MAT 540 Quantiative Methods Strayer wk 7
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Tenth Edition
Bernard W. Taylor III Virginia Polytechnic Institute and State University
Prentice Hall Upper Saddle River, New Jersey 07458
Introduction to Management Science
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
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Taylor, Bernard W. Introduction to management science / Bernard W. Taylor III.—10th ed.
p. cm. Includes bibliographical references and index. ISBN-13: 978-0-13-606436-7 (alk. paper) ISBN-10: 0-13-606436-1 (alk. paper)
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
1
C h a p t e r 1
Management Science
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
2 Chapter 1 Management Science
Management science is a scientific approach to solving
management problems.
Management science is the application of a scientific approach to solving managementproblems in order to help managers make better decisions. As implied by this defini- tion, management science encompasses a number of mathematically oriented techniques that have either been developed within the field of management science or been adapted from other disciplines, such as the natural sciences, mathematics, statistics, and engineer- ing. This text provides an introduction to the techniques that make up management science and demonstrates their applications to management problems.
Management science is a recognized and established discipline in business. The applica- tions of management science techniques are widespread, and they have been frequently credited with increasing the efficiency and productivity of business firms. In various sur- veys of businesses, many indicate that they use management science techniques, and most rate the results to be very good. Management science (also referred to as operations research, quantitative methods, quantitative analysis, and decision sciences) is part of the fundamental curriculum of most programs in business.
As you proceed through the various management science models and techniques con- tained in this text, you should remember several things. First, most of the examples pre- sented in this text are for business organizations because businesses represent the main users of management science. However, management science techniques can be applied to solve problems in different types of organizations, including services, government, mili- tary, business and industry, and health care.
Second, in this text all of the modeling techniques and solution methods are mathemat- ically based. In some instances the manual, mathematical solution approach is shown because it helps one understand how the modeling techniques are applied to different problems. However, a computer solution is possible for each of the modeling techniques in this text, and in many cases the computer solution is emphasized. The more detailed math- ematical solution procedures for many of the modeling techniques are included as supple- mental modules on the companion Web site for this text.
Finally, as the various management science techniques are presented, keep in mind that management science is more than just a collection of techniques. Management science also involves the philosophy of approaching a problem in a logical manner (i.e., a scientific approach). The logical, consistent, and systematic approach to problem solving can be as useful (and valuable) as the knowledge of the mechanics of the mathematical techniques themselves. This understanding is especially important for those readers who do not always see the immediate benefit of studying mathematically oriented disciplines such as manage- ment science.
The Management Science Approach to Problem Solving
As indicated in the previous section, management science encompasses a logical, systematic approach to problem solving, which closely parallels what is known as the scientific method for attacking problems. This approach, as shown in Figure 1.1, follows a generally recognized and ordered series of steps: (1) observation, (2) definition of the problem, (3) model construction, (4) model solution, and (5) implementation of solution results. We will analyze each of these steps individually.
Observation The first step in the management science process is the identification of a problem that exists in the system (organization). The system must be continuously and closely observed
Management science can be used in a variety of organizations to
solve many different types of problems.
Management science encompasses a logical approach to problem
solving.
The steps of the scientific method are (1) observation, (2) problem
definition, (3) model construction, (4) model solution, and
(5) implementation.
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
so that problems can be identified as soon as they occur or are anticipated. Problems are not always the result of a crisis that must be reacted to but, instead, frequently involve an anticipatory or planning situation. The person who normally identifies a problem is the manager because managers work in places where problems might occur. However, prob- lems can often be identified by a management scientist, a person skilled in the techniques of management science and trained to identify problems, who has been hired specifically to solve problems using management science techniques.
Definition of the Problem Once it has been determined that a problem exists, the problem must be clearly and con- cisely defined. Improperly defining a problem can easily result in no solution or an inap- propriate solution. Therefore, the limits of the problem and the degree to which it pervades other units of the organization must be included in the problem definition. Because the existence of a problem implies that the objectives of the firm are not being met in some way, the goals (or objectives) of the organization must also be clearly defined. A stated objective helps to focus attention on what the problem actually is.
Model Construction A management science model is an abstract representation of an existing problem situa- tion. It can be in the form of a graph or chart, but most frequently a management science model consists of a set of mathematical relationships. These mathematical relationships are made up of numbers and symbols.
As an example, consider a business firm that sells a product. The product costs $5 to produce and sells for $20. A model that computes the total profit that will accrue from the items sold is
In this equation x represents the number of units of the product that are sold, and Z rep- resents the total profit that results from the sale of the product. The symbols x and Z are variables. The term variable is used because no set numeric value has been specified for these items. The number of units sold, x, and the profit, Z, can be any amount (within limits); they can vary. These two variables can be further distinguished. Z is a dependent variable because
Z = $20x - 5x
The Management Science Approach to Problem Solving 3
Management science techniques
Observation
Problem definition
Model construction
Solution
Feedback
Information
Implementation
Figure 1.1
The management science process
A management scientist is a person skilled in the application of
management science techniques.
A variable is a symbol used to represent an item that can take on
any value.
A model is an abstract mathematical representation of a
problem situation.
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
4 Chapter 1 Management Science
its value is dependent on the number of units sold; x is an independent variable because the number of units sold is not dependent on anything else (in this equation).
The numbers $20 and $5 in the equation are referred to as parameters. Parameters are constant values that are generally coefficients of the variables (symbols) in an equation. Parameters usually remain constant during the process of solving a specific problem. The parameter values are derived from data (i.e., pieces of information) from the problem envi- ronment. Sometimes the data are readily available and quite accurate. For example, pre- sumably the selling price of $20 and product cost of $5 could be obtained from the firm’s accounting department and would be very accurate. However, sometimes data are not as readily available to the manager or firm, and the parameters must be either estimated or based on a combination of the available data and estimates. In such cases, the model is only as accurate as the data used in constructing the model.
The equation as a whole is known as a functional relationship (also called function and relationship). The term is derived from the fact that profit, Z, is a function of the number of units sold, x, and the equation relates profit to units sold.
Because only one functional relationship exists in this example, it is also the model. In this case the relationship is a model of the determination of profit for the firm. However, this model does not really replicate a problem. Therefore, we will expand our example to create a problem situation.
Let us assume that the product is made from steel and that the business firm has 100 pounds of steel available. If it takes 4 pounds of steel to make each unit of the product, we can develop an additional mathematical relationship to represent steel usage:
This equation indicates that for every unit produced, 4 of the available 100 pounds of steel will be used. Now our model consists of two relationships:
We say that the profit equation in this new model is an objective function, and the resource equation is a constraint. In other words, the objective of the firm is to achieve as much profit, Z, as possible, but the firm is constrained from achieving an infinite profit by the limited amount of steel available. To signify this distinction between the two relation- ships in this model, we will add the following notations:
subject to
This model now represents the manager’s problem of determining the number of units to produce. You will recall that we defined the number of units to be produced as x. Thus, when we determine the value of x, it represents a potential (or recommended) decision for the manager. Therefore, x is also known as a decision variable. The next step in the manage- ment science process is to solve the model to determine the value of the decision variable.
Model Solution Once models have been constructed in management science, they are solved using the management science techniques presented in this text. A management science solution technique usually applies to a specific type of model. Thus, the model type and solution method are both part of the management science technique. We are able to say that a model
4x = 100
maximize Z = $20x - 5x
4x = 100 Z = $20x - 5x
4x = 100 lb. of steel
Parameters are known, constant values that are often coefficients of
variables in equations.
Data are pieces of information from the problem environment.
A model is a functional relationship that includes
variables, parameters, and equations.
A management science technique usually applies to a specific
model type.
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The Management Science Approach to Problem Solving 5
is solved because the model represents a problem. When we refer to model solution, we also mean problem solution.
For the example model developed in the previous section,
subject to
the solution technique is simple algebra. Solving the constraint equation for x, we have
Substituting the value of 25 for x into the profit function results in the total profit:
Thus, if the manager decides to produce 25 units of the product and all 25 units sell, the business firm will receive $375 in profit. Note, however, that the value of the decision vari- able does not constitute an actual decision; rather, it is information that serves as a recom- mendation or guideline, helping the manager make a decision.
= $375 = 20(25) - 5(25)
Z = $20x - 5x
x = 25 units x = 100>4
4x = 100
4x = 100
maximize Z = $20x - 5x
Ti m e O u t for Pioneers in Management Science Throughout this text TIME OUT boxes introduce you to the individuals who developed the various techniques that are described in the chapters. This will provide a historical per- spective on the development of the field of management science. In this first instance we will briefly outline the develop- ment of management science.
Although a number of the mathematical techniques that make up management science date to the turn of the twentieth century or before, the field of management science itself can trace its beginnings to military operations research (OR) groups formed during World War II in Great Britain circa 1939. These OR groups typically consisted of a team of about a dozen individuals from different fields of science, mathematics, and the military, brought together to find solutions to military- related problems. One of the most famous of these groups— called “Blackett’s circus” after its leader, Nobel laureate P. M. S. Blackett of the University of Manchester and a former naval officer—included three physiologists, two mathematical physi- cists, one astrophysicist, one general physicist, two mathemati- cians, an Army officer, and a surveyor. Blackett’s group and the other OR teams made significant contributions in improving Britain’s early-warning radar system (which was instrumental in their victory in the Battle of Britain), aircraft gunnery, anti- submarine warfare, civilian defense, convoy size determination, and bombing raids over Germany.
The successes achieved by the British OR groups were observed by two Americans working for the U.S. military, Dr. James B. Conant and Dr. Vannevar Bush, who recommended that OR teams be established in the U.S. branches of the military. Subsequently, both the Air Force and Navy created OR groups.
After World War II the contributions of the OR groups were considered so valuable that the Army, Air Force, and Navy set up various agencies to continue research of military problems. Two of the more famous agencies were the Navy’s Operations Evaluation Group at MIT and Project RAND, established by the Air Force to study aerial warfare. Many of the individuals who developed operations research and management science techniques did so while working at one of these agencies after World War II or as a result of their work there.
As the war ended and the mathematical models and tech- niques that were kept secret during the war began to be released, there was a natural inclination to test their applicabil- ity to business problems. At the same time, various consulting firms were established to apply these techniques to industrial and business problems, and courses in the use of quantitative techniques for business management began to surface in American universities. In the early 1950s the use of these quan- titative techniques to solve management problems became known as management science, and it was popularized by a book of that name by Stafford Beer of Great Britain.
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6 Chapter 1 Management Science
Some management science techniques do not generate an answer or a recommended decision. Instead, they provide descriptive results: results that describe the system being modeled. For example, suppose the business firm in our example desires to know the aver- age number of units sold each month during a year. The monthly data (i.e., sales) for the past year are as follows:
M a n a g e m e n t S c i e n c e A p p l i c a t i o n Management Science at Taco Bell
Taco Bell, an international fast-food chain with annual salesof approximately $4.6 billion, operates more than 6,500 locations worldwide. In the fast-food business the operating objective is, in general, to provide quality food, good service, and a clean environment. Although Taco Bell sees these three attributes as equally important, good service, as measured by its speed, has the greatest impact on revenues.
The 3-hour lunch period 11:00 A.M. to 2:00 P.M. accounts for 52% of Taco Bell’s daily sales. Most fast-food restaurants have lines of waiting customers during this period, and so speed of service determines sales capacity. If service time decreases, sales capacity increases, and vice versa. However, as speed of service increases, labor costs also increase. Because very few food items can be pre- pared in advance and inventoried, products must be prepared when they are ordered, making food preparation very labor inten- sive. Thus, speed of service depends on labor availability.
Taco Bell research studies showed that when customers are in line up to 5 minutes only, their perception of that waiting time is only a few minutes. However, after waiting time exceeds 5 minutes, customer perception of that waiting time increases exponentially. The longer the perceived waiting time, the more likely the customer is to leave the restaurant without ordering. The company determined that a 3-minute average waiting time would result in only 2.5% of customers leaving. The company believed this was an acceptable level of attrition, and it estab- lished this waiting time as its service goal.
To achieve this goal Taco Bell developed a labor-management system based on an integrated set of management science mod- els to forecast customer traffic for every 15-minute interval
during the day and to schedule employees accordingly to meet customer demand. This labor-management system includes a forecasting model to predict customer transactions; a simulation model to determine labor requirements based on these trans- actions; and an integer programming model to schedule employees and minimize payroll. From 1993 through 1997 the labor-management system using these models saved Taco Bell over $53 million.
Source: J. Heuter and W. Swart, “An Integrated Labor- Management System for Taco Bell,” Interfaces 28, no. 1 (January– February 1998): 75–91.
Month Sales Month Sales
January 30 July 35
February 40 August 50
March 25 September 60
April 60 October 40
May 30 November 35
June 25 December 50
Total 480 units
A management science solution can be either a recommended
decision or information that helps a manager make a decision.
Monthly sales average 40 units . This result is not a decision; it is informa- tion that describes what is happening in the system. The results of the management science
(480 , 12)
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Model Building: Break-Even Analysis 7
techniques in this text are examples of the two types shown in this section: (1) solutions/ decisions and (2) descriptive results.
Implementation The final step in the management science process for problem solving described in Figure 1.1 is implementation. Implementation is the actual use of the model once it has been developed or the solution to the problem the model was developed to solve. This is a critical but often overlooked step in the process. It is not always a given that once a model is developed or a solution found, it is automatically used. Frequently the person responsible for putting the model or solution to use is not the same person who developed the model, and thus the user may not fully understand how the model works or exactly what it is sup- posed to do. Individuals are also sometimes hesitant to change the normal way they do things or to try new things. In this situation the model and solution may get pushed to the side or ignored altogether if they are not carefully explained and their benefit fully demon- strated. If the management science model and solution are not implemented, then the effort and resources used in their development have been wasted.
Model Building: Break-Even Analysis
In the previous section we gave a brief, general description of how management science models are formulated and solved, using a simple algebraic example. In this section we will continue to explore the process of building and solving management science models, using break-even analysis, also called profit analysis. Break-even analysis is a good topic to expand our discussion of model building and solution because it is straightforward, rela- tively familiar to most people, and not overly complex. In addition, it provides a convenient means to demonstrate the different ways management science models can be solved— mathematically (by hand), graphically, and with a computer.
The purpose of break-even analysis is to determine the number of units of a product (i.e., the volume) to sell or produce that will equate total revenue with total cost. The point where total revenue equals total cost is called the break-even point, and at this point profit is zero. The break-even point gives a manager a point of reference in determining how many units will be needed to ensure a profit.
Components of Break-Even Analysis The three components of break-even analysis are volume, cost, and profit. Volume is the level of sales or production by a company. It can be expressed as the number of units (i.e., quantity) produced and sold, as the dollar volume of sales, or as a percentage of total capac- ity available.
Two type of costs are typically incurred in the production of a product: fixed costs and variable costs. Fixed costs are generally independent of the volume of units produced and sold. That is, fixed costs remain constant, regardless of how many units of product are pro- duced within a given range. Fixed costs can include such items as rent on plant and equip- ment, taxes, staff and management salaries, insurance, advertising, depreciation, heat and light, and plant maintenance. Taken together, these items result in total fixed costs.
Variable costs are determined on a per-unit basis. Thus, total variable costs depend on the number of units produced. Variable costs include such items as raw materials and resources, direct labor, packaging, material handling, and freight.
Fixed costs are independent of volume and remain constant.
Variable costs depend on the number of items produced.
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8 Chapter 1 Management Science
Total variable costs are a function of the volume and the variable cost per unit. This relationship can be expressed mathematically as
where and (number of units) sold. The total cost of an operation is computed by summing total fixed cost and total vari-
able cost, as follows:
or
where . As an example, consider Western Clothing Company, which produces denim jeans. The
company incurs the following monthly costs to produce denim jeans:
If we arbitrarily let the monthly sales volume, v, equal 400 pairs of denim jeans, the total cost is
The third component in our break-even model is profit. Profit is the difference between total revenue and total cost. Total revenue is the volume multiplied by the price per unit,
where . For our clothing company example, if denim jeans sell for $23 per pair and we sell
400 pairs per month, then the total monthly revenue is
Now that we have developed relationships for total revenue and total cost, profit (Z) can be computed as follows:
Computing the Break-Even Point For our clothing company example, we have determined total revenue and total cost to be $9,200 and $13,200, respectively. With these values, there is no profit but, instead, a loss of $4,000:
We can verify this result by using our total profit formula,
and the values , , , and :
= - $4,000 = $9,200 - 10,000 - 3,200 = $(400)(23) - 10,000 - (400)(8)
Z = vp - cf - vcv
cv = $8cf = $10,000p = $23v = 400
Z = vp - cf - vcv
total profit = total revenue - total cost = $9,200 - 13,200 = - $4,000
= vp - cf - vcv Z = vp - (cf + vcv)
total profit = total revenue - total cost
total revenue = vp = (400)(23) = $9,200
p = price per unit
total revenue = vp
TC = cf + vcv = $10,000 + (400)(8) = $13,200
variable cost = cv = $8 per pair fixed cost = cf = $10,000
cf = fixed cost
TC = cf + vcv
total cost = total fixed cost + total variable cost
v = volumecv = variable cost per unit
total variable cost = vcv
Total cost (TC) equals the fixed cost plus the variable cost
per unit multiplied by volume (v).
(cv) (cf )
Profit is the difference between total revenue (volume multiplied
by price) and total cost.
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Model Building: Break-Even Analysis 9
Obviously, the clothing company does not want to operate with a monthly loss of $4,000 because doing so might eventually result in bankruptcy. If we assume that price is static because of market conditions and that fixed costs and the variable cost per unit are not subject to change, then the only part of our model that can be varied is volume. Using the modeling terms we developed earlier in this chapter, price, fixed costs, and variable costs are parameters, whereas the volume, v, is a decision variable. In break-even analysis we want to compute the value of v that will result in zero profit.
At the break-even point, where total revenue equals total cost, the profit, Z, equals zero. Thus, if we let profit, Z, equal zero in our total profit equation and solve for v, we can deter- mine the break-even volume:
In other words, if the company produces and sells 666.7 pairs of jeans, the profit (and loss) will be zero and the company will break even. This gives the company a point of refer- ence from which to determine how many pairs of jeans it needs to produce and sell in order to gain a profit (subject to any capacity limitations). For example, a sales volume of 800 pairs of denim jeans will result in the following monthly profit:
In general, the break-even volume can be determined using the following formula:
For our example,
Graphical Solution It is possible to represent many of the management science models in this text graphically and use these graphical models to solve problems. Graphical models also have the advan- tage of providing a “picture” of the model that can sometimes help us understand the mod- eling process better than the mathematics alone can. We can easily graph the break-even model for our Western Clothing Company example because the functions for total cost and total revenue are linear. That means we can graph each relationship as a straight line on a set of coordinates, as shown in Figure 1.2.
In Figure 1.2, the fixed cost, , has a constant value of $10,000, regardless of the volume. The total cost line, TC, represents the sum of variable cost and fixed cost. The total cost line
cf
= 666.7 pairs of jeans
= 10,000
23 - 8
v = cf
p - cv
v = cf
p - cv
v(p - cv) = cf 0 = v(p - cv) - cf Z = vp - cf - vcv
= $(800)(23) - 10,000 - (800)(8) = $2,000 Z = vp - cf - vcv
v = 666.7 pairs of jeans 15v = 10,000
0 = 23v - 10,000 - 8v 0 = v(23) - 10,000 - v(8) Z = vp - cf - vcv
The break-even point is the volume (v) that equates total revenue with total cost where
profit is zero.
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10 Chapter 1 Management Science
10
20
30
40
50
200 0
400 600 800 1,000 1,200 1,400 1,600 Volume, v
Total cost
R e
ve n
u e
, co
st ,
a n
d p
ro fit
( $
1 ,0
0 0
s)
Variable cost
Fixed cost
Total revenue
Loss
Profit
Break-even point
Figure 1.2
Break-even model
increases because variable cost increases as the volume increases. The total revenue line also increases as volume increases, but at a faster rate than total cost. The point where these two lines intersect indicates that total revenue equals total cost. The volume, v, that corresponds to this point is the break-even volume. The break-even volume in Figure 1.2 is 666.7 pairs of denim jeans.
Sensitivity Analysis We have now developed a general relationship for determining the break-even volume, which was the objective of our modeling process. This relationship enables us to see how the level of profit (and loss) is directly affected by changes in volume. However, when we developed this model, we assumed that our parameters, fixed and variable costs and price, were constant. In reality such parameters are frequently uncertain and can rarely be assumed to be constant, and changes in any of the parameters can affect the model solu- tion. The study of changes on a management science model is called sensitivity analysis— that is, seeing how sensitive the model is to changes.
Sensitivity analysis can be performed on all management science models in one form or another. In fact, sometimes companies develop models for the primary purpose of experi- mentation to see how the model will react to different changes the company is contemplat- ing or that management might expect to occur in the future. As a demonstration of how sensitivity analysis works, we will look at the effects of some changes on our break-even model.
The first thing we will analyze is price. As an example, we will increase the price for denim jeans from $23 to $30. As expected, this increases the total revenue, and it therefore reduces the break-even point from 666.7 pairs of jeans to 454.5 pairs of jeans:
The effect of the price change on break-even volume is illustrated in Figure 1.3. Although a decision to increase price looks inviting from a strictly analytical point of
view, it must be remembered that the lower break-even volume and higher profit are possible but not guaranteed. A higher price can make it more difficult to sell the product. Thus, a change in price often must be accompanied by corresponding increases in costs, such as those for advertising, packaging, and possibly production (to enhance quality). However, even such direct changes as these may have little effect on product demand
= 10,000
30 - 8 = 454.5 pairs of denim jeans
v = cf
p - cv
In general, an increase in price lowers the break-even point, all
other things held constant.
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because price is often sensitive to numerous factors, such as the type of market, monopo- listic elements, and product differentiation.
When we increased price, we mentioned the possibility of raising the quality of the product to offset a potential loss of sales due to the price increase. For example, suppose the stitching on the denim jeans is changed to make the jeans more attractive and stronger. This change results in an increase in variable costs of $4 per pair of jeans, thus raising the variable cost per unit, , to $12 per pair. This change (in conjunction with our previous price change to $30) results in a new break-even volume:
This new break-even volume and the change in the total cost line that occurs as a result of the variable cost change are shown in Figure 1.4.
= 10,000
30 - 12 = 555.5 pairs of denim jeans
v = cf
p - cv
cv
Model Building: Break-Even Analysis 11
10
20
30
40
50
200 0
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st ,
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ro fit
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0 0
s)
New total revenue
Fixed cost
Old total revenue
Old B-E point
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Figure 1.3
Break-even model with an increase in price
In general, an increase in variable costs will decrease the break-even
point, all other things held constant.
10
20
30
40
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400 600 800 1,000 1,200 1,400 1,600 Volume, v
Old total cost
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n u e ,
co st
, a n d p
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s)
Total revenue
Fixed cost
New total cost
New B-E point
Old B-E point
Figure 1.4
Break-even model with an increase in variable cost
Next let’s consider an increase in advertising expenditures to offset the potential loss in sales resulting from a price increase. An increase in advertising expenditures is an addition to fixed costs. For example, if the clothing company increases its monthly advertising bud- get by $3,000, then the total fixed cost, , becomes $13,000. Using this fixed cost, as well ascf
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400 600 800 1,000 1,200 1,400 1,600 Volume, v
Old total cost
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a n
d p
ro fit
( $
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Total revenue
Old fixed cost
New total cost
New fixed cost
New B-E point
Old B-E point
Figure 1.5
Break-even model with a change in fixed cost
the increased variable cost per unit of $12 and the increased price of $30, we compute the break-even volume as follows:
This new break-even volume, representing changes in price, fixed costs, and variable costs, is illustrated in Figure 1.5. Notice that the break-even volume is now higher than the origi- nal volume of 666.7 pairs of jeans, as a result of the increased costs necessary to offset the potential loss in sales. This indicates the necessity to analyze the effect of a change in one of the break-even components on the whole break-even model. In other words, generally it is not sufficient to consider a change in one model component without considering the over- all effect.
= 722.2 pairs of denim jeans
= 13,000
30 - 12
v = cf
p - cv
In general, an increase in fixed costs will increase the break-even
point, all other things held constant.
Computer Solution
Throughout the text we will demonstrate how to solve management science models on the computer by using Excel spreadsheets and QM for Windows, a general-purpose quantita- tive methods software package by Howard Weiss. QM for Windows has program modules to solve almost every type of management science problem you will encounter in this book. There are a number of similar quantitative methods software packages available on the market, with characteristics and capabilities similar to those of QM for Windows. In most cases you simply input problem data (i.e., model parameters) into a model template, click on a solve button, and the solution appears in a Windows format. QM for Windows is included on the companion Web site for this text.
Spreadsheets are not always easy to use, and you cannot conveniently solve every type of management science model by using a spreadsheet. Most of the time you must not only input the model parameters but also set up the model mathematics, including formulas, as well as your own model template with headings to display your solution output. However, spreadsheets provide a powerful reporting tool in which you can present your model and results in any format you choose. Spreadsheets such as Excel have become almost
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Computer Solution 13
universally available to anyone who owns a computer. In addition, spreadsheets have become very popular as a teaching tool because they tend to guide the student through a modeling procedure, and they can be interesting and fun to use. However, because spread- sheets are somewhat more difficult to set up and apply than is QM for Windows, we will spend more time explaining their use to solve various types of problems in this text.
One of the difficult aspects of using spreadsheets to solve management science problems is setting up a spreadsheet with some of the more complex models and formulas. For the most complex models in the text we will show how to use Excel QM, a supplemental spreadsheet macro that is included on the companion Web site for this text. A macro is a template or an overlay that already has the model format with the necessary formulas set up on the spreadsheet so that the user only has to input the model parameters. We will demonstrate Excel QM in six chapters, including this chapter, Chapter 6 (“Transporta- tion, Transshipment, and Assignment Problems”), Chapter 12 (“Decision Analysis”), Chapter 13 (“Queuing Analysis”), Chapter 15 (“Forecasting”), and Chapter 16 (“Inventory Management”).
Later in this text we will also demonstrate two spreadsheet add-ins, TreePlan and Crystal Ball, which are also included on the companion Web site for this text. TreePlan is a program for setting up and solving decision trees that we use in Chapter 12 (“Decision Analysis”), whereas Crystal Ball is a simulation package that we use in Chapter 14 (“Simulation”). Also, in Chapter 8 (“Project Management”) we will demonstrate Microsoft Project.
In this section we will demonstrate how to use Excel, Excel QM, and QM for Windows, using our break-even model example for Western Clothing Company.
Excel Spreadsheets To solve the break-even model using Excel, you must set up a spreadsheet with headings to identify your model parameters and variables and then input the appropriate mathemati- cal formulas into the cells where you want to display your solution. Exhibit 1.1 shows the spreadsheet for the Western Clothing Company example. Setting up the different headings to describe the parameters and the solution is not difficult, but it does require that you know your way around Excel a little. Appendix B provides a brief tutorial titled “Setting Up and Editing a Spreadsheet” for solving management science problems.
Formula for v, break-even point
Exhibit 1.1
Notice that cell D10 contains the break-even formula, which is displayed on the toolbar near the top of the screen. The fixed cost of $10,000 is typed in cell D4, the variable cost of $8 is in cell D6, and the price of $23 is in cell D8.
As we present more complex models and problems in the chapters to come, the spread- sheets we will develop to solve these problems will become more involved and will enable us to demonstrate different features of Excel and spreadsheet modeling.
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14 Chapter 1 Management Science
Exhibit 1.2
Enter model parameters in cells B10:B13.
Exhibit 1.3
The Excel QM Macro for Spreadsheets Excel QM is included on the companion Web site for this text. You can install Excel QM onto your computer by following a brief series of steps displayed when the program is first accessed.
After Excel is started, Excel QM is normally accessed from the computer’s program files, where it is usually loaded. When Excel QM is activated, “Add-Ins” will appear at the top of the spreadsheet (as indicated in Exhibit 1.2). Clicking on “Excel QM” will pull down a menu of the topics in Excel QM, one of which is break-even analysis. Clicking on “Break- Even Analysis” will result in the window for spreadsheet initialization shown in Exhibit 1.2. Every Excel QM macro listed on the menu will start with a “Spreadsheet Initialization” window similar to this one.
In the window in Exhibit 1.2 you can enter a spreadsheet title and choose under “Options” whether you also want volume analysis and a graph. Clicking on “OK” will result in the spreadsheet shown in Exhibit 1.3. The first step is to input the values for the Western Clothing Company example in cells B10 to B13, as shown in Exhibit 1.3. The spreadsheet shows the break-even volume in cell B17. However, notice that we have also chosen to per- form some volume analysis by entering a hypothetical volume of 800 units in cell B13, which results in the volume analysis in cells B20 to B23.
QM for Windows You begin using QM for Windows by clicking on the “Module” button on the toolbar at the top of the main window that appears when you start the program. This will pull down a window with a list of all the model solution modules available in QM for Windows.
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Management Science Modeling Techniques 15
Exhibit 1.4
Clicking on the “Break-even Analysis” module will access a new screen for typing in the problem title. Clicking again will access a screen with input cells for the model parame- ters—that is, fixed cost, variable cost, and price (or revenue). Next, clicking on the “Solve” button at the top of the screen will provide the solution to the Western Clothing Company example, as shown in Exhibit 1.4.
Exhibit 1.5
You can also get the graphical model and solution for this problem by clicking on “Win- dow” at the top of the solution screen and selecting the menu item for a graph of the prob- lem. The break-even graph for the Western Clothing example is shown in Exhibit 1.5.
Management Science Modeling Techniques
This text focuses primarily on two of the five steps of the management science process described in Figure 1.1—model construction and solution. These are the two steps that use the management science technique. In a textbook, it is difficult to show how an unstructured real-world problem is identified and defined because the problem must be written out.
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16 Chapter 1 Management Science
Management science techniques
Probabilistic techniques
Linear mathematical programming
Linear programming models Graphical analysis Sensitivity analysis Transportation, transshipment, and assignment Integer linear programming Goal programming
Decision analysis
Probability and statistics
Queuing
Network
Text
techniques
Network flow Project
management (CPM/PERT)
Other techniques
Forecasting Simulation
Inventory
Analytical hierarchy process (AHP) Nonlinear programming
Companion Web Site
Branch and bound Markov analysis Game theory
method
Simplex method Transportation
and assignment solution methods
Nonlinear programming
Figure 1.6
Classification of management science techniques
However, once a problem statement has been given, we can show how a model is con- structed and a solution is derived. The techniques presented in this text can be loosely classified into four categories, as shown in Figure 1.6.
Linear Mathematical Programming Techniques Chapters 2 through 6 and 9 present techniques that together make up linear mathematical programming. (The first example used to demonstrate model construction earlier in this chapter is a very rudimentary linear programming model.) The term programming used to identify this technique does not refer to computer programming but rather to a predeter- mined set of mathematical steps used to solve a problem. This particular class of techniques holds a predominant position in this text because it includes some of the more frequently used and popular techniques in management science.
In general, linear programming models help managers determine solutions (i.e., make decisions) for problems that will achieve some objective in which there are restrictions, such as limited resources or a recipe or perhaps production guidelines. For example, you could actually develop a linear programming model to help determine a breakfast menu for yourself that would meet dietary guidelines you may have set, such as number of calo- ries, fat content, and vitamin level, while minimizing the cost of the breakfast. Manufactur- ing companies develop linear programming models to help decide how many units of different products they should produce to maximize their profit (or minimize their cost), given scarce resources such as capital, labor, and facilities.
Six chapters in this text are devoted to this topic because there are several variations of linear programming models that can be applied to specific types of problems. Chapter 4 is devoted entirely to describing example linear programming models for several different types of problem scenarios. Chapter 6, for example, focuses on one particular type of linear programming application for transportation, transshipment, and assignment problems. An example of a transportation problem is a manager trying to determine the lowest-cost routes to use to ship goods from several sources (such as plants or warehouses) to several destinations (such as retail stores), given that each source may have limited goods available and each destination may have limited demand for the goods. Also, Chapter 9 includes the topic of goal programming, which is a form of linear programming that addresses prob- lems with more than one objective or goal.
As mentioned previously in this chapter, some of the more mathematical topics in the text are included as supplementary modules on the companion Web site for the text. Among
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Management Science Modeling Techniques 17
the linear programming topics included on the companion Web site are modules on the simplex method; the transportation, transshipment, and assignment solution methods; and the branch and bound solution method for integer programming models. Also included on the companion Web site are modules on nonlinear programming, game theory, and Markov analysis.
Probabilistic Techniques Probabilistic techniques are presented in Chapters 11 through 13. These techniques are distinguished from mathematical programming techniques in that the results are proba- bilistic. Mathematical programming techniques assume that all parameters in the models are known with certainty. Therefore, the solution results are assumed to be known with cer- tainty, with no probability that other solutions might exist. A technique that assumes certainty in its solution is referred to as deterministic. In contrast, the results from a prob- abilistic technique do contain uncertainty, with some possibility that alternative solutions might exist. In the model solution presented earlier in this chapter, the result of the first example ( units to produce) is deterministic, whereas the result of the second exam- ple (estimating an average of 40 units sold each month) is probabilistic.
An example of a probabilistic technique is decision analysis, the subject of Chapter 12. In decision analysis it is shown how to select among several different decision alternatives, given uncertain (i.e., probabilistic) future conditions. For example, a developer may want to decide whether to build a shopping mall, build an office complex, build condominiums, or not build anything at all, given future economic conditions that might be good, fair, or poor, each with a probability of occurrence. Chapter 13, on queuing analysis, presents probabilistic techniques for analyzing waiting lines that might occur, for example, at the grocery store, at a bank, or at a movie. The results of waiting line analysis are statistical averages showing, among other things, the average number of customers in line waiting to be served or the average time a customer might have to wait for service.
Network Techniques Networks, the topic of Chapters 7 and 8, consist of models that are represented as diagrams rather than as strictly mathematical relationships. As such, these models offer a pictorial representation of the system under analysis. These models represent either probabilistic or deterministic systems.
For example, in shortest-route problems, one of the topics in Chapter 7 (“Network Flow Models”), a network diagram can be drawn to help a manager determine the shortest route among a number of different routes from a source to a destination. For example, you could use this technique to determine the shortest or quickest car route from St. Louis to Daytona Beach for a spring break vacation. In Chapter 8 (“Project Management”), a network is drawn that shows the relationships of all the tasks and activities for a project, such as build- ing a house or developing a new computer system. This type of network can help a man- ager plan the best way to accomplish each of the tasks in the project so that it will take the shortest amount of time possible. You could use this type of technique to plan for a concert or an intramural volleyball tournament on your campus.
Other Techniques Some topics in the text are not easily categorized; they may overlap several categories, or they may be unique. The analytical hierarchy process (AHP) in Chapter 9 is such a topic that is not easily classified. It is a mathematical technique for helping the decision maker choose between several alternative decisions, given more than one objective; however, it is not a form of linear
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programming, as is goal programming, the shared topic in Chapter 9, on multicriteria deci- sion making. The structure of the mathematical models for nonlinear programming prob- lems in Chapter 10 is similar to the linear programming problems in Chapters 2 through 6; however, the mathematical equations and functions in nonlinear programming can be non- linear instead of linear, thus requiring the use of calculus to solve them. Simulation, the sub- ject of Chapter 14, is probably the single most unique topic in the text. It has the capability to solve probabilistic and deterministic problems and is often the technique of last resort when no other management science technique will work. In simulation, a mathematical model is constructed (typically using a computer) that replicates a real-world system under analysis, and then that simulation model is used to solve problems in the “simulated” real-world system. For example, with simulation you could build a model to simulate the traffic patterns of vehicles at a busy intersection to determine how to set the traffic light signals.
Forecasting, the subject of Chapter 15, and inventory management, in Chapter 16, are topics traditionally considered to be part of the field of operations management. However, because they are both important business functions that also rely heavily on quantitative models for their analysis, they are typically considered important topics in the study of management science as well. Both topics also include probabilistic as well as deterministic aspects. In Chapter 15 we will look at several different quantitative models that help man- agers predict what the future demand for products and services will look like. In general, historical sales and demand data are used to build a mathematical function or formula that can be used to estimate product demand in the future. In Chapter 16 we will look at several different quantitative models that help organizations determine how much inventory to keep on hand in order to minimize inventory costs, which can be significant.
Business Usage of Management Science Techniques
Not all management science techniques are equally useful or equally used by business firms and other organizations. Some techniques are used quite frequently by business practition- ers and managers; others are used less often. The most frequently used techniques are linear and integer programming, simulation, network analysis (including critical path method/project evaluation and review technique [CPM/PERT]), inventory control, deci- sion analysis, and queuing theory, as well as probability and statistics. An attempt has been made in this text to provide a comprehensive treatment of all the topics generally consid- ered within the field of management science, regardless of how frequently they are used. Although some topics may have limited direct applicability, their study can reveal informa- tive and unique means of approaching a problem and can often enhance one’s understand- ing of the decision-making process.
The variety and breadth of management science applications and of the potential for applying management science, not only in business and industry but also in government, health care, and service organizations, are extensive. Areas of application include project planning, capital budgeting, production planning, inventory analysis, scheduling, market- ing planning, quality control, plant location, maintenance policy, personnel management, and product demand forecasting, among others. In this text the applicability of manage- ment science to a variety of problem areas is demonstrated via individual chapter examples and the problems that accompany each chapter.
A small portion of the thousands of applications of management science that occur each year are recorded in various academic and professional journals. Frequently, these journal articles are as complex as the applications themselves and are very difficult to read. However, one particular journal, Interfaces, is devoted specifically to the application of management
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Business Usage of Management Science Techniques 19
science and is written not just for college professors but for businesspeople, practitioners, and students as well. Interfaces is published by INFORMS (Institute for Operations Research and Management Sciences), an international professional organization whose members include college professors, businesspeople, scientists, students, and a variety of professional people interested in the practice and application of management science and operations research.
Interfaces regularly publishes articles that report on the application of management science to a wide variety of problems. The chapters that follow present examples of appli- cations of management science from Interfaces and other professional journals. These examples, as presented here, do not detail the actual models and the model components. Instead, they briefly indicate the type of problem the company or organization faced, the objective of the solution approach developed to solve the problem, and the benefits derived from the model or technique (i.e., what was accomplished). The interested reader who desires more detailed information about these and other management science applications is encouraged to go to the library and peruse Interfaces and the many other journals that contain articles on the application of management science.
M a n a g e m e n t S c i e n c e A p p l i c a t i o n Employee Scheduling with Management Science
Management science models continue to be applied tothousands of operational problems in companies and organizations around the world. One of the largest areas of application of management science models is employee sched- uling, which is often generally referred to as the capacity- planning problem. The overall problem, and the subsequent model development, involves a staffing component—that is, how many, when, and where employees are needed—and a scheduling component, which determines when and where each employee is assigned to work. This problem is generally solved in two parts. The staffing problem is solved first using such management science modeling techniques as demand forecast- ing (Chapter 15), queuing analysis (Chapter 13), simulation (Chapter 14), and/or inventory management (Chapter 16). Once a staffing plan has been developed, a schedule is con- structed that assigns employees to specific jobs or locations at specific times, using a variety of possible management science techniques including various forms of linear programming (Chapters 2 through 6 and 10) and simulation (Chapter 14). Often, the development of these models for staffing and scheduling is part of a focused project management effort (Chapter 8). A specific example of this type of problem is the nurse scheduling problem, in which a specific number of nurses in a hospital must be assigned to days and shifts over a specified time period. The solution objective of this problem is generally to minimize the number of nurses in order to avoid waste (and save money) while making sure there is adequate service for patient care. A good solution must also be able to satisfy hospi- tal, legislative, and union policies and maintain nurse and patient morale and retention. In general, this is called a constrained optimization problem, in other words, trying to
optimize one or more objectives that are subject to various constraints. Many solutions have been developed for the employee scheduling problem—and the nurse scheduling prob- lem specifically—during the past 40 years (basically since high- speed computers became available). In a recent survey it was estimated that more than 570 systems were currently available from system software development vendors for the nurse sched- uling problem alone. These systems employed a variety of dif- ferent mathematical modeling solution approaches using the various management science techniques presented in this text.
Source: D. Kellogg and S. Walczak, “Nurse Scheduling: From Academia to Implementation or Not?” Interfaces 37, no. 4 (July–August 2007): 355–69.
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Decision maker
User interface
Internet/e-business
Databases Management science models
Figure 1.7
A decision support system
Management Science Models in Decision Support Systems
Historically, management science models have been applied to the solution of specific types of problems; for example, a waiting line model is used to analyze a specific waiting line sys- tem at a store or bank. However, the evolution of computer and information technology has enabled the development of expansive computer systems that combine several man- agement science models and solution techniques in order to address more complex, interrelated organizational problems. A decision support system (DSS) is a computer- based system that helps decision makers address complex problems that cut across different parts of an organization and operations.
A DSS is normally interactive, combining various databases and different management science models and solution techniques with a user interface that enables the decision maker to ask questions and receive answers. In its simplest form any computer-based soft- ware program that helps a decision maker make a decision can be referred to as a DSS. For example, an Excel spreadsheet like the one shown for break-even analysis in Exhibit 1.1 or the QM for Windows model shown in Exhibit 1.4 can realistically be called a DSS. Alterna- tively, enterprisewide DSSs can encompass many different types of models and large data warehouses, and they can serve many decision makers in an organization. They can provide decision makers with interrelated information and analyses about almost anything in a company.
Figure 1.7 illustrates the basic structure of a DSS with a database component, a model- ing component, and a user interface with the decision maker. As noted earlier, a DSS can be small and singular, with one analytical model linked to a database, or it can be very large and complex, linking many models and large databases. A DSS can be primarily a data- oriented system, or it can be a model-oriented system. A new type of DSS, called an online analytical processing system, or OLAP, focuses on the use of analytical techniques such as management science models and statistics for decision making. A desktop DSS for a single user can be a spreadsheet program such as Excel to develop specific solutions to individual problems. Exhibit 1.1 includes all the components of a DSS—cost, volume, and price data, a break-even model, and the opportunity for the user to manipulate the data and see the results (i.e., a user interface). Expert Choice is another example of a desktop DSS that uses the analytical hierarchy process (AHP) described in Chapter 9 to structure complex problems by establishing decision criteria, developing priorities, and ranking decision alternatives.
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On the other end of the DSS spectrum, an enterprise resource planning (ERP) system is software that can connect the components and functions of an entire company. It can transform data, such as individual daily sales, directly into information that supports immediate decisions in other parts of the company, such as ordering, manufacturing, inventory, and distribution. A large-scale DSS such as an ERP system in a company might include a forecasting model (Chapter 15) to analyze sales data and help determine future product demand; an inventory model (Chapter 16) to determine how much inventory to keep on hand; a linear programming model (Chapters 2–5) to determine how much mate- rial to order and product to produce, and when to produce it; a transportation model (Chapter 6) to determine the most cost-effective method of distributing a product to cus- tomers; and a network flow model (Chapter 7) to determine the best delivery routes. All these different management science models and the data necessary to support them can be linked in a single enterprisewide DSS that can provide many decisions to many different decision makers.
In addition to helping managers answer specific questions and make decisions, a DSS may be most useful in answering What if? questions and performing sensitivity analysis. In other words, a DSS provides a computer-based laboratory to perform experiments.
M a n a g e m e n t S c i e n c e A p p l i c a t i o n A Decision Support System for Aluminum Can Production at Coors
Valley Metal Container (VMC), a joint venture betweenCoors Brewing Company and American National Can, operates the largest single facility for aluminum can produc- tion in the world in Golden, Colorado. The plant, encompass- ing six production lines, manufactures over 4 billion cans per year for seven Coors beer labels produced at Coors breweries in Colorado, Virginia, and Tennessee. The breweries’ weekly pro- duction schedules are unpredictable, and if sufficient cans are not available, brewery fill lines can be shut down, at a cost of $65 per minute. In order to cope with variable brewery demand, VMC builds up inventories of cans for all seven Coors labels during winter and spring, when beer demand is lower, in order to meet higher demand in the summer. Each production line at the can production plant produces multiple labels, and when a line is switched from one label to another, a label change occurs. Finished cans go into either short-term inven- tory (in trailers), from which they are shipped to a brewery within a day, or two types of longer-term inventory, in which cans are stored on pallets for future delivery.
VMC developed a DSS to determine the weekly production schedule for cans that would meet brewery demand while mini- mizing the number of costly label changes and associated inven- tory costs. The DSS includes an Excel-based user interface for data entry. The Excel spreadsheets are linked to a linear program- ming model that develops the optimal production schedule that minimizes the costs associated with label changes while meeting demand. The spreadsheet format allows data to be entered and adjusted easily on one worksheet, and the production schedule is
developed on another worksheet. The production schedule is then reported on another specially formatted spreadsheet, which the user can view, print, and edit. Cost savings with this DSS average about $3,000 per week and annual savings of over $160,000.
Source: E. Katok and D. Ott, “Using Mixed-Integer Program- ming to Reduce Label Changes in the Coors Aluminum Can Plant,” Interfaces 30, no. 2 (March–April 2000): 1–12.
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By linking various management science models together with different databases, a user can change a parameter in one model related to one company function and see what the effect will be in a model related to a different operation in the company. For example, by changing the data in a forecasting model, a manager could see the impact of a hypothetical change in product demand on the production schedule, as determined by a linear programming model.
Advances in information and computer technology have provided the opportunity to apply management science models to a broad array of complex organizational problems by linking different models to databases in a DSS. These advances have also made the applica- tion of management science models more readily available to individual users in the form of desktop DSSs that can help managers make better decisions relative to their day-to-day operations. In the future it will undoubtedly become even easier to apply management science to the solution of problems with the development of newer software, and manage- ment science will become even more important and pervasive as an aid to decision makers as managers are linked within companies with sophisticated computer systems and to other companies via the Internet.
Many companies now interface with new types of DSS over the Internet. In e-business applications, companies can link to other business units around the world through com- puter systems called intranets, with other companies through systems called extranets, and over the Internet. For example, electronic data interchange (EDI) and point-of-sale data (through bar codes) can provide companies with instantaneous records of business trans- actions and sales at retail stores that are immediately entered into a company’s DSS to update inventory and production scheduling, using management science models. Internet transportation exchanges enable companies to arrange cost-effective transportation of their products at Web sites that match shipping loads with available trucks at the lowest cost and fastest delivery speed, using sophisticated management science models.
Summary
In the chapters that follow, the model construction and solutions that constitute each man-agement science technique are presented in detail and illustrated with examples. In fact, the primary method of presenting the techniques is through examples. Thus, the text offers you a broad spectrum of knowledge of the mechanics of management science techniques and the types of problems to which these techniques are applied. However, the ultimate test of a management scientist or a manager who uses management science techniques is the ability to transfer textbook knowledge to the business world. In such instances there is an art to the application of management science, but it is an art predicated on practical experience and sound textbook knowledge. Providing the first of these necessities is beyond the scope of textbooks; providing the second is the objective of this text.
References
Ackoff, Russell L., and Sasieni, Maurice W. Fundamentals of Opera-
tions Research. New York: John Wiley & Sons, 1968.
Beer, Stafford. Management Sciences: The Business Use of Operations
Research. New York: Doubleday, 1967.
Churchman, C. W., Ackoff, R. L., and Arnoff, E. L. Introduction to
Operations Research. New York: John Wiley & Sons, 1957.
Fabrycky, W. J., and Torgersen, P. E. Operations Economy: Industrial
Applications of Operations Research. Upper Saddle River, NJ:
Prentice Hall, 1966.
Hillier, F. S., and Lieberman, G. J. Operations Research, 4th ed. San
Francisco: Holden-Day, 1987.
Management science is an art.
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Problems 23
Problems
1. The Willow Furniture Company produces tables. The fixed monthly cost of production is $8,000, and the variable cost per table is $65. The tables sell for $180 apiece.
a. For a monthly volume of 300 tables, determine the total cost, total revenue, and profit. b. Determine the monthly break-even volume for the Willow Furniture Company.
2. The Retread Tire Company recaps tires. The fixed annual cost of the recapping operation is $60,000. The variable cost of recapping a tire is $9. The company charges $25 to recap a tire.
a. For an annual volume of 12,000 tires, determine the total cost, total revenue, and profit. b. Determine the annual break-even volume for the Retread Tire Company operation.
3. The Rolling Creek Textile Mill produces denim. The fixed monthly cost is $21,000, and the variable cost per yard of denim is $0.45. The mill sells a yard of denim for $1.30.
a. For a monthly volume of 18,000 yards of denim, determine the total cost, total revenue, and profit.
b. Determine the annual break-even volume for the Rolling Creek Textile Mill.
4. Evergreen Fertilizer Company produces fertilizer. The company’s fixed monthly cost is $25,000, and its variable cost per pound of fertilizer is $0.15. Evergreen sells the fertilizer for $0.40 per pound. Determine the monthly break-even volume for the company.
5. Graphically illustrate the break-even volume for the Retread Tire Company determined in Problem 2.
6. Graphically illustrate the break-even volume for the Evergreen Fertilizer Company determined in Problem 4.
7. Andy Mendoza makes handcrafted dolls, which he sells at craft fairs. He is considering mass- producing the dolls to sell in stores. He estimates that the initial investment for plant and equip- ment will be $25,000, whereas labor, material, packaging, and shipping will be about $10 per doll. If the dolls are sold for $30 each, what sales volume is necessary for Andy to break even?
8. If the maximum operating capacity of the Retread Tire Company, as described in Problem 2, is 8,000 tires annually, determine the break-even volume as a percentage of that capacity.
9. If the maximum operating capacity of the Rolling Creek Textile Mill described in Problem 3 is 25,000 yards of denim per month, determine the break-even volume as a percentage of capacity.
10. If the maximum operating capacity of Evergreen Fertilizer Company described in Problem 4 is 120,000 pounds of fertilizer per month, determine the break-even volume as a percentage of capacity.
11. If the Retread Tire Company in Problem 2 changes its pricing for recapping a tire from $25 to $31, what effect will the change have on the break-even volume?
12. If Evergreen Fertilizer Company in Problem 4 changes the price of its fertilizer from $0.40 per pound to $0.60 per pound, what effect will the change have on the break-even volume?
13. If Evergreen Fertilizer Company changes its production process to add a weed killer to the fertilizer in order to increase sales, the variable cost per pound will increase from $0.15 to $0.22. What effect will this change have on the break-even volume computed in Problem 12?
Taha, Hamdy A. Operations Research, An Introduction, 4th ed. New
York: Macmillan, 1987.
Teichroew, P. An Introduction to Management Science. New York:
John Wiley & Sons, 1964.
Wagner, Harvey M. Principles of Management Science. Upper Saddle
River, NJ: Prentice Hall, 1975.
———. Principles of Operations Research. 2nd ed. Upper Saddle
River, NJ: Prentice Hall, 1975.
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24 Chapter 1 Management Science
14. If Evergreen Fertilizer Company increases its advertising expenditures by $14,000 per year, what effect will the increase have on the break-even volume computed in Problem 13?
15. Pastureland Dairy makes cheese, which it sells at local supermarkets. The fixed monthly cost of production is $4,000, and the variable cost per pound of cheese is $0.21. The cheese sells for $0.75 per pound; however, the dairy is considering raising the price to $0.95 per pound. The dairy currently produces and sells 9,000 pounds of cheese per month, but if it raises its price per pound, sales will decrease to 5,700 pounds per month. Should the dairy raise the price?
16. For the doll-manufacturing enterprise described in Problem 7, Andy Mendoza has determined that $10,000 worth of advertising will increase sales volume by 400 dolls. Should he spend the extra amount for advertising?
17. Andy Mendoza in Problem 7 is concerned that the demand for his dolls will not exceed the break-even point. He believes he can reduce his initial investment by purchasing used sewing machines and fewer machines. This will reduce his initial investment from $25,000 to $17,000. However, it will also require his employees to work more slowly and perform more operations by hand, thus increasing variable cost from $10 to $14 per doll. Will these changes reduce his break-even point?
18. The General Store at State University is an auxiliary bookstore located near the dormitories that sells academic supplies, toiletries, sweatshirts and T-shirts, magazines, packaged food items, and canned soft drinks and fruit drinks. The manager of the store has noticed that several pizza delivery services near campus make frequent deliveries. The manager is therefore considering selling pizza at the store. She could buy premade frozen pizzas and heat them in an oven. The cost of the oven and freezer would be $27,000. The frozen pizzas cost $3.75 each to buy from a distributor and to prepare (including labor and a box). To be competitive with the local delivery services, the manager believes she should sell the pizzas for $8.95 apiece. The manager needs to write up a proposal for the university’s director of auxiliary services.
a. Determine how many pizzas would have to be sold to break even. b. If The General Store sells 20 pizzas per day, how many days would it take to break even? c. The manager of the store anticipates that once the local pizza delivery services start
losing business, they will react by cutting prices. If after a month (30 days) the manager has to lower the price of a pizza to $7.95 to keep demand at 20 pizzas per day, as she expects, what will the new break-even point be, and how long will it take the store to break even?
19. Kim Davis has decided to purchase a cellular phone, but she is unsure about which rate plan to select. The “regular” plan charges a fixed fee of $55 per month for 1,000 minutes of airtime plus $0.33 per minute for any time over 1,000 minutes. The “executive” plan charges a fixed fee of $100 per month for 1,200 minutes of airtime plus $0.25 per minute over 1,200 minutes.
a. If Kim expects to use the phone for 21 hours per month, which plan should she select? b. At what level of use would Kim be indifferent between the two plans?
20. Annie McCoy, a student at Tech, plans to open a hot dog stand inside Tech’s football stadium dur- ing home games. There are seven home games scheduled for the upcoming season. She must pay the Tech athletic department a vendor’s fee of $3,000 for the season. Her stand and other equip- ment will cost her $4,500 for the season. She estimates that each hot dog she sells will cost her $0.35. She has talked to friends at other universities who sell hot dogs at games. Based on their information and the athletic department’s forecast that each game will sell out, she anticipates that she will sell approximately 2,000 hot dogs during each game.
a. What price should she charge for a hot dog in order to break even? b. What factors might occur during the season that would alter the volume sold and thus the
break-even price Annie might charge?
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Problems 25
c. What price would you suggest that Annie charge for a hot dog to provide her with a reasonable profit while remaining competitive with other food vendors?
21. Molly Dymond and Kathleen Taylor are considering the possibility of teaching swimming to kids during the summer. A local swim club opens its pool at noon each day, so it is available to rent dur- ing the morning. The cost of renting the pool during the 10-week period for which Molly and Kathleen would need it is $1,700. The pool would also charge Molly and Kathleen an admission, towel service, and life guarding fee of $7 per pupil, and Molly and Kathleen estimate an additional $5 cost per student to hire several assistants. Molly and Kathleen plan to charge $75 per student for the 10-week swimming class.
a. How many pupils do Molly and Kathleen need to enroll in their class to break even? b. If Molly and Kathleen want to make a profit of $5,000 for the summer, how many pupils do
they need to enroll? c. Molly and Kathleen estimate that they might not be able to enroll more than 60 pupils. If
they enroll this many pupils, how much would they need to charge per pupil in order to realize their profit goal of $5,000?
22. The College of Business at Tech is planning to begin an online MBA program. The initial start-up cost for computing equipment, facilities, course development, and staff recruitment and develop- ment is $350,000. The college plans to charge tuition of $18,000 per student per year. However, the university administration will charge the college $12,000 per student for the first 100 students enrolled each year for administrative costs and its share of the tuition payments.
a. How many students does the college need to enroll in the first year to break even? b. If the college can enroll 75 students the first year, how much profit will it make? c. The college believes it can increase tuition to $24,000, but doing so would reduce
enrollment to 35. Should the college consider doing this?
23. The Star Youth Soccer Club helps to support its 20 boys’ and girls’ teams financially, primarily through the payment of coaches. The club puts on a tournament each fall to help pay its expenses. The cost of putting on the tournament is $8,000, mainly for development, printing, and mailing of the tournament brochures. The tournament entry fee is $400 per team. For every team that enters, it costs the club about $75 to pay referees for the three-game minimum each team is guaranteed. If the club needs to clear $60,000 from the tournament, how many teams should it invite?
24. In the example used to demonstrate model construction in this chapter (p. 4), a firm sells a prod- uct, x, for $20 that costs $5 to make, it has 100 pounds of steel to make the product, and it takes 4 pounds of steel to make each unit. The model that was constructed is
subject to
Now suppose that there is a second product, y, that has a profit of $10 and requires 2 pounds of steel to make, such that the model becomes
subject to
Can you determine a solution to this new model that will achieve the objective? Explain your answer.
25. Consider a model in which two products, x and y, are produced. There are 100 pounds of material and 80 hours of labor available. It requires 2 pounds of material and 1 hour of labor to produce a
4x + 2y = 100
maximize Z = 15x + 10y
4x = 100
maximize Z = 15x
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26 Chapter 1 Management Science
Month 1 2 3 4 5 6
Site visits 6,300 10,200 14,700 18,500 25,100 30,500
Registers staffed 1 2 3 4 5 6 7 8
Waiting time (mins) 20.0 14.0 9.0 4.0 1.7 1.0 0.5 0.1
unit of x, and 4 pounds of material and 5 hours of labor to produce a unit of y. The profit for x is $30 per unit, and the profit for y is $50 per unit. If we want to know how many units of x and y to produce to maximize profit, the model is
subject to
Determine the solution to this problem and explain your answer.
26. The Easy Drive Car Rental Agency needs 500 new cars in its Nashville operation and 300 new cars in Jacksonville, and it currently has 400 new cars in both Atlanta and Birmingham. It costs $30 to move a car from Atlanta to Nashville, $70 to move a car from Atlanta to Jacksonville, $40 to move a car from Birmingham to Nashville, and $60 to move a car from Birmingham to Jacksonville. The agency wants to determine how many cars should be transported from the agencies in Atlanta and Birmingham to the agencies in Nashville and Jacksonville in order to meet demand while minimiz- ing the transport costs. Develop a mathematical model for this problem and use logic to determine a solution.
27. Ed Norris has developed a Web site for his used textbook business at State University. To sell adver- tising he needs to forecast the number of site visits he expects in the future. For the past 6 months he has had the following number of site visits:
x + 5y = 80 2x + 4y = 100
maximize Z = 30x + 50y
Determine a forecast for Ed to use for month 7 and explain the logic used to develop your forecast.
28. When Marie McCoy wakes up on Saturday morning, she remembers that she promised the PTA she would make some cakes and/or homemade bread for its bake sale that afternoon. However, she does not have time to go to the store and get ingredients, and she has only a short time to bake things in her oven. Because cakes and breads require different baking temperatures, she cannot bake them simultaneously, and she has only 3 hours available to bake. A cake requires 3 cups of flour, and a loaf of bread requires 8 cups; Marie has 20 cups of flour. A cake requires 45 minutes to bake, and a loaf of bread requires 30 minutes. The PTA will sell a cake for $10 and a loaf of bread for $6. Marie wants to decide how many cakes and loaves of bread she should make. Identify all the possible solutions to this problem (i.e., combinations of cakes and loaves of bread Marie has the time and flour to bake) and select the best one.
29. The local Food King grocery store has eight possible checkout stations with registers. On Saturday mornings customer traffic is relatively steady from 8 A.M. to noon. The store manager would like to determine how many checkout stations to staff during this time period. The manager knows from information provided by the store’s national office that each minute past 3 minutes a customer must wait in line costs the store on average $50 in ill will and lost sales. Alternatively, each addi- tional checkout station the store operates on Saturday morning costs the store $60 in salary and benefits. The following table shows the waiting time for the different staff levels.
How many registers should the store staff and why?
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Case Problems 27
Staunton
Charlottesville
Front Royal
Roanoke
Danville
4
3
2
5
7
8
9
6
1
Washington, DC
Richmond Lynchburg
76
85
53
88 65 137
117 24
61 72
31
97
106
67
Petersburg
30. A furniture manufacturer in Roanoke, Virginia, must deliver a tractor trailer load of furniture to a retail store in Washington, DC. There are a number of different routes the truck can take from Roanoke to DC, as shown in the following road network, with the distance for each segment shown in miles.
Case Problem
The Clean Clothes Corner Laundry
When Molly Lai purchased the Clean Clothes Corner Laun-dry, she thought that because it was in a good location near several high-income neighborhoods, she would automatically generate good business if she improved the laundry’s physical appearance. Thus, she initially invested a lot of her cash reserves in remodeling the exterior and interior of the laundry. However, she just about broke even in the year following her acquisition of the laundry, which she didn’t feel was a sufficient return, given how hard she had worked. Molly didn’t realize that the dry-cleaning business is very competitive and that success is based more on price and quality service, including quickness of service, than on the laundry’s appearance.
In order to improve her service, Molly is considering purchas- ing new dry-cleaning equipment, including a pressing machine that could substantially increase the speed at which she can dry- clean clothes and improve their appearance. The new machinery costs $16,200 installed and can clean 40 clothes items per hour (or 320 items per day). Molly estimates her variable costs to be $0.25
per item dry-cleaned, which will not change if she purchases the new equipment. Her current fixed costs are $1,700 per month. She charges customers $1.10 per clothing item.
A. What is Molly’s current monthly volume? B. If Molly purchases the new equipment, how many additional
items will she have to dry-clean each month to break even? C. Molly estimates that with the new equipment she can
increase her volume to 4,300 items per month. What monthly profit would she realize with that level of business during the next 3 years? After 3 years?
D. Molly believes that if she doesn’t buy the new equipment but lowers her price to $0.99 per item, she will increase her business volume. If she lowers her price, what will her new break-even volume be? If her price reduction results in a monthly volume of 3,800 items, what will her monthly profit be?
E. Molly estimates that if she purchases the new equipment and lowers her price to $0.99 per item, her volume will increase to about 4,700 units per month. Based on the local market, that is the largest volume she can realistically expect. What should Molly do?
Determine the shortest route the truck can take from Roanoke to Washington, DC.
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28 Chapter 1 Management Science
Case Problem
Constructing a Downtown Parking Lot in Draper
The town of Draper, with a population of 20,000, sits adjacent toState University, which has an enrollment of 27,000 students. Downtown Draper merchants have long complained about the lack of parking available to their customers. This is one primary reason for the steady migration of downtown businesses to a mall several miles outside town. The local chamber of commerce has finally convinced the town council to consider the construction of a new multilevel indoor parking facility downtown. Kelly Mattingly, the town’s public works director, has developed plans for a facility that would cost $4.5 million to construct. To pay for the project, the town would sell municipal bonds with a duration of 30 years at 8%
interest. Kelly also estimates that five employees would be required to operate the lot on a daily basis, at a total annual cost of $140,000. It is estimated that each car that enters the lot would park for an average of 2.5 hours and pay an average fee of $3.20. Further, it is estimated that each car that parks in the lot would (on average) cost the town $0.60 in annual maintenance for cleaning and repairs to the facility. Most of the downtown businesses (which include a number of restaurants) are open 7 days per week.
A. Using break-even analysis, determine the number of cars that would have to park in the lot on an annual basis to pay off the project in the 30-year time frame.
B. From the results in (A), determine the approximate num- ber of cars that would have to park in the lot on a daily basis. Does this seem to be a reasonable number to achieve, given the size of the town and college population?
Case Problem
The Ocobee River Rafting Company
Vicki Smith, Penny Miller, and Darryl Davis are students at StateUniversity. In the summer they often go rafting with other stu- dents down the Ocobee River in the nearby Blue Ridge Mountain foothills. The river has a number of minor rapids but is not gener- ally dangerous. The students’ rafts basically consist of large rubber tubes, sometimes joined together with ski rope. They have noticed that a number of students who come to the river don’t have rubber rafts and often ask to borrow theirs, which can be very annoying. In discussing this nuisance, it occurred to Vicki, Penny, and Darryl that the problem might provide an opportunity to make some extra money. They considered starting a new enterprise, the Ocobee River Rafting Company, to sell rubber rafts at the river. They determined that their initial investment would be about $3,000 to rent a small parcel of land next to the river on which to make and sell the rafts; to purchase a tent to operate out of; and to buy some small equip- ment such as air pumps and a rope cutter. They estimated that the labor and material cost per raft will be about $12, including the purchase and shipping costs for the rubber tubes and rope. They plan to sell the rafts for $20 apiece, which they think is about the maximum price students will pay for a preassembled raft.
Soon after they determined these cost estimates, the newly formed company learned about another rafting company in North Carolina that was doing essentially what they planned to
do. Vicki got in touch with one of the operators of that company, and he told her the company would be willing to supply rafts to the Ocobee River Rafting Company for an initial fixed fee of $9,000 plus $8 per raft, including shipping. (The Ocobee River Rafting Company would still have to rent the parcel of riverside land and tent for $1,000.) The rafts would already be inflated and assembled. This alternative appealed to Vicki, Penny, and Darryl because it would reduce the amount of time they would have to work pumping up the tubes and putting the rafts together, and it would increase time for their schoolwork.
Although the students prefer the alternative of purchasing the rafts from the North Carolina company, they are concerned about the large initial cost and worried about whether they will lose money. Of course, Vicki, Penny, and Darryl realize that their profit, if any, will be determined by how many rafts they sell. As such, they believe that they first need to determine how many rafts they must sell with each alternative in order to make a profit and which alternative would be best given different levels of demand. Furthermore, Penny has conducted a brief sample survey of people at the river and estimates that demand for rafts for the summer will be around 1,000 rafts.
Perform an analysis for the Ocobee River Rafting Company to determine which alternative would be best for different levels of demand. Indicate which alternative should be selected if demand is approximately 1,000 rafts and how much profit the company would make.
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488
C h a p t e r 11
Probability and Statistics
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Types of Probability 489
The techniques presented in Chapters 2 through 10 are typically thought of asdeterministic; that is, they are not subject to uncertainty or variation. With determin- istic techniques, the assumption is that conditions of complete certainty and perfect knowledge of the future exist. In the linear programming models presented in previous chapters, the various parameters of the models and the model results were assumed to be known with certainty. In the model constraints, we did not say that a bowl would require 4 pounds of clay “70% of the time.” We specifically stated that each bowl would require exactly 4 pounds of clay (i.e., there was no uncertainty in our problem statement). Similarly, the solutions we derived for the linear programming models contained no vari- ation or uncertainty. It was assumed that the results of the model would occur in the future, without any degree of doubt or chance.
In contrast, many of the techniques in management science do reflect uncertain infor- mation and result in uncertain solutions. These techniques are said to be probabilistic. This means that there can be more than one outcome or result to a model and that there is some doubt about which outcome will occur. The solutions generated by these techniques have a probability of occurrence. They may be in the form of averages; the actual values that occur will vary over time.
Many of the upcoming chapters in this text present probabilistic techniques. The pre- sentation of these techniques requires that the reader have a fundamental understanding of probability. Thus, the purpose of this chapter is to provide an overview of the fundamen- tals, properties, and terminology of probability and statistics.
Types of Probability
Two basic types of probability can be defined: objective probability and subjective probabil- ity. First, we will consider what constitutes objective probability.
Objective Probability Consider a referee’s flipping a coin before a football game to determine which team will kick off and which team will receive. Before the referee tosses the coin, both team captains know that they have a .50 (or 50%) probability (or chance) of winning the toss. None of the onlookers in the stands or anywhere else would argue that the probability of a head or a tail was not .50. In this example, the probability of .50 that either a head or a tail will occur when a coin is tossed is called an objective probability. More specifically, it is referred to as a classical or a priori (prior to the occurrence) probability, one of the two types of objective probabilities.
A classical, or a priori, probability can be defined as follows: Given a set of outcomes for an activity (such as a head or a tail when a coin is tossed), the probability of a specific (desired) outcome (such as a head) is the ratio of the number of specific outcomes to the total number of outcomes. For example, in our coin-tossing example, the probability of a head is the ratio of the number of specific outcomes (a head) to the total number of out- comes (a head and a tail), or 1/2. Similarly, the probability of drawing an ace from a deck of 52 cards would be found by dividing 4 (the number of aces) by 52 (the total number of cards in a deck) to get 1/13. If we spin a roulette wheel with 50 red numbers and 50 black num- bers, the probability of the wheel’s landing on a red number is 50 divided by 100, or 1/2.
These examples are referred to as a priori probabilities because we can state the proba- bilities prior to the actual occurrence of the activity (i.e., ahead of time). This is because we know (or assume we know) the number of specific outcomes and total outcomes prior to the occurrence of the activity. For example, we know that a deck of cards consists of four
Deterministic techniques assume that no uncertainty exists in model
parameters.
Probabilistic techniques include uncertainty and assume that there
can be more than one model solution.
Objective probabilities that can be stated prior to the occurrence of an
event are classical, or a priori, probabilities.
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490 Chapter 11 Probability and Statistics
aces and 52 total cards before we draw a card from the deck and that a coin contains one head and one tail before we toss it. These probabilities are also known as classical probabil- ities because some of the earliest references in history to probabilities were related to games of chance, to which (as the preceding examples show) these probabilities readily apply.
The second type of objective probability is referred to as relative frequency probability. This type of objective probability indicates the relative frequency with which a specific outcome has been observed to occur in the long run. It is based on the observation of past occurrences. For example, suppose that over the past 4 years, 3,000 business students have taken the introductory management science course at State University, and 300 of them have made an A in the course. The relative frequency probability of making an A in man- agement science would be 300/3,000 or .10. Whereas in the case of a classical probability we indicate a probability before an activity (such as tossing a coin) takes place, in the case of a relative frequency we determine the probability after observing, for example, what 3,000 students have done in the past.
The relative frequency definition of probability is more general and widely accepted than the classical definition. Actually, the relative frequency definition can encompass the classical case. For example, if we flip a coin many times, in the long run the relative fre- quency of a head’s occurring will be .50. If, however, you tossed a coin 10 times, it is con- ceivable that you would get 10 consecutive heads. Thus, the relative frequency (probability) of a head would be 1.0. This illustrates one of the key characteristics of a relative frequency probability: The relative frequency probability becomes more accurate as the total number of observations of the activity increases. If a coin were tossed about 350 times, the relative frequency would approach 1/2 (assuming a fair coin).
Subjective Probability When relative frequencies are not available, a probability is often determined anyway. In these cases a person must rely on personal belief, experience, and knowledge of the situa- tion to develop a probability estimate. A probability estimate that is not based on prior or past evidence is a subjective probability. For example, when a meteorologist forecasts a “60% chance of rain tomorrow,” the .60 probability is usually based on the meteorologist’s experience and expert analysis of the weather conditions. In other words, the meteorolo- gist is not saying that these exact weather conditions have occurred 1,000 times in the past and on 600 occasions it has rained, thus there is a 60% probability of rain. Likewise, when a sportswriter says that a football team has an 80% chance of winning, it is usually not because the team has won 8 of its 10 previous games. The prediction is judgmental, based on the sportswriter’s knowledge of the teams involved, the playing conditions, and so forth. If the sportswriter had based the probability estimate solely on the team’s relative frequency of winning, then it would have been an objective probability. However, once the relative frequency probability becomes colored by personal belief, it is subjective.
Subjective probability estimates are frequently used in making business decisions. For example, suppose the manager of Beaver Creek Pottery Company (referred to in Chapters 2 and 3) is thinking about producing plates in addition to the bowls and mugs it already produces. In making this decision, the manager will determine the chances of the new product’s being successful and returning a profit. Although the manager can use personal knowledge of the market and judgment to determine a probability of success, direct relative frequency evidence is not generally available. The manager cannot observe the frequency with which the introduction of this new product was successful in the past. Thus, the man- ager must make a subjective probability estimate.
This type of subjective probability analysis is common in the business world. Decision makers frequently must consider their chances for success or failure, the probability of
Subjective probability is an estimate based on personal belief,
experience, or knowledge of a situation.
Objective probabilities that are stated after the outcomes of an
event have been observed are relative frequency probabilities.
Relative frequency is the more widely used definition of objective
probability.
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Fundamentals of Probability 491
achieving a certain market share or profit, the probability of a level of demand, and the like without the benefit of relative frequency probabilities based on past observations. Although there may not be a consensus as to the accuracy of a subjective estimate, as there is with an objective probability (e.g., everyone is sure there is a .50 probability of getting a head when a coin is tossed), subjective probability analysis is often the only means avail- able for making probabilistic estimates, and it is frequently used.
A brief note of caution must be made regarding the use of subjective probabilities. Different people will often arrive at different subjective probabilities, whereas everyone should arrive at the same objective probability, given the same numbers and correct calcu- lations. Therefore, when a probabilistic analysis is to be made of a situation, the use of objective probability will provide more consistent results. In the material on probability in the remainder of this chapter, we will use objective probabilities unless the text indicates otherwise.
Fundamentals of Probability
Let us return to our example of a referee’s tossing a coin prior to a football game. In the ter- minology of probability, the coin toss is referred to as an experiment. An experiment is an activity (such as tossing a coin) that results in one of the several possible outcomes. Our coin-tossing experiment can result in either one of two outcomes, which are referred to as events: a head or a tail. The probabilities associated with each event in our experiment follow:
Event Probability Head .50 Tail .50
1.00
This simple example highlights two of the fundamental characteristics of probability. First, the probability of an event is always greater than or equal to zero and less than or equal to one [i.e., (event) ]. In our coin-tossing example, each event has a probability of .50, which is in the range of 0 to 1.0. Second, the probability of all the events included in an experiment must sum to one. Notice in our example that the probability of each of the two events is .50, and the sum of these two probabilities is 1.0.
The specific example of tossing a coin also exhibits a third characteristic: The events in a set of events are mutually exclusive. The events in an experiment are mutually exclusive if only one of them can occur at a time. In the context of our experiment, the term mutually exclusive means that any time the coin is tossed, only one of the two events can take place— either a head or a tail can occur, but not both. Consider a customer who enters a store to shop for shoes. The store manager estimates that there is a .60 probability that the customer will buy a pair of shoes and a .40 probability that the customer will not buy a pair of shoes. These two events are mutually exclusive because it is impossible to buy shoes and not buy shoes at the same time. In general, events are mutually exclusive if only one of the events can occur, but not both.
Because the events in our example of obtaining a head or tail are mutually exclusive, we can infer that the probabilities of mutually exclusive events sum to 1.0. Also, the probabili- ties of mutually exclusive events can be added. The following example will demonstrate these fundamental characteristics of probability.
The staff of the dean of the business school at State University has analyzed the records of the 3,000 students who received a grade in management science during the past 4 years.
… 1.00 … P
An experiment is an activity that results in one of several possible
outcomes.
Two fundamentals of probability: (events) and the
probabilities of all events in an experiment sum to one.
… 1.0,0 … P
The events in an experiment are mutually exclusive if only one
can occur at a time.
The probabilities of mutually exclusive events sum to one.
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492 Chapter 11 Probability and Statistics
The dean wants to know the number of students who made each grade (A, B, C, D, or F) in the course. The dean’s staff has developed the following table of information:
Event Grade Number of Students Relative Frequency Probability
A 300 300/3,000 .10
B 600 600/3,000 .20
C 1,500 1,500/3,000 .50
D 450 450/3,000 .15
F 150 150/3,000 .05
3,000 1.00
This example demonstrates several of the characteristics of probability. First, the data (numerical information) in the second column show how the students are distributed across the different grades (events). The third column shows the relative frequency with which each event occurred for the 3,000 observations. In other words, the relative frequency of a student’s making a C is 1,500/3,000, which also means that the probability of selecting a student who had obtained a C at random from those students who took management science in the past 4 years is .50.
This information, organized according to the events in the experiment, is called a frequency distribution. The list of the corresponding probabilities for each event in the last column is referred to as a probability distribution.
All the events in this example are mutually exclusive; it is not possible for two or more of these events to occur at the same time. A student can make only one grade in the course, not two or more grades. As indicated previously, mutually exclusive probabilities of an experiment can be summed to equal one. There are five mutually exclusive events in this experiment, the probabilities of which (.10, .20, .50, .15, and .05) sum to one.
This example exhibits another characteristic of probability: Because the five events in the example are all that can occur (i.e., no other grade in the course is possible), the experiment is said to be collectively exhaustive. Likewise, the coin-tossing experiment is collectively exhaus- tive because the only two events that can occur are a head and a tail. In general, when a set of events includes all the events that can possibly occur, the set is said to be collectively exhaustive.
The probability of a single event occurring, such as a student receiving an A in a course, is represented symbolically as P(A). This probability is called the marginal probability in the terminology of probability. For our example, the marginal probability of a student’s getting an A in management science is
For mutually exclusive events, it is possible to determine the probability that one or the other of several events will occur. This is done by summing the individual marginal prob- abilities of the events. For example, the probability of a student receiving an A or a B is determined as follows:
In other words, 300 students received an A and 600 students received a B; thus, the num- ber of students who received an A or a B is 900. Dividing 900 students who received an A or a B by the total number of students, 3,000, yields the probability of a student’s receiving an A or a B [i.e., ].
Mutually exclusive events can be shown pictorially with a Venn diagram. Figure 11.1 shows a Venn diagram for the mutually exclusive events A and B in our example.
P1A or B2 = .30
P(A or B) = P(A) + P1B2 = .10 + .20 = .30
P1A2 = .10
A frequency distribution is an organization of numeric data
about the events in an experiment.
A probability distribution shows the probability of occurrence of all
events in an experiment.
A set of events is collectively exhaustive when it includes all the events that can occur in an
experiment.
A Venn diagram visually displays mutually exclusive and
non–mutually exclusive events.
A marginal probability is the probability of a single event
occurring.
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Fundamentals of Probability 493
Now let us consider a case in which two events are not mutually exclusive. In this case the probability that A or B or both will occur is expressed as
where the term P(AB), referred to as the joint probability of A and B, is the probability that both A and B will occur. For mutually exclusive events, this term has to equal zero because both events cannot occur together. Thus, for mutually exclusive events our formula becomes
which is the same formula we developed previously for mutually exclusive events.
= P(A) + P1B2 = P(A) + P(B) - 0
P(A or B) = P(A) + P(B) - P(AB)
P(A or B) = P(A) + P(B) - P1AB2 A joint probability is the
probability that two or more events that are not mutually
exclusive can occur simultaneously.
A B
Figure 11.1
Venn diagram for mutually exclusive events
M a n a g e m e n t S c i e n c e A p p l i c a t i o n Treasure Hunting with Probability and Statistics
In 1857 the SS Central America, a wooden-hulled steamship,sank during a hurricane approximately 20 miles off the coast of Charleston, South Carolina, in the Atlantic Ocean. A total of 425 passengers and crew members lost their lives, and approx- imately $400 million in gold bars and coins sank to the ocean bottom approximately 8,000 feet (i.e., 1.5 miles) below the sur- face. This shipwreck was the most famous of its time, causing a financial panic in the United States and a recommendation from insurance investigators that new ships be built with iron hulls and watertight compartments.
In 1985 the Columbus-America Discovery Group was formed to locate and recover the remains of the SS Central America. A search plan was developed, using statistical tech- niques and Monte Carlo simulation. From the available histor- ical information about the wreck, three scenarios were con- structed, describing different search areas based on different accounts from nearby ships and witnesses. Using historical sta- tistical data on weather, winds and currents in the area, and estimates of uncertainties in celestial navigation, probability distributions were developed for different possible sites where the ship might have sunk. Using Monte Carlo simulation, a map was configured for each scenario, consisting of cells repre- senting locations. Each cell was assigned a probability that the ship sank in that location. The probability maps for the three scenarios were combined into a composite probability map, which provided a guide for searching the ocean bottom using
sonar. The search consisted of long, straight paths that concen- trated on the high-probability areas. Based on the results of the sonar search of 1989, the discovery group recovered 1 ton of gold bars and coins from the wreck. In 1993 the total gold recovered from the wreck was valued at $21 million. However, because of the historical significance of all recovered items, they were expected to bring a much higher return when sold.
Source: L. D. Stone, “Search for the SS Central America: Mathematical Treasure Hunting,” Interfaces 22, no. 1 (January– February 1992): 32–54.
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494 Chapter 11 Probability and Statistics
The following example will illustrate the case in which two events are not mutually exclusive. Suppose it has been determined that 40% of all students in the school of busi- ness are at present taking management, and 30% of all the students are taking finance. Also, it has been determined that 10% take both subjects. Thus, our probabilities are
The probability of a student’s taking one or the other or both of the courses is deter- mined as follows:
Observing this formulation closely, we can see why the joint probability, P(MF), was sub- tracted out. The 40% of the students who were taking management also included those stu- dents taking both courses. Likewise, the 30% of the students taking finance also included those students taking both courses. Thus, if we add the two marginal probabilities, we are double-counting the percentage of students taking both courses. By subtracting out one of these probabilities (that we added in twice), we derive the correct probability.
Figure 11.2 contains a Venn diagram that shows the two events, M and F, that are not mutually exclusive, and the joint event, MF.
P(M or F) = P(M) + P(F) - P1MF2 = .40 + .30 - .10 = .60
P1MF2 = .10 P(F) = .30
P(M) = .40
An alternative way to construct a probability distribution is to add the probability of an event to the sum of all previously listed probabilities in a probability distribution. Such a list is referred to as a cumulative probability distribution. The cumulative probability dis- tribution for our management science grade example is as follows:
Event Grade Probability Cumulative Probability
A .10 .10
B .20 .30
C .50 .80
D .15 .95
F .05 1.00
1.00
The value of a cumulative probability distribution is that it organizes the event proba- bilities in a way that makes it easier to answer certain questions about the probabilities. For example, if we want to know the probability that a student will get a grade of C or higher, we can add the probabilities of the mutually exclusive events A, B, and C:
Or we can look directly at the cumulative probability distribution and see that the prob- ability of a C and the events preceding it in the distribution (A and B) equals .80. Alternatively, if we want to know the probability of a grade lower than C, we can subtract the cumulative probability of a C from 1.00 (i.e., ).1.00 - .80 = .20
P(A or B or C) = P(A) + P(B) + P1C2 = .10 + .20 + .50 = .80
M F
MF
Figure 11.2
Venn diagram for non–mutually exclusive events and the joint
event
In a cumulative probability distribution the probability of an
event is added to the sum of all previously listed probabilities in a
distribution.
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Statistical Independence and Dependence 495
Statistical Independence and Dependence
Statistically, events are either independent or dependent. If the occurrence of one event does not affect the probability of the occurrence of another event, the events are independent. Conversely, if the occurrence of one event affects the probability of the occur- rence of another event, the events are dependent. We will first turn our attention to a dis- cussion of independent events.
Independent Events When we toss a coin, the two events—getting a head and getting a tail—are independent. If we get a head on the first toss, this result has absolutely no effect on the probability of getting a head or a tail on the next toss. The probability of getting either a head or a tail will still be .50, regardless of the outcomes of previous tosses. In other words, the two events are independent events.
When events are independent, it is possible to determine the probability that both events will occur in succession by multiplying the probabilities of each event. For example, what is the probability of getting a head on the first toss and a tail on the second toss? The answer is
P(HT) = P(H) • P(T) where
probability of a head
probability of a tail
joint probability of a head and a tail
Therefore,
P(HT) = P(H) • P(T) = (.5)(.5) = .25
As we indicated previously, the probability of both events occurring, P(HT), is referred to as the joint probability.
Another property of independent events relates to conditional probabilities. A condi- tional probability is the probability that event A will occur given that event B has already occurred. This relationship is expressed symbolically as
The term in parentheses, “A slash B,” means “A, given the occurrence of B.” Thus, the entire term is interpreted as the probability that A will occur, given that B has already occurred. If A and B are independent events, then
In words, this result says that if A and B are independent, then the probability of A, given the occurrence of event B, is simply equal to the probability of A. Because the events are independent of each other, the occurrence of event B will have no effect on the occurrence of A. Therefore, the probability of A is in no way dependent on the occurrence of B.
In summary, if events A and B are independent, the following two properties hold:
1. P(AB) = P(A) • P(B) 2. P(A|B) = P(A)
P(A|B) = P1A2
P1A|B2
P1A|B2
P1HT2 = P1T2 = P1H2 =
The probability that independent events will occur in succession is
computed by multiplying the probabilities of each event.
A conditional probability is the probability that an event will
occur, given that another event has already occurred.
A succession of events that do not affect each other are independent
events.
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496 Chapter 11 Probability and Statistics
Probability Trees Consider an example in which a coin is tossed three consecutive times. The possible out- comes of this example can be illustrated by using a probability tree, as shown in Figure 11.3.
The probability tree in Figure 11.3 demonstrates the probabilities of the various occur- rences, given three tosses of a coin. Notice that at each toss, the probability of either event’s occurring remains the same: Thus, the events are independent. Next, the joint probabilities that events will occur in succession are computed by multiplying the probabilities of all the events. For example, the probability of getting a head on the first toss, a tail on the second, and a tail on the third is .125:
P(HTT) = P(H) • P(T) • P(T) = (.5)(.5)(.5) = .125
However, do not confuse the results in the probability tree with conditional probabili- ties. The probability that a head and then two tails will occur on three consecutive tosses is computed prior to any tosses. If the first two tosses have already occurred, the probability of getting a tail on the third toss is still .5:
The Binomial Distribution Some additional information can be drawn from the probability tree of our example. For instance, what is the probability of achieving exactly two tails on three tosses? The answer can be found by observing the instances in which two tails occurred. It can be seen that two tails in three tosses occurred three times, each time with a probability of .125. Thus, the probability of getting exactly two tails in three tosses is the sum of these three probabilities, or .375. The use of a probability tree can become very cumbersome, especially if we are considering an example with 20 tosses. However, the example of tossing a coin exhibits cer- tain properties that enable us to define it as a Bernoulli process. The properties of a Bernoulli process follow:
1. There are two possible outcomes for each trial (i.e., each toss of a coin). Outcomes can be success or failure, yes or no, heads or tails, good or bad, and so on.
2. The probability of the outcomes remains constant over time. In other words, the probability of getting a head on a coin toss remains the same, regardless of the num- ber of tosses.
P1T|HT2 = P1T2 = .5
P(H) = P1T2 = .5.
P(H) = .5
P(T) = .5
P(H) = .5
P(T) = .5
P(T) = .5
P(H) = .5
P(H) = .5
P(T) = .5
P(H) = .5
P(T) = .5
P(H) = .5
P(T) = .5
P(H) = .5
P(T) = .5 .125 (TTT)
.125 (THT)
.125 (TTH)
.125 (HTT)
.125 (THH)
.125 (HHT)
.125 (HTH)
.125 (HHH)
.25
.25
.25
.25
.5
.5
Figure 11.3
Probability tree for coin-tossing example
Properties of a Bernoulli process.
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Statistical Independence and Dependence 497
3. The outcomes of the trials are independent. The fact that we get a head on the first toss does not affect the probabilities on subsequent tosses.
4. The number of trials is discrete and an integer. The term discrete indicates values that are countable and, thus, usually an integer—for example, 1 car or 2 people rather than 1.34 cars or 2.51 people. There are 1, 2, 3, 4, 5, . . . tosses of the coin, not 3.36 tosses.
Given the properties of a Bernoulli process, a binomial distribution function can be used to determine the probability of a number of successes in n trials. The binomial distri- bution is an example of a discrete distribution because the value of the distribution (the number of successes) is discrete, as is the number of trials. The formula for the binomial distribution is
where probability of a success
probability of a failure number of trials number of successes in n trials
The terms n!, (n – r)!, and r! are called factorials. Factorials are computed using the formula
m! = m(m – 1)(m – 2)(m – 3) . . . (2)(1)
The factorial 0! always equals one. The binomial distribution formula may look complicated, but using it is not difficult.
For example, suppose we want to determine the probability of getting exactly two tails in three tosses of a coin. For this example, getting a tail is a success because it is the object of the analysis. The probability of a tail, p, equals .5; therefore, The number of tosses, n, is 3, and the number of tails, r, is 2. Substituting these values into the binomial formula will result in the probability of two tails in three coin tosses:
Notice that this is the same result achieved by using a probability tree in the previous section.
Now let us consider an example of more practical interest. An electrical manufacturer produces microchips. The microchips are inspected at the end of the production process at a quality control station. Out of every batch of microchips, four are randomly selected and tested for defects. Given that 20% of all transistors are defective, what is the probability that each batch of microchips will contain exactly two defective microchips?
The two possible outcomes in this example are a good microchip and a defective microchip. Because defective microchips are the object of our analysis, a defective item is a
P1r = 22 = .375 =
6
2 (.125)
= (3 # 2 # 1) (2 # 1)(1) (.25)(.5)
P(2 tails) = P(r = 2) = 3!
2!(3 - 2)! (.5)2(.5)3 - 2
q = 1 - .5 = .5.
r = n = q = 1 - p = p =
P1r2 = n! r!1n - r2! p
rqn - r
A binomial distribution indicates the probability of r successes in n
trials.
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498 Chapter 11 Probability and Statistics
success. The probability of a success is the probability of a defective microchip, or The number of trials, n, equals 4. Now let us substitute these values into the binomial formula:
Thus, the probability of getting exactly two defective items out of four microchips is .1536. Now, let us alter this problem to make it even more realistic. The manager has deter-
mined that four microchips from every large batch should be tested for quality. If two or more defective microchips are found, the whole batch will be rejected. The manager wants to know the probability of rejecting an entire batch of microchips, if, in fact, the batch has 20% defective items.
From our previous use of the binomial distribution, we know that it gives us the prob- ability of an exact number of integer successes. Thus, if we want the probability of two or more defective items, it is necessary to compute the probability of two, three, and four defective items:
Substituting the values and and 4 into the binomial dis- tribution results in the probability of two or more defective items:
Thus, the probability that a batch of microchips will be rejected due to poor quality is .1808.
Notice that the collectively exhaustive set of events for this example is 0, 1, 2, 3, and 4 defective transistors. Because the sum of the probabilities of a collectively exhaustive set of events equals 1.0,
Recall that the results of the immediately preceding example show that
Given this result, we can compute the probability of “less than two defectives” as follows:
Although our examples included very small values for n and r that enabled us to work out the examples by hand, problems containing larger values for n and r can be solved eas- ily by using an electronic calculator.
= .8192 = 1.0 - .1808 = 1.0 - [P1r = 22 + P1r = 32 + P1r = 42]
P1r 6 22 = P(r = 0) + P(r = 1)
P1r = 22 + P1r = 32 + P1r = 42 = .1808
P1r = 0, 1, 2, 3, 42 = P(r = 0) + P(r = 1) + P(r = 2) + P(r = 3) + P(r = 4) = 1.0
= .1808 = .1536 + .0256 + .0016
P1r Ú 22 = 4! 2!14 - 22!1.22
21.822 + 4! 3!14 - 32!1.22
31.821 + 4! 4!14 - 42!1.22
41.820
r = 2, 3,p = .2, n = 4, q = .8,
P1r Ú 22 = P1r = 22 + P1r = 32 + P1r = 42
= .1536
= 24
4 1.02562
= (4 # 3 # 2 # 1) (2 # 1)(2 # 1)(.04)(.64)
P(r = defectives) = 4!
2!(4 - 2)! (.2)2(.8)2
p = .2.
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Statistical Independence and Dependence 499
Dependent Events As stated earlier, if the occurrence of one event affects the probability of the occurrence of another event, the events are dependent events. The following example illustrates depen- dent events.
Two buckets contain a number of colored balls each. Bucket 1 contains two red balls and four white balls, and bucket 2 contains one blue ball and five red balls. A coin is tossed. If a head results, a ball is drawn out of bucket 1. If a tail results, a ball is drawn from bucket 2. These events are illustrated in Figure 11.4.
R R W
W W W
B R R
R R R
Flip a coin
Heads Tails
Bucket 1 Bucket 2
Figure 11.4
Dependent events
R R W
W W W
W R R
R R R
Bucket 1 Bucket 2
Flip a coin
Heads Tails
Figure 11.5
Another set of dependent events
Red
Red
White
White
Bucket 1
Bucket 2
Head
Tail
Flip a coin
Figure 11.6
Probability tree for dependent events
In this example the probability of drawing a blue ball is clearly dependent on whether a head or a tail occurs on the coin toss. If a tail occurs, there is a 1/6 chance of drawing a blue ball from bucket 2. However, if a head results, there is no possibility of drawing a blue ball from bucket 1. In other words, the probability of the event “drawing a blue ball” is depen- dent on the event “flipping a coin.”
Like statistically independent events, dependent events exhibit certain defining proper- ties. In order to describe these properties, we will alter our previous example slightly, so that bucket 2 contains one white ball and five red balls. Our new example is shown in Figure 11.5. The outcomes that can result from the events illustrated in Figure 11.5 are shown in Figure 11.6. When the coin is flipped, one of two outcomes is possible, a head or a tail. The probability of getting a head is .50, and the probability of getting a tail is .50:
P1T2 = .50 P1H2 = .50
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500 Chapter 11 Probability and Statistics
As indicated previously, these probabilities are referred to as marginal probabilities. They are also unconditional probabilities because they are the probabilities of the occur- rence of a single event and are not conditional on the occurrence of any other event(s). They are the same as the probabilities of independent events defined earlier, and like those of independent events, the marginal probabilities of a collectively exhaustive set of events sum to one.
Once the coin has been tossed and a head or tail has resulted, a ball is drawn from one of the buckets. If a head results, a ball is drawn from bucket 1; there is a 2/6, or .33, proba- bility of drawing a red ball and a 4/6, or .67, probability of drawing a white ball. If a tail resulted, a ball is drawn from bucket 2; there is a 5/6, or .83, probability of drawing a red ball and a 1/6, or .17, probability of drawing a white ball. These probabilities of drawing red or white balls are called conditional probabilities because they are conditional on the outcome of the event of tossing a coin. Symbolically, these conditional probabilities are expressed as follows:
The first term, which can be expressed verbally as “the probability of drawing a red ball, given that a head results from the coin toss,” equals .33. The other conditional probabilities are expressed similarly.
Conditional probabilities can also be defined by the following mathematical relation- ship. Given two dependent events A and B,
the term P(AB) is the joint probability of the two events, as noted previously. This rela- tionship can be manipulated by multiplying both sides by P(B), to yield
Thus, the joint probability can be determined by multiplying the conditional probability of A by the marginal probability of B.
Recall from our previous discussion of independent events that
Substituting this result into the relationship for a conditional probability yields
P1A|B2 = P1A) # (B2
P1B2
P1AB2 = P(A) # P1B2
P1A|B2 # P1B2 = P1AB2
P1A|B2 = P1AB2 P1B2
P1W|T2 = .17 P(R|T) = .83
P(W|H) = .67 P(R|H) = .33
which is consistent with the property for independent events. Returning to our example, the joint events are the occurrence of a head and a red ball,
a head and a white ball, a tail and a red ball, and a tail and a white ball. The probabilities of these joint events are as follows:
P1WT2 = P1W|T2 # P1T2 = 1.1721.52 = .085 P1RT2 = P1R|T2 # P1T2 = 1.8321.52 = .415
P1WH2 = P1W|H2 # P1H2 = 1.6721.52 = .335 P1RH2 = P1R|H2 # P1H2 = 1.3321.52 = .165
= P1A2
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Statistical Independence and Dependence 501
The marginal, conditional, and joint probabilities for this example are summarized in Figure 11.7. Table 11.1 is a joint probability table, which summarizes the joint probabilities for the example.
Bayesian Analysis The concept of conditional probability given statistical dependence forms the necessary foundation for an area of probability known as Bayesian analysis. The technique is named after Thomas Bayes, an eighteenth-century clergyman who pioneered this area of analysis.
The basic principle of Bayesian analysis is that additional information (if available) can sometimes enable one to alter (improve) the marginal probabilities of the occurrence of an event. The altered probabilities are referred to as revised, or posterior, probabilities.
The concept of posterior probabilities will be illustrated using an example. A produc- tion manager for a manufacturing firm is supervising the machine setup for the produc- tion of a product. The machine operator sets up the machine. If the machine is set up cor- rectly, there is a 10% chance that an item produced on the machine will be defective; if the machine is set up incorrectly, there is a 40% chance that an item will be defective. The pro- duction manager knows from past experience that there is a .50 probability that a machine will be set up correctly or incorrectly by an operator. In order to reduce the chance that an item produced on the machine will be defective, the manager has decided that the opera- tor should produce a sample item. The manager wants to know the probability that the machine has been set up incorrectly if the sample item turns out to be defective.
The probabilities given in this problem statement can be summarized as follows:
where correct incorrect defective D =
IC = C =
P1IC2 = .50 P1D|IC2 = .40 P1C2 = .50 P1D|C2 = .10
In Bayesian analysis, additional information is used to alter the
marginal probability of the occurrence of an event.
A posterior probability is the altered marginal probability of an
event, based on additional information.
Red
Red
White
White
Bucket 1
Bucket 2
Head P(H) = .50
Flip a coin
Tail P(T) = .50
P(R � H) = .33
P(W � T) = .17
P(W � H) = .67 P(R � T) = .83
P(RH)
P(WH) = .335
P(RT)
P(WT)
= .165
= .415
= .085
Conditional probabilities
Marginal probabilities
Joint probabilities
Figure 11.7
Probability tree with marginal, conditional, and joint
probabilities
Draw a Ball Marginal
Flip a Coin Red White Probabilities
Head P(RH) = .165 P(WH) = .335 P(H) = .50 Tail P(RT) = .415 P(WT) = .085 P(T) = .50
Marginal P(R) = .580 P(W) = .420 1.00 probabilities
Table 11.1 Joint probability table
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502 Chapter 11 Probability and Statistics
The posterior probability for our example is the conditional probability that the machine has been set up incorrectly, given that the sample item proves to be defective, or
In Bayesian analysis, once we are given the initial marginal and conditional prob- abilities, we can compute the posterior probability by using Bayes’s rule, as follows:
Previously, the manager knew that there was a 50% chance that the machine was set up incorrectly. Now, after producing and testing a sample item, the manager knows that if it is defective, there is an .80 probability that the machine was set up incorrectly. Thus, by gath- ering some additional information, the manager can revise the estimate of the probability that the machine was set up correctly. This will obviously improve decision making by allowing the manager to make a more informed decision about whether to have the machine set up again.
In general, given two events, A and B, and a third event, C, that is conditionally depen- dent on A and B, Bayes’s rule can be written as
Expected Value
It is often possible to assign numeric values to the various outcomes that can result from an experiment. When the values of the variables occur in no particular order or sequence, the variables are referred to as random variables. Every possible value of a variable has a probability of occurrence associated with it. For example, if a coin is tossed three times, the number of heads obtained is a random variable. The possible values of the random vari- able are 0, 1, 2, and 3 heads. The values of the variable are random because there is no way of predicting which value (0, 1, 2, or 3) will result when the coin is tossed three times. If three tosses are made several times, the values (i.e., numbers of heads) that will result will have no sequence or pattern; they will be random.
Like the variables defined in previous chapters in this text, random variables are typi- cally represented symbolically by a letter, such as x, y, or z. Consider a vendor who sells hot dogs outside a building every day. If the number of hot dogs the vendor sells is defined as the random variable x, then x will equal 0, 1, 2, 3, 4, . . . hot dogs sold daily.
Although the exact values of the random variables in the foregoing examples are not known prior to the event, it is possible to assign a probability to the occurrence of the possible values that can result. Consider a production operation in which a machine breaks down periodically. From experience it has been determined that the machine will break down 0, 1, 2, 3, or 4 times per month. Although managers do not know the exact number of breakdowns (x) that will occur each month, they can determine the relative
P1A|C2 = P1C|A2P1A2 P1C|A2P1A2 + P1C|B2P1B2
= .80
= 1.4021.502
1.4021.502 + 1.1021.502
P1IC|D2 = P1D|IC2P1IC2 P1D|IC2P1IC2 + P1D|C2P1C2
P1IC|D2. Bayes’s rule is a formula for
computing the posterior probability given marginal and
conditional probabilities.
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Expected Value 503
frequency probability of each number of breakdowns P(x). These probabilities are as follows:
x P(x) 0 .10 1 .20 2 .30 3 .25 4 .15
1.00
These probability values taken together form a probability distribution. That is, the prob- abilities are distributed over the range of possible values of the random variable x.
The expected value of the random variable (the number of breakdowns in any given month) is computed by multiplying each value of the random variable by its probability of occurrence and summing these products.
For our example, the expected number of breakdowns per month is computed as follows:
This means that, on the average, management can expect 2.15 breakdowns every month. The expected value is often referred to as the weighted average, or mean, of the proba-
bility distribution and is a measure of central tendency of the distribution. In addition to knowing the mean, it is often desirable to know how the values are dispersed (or scattered) around the mean. A measure of dispersion is the variance, which is computed as follows:
1. Square the difference between each value and the expected value. 2. Multiply these resulting amounts by the probability of each value. 3. Sum the values compiled in step 2.
The general formula for computing the variance, which we will designate as is
The variance for the machine breakdown example is computed as follows:
xi P(xi) xi – E(x) [xi – E(x)] 2 [xi – E(x)]
2 • P(xi) 0 .10 –2.15 4.62 .462 1 .20 –1.15 1.32 .264 2 .30 –0.15 .02 .006 3 .25 0.85 .72 .180 4 .15 1.85 3.42 .513
1.00 1.425
The standard deviation is another widely recognized measure of dispersion. It is desig- nated symbolically as and is computed by taking the square root of the variance, as follows:
= 1.19 breakdowns per month s = 21.425
s
s2 = 1.425 breakdowns per month
1s22 s2 = a
n
i = 1 3xi - E1xi242P1xi2
s2,
= 2.15 breakdowns = 0 + .20 + .60 + .75 + .60
E1x2 = 1021.102 + 1121.202 + 1221.302 + 1321.252 + 1421.152
The expected value of a random variable is computed by
multiplying each possible value of the variable by its probability and
summing these products.
The expected value is the mean of the probability distribution of the
random variable.
Variance is a measure of the dispersion of random variable
values about the expected value, or mean.
The standard deviation is computed by taking the square
root of the variance.
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504 Chapter 11 Probability and Statistics
M a n a g e m e n t S c i e n c e A p p l i c a t i o n A Probability Model for Analyzing Coast Guard Patrol Effectiveness
Aprimary responsibility of the U.S. Coast Guard is to mon-itor and protect the maritime borders of the United States. One measure of Coast Guard effectiveness in meeting this responsibility is determined by a probability model that indi- cates the probability that the Coast Guard will intercept a ran- domly selected target vessel attempting to penetrate a Coast Guard patrol perimeter. The probability P(I), the probability of intercepting the target vessel, is computed according to the formula
is the probability that a target vessel will be inter- cepted (I), given that it is detected (D). It is a constant value estimated from Coast Guard historical data. is the probability that a target vessel is detected, given that a Coast Guard vessel is available (A). This value is computed using a submodel based on the patrol area and speed of the Coast Guard vessel and the speed of the target vessel. is the probability that the Coast Guard vessel is available, given that it is on the scene (O) in the patrol area. P(O) is the probability that a Coast Guard vessel is on the scene in the patrol area, and currently it is assigned a constant value of one. The Coast Guard used this model to perform sensitivity analysis and ana- lyze the effect of different parameter changes on effectiveness.
P1A|O2
P1D|A2 P1I|D2
P1I2 = P1I|D2P1D|A2P1A|O2P1O2
An example might be testing the impact of a change in the patrol area or vessel speed.
Source: S. O. Kimbrough, J. R. Oliver, and C. W. Pritchett, “On Post-Evaluation Analysis: Candle-Lighting and Surrogate Models,” Interfaces 23, no. 3 (May–June 1993): 17–28.
A small standard deviation or variance, relative to the expected value, or mean, indicates that most of the values of the random variable distribution are bunched close to the expected value. Conversely, a large relative value for the measures of dispersion indicates that the values of the random variable are widely dispersed from the expected value.
The Normal Distribution
Previously, a discrete value was defined as a value that is countable (and usually an integer). A random variable is discrete if the values it can equal are finite and countable. The prob- ability distributions we have encountered thus far have been discrete distributions. In other words, the values of the random variables that made up these discrete distributions were always finite (for example, in the preceding example, there were five possible values of the random variable breakdowns per month). Because every value of the random variable had a unique probability of occurrence, the discrete probability distribution consisted of all the (finite) values of a random variable and their associated probabilities.
In contrast, a continuous random variable can take on an infinite number of values within some interval. This is because continuous random variables have values that are not specifi- cally countable and are often fractional. The distinction between discrete and continuous ran- dom variables is sometimes made by saying that discrete relates to things that can be counted
A continuous random variable can take on an infinite number of
values within an interval, or range.
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The Normal Distribution 505
and continuous relates to things that are measured. For example, a load of oil being transported by tanker may consist of not just 1 million or 2 million barrels but 1.35 million barrels. If the range of the random variable is between 1 million and 2 million barrels, then there is an infinite number of possible (fractional) values between 1 million and 2 million barrels, even though the value 1.35 million corresponds to a discrete value of 1,350,000 barrels of oil. No matter how small an interval exists between two values in the distribution, there is always at least one value—and, in fact, an infinite number of values—between the two values.
Because a continuous random variable can take on an extremely large or infinite num- ber of values, assigning a unique probability to every value of the random variable would require an infinite (or very large) number of probabilities, each of which would be infi- nitely small. Therefore, we cannot assign a unique probability to each value of the contin- uous random variable, as we did in a discrete probability distribution. In a continuous dis- tribution, we can refer only to the probability that a value of the random variable is within some range. For example, we can determine the probability that between 1.35 million and 1.40 million barrels of oil are transported, or the probability that fewer than or more than 1.35 million barrels are shipped, but we cannot determine the probability that exactly 1.35 million barrels of oil are transported.
One of the most frequently used continuous probability distributions is the normal dis- tribution, which is a continuous curve in the shape of a bell (i.e., it is symmetrical). The normal distribution is a popular continuous distribution because it has certain mathemat- ical properties that make it easy to work with, and it is a reasonable approximation of the continuous probability distributions of a number of natural phenomena. Figure 11.8 is an illustration of the normal distribution.
The normal distribution is a continuous probability distribution that is symmetrical on both sides of
the mean—that is, it is shaped like a bell.
In a continuous distribution the random variables can equal an
infinite number of values within an interval.
The center of a normal distribution is its mean, μ.
Area = .50 Area = .50
− ∞ + ∞μ = Mean x
Figure 11.8
The normal curve
The fact that the normal distribution is a continuous curve reflects the fact that it con- sists of an infinite or extremely large number of points (on the curve). The bell-shaped curve can be flatter or taller, depending on the degree to which the values of the random variable are dispersed from the center of the distribution. The center of the normal distri- bution is referred to as the mean and it is analogous to the average of the distribution.
Notice that the two ends (or tails) of the distribution in Figure 11.8 extend from to . In reality, random variables do not often take on values over an infinite range. Therefore, when the normal distribution is applied, it actually approximates the distribu- tion of a random variable with finite limits. The area under the normal curve represents probability. The entire area under the curve equals 1.0 because the sum of the probabilities of all values of a random variable in a probability distribution must equal 1.0. Fifty percent of the curve lies to the right of the mean, and 50% lies to the left. Thus, the probability that a random variable x will have a value greater (or less) than the mean is .50.
As an example of the application of the normal distribution, consider the Armor Carpet Store, which sells Super Shag carpet. From several years of sales records, store management has determined that the mean number of yards of Super Shag demanded by customers during a week is 4,200 yards, and the standard deviation is 1,400 yards. It is necessary to
+ q - q
1m2, The area under the normal curve
represents probability, and the total area under the curve
sums to one.
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506 Chapter 11 Probability and Statistics
know both the mean and the standard deviation to perform a probabilistic analysis using the normal distribution. The store management assumes that the continuous random vari- able, yards of carpet demanded per week, is normally distributed (i.e., the values of the ran- dom variable have approximately the shape of the normal curve). The mean of the normal distribution is represented by the symbol μ, and the standard deviation is represented by the symbol
The store manager wants to know the probability that the demand for Super Shag in the upcoming week will exceed 6,000 yards. The normal curve for this example is shown in Figure 11.9. The probability that x (the number of yards of carpet) will be equal to or greater than 6,000 is expressed as
which corresponds to the area under the normal curve to the right of the value 6,000 because the area under the curve (in Figure 11.9) represents probability. In a normal distribution, area
P1x Ú 6,0002
σ = 1,400 yd. μ = 4,200 yd.
σ:
or probability is measured by determining the number of standard deviations the value of the ran- dom variable x is from the mean. The number of standard deviations a value is from the mean is represented by Z and is computed using the formula
The number of standard deviations a value is from the mean gives us a consistent standard of measure for all normal distributions. In our example, the units of measure are yards; in other problems, the units of measure may be pounds, hours, feet, or tons. By con- verting these various units of measure into a common measure (number of standard devi- ations), we create a standard that is the same for all normal distributions.
Actually, the standard form of the normal distribution has a mean of zero (μ � 0) and a standard deviation of one The value Z enables us to convert this scale of mea- sure into whatever scale our problem requires.
Figure 11.10 shows the standard normal distribution, with our example distribution of carpet demand above it. This illustrates the conversion of the scale of measure along the horizontal axis from yards to number of standard deviations.
The horizontal axis along the bottom of Figure 11.10 corresponds to the standard nor- mal distribution. Notice that the area under the normal curve between and rep- resents 68% of the total area under the normal curve, or a probability of .68. Now look at the horizontal axis corresponding to yards in our example. Given that the standard deviation is 1,400 yards, the area between (2,800 yards) and (5,600 yards) is also 68% of the1s- 1s
1s- 1s
1s = 12.
Z = x - m s
μ = 4,200 σ = 1,400
x (yards) 0 6,000
Figure 11.9
The normal distribution for carpet demand
The area under a normal curve is measured by determining the
number of standard deviations the value of a random variable is from
the mean.
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The Normal Distribution 507
total area under the curve. Thus, if we measure distance along the horizontal axis in terms of the number of standard deviations, we will determine the same probability, no matter what the units of measure are. The formula for Z makes this conversion for us.
Returning to our example, recall that the manager of the carpet store wants to know the probability that the demand for Super Shag in the upcoming week will be 6,000 yards or more. Substituting the values and yards into our formula for Z, we can determine the number of standard deviations the value 6,000 is from the mean:
The value is 1.29 standard deviations from the mean, as shown in Figure 11.11.x = 6,000
= 1.29 standard deviations
= 6,000 - 4,200
1,400
Z = x - m s
s = 1,400x = 6,000, m = 4,200,
The area under the standard normal curve for values of Z has been computed and is dis- played in easily accessible normal tables. Table A.1 in Appendix A is such a table. It shows that
standard deviations corresponds to an area, or probability, of .4015. However, this is the area between and because what was measured was the area within 1.29 standard deviations of the mean. Recall, though, that 50% of the area under the curve lies to the right of the mean. Thus, we can subtract .4015 from .5000 to get the area to the right of
This means that there is a .0985 (or 9.85%) probability that the demand for carpet next week will be 6,000 yards or more.
= .0985 P1x Ú 6,0002 = .5000 - .4015
x = 6,000:
x = 6,000m = 4,200 Z = 1.29
0 1,400 2,800 8,4007,0005,600μ = 4,200
− 3σ − 2σ − 1σ 1σ 2σ 3σμ = 0
x
x
68%
68%
Figure 11.10
The standard normal distribution
μ = 4,200 σ = 1,400
x (yards) 0 6,000
.0985.4015
1.29σ
Figure 11.11
Determination of the Z value
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508 Chapter 11 Probability and Statistics
Now suppose that the carpet store manager wishes to consider two additional questions: (1) What is the probability that demand for carpet will be 5,000 yards or less? (2) What is the probability that the demand for carpet will be between 3,000 yards and 5,000 yards? We will consider each of these questions separately.
First, we want to determine The area representing this probability is shown in Figure 11.12. The area to the left of the mean in Figure 11.12 equals .50. That leaves only the area between and to be determined. The number of standard deviations is from the mean is
= .57 standard deviation
= 800
1,400
= 5,000 - 4,200
1,400
Z = x - m s
x = 5,000 x = 5,000m = 4,200
P1x … 5,0002.
The value corresponds to a probability of .2157 in Table A.1 in Appendix A. Thus, the area between 4,200 and 5,000 in Figure 11.12 is .2157. To find our desired prob- ability, we simply add this amount to .5000:
Next, we want to determine The area representing this proba- bility is shown in Figure 11.13. The shaded area in Figure 11.13 is computed by finding two
P13,000 … x … 5,0002. P1x … 5,0002 = .5000 + .2157 = .7157
Z = .57
μ = 4,200 σ = 1,400
x (yards) 0
5,000
.2157.5000
.57σ
Figure 11.12
Normal distribution for P1x … 5,000 yards2
μ = 4,200 σ = 1,400
x (yards) 0
5,000
.2157.3051
.86σ
3,000
Figure 11.13
Normal distribution with P (3,000 yards ≤ x ≤ 5,000 yards)
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The Normal Distribution 509
areas—the area between and and the area between and —and summing them. We already computed the area between 4,200 and 5,000
in the previous example and found it to be .2157. The area between and is found by determining how many standard deviations x = 3,000 is from the
mean:
The minus sign is ignored when we find the area corresponding to the Z value of .86 in Table A.1. This value is .3051. We now find our probability by summing .2157 and .3051:
The normal distribution, although applied frequently in probability analysis, is just one of a number of continuous probability distributions. In subsequent chapters, other con- tinuous distributions will be identified. Being acquainted with the normal distribution will make the use of these other distributions much easier.
Sample Mean and Variance The mean and variance we described in the previous section are actually more correctly referred to as the population mean and variance The population mean is the mean of the entire set of possible measurements or set of data being analyzed, and the population variance is defined similarly. Although we provided numeric values of the population mean and variance (or standard deviation) in our examples in the previous section, it is more likely that a sample mean, x–,would be used to estimate the population mean, and a sample variance, would be used to estimate the population variance. The computation of a true population mean and variance is usually too time-consuming and costly, or the entire population of data may not be available. Instead, a sample, which is a smaller subset of the population, is used. The mean and variance of this sample can then be used as an estimate of the population mean and variance if the population is assumed to be normally distributed. In other words, a smaller set of sample data is used to make generalizations or estimates, referred to as inferences, about the whole (popula- tion) set of data.
For example, in the previous section we indicated that the mean weekly demand for Super Shag carpet was 4,200 yards and the standard deviation was 1,400; thus and However, we noted that the management of the Armor Carpet Store deter- mined the mean and standard deviation from “several years of sales records.” Thus, the mean and standard deviation were actually a sample mean and standard deviation, based on several years of sample data from a normal distribution. In other words, we used a sample mean and sample standard deviation to estimate the population mean and stan- dard deviation. The population mean and standard deviation would have actually been computed from all weekly demand since the company started, for some extended period of time.
s = 1,400. m = 4,200
s2,
1s22.1m2
= .5208 P13,000 … x … 5,0002 = .2157 + .3051
= - .86
= - 1,200 1,400
Z = 3,000 - 4,200
1,400
m = 4,200 x = 3,000
x = 5,000 m = 4,200m = 4,200x = 3,000
The population mean and variance are for the entire set of
data being analyzed.
A sample mean and variance are derived from a subset of the
population data and are used to make inferences about the
population.
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510 Chapter 11 Probability and Statistics
The sample mean, x–, is computed using the following formula:
The sample variance, is computed as follows:
or, in shortcut form:
The sample standard deviation is simply the square root of the variance:
Let us consider our example for Armor Carpet Store again, except now we will use a sample of 10 weeks’ demand data for Super Shag carpet, as follows:
Week i Demand xi
1 2,900
2 5,400
3 3,100
4 4,700
5 3,800
6 4,300
7 6,800
8 2,900
9 3,600
10 4,500
Σ = 42,000
The sample mean is computed as
= 42,000
10 = 4,200 yd.
x = a n
i=1 xi
n
s = 2s2
s2 = a n
i=1 x 2i -
¢a n
i=1 x i≤
2
n
n - 1
s2 = a n
i=1 1xi - x22 n - 1
s2,
x = a n
i=1 xi
n
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The Normal Distribution 511
The sample variance is computed as
The sample standard deviation, s, is
These values are very close to the mean and standard deviation we originally estimated in this example in the previous section (i.e., yd., yd.). In general, the accuracy of the sample depends on two factors: the sample size (n) and the variation in the data. The larger the sample, the more accurate the sample statistics will be in estimat- ing the population statistics. Also, the more variable the data are, the less accurate the sample statistics will be as estimates of the population statistics.
The Chi-Square Test for Normality Several of the quantitative techniques that are presented in the remainder of this text include probabilistic data and parameters and statistical analysis. In many cases the problem data are assumed (or stated) to be normally distributed, with a mean and stan- dard deviation, which enable statistical analysis to be performed based on the normal distribution. However, in reality it can never be simply assumed that data are nor- mally distributed or in fact reflect any probability distribution. Frequently, a statistical test must be performed to determine the exact distribution (if any) to which the data conform.
The chi-square test is one such statistical test to see if observed data fit a particular probability distribution, including the normal distribution. The chi-square test compares the actual frequency distribution for a set of data with a theoretical fre- quency distribution that would be expected to occur for a specific distribution. This is also referred to as testing the goodness-of-fit of a set of data to a specific probability distribution.
To perform a chi-square goodness-of-fit test, the actual number of frequencies in each class or the range of a frequency distribution is compared to the theoretical frequencies that should occur in each class if the data followed a particular distribution. These numeric differences between the actual and theoretical values in each class are used in a formula to compute a test statistic, or number, which is then compared to a number from a chi- square table called a critical value. If the computed test statistic is greater than the tabular critical value, then the data do not follow the distribution being tested; if the critical value is greater than the computed test statistic, the distribution does exist. In statistical terminology this is referred to as testing the hypothesis that the data are, for example,
x2 x2
1x22
s = 1,400m = 4,200
= 1,232 yd. = 21,517,777
s = 2s2
s2 = 1,517,777
= (190,060,000) -
11,764,000,0002 10
9
s2 = a n
i=1 x 2i -
aa n
i=1 xib
2
n
n - 1
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512 Chapter 11 Probability and Statistics
normally distributed. If the test statistic is greater than the tabular critical value, the hypothesis that the data fits the hypothesized distribution is rejected, and the distri- bution does not exist; otherwise, it is accepted.
To demonstrate how to apply the chi-square test for a normal distribution, we will use an expanded version of our Armor Carpet Store example used in the previous section. To test the distribution, we need a lot more data than the 10 weeks of demand we used previ- ously. Instead, we will now assume that we have collected a sample of 200 weeks of demand for Super Shag carpet and that these data have been grouped according to the following frequency distribution:
Range, Frequency Weekly Demand (yd.) (weeks)
0 < 1,000 2
1,000 < 2,000 5
2,000 < 3,000 22
3,000 < 4,000 50
4,000 < 5,000 62
5,000 < 6,000 40
6,000 < 7,000 15
7,000 < 8,000 3
8,000+ 1
200
Because we haven’t provided the actual data, we are also going to assume that the sample mean (x–) equals 4,200 yards and the sample standard deviation (s) equals 1,232 yards, although normally these values would be computed directly from the data as in the previ- ous section.
The first step in performing the chi-square test is to determine the number of observa- tions that should be in each frequency range, if the distribution is normal. We start by deter- mining the area (or probability) that should be in each class, using the sample mean and standard deviation. Figure 11.14 shows the theoretical normal distribution, with the area in each range. The area of probability for each range is computed using the normal probability
1Ho2 x2
.0636 � .2422 � .3058
.0047 1
.0320 .1293 .2704 .1857 .0605 .0106 .0010 2 3 4
x � 4.2 5 6 7 8 (1,000 yd)
Figure 11.14
The theoretical normal distribution
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The Normal Distribution 513
table (Table A.1 in Appendix A), as demonstrated earlier in this chapter. For example, the area less than 1,000 yards is determined by computing the Z statistic for
This corresponds to a normal table value of .4953. This is the area from 1,000 to the sam- ple mean (4,200). Subtracting this value from .5000 results in the area less than 1,000, or
The area in the range of 1,000 to 2,000 yards is computed by sub- tracting the area from 2,000 to the mean (.4633) from the area from 1,000 to the mean (.4953), or The Z values for these ranges and all the range areas in Figure 11.14 are shown in Table 11.2.
.4953 - .4633 = .0320.
.5000 - .4953 = .0047.
= - 2.60
= 1,000 - 4,200
1,232
Z = x - x
s
x = 1,000:
Area: Range Normal Range Z x → x– Area Frequency (n = 200)
0 < 1,000 — .5000 .0047 0.94 1,000 < 2,000 –2.60 .4953 .0320 6.40 2,000 < 3,000 –1.79 .4633 .1293 25.86 3,000 < 4,000 –.97 .3340 .2704 54.08
4,000 < 5,000 –.16 .0636
.3058 61.16 .65 .2422
5,000 < 6,000 1.46 .4279 .1857 37.14 6,000 < 7,000 2.27 .4884 .0605 12.10 7,000 < 8,000 3.08 .4990 .0106 2.12 8,000+ — .5000 .0010 0.20
{ }
Table 11.2 The determination of the
theoretical range frequencies
Notice that the area for the range that includes the mean between 4,000 and 5,000 yards is determined by adding the two areas to the immediate right and left of the mean,
The next step is to compute the theoretical frequency for each range by multiplying the area in each range by For example, the frequency in the range from 0 to 1,000 is and the frequency for the range from 1,000 to 2,000 is
These and the remaining theoretical frequencies are shown in the last column in Table 11.2.
Next we must compare these theoretical frequencies with the actual frequencies in each range, using the following chi-square test statistic:
where observed frequency theoretical frequency the number of classes or ranges the number of estimated parameters degrees of freedomk - p - 1 =
p = k = ft = fo =
χ 2 1
2
k p
o t
tk
f f
f− − =
−( )∑
1.0320212002 = 6.40. 1.0047212002 = 0.94,
n = 200.
.0636 + .2422 = .3058.
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However, before we can apply this formula, we must make an important adjustment. Before we can apply the chi-square test, each range must include at least five theoretical observations. Thus, we need to combine some of the ranges so that they will contain at least five theoretical observations. For our distribution we can accomplish this by combining the two lower class ranges (0–1,000 and 1,000–2,000) and the three higher ranges
. This results in a revised frequency distribution with six frequency classes, as shown in Table 11.3. 7,000; 7,000 - 8,000; and 8,000 +2
16,000 -
The completed chi-square test statistic is shown in the last column in Table 11.3 and is computed as follows:
Range, Observed Theoretical Weekly Demand Frequency fo Frequency ft ( fo – ft)
2 ( fo – ft ) 2/ft
0 < 2,000 7 7.34 .12 .016 2,000 < 3,000 22 25.86 14.90 .576 3,000 < 4,000 50 54.08 16.64 .308 4,000 < 5,000 62 61.16 .71 .012 5,000 < 6,000 40 37.14 8.18 .220 6,000+ 19 14.42 21.00 1.456
2.588
Table 11.3 Computation of χ2 test statistic
χ 2 2
6
2 588
= −⎛⎝
⎞ ⎠
=
∑ f f
f
o t
t
.
Next we must compare this test statistic with a critical value obtained from the chi- square table (Table A.2) in Appendix A. The degrees of freedom for the critical value are �
where k is the number of frequency classes, or 6; and p is the number of para- meters that were estimated for the distribution, which in this case is 2, the sample mean and the sample standard deviation. Thus,
Using a level of significance (degree of confidence) of .05 (i.e., ), from Table A.2:
Because 7.815>2.588, we accept the hypothesis that the distribution is normal. If the χ2 value of 7.815 were less than the computed χ2 test statistic, the distribution would not be considered normal.
Statistical Analysis with Excel QM for Windows does not have statistical program modules. Therefore, to perform statis- tical analysis, and specifically to compute the mean and standard deviation from sample data, we must rely on Excel. Exhibit 11.1 shows the Excel spreadsheet for our Armor Carpet Store example. The average demand for the sample data (4,200) is computed in cell C16, using the formula �AVERAGE(C4:C13), which is also shown on the formula bar at the top of the spreadsheet. Cell C17 contains the sample standard deviation (1,231.98), computed by using the formula �STDEV(C4:C13).
x2.05,3 = 7.815
a = .05
= 3 degrees of freedom k - p - 1 = 6 - 2 - 1
k - p - 1,
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The Normal Distribution 515
A statistical analysis of the sample data can also be obtained by using the “Data Analysis” option from the “Data” tab at the top of the spreadsheet. (If this option is not available on your “Data” menu, select the “Add-Ins” option from the “Excel options” menu and then select the “Analysis ToolPak” option. This will insert the “Data Analysis” option on your “Data” toolbar when you return to it.) Selecting the “Data Analysis” option from the menu will result in the “Data Analysis” window shown in Exhibit 11.1.
Select “Descriptive Statistics” from this window. This will result in the dialog window titled Descriptive Statistics shown in Exhibit 11.2. This window, completed as shown,
=STDEV(C4:C13) =AVERAGE(C4:C13)
Click on “Data” tab on toolbar; then on “Data Analysis”; then select “Descriptive Statistics.”
Table generated from “Data Analysis”
Exhibit 11.1
Specifies location of statistical summary on spreadsheet.
Indicates that the first row of data (in C3) is a label.
Cells with data
Exhibit 11.2
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516 Chapter 11 Probability and Statistics
results in summary statistics for the demand data in our carpet store example. Notice that the input range we entered, C3:C13, includes the “Demand” heading on the spreadsheet in cell C3, which we acknowledge by checking the “Labels in first row” box. This will result in our statistical summary being labeled “Demand” on the spreadsheet, as shown in cell E3 in Exhibit 11.1. Notice that we indicated where we wanted to locate the summary statistics on our spreadsheet by typing E3 in the “Output Range” window. We obtained the summary statistics by checking the “Summary statistics” box at the bottom of the screen.
Summary
The field of probability and statistics is quite large and complex, and it contains muchmore than has been presented in this chapter. This chapter presented the basic prin- ciples and fundamentals of probability. The primary purpose of this brief overview was to prepare the reader for other material in the book. The topics of decision analysis (Chapter 12), simulation (Chapter 14), probabilistic inventory models (Chapter 16), and, to a certain extent, project management (Chapter 8) are probabilistic in nature and require an understanding of the fundamentals of probability.
References
Chou, Ya-lun. Statistical Analysis, 2nd ed. New York: Holt, Rinehart
& Winston, 1975.
Cramer, H. The Elements of Probability Theory and Some of Its
Applications, 2nd ed. New York: John Wiley & Sons, 1973.
Dixon, W. J., and Massey, F. J. Introduction to Statistical Analysis, 4th
ed. New York: McGraw-Hill, 1983.
Hays, W. L., and Winkler, R. L. Statistics: Probability, Inference,
and Decision, 2nd ed. New York: Holt, Rinehart & Winston,
1975.
Mendenhall, W., Reinmuth, J. E., Beaver, R., and Duhan, D. Statistics
for Management and Economics, 5th ed. North Scituate, MA:
Duxbury Press, 1986.
Neter, J., Wasserman, W., and Whitmore, G. A. Applied Statistics, 3rd
ed. Boston: Allyn & Bacon, 1988.
Sasaki, K. Statistics for Modern Business Decision Making. Belmont,
CA: Wadsworth, 1969.
Spurr, W. A., and Bonini, C. P. Statistical Analysis for Business
Decisions. Homewood, IL: Richard D. Irwin, 1973.
The following example will illustrate the solution of a problem involving a normal probability.
Problem Statement
The Radcliffe Chemical Company and Arsenal produces explosives for the U.S. Army. Because of the nature of its products, the company devotes strict attention to safety, which is also scrutinized by the federal government. Historical records show that the annual number of property damage and personal injury accidents is normally distrib- uted, with a mean of 8.3 accidents and a standard deviation of 1.8 accidents.
A. What is the probability that the company will have fewer than 5 accidents next year? more than 10?
B. The government will fine the company $200,000 if the number of accidents exceeds 12 in a 1-year period. What average annual fine can the company expect?
Example Problem Solution
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Example Problem Solution 517
Solution
Step 1: Set Up the Normal Distribution
x (accidents)105 � = 8.3 � = 1.8
Step 2: Solve Part A
From Table A.1 in Appendix A we see that corresponds to a probability of .4664; thus,
From Table A.1 in Appendix A we see that corresponds to a probability of .3264; thus,
P1x Ú 102 = .5000 - .3264 Z = .94
= .94
= 10 - 8.3
1.8
Z = x - m s
P1x Ú 10 accidents2 = .0336
P(x … 5) = .5000 - .4664
Z = - 1.83
= - 1.83
= 5 - 8.3
1.8
Z = x - m s
P1x … 5 accidents2
Step 3: Solve Part B
= 2.06
= 12 - 8.3
1.8
Z = x - m s
P(x Ú 12 accidents)
= .1736
From Table A.1 in Appendix A we see that corresponds to a probability of .4803; thus,
= .0197 P1x Ú 122 = .5000 - .4803
Z = 2.06
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518 Chapter 11 Probability and Statistics
Therefore, the company’s expected annual fine is
Problems
1. Indicate which of the following probabilities are objective and which are subjective. (Note that in some cases, the probabilities may not be entirely one or the other.)
a. The probability of snow tomorrow b. The probability of catching a fish c. The probability of the prime interest rate rising in the coming year d. The probability that the Cincinnati Reds will win the World Series e. The probability that demand for a product will be a specific amount next month f. The probability that a political candidate will win an election g. The probability that a machine will break down h. The probability of being dealt four aces in a poker hand
2. A gambler in Las Vegas is cutting a deck of cards for $1,000. What is the probability that the card for the gambler will be the following?
a. A face card b. A queen c. A spade d. A jack of spades
3. Downhill Ski Resort in Colorado has accumulated information from records of the past 30 winters regarding the measurable snowfall. This information is as follows:
Snowfall (in.) Frequency
0–19 2 20–29 7 30–39 8 40–49 8 50+ 5
30
a. Determine the probability of each event in this frequency distribution. b. Are all the events in this distribution mutually exclusive? Explain.
4. Employees in the textile industry can be segmented as follows:
Employees Number
Female and union 12,000 Female and nonunion 25,000 Male and union 21,000 Male and nonunion 42,000
a. Determine the probability of each event in this distribution. b. Are the events in this distribution mutually exclusive? Explain. c. What is the probability that an employee is male? d. Is this experiment collectively exhaustive? Explain.
= $3,940 $200,000 # P(x Ú 12) = ($200,000)1.01972
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Problems 519
5. The quality control process at a manufacturing plant requires that each lot of finished units be sampled for defective items. Twenty units from each lot are inspected. If five or more defective units are found, the lot is rejected. If a lot is known to contain 10% defective items, what is the proba- bility that the lot will be rejected? accepted?
6. A manufacturing company has 10 machines in continuous operation during a workday. The prob- ability that an individual machine will break down during the day is .10. Determine the probabil- ity that during any given day 3 machines will break down.
7. A polling firm is taking a survey regarding a proposed new law. Of the voters polled, 30% are in favor of the law. If 10 people are surveyed, what is the probability that 4 will indicate that they are opposed to the passage of the new law?
8. An automobile manufacturer has discovered that 20% of all the transmissions it installed in a par- ticular style of truck one year are defective. It has contacted the owners of these vehicles and asked them to return their trucks to the dealer to check the transmission. The Friendly Auto Mart sold seven of these trucks and has two of the new transmissions in stock. What is the probability that the auto dealer will need to order more new transmissions?
9. A new county hospital is attempting to determine whether it needs to add a particular specialist to its staff. Five percent of the general hospital population in the county contracts the illness the spe- cialist would treat. If 12 patients check into the hospital in a day, what is the probability that 4 or more will have the illness?
10. A large research hospital has accumulated statistical data on its patients for an extended period. Researchers have determined that patients who are smokers have an 18% chance of contracting a serious illness such as heart disease, cancer, or emphysema, whereas there is only a .06 probability that a nonsmoker will contract a serious illness. From hospital records, the researchers know that 23% of all hospital patients are smokers, while 77% are nonsmokers. For planning purposes, the hospital physician staff would like to know the probability that a given patient is a smoker if the patient has a serious illness.
11. Two law firms in a community handle all the cases dealing with consumer suits against companies in the area. The Abercrombie firm takes 40% of all suits, and the Olson firm handles the other 60%. The Abercrombie firm wins 70% of its cases, and the Olson firm wins 60% of its cases.
a. Develop a probability tree showing all marginal, conditional, and joint probabilities. b. Develop a joint probability table. c. Using Bayes’s rule, determine the probability that the Olson firm handled a particular case,
given that the case was won.
12. The Senate consists of 100 senators, of whom 34 are Republicans and 66 are Democrats. A bill to increase defense appropriations is before the Senate. Thirty-five percent of the Democrats and 70% of the Republicans favor the bill. The bill needs a simple majority to pass. Using a probability tree, determine the probability that the bill will pass.
13. A retail outlet receives radios from three electrical appliance companies. The outlet receives 20% of its radios from A, 40% from B, and 40% from C. The probability of receiving a defective radio from A is .01; from B, .02; and from C, .08.
a. Develop a probability tree showing all marginal, conditional, and joint probabilities. b. Develop a joint probability table. c. What is the probability that a defective radio returned to the retail store came from
company B?
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520 Chapter 11 Probability and Statistics
14. A metropolitan school system consists of three districts—north, south, and central. The north dis- trict contains 25% of all students, the south district contains 40%, and the central district contains 35%. A minimum-competency test was given to all students; 10% of the north district students failed, 15% of the south district students failed, and 5% of the central district students failed.
a. Develop a probability tree showing all marginal, conditional, and joint probabilities. b. Develop a joint probability table. c. What is the probability that a student selected at random failed the test?
15. A service station owner sells Goodroad tires, which are ordered from a local tire distributor. The dis- tributor receives tires from two plants, A and B. When the owner of the service station receives an order from the distributor, there is a .50 probability that the order consists of tires from plant A or plant B. However, the distributor will not tell the owner which plant the tires come from. The owner knows that 20% of all tires produced at plant A are defective, whereas only 10% of the tires produced at plant B are defective. When an order arrives at the station, the owner is allowed to inspect it briefly. The owner takes this opportunity to inspect one tire to see if it is defective. If the owner believes the tire came from plant A, the order will be sent back. Using Bayes’s rule, determine the posterior prob- ability that a tire is from plant A, given that the owner finds that it is defective.
16. A metropolitan school system consists of two districts, east and west. The east district contains 35% of all students, and the west district contains the other 65%. A vocational aptitude test was given to all students; 10% of the east district students failed, and 25% of the west district students failed. Given that a student failed the test, what is the posterior probability that the student came from the east district?
17. The Ramshead Pub sells a large quantity of beer every Saturday. From past sales records, the pub has determined the following probabilities for sales:
Barrels Probability
6 .10 7 .20 8 .40 9 .25
10 .05 1.00
Compute the expected number of barrels that will be sold on Saturday.
18. The following probabilities for grades in management science have been determined based on past records:
Grade Probability
A .10 B .30 C .40 D .10 F .10
1.00
The grades are assigned on a 4.0 scale, where an A is a 4.0, a B a 3.0, and so on. Determine the expected grade and variance for the course.
19. A market in Boston orders oranges from Florida. The oranges are shipped to Boston from Florida by either railroad, truck, or airplane; an order can take 1, 2, 3, or 4 days to arrive in Boston once it
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Problems 521
is placed. The following probabilities have been assigned to the number of days it takes to receive an order once it is placed (referred to as lead time):
Lead Time Probability
1 .20 2 .50 3 .20 4 .10
1.00
Compute the expected number of days it takes to receive an order and the standard deviation.
20. An investment firm is considering two alternative investments, A and B, under two possible future sets of economic conditions, good and poor. There is a .60 probability of good economic condi- tions occurring and a .40 probability of poor economic conditions occurring. The expected gains and losses under each economic type of conditions are shown in the following table:
Economic Conditions
Investment Good Poor
A $900,000 –$800,000 B 120,000 70,000
Using the expected value of each investment alternative, determine which should be selected.
21. An investor is considering two investments, an office building and bonds. The possible returns from each investment and their probabilities are as follows:
Office Building Bonds
Return Probability Return Probability
$50,000 .30 $30,000 .60 60,000 .20 40,000 .40 80,000 .10 1.00 10,000 .30
0 .10 1.00
Using expected value and standard deviation as a basis for comparison, discuss which of the two investments should be selected.
22. The Jefferson High School Band Booster Club has organized a raffle. The prize is a $6,000 car. Two thousand tickets to the raffle are to be sold at $1 apiece. If a person purchases four tickets, what will be the expected value of the tickets?
23. The time interval between machine breakdowns in a manufacturing firm is defined according to the following probability distribution:
Time Interval (hr.) Probability
1 .15 2 .20 3 .40 4 .25
1.00
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522 Chapter 11 Probability and Statistics
Determine the cumulative probability distribution and compute the expected time between machine breakdowns.
24. The life of an electronic transistor is normally distributed, with a mean of 500 hours and a stan- dard deviation of 80 hours. Determine the probability that a transistor will last for more than 400 hours.
25. The grade point average of students at a university is normally distributed, with a mean of 2.6 and a standard deviation of 0.6. A recruiter for a company is interviewing students for summer employ- ment. What percentage of the students will have a grade point average of 3.5 or greater?
26. The weight of bags of fertilizer is normally distributed, with a mean of 50 pounds and a standard deviation of 6 pounds. What is the probability that a bag of fertilizer will weigh between 45 and 55 pounds?
27. The monthly demand for a product is normally distributed, with a mean of 700 units and a stan- dard deviation of 200 units. What is the probability that demand will be greater than 900 units in a given month?
28. The Polo Development Firm is building a shopping center. It has informed renters that their rental spaces will be ready for occupancy in 19 months. If the expected time until the shopping center is completed is estimated to be 14 months, with a standard deviation of 4 months, what is the prob- ability that the renters will not be able to occupy in 19 months?
29. A warehouse distributor of carpet keeps 6,000 yards of deluxe shag carpet in stock during a month. The average demand for carpet from the stores that purchase from the distributor is 4,500 yards per month, with a standard deviation of 900 yards. What is the probability that a customer’s order will not be met during a month? (This situation is referred to as a stockout.)
30. The manager of the local National Video Store sells videocassette recorders at discount prices. If the store does not have a video recorder in stock when a customer wants to buy one, it will lose the sale because the customer will purchase a recorder from one of the many local competitors. The prob- lem is that the cost of renting warehouse space to keep enough recorders in inventory to meet all demand is excessively high. The manager has determined that if 90% of customer demand for recorders can be met, then the combined cost of lost sales and inventory will be minimized. The manager has estimated that monthly demand for recorders is normally distributed, with a mean of 180 recorders and a standard deviation of 60. Determine the number of recorders the manager should order each month to meet 90% of customer demand.
31. The owner of Western Clothing Company has determined that the company must sell 670 pairs of denim jeans each month to break even (i.e., to reach the point where total revenue equals total cost). The company’s marketing department has estimated that monthly demand is normally dis- tributed, with a mean of 805 pairs of jeans and a standard deviation of 207 pairs. What is the prob- ability that the company will make a profit each month?
32. Lauren Moore, a professor in management science, is computing her final grades for her introduc- tory management science class. The average final grade is a 63, with a standard deviation of 10. Professor Moore wants to curve the final grades according to a normal distribution so that 10% of the grades are Fs, 20% are Ds, 40% are Cs, 20% are Bs, and 10% are As. Determine the numeric grades that conform to the curve Professor Moore wants to establish.
33. The SAT scores of all freshmen accepted at State University are normally distributed, with a mean of 1,050 and a standard deviation of 120. The College of Business at State University has accepted 620 of these freshmen into the college. All students in the college who score over 1,200 are eligible for merit scholarships. How many students can the college administration expect to be eligible for merit scholarships?
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Problems 523
34. Erin Richards is a junior at Central High School, and she has talked to her guidance counselor about her chances of being admitted to Tech after her graduation. The guidance counselor has told her that Tech generally accepts only those applicants who graduate in the top 10% of their high school class. The average grade point average of the last four senior classes has been 2.67, with a standard deviation of 0.58. What GPA will Erin have to achieve to be in the top 10% of her class?
35. The associate dean in the college of business at Tech is going to purchase a new copying machine for the college. One model he is considering is the Zerox X10. The sales representative has told him that this model will make an average of 125,000 copies, with a standard deviation of 40,000 copies, before breaking down. What is the probability that the copier will make 200,000 copies before breaking down?
36. The Palace Hotel believes its customers may be waiting too long for room service. The hotel oper- ations manager knows that the time for room service orders is normally distributed, and he sam- pled 10 room service orders during a 3-day period and timed each (in minutes), as follows:
23 23 15 12 26 16 19 18 30 25
The operations manager believes that only 10% of the room service orders should take longer than 25 minutes if the hotel has good customer service. Does the hotel room service meet this goal?
37. Agnes Hammer is a senior majoring in management science. She has been interviewing with sev- eral companies for a job when she graduates, and she is curious about what starting salary offers she might receive. There are 140 seniors in the graduating class for her major, and more than half have received job offers. She asked 12 of her classmates at random what their annual starting salary offers were, and she received the following responses:
$28,500 $35,500 32,600 36,000 34,000 25,700 27,500 29,000 24,600 31,500 34,500 26,800
Assume that starting salaries are normally distributed. Compute the mean and standard deviation for these data and determine the probability that Agnes will receive a salary offer of less than $27,000.
38. The owner of Gilley’s Ice Cream Parlor has noticed that she sells more ice cream on hotter days during the summer, especially on days when the temperature is 85° or higher. To plan how much ice cream to stock, she would like to know the average daily high temperature for the summer months of July and August. Assuming that daily temperatures are normally distributed, she has gathered the following data for the high temperature for 20 days from a local almanac:
86° 92° 85 94 78 83 91 81 90 84 92 76 83 78 80 78 69 85 74 90
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Compute the mean and standard deviation for these data and determine the expected number of days in July and August that the high temperature will be 85° or greater.
39. The state of Virginia has implemented a Standard of Learning (SOL) test that all public school stu- dents must pass before they can graduate from high school. A passing grade is 75. Montgomery County High School administrators want to gauge how well their students might do on the SOL test, but they don’t want to take the time to test the whole student population. Instead, they selected 20 students at random and gave them the test. The results are as follows:
83 79 56 93 48 92 37 45 72 71 92 71 66 83 81 80 58 95 67 78
Assume that SOL test scores are normally distributed. Compute the mean and standard deviation for these data and determine the probability that a student at the high school will pass the test.
40. The department of management science at Tech has sampled 250 of its majors and compiled the following frequency distribution of grade point averages (on a 4.0 scale) for the previous semester:
GPA Frequency
0 < 0.5 1 0.5 < 1.0 4 1.0 < 1.5 20 1.5 < 2.0 35 2.0 < 2.5 67 2.5 < 3.0 58 3.0 < 3.5 47 3.5 < 4.0 18
250
The sample mean (x–) for this distribution is 2.5, and the sample standard deviation (s) is 0.72. Determine whether the student GPAs are normally distributed, using a .05 level of significance (i.e., ).
41. Geo-net, a cellular phone company, has collected the following frequency distribution for the length of calls outside its normal customer roaming area:
Length (min.) Frequency
0 < 5 26 5 < 10 75
10 < 15 139 15 < 20 105 20 < 25 37 25+ 18
400
The sample mean (x–) for this distribution is 14.3 minutes, and the sample standard deviation is 3.7 minutes. Determine whether these data are normally distributed 1a = .052.
a = .05
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Case Problem 525
Case Problem
Valley Swim Club
The Valley Swim Club has 300 stockholders, each holding oneshare of stock in the club. A share of club stock allows the shareholder’s family to use the club’s heated outdoor pool during the summer, upon payment of annual membership dues of $175. The club has not issued any new stock in years, and only a few of the existing shares come up for sale each year. The board of direc- tors administers the sale of all stock. When a shareholder wants to sell, he or she turns the stock in to the board, which sells it to the person at the top of the waiting list. For the past few years, the length of the waiting list has remained relatively steady, at approx- imately 20 names.
However, during the past winter, two events occurred that have increased the demand for shares in the club. The winter was especially severe, and subzero weather and heavy ice storms caused both the town and the county pools to buckle and crack. The problems were not discovered until maintenance crews began to ready the pools for the summer, and repairs cannot be com- pleted until the fall. Also during the winter, the manager of the local country club had an argument with her board of directors and one night burned down the clubhouse. Although the pool itself was not damaged, the dressing room facilities, showers, and snack bar were destroyed. As a result of these two events, the Valley Swim Club was inundated with applications to purchase shares. The waiting list suddenly grew to 250 people as the sum- mer approached.
The board of directors of the swim club had refrained from issuing new shares in the past because there was never a very great demand, and the demand that did exist was usually absorbed within a year by stock turnover. In addition, the board has a real concern about overcrowding. It seemed like the present member- ship was about right, and there were very few complaints about overcrowding, except on holidays like Memorial Day and the Fourth of July. However, at a recent board meeting, a number of new applicants had attended and asked the board to issue new shares. In addition, a number of current shareholders suggested that this might be an opportunity for the club to raise some capi- tal for needed repairs and to improve some of the existing facili- ties. This was tempting to the board. Although it had set the share price at $500 in the past, the board could set it at a much higher level now. In addition, any new shares sold would result in almost total profit because the manager, lifeguard, and maintenance costs had already been budgeted for the summer and would not increase with additional members.
Before the board of directors could make a decision on whether to sell more shares and, if so, how many, the board members felt
they needed more information. Specifically, they would like to know the average number of people (family members, guests, etc.) that might use the pool each day during the summer. They would also like to know the number of days they could expect more than 500 people to use the pool from June through August, given the current number of shares.
The board of directors has the following daily attendance records for June through August from the previous summer; it thinks the figures would provide accurate estimates for the upcom- ing summer:
139 380 193 399 177 238
273 367 378 197 161 224
172 359 461 273 308 368
275 463 242 213 256 541
337 578 177 303 391 235
402 287 245 262 400 218
487 247 390 447 224 271
198 356 284 399 239 259
310 322 417 275 274 232
347 419 474 241 205 317
393 516 194 190 361 369
421 478 207 243 411 361
595 303 215 277 419
497 223 304 241 258
341 315 331 384 130
291 258 407 246 195
The board has developed the following criteria for making a deci- sion on whether to issue new shares:
1. The expected number of days on which attendance would exceed 500 should be no more than 5 with the current membership.
2. The current average daily attendance should be no more than 320.
3. The average daily weekend (Saturday and Sunday) attendance should be no more than 500. (Weekend atten- dance is every sixth and seventh entry in each progression of seven entries in the preceding data.)
If these criteria are met, the club will issue one new share, at a price of $1,000, for every two average attendees between the cur- rent daily average and an upper limit of 400.
Should the club issue new shares? If so, how many will it issue, and how much additional revenue will it realize?
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C h a p t e r 12
Decision Analysis
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Components of Decision Making 527
In the previous chapters dealing with linear programming, models were formulated andsolved in order to aid the manager in making a decision. The solutions to the models were represented by values for decision variables. However, these linear programming mod- els were all formulated under the assumption that certainty existed. In other words, it was assumed that all the model coefficients, constraint values, and solution values were known with certainty and did not vary.
In actual practice, however, many decision-making situations occur under conditions of uncertainty. For example, the demand for a product may be not 100 units next week, but 50 or 200 units, depending on the state of the market (which is uncertain). Several decision-making techniques are available to aid the decision maker in dealing with this type of decision situation in which there is uncertainty.
Decision situations can be categorized into two classes: situations in which probabilities cannot be assigned to future occurrences and situations in which probabilities can be assigned. In this chapter we will discuss each of these classes of decision situations sepa- rately and demonstrate the decision-making criterion most commonly associated with each. Decision situations in which there are two or more decision makers who are in com- petition with each other are the subject of game theory, a topic included on the CD that accompanies this text.
Components of Decision Making
A decision-making situation includes several components—the decisions themselves and the actual events that may occur in the future, known as states of nature. At the time a decision is made, the decision maker is uncertain which states of nature will occur in the future and has no control over them.
Suppose a distribution company is considering purchasing a computer to increase the number of orders it can process and thus increase its business. If economic conditions remain good, the company will realize a large increase in profit; however, if the economy takes a downturn, the company will lose money. In this decision situation, the possible decisions are to purchase the computer and to not purchase the computer. The states of nature are good economic conditions and bad economic conditions. The state of nature that occurs will determine the outcome of the decision, and it is obvious that the decision maker has no control over which state will occur.
As another example, consider a concessions vendor who must decide whether to stock coffee for the concession stands at a football game in November. If the weather is cold, most of the coffee will be sold, but if the weather is warm, very little coffee will be sold. The deci- sion is to order or not to order coffee, and the states of nature are warm and cold weather.
To facilitate the analysis of these types of decision situations so that the best decisions result, they are organized into payoff tables. In general, a payoff table is a means of orga- nizing and illustrating the payoffs from the different decisions, given the various states of nature in a decision problem. A payoff table is constructed as shown in Table 12.1.
Each decision, 1 or 2, in Table 12.1 will result in an outcome, or payoff, for the particular state of nature that will occur in the future. Payoffs are typically expressed in terms of profit
The two categories of decision situations are probabilities that
can be assigned to future occurrences and probabilities that
cannot be assigned.
A state of nature is an actual event that may occur in the future.
Using a payoff table is a means of organizing a decision situation,
including the payoffs from different decisions, given the
various states of nature.
State of Nature
Decision a b
1 Payoff 1a Payoff 1b 2 Payoff 2a Payoff 2b
Table 12.1 Payoff table
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528 Chapter 12 Decision Analysis
revenues, or cost (although they can be expressed in terms of a variety of values). For example, if decision 1 is to purchase a computer and state of nature a is good economic conditions, payoff 1a could be $100,000 in profit.
It is often possible to assign probabilities to the states of nature to aid the decision maker in selecting the decision that has the best outcome. However, in some cases the decision maker is not able to assign probabilities, and it is this type of decision-making situation that we will address first.
EconomicreportE con
omi c
rep ort
?
Apartment building
Office building
Warehouse
Figure 12.1
Decision situation with real estate investment alternatives
Decision Making without Probabilities
The following example will illustrate the development of a payoff table without probabilities. An investor is to purchase one of three types of real estate, as illustrated in Figure 12.1. The investor must decide among an apartment building, an office building, and a warehouse. The
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Decision Making without Probabilities 529
future states of nature that will determine how much profit the investor will make are good economic conditions and poor economic conditions. The profits that will result from each decision in the event of each state of nature are shown in Table 12.2.
Decision-Making Criteria Once the decision situation has been organized into a payoff table, several criteria are avail- able for making the actual decision. These decision criteria, which will be presented in this section, include maximax, maximin, minimax regret, Hurwicz, and equal likelihood. On occasion these criteria will result in the same decision; however, often they will yield dif- ferent decisions. The decision maker must select the criterion or combination of criteria that best suits his or her needs.
The Maximax Criterion With the maximax criterion, the decision maker selects the decision that will result in the maximum of the maximum payoffs. (In fact, this is how this criterion derives its name—a maximum of a maximum.) The maximax criterion is very optimistic. The decision maker assumes that the most favorable state of nature for each decision alternative will occur. Thus, for example, using this criterion, the investor would optimistically assume that good economic conditions will prevail in the future.
The maximax criterion is applied in Table 12.3. The decision maker first selects the maximum payoff for each decision. Notice that all three maximum payoffs occur under good economic conditions. Of the three maximum payoffs—$50,000, $100,000, and $30,000—the maximum is $100,000; thus, the corresponding decision is to purchase the office building.
Although the decision to purchase an office building will result in the largest payoff ($100,000), such a decision completely ignores the possibility of a potential loss of $40,000. The decision maker who uses the maximax criterion assumes a very optimistic future with respect to the state of nature.
State of Nature
Decision Good Economic Poor Economic (Purchase) Conditions Conditions
Apartment building $ 50,000 $ 30,000 Office building 100,000 �40,000 Warehouse 30,000 10,000
Table 12.2 Payoff table for the real estate
investments
State of Nature
Decision Good Economic Poor Economic (Purchase) Conditions Conditions
Apartment building $ 50,000 $ 30,000 Office building 100,000 �40,000 Warehouse 30,000 10,000
Maximum payoff
Table 12.3 Payoff table illustrating a
maximax decision
The maximax criterion results in the maximum of the maximum
payoffs.
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530 Chapter 12 Decision Analysis
Before the next criterion is presented, it should be pointed out that the maximax deci- sion rule as presented here deals with profit. However, if the payoff table consisted of costs, the opposite selection would be indicated: the minimum of the minimum costs, or a minimin criterion. For the subsequent decision criteria we encounter, the same logic in the case of costs can be used.
The Maximin Criterion In contrast with the maximax criterion, which is very optimistic, the maximin criterion is pessimistic. With the maximin criterion, the decision maker selects the decision that will reflect the maximum of the minimum payoffs. For each decision alternative, the decision maker assumes that the minimum payoff will occur. Of these minimum payoffs, the maxi- mum is selected. The maximin criterion for our investment example is demonstrated in Table 12.4.
The maximin criterion results in the maximum of the minimum
payoff.
M a n a g e m e n t S c i e n c e A p p l i c a t i o n Decision Analysis at DuPont
DuPont has used decision analysis extensively since themid-1960s to create, evaluate, and implement strategic alternatives within 10 of its businesses, each worldwide in scope. As an example, one of these DuPont businesses was experiencing declining financial performance due to eroding prices and loss of market share to European and Japanese com- petitors. The business identified three possible strategies to evaluate from a number of alternatives. These strategies were (1) to continue with the current strategy, (2) to reestablish product leadership by strengthening new product development efforts and product differentiation, and (3) to establish a low- cost market position by closing a plant and streamlining the product line to improve production efficiency. Uncertainties in the evaluation process derived from competitor strategies and from market size, share, and prices. The decision criterion was net present value for each of the three strategies, calculated using a Super Tree spreadsheet model. The ultimate decision was to strengthen product development and product differen- tiation. The selected strategy resulted in an approximate increase in value of $175 million.
Source: F. Krumm and C. Rolle, “Management and Application of Decision and Risk Analysis in DuPont,” Interfaces 22, no. 6 (November–December 1992): 84–93.
State of Nature
Decision Good Economic Poor Economic (Purchase) Conditions Conditions
Apartment building $ 50,000 $ 30,000 Office building 100,000 �40,000 Maximum payoff Warehouse 30,000 10,000
Table 12.4 Payoff table illustrating a
maximin decision
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Decision Making without Probabilities 531
The minimum payoffs for our example are and $10,000. The maximum of these three payoffs is $30,000; thus, the decision arrived at by using the maximin criterion would be to purchase the apartment building. This decision is relatively conservative because the alternatives considered include only the worst outcomes that could occur. The decision to purchase the office building as determined by the maximax criterion includes the possibility of a large loss The worst that can occur from the decision to purchase the apart- ment building, however, is a gain of $30,000. On the other hand, the largest possible gain from purchasing the apartment building is much less than that of purchasing the office building (i.e., $50,000 vs. $100,000).
If Table 12.4 contained costs instead of profits as the payoffs, the conservative approach would be to select the maximum cost for each decision. Then the decision that resulted in the minimum of these costs would be selected.
The Minimax Regret Criterion In our example, suppose the investor decided to purchase the warehouse, only to discover that economic conditions in the future were better than expected. Naturally, the investor would be disappointed that she had not purchased the office building because it would have resulted in the largest payoff ($100,000) under good economic conditions. In fact, the investor would regret the decision to purchase the warehouse, and the degree of regret would be $70,000, the difference between the payoff for the investor’s choice and the best choice.
This brief example demonstrates the principle underlying the decision criterion known as minimax regret criterion. With this decision criterion, the decision maker attempts to avoid regret by selecting the decision alternative that minimizes the maximum regret.
To use the minimax regret criterion, a decision maker first selects the maximum payoff under each state of nature. For our example, the maximum payoff under good economic conditions is $100,000, and the maximum payoff under poor economic conditions is $30,000. All other payoffs under the respective states of nature are subtracted from these amounts, as follows:
Good Economic Conditions Poor Economic Conditions $100,000 � 50,000 = $50,000 $30,000 � 30,000 = $0 $100,000 � 100,000 = $0 $30,000 � (�40,000) = $70,000 $100,000 � 30,000 = $70,000 $30,000 � 10,000 = $20,000
These values represent the regret that the decision maker would experience if a decision were made that resulted in less than the maximum payoff. The values are summarized in a modified version of the payoff table known as a regret table, shown in Table 12.5. (Such a table is sometimes referred to as an opportunity loss table, in which case the term opportunity loss is synonymous with regret.)
1- $40,0002.
$30,000, - $40,000,
State of Nature
Decision Good Economic Poor Economic (Purchase) Conditions Conditions
Apartment building $50,000 $ 0 Office building 0 70,000 Warehouse 70,000 20,000
Table 12.5 Regret table
Regret is the difference between the payoff from the best decision
and all other decision payoffs.
The minimax regret criterion minimizes the maximum regret.
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532 Chapter 12 Decision Analysis
To make the decision according to the minimax regret criterion, the maximum regret for each decision must be determined. The decision corresponding to the minimum of these regret values is then selected. This process is illustrated in Table 12.6.
According to the minimax regret criterion, the decision should be to purchase the apart- ment building rather than the office building or the warehouse. This particular decision is based on the philosophy that the investor will experience the least amount of regret by pur- chasing the apartment building. In other words, if the investor purchased either the office building or the warehouse, $70,000 worth of regret could result; however, the purchase of the apartment building will result in, at most, $50,000 in regret.
The Hurwicz Criterion The Hurwicz criterion strikes a compromise between the maximax and maximin crite- ria. The principle underlying this decision criterion is that the decision maker is neither totally optimistic (as the maximax criterion assumes) nor totally pessimistic (as the max- imin criterion assumes). With the Hurwicz criterion, the decision payoffs are weighted by a coefficient of optimism, a measure of the decision maker’s optimism. The coefficient of optimism, which we will define as is between zero and one (i.e., ). If
then the decision maker is said to be completely optimistic; if then the decision maker is completely pessimistic. (Given this definition, if is the coefficient of optimism, is the coefficient of pessimism.)
The Hurwicz criterion requires that for each decision alternative, the maximum payoff be multiplied by and the minimum payoff be multiplied by For our investment example, if equals .4 (i.e., the investor is slightly pessimistic), and the fol- lowing values will result:
Decision Values Apartment building $ 50,000(.4) + 30,000(.6) = $38,000 Office building $100,000(.4) � 40,000(.6) = $16,000 Warehouse $ 30,000(.4) + 10,000(.6) = $18,000
The Hurwicz criterion specifies selection of the decision alternative corresponding to the maximum weighted value, which is $38,000 for this example. Thus, the decision would be to purchase the apartment building.
It should be pointed out that when the Hurwicz criterion is actually the maximin criterion; when it is the maximax criterion. A limitation of the Hurwicz criterion is the fact that must be determined by the decision maker. It can be quite difficult for a decision maker to accurately determine his or her degree of optimism. Regardless of how the decision maker determines , it is still a completely subjective measure of the decisiona
a
a = 1.0, a = 0,
1 - a = .6,a 1 - a.a
1 - a a
a = 0,a = 1.0, 0 … a … 1.0a,
State of Nature
Decision Good Economic Poor Economic (Purchase) Conditions Conditions
Apartment building $50,000 $ 0 Office building 0 70,000 Warehouse 70,000 20,000
The minimum regret value
Table 12.6 Regret table illustrating the
minimax regret decision
The Hurwicz criterion is a compromise between the maximax
and maximin criteria.
The coefficient of optimism, is a measure of the decision maker’s
optimism.
a,
The Hurwicz criterion multiplies the best payoff by the coefficient of optimism, and the worst payoff by for each decision, and
the best result is selected. 1 - a,
a,
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Decision Making without Probabilities 533
maker’s degree of optimism. Therefore, the Hurwicz criterion is a completely subjective decision-making criterion.
The Equal Likelihood Criterion When the maximax criterion is applied to a decision situation, the decision maker implic- itly assumes that the most favorable state of nature for each decision will occur. Alternatively, when the maximin criterion is applied, the least favorable states of nature are assumed. The equal likelihood, or LaPlace, criterion weights each state of nature equally, thus assuming that the states of nature are equally likely to occur.
Because there are two states of nature in our example, we assign a weight of .50 to each one. Next, we multiply these weights by each payoff for each decision:
Decision Values Apartment building $ 50,000(.50) + 30,000(.50) = $40,000 Office building $100,000(.50) � 40,000(.50) = $30,000 Warehouse $ 30,000(.50) + 10,000(.50) = $20,000
As with the Hurwicz criterion, we select the decision that has the maximum of these weighted values. Because $40,000 is the highest weighted value, the investor’s decision would be to purchase the apartment building.
In applying the equal likelihood criterion, we are assuming a 50% chance, or .50 proba- bility, that either state of nature will occur. Using this same basic logic, it is possible to weight the states of nature differently (i.e., unequally) in many decision problems. In other words, different probabilities can be assigned to each state of nature, indicating that one state is more likely to occur than another. The application of different probabilities to the states of nature is the principle behind the decision criteria to be presented in the section on expected value.
Summary of Criteria Results The decisions indicated by the decision criteria examined so far can be summarized as follows:
Criterion Decision (Purchase) Maximax Office building Maximin Apartment building Minimax regret Apartment building Hurwicz Apartment building Equal likelihood Apartment building
The decision to purchase the apartment building was designated most often by the var- ious decision criteria. Notice that the decision to purchase the warehouse was never indi- cated by any criterion. This is because the payoffs for an apartment building, under either set of future economic conditions, are always better than the payoffs for a warehouse. Thus, given any situation with these two alternatives (and any other choice, such as purchasing the office building), the decision to purchase an apartment building will always be made over the decision to purchase a warehouse. In fact, the warehouse decision alternative could have been eliminated from consideration under each of our criteria. The alternative of pur- chasing a warehouse is said to be dominated by the alternative of purchasing an apartment building. In general, dominated decision alternatives can be removed from the payoff table and not considered when the various decision-making criteria are applied. This reduces the complexity of the decision analysis somewhat. However, in our discussions throughout this chapter of the application of decision criteria, we will leave the dominated alternative in the payoff table for demonstration purposes.
A dominant decision is one that has a better payoff than another
decision under each state of nature.
The equal likelihood criterion multiplies the decision payoff for each state of nature by an equal
weight.
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534 Chapter 12 Decision Analysis
The use of several decision criteria often results in a mix of decisions, with no one deci- sion being selected more than the others. The criterion or collection of criteria used and the resulting decision depend on the characteristics and philosophy of the decision maker. For example, the extremely optimistic decision maker might eschew the majority of the foregoing results and make the decision to purchase the office building because the maxi- max criterion most closely reflects his or her personal decision-making philosophy.
Solution of Decision-Making Problems Without Probabilities with QM for Windows QM for Windows includes a module to solve decision analysis problems. QM for Windows will be used to illustrate the use of the maximax, maximin, minimax regret, equal likelihood, and Hurwicz criteria for the real estate problem considered in this section. The problem data are input very easily. A summary of the input and solution output for the maximax, max- imin, and Hurwicz criteria is shown in Exhibit 12.1. The decision with the equal likelihood criterion can be determined by using an alpha value for the Hurwicz criterion equal to the equal likelihood weight, which is .5 for our real estate investment example. The solution out- put with alpha equal to .5 is shown in Exhibit 12.2. The decision with the minimax regret criterion is shown in Exhibit 12.3.
Exhibit 12.1
Equal likelihood weight
Exhibit 12.2
Exhibit 12.3
The appropriate criterion is dependent on the risk
personality and philosophy of the decision maker.
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Decision Making with Probabilities 535
Solution of Decision-Making Problems Without Probabilities with Excel Excel can also be used to solve decision analysis problems using the decision-making crite- ria presented in this section. Exhibit 12.4 illustrates the application of the maximax, mini- max, minimax regret, Hurwicz, and equal likelihood criteria for our real estate investment example.
In cell E7 the formula �MAX(C7,D7) selects the maximum payoff outcome for the decision to purchase the apartment building. Next, in cell C11 the maximum of the maxi- mum payoffs is determined with the formula �MAX(E7:E9). The maximin decision is determined similarly.
In the regret table in Exhibit 12.4, in cell C18 the formula �MAX(C7:C9) – C7 com- putes the regret for the apartment building decision under good economic conditions, and then the maximum regret for the apartment building is determined in cell E18, using the formula �MAX(C18,D18). The minimax regret value is determined in cell C22 with the formula �MIN(E18:E20).
The Hurwicz and equal likelihood decisions are determined using their respective for- mulas in cells C27:C29 and C32:C34.
=MIN(C7,D7)
=C7*0.5+D7*0.5
=C7*C25+D7*C26
=MAX(C18,D18)
=MAX(E7,E9)
=MAX(C7:C9)–C9
=MAX(F7:F9)
Exhibit 12.4
Decision Making with Probabilities
The decision-making criteria just presented were based on the assumption that no infor- mation regarding the likelihood of the states of nature was available. Thus, no probabilities of occurrence were assigned to the states of nature, except in the case of the equal likelihood criterion. In that case, by assuming that each state of nature was equally likely and assign- ing a weight of .50 to each state of nature in our example, we were implicitly assigning a probability of .50 to the occurrence of each state of nature.
It is often possible for the decision maker to know enough about the future states of nature to assign probabilities to their occurrence. Given that probabilities can be assigned,
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536 Chapter 12 Decision Analysis
several decision criteria are available to aid the decision maker. We will consider two of these criteria: expected value and expected opportunity loss (although several others, includ- ing the maximum likelihood criterion, are available).
Expected Value To apply the concept of expected value as a decision-making criterion, the decision maker must first estimate the probability of occurrence of each state of nature. Once these esti- mates have been made, the expected value for each decision alternative can be computed. The expected value is computed by multiplying each outcome (of a decision) by the prob- ability of its occurrence and then summing these products. The expected value of a ran- dom variable x, written symbolically as is computed as follows:
where of values of the random variable x
Using our real estate investment example, let us suppose that, based on several economic forecasts, the investor is able to estimate a .60 probability that good economic conditions will prevail and a .40 probability that poor economic conditions will prevail. This new information is shown in Table 12.7.
n = number
EV(x) = a n
i = 1 xi P(xi)
EV1x2,
Expected value is computed by multiplying each decision outcome
under each state of nature by the probability of its occurrence.
State of Nature
Good Economic Poor Economic
Decision Conditions Conditions (Purchase) .60 .40
Apartment building $ 50,000 $ 30,000 Office building 100,000 �40,000 Warehouse 30,000 10,000
Table 12.7 Payoff table with probabilities
for states of nature
The expected value (EV) for each decision is computed as follows:
The best decision is the one with the greatest expected value. Because the greatest expected value is $44,000, the best decision is to purchase the office building. This does not mean that $44,000 will result if the investor purchases the office building; rather, it is assumed that one of the payoff values will result (either $100,000 or ). The expected value means that if this decision situation occurred a large number of times, an average payoff of $44,000 would result. Alternatively, if the payoffs were in terms of costs, the best decision would be the one with the lowest expected value.
Expected Opportunity Loss A decision criterion closely related to expected value is expected opportunity loss. To use this criterion, we multiply the probabilities by the regret (i.e., opportunity loss) for each decision outcome rather than multiplying the decision outcomes by the probabilities of their occurrence, as we did for expected monetary value.
- $40,000
EV(warehouse) = $30,000(.60) + 10,0001.402 = $22,000 EV(office) = $100,000(.60) – 40,000(.40) = $44,000
EV(apartment) = $50,000(.60) + 30,000(.40) = $42,000
Expected opportunity loss is the expected value of the regret for
each decision.
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Decision Making with Probabilities 537
The concept of regret was introduced in our discussion of the minimax regret criterion. The regret values for each decision outcome in our example were shown in Table 12.6. These values are repeated in Table 12.8, with the addition of the probabilities of occurrence for each state of nature.
Exhibit 12.5
The expected opportunity loss (EOL) for each decision is computed as follows:
As with the minimax regret criterion, the best decision results from minimizing the regret, or, in this case, minimizing the expected regret or opportunity loss. Because $28,000 is the minimum expected regret, the decision is to purchase the office building.
Notice that the decisions recommended by the expected value and expected opportunity loss criteria were the same—to purchase the office building. This is not a coincidence because these two methods always result in the same decision. Thus, it is repetitious to apply both methods to a decision situation when one of the two will suffice.
In addition, note that the decisions from the expected value and expected opportunity loss criteria are totally dependent on the probability estimates determined by the decision maker. Thus, if inaccurate probabilities are used, erroneous decisions will result. It is there- fore important that the decision maker be as accurate as possible in determining the prob- ability of each state of nature.
Solution of Expected Value Problems with QM for Windows QM for Windows not only solves decision analysis problems without probabilities but also has the capability to solve problems using the expected value criterion. A summary of the input data and the solution output for our real estate example is shown in Exhibit 12.5. Notice that the expected value results are included in the third column of this solution screen.
EOL(warehouse) = $70,000(.60) + 20,0001.402 = $50,000 EOL(office) = $0(.60) + 70,000(.40) = $28,000
EOL(apartment) = $50,000(.60) + 0(.40) = $30,000
Table 12.8 Regret (opportunity loss) table with probabilities for states of
nature
State of Nature
Good Economic Poor Economic
Decision Conditions Conditions (Purchase) .60 .40
Apartment building $50,000 $ 0 Office building 0 70,000 Warehouse 70,000 20,000
The expected value and expected opportunity loss criteria result in
the same decision.
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538 Chapter 12 Decision Analysis
Expected value for apartment building
Exhibit 12.6
Click on “Add-Ins” to access the “Excel QM” menu
Exhibit 12.7
Solution of Expected Value Problems with Excel and Excel QM This type of expected value problem can also be solved by using an Excel spreadsheet. Exhibit 12.6 shows our real estate investment example set up in a spreadsheet format. Cells E7, E8, and E9 include the expected value formulas for this example. The expected value formula for the first decision, purchasing the apartment building, is embedded in cell E7 and is shown on the formula bar at the top of the spreadsheet.
Excel QM is a set of spreadsheet macros that is included on the companion Web site that accompanies this text, and it has a macro to solve decision analysis problems. Once activated, clicking on “Decision Analysis” will result in a Spreadsheet Initialization window. After enter- ing several problem parameters, including the number of decisions and states of nature, and then clicking on “OK,” the spreadsheet shown in Exhibit 12.7 will result. Initially, this spread- sheet contains example values in cells B8:C11. Exhibit 12.7 shows the spreadsheet with our problem data already typed in. The results are computed automatically as the data are entered, using the cell formulas already embedded in the macro.
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Decision Making with Probabilities 539
Expected Value of Perfect Information It is often possible to purchase additional information regarding future events and thus make a better decision. For example, a real estate investor could hire an economic fore- caster to perform an analysis of the economy to more accurately determine which eco- nomic condition will occur in the future. However, the investor (or any decision maker) would be foolish to pay more for this information than he or she stands to gain in extra profit from having the information. That is, the information has some maximum value that represents the limit of what the decision maker would be willing to spend. This value of information can be computed as an expected value—hence its name, the expected value of perfect information (also referred to as EVPI).
To compute the expected value of perfect information, we first look at the decisions under each state of nature. If we could obtain information that assured us which state of nature was going to occur (i.e., perfect information), we could select the best decision for that state of nature. For example, in our real estate investment example, if we know for sure that good economic conditions will prevail, then we will decide to purchase the office building. Similarly, if we know for sure that poor economic conditions will occur, then we will decide to purchase the apartment building. These hypothetical “perfect” decisions are summarized in Table 12.9.
The expected value of perfect information is the maximum
amount a decision maker would pay for additional information.
The probabilities of each state of nature (i.e., .60 and .40) tell us that good economic conditions will prevail 60% of the time and poor economic conditions will prevail 40% of the time (if this decision situation is repeated many times). In other words, even though perfect information enables the investor to make the right decision, each state of nature will occur only a certain portion of the time. Thus, each of the decision outcomes obtained using perfect information must be weighted by its respective probability:
The amount $72,000 is the expected value of the decision, given perfect information, not the expected value of perfect information. The expected value of perfect information is the maximum amount that would be paid to gain information that would result in a decision better than the one made without perfect information. Recall that the expected value deci- sion without perfect information was to purchase an office building, and the expected value was computed as
The expected value of perfect information is computed by subtracting the expected value without perfect information ($44,000) from the expected value given perfect infor- mation ($72,000):
EVPI = $72,000 - 44,000 = $28,000
EV(office) = $100,000(.60) - 40,0001.402 = $44,000
$100,000(.60) + 30,0001.402 = $72,000
State of Nature
Good Economic Poor Economic
Decision Conditions Conditions (Purchase) .60 .40
Apartment building $ 50,000 $ 30,000 Office building 100,000 �40,000 Warehouse 30,000 10,000
Table 12.9 Payoff table with decisions,
given perfect information
EVPI equals the expected value, given perfect information, minus
the expected value without perfect information.
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540 Chapter 12 Decision Analysis
The expected value, given perfect information, in Cell F12
=MAX(E7:E9)
=F12–F11
Exhibit 12.8
The expected value of perfect information, $28,000, is the maximum amount that the investor would pay to purchase perfect information from some other source, such as an economic forecaster. Of course, perfect information is rare and usually unobtainable. Typically, the decision maker would be willing to pay some amount less than $28,000, depending on how accurate (i.e., close to perfection) the decision maker believed the infor- mation was.
It is interesting to note that the expected value of perfect information, $28,000 for our example, is the same as the expected opportunity loss (EOL) for the decision selected, using this later criterion:
This will always be the case, and logically so, because regret reflects the difference between the best decision under a state of nature and the decision actually made. This is actu- ally the same thing determined by the expected value of perfect information.
Excel QM for decision analysis computes the expected value of perfect information, as shown in cell E17 at the bottom of the spreadsheet in Exhibit 12.7. The expected value of perfect information can also be determined by using Excel. Exhibit 12.8 shows the EVPI for our real estate investment example.
EOL(office) = $0(.60) + 70,0001.402 = $28,000
The expected value of perfect information equals the expected
opportunity loss for the best decision.
Decision Trees Another useful technique for analyzing a decision situation is using a decision tree. A deci- sion tree is a graphical diagram consisting of nodes and branches. In a decision tree the user computes the expected value of each outcome and makes a decision based on these expected values. The primary benefit of a decision tree is that it provides an illustration (or picture) of the decision-making process. This makes it easier to correctly compute the nec- essary expected values and to understand the process of making the decision.
We will use our example of the real estate investor to demonstrate the fundamentals of decision tree analysis. The various decisions, probabilities, and outcomes of this example, initially presented in Table 12.7, are repeated in Table 12.10. The decision tree for this example is shown in Figure 12.2.
A decision tree is a diagram consisting of square decision
nodes, circle probability nodes, and branches representing
decision alternatives.
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The circles (●) and the square (■) in Figure 12.2 are referred to as nodes. The square is a decision node, and the branches emanating from a decision node reflect the alternative decisions possible at that point. For example, in Figure 12.2, node 1 signifies a decision to purchase an apartment building, an office building, or a warehouse. The circles are proba- bility, or event, nodes, and the branches emanating from them indicate the states of nature that can occur: good economic conditions or poor economic conditions.
The decision tree represents the sequence of events in a decision situation. First, one of the three decision choices is selected at node 1. Depending on the branch selected, the deci- sion maker arrives at probability node 2, 3, or 4, where one of the states of nature will pre- vail, resulting in one of six possible payoffs.
Determining the best decision by using a decision tree involves computing the expected value at each probability node. This is accomplished by starting with the final outcomes (pay- offs) and working backward through the decision tree toward node 1. First, the expected value of the payoffs is computed at each probability node:
These values are now shown as the expected payoffs from each of the three branches emanating from node 1 in Figure 12.3. Each of these three expected values at nodes 2, 3, and 4 is the outcome of a possible decision that can occur at node 1. Moving toward node 1,
EV(node 4) = .60($30,000) + .401$10,0002 = $22,000 EV(node 3) = .60($100,000) + .40( - $40,000) = $44,000 EV(node 2) = .60($50,000) + .40($30,000) = $42,000
2
3
4
1
$50,000
$30,000
Good economic conditions (.60)
Poor economic conditions (.40)
$100,000
–$40,000
Good economic conditions (.60)
Poor economic conditions (.40)
$30,000
$10,000
Good economic conditions (.60)
Poor economic conditions (.40)
Apartment building
Office building
Warehouse
Purchase
Figure 12.2
Decision tree for real estate investment example
State of Nature
Good Economic Poor Economic
Decision Conditions Conditions (Purchase) .60 .40
Apartment building $ 50,000 $ 30,000 Office building 100,000 �40,000 Warehouse 30,000 10,000
Table 12.10 Payoff table for real estate
investment example
The expected value is computed at each probability node.
Branches with the greatest expected value are selected.
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2
3
4
1
$50,000
$30,000
Good economic conditions (.60)$42,000
Poor economic conditions (.40)
$100,000
–$40,000
Good economic conditions (.60)$44,000
Poor economic conditions (.40)
$30,000
$10,000
Good economic conditions (.60)$22,000
Poor economic conditions (.40)
Apartment building
Office building
Warehouse
Purchase
Figure 12.3
Decision tree with expected value at probability nodes
we select the branch that comes from the probability node with the highest expected payoff. In Figure 12.3, the branch corresponding to the highest payoff, $44,000, is from node 1 to node 3. This branch represents the decision to purchase the office building. The decision to purchase the office building, with an expected payoff of $44,000, is the same result we achieved earlier by using the expected value criterion. In fact, when only one decision is to be made (i.e., there is not a series of decisions), the decision tree will always yield the same decision and expected payoff as the expected value criterion. As a result, in these decision situations a decision tree is not very useful. However, when a sequence or series of decisions is required, a decision tree can be very useful.
Decision Trees with QM for Windows In QM for Windows, when the “Decision Analysis” module is accessed and you click on “New” to input a new problem, a menu appears that allows you to select either “Decision Tables” or “Decision Trees.” Exhibit 12.9 shows the Decision Tree Results solution screen for our real estate investment example. Notice that QM for Windows requires that all nodes
Exhibit 12.9
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be numbered, including the end, or terminal, nodes, which are shown as solid dots in Figure 12.3. For QM for Windows, we have numbered these end nodes 5 through 10.
Decision Trees with Excel and TreePlan TreePlan is an Excel add-in program developed by Michael Middleton to construct and solve decision trees in an Excel spreadsheet format. Although Excel has the graphical and computational capability to develop decision trees, it is a difficult and slow process. TreePlan is basically a decision tree template that greatly simplifies the process of setting up a decision tree in Excel.
The first step in using TreePlan is to gain access to it. The best way to go about this is to copy the TreePlan add-in file, TreePlan.xla, from the companion Web site accompanying this text onto your hard drive and then add it to the “Data/Add-Ins” menu that you access at the top of your spreadsheet screen. Once you have added TreePlan to the “Add-Ins” menu, you can invoke it by clicking on the “Decision Trees” menu item.
We will demonstrate how to use TreePlan with our real estate investment example shown in Figure 12.3. The first step in using TreePlan is to generate a new tree on which to begin work. Exhibit 12.10 shows a new tree that we generated by positioning the cursor on cell B1 and then invoking the “Add-ins” menu and clicking on “Decision Trees.” This results in a menu from which we click on “New Tree,” which creates the decision tree shown in Exhibit 12.10.
The decision tree in Exhibit 12.10 uses the normal nodal convention we used in creat- ing the decision trees in Figures 12.2 and 12.3—squares for decision nodes and circles for probability nodes (which TreePlan calls event nodes). However, this decision tree is only a starting point or template that we need to expand to replicate our example decision tree in Figure 12.3.
In Figure 12.3, three branches emanate from the first decision node, reflecting the three investment decisions in our example. To create a third branch using TreePlan, click on the decision node in cell B5 in Exhibit 12.10 and then invoke TreePlan from the “Add-Ins” menu. A window will appear, with several menu items, including “Add a Branch.” Select this menu item and click on “OK.” This will create a third branch on our decision tree, as shown in Exhibit 12.11.
Invoke TreePlan from the “Add Ins” menu.
To create another branch, click B5, then the “Decision Tree” menu, and select “Add a Branch” from the menu.
Exhibit 12.10
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Add numeric ($) values in these cells in column H.
These cells contain decision tree formulas; do not type in these cells in columns E and I.
Exhibit 12.12
Click on cell F3 to add event nodes.
Exhibit 12.11
Next we need to expand our decision tree in Exhibit 12.11 by adding probability nodes (2, 3, and 4 in Figure 12.3) and branches from these nodes for our example. To add a new node, click on the end node in cell F3 in Exhibit 12.11 and then invoke TreePlan from the “Add-Ins” menu. From the menu window that appears, select “Change to Event Node” and then select “Two Branches” from the same menu and click on “OK.” This process must be repeated two more times for the other two end nodes (in cells F8 and F13) to create our three probability nodes. The resulting decision tree is shown in Exhibit 12.12, with the new probability nodes at cells F5, F15, and F25 and with accompanying branches.
The next step is to edit the decision tree labels and add the numeric data from our example. Generic descriptive labels are shown above each branch in Exhibit 12.12—for example, “Decision 1” in cell D4 and “Event 4” in cell H2. We edit the labels the same way we would edit
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Decision Making with Probabilities 545
any spreadsheet. For example, if we click on cell D4, we can type in “Apartment Building” in place of “Decision 1,” reflecting the decision corresponding to this branch in our example, as shown in Figure 12.3. We can change the other labels on the other branches the same way. The decision tree with the edited labels corresponding to our example is shown in Exhibit 12.13.
Looking back to Exhibit 12.12 for a moment, focus on the two 0 values below each branch—for example, in cells D6 and E6 and in cells H9 and I4. The first 0 cell is where we type in the numeric value (i.e., $ amount) for that branch. For our example, we would type in 50,000 in cell H4, 30,000 in H9, 100,000 in H14, and so on. These values are shown on the decision tree in Exhibit 12.13. Likewise, we would type in the probabilities for the branches in the cells just above the branch—H1, H6, H11, and so on. For example, we would type in 0.60 in cell H1 and 0.40 in cell H6. These probabilities are also shown in Exhibit 12.13. However, we need to be very careful not to type anything into the second 0 branch cell—for example, E6, I4, I9, E16, I14, I19, and so on. These cells automatically con- tain the decision tree formulas that compute the expected values at each node and select the best decision branches, so we do not want to type anything in these cells that would eliminate these formulas. For example, in Exhibit 12.12 the formula in cell E6 is shown on the formula bar at the top of the screen. This is the expected value for that probability node.
The expected value for this decision tree and our example, $44,000, is shown in cell A16 in Exhibit 12.13.
Sequential Decision Trees As noted earlier, when a decision situation requires only a single decision, an expected value payoff table will yield the same result as a decision tree. However, a payoff table is usually limited to a single decision situation, as in our real estate investment example. If a decision situation requires a series of decisions, then a payoff table cannot be created, and a decision tree becomes the best method for decision analysis.
To demonstrate the use of a decision tree for a sequence of decisions, we will alter our real estate investment example to encompass a 10-year period during which several decisions must be made. In this new example, the first decision facing the investor is whether to pur- chase an apartment building or land. If the investor purchases the apartment building, two
Exhibit 12.13
A sequential decision tree illustrates a situation requiring a
series of decisions.
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M a n a g e m e n t S c i e n c e A p p l i c a t i o n Evaluating Electric Generator Maintenance Schedules Using Decision Tree Analysis
Electric utility companies plan annual outage periodsfor preventive maintenance on generators. The outages are typically part of 5- to-20-year master schedules. However, at Entergy Electric Systems these scheduled outages were tradi- tionally based on averages that did not reflect short-term fluctuations in demand due to breakdowns and bad weather conditions. The master schedule had to be reviewed each week by outage planners who relied on their experience to determine whether the schedule needed to be changed. A user-friendly software system was developed to assist planners at Entergy Electric Systems in making changes in their schedule. The system is based on decision tree analysis.
Each week in the master schedule is represented by a decision tree that is based on changes in customer demand, unexpected generator breakdowns, and delays in returning generators from planned outages. The numeric outcome of the decision tree is the average reserve margin of megawatts (MW) for a specific week. This value enables planners to determine whether customer demand will be met and whether the maintenance schedule planned for the week is acceptable. The planner’s objective is to avoid negative power reserves by making changes in the generators’ maintenance schedule. The branches of the decision tree, their probabilities of occurrence, and the branch MW values are based on historical data. The system has enabled Entergy Electric Systems to isolate high- risk weeks and to develop timely maintenance schedules on
short notice. The new computerized system has reduced the maintenance schedules review time for as many as 260 weeks from several days to less than an hour.
Source: H. A. Taha and H. M. Wolf, “Evaluation of Generator Maintenance Schedules at Entergy Electric Systems,” Interfaces 26, no. 4 (July–August 1996): 56–65.
states of nature are possible: Either the population of the town will grow (with a probabil- ity of .60) or the population will not grow (with a probability of .40). Either state of nature will result in a payoff. On the other hand, if the investor chooses to purchase land, 3 years in the future another decision will have to be made regarding the development of the land. The decision tree for this example, shown in Figure 12.4, contains all the pertinent data, including decisions, states of nature, probabilities, and payoffs.
At decision node 1 in Figure 12.4, the decision choices are to purchase an apartment building or to purchase land. Notice that the cost of each venture ($800,000 and $200,000, respectively) is shown in parentheses. If the apartment building is purchased, two states of nature are possible at probability node 2: The town may exhibit population growth, with a probability of .60, or there may be no population growth or a decline, with a probability of .40. If the population grows, the investor will achieve a payoff of $2,000,000 over a 10-year period. (Note that this whole decision situation encompasses a 10-year time span.) However, if no population growth occurs, a payoff of only $225,000 will result.
If the decision is to purchase land, two states of nature are possible at probability node 3. These two states of nature and their probabilities are identical to those at node 2; however, the payoffs are different. If population growth occurs for a 3-year period, no payoff will occur, but the investor will make another decision at node 4 regarding development of the land. At that point, either apartments will be built, at a cost of $800,000, or the land will be sold, with a pay- off of $450,000. Notice that the decision situation at node 4 can occur only if population
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growth occurs first. If no population growth occurs at node 3, there is no payoff, and another decision situation becomes necessary at node 5: The land can be developed commercially at a cost of $600,000, or the land can be sold for $210,000. (Notice that the sale of the land results in less profit if there is no population growth than if there is population growth.)
If the decision at decision node 4 is to build apartments, two states of nature are possi- ble: The population may grow, with a conditional probability of .80, or there may be no population growth, with a conditional probability of .20. The probability of population growth is higher (and the probability of no growth is lower) than before because there has already been population growth for the first 3 years, as shown by the branch from node 3 to node 4. The payoffs for these two states of nature at the end of the 10-year period are $3,000,000 and $700,000, respectively, as shown in Figure 12.4.
If the investor decides to develop the land commercially at node 5, then two states of nature can occur: Population growth can occur, with a probability of .30 and an eventual payoff of $2,300,000, or no population growth can occur, with a probability of .70 and a payoff of $1,000,000. The probability of population growth is low (i.e., .30) because there has already been no population growth, as shown by the branch from node 3 to node 5.
This decision situation encompasses several sequential decisions that can be analyzed by using the decision tree approach outlined in our earlier (simpler) example. As before, we start at the end of the decision tree and work backward toward a decision at node 1.
First, we must compute the expected values at nodes 6 and 7:
These expected values (and all other nodal values) are shown in boxes in Figure 12.5. At decision nodes 4 and 5, we must make a decision. As with a normal payoff table, we
make the decision that results in the greatest expected value. At node 4 we have a choice between two values: $1,740,000, the value derived by subtracting the cost of building an apartment building ($800,000) from the expected payoff of $2,540,000, or $450,000, the
EV(node 7) = .30($2,300,000) + .701$1,000,0002 = $1,390,000 EV(node 6) = .80($3,000,000) + .20($700,000) = $2,540,000
2
6
7
4
5
3
1
$2,000,000
Population growthBuild
apartments (–$800,000)
$225,000
$450,000
$3,000,000
$700,000
Population growth
.80
No population growth
.20
Sell land
Population growthDevelop
commercially (–$600,000)
$210,000
$2,300,000
$1,000,000
.30
No population growth
.70
Sell land
No population growth (3 years, $0 payoff)
Population growth (3 years, $0 payoff)
.60
.40
Purchase land (–$200,000)
Purchase apartment building (–$800,000) No population
growth
.60
.40
Figure 12.4
Sequential decision tree
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2
6
7
4
5
3
1
$2,000,000
Population growthBuild
apartments (–$800,000)
$225,000
$450,000
$3,000,000
$700,000
Population growth
.80
No population growth
.20
Sell land
Population growthDevelop
commercially (–$600,000)
$210,000
$2,300,000
$1,000,000
.30
No population growth
.70
Sell land
No population growth (3 years, $0 payoff)
Population growth (3 years, $0 payoff)
.60
.40
Purchase land (–$200,000)
Purchase apartment building (–$800,000)
$1,160,000
$1,290,000
$1,360,000
$790,000
$1,740,000
$2,540,000
$1,390,000
No population growth
.60
.40
Figure 12.5
Sequential decision tree with nodal
expected values
expected value of selling the land computed with a probability of 1.0. The decision is to build the apartment building, and the value at node 4 is $1,740,000.
This same process is repeated at node 5. The decisions at node 5 result in payoffs of $790,000 (i.e., ) and $210,000. Because the value $790,000 is higher, the decision is to develop the land commercially.
Next, we must compute the expected values at nodes 2 and 3:
(Note that the expected value for node 3 is computed from the decision values previously determined at nodes 4 and 5.)
Now we must make the final decision for node 1. As before, we select the decision with the greatest expected value after the cost of each decision is subtracted out:
Because the highest net expected value is $1,160,000, the decision is to purchase land, and the payoff of the decision is $1,160,000.
This example demonstrates the usefulness of decision trees for decision analysis. A deci- sion tree allows the decision maker to see the logic of decision making because it provides a picture of the decision process. Decision trees can be used for decision problems more complex than the preceding example without too much difficulty.
Sequential Decision Tree Analysis with QM for Windows We have already demonstrated the capability of QM for Windows to perform decision tree analysis. For the sequential decision tree example described in the preceding section and
land: $1,360,000 - 200,000 = $1,160,000 apartment building: $1,290,000 - 800,000 = $490,000
EV(node 3) = .60($1,740,000) + .401$790,0002 = $1,360,000 EV(node 2) = .60($2,000,000) + .40($225,000) = $1,290,000
$1,390,000 - 600,000 = $790,000
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illustrated in Figures 12.4 and 12.5, the program input and solution screen for QM for Windows is shown in Exhibit 12.14.
Notice that the expected value for the decision tree, $1,160,000, is given in the first row of the last column on the solution screen in Exhibit 12.14.
Sequential Decision Tree Analysis with Excel and TreePlan The sequential decision tree example shown in Figure 12.5, developed and solved by using TreePlan, is shown in Exhibit 12.15. Although this TreePlan decision tree is larger than the one we previously developed in Exhibit 12.13, it was accomplished in exactly the same way.
Expected value for the decision tree
Exhibit 12.14
Exhibit 12.15
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Decision Analysis with Additional Information
Earlier in this chapter we discussed the concept of the expected value of perfect information. We noted that if perfect information could be obtained regarding which state of nature would occur in the future, the decision maker could obviously make better decisions. Although perfect information about the future is rare, it is often possible to gain some amount of additional (imperfect) information that will improve decisions.
In this section we will present a process for using additional information in the decision-making process by applying Bayesian analysis. We will demonstrate this process using the real estate investment example employed throughout this chapter. Let’s review this example briefly: A real estate investor is considering three alternative investments, which will occur under one of the two possible economic conditions (states of nature) shown in Table 12.11.
Recall that, using the expected value criterion, we found the best decision to be the pur- chase of the office building, with an expected value of $44,000. We also computed the expected value of perfect information to be $28,000. Therefore, the investor would be will- ing to pay up to $28,000 for information about the states of nature, depending on how close to perfect the information was.
Now suppose that the investor has decided to hire a professional economic analyst who will provide additional information about future economic conditions. The analyst is con- stantly researching the economy, and the results of this research are what the investor will be purchasing.
The economic analyst will provide the investor with a report predicting one of two out- comes. The report will be either positive, indicating that good economic conditions are most likely to prevail in the future, or negative, indicating that poor economic conditions will probably occur. Based on the analyst’s past record in forecasting future economic con- ditions, the investor has determined conditional probabilities of the different report out- comes, given the occurrence of each state of nature in the future. We will use the following notations to express these conditional probabilities:
N = negative economic report P = positive economic report p = poor economic conditions g = good economic conditions
In Bayesian analysis additional information is used to alter the
marginal probability of the occurrence of an event.
State of Nature
Good Economic Poor Economic
Decision Conditions Conditions (Purchase) .60 .40
Apartment building $ 50,000 $ 30,000 Office building 100,000 �40,000 Warehouse 30,000 10,000
Table 12.11 Payoff table for the real estate
investment example
A conditional probability is the probability that an event will
occur, given that another event has already occurred.
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M a n a g e m e n t S c i e n c e A p p l i c a t i o n Decision Analysis in the Electric Power Industry
Oglethorpe Power Corporation (OPC), a wholesale powergeneration and transmission cooperative, provides power to 34 consumer-owned distribution cooperatives in Georgia, representing 20% of the power in the state. Georgia Power Company meets the remaining power demand in Georgia. In general, Georgia has the ability to produce surplus power, whereas Florida, because of its rapidly growing population, must buy power from outside the state to meet its demand. In 1990, OPC learned that Florida Power Corporation wanted to add a transmission line to Georgia capable of transmitting an additional 1,000 megawatts (MW). OPC had to decide whether to add this additional capacity at a cost of around $100 million, with annual savings of around $20 million or more. OPC used decision analysis and, specifically, decision trees to help make its ultimate decision. Decision analysis has become a very pop- ular form of quantitative analysis in the electric power indus- try in the United States. OPC’s decision tree analysis included a series of decisions combined with uncertainties. The initial decision alternatives in the tree were whether to build a new transmission line alone, build it in a joint venture with Georgia Power, or not build a new line at all. Subsequent decisions included whether to upgrade existing facilities to meet Florida’s needs plus control of the new facilities. Uncertainties included construction costs, Florida’s demand for power, competition from other power sellers to Florida, and the share of Florida power for which OPC would be able to contract. The full deci- sion tree for this decision problem, including all combinations of decision nodes and probability nodes, includes almost 8,000 possible decision paths. The decision tree helped OPC
understand its decision process better and enabled it to approach the problem so that it would understand its compet- itive situation with Florida Power Corporation to a greater extent before making a final decision.
Source: A. Borison, “Oglethorpe Power Corporation Decides about Investing in a Major Transmission System,” Interfaces 25, no. 2 (March–April 1995): 25–36.
The conditional probability of each report outcome, given the occurrence of each state of nature, follows:
For example, if the future economic conditions are, in fact, good (g), the probability that a positive report (P) will have been given by the analyst, is .80. The other three con- ditional probabilities can be interpreted similarly. Notice that these probabilities indicate that the analyst is a relatively accurate forecaster of future economic conditions.
The investor now has quite a bit of probabilistic information available—not only the conditional probabilities of the report but also the prior probabilities that each state of nature will occur. These prior probabilities that good or poor economic conditions will occur in the future are
P1p2 = .40 P(g) = .60
P1P|g2, P1N|p2 = .90 P(P|p) = .10 P(N|g) = .20 P(P|g) = .80
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Given the conditional probabilities, the prior probabilities can be revised to form posterior probabilities by means of Bayes’s rule. If we know the conditional probability that a positive report was presented, given that good economic conditions prevail, the posterior probability of good economic conditions, given a positive report, can be determined using Bayes’s rule, as follows:
The prior probability that good economic conditions will occur in the future is .60. However, by obtaining the additional information of a positive report from the analyst, the investor can revise the prior probability of good conditions to a .923 probability that good economic conditions will occur. The remaining posterior (revised) probabilities are
Decision Trees with Posterior Probabilities The original decision tree analysis of the real estate investment example is shown in Figures 12.2 and 12.3. Using these decision trees, we determined that the appropriate decision was the purchase of an office building, with an expected value of $44,000. However, if the investor hires an economic analyst, the decision regarding which piece of real estate to invest in will not be made until after the analyst presents the report. This creates an additional stage in the decision-making process, which is shown in the decision tree in Figure 12.6. This decision tree differs in two respects from the decision trees in Figures 12.2 and 12.3. The first differ- ence is that there are two new branches at the beginning of the decision tree that represent the two report outcomes. Notice, however, that given either report outcome, the decision alternatives, the possible states of nature, and the payoffs are the same as those in Figures 12.2 and 12.3.
The second difference is that the probabilities of each state of nature are no longer the prior probabilities given in Figure 12.2; instead, they are the revised posterior probabilities computed in the previous section, using Bayes’s rule. If the economic analyst issues a pos- itive report, then the upper branch in Figure 12.6 (from node 1 to node 2) will be taken. If an apartment building is purchased (the branch from node 2 to node 4), the probability of good economic conditions is .923, whereas the probability of poor conditions is .077. These are the revised posterior probabilities of the economic conditions, given a positive report. However, before we can perform an expected value analysis using this decision tree, one more piece of probabilistic information must be determined—the initial branch probabil- ities of positive and negative economic reports.
The probability of a positive report, P(P), and of a negative report, P(N), can be deter- mined according to the following logic. The probability that two dependent events, A and B, will both occur is
If event A is a positive report and event B is good economic conditions, then according to the preceding formula,
P(Pg) = P(P|g)P1g2
P(AB) = P(A|B)P1B2
P1p|N2 = .750 P(p|P) = .077 P(g|N) = .250
= .923
= (.80)(.60)
(.80)(.60) + (.10)1.402
P(g|P) = P(P|g)P(g)
P(P|g)P(g) + P(P|p)P(p)
P1g|P2, P1P|g2,
A posterior probability is the altered marginal probability of an
event, based on additional information.
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Decision Analysis with Additional Information 553
2
9
3 8
7
6
5
4
1
Warehouse
Office
building
Apartment building
Positive report
P(g � N) = .250 $30,000
$10,000
Warehouse
P(p � N) = .750
P(g � N) = .250 $100,000
–$40,000P(p � N) = .750
P(g � N) = .250 $50,000
$30,000P(p � N) = .750
P(g � P) = .923 $30,000
$10,000P(p � P) = .077
P(g � P) = .923 $100,000
–$40,000P(p � P) = .077
P(g � P) = .923 $50,000
$30,000P(p � P) = .077
Office
building
Apartment building
Negative report
Figure 12.6
Decision tree with posterior probabilities
M a n a g e m e n t S c i e n c e A p p l i c a t i o n Discount Fare Allocation at American Airlines
The management of its reservation system, referred to asyield management, is a significant factor in American Airlines’s profitability. At American Airlines, yield management consists of three functions: overbooking, discount fare alloca- tion, and traffic management. Overbooking is the practice of selling more reservations for a flight than there are available seats, to compensate for no-shows and cancellations. Discount allocation is the process of determining the number of dis- count fares to make available on a flight. Too few discount fares will leave a flight with empty seats, while too many will limit the number of flights to schedule to maximize revenues.
The process of determining the allocation of discount fares for a flight is essentially a decision analysis model with a decision tree. The initial decision is to accept or reject a discount fare request. If the initial decision is to reject the request, there is a probability, p, that the seat will eventually be sold at full fare and a probability, that the seat will not be sold at all. The esti- mate of p is updated several times before flight departure, depending on the forecast for seat demand and the number of available seats remaining. If the expected value of the decision to reject—that is, EV (reject discount fare) ($full fare)—is greater than the discount fare, then the request is rejected.
= 1p2
1 - p,
Source: B. Smith, J. Leimkuhler, and R. Darrow, “Yield Management at American Airlines,” Interfaces 22, no. 1 (January– February 1992): 8–31.
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554 Chapter 12 Decision Analysis
We can also determine the probability of a positive report and poor economic conditions the same way:
Next, we consider the two probabilities and also called joint probabilities. These are, respectively, the probability of a positive report and good economic conditions and the probability of a positive report and poor economic conditions. These two sets of occurrences are mutually exclusive because both good and poor economic conditions can- not occur simultaneously in the immediate future. Conditions will be either good or poor, but not both. To determine the probability of a positive report, we add the mutually exclu- sive probabilities of a positive report with good economic conditions and a positive report with poor economic conditions, as follows:
Now, if we substitute into this formula the relationships for and determined earlier, we have
You might notice that the right-hand side of this equation is the denominator of the Bayesian formula we used to compute in the previous section. Using the condi- tional and prior probabilities that have already been established, we can determine that the probability of a positive report is
Similarly, the probability of a negative report is
and are also referred to as marginal probabilities. Now we have all the information needed to perform a decision tree analysis. The deci-
sion tree analysis for our example is shown in Figure 12.7. To see how the decision tree analysis is conducted, consider the result at node 4 first. The value $48,460 is the expected value of the purchase of an apartment building, given both states of nature. This expected value is computed as follows:
The expected values at nodes 5, 6, 7, 8, and 9 are computed similarly. The investor will actually make the decision about the investment at nodes 2 and 3. It is
assumed that the investor will make the best decision in each case. Thus, the decision at node 2 will be to purchase an office building, with an expected value of $89,220; the deci- sion at node 3 will be to purchase an apartment building, with an expected value of $35,000. These two results at nodes 2 and 3 are referred to as decision strategies. They rep- resent a plan of decisions to be made, given either a positive or a negative report from the economic analyst.
The final step in the decision tree analysis is to compute the expected value of the deci- sion strategy, given that an economic analysis is performed. This expected value, shown as $63,194 at node 1 in Figure 12.7, is computed as follows:
This amount, $63,194, is the expected value of the investor’s decision strategy, given that a report forecasting future economic condition is generated by the economic analyst.
= $63,194 EV(strategy) = $89,220(.52) + 35,0001.482
EV(apartment building) = $50,000(.923) + 30,0001.0772 = $48,460
P1N2P1P2 P(N) = P(N|g)P(g) + P(N|p)P(p) = (.20)(.60) + (.90)1.402 = .48
P(P) = P(P|g)P(g) + P(P|p)P(p) = (.80)(.60) + (.10)1.402 = .52
P1g|P2 P(P) = P(P|g)P(g) + P(P|p)P1p2
P1Pp2P1Pg2 P(P) = P(Pg) + P1Pp2
P1Pp2,P1Pg2 P(Pp) = P(P|p)P1p2
Events are mutually exclusive if only one can occur at a time.
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Decision Analysis with Additional Information 555
Computing Posterior Probabilities with Tables One of the difficulties that can occur with decision analysis with additional information is that as the size of the problem increases (i.e., as we add more decision alternatives and states of nature), the application of Bayes’s rule to compute the posterior probabilities becomes more complex. In such cases, the posterior probabilities can be computed by using tables. This tabular approach will be demonstrated with our real estate investment example. The table for computing posterior probabilities for a positive report and is initially set up as shown in Table 12.12.
P1P2
2
9
3 8
7
6
5
4
1
Warehouse
Office
building
Apartment building
$89,220
Positive report
P(g � N) = .250 $30,000
$10,000
Warehouse
P(p � N) = .750
P(g � N) = .250 $100,000
–$40,000P(p � N) = .750
P(g � N) = .250 $50,000
$30,000P(p � N) = .750
P(g � P) = .923 $30,000
$10,000P(p � P) = .077
P(g � P) = .923 $100,000
–$40,000P(p � P) = .077
P(g � P) = .923 $50,000
$30,000
–$5,000
$35,000
$28,460
$89,220
$48,460
$15,000
Office
building
Apartment building
$35,000
Negative report
P(P) = .52
P(N) = .48
$63,194
P(p � P) = .077
Figure 12.7
Decision tree analysis for real estate investment example
The posterior probabilities for either state of nature (good or poor economic condi- tions), given a negative report, are computed similarly.
(4) Prior
Probability × (5) (1) (2) (3) Conditional Posterior
State Prior Conditional Probability: Probability: of Nature Probability Probability (2) × (3) (4) ÷ Σ(4)
Good conditions P(g) = .60 P(P|g) = .80 P(Pg) = .48
Poor conditions P(p) = .40 P(P|p) = .10 P1p|P2 = .04 .52
= .077P(Pp) = .04
a = P1P2 = .52
P1g|P2 = .48 .52
= .923
Table 12.12 Computation of posterior
probabilities
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556 Chapter 12 Decision Analysis
No matter how large the decision analysis, the steps of this tabular approach can be fol- lowed the same way as in this relatively small problem. This approach is more systematic than the direct application of Bayes’s “rule,” making it easier to compute the posterior probabilities for larger problems.
Computing Posterior Probabilities with Excel The posterior probabilities computed in Table 12.12 can also be computed by using Excel. Exhibit 12.16 shows Table 12.12 set up in an Excel spreadsheet format, as well as the table for computing P1N2.
The Expected Value of Sample Information Recall that we computed the expected value of our real estate investment example to be $44,000 when we did not have any additional information. After obtaining the additional information provided by the economic analyst, we computed an expected value of $63,194, using the decision tree in Figure 12.7. The difference between these two expected values is called the expected value of sample information (EVSI), and it is computed as follows:
For our example, the expected value of sample information is
This means that the real estate investor would be willing to pay the economic analyst up to $19,194 for an economic report that forecasted future economic conditions.
After we computed the expected value of the investment without additional informa- tion, we computed the expected value of perfect information, which equaled $28,000. However, the expected value of the sample information was only $19,194. This is a logical result because it is rare that absolutely perfect information can be determined. Because the additional information that is obtained is less than perfect, it will be worth less to the deci- sion maker. We can determine how close to perfect our sample information is by comput- ing the efficiency of sample information as follows:
Thus, the analyst’s economic report is viewed by the investor to be 69% as efficient as per- fect information. In general, a high efficiency rating indicates that the information is very good, or close to being perfect information, and a low rating indicates that the additional
efficiency = EVSI , EVPI = $19,194 / 28,000 = .69
EVSI = $63,194 - 44,000 = $19,194
EVSI = EVwith information - EVwithout information
Exhibit 12.16
The expected value of sample information is the difference
between the expected value with and without additional
information.
The efficiency of sample information is the ratio of
the expected value of sample information to the expected value
of perfect information.
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Utility 557
information is not very good. For our example, the efficiency of .69 is relatively high; thus, it is doubtful that the investor would seek additional information from an alternative source. (However, this is usually dependent on how much money the decision maker has available to purchase information.) If the efficiency had been lower, however, the investor might seek additional information elsewhere.
Utility
All the decision-making criteria presented so far in this chapter have been based on mon- etary value. In other words, decisions have been based on the potential dollar payoffs of the alternatives. However, there are certain decision situations in which individuals do not make decisions based on the expected dollar gain or loss.
M a n a g e m e n t S c i e n c e A p p l i c a t i o n Scheduling Refueling at the Indian Point 3 Nuclear Power Plant
The New York Power Authority (NYPA) owns and operates 12 power projects that provide roughly one-fourth of all the electricity used in New York State. Approximately 20% of NYPA’s electrical power (supplying Westchester County and New York City) is generated by the Indian Point 3 (IP3) nuclear power plant, located on the Hudson River. IP3 withdraws 840,000 gallons of water per minute from the river for cooling steam and then returns it to the river. When water is withdrawn from the river, fish, especially small fish and eggs, do not always survive as they pass through the cooling system. The NYPA can reduce the nega- tive effects by scheduling plant shutdowns to refuel IP3 when fish eggs and small fish are most abundant in the Hudson River. In the past, NYPA developed a refueling schedule according to a 10-year planning horizon, although unforeseen events often altered this schedule. There is uncertainty in the future about possible dereg- ulation of the electric utility industry and its effect on the price of replacement power during refueling outages. The NYPA must consider these uncertainties in its attempt to provide low-cost power and minimize the environmental effects of refueling out- ages. The NYPA developed a decision analysis model to compare alternative strategies for refueling that balanced fish protection, the cost of buying fuel, and the uncertainties of deregulation.
The key decisions of the model were the times of year that refueling outages should occur, which affects the level of fish protection, and the amount of fuel that should be ordered for the nuclear reactor core that allows for operation for a target number of days, which affects the cost of the operation. The cost consideration comprises three objectives: minimizing the cost of replacement electricity when IP3 is shut down, minimizing the amount of unused fuel at the end of an operating cycle, and minimizing the time that IP3 would operate at less than full power before refueling. The objective of fish protection was to minimize the sum of the average percentage reduction in the mortality rate caused by water removal at IP3 for five types of
fish over the 10-year planning horizon. The three major uncer- tainties in the model were the cost of refueling, how long it takes IP3 to refuel and how well it operates, and when New York State is likely to deregulate the electric utility industry. The decision analysis model considered five strategies based on different schedules that reflected different time windows when fish were most vulnerable. The model showed that no strategy simultane- ously minimized refueling cost while minimizing fish mortality rates. The strategy that was selected restricted the starting date for refueling to the third week in May, when fish protection would meet accepted standards, at a cost savings of $10 million over the previous refueling schedule. The NYPA used the deci- sion analysis model to develop its refueling schedule for the 10-year period from 1999 to 2008.
Source: D. J. Dunning, S. Lockfort, Q. E. Ross, P. C. Beccue, and J. S. Stonebraker, “New York Power Authority Uses Decision Analysis to Schedule Refueling of Its Indian Point 3 Nuclear Power Plant,” Interfaces 31, no. 5 (September–October 2001): 121–35.
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558 Chapter 12 Decision Analysis
For example, consider an individual who purchases automobile insurance. The decisions are to purchase and to not purchase, and the states of nature are an accident and no accident. The payoff table for this decision situation, including probabilities, is shown in Table 12.13.
The dollar outcomes in Table 12.13 are the costs associated with each outcome. The insurance costs $500 whether there is an accident or no accident. If the insurance is not purchased and there is no accident, then there is no cost at all. However, if an accident does occur, the individual will incur a cost of $10,000.
The expected cost (EC) for each decision is
Because the lower expected cost is $80, the decision should be not to purchase insurance. However, people almost always purchase insurance (even when they are not legally required to do so). This is true of all types of insurance, such as accident, life, or fire.
Why do people shun the greater expected dollar outcome in this type of situation? The answer is that people want to avoid a ruinous or painful situation. When faced with a rel- atively small dollar cost versus a disaster, people typically pay the small cost to avert the dis- aster. People who display this characteristic are referred to as risk averters because they avoid risky situations.
Alternatively, people who go to the track to wager on horse races, travel to Atlantic City to play roulette, or speculate in the commodities market decide to take risks even though the greatest expected value would occur if they simply held on to the money. These people shun the greater expected value accruing from a sure thing (keeping their money) in order to take a chance on receiving a “bonanza.” Such people are referred to as risk takers.
For both risk averters and risk takers (as well as those who are indifferent to risk), the decision criterion is something other than the expected dollar outcome. This alternative criterion is known as utility. Utility is a measure of the satisfaction derived from money. In our examples of risk averters and risk takers presented earlier, the utility derived from their decisions exceeded the expected dollar value. For example, the utility to the average deci- sion maker of having insurance is much greater than the utility of not having insurance.
As another example, consider two people, each of whom is offered $100,000 to perform some particularly difficult and strenuous task. One individual has an annual income of $10,000; the other individual is a multimillionaire. It is reasonable to assume that the aver- age person with an annual income of only $10,000 would leap at the opportunity to earn $100,000, whereas the multimillionaire would reject the offer. Obviously, $100,000 has more utility (i.e., value) for one individual than for the other.
In general, the same incremental amount of money does not have the same intrinsic value to every person. For individuals with a great deal of wealth, more money does not usu- ally have as much intrinsic value as it does for individuals who have little money. In other words, although the dollar value is the same, the value as measured by utility is different, depending on how much wealth a person has. Thus, utility in this case is a measure of the pleasure or satisfaction an individual would receive from an incremental increase in wealth.
EC(no insurance) = .992($0) + .0081$10,0002 = $80 EC(insurance) = .992($500) + .008($500) = $500
State of Nature
No Accident Accident
Decision .992 .008
Purchase insurance $500 $ 500 Do not purchase insurance 0 10,000
Table 12.13 Payoff table for auto insurance
example
People who forgo a high expected value to avoid a disaster with a
low probability are risk averters.
People who take a chance on a bonanza with a very low
probability of occurrence in lieu of a sure thing are risk takers.
Utility is a measure of personal satisfaction derived from money.
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Summary
The purpose of this chapter was to demonstrate the concepts and fundamentals ofdecision making when uncertainty exists. Within this context, several decision-mak- ing criteria were presented. The maximax, maximin, minimax regret, equal likelihood, and Hurwicz decision criteria were demonstrated for cases in which probabilities could not be attached to the occurrence of outcomes. The expected value criterion and deci- sion trees were discussed for cases in which probabilities could be assigned to the states of nature of a decision situation.
All the decision criteria presented in this chapter were demonstrated via rather simpli- fied examples; actual decision-making situations are usually more complex. Nevertheless, the process of analyzing decisions presented in this chapter is the logical method that most decision makers follow to make a decision.
Example Problem Solution 559
In some decision situations, decision makers attempt to assign a subjective value to utility. This value is typically measured in terms of units called utiles. For example, the $100,000 offered to the two individuals may have a utility value of 100 utiles to the person with a low income and 0 utiles to the multimillionaire.
In our automobile insurance example, the expected utility of purchasing insurance could be 1,000 utiles, and the expected utility of not purchasing insurance only 1 utile. These util- ity values are completely reversed from the expected monetary values computed from Table 12.13, which explains the decision to purchase insurance.
As might be expected, it is usually very difficult to measure utility and the number of utiles derived from a decision outcome. The process is a very subjective one in which the decision maker’s psychological preferences must be determined. Thus, although the concept of utility is realistic and often portrays actual decision-making criteria more accurately than does expected monetary value, its application is difficult and, as such, somewhat limited.
Utiles are units of subjective measures of utility.
References
Baumol, W. J. Economic Theory and Operations Analysis, 4th ed.
Upper Saddle River, NJ: Prentice Hall, 1977.
Bell, D.“Bidding for the S.S. Kuniang.” Interfaces 14, no. 2 (March–April
1984): 17–23.
Dorfman, R., Samuelson, P. A., and Solow, R. M. Linear Programming
and Economic Analysis. New York: McGraw-Hill, 1958.
Holloway, C. A. Decision Making Under Uncertainty. Upper Saddle
River, NJ: Prentice Hall, 1979.
Howard, R. A. “An Assessment of Decision Analysis.” Operations
Research 28, no. 1 (January–February 1980): 4–27.
Keeney, R. L. “Decision Analysis: An Overview.” Operations Research
30, no. 5 (September–October 1982): 803–38.
Luce, R. D., and Raiffa, H. Games and Decisions. New York: John
Wiley & Sons, 1957.
Von Neumann, J., and Morgenstern, O. Theory of Games and
Economic Behavior, 3rd ed. Princeton, NJ: Princeton University
Press, 1953.
Williams, J. D. The Compleat Strategyst, rev. ed. New York: McGraw-
Hill, 1966.
The following example will illustrate the solution procedure for a decision analysis problem.
Problem Statement
T. Bone Puckett, a corporate raider, has acquired a textile company and is contemplating the future of one of its major plants, located in South Carolina. Three alternative deci- sions are being considered: (1) expand the plant and produce lightweight, durable
Example Problem Solution
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560 Chapter 12 Decision Analysis
materials for possible sales to the military, a market with little foreign competition; (2) maintain the status quo at the plant, continuing production of textile goods that are subject to heavy foreign competition; or (3) sell the plant now. If one of the first two alternatives is chosen, the plant will still be sold at the end of a year. The amount of profit that could be earned by selling the plant in a year depends on foreign market con- ditions, including the status of a trade embargo bill in Congress. The following payoff table describes this decision situation:
State of Nature
Good Foreign Poor Foreign Decision Competitive Conditions Competitive Conditions
Expand $ 800,000 $ 500,000 Maintain status quo 1,300,000 �150,000 Sell now 320,000 320,000
A. Determine the best decision by using the following decision criteria: 1. Maximax 2. Maximin 3. Minimax regret 4. Hurwicz 5. Equal likelihood
B. Assume that it is now possible to estimate a probability of .70 that good foreign competitive conditions will exist and a probability of .30 that poor conditions will exist. Determine the best decision by using expected value and expected opportunity loss.
C. Compute the expected value of perfect information. D. Develop a decision tree, with expected values at the probability nodes. E. T. Bone Puckett has hired a consulting firm to provide a report on future political
and market situations. The report will be positive (P) or negative (N), indicating either a good (g) or poor (p) future foreign competitive situation. The conditional probability of each report outcome, given each state of nature, is
Determine the posterior probabilities by using Bayes’s rule. F. Perform a decision tree analysis by using the posterior probability obtained in (E).
Solution
Step 1 (part A): Determine Decisions Without Probabilities
Maximax:
Expand $ 800,000 Status quo 1,300,000 ← Maximum Sell 320,000
Decision: Maintain status quo.
P1N|p2 = .80 P(P|p) = .20 P(N|g) = .30 P(P|g) = .70
1a = .32
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Example Problem Solution 561
Maximin: Expand $ 500,000 ← Maximum Status quo �150,000
Sell 320,000
Decision: Expand.
Minimax regret: Expand $500,000 ← Minimum Status quo 650,000
Sell 980,000
Decision: Expand.
Hurwicz Expand $800,000(.3) + 500,000(.7) = $590,000
Status quo $1,300,000(.3) � 150,000(.7) = $285,000
Sell $320,000(.3) + 320,000(.7) = $320,000
Decision: Expand.
Equal likelihood: Expand $800,000(.50) + 500,000(.50) = $650,000
Status quo $1,300,000(.50) � 150,000(.50) = $575,000
Sell $320,000(.50) + 320,000(.50) = $320,000
Decision: Expand.
Step 2 (part B): Determine Decisions with EV and EOL
Expected value: Expand $800,000(.70) + 500,000(.30) = $710,000
Status quo $1,300,000(.70) � 150,000(.30) = $865,000
Sell $320,000(.70) + 320,000(.30) = $320,000
Decision: Maintain status quo.
Expected opportunity loss: Expand $500,000(.70) + 0(.30) = $350,000
Status quo $0(.70) + 650,000(.30) = $195,000
Sell $980,000(.70) + 180,000(.30) = $740,000
Decision: Maintain status quo.
Step 3 (part C): Compute EVPI
= $1,060,000 expected value given perfect information = 1,300,000(.70) + 500,000(.30)
1a = .32:
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562 Chapter 12 Decision Analysis
Step 4 (part D): Develop a Decision Tree
EVPI = $1,060,000 - 865,000 = $195,000 = $865,000
expected value without perfect information = $1,300,000(.70) - 150,0001.302
Step 5 (part E): Determine Posterior Probabilities
P1p|N2 = .533 = .467
= (.30)(.70)
(.30)(.70) + (.80)(.30)
P(g|N) = P(N|g)P(g)
P(N|g)P(g) + P(N|p)P(p)
P(p|P) = .109 = .891
= (.70)(.70)
(.70)(.70) + (.20)(.30)
P(g|P) = P(P|g)P(g)
P(P|g)P(g) + P(P|p)P(p)
Expand
$800,000
$500,000
$1,300,000
$320,000
$320,000
–$150,000
Good conditions (.7)
Poor conditions (.3)
Good conditions (.7)
Good conditions (.7)
Poor conditions (.3)
Poor conditions (.3)
Maintain status quo
Sell
$320,000
$865,000 $865,000
$710,000
1
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems
1. A farmer in Iowa is considering either leasing some extra land or investing in savings certificates at the local bank. If weather conditions are good next year, the extra land will give the farmer an excel- lent harvest. However, if weather conditions are bad, the farmer will lose money. The savings cer- tificates will result in the same return, regardless of the weather conditions. The return for each investment, given each type of weather condition, is shown in the following payoff table:
Weather
Decision Good Bad
Lease land $90,000 $�40,000 Buy savings certificate 10,000 10,000
Select the best decision, using the following decision criteria: a. Maximax b. Maximin
Problems 563
Step 6 (part F): Perform Decision Tree Analysis with Posterior Probabilities
$767,300
$1,141,950
$1,141,950
$916,117.50 $320,000
$640,100
$527,150
$640,100
$320,000
1
3
2
4
5
6
7
8
9
Positive report
Expand
Sell
Status quo
Expand
$800,000
$500,000
$1,300,000
$320,000
$320,000
–$150,000
P(g P) = .891
P(p P) = .109
P(p P) = .109
P(g N) = .467
P(p N) = .533
P(p N) = .533
P(g N) = .467
P(g N) = .467
P(p N) = .533
P(g P) = .891
P(g P) = .891
P(p P) = .109
Sell
Status quo
Negative report
P(N) = .45
P(P) = .55
$800,000
$500,000
$1,300,000
$320,000
$320,000
–$150,000
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
564 Chapter 12 Decision Analysis
2. The owner of the Burger Doodle Restaurant is considering two ways to expand operations: open a drive-up window or serve breakfast. The increase in profits resulting from these proposed expan- sions depends on whether a competitor opens a franchise down the street. The possible profits from each expansion in operations, given both future competitive situations, are shown in the fol- lowing payoff table:
Competitor
Decision Open Not Open
Drive-up window $�6,000 $20,000 Breakfast 4,000 8,000
Select the best decision, using the following decision criteria. a. Maximax b. Maximin
3. Stevie Stone, a bellhop at the Royal Sundown Hotel in Atlanta, has been offered a management position. Although accepting the offer would assure him a job if there was a recession, if good eco- nomic conditions prevailed, he would actually make less money as a manager than as a bellhop (because of the large tips he gets as a bellhop). His salary during the next 5 years for each job, given each future economic condition, is shown in the following payoff table:
Economic Conditions
Decision Good Recession
Bellhop $120,000 $60,000 Manager 85,000 85,000
Select the best decision, using the following decision criteria. a. Minimax regret b. Hurwicz c. Equal likelihood
4. Brooke Bentley, a student in business administration, is trying to decide which management sci- ence course to take next quarter—I, II, or III. “Steamboat” Fulton, “Death” Ray, and “Sadistic” Scott are the three management science professors who teach the courses. Brooke does not know who will teach what course. Brooke can expect a different grade in each of the courses, depending on who teaches it next quarter, as shown in the following payoff table:
Professor
Course Fulton Ray Scott
I B D D II C B F III F A C
Determine the best course to take next quarter, using the following criteria. a. Maximax b. Maximin
1a = .42
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 565
5. A farmer in Georgia must decide which crop to plant next year on his land: corn, peanuts, or soybeans. The return from each crop will be determined by whether a new trade bill with Russia passes the Senate. The profit the farmer will realize from each crop, given the two possible results on the trade bill, is shown in the following payoff table:
Trade Bill
Crop Pass Fail
Corn $35,000 $ 8,000 Peanuts 18,000 12,000 Soybeans 22,000 20,000
Determine the best crop to plant, using the following decision criteria. a. Maximax b. Maximin c. Minimax regret d. Hurwicz e. Equal likelihood
6. A company must decide now which of three products to make next year to plan and order proper materials. The cost per unit of producing each product will be determined by whether a new union labor contract passes or fails. The cost per unit for each product, given each contract result, is shown in the following payoff table:
Contract Outcome
Product Pass Fail
1 $7.50 $6.00 2 4.00 7.00 3 6.50 3.00
Determine which product should be produced, using the following decision criteria. a. Minimin b. Minimax
7. The owner of the Columbia Construction Company must decide between building a housing development, constructing a shopping center, and leasing all the company’s equipment to another company. The profit that will result from each alternative will be determined by whether material costs remain stable or increase. The profit from each alternative, given the two possibilities for material costs, is shown in the following payoff table:
Material Costs
Decision Stable Increase
Houses $ 70,000 $30,000 Shopping center 105,000 20,000 Leasing 40,000 40,000
Determine the best decision, using the following decision criteria. a. Maximax b. Maximin
1a = .32
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
566 Chapter 12 Decision Analysis
c. Minimax regret d. Hurwicz e. Equal likelihood
8. A local real estate investor in Orlando is considering three alternative investments: a motel, a restaurant, or a theater. Profits from the motel or restaurant will be affected by the availability of gasoline and the number of tourists; profits from the theater will be relatively stable under any conditions. The following payoff table shows the profit or loss that could result from each investment:
Gasoline Availability
Investment Shortage Stable Supply Surplus
Motel $�8,000 $15,000 $20,000 Restaurant 2,000 8,000 6,000 Theater 6,000 6,000 5,000
Determine the best investment, using the following decision criteria. a. Maximax b. Maximin c. Minimax regret d. Hurwicz e. Equal likelihood
9. A television network is attempting to decide during the summer which of the following three foot- ball games to televise on the Saturday following Thanksgiving Day: Alabama versus Auburn, Georgia versus Georgia Tech, or Army versus Navy. The estimated viewer ratings (millions of homes) for the games depend on the win–loss records of the six teams, as shown in the following payoff table:
Number of Viewers (1,000,000s)
One Team Has Both Teams Winning Record; Both Teams
Have Winning One Team Has Have Losing Game Records Losing Record Records
Alabama vs. Auburn 10.2 7.3 5.4 Georgia vs. Georgia Tech 9.6 8.1 4.8 Army vs. Navy 12.5 6.5 3.2
Determine the best game to televise, using the following decision criteria. a. Maximax b. Maximin c. Equal likelihood
10. Ann Tyler has come into an inheritance from her grandparents. She is attempting to decide among several investment alternatives. The return after 1 year is primarily dependent on the interest rate during the next year. The rate is currently 7%, and Ann anticipates that it will stay the same or go
1a = .42
1a = .22
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 567
up or down by at most two points. The various investment alternatives plus their returns ($10,000s), given the interest rate changes, are shown in the following table:
Interest Rate
Investments 5% 6% 7% 8% 9%
Money market fund 2 3.1 4 4.3 5 Stock growth fund �3 �2 2.5 4 6 Bond fund 6 5 3 3 2 Government fund 4 3.6 3.2 3 2.8 Risk fund �9 �4.5 1.2 8.3 14.7 Savings bonds 3 3 3.2 3.4 3.5
Determine the best investment, using the following decision criteria. a. Maximax b. Maximin c. Equal likelihood
11. The Tech football coaching staff has six basic offensive plays it runs every game. Tech has an upcoming game against State on Saturday, and the Tech coaches know that State employs five dif- ferent defenses. The coaches have estimated the number of yards Tech will gain with each play against each defense, as shown in the following payoff table:
Defense
Play 54 63 Wide Tackle Nickel Blitz
Off tackle 3 �2 9 7 �1 Option �1 8 �2 9 12 Toss sweep 6 16 �5 3 14 Draw �2 4 3 10 �3 Pass 8 20 12 �7 �8 Screen �5 �2 8 3 16
a. If the coaches employ an offensive game plan, they will use the maximax criterion. What will be their best play?
b. If the coaches employ a defensive plan, they will use the maximin criterion. What will be their best play?
c. What will be their best offensive play if State is equally likely to use any of its five defenses?
12. Microcomp is a U.S.-based manufacturer of personal computers. It is planning to build a new manufacturing and distribution facility in either South Korea, China, Taiwan, the Philippines, or Mexico. It will take approximately 5 years to build the necessary infrastructure (roads, etc.), construct the new facility, and put it into operation. The eventual cost of the facility will differ between countries and will even vary within countries depending on the financial, labor, and polit- ical climate, including monetary exchange rates. The company has estimated the facility cost (in $1,000,000s) in each country under three different future economic and political climates, as follows:
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
568 Chapter 12 Decision Analysis
Economic/Political Climate
Country Decline Same Improve
South Korea 21.7 19.1 15.2 China 19.0 18.5 17.6 Taiwan 19.2 17.1 14.9 Philippines 22.5 16.8 13.8 Mexico 25.0 21.2 12.5
Determine the best decision, using the following decision criteria. a. Minimin b. Minimax c. Hurwicz d. Equal likelihood
13. Place-Plus, a real estate development firm, is considering several alternative development projects. These include building and leasing an office park, purchasing a parcel of land and building an office building to rent, buying and leasing a warehouse, building a strip mall, and building and selling con- dominiums. The financial success of these projects depends on interest rate movement in the next 5 years. The various development projects and their 5-year financial return (in $1,000,000s) given that interest rates will decline, remain stable, or increase, are shown in the following payoff table:
Interest Rate
Project Decline Stable Increase
Office park $0.5 $1.7 $4.5 Office building 1.5 1.9 2.5 Warehouse 1.7 1.4 1.0 Mall 0.7 2.4 3.6 Condominiums 3.2 1.5 0.6
Determine the best investment, using the following decision criteria. a. Maximax b. Maximin c. Equal likelihood d. Hurwicz
14. The Oakland Bombers professional basketball team just missed making the playoffs last season and believes it needs to sign only one very good free agent to make the playoffs next season. The team is considering four players: Barry Byrd, Rayneal O’Neil, Marvin Johnson, and Michael Gordan. Each player differs according to position, ability, and attractiveness to fans. The payoffs (in $1,000,000s) to the team for each player, based on the contract, profits from attendance, and team product sales for several different season outcomes, are provided in the following table:
Season Outcome
Makes Player Loser Competitive Playoffs
Byrd $�3.2 $ 1.3 4.4 O’Neil �5.1 1.8 6.3 Johnson �2.7 0.7 5.8 Gordan �6.3 �1.6 9.6
1a = .32
1a = .42
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 569
Determine the best decision, using the following decision criteria. a. Maximax b. Maximin c. Hurwicz d. Equal likelihood
15. A machine shop owner is attempting to decide whether to purchase a new drill press, a lathe, or a grinder. The return from each will be determined by whether the company succeeds in getting a government military contract. The profit or loss from each purchase and the probabilities associ- ated with each contract outcome are shown in the following payoff table:
Contract No Contract Purchase .40 .60
Drill press $40,000 $�8,000 Lathe 20,000 4,000 Grinder 12,000 10,000
Compute the expected value for each purchase and select the best one.
16. A concessions manager at the Tech versus A&M football game must decide whether to have the vendors sell sun visors or umbrellas. There is a 30% chance of rain, a 15% chance of overcast skies, and a 55% chance of sunshine, according to the weather forecast in College Junction, where the game is to be held. The manager estimates that the following profits will result from each decision, given each set of weather conditions:
Weather Conditions
Decision Rain Overcast Sunshine .30 .15 .55
Sun visors $�500 $�200 $1,500 Umbrellas 2,000 0 �900
a. Compute the expected value for each decision and select the best one. b. Develop the opportunity loss table and compute the expected opportunity loss for each decision.
17. Allen Abbott has a wide-curving, uphill driveway leading to his garage. When there is a heavy snow, Allen hires a local carpenter, who shovels snow on the side in the winter, to shovel his driveway. The snow shoveler charges $30 to shovel the driveway. Following is a probability distribution of the number of heavy snows each winter:
Heavy Snows Probability
1 .13 2 .18 3 .26 4 .23 5 .10 6 .07 7 .03
1.00
1a = .602
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Allen is considering purchasing a new self-propelled snowblower for $625 that would allow him, his wife, or his children to clear the driveway after a snow. Discuss what you think Allen’s decision should be and why.
18. The Miramar Company is going to introduce one of three new products: a widget, a hummer, or a nimnot. The market conditions (favorable, stable, or unfavorable) will determine the profit or loss the company realizes, as shown in the following payoff table:
Market Conditions
Favorable Stable Unfavorable Product .2 .7 .1
Widget $120,000 $70,000 $�30,000 Hummer 60,000 40,000 20,000 Nimnot 35,000 30,000 30,000
a. Compute the expected value for each decision and select the best one. b. Develop the opportunity loss table and compute the expected opportunity loss for each
product. c. Determine how much the firm would be willing to pay to a market research firm to gain
better information about future market conditions.
19. The financial success of the Downhill Ski Resort in the Blue Ridge Mountains is dependent on the amount of snowfall during the winter months. If the snowfall averages more than 40 inches, the resort will be successful; if the snowfall is between 20 and 40 inches, the resort will receive a mod- erate financial return; and if snowfall averages less than 20 inches, the resort will suffer a financial loss. The financial return and probability, given each level of snowfall, follow:
Snowfall Level (in.) Financial Return
>40, .4 $ 120,000 20–40, .2 40,000 <20, .4 �40,000
A large hotel chain has offered to lease the resort for the winter for $40,000. Compute the expected value to determine whether the resort should operate or lease. Explain your answer.
20. An investor must decide between two alternative investments—stocks and bonds. The return for each investment, given two future economic conditions, is shown in the following payoff table:
Economic Conditions
Investment Good Bad
Stocks $10,000 $�4,000 Bonds 7,000 2,000
What probability for each economic condition would make the investor indifferent to the choice between stocks and bonds?
570 Chapter 12 Decision Analysis
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 571
21. In Problem 10, Ann Tyler, with the help of a financial newsletter and some library research, has been able to assign probabilities to each of the possible interest rates during the next year, as follows:
Interest Rate (%) Probability
5 .2 6 .3 7 .3 8 .1 9 .1
Using expected value, determine her best investment decision.
22. In Problem 11 the Tech coaches have reviewed game films and have determined the following prob- abilities that State will use each of its defenses:
Defense Probability
54 .40 63 .10 Wide tackle .20 Nickel .20 Blitz .10
a. Using expected value, rank Tech’s plays from best to worst. b. During the actual game, a situation arises in which Tech has a third down and 10 yards
to go, and the coaches are 60% certain State will blitz, with a 10% chance of any of the other four defenses. What play should Tech run? Is it likely the team will make the first down?
23. A global economist hired by Microcomp, the U.S.-based computer manufacturer in Problem 12, estimates that the probability that the economic and political climate overseas and in Mexico will decline during the next 5 years is .40, the probability that it will remain approximately the same is .50, and the probability that it will improve is .10. Determine the best country to construct the new facility in and the expected value of perfect information.
24. In Problem 13 the Place-Plus real estate development firm has hired an economist to assign a prob- ability to each direction interest rates may take over the next 5 years. The economist has determined that there is a .50 probability that interest rates will decline, a .40 probability that rates will remain stable, and a .10 probability that rates will increase.
a. Using expected value, determine the best project. b. Determine the expected value of perfect information.
25. Fenton and Farrah Friendly, husband-and-wife car dealers, are soon going to open a new dealer- ship. They have three offers: from a foreign compact car company, from a U.S.-producer of full- sized cars, and from a truck company. The success of each type of dealership will depend on how
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
572 Chapter 12 Decision Analysis
much gasoline is going to be available during the next few years. The profit from each type of deal- ership, given the availability of gas, is shown in the following payoff table:
Gasoline Availability
Dealership Shortage Surplus .6 .4
Compact cars $ 300,000 $150,000 Full-sized cars �100,000 600,000 Trucks 120,000 170,000
Determine which type of dealership the couple should purchase.
26. The Steak and Chop Butcher Shop purchases steak from a local meatpacking house. The meat is purchased on Monday at $2.00 per pound, and the shop sells the steak for $3.00 per pound. Any steak left over at the end of the week is sold to a local zoo for $.50 per pound. The possible demands for steak and the probability of each are shown in the following table:
Demand (lb.) Probability
20 .10 21 .20 22 .30 23 .30 24 .10
1.00
The shop must decide how much steak to order in a week. Construct a payoff table for this deci- sion situation and determine the amount of steak that should be ordered, using expected value.
27. The Loebuck Grocery must decide how many cases of milk to stock each week to meet demand. The probability distribution of demand during a week is shown in the following table:
Demand (cases) Probability
15 .20 16 .25 17 .40 18 .15
1.00
Each case costs the grocer $10 and sells for $12. Unsold cases are sold to a local farmer (who mixes the milk with feed for livestock) for $2 per case. If there is a shortage, the grocer considers the cost of customer ill will and lost profit to be $4 per case. The grocer must decide how many cases of milk to order each week.
a. Construct the payoff table for this decision situation. b. Compute the expected value of each alternative amount of milk that could be stocked and
select the best decision. c. Construct the opportunity loss table and determine the best decision. d. Compute the expected value of perfect information.
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 573
28. The manager of the greeting card section of Mazey’s department store is considering her order for a particular line of Christmas cards. The cost of each box of cards is $3; each box will be sold for $5 during the Christmas season. After Christmas, the cards will be sold for $2 a box. The card section manager believes that all leftover cards can be sold at that price. The estimated demand during the Christmas season for the line of Christmas cards, with associated probabilities, is as follows:
Demand (boxes) Probability
25 .10 26 .15 27 .30 28 .20 29 .15 30 .10
a. Develop the payoff table for this decision situation. b. Compute the expected value for each alternative and identify the best decision. c. Compute the expected value of perfect information.
29. The Palm Garden Greenhouse specializes in raising carnations that are sold to florists. Carnations are sold for $3.00 per dozen; the cost of growing the carnations and distributing them to the florists is $2.00 per dozen. Any carnations left at the end of the day are sold to local restaurants and hotels for $0.75 per dozen. The estimated cost of customer ill will if demand is not met is $1.00 per dozen. The expected daily demand (in dozens) for the carnations is as follows:
Daily Demand Probability
20 .05 22 .10 24 .25 26 .30 28 .20 30 .10
1.00
a. Develop the payoff table for this decision situation. b. Compute the expected value of each alternative number of (dozens of ) carnations that
could be stocked and select the best decision. c. Construct the opportunity loss table and determine the best decision. d. Compute the expected value of perfect information.
30. Assume that the probabilities of demand in Problem 28 are no longer valid; the decision situation is now one without probabilities. Determine the best number of cards to stock, using the following decision criteria.
a. Maximin b. Maximax c. Hurwicz d. Minimax regret
1a = .42
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
574 Chapter 12 Decision Analysis
31. In Problem 14, the Bombers’ management has determined the following probabilities of the occur- rence of each future season outcome for each player:
Probability
Player Loser Competitive Makes Playoffs
Byrd .15 .55 .30 O’Neil .18 .26 .56 Johnson .21 .32 .47 Gordan .30 .25 .45
Compute the expected value for each player and indicate which player the team should try to sign.
32. The director of career advising at Orange Community College wants to use decision analysis to provide information to help students decide which 2-year degree program they should pursue. The director has set up the following payoff table for six of the most popular and successful degree pro- grams at OCC that shows the estimated 5-year gross income ($) from each degree for four future economic conditions:
Economic Conditions
Degree Program Recession Average Good Robust
Graphic design 145,000 175,000 220,000 260,000 Nursing 150,000 180,000 205,000 215,000 Real estate 115,000 165,000 220,000 320,000 Medical technology 130,000 180,000 210,000 280,000 Culinary technology 115,000 145,000 235,000 305,000 Computer information
technology 125,000 150,000 190,000 250,000
Determine the best degree program in terms of projected income, using the following decision criteria:
a. Maximax b. Maximin c. Equal likelihood d. Hurwicz
33. In Problem 32 the director of career advising at Orange Community College has paid a small fee to a local investment firm to indicate a probability for each future economic condition over the next 5 years. The firm estimates that there is a .20 probability of a recession, a .40 probability that the economy will be average, a .30 probability that the economy will be good, and a .10 probability that it will be robust. Using expected value determine the best degree program in terms of projected income. If you were the director of career advising, which degree program would you recommend?
34. The Blue Sox American League baseball team is going to enter the free-agent market over the win- ter to sign a new starting pitcher. They are considering five prospects who will enter the free-agent market. All five pitchers are in their mid-20s, have been in the major leagues for approximately 5 years, and have been relatively successful. The team’s general manager has compiled a lot of infor- mation about the pitchers from scouting reports and their playing histories since high school. He has developed a chart projecting how many wins each pitcher will likely have during the next
1a = .502
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 575
10 years given three possible future states of nature: the pitchers will be relatively injury free, they will have a normal career with injuries, or they will have excessive injuries, as shown in the follow- ing payoff table:
Physical Condition
Pitcher No injuries Normal Excessive injuries
Jose Diaz 153 122 76 Jerry Damon 173 135 46 Frank Thompson 133 115 88 Derek Rodriguez 105 101 95 Ken Griffin 127 98 75
Determine the best pitcher to sign, using the following decision criteria: a. Maximax b. Maximin c. Equal likelihood d. Hurwicz
35. In Problem 34 the Blue Sox general manager has asked a superscout to assign a probability to each of the three states of nature for the pitchers during the next 10 years. The scout estimates there is a .10 probability that these pitchers at this stage of their careers will have no injuries, a .60 proba- bility that they will have a career with the normal number of injuries, and a .30 probability that they will have excessive injuries.
a. Using expected value, determine the best pitcher to sign. b. Given the following 10-year contract price for each pitcher (in $millions), which would you
recommend signing?
Jose Diaz $ 97.3 Jerry Damon $121.5 Frank Thompson $ 73.5 Derek Rodriguez $103.4 Ken Griffin $ 85.7
c. Suppose that the general manager asked the superscout to determine the probabilities of each state of nature for each individual pitcher, and the results were as follows:
Physical Condition
Pitcher No injuries Normal Excessive injuries
Jose Diaz 0.30 0.45 0.25 Jerry Damon 0.13 0.47 0.40 Frank Thompson 0.45 0.35 0.20 Derek Rodriguez 0.50 0.40 0.10 Ken Griffin 0.15 0.60 0.25
Determine the expected number of wins for each pitcher, and combined with the contract price in part b, indicate which pitcher you would recommend signing.
36. Construct a decision tree for the decision situation described in Problem 25 and indicate the best decision.
1a = .352
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576 Chapter 12 Decision Analysis
37. Given the following sequential decision tree, determine which is the optimal investment, A or B:
4
7
2
1
5
8
6
9
3
.40 $300,000
$60,000
$45,000
$75,000
.60
(–$20,000)
.30
.70 .20 $200,000
$70,000
$60,000
.80
(–$17,000)
.35 $105,000
$40,000
$55,000
$80,000
.65
(–$9,000)
.55
.30
.15
Investment A (–$70,000)
Investment B (–$50,000)
38. The management of First American Bank was concerned about the potential loss that might occur in the event of a physical catastrophe such as a power failure or a fire. The bank estimated that the loss from one of these incidents could be as much as $100 million, including losses due to inter- rupted service and customer relations. One project the bank is considering is the installation of an emergency power generator at its operations headquarters. The cost of the emergency generator is $800,000, and if it is installed, no losses from this type of incident will be incurred. However, if the generator is not installed, there is a 10% chance that a power outage will occur during the next year. If there is an outage, there is a .05 probability that the resulting losses will be very large, or approx- imately $80 million in lost earnings. Alternatively, it is estimated that there is a .95 probability of only slight losses of around $1 million. Using decision tree analysis, determine whether the bank should install the new power generator.
39. The Americo Oil Company is considering making a bid for a shale oil development contract to be awarded by the federal government. The company has decided to bid $112 million. The company estimates that it has a 60% chance of winning the contract with this bid. If the firm wins the con- tract, it can choose one of three methods for getting the oil from the shale. It can develop a new method for oil extraction, use an existing (inefficient) process, or subcontract the processing to a number of smaller companies once the shale has been excavated. The results from these alterna- tives are as follows:
Develop new process:
Outcomes Probability Profit ($1,000,000s)
Great success .30 $ 600 Moderate success .60 300 Failure .10 �100
Use present process:
Outcomes Probability Profit ($1,000,000s)
Great success .50 $ 300 Moderate success .30 200 Failure .20 �40
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 577
Subcontract:
Outcome Probability Profit ($1,000,000s)
Moderate success 1.00 250
The cost of preparing the contract proposal is $2 million. If the company does not make a bid, it will invest in an alternative venture with a guaranteed profit of $30 million. Construct a sequen- tial decision tree for this decision situation and determine whether the company should make a bid.
40. The machine shop owner in Problem 15 is considering hiring a military consultant to ascertain whether the shop will get the government contract. The consultant is a former military officer who uses various personal contacts to find out such information. By talking to other shop owners who have hired the consultant, the owner has estimated a .70 probability that the consultant would pre- sent a favorable report, given that the contract is awarded to the shop, and a .80 probability that the consultant would present an unfavorable report, given that the contract is not awarded. Using deci- sion tree analysis, determine the decision strategy the owner should follow, the expected value of this strategy, and the maximum fee the owner should pay the consultant.
41. The Miramar Company in Problem 18 is considering contracting with a market research firm to do a survey to determine future market conditions. The results of the survey will indicate either positive or negative market conditions. There is a .60 probability of a positive report, given favor- able conditions; a .30 probability of a positive report, given stable conditions; and a .10 probabili- ty of a positive report, given unfavorable conditions. There is a .90 probability of a negative report, given unfavorable conditions; a .70 probability, given stable conditions; and a .40 probability, given favorable conditions. Using decision tree analysis and posterior probability tables, determine the decision strategy the company should follow, the expected value of the strategy, and the maximum amount the company should pay the market research firm for the survey results.
42. The Friendlys in Problem 25 are considering hiring a petroleum analyst to determine the future availability of gasoline. The analyst will report that either a shortage or a surplus will occur. The probability that the analyst will indicate a shortage, given that a shortage actually occurs is .90; the probability that the analyst will indicate a surplus, given that a surplus actually occurs is .70.
a. Determine the decision strategy the Friendlys should follow, the expected value of this strategy, and the maximum amount the Friendlys should pay for the analyst’s services.
b. Compute the efficiency of the sample information for the Friendly car dealership.
43. Jeffrey Mogul is a Hollywood film producer, and he is currently evaluating a script by a new screen- writer and director, Betty Jo Thurston. Jeffrey knows that the probability of a film by a new direc- tor being a success is about .10 and that the probability it will flop is .90. The studio accounting department estimates that if this film is a hit, it will make $25 million in profit, whereas if it is a box office failure, it will lose $8 million. Jeffrey would like to hire noted film critic Dick Roper to read the script and assess its chances of success. Roper is generally able to correctly predict a suc- cessful film 70% of the time and correctly predict an unsuccessful film 80% of the time. Roper wants a fee of $1 million. Determine whether Roper should be hired, the strategy Mogul should fol- low if Roper is hired, and the expected value.
44. Tech is playing State in the last conference game of the season. Tech is trailing State 21 to 14, with 7 seconds left in the game, when Tech scores a touchdown. Still trailing 21 to 20, Tech can either go for 2 points and win or go for 1 point to send the game into overtime. The conference champi- onship will be determined by the outcome of this game. If Tech wins, it will go to the Sugar Bowl, with a payoff of $7.2 million; if it loses, it will go to the Gator Bowl, with a payoff of $1.7 million. If Tech goes for 2 points, there is a 33% chance it will be successful and win (and a 67% chance it will fail and lose). If it goes for 1 point, there is a 0.98 probability of success and a tie and a 0.02
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probability of failure. If the teams tie, they will play overtime, during which Tech believes it has only a 20% chance of winning because of fatigue.
a. Use decision tree analysis to determine whether Tech should go for 1 or 2 points. b. What would Tech’s probability of winning the game in overtime have to be to make Tech
indifferent to going for either 1 or 2 points?
45. Jay Seago is suing the manufacturer of his car for $3.5 million because of a defect that he believes caused him to have an accident. The accident kept him out of work for a year. The company has offered him a settlement of $700,000, of which Jay would receive $600,000 after attorneys’ fees. His attorney has advised him that he has a 50% chance of winning his case. If he loses, he will incur attorneys’ fees and court costs of $75,000. If he wins, he is not guaranteed his full requested settle- ment. His attorney believes that there is a 50% chance he could receive the full settlement, in which case Jay would realize $2 million after his attorney takes her cut, and a 50% chance that the jury will award him a lesser amount of $1 million, of which Jay would get $500,000.
Using decision tree analysis, decide whether Jay should proceed with his lawsuit against the manufacturer.
46. Tech has three health care plans for its faculty and staff to choose from, as follows:
Plan 1—monthly cost of $32, with a $500 deductible; the participants pay the first $500 of medical costs for the year; the insurer pays 90% of all remaining expenses.
Plan 2—monthly cost of $5 but a deductible of $1,200, with the insurer paying 90% of medical expenses after the insurer pays the first $1,200 in a year.
Plan 3—monthly cost of $24, with no deductible; the participants pay 30% of all expenses, with the remainder paid by the insurer.
Tracy McCoy, an administrative assistant in the management science department, estimates that her annual medical expenses are defined by the following probability distribution:
Annual Medical Expenses Probability
$ 100 .15 500 .30
1,500 .35 3,000 .10 5,000 .05
10,000 .05
Determine which medical plan Tracy should select.
47. The Valley Wine Company purchases grapes from one of two nearby growers each season to pro- duce a particular red wine. It purchases enough grapes to produce 3,000 bottles of the wine. Each grower supplies a certain portion of poor-quality grapes, resulting in a percentage of bottles being used as fillers for cheaper table wines, according to the following probability distribution:
Probability of Percentage Defective
Defective (%) Grower A Grower B
2 .15 .30 4 .20 .30 6 .25 .20 8 .30 .10
10 .10 .10
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Problems 579
The two growers charge different prices for their grapes and, because of differences in taste, the company charges different prices for its wine, depending on which grapes it uses. Following is the annual profit from the wine produced from each grower’s grapes for each percentage defective:
Profit
Defective (%) Grower A Grower B
2 $44,200 $42,600 4 40,200 40,300 6 36,200 38,000 8 32,200 35,700
10 28,200 33,400
Use decision tree analysis to determine from which grower the company should purchase grapes.
48. Kroft Food Products is attempting to decide whether it should introduce a new line of salad dress- ings called Special Choices. The company can test market the salad dressings in selected geographic areas or bypass the test market and introduce the product nationally. The cost of the test market is $150,000. If the company conducts the test market, it must wait to see the results before deciding whether to introduce the salad dressings nationally. The probability of a positive test market result is estimated to be 0.6. Alternatively, the company can decide not to conduct the test market and go ahead and make the decision to introduce the dressings or not. If the salad dressings are intro- duced nationally and are a success, the company estimates that it will realize an annual profit of $1.6 million, whereas if the dressings fail, it will incur a loss of $700,000. The company believes the probability of success for the salad dressings is 0.50 if they are introduced without the test market. If the company does conduct the test market and it is positive, then the probability of successfully introducing the salad dressings increases to 0.8. If the test market is negative and the company introduces the salad dressings anyway, the probability of success drops to 0.30.
Using decision tree analysis, determine whether the company should conduct the test market.
49. In Problem 48, determine the expected value of sample information (EVSI) (i.e., the test market value) and the expected value of perfect information (EVPI).
50. Ellie Daniels has $200,000 and is considering three mutual funds for investment—a global fund, an index fund, and an Internet stock fund. During the first year of investment, Ellie estimates that there is a .70 probability that the market will go up and a .30 probability that the market will go down. Following are the returns on her $200,000 investment at the end of the year under each mar- ket condition:
Market Conditions
Fund Up Down
Global $25,000 $ �8,000 Index 35,000 5,000 Internet 60,000 �35,000
At the end of the first year, Ellie will either reinvest the entire amount plus the return or sell and take the profit or loss. If she reinvests, she estimates that there is a .60 probability the market will go up and a .40 probability the market will go down. If Ellie reinvests in the global fund after it has gone up, her return on her initial $200,000 investment plus her $25,000 return after 1 year will be
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$45,000. If the market goes down, her loss will be $15,000. If she reinvests after the market has gone down, her return will be $34,000, and her loss will be $17,000. If Ellie reinvests in the index fund after the market has gone up, after 2 years her return will be $65,000 if the market continues upward, but only $5,000 if the market goes down. Her return will be $55,000 if she reinvests and the market reverses itself and goes up after initially going down, and it will be $5,000 if the market continues to go down. If Ellie invests in the Internet fund, she will make $60,000 if the market goes up, but she will lose $35,000 if it goes down. If she reinvests as the market continues upward, she will make an additional $100,000; but if the market reverses and goes down, she will lose $70,000. If she reinvests after the market has initially gone down, she will make $65,000, but if the market continues to go down, she will lose an additional $75,000.
Using decision tree analysis, determine which fund Ellie should invest in and its expected value.
51. Blue Ridge Power and Light is an electric utility company with a large fleet of vehicles, including automobiles, light trucks, and construction equipment. The company is evaluating four alternative strategies for maintaining its vehicles at the lowest cost: (1) do no preventive maintenance at all and repair vehicle components when they fail; (2) take oil samples at regular intervals and perform whatever preventive maintenance is indicated by the oil analysis; (3) change the vehicle oil on a reg- ular basis and perform repairs when needed; (4) change the oil at regular intervals, take oil samples regularly, and perform maintenance repairs as indicated by the sample analysis.
For autos and light trucks, strategy 1 (no preventive maintenance) costs nothing to implement and results in two possible outcomes: There is a .10 probability that a defective component will occur, requiring emergency maintenance at a cost of $1,200, or there is a .90 probability that no defects will occur and no maintenance will be necessary.
Strategy 2 (take oil samples) costs $20 to implement (i.e., take a sample), and there is a .10 prob- ability that there will be a defective part and .90 probability that there will not be a defect. If there is actually a defective part, there is a .70 probability that the sample will correctly identify it, result- ing in preventive maintenance at a cost of $500. However, there is a .30 probability that the sample will not identify the defect and indicate that everything is okay, resulting in emergency mainte- nance later at a cost of $1,200. On the other hand, if there are actually no defects, there is a .20 prob- ability that the sample will erroneously indicate that there is a defect, resulting in unnecessary maintenance at a cost of $250. There is an .80 probability that the sample will correctly indicate that there are no defects, resulting in no maintenance and no costs.
Strategy 3 (changing the oil regularly) costs $14.80 to implement and has two outcomes: a .04 probability of a defective component, which will require emergency maintenance at a cost of $1,200, and a .96 probability that no defects will occur, resulting in no maintenance and no cost.
Strategy 4 (changing the oil and sampling) costs $34.80 to implement and results in the same probabilities of defects and no defects as strategy 3. If there is a defective component, there is a .70 probability that the sample will detect it and $500 in preventive maintenance costs will be incurred. Alternatively, there is a .30 probability that the sample will not detect the defect, resulting in emer- gency maintenance at a cost of $1,200. If there is no defect, there is a .20 probability that the sam- ple will indicate that there is a defect, resulting in an unnecessary maintenance cost of $250, and there is an .80 probability that the sample will correctly indicate no defects, resulting in no cost.
Develop a decision strategy for Blue Ridge Power and Light and indicate the expected value of this strategy.1
52. In Problem 51, the decision analysis is for automobiles and light trucks. Blue Ridge Power and Light would like to reformulate the problem for its heavy construction equipment. Emergency maintenance is much more expensive for heavy equipment, costing $15,000. Required preventive
580 Chapter 12 Decision Analysis
1This problem is based on J. Mellichamp, D. Miller, and O.-J. Kwon, “The Southern Company Uses a Probability Model for Cost Justification of Oil Sample Analysis,” Interfaces 23, no. 3 (May–June 1993): 118–24.
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maintenance costs $2,000, and unnecessary maintenance costs $1,200. The cost of an oil change is $100, and the cost of taking an oil sample and analyzing it is $30. All the probabilities remain the same. Determine the strategy Blue Ridge Power and Light should use for its heavy equipment.
53. In Problem 14, the management of the Oakland Bombers is considering hiring superscout Jerry McGuire to evaluate the team’s chances for the coming season. McGuire will evaluate the team, assuming that it will sign one of the four free agents. The team’s management has determined the probability that the team will have a losing record with any of the free agents to be .21 by averag- ing the probabilities of losing for the four free agents in Problem 31. The probability that the team will have a competitive season but not make the playoffs is developed similarly, and it is .35. The probability that the team will make the playoffs is .44. The probability that McGuire will correctly predict that the team will have a losing season is .75, whereas the probability that he will predict a competitive season, given that it has a losing season, is .15, and the probability that he will incor- rectly predict a playoff season, given that the team has a losing season, is .10. The probability that he will successfully predict a competitive season is .80, whereas the probability that he will incor- rectly predict a losing season, given that the team is competitive, is .10, and the probability that he will incorrectly predict a playoff season, given the team has a competitive season, is .10. The prob- ability that he will correctly predict a playoff season is .85, whereas the probability that he will incorrectly predict a losing season, given that the team makes the playoffs, is .05, and the probabil- ity that he will predict a competitive season, given the team makes the playoffs, is .10. Using deci- sion tree analysis and posterior probabilities, determine the decision strategy the team should fol- low, the expected value of the strategy, and the maximum amount the team should pay for Jerry McGuire’s predictions.
54. The Place-Plus real estate development firm in Problem 24 is dissatisfied with the economist’s esti- mate of the probabilities of future interest rate movement, so it is considering having a financial consulting firm provide a report on future interest rates. The consulting firm is able to cite a track record which shows that 80% of the time when interest rates declined, it had predicted they would, whereas 10% of the time when interest rates declined, the firm had predicted they would remain stable and 10% of the time it had predicted they would increase. The firm has been correct 70% of the time when rates have remained stable, whereas 10% of the time it has incorrectly predicted that rates would decrease, and 20% of the time it has incorrectly predicted that rates would increase. The firm has correctly predicted that interest rates would increase 90% of the time and incorrectly predicted rates would decrease 2% and remain stable 8% of the time. Assuming that the consult- ing firm could supply an accurate report, determine how much Place-Plus should be willing to pay the consulting firm and how efficient the information will be.
55. A young couple has $5,000 to invest in either savings bonds or a real estate deal. The expected return on each investment, given good and bad economic conditions, is shown in the following payoff table:
Economic Conditions
Investment Good Bad .6 .4
Savings bonds $ 1,000 $ 1,000 Real estate 10,000 �2,000
The expected value of investing in savings bonds is $1,000, and the expected value of the real estate investment is $5,200. However, the couple decides to invest in savings bonds. Explain the couple’s decision in terms of the utility they might associate with each investment.
Problems 581
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56. Annie Hays recently sold a condominium she had bought and lived in while she was a college student over 15 years ago. She received $200,000 for the condominium and is considering two investment alternatives. Annie can invest the entire amount in a bank money market for 1 year at 8% interest and thus receive $162,000 at the end of a year, or she can invest in a speculative oil exploration project with a 50–50 chance of doubling her investment at the end of the year or losing everything.
a. Determine which alternative Annie should invest in, according to the expected value criterion, and indicate whether that is the alternative you would also select.
b. Suppose Annie determined that the probability of success for the oil exploration investment would have to be .80 before she would be indifferent between the two alternatives. In other words, if the probability of success for the oil exploration investment was less than .80, Annie would invest in the money market, but if the probability of success was greater than .80, she would invest in the oil exploration. In this case, .80 is the utility value of the $162,000 safe return. Determine the preferred alternative, according to the expected utility value.
582 Chapter 12 Decision Analysis
Case Problem
Steeley Associates Versus Concord Falls
Steeley Associates, Inc., a property development firm, pur-chased an old house near the town square in Concord Falls, where State University is located. The old house was built in the mid-1800s, and Steeley Associates restored it. For almost a decade, Steeley has leased it to the university for academic office space. The house is located on a wide lawn and has become a town landmark.
However, in 2008, the lease with the university expired, and Steeley Associates decided to build high-density student apart- ments on the site, using all the open space. The community was outraged and objected to the town council. The legal counsel for the town spoke with a representative from Steeley and hinted that if Steeley requested a permit, the town would probably reject it. Steeley had reviewed the town building code and felt confident that its plan was within the guidelines, but that did not necessarily mean that it could win a lawsuit against the town to force the town to grant a permit.
The principals at Steeley Associates held a series of meetings to review their alternatives. They decided that they had three options: They could request the permit, they could sell the prop- erty, or they could request a permit for a low-density office build- ing, which the town had indicated it would not fight. Regarding the last two options, if Steeley sells the house and property, it thinks it can get $900,000. If it builds a new office building, its return will depend on town business growth in the future. It feels that there is a 70% chance of future growth, in which case Steeley will see a return of $1.3 million (over a 10-year planning horizon); if no growth (or erosion) occurs, it will make only $200,000.
If Steeley requests a permit for the apartments, a host of good and bad outcomes are possible. The immediate good outcome is approval of its permit, which it estimates will result in a return of $3 million. However, Steeley gives that result only a 10% chance that it will occur. Alternatively, Steeley thinks there is a 90% chance that the town will reject its application, which will result in another set of decisions.
Steeley can sell the property at that point. However, the rejection of the permit will undoubtedly decrease the value to potential buyers, and Steeley estimates that it will get only $700,000. Alternatively, it can construct the office building and face the same potential outcomes it did earlier, namely, a 30% chance of no town growth and a $200,000 return or a 70% chance of growth with a return of $1.3 million. A third option is to sue the town. On the surface, Steeley’s case looks good, but the town building code is vague, and a sympathetic judge could throw out its suit. Whether or not it wins, Steeley estimates its possible legal fees to be $300,000, and it feels it has only a 40% chance of win- ning. However, if Steeley does win, it estimates that the award will be approximately $1 million, and it will also get its $3 million return for building the apartments. Steeley also estimates that there is a 10% chance that the suit could linger on in the courts for such a long time that any future return would be negated dur- ing its planning horizon, and it would incur an additional $200,000 in legal fees.
If Steeley loses the suit, it will then be faced with the same options of selling the property or constructing an office building. However, if the suit is carried this far into the future, it feels that the selling price it can ask will be somewhat dependent on the town’s growth prospects at that time, which it feels it can estimate at only 50–50. If the town is in a growth mode that far in the
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Case Problems 583
future, Steeley thinks that $900,000 is a conservative estimate of the potential sale price, whereas if the town is not growing, it thinks $500,000 is a more likely estimate. Finally, if Steeley constructs the office building, it feels that the chance of town growth is 50%, in which case the return will be only $1.2 million. If no growth occurs, it conservatively estimates only a $100,000 return.
A. Perform a decision tree analysis of Steeley Associates’s deci- sion situation, using expected value, and indicate the appropriate decision with these criteria.
B. Indicate the decision you would make and explain your reasons.
Case Problem
Transformer Replacement at Mountain States Electric Service
Mountain States Electric Service is an electrical utility com-pany serving several states in the Rocky Mountains region. It is considering replacing some of its equipment at a generating substation and is attempting to decide whether it should replace an older, existing PCB transformer. (PCB is a toxic chemical known formally as polychlorinated biphenyl.) Even though the PCB generator meets all current regulations, if an incident occurred, such as a fire, and PCB contamination caused harm either to neighboring businesses or farms or to the environment, the company would be liable for damages. Recent court cases have shown that simply meeting utility regulations does not relieve a utility of liability if an incident causes harm to others. Also, courts have been awarding large damages to individuals and businesses harmed by hazardous incidents.
If the utility replaces the PCB transformer, no PCB incidents will occur, and the only cost will be the cost of the transformer, $85,000. Alternatively, if the company decides to keep the exist- ing PCB transformer, then management estimates that there is a 50–50 chance of there being a high likelihood of an incident or a low likelihood of an incident. For the case in which there is a
high likelihood that an incident will occur, there is a .004 prob- ability that a fire will occur sometime during the remaining life of the transformer and a .996 probability that no fire will occur. If a fire occurs, there is a .20 probability that it will be bad and the utility will incur a very high cost of approximately $90 mil- lion for the cleanup, whereas there is an .80 probability that the fire will be minor and cleanup can be accomplished at a low cost of approximately $8 million. If no fire occurs, no cleanup costs will occur. For the case in which there is a low likelihood that an incident will occur, there is a .001 probability that a fire will occur during the life of the existing transformer and a .999 probability that a fire will not occur. If a fire does occur, the same probabilities exist for the incidence of high and low cleanup costs, as well as the same cleanup costs, as indicated for the previous case. Similarly, if no fire occurs, there is no cleanup cost.
Perform a decision tree analysis of this problem for Mountain States Electric Service and indicate the recommended solution. Is this the decision you believe the company should make? Explain your reasons.
This case was adapted from W. Balson, J. Welsh, and D. Wilson, “Using Decision Analysis and Risk Analysis to Manage Utility Environmental Risk,” Interfaces 22, no. 6 (November–December 1992): 126–39.
Case Problem
The Carolina Cougars
The Carolina Cougars is a major league baseball expansion teambeginning its third year of operation. The team had losing records in each of its first 2 years and finished near the bottom of its division. However, the team was young and generally competi- tive. The team’s general manager, Frank Lane, and manager, Biff Diamond, believe that with a few additional good players, the Cougars can become a contender for the division title and perhaps even for the pennant. They have prepared several proposals for free- agent acquisitions to present to the team’s owner, Bruce Wayne.
Under one proposal the team would sign several good avail- able free agents, including two pitchers, a good fielding shortstop, and two power-hitting outfielders for $52 million in bonuses and annual salary. The second proposal is less ambitious, costing $20 million to sign a relief pitcher, a solid, good-hitting infielder, and one power-hitting outfielder. The final proposal would be to stand pat with the current team and continue to develop.
General Manager Lane wants to lay out a possible season scenario for the owner so he can assess the long-run ramifications of each decision strategy. Because the only thing the owner under- stands is money, Frank wants this analysis to be quantitative, indi- cating the money to be made or lost from each strategy. To help
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develop this analysis, Frank has hired his kids, Penny and Nathan, both management science graduates from Tech.
Penny and Nathan analyzed league data for the previous five seasons for attendance trends, logo sales (i.e., clothing, souvenirs, hats, etc.), player sales and trades, and revenues. In addition, they interviewed several other owners, general managers, and league officials. They also analyzed the free agents that the team was con- sidering signing.
Based on their analysis, Penny and Nathan feel that if the Cougars do not invest in any free agents, the team will have a 25% chance of contending for the division title and a 75% chance of being out of contention most of the season. If the team is a con- tender, there is a .70 probability that attendance will increase as the season progresses and the team will have high attendance levels (between 1.5 million and 2.0 million) with profits of $170 million from ticket sales, concessions, advertising sales, TV and radio sales, and logo sales. They estimate a .25 probability that the team’s attendance will be mediocre (between 1.0 million and 1.5 million) with profits of $115 million and a .05 probability that the team will suffer low attendance (less than 1.0 million) with profit of $90 million. If the team is not a contender, Penny and Nathan estimate that there is .05 probability of high attendance with profits of $95 million, a .20 probability of medium atten- dance with profits of $55 million, and a .75 probability of low attendance with profits of $30 million.
If the team marginally invests in free agents at a cost of $20 million, there is a 50–50 chance it will be a contender. If it is a con- tender, then later in the season it can either stand pat with its existing roster or buy or trade for players that could improve the team’s chances of winning the division. If the team stands pat, there is a .75 probability that attendance will be high and profits will be $195 million. There is a .20 probability that attendance will be mediocre with profits of $160 million and a .05 probability of low attendance and profits of $120 million. Alternatively, if the team decides to buy or trade for players, it will cost $8 million, and the probability of high attendance with profits of $200 million will be .80. The probability of mediocre attendance with $170 mil- lion in profits will be .15, and there will be a .05 probability of low attendance, with profits of $125 million.
If the team is not in contention, then it will either stand pat or sell some of its players, earning approximately $8 million in profit. If the team stands pat, there is a .12 probability of high attendance, with profits of $110 million; a .28 probability of mediocre attendance, with profits of $65 million; and a .60 prob- ability of low attendance, with profits of $40 million. If the team sells players, the fans will likely lose interest at an even faster rate, and the probability of high attendance with profits of $100 mil- lion will drop to .08, the probability of mediocre attendance with profits of $60 million will be .22, and the probability of low atten- dance with profits of $35 million will be .70.
The most ambitious free-agent strategy will increase the team’s chances of being a contender to 65%. This strategy will also excite the fans most during the off-season and boost ticket sales and advertising and logo sales early in the year. If the team does con- tend for the division title, then later in the season it will have to decide whether to invest in more players. If the Cougars stand pat, the probability of high attendance with profits of $210 million will be .80, the probability of mediocre attendance with profits of $170 million will be .15, and the probability of low attendance with profits of $125 million will be .05. If the team buys players at a cost of $10 million, then the probability of having high atten- dance with profits of $220 million will increase to .83, the proba- bility of mediocre attendance with profits of $175 million will be .12, and the probability of low attendance with profits of $130 million will be .05.
If the team is not in contention, it will either sell some players’ contracts later in the season for profits of around $12 million or stand pat. If it stays with its roster, the probability of high atten- dance with profits of $110 million will be .15, the probability of mediocre attendance with profits of $70 million will be .30, and the probability of low attendance with profits of $50 million will be .55. If the team sells players late in the season, there will be a .10 probability of high attendance with profits of $105 million, a .30 probability of mediocre attendance with profits of $65 mil- lion, and a .60 probability of low attendance with profits of $45 million.
Assist Penny and Nathan in determining the best strategy to follow and its expected value.
Case Problem
Evaluating R&D Projects at WestCom Systems Products Company
WestCom Systems Products Company develops computersystems and software products for commercial sale. Each year it considers and evaluates a number of different R&D pro- jects to undertake. It develops a road map for each project, in the form of a standardized decision tree that identifies the different decision points in the R&D process from the initial decision to
invest in a project’s development through the actual commercial- ization of the final product.
The first decision point in the R&D process is whether to fund a proposed project for 1 year. If the decision is no, then there is no resulting cost; if the decision is yes, then the project proceeds at an incremental cost to the company. The company establishes specific short-term, early technical milestones for its projects after 1 year. If the early milestones are achieved, the project pro- ceeds to the next phase of project development; if the milestones are not achieved, the project is abandoned. In its planning
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Case Problems 585
process the company develops probability estimates of achieving and not achieving the early milestones. If the early milestones are achieved, the project is funded for further development during an extended time frame specific to a project. At the end of this time frame, a project is evaluated according to a second set of (later) technical milestones. Again, the company attaches proba- bility estimates for achieving and not achieving these later milestones. If the later milestones are not achieved, the project is abandoned.
If the later milestones are achieved, technical uncertainties and problems have been overcome, and the company next assesses the project’s ability to meet its strategic business objectives. At this stage, the company wants to know if the eventual product coin- cides with the company’s competencies and whether there appears to be an eventual, clear market for the product. It invests in a product “prelaunch” to ascertain the answers to these ques- tions. The outcomes of the prelaunch are that either there is a
strategic fit or there is not, and the company assigns probability estimates to each of these two possible outcomes. If there is not a strategic fit at this point, the project is abandoned and the com- pany loses its investment in the prelaunch process. If it is deter- mined that there is a strategic fit, then three possible decisions result: (1) The company can invest in the product’s launch, and a successful or unsuccessful outcome will result, each with an esti- mated probability of occurrence; (2) the company can delay the product’s launch and at a later date decide whether to launch or abandon; and (3) if it launches later, the outcomes are success or failure, each with an estimated probability of occurrence. Also, if the product launch is delayed, there is always a likelihood that the technology will become obsolete or dated in the near future, which tends to reduce the expected return.
The following table provides the various costs, event probabil- ities, and investment outcomes for five projects the company is considering:
Project
Decision Outcomes/Event 1 2 3 4 5
Fund—1 year $ 200,000 $ 350,000 $ 170,000 $ 230,000 $ 400,000 P(Early milestones—yes) .70 .67 .82 .60 .75 P(Early milestones—no) .30 .33 .18 .40 .25 Long-term funding $ 650,000 780,000 450,000 300,000 450,000 P(Late milestones—yes) .60 .56 .65 .70 .72 P(Late milestones—no) .40 .44 .35 .30 .28 Prelaunch funding $ 300,000 450,000 400,000 500,000 270,000 P(Strategic fit—yes) .80 .75 .83 .67 .65 P(Strategic fit—no) .20 .25 .17 .33 .35 P(Invest—success) .60 .65 .70 .75 .80 P(Invest—failure) .40 .35 .30 .25 .20 P(Delay—success) .80 .70 .65 .80 .85 P(Delay—failure) .20 .30 .35 .20 .15 Invest—success $ 7,300,000 8,000,000 4,500,000 5,200,000 3,800,000 Invest—failure �2,000,000 �3,500,000 �1,500,000 �2,100,000 �900,000 Delay—success 4,500,000 6,000,000 3,300,000 2,500,000 2,700,000 Delay—failure �1,300,000 �4,000,000 �800,000 �1,100,000 �900,000
Determine the expected value for each project and then rank the projects accordingly for the company to consider.
This case was adapted from R. K. Perdue, W. J. McAllister, P. V. King, and B. G. Berkey, “Valuation of R and D Projects Using Options Pricing and Decision Analysis Models,” Interfaces 29, no. 6 (November–December 1999): 57–74.
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627
C h a p t e r 14
Simulation
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The Monte Carlo Process
One characteristic of some systems that makes them difficult to solve analytically is that they consist of random variables represented by probability distributions. Thus, a large proportion of the applications of simulations are for probabilistic models.
The term Monte Carlo has become synonymous with probabilistic simulation in recent years. However, the Monte Carlo technique can be more narrowly defined as a technique for selecting numbers randomly from a probability distribution (i.e., “sampling”) for use in a trial (computer) run of a simulation. The Monte Carlo technique is not a type of simu- lation model but rather a mathematical process used within a simulation.
The name Monte Carlo is appropriate because the basic principle behind the process is the same as in the operation of a gambling casino in Monaco. In Monaco such devices as roulette wheels, dice, and playing cards are used. These devices produce numbered results at random from well-defined populations. For example, a 7 resulting from thrown dice is a random value from a population of 11 possible numbers (i.e., 2 through 12). This same process is employed, in principle, in the Monte Carlo process used in simulation models.
The Use of Random Numbers The Monte Carlo process of selecting random numbers according to a probability distri- bution will be demonstrated using the following example. The manager of ComputerWorld, a store that sells computers and related equipment, is attempting to determine how many laptop PCs the store should order each week. A primary consideration in this decision
628 Chapter 14 Simulation
Monte Carlo is a technique for selecting numbers randomly from
a probability distribution.
Simulation represents a major divergence from the topics presented in the previouschapters of this text. Previous topics usually dealt with mathematical models and for- mulas that could be applied to certain types of problems. The solution approaches to these problems were, for the most part, analytical. However, not all real-world problems can be solved by applying a specific type of technique and then performing the calculations. Some problem situations are too complex to be represented by the concise techniques presented so far in this text. In such cases, simulation is an alternative form of analysis.
Analogue simulation is a form of simulation that is familiar to most people. In ana- logue simulation, an original physical system is replaced by an analogous physical system that is easier to manipulate. Much of the experimentation in staffed spaceflight was con- ducted using physical simulation that re-created the conditions of space. For example, con- ditions of weightlessness were simulated using rooms filled with water. Other examples include wind tunnels that simulate the conditions of flight and treadmills that simulate automobile tire wear in a laboratory instead of on the road.
This chapter is concerned with an alternative type of simulation, computer mathematical simulation. In this form of simulation, systems are replicated with mathematical models, which are analyzed using a computer. This form of simulation has become very popular and has been applied to a wide variety of business problems. One reason for its popularity is that it offers a means of analyzing very complex systems that cannot be analyzed by using the other management science techniques in this text. However, because such complex systems are beyond the scope of this text, we will not present actual simulation models; instead, we will present simplified simulation models of systems that can also be analyzed analytically. We will begin with one of the simplest forms of simulation models, which encompasses the Monte Carlo process for simulating random variables.
The Monte Carlo process is analogous to gambling devices.
In computer mathematical simulation, a system is replicated
with a mathematical model that is analyzed by using the computer.
Analogue simulation replaces a physical system with an analogous
physical system that is easier to manipulate.
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The Monte Carlo Process 629
20%
40%
20%
10%
10%
Spin
x = 1
x = 4
x = 3
x = 2
x = 0
Figure 14.1
A roulette wheel for demand
is the average number of laptop computers that the store will sell each week and the aver- age weekly revenue generated from the sale of laptop PCs. A laptop sells for $4,300. The number of laptops demanded each week is a random variable (which we will define as x) that ranges from 0 to 4. From past sales records, the manager has determined the frequency of demand for laptop PCs for the past 100 weeks. From this frequency distribution, a prob- ability distribution of demand can be developed, as shown in Table 14.1.
Because the surface area on the roulette wheel is partitioned according to the probabil- ity of each weekly demand value, the wheel replicates the probability distribution for demand if the values of demand occur in a random manner. To simulate demand for 1 week, the manager spins the wheel; the segment at which the wheel stops indicates demand for 1 week. Over a period of weeks (i.e., many spins of the wheel), the frequency with which demand values occur will approximate the probability distribution, P(x). This method of generating values of a variable, x, by randomly selecting from the probability distribution—the wheel—is the Monte Carlo process.
By spinning the wheel, the manager artificially reconstructs the purchase of PCs during a week. In this reconstruction, a long period of real time (i.e., a number of weeks) is repre- sented by a short period of simulated time (i.e., several spins of the wheel).
Now let us slightly reconstruct the roulette wheel. In addition to partitioning the wheel into segments corresponding to the probability of demand, we will put numbers along the outer rim, as on a real roulette wheel. This reconstructed roulette wheel is shown in Figure 14.2.
There are 100 numbers from 0 to 99 on the outer rim of the wheel, and they have been partitioned according to the probability of each demand value. For example, 20 numbers
The purpose of the Monte Carlo process is to generate the random variable, demand, by sampling from the probability distribution, P(x). The demand per week can be randomly generated according to the probability distribution by spinning a wheel that is partitioned into segments corresponding to the probabilities, as shown in Figure 14.1.
Table 14.1 Probability distribution of
demand for laptop PCs
PCs Demanded Frequency Probability per Week of Demand of Demand, P(x)
0 20 .20 1 40 .40 2 20 .20 3 10 .10 4 10 .10
100 1.00
In the Monte Carlo process, values for a random variable are
generated by sampling from a probability distribution.
A long period of real time is represented by a short period of
simulated time.
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630 Chapter 14 Simulation
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0999897
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57 56 55 54 53 52 51 50 49 48 47
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37
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30 29
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82 81
80 79
13 12
11 10
9 8
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54 321
x = 1
x = 0
x = 4
x = 3
x = 2
Figure 14.2
Numbered roulette wheel
from 0 to 19 (i.e., 20% of the total 100 numbers) correspond to a demand of no (0) PCs. Now we can determine the value of demand by seeing which number the wheel stops at as well as by looking at the segment of the wheel.
When the manager spins this new wheel, the actual demand for PCs will be determined by a number. For example, if the number 71 comes up on a spin, the demand is 2 PCs per week; the number 30 indicates a demand of 1. Because the manager does not know which number will come up prior to the spin and there is an equal chance that any of the 100 numbers will occur, the numbers occur at random; that is, they are random numbers.
Obviously, it is not generally practical to generate weekly demand for PCs by spinning a wheel. Alternatively, the process of spinning a wheel can be replicated by using random numbers alone.
First, we will transfer the ranges of random numbers for each demand value from the roulette wheel to a table, as in Table 14.2. Next, instead of spinning the wheel to get a random
Demand, x Ranges of Random Numbers, r
0 0–19 1 20–59 r = 39 2 60–79 3 80–89 4 90–99
Table 14.2 Generating demand from
random numbers
Random numbers are numbers equally likely to be chosen from a
large population of numbers.
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The Monte Carlo Process 631
number, we will select a random number from Table 14.3, which is referred to as a random number table. (These random numbers have been generated by computer so that they are all equally likely to occur, just as if we had spun a wheel. The development of random num- bers is discussed in more detail later in this chapter.) As an example, let us select the num- ber 39, the first entry in Table 14.3. Looking again at Table 14.2, we can see that the random number 39 falls in the range 20–59, which corresponds to a weekly demand of 1 laptop PC.
39 65 76 45 45 19 90 69 64 61 20 26 36 31 62 58 24 97 14 97 95 06 70 99 00 73 71 23 70 90 65 97 60 12 11 31 56 34 19 19 47 83 75 51 33 30 62 38 20 46 72 18 47 33 84 51 67 47 97 19 98 40 07 17 66 23 05 09 51 80 59 78 11 52 49 75 12 25 69 17 17 95 21 78 58 24 33 45 77 48 69 81 84 09 29 93 22 70 45 80 37 17 79 88 74 63 52 06 34 30 01 31 60 10 27 35 07 79 71 53 28 99 52 01 41
02 48 08 16 94 85 53 83 29 95 56 27 09 24 43 21 78 55 09 82 72 61 88 73 61 87 89 15 70 07 37 79 49 12 38 48 13 93 55 96 41 92 45 71 51 09 18 25 58 94 98 18 71 70 15 89 09 39 59 24 00 06 41 41 20 14 36 59 25 47 54 45 17 24 89 10 83 58 07 04 76 62 16 48 68 58 76 17 14 86 59 53 11 52 21 66 04 18 72 87 47 08 56 37 31 71 82 13 50 41 27 55 10 24 92 28 04 67 53 44 95 23 00 84 47
93 90 31 03 07 34 18 04 52 35 74 13 39 35 22 68 95 23 92 35 36 63 70 35 33 21 05 11 47 99 11 20 99 45 18 76 51 94 84 86 13 79 93 37 55 98 16 04 41 67 95 89 94 06 97 27 37 83 28 71 79 57 95 13 91 09 61 87 25 21 56 20 11 32 44 97 18 31 55 73 10 65 81 92 59 77 31 61 95 46 20 44 90 32 64 26 99 76 75 63 69 08 88 86 13 59 71 74 17 32 48 38 75 93 29 73 37 32 04 05 60 82 29 20 25
41 26 10 25 03 87 63 93 95 17 81 83 83 04 49 77 45 85 50 51 79 88 01 97 30 91 47 14 63 62 08 61 74 51 69 92 79 43 89 79 29 18 94 51 23 14 85 11 47 23 80 94 54 18 47 08 52 85 08 40 48 40 35 94 22 72 65 71 08 86 50 03 42 99 36 67 06 77 63 99 89 85 84 46 06 64 71 06 21 66 89 37 20 70 01 61 65 70 22 12 59 72 24 13 75 42 29 72 23 19 06 94 76 10 08 81 30 15 39 14 81 33 17 16 33
63 62 06 34 41 79 53 36 02 95 94 61 09 43 62 20 21 14 68 86 84 95 48 46 45 78 47 23 53 90 79 93 96 38 63 34 85 52 05 09 85 43 01 72 73 14 93 87 81 40 87 68 62 15 43 97 48 72 66 48 53 16 71 13 81 59 97 50 99 52 24 62 20 42 31 47 60 92 10 77 26 97 05 73 51 88 46 38 03 58 72 68 49 29 31 75 70 16 08 24 56 88 87 59 41 06 87 37 78 48 65 88 69 58 39 88 02 84 27 83 85 81 56 39 38
22 17 68 65 84 87 02 22 57 51 68 69 80 95 44 11 29 01 95 80 49 34 35 36 47 19 36 27 59 46 39 77 32 77 09 79 57 92 36 59 89 74 39 82 15 08 58 94 34 74 16 77 23 02 77 28 06 24 25 93 22 45 44 84 11 87 80 61 65 31 09 71 91 74 25 78 43 76 71 61 97 67 63 99 61 30 45 67 93 82 59 73 19 85 23 53 33 65 97 21 03 28 28 26 08 69 30 16 09 05 53 58 47 70 93 66 56 45 65 79 45 56 20 19 47
04 31 17 21 56 33 73 99 19 87 26 72 39 27 67 53 77 57 68 93 60 61 97 22 61 61 06 98 03 91 87 14 77 43 96 43 00 65 98 50 45 60 33 01 07 98 99 46 50 47 23 68 35 26 00 99 53 93 61 28 52 70 05 48 34 56 65 05 61 86 90 92 10 70 80 15 39 25 70 99 93 86 52 77 65 15 33 59 05 28 22 87 26 07 47 86 96 98 29 06 58 71 96 30 24 18 46 23 34 27 85 13 99 24 44 49 18 09 79 49 74 16 32 23 02
93 22 53 64 39 07 10 63 76 35 87 03 04 79 88 08 13 13 85 51 55 34 57 72 69 78 76 58 54 74 92 38 70 96 92 52 06 79 79 45 82 63 18 27 44 69 66 92 19 09 61 81 31 96 82 00 57 25 60 59 46 72 60 18 77 55 66 12 62 11 08 99 55 64 57 42 88 07 10 05 24 98 65 63 21 47 21 61 88 32 27 80 30 21 60 10 92 35 36 12 77 94 30 05 39 28 10 99 00 27 12 73 73 99 12 49 99 57 94 82 96 88 57 17 91
Table 14.3 Random number table
By repeating this process of selecting random numbers from Table 14.3 (starting any- where in the table and moving in any direction but not repeating the same sequence) and then determining weekly demand from the random number, we can simulate demand for a period of time. For example, Table 14.4 shows demand for a period of 15 consecutive weeks.
Random numbers are equally likely to occur.
In a random number table, the random numbers are derived from
some artificial process, like a computer program.
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632 Chapter 14 Simulation
From Table 14.4 the manager can compute the estimated average weekly demand and revenue:
The manager can then use this information to help determine the number of PCs to order each week.
Although this example is convenient for illustrating how simulation works, the average demand could have more appropriately been calculated analytically using the formula for expected value. The expected value or average for weekly demand can be computed analyt- ically from the probability distribution, P(x):
where
demand value i probability of demand the number of different demand values
Therefore,
The analytical result of 1.5 PCs is close to the simulated result of 2.07 PCs, but clearly there is some difference. The margin of difference (0.57 PC) between the simulated value and the analytical value is a result of the number of periods over which the simulation was conducted. The results of any simulation study are subject to the number of times the sim- ulation occurred (i.e., the number of trials). Thus, the more periods for which the simula- tion is conducted, the more accurate the result. For example, if demand was simulated for
= 1.5 PCs per week EV(x) = (.20)(0) + (.40)(1) + (.20)(2) + (.10)(3) + (.10)(4)
n = P(xi) =
xi =
EV(x) = a n
i = 1 P(xi)x
estimated average revenue = $133,300
15 = $8,886.67
estimated average demand = 31
15 = 2.07 laptop PCs per week
Simulation results will not equal analytical results unless enough
trials of the simulation have been conducted to reach steady state.
Week r Demand, x Revenue
1 39 1 $ 4,300 2 73 2 8,600 3 72 2 8,600 4 75 2 8,600 5 37 1 4,300 6 02 0 0 7 87 3 12,900 8 98 4 17,200 9 10 0 0
10 47 1 4,300 11 93 4 17,200 12 21 1 4,300 13 95 4 17,200 14 97 4 17,200 15 69 2 8,600
Σ = 31 $133,300
Table 14.4 Randomly generated demand
for 15 weeks
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1,000 weeks, in all likelihood an average value exactly equal to the analytical value (1.5 lap- top PCs per week) would result.
Once a simulation has been repeated enough times that it reaches an average result that remains constant, this result is analogous to the steady-state result, a concept we discussed previously in our presentation of queuing. For this example, 1.5 PCs is the long-run aver- age or steady-state result, but we have seen that the simulation might have to be repeated more than 15 times (i.e., weeks) before this result is reached.
Comparing our simulated result with the analytical (expected value) result for this example points out one of the problems that can occur with simulation. It is often difficult to validate the results of a simulation model—that is, to make sure that the true steady-state average result has been reached. In this case we were able to compare the simulated result with the expected value (which is the true steady-state result), and we found there was a slight difference. We logically deduced that the 15 trials of the simulation were not suffi- cient to determine the steady-state average. However, simulation most often is employed whenever analytical analysis is not possible (this is one of the reasons that simulation is generally useful). In these cases, there is no analytical standard of comparison, and valida- tion of the results becomes more difficult. We will discuss this problem of validation in more detail later in the chapter.
Computer Simulation with Excel Spreadsheets 633
The mathematics of the Monte Carlo method have been known for years; the British mathematician Lord Kelvin used the technique in a paper in 1901. However, it was formally identified and given this name by the Hungarian mathemati- cian John Von Neumann while working on the Los Alamos atomic bomb project during World War II. During this project, physicists confronted a problem in determining how far neu- trons would travel through various materials (i.e., neutron diffusion in fissile material). The Monte Carlo process was
suggested to Von Neumann by a colleague at Los Alamos, Stanislas Ulam, as a means to solve this problem—that is, by selecting random numbers to represent the random actions of neutrons. However, the Monte Carlo method as used in simu- lation did not gain widespread popularity until the develop- ment of the modern electronic computer after the war. Interestingly, this remarkable man, John Von Neumann, is credited with being the key figure in the development of the computer.
T I M E O U T for John Von Neumann
It is often difficult to validate that the results of a simulation truly
replicate reality.
Computer Simulation with Excel Spreadsheets
The simulation we performed manually for the ComputerWorld example was not too dif- ficult. However, if we had performed the simulation for 1,000 weeks, it would have taken several hours. On the other hand, this simulation could be done on a computer in several seconds. Also, our simulation example was not very complex. As simulation models get progressively more complex, it becomes virtually impossible to perform them manually, thus making the computer a necessity.
Although we will not develop a simulation model in a computer language for this exam- ple, we will demonstrate how a computerized simulation model is developed by using Excel spreadsheets.
The first step in developing a simulation model is to generate a random number, r. Numerous subroutines that are available on practically every computer system generate random numbers. Most are quite easy to use and require the insertion of only a few state- ments in a program. These random numbers are generated by mathematical processes as opposed to a physical process, such as spinning a roulette wheel. For this reason, they are
Simulations are normally done on the computer.
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referred to as pseudorandom numbers. It should be apparent from the previous discus- sion that random numbers play a very important part in a probabilistic simulation. Some of the random numbers we used came from Table 14.3, a table of random numbers. However, random numbers do not come just from tables, and their generation is not as simple as one might initially think. If random numbers are not truly random, the validity of simulation results can be significantly affected.
The random numbers in Table 14.3 were generated by using a numerical technique. Thus, they are not true random numbers but pseudorandom numbers. True random num- bers can be produced only by a physical process, such as spinning a roulette wheel over and over. However, a physical process, such as spinning a roulette wheel, cannot be conveniently employed in a computerized simulation model. Thus, there is a need for a numerical method that artificially creates random numbers.
To truly reflect the system being simulated, the artificially created random numbers must have the following characteristics:
1. The random numbers must be uniformly distributed. This means that each random number in the interval of random numbers (i.e., 0 to 1 or 0 to 100) has an equal chance of being selected. If this condition is not met, then the simulation results will be biased by the random numbers that have a more likely chance of being selected.
2. The numerical technique for generating random numbers should be efficient. This means that the random numbers should not degenerate into constant values or recy- cle too frequently.
3. The sequence of random numbers should not reflect any pattern. For example, the sequence of numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, and so on, although uniform, is not random.
Random numbers between 0 and 1 can be generated in Excel by entering the formula in a cell. The random numbers generated by this formula include all the neces-
sary characteristics for randomness and uniformity that we discussed earlier. Exhibit 14.1 is an Excel spreadsheet with 100 random numbers generated by entering the formula
in cell A3 and copying to the cells in the range A3:J12. Recall that you can copy things in a range of cells in two ways. You can first cover cells A3:J12 with the cursor and then type the formula into cell A3. Then you press the “Ctrl” and “Enter” keys simultaneously. Alternatively, you can type in cell A3, copy this cell (using the right mouse button), then cover cells A3:J12 with the cursor, and (again with the right mouse button) paste this formula in these cells.
= RAND() = RAND()
= RAND()
= RAND()
634 Chapter 14 Simulation
Random numbers generated by a mathematical process instead
of a physical process are pseudorandom numbers.
Random numbers are typically generated on the computer by using a numerical technique.
A table of random numbers must be uniform, efficiently generated,
and absent of patterns.
If you attempt to replicate this spreadsheet, you will generate different random numbers from those shown in Exhibit 14.1. Every time you generate random numbers, they will be different. In fact, any time you recalculate anything on your spreadsheet, the random num- bers will change. You can see this by pressing the F9 key and observing that all the random
Exhibit 14.1
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Computer Simulation with Excel Spreadsheets 635
numbers change. However, sometimes it is useful in a simulation model to be able to use the same set (or stream) of random numbers over and over. You can freeze the random numbers you are using on your spreadsheet by first covering the cells with random num- bers in them with the cursor—for example, cells A3:J12 in Exhibit 14.1. Next, you copy these cells (using the right mouse button); then you click on the “Edit” menu at the top of your spreadsheet and select “Paste Special” from this menu. Next, you select the “Values” option and click on “OK.” This procedure pastes a copy of the numbers in these cells over the same cells with formulas in them, thus freezing the numbers in place.
Notice one more thing from Exhibit 14.1: The random numbers are all between 0 and 1, whereas the random numbers in Table 14.3 are whole numbers between 0 and 100. We used whole random numbers previously for illustrative purposes; however, computer programs such as Excel generally provide random numbers between 0 and 1.
Now we are ready to duplicate our example simulation model for the ComputerWorld store by using Excel. The spreadsheet in Exhibit 14.2 includes the simulation model origi- nally developed in Table 14.4.
= RAND()
Enter this formula in G6 and copy it to G7:G20.
Enter = 4300*G6 in H6 and copy it to H7:H20.
Generate random numbers for cells F6 : F20 with the formula = RAND() in F6 and copy them to F7:F20.
= AVERAGE(G6:G20)
Exhibit 14.2
Note that the probability distribution for the weekly demand for laptops has been entered in cells A6:C10. Also notice that we have entered a new set of cumulative proba- bility values in column B. We generated these cumulative probabilities by first entering 0 in cell B6, then entering the formula in cell B7, and copying this formula to cells B8:B10. This cumulative probability creates a range of random numbers for each demand value. For example, any random number less than 0.20 will result in a demand value of 0, and any random number greater than 0.20 but less than 0.60 will result in a demand value of 1, and so on. (Notice that there is no value of 1.00 in cell B11; the last demand value, 4, will be selected for any random number equal to or greater than .90.)
Random numbers are generated in cells F6:F20 by entering the formula in cell F6 and copying it to the range of cells in F7:F20.
= RAND()
= A6 + B6
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636 Chapter 14 Simulation
Now we need to be able to generate demand values for each of these random numbers in column F. We accomplish this by first covering the cumulative probabilities and the demand values in cells B6:C10 with the cursor. Then we give this range of cells the name “Lookup.” This can be done by typing “Lookup” directly on the formula bar in place of B6 or by clicking on the “Insert” button at the top of the spreadsheet and selecting “Name” and “Define” and then entering the name “Lookup.” This has the effect of creating a table called “Lookup” with the ranges of random numbers and associated demand values in it. Next, we enter the formula in cell G6 and copy it to the cells in the range G7:G20. This formula will compare the random numbers in column F with the cumulative probabilities in B6:B10 and generate the correct demand value from cells C6:C10.
Once the demand values have been generated in column G, we can determine the weekly revenue values by entering the formula in H6 and copying it to cells H7:H20.
Average weekly demand is computed in cell C13 by using the formula and the average weekly revenue is computed by entering a similar formula in
cell C14. Notice that the average weekly demand value of 1.53 in Exhibit 14.2 is different from the
simulation result (2.07) we obtained from Table 14.4. This is because we used a different stream of random numbers. As mentioned previously, to acquire an average closer to the true steady-state value, the simulation probably needs to include more repetitions than 15 weeks. As an example, Exhibit 14.3 simulates demand for 100 weeks. The window has been “frozen” at row 16 and scrolled up to show the first 10 weeks and the last 6 weeks on the screen in Exhibit 14.3.
(G6:G20), = AVERAGE
= 4300*G6
= VLOOKUP(F6,Lookup,2)
Decision Making with Simulation In our previous example, the manager of ComputerWorld acquired some useful informa- tion about the weekly demand and revenue for laptops that would be helpful in making a decision about how many laptops would be needed each week to meet demand. However, this example did not lead directly to a decision. Next, we will expand our ComputerWorld store example so that a possible decision will result.
Spreadsheet “frozen” at row 16 to show first 10 weeks and last 6.
Click on “View” tab, then on “Freeze Panes.“
Exhibit 14.3
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Computer Simulation with Excel Spreadsheets 637
From the simulation in Exhibit 14.3 the manager of the store knows that the average weekly demand for laptop PCs will be approximately 1.49; however, the manager cannot order 1.49 laptops each week. Because fractional laptops are not possible, either one or two must be ordered. Thus, the manager wants to repeat the earlier simulation with two possi- ble order sizes, 1 and 2. The manager also wants to include some additional information in the model that will affect the decision.
If too few laptops are on hand to meet demand during the week, then not only will there be a loss of revenue, but there will also be a shortage cost of $500 per unit incurred because the customer will be unhappy. However, each laptop still in stock at the end of each week that has not been sold will incur an inventory or storage cost of $50. Thus, it costs the store money to have either too few or too many laptops on hand each week. Given this scenario, the manager wants to order either one or two laptops, depending on which order size will result in the greatest average weekly revenue.
Exhibit 14.4 shows the Excel spreadsheet for this revised example. The simulation is for 100 weeks. The columns labeled “1,” “2,” and “4” for “Week,” “RN,” and “Demand” were constructed similarly to the model in Exhibit 14.3. The array of cells B6:C10 was given the name “Lookup,” and the formula was entered in cell H6 and copied to cells H7:H105.
= VLOOKUP(F6,Lookup,2)
This formula is entered in G7 and copied to G8 : G105.
Shortages are computed by entering = MIN(G6 � H6,0) in I6 and copying to I7 : I105.
= G6*50 is entered in cell L6 and copied to L7 : L105.
= Vlookup(F6,LOOKUP,2) is entered in H6 and copied to H7:H105.
Exhibit 14.4
The simulation in Exhibit 14.4 is for an order size of one laptop each week. The “Inventory” column (3) keeps track of the amount of inventory available each week—the one laptop that comes in on order plus any laptops carried over from the previous week. The cumulative inventory is computed each week by entering the formula in cell G7 and copying it to cells G8:G105. This formula adds the one laptop on order to either the number left over from the previous week (G6 H6) or 0, if there were not enough laptops on hand to meet demand. It does not allow for negative inventory levels, called back orders. In
-
= 1 + MAX(G6 - H6,0)
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638 Chapter 14 Simulation
other words, if a sale cannot be made owing to a shortage, then it is gone. The inventory val- ues in column 3 are eventually multiplied by the inventory cost of $50 per unit in column 8, using the formula .
If there is a shortage, it is recorded in column 5, labeled “Shortage.” The shortage is com- puted by entering the formula in cell I6 and copying it to cells I7:I105. Shortage costs are computed in column 7 by multiplying the shortage values in column 5 by $500 by entering the formula in cell K6 and copying it to cells K7:K105.
Weekly revenues are computed in column 6 by entering the formula in cell J6 and copying it to cells J7:J105. In other words, the revenue is deter-
mined by either the inventory level in column 3 or the demand in column 4, whichever is smaller.
Total weekly revenue is computed in column 9 by subtracting shortage costs and inven- tory costs from revenue by entering the formula J6-K6-L6 in cell M6 and copying it to cells M7:M105.
The average weekly demand, 1.50, is shown in cell C13. The average weekly revenue, $3,875, is computed in cell C14.
Next, we must repeat this same simulation for an order size of two laptops each week. The spreadsheet for an order size of two is shown in Exhibit 14.5. Notice that the only actual difference is the use of a new formula to compute the weekly inventory level in col- umn 3. This formula in cell G7, reflecting two laptops ordered each week, is shown on the formula bar at the top of the spreadsheet.
=
MIN(H6,G6) = 4300*
= I6*500
= MIN(G6 - H6,0)
= G6*50
This second simulation, in Exhibit 14.5, results in an average weekly demand of 1.50 lap- tops and an average weekly total revenue of $4,927.50. This is higher than the total weekly revenue of $3,875 achieved in the first simulation run in Exhibit 14.4, even though the store would incur significantly higher inventory costs. Thus, the correct decision—based on weekly revenue—would be to order two laptops per week. However, there are probably additional aspects of this problem the manager would want to consider in the decision-making process, such as the increasingly high inventory levels as the simulation progresses. For example, there may not be enough storage space to accommodate this much inventory. Such questions as this and others can also be analyzed with simulation. In fact, one of the main attributes of simulation is its usefulness as a model for experimenting, called what-if? analysis.
New formula for two laptops ordered per week
Exhibit 14.5
What-if? analysis is a form of model experimentation using a
computer to ascertain the results of making changes in a model.
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Simulation of a Queuing System 639
This example briefly demonstrates how simulation can be used to make a decision (i.e., to “optimize”). In this example we experimented with two order sizes and determined the one that resulted in the greater revenue. The same basic modeling principles can be used to solve larger problems with hundreds of possible order sizes and a probability distribu- tion for demand with many more values plus variable lead times (i.e., the time it takes to receive an order), the ability to back-order, and other complicating factors. These factors make the simulation model larger and more complex, but such models are frequently developed and used in business.
Simulation of a Queuing System
To demonstrate the simulation of a queuing system, we will use an example. Burlingham Mills produces denim, and one of the main steps in the production process is dyeing the cotton yarn that is subsequently woven into denim cloth. The yarn is dyed in a large con- crete vat like a narrow swimming pool. The yarn is strung over a series of rollers so that it passes through the vat, up over a set of rollers, back down into the vat, back up and over another set of rollers, and so on. Yarn is dyed in batches that arrive at the dyeing vat in 1-, 2-, 3-, or 4-day intervals, according to the probability distribution shown in Table 14.5. Once a batch of yarn arrives at the dyeing facility, it takes either 0.5, 1.0, or 2.0 days to com- plete the dyeing process, according to the probability distribution shown in Table 14.6.
Service Time Probability, Cumulative Random Number (days), y P(y) Probability Range, r2
0.5 .20 .20 1–20 1.0 .50 .70 21–70 2.0 .30 1.00 71–99, 00
Table 14.5 Distribution of arrival intervals
Table 14.6 Distribution of service times
Arrival Interval Probability, Cumulative Random Number (days), x P(x) Probability Range, r1
1.0 .20 .20 1–20 2.0 .40 .60 21–60 3.0 .30 .90 61–90 4.0 .10 1.00 91–99, 00
Table 14.5 defines the interarrival time, or how often batches arrive at the dyeing facil- ity. For example, there is a .20 probability that a batch will arrive 1 day after the previous batch. Table 14.6 defines the service time for a batch. Notice that cumulative probabilities have been included in Tables 14.5 and 14.6. The cumulative probability provides a means for determining the ranges of random numbers associated with each probability, as we demonstrated with our Excel example in the previous section. For example, in Table 14.5 the first random number range for r1 is from 1 to 20, which corresponds to the cumulative probability of .20. The second range of random numbers is from 21 to 60, which corre- sponds to a cumulative probability of .60. Although the cumulative probability goes up to 1.00, Table 14.3 contains only random number values from 0 to 99. Thus, the number 0 is used in place of 100 in the last random number range of each table.
Developing the cumulative probability distribution helps to
determine random number ranges.
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640 Chapter 14 Simulation
Table 14.7 illustrates the simulation of 10 batch arrivals at the dyeing vat.
The manual simulation process illustrated in Table 14.7 can be interpreted as follows:
1. Batch 1 arrives at time 0, which is recorded on an arrival clock. Because there are no batches in the system, batch 1 approaches the dyeing facility immediately, also at time 0. The waiting time is 0.
2. Next, a random number, is selected from the second column in Table 14.3. From Table 14.6 we see that the random number 65 results in a service time, y, of 1 day. After leaving the dyeing vat, the batch departs at time 1 day, having been in the queuing system a total of 1 day.
3. The next random number, is selected from Table 14.3, which specifies that batch 2 arrives 3 days after batch 1, or at time 3.0, as shown on the arrival clock. Because batch 1 departed the service facility at time 1.0, batch 2 can be served imme- diately and thus incurs no waiting time.
4. The next random number, is selected from Table 14.3, which indicates that batch 2 will spend 0.5 day being served and will depart at time 3.5.
This process of selecting random numbers and generating arrival intervals and service times continues until 10 batch arrivals have been simulated, as shown in Table 14.7.
Once the simulation is complete, we can compute operating characteristics from the simulation results, as follows:
However, as in our previous example, these results must be viewed with skepticism. Ten trials of the system do not ensure steady-state results. In general, we can expect a consider- able difference between the true average values and the values estimated from only 10 ran- dom draws. One reason is that we cannot be sure that the random numbers we selected in this example replicated the actual probability distributions because we used so few random numbers. The stream of random numbers that was used might have had a preponderance of high or low values, thus biasing the final model results. For example, of nine arrivals, five had interarrival times of 1 day. This corresponds to a probability of .55 (i.e., 5/9); however,
average time in the system = 24.5 days
10 batches = 2.45 days per batch
average waiting time = 12.5 days
10 batches = 1.25 days per batch
r2 = 18,
r1 = 71,
r2 = 65,
Table 14.7 Simulation at the Burlingham Mills Dyeing Facility
Arrival Enter Interval, Arrival Facility Waiting Service Departure Time in
Batch r1 x Clock Clock Time r2 Time, y Clock System
1 — — 0.0 0.0 0.0 65 1.0 1.0 1.0 2 71 3.0 3.0 3.0 0.0 18 0.5 3.5 0.5 3 12 1.0 4.0 4.0 0.0 17 0.5 4.5 0.5 4 48 2.0 6.0 6.0 0.0 89 2.0 8.0 2.0 5 18 1.0 7.0 8.0 1.0 83 2.0 10.0 3.0 6 08 1.0 8.0 10.0 2.0 90 2.0 12.0 4.0 7 05 1.0 9.0 12.0 3.0 89 2.0 14.0 5.0 8 18 1.0 10.0 14.0 4.0 08 0.5 14.5 4.5 9 26 2.0 12.0 14.5 2.5 47 1.0 15.5 3.5
10 94 4.0 16.0 16.0 0.0 06 0.5 16.5 0.5 12.5 24.5
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Simulation of a Queuing System 641
the actual probability of an arrival interval of 1 day is .20 (from Table 14.5). This excessive number of short interarrival times (caused by the sequence of random numbers) has prob- ably artificially inflated the operating statistics of the system.
As the number of random trials is increased, the probabilities in the simulation will more closely conform to the actual probability distributions. That is, if we simulated the queuing system for 1,000 arrivals, we could more reasonably expect that 20% of the arrivals would have an interarrival time of 1 day.
An additional factor that can affect simulation results is the starting conditions. If we start our queuing system with no batches in the system, we must simulate a length of time before the system replicates normal operating conditions. In this example, it is logical to start simulating at the time the vat starts operating in the morning, especially if we simulate an entire working day. Some queuing systems, however, start with items already in the sys- tem. For example, the dyeing facility might logically start each day with partially completed batches from the previous day. In this case, it is necessary to begin the simulation with batches already in the system.
By adding a second random variable to a simulation model, such as the one just shown, we increase the complexity and therefore the manual operations. To simulate the example in Table 14.7 manually for 1,000 trials would require several hours. It would be far prefer- able to perform this type of simulation on a computer. A number of mathematical com- putations would be required to determine the various column values of Table 14.7, as we will demonstrate in the Excel spreadsheet simulation of this example in the next section.
Computer Simulation of the Queuing Example with Excel The simulation of the dyeing process at Burlingham Mills shown in Table 14.7 can also be done in Excel. Exhibit 14.6 shows the spreadsheet simulation model for this example.
This formula is entered in D15 and copied to D16 : D23.
= AVERAGE(G14 : G23)Arrival times are generated by entering = E14 + D15 in E15 and copying it to E16 : E23.
Copy = VLOOKUP(H14,Lookup2,2) to I14 : I23.
Clock time is generated by entering = MAX(E15,J14) in F15 and copying it to F16 : F23.
Exhibit 14.6
A factor that can affect simulation results is the starting conditions.
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The stream of random numbers in column C is generated by the formula used in the ComputerWorld examples in Exhibits 14.2 through 14.5. (Notice that for this computer simulation, we have changed our random numbers from whole numbers to numbers between 0.0 and 1.0.) The arrival times are generated from the cumulative prob- ability distribution of arrival intervals in cells B6:C9. This array of cells is renamed “Lookup1” because there are two probability distributions in the model. The formula
is entered in cell D15 and copied to cells D16:D23 to gen- erate the arrival times in column D. The arrival clock times are computed by entering the formula in cell E15 and copying it to cells E16:E23.
A batch of yarn can enter the dyeing facility as soon as it arrives (in column E) if there is not another batch being dyed or as soon as any batches being dyed or waiting to be dyed have departed the facility (column J). This clock time is computed by entering the formula
in cell F15 and copying it to cells F16:F23. The waiting time is computed with the formula copied in cells G14:G23.
A second set of random numbers is generated in column H by using the RAND() func- tion. The service times are generated in column I from the cumulative probability distrib- ution in cells H6:I8, using the “Lookup” function again. In this case the array of cells in H6:I8 is named “Lookup2,” and the service times in column I are generated by copying the formula in cells I14:I23. The departure times in column J= VLOOKUP(H14, Lookup2,2)
= F14 - E14, = MAX(E15, J14)
= E14 + D15
= VLOOKUP(C15, Lookup1,2)
= RAND(),
642 Chapter 14 Simulation
M a n a g e m e n t S c i e n c e A p p l i c a t i o n Planning for Health Emergencies Created by Terrorist Attacks
In today’s world the threat of terrorist attacks cannot be under-estimated or discounted, especially in densely populated or strategically important areas of the country, like Washington, DC. Because such attacks could take the form of outbreaks of contagious diseases, public health departments have been prompted to plan how to respond to such health disasters. If ter- rorists released a lethal virus like smallpox into the general population, everyone would have to be vaccinated within a few days. This would require a massive response including the mass- dispensing of vaccinations through clinics staffed with the nec- essary people to perform the tasks. For example, Montgomery County, Maryland, which borders the District of Columbia, would need to vaccinate nearly a million people in just a few days. To vaccinate such a large number of people would require mass-vaccination clinics throughout the county. Any type of planning model would need to include the number of people to train beforehand, the clinics’ capacities, and the time residents would spend at a clinic (i.e., time in the system). A simulation study was used to evaluate alternative mass-vaccination clinic designs—specifically, clinic operations, clinic capacity, and resi- dent time in the system. Model data for arrivals, the time residents stayed at each station, and the total time at the clinic was derived from a full-scale test exercise at a mock-up clinic with actual workers and volunteers. The simulation model included animation so that the user could visualize residents moving through the clinic. The simulation study provided
guidelines on how many staff members would be required (fewer than estimated) and how to manage bus arrivals. The simulation study provided input into additional quantitative models for capacity planning, clinic layout, overall clinic oper- ations, and the flow of residents through the clinic.
Source: K. Aaby, J. Herrmann, C. Jordan, M. Treadwell, and K. Wood, “Montgomery County’s Public Health Service Uses Operations Research to Plan Emergency Mass Dispensing and Vaccination Clinics,” Interfaces 36, no. 6 (November–December 2006): 569–79.
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Continuous Probability Distributions 643
Continuous Probability Distributions
In the first example in this chapter, ComputerWorld’s store manager considered a proba- bility distribution of discrete demand values. In the queuing example, the probability dis- tributions were for discrete interarrival times and service times. However, applications of simulation models reflecting continuous distributions are more common than those of models employing discrete distributions.
We have concentrated on examples with discrete distributions because with a discrete distribution, the ranges of random numbers can be explicitly determined and are thus eas- ier to illustrate. When random numbers are being selected according to a continuous prob- ability distribution, a continuous function must be used. For example, consider the following continuous probability function, f(x), for time (minutes), x :
The area under the curve, f(x), represents the probability of the occurrence of the random variable x. Therefore, the area under the curve must equal 1.0 because the sum of all prob- abilities of the occurrence of a random variable must equal 1.0. By computing the area under the curve from 0 to any value of the random variable x, we can determine the cumu- lative probability of that value of x, as follows:
Cumulative probabilities are analogous to the discrete ranges of random numbers we used in previous examples. Thus, we let this function, F(x), equal the random number r,
and solve for x,
By generating a random number, r, and substituting it into this function, we determine a value for x, “time.” (However, for a continuous function, the range of random numbers must be between zero and one to correspond to probabilities between 0.0 and 1.00.) For example, if then
x = 42.25 = 2 min. r = .25,
x = 42r
r = x2
16
F(x) = x2
16
F(x) = L x
0
x
8 dx =
1
8 L x
0 xdx =
1
8 a1
2 x2b d
x
0
f (x) = x
8 , 0 … x … 4
are determined by using the formula copied in cells J14:J23, and the “Time in System” values are computed by using the formula copied in cells K14:K23.
The operating statistic, average waiting time, is computed by using the formula in cell G26, and the average time in the system is computed with
a similar formula in cell L26. Notice that both the average waiting time of 0.4 day and the average time in the system of 1.9 days are significantly lower than the simulation conducted in Table 14.7, as we speculated they might be.
= AVERAGE(G14:G23)
= J14 - E14, = F14 + I14
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644 Chapter 14 Simulation
The purpose of briefly presenting this example is to demonstrate the difference between discrete and continuous functions. This continuous function is relatively simple; as functions become more complex, it becomes more difficult to develop the equation for determining the random variable x from r. Even this simple example required some calculus, and developing more complex models would require a higher level of mathematics.
Simulation of a Machine Breakdown and Maintenance System In this example we will demonstrate the use of a continuous probability distribution. The Bigelow Manufacturing Company produces a product on a number of machines. The elapsed time between breakdowns of the machines is defined by the following continuous probability distribution:
where
weeks between machine breakdowns
As indicated in the previous section on continuous probability distributions, the equa- tion for generating x , given the random number is
When a machine breaks down, it must be repaired, and it takes either 1, 2, or 3 days for the repair to be completed, according to the discrete probability distribution shown in Table 14.8. Every time a machine breaks down, the cost to the company is an estimated $2,000 per day in lost production until the machine is repaired.
x = 42r1 r1,
x =
f (x) = x
8 , 0 … x … 4 weeks
A continuous probability distribution of the time between
machine breakdowns.
Machine Repair Probability of Cumulative Random Number Time, y (days) Repair Time, P(y) Probability Range, r2
1 .15 .15 0.00–.15 2 .55 .70 .16–.70 3 .30 1.00 .71–1.00
Table 14.8 Probability distribution of
machine repair time
The company would like to know whether it should implement a machine maintenance program at a cost of $20,000 per year that would reduce the frequency of breakdowns and thus the time for repair. The maintenance program would result in the following continu- ous probability function for time between breakdowns:
where
weeks between machine breakdowns
The equation for generating x, given the random number for this probability distribution is
x = 62r1
r1,
x =
f (x) = x>18, 0 … x … 6 weeks
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Continuous Probability Distributions 645
The reduced repair time resulting from the maintenance program is defined by the discrete probability distribution shown in Table 14.9.
Machine Repair Probability of Cumulative Random Number Time, y (days) Repair Time, P(y) Probability Range, r2
1 .40 .40 0.00–.40 2 .50 .90 .41–.90 3 .10 1.00 .91–1.00
Time Between Breakdowns, Repair Time, Cost, Cumulative Time,
Breakdowns r1 x (weeks) r2 y (days) $2,000y Σx (weeks)
1 .45 2.68 .19 2 $ 4,000 2.68 2 .90 3.80 .65 2 4,000 6.48 3 .84 3.67 .51 2 4,000 10.15 4 .17 1.65 .17 2 4,000 11.80 5 .74 3.44 .63 2 4,000 15.24 6 .94 3.88 .85 3 6,000 19.12 7 .07 1.06 .37 2 4,000 20.18 8 .15 1.55 .89 3 6,000 21.73 9 .04 0.80 .76 3 6,000 22.53
10 .31 2.23 .71 3 6,000 24.76 11 .07 1.06 .34 2 4,000 25.82 12 .99 3.98 .11 1 2,000 29.80 13 .97 3.94 .27 2 4,000 33.74 14 .73 3.42 .10 1 2,000 37.16 15 .13 1.44 .59 2 4,000 38.60 16 .03 0.70 .87 3 6,000 39.30 17 .62 3.15 .08 1 2,000 42.45 18 .47 2.74 .08 1 2,000 45.19 19 .99 3.98 .89 3 6,000 49.17 20 .75 3.46 .42 2 4,000 52.63
$84,000
Table 14.10 Simulation of machine
breakdowns and repair times
Table 14.9 Revised probability distribution of machine repair time with the
maintenance program
To solve this problem, we must first simulate the existing system to determine an esti- mate of the average annual repair costs. Then we must simulate the system with the main- tenance program installed to see what the average annual repair costs will be with the maintenance program. We will then compare the average annual repair cost with and with- out the maintenance program and compute the difference, which will be the average annual savings in repair costs with the maintenance program. If this savings is more than the annual cost of the maintenance program ($20,000), we will recommend that it be implemented; if it is less, we will recommend that it not be implemented.
First, we will manually simulate the existing breakdown and repair system without the maintenance program, to see how the simulation model is developed. Table 14.10 illus- trates the simulation of machine breakdowns and repair for 20 breakdowns that occur over a period of approximately 1 year (i.e., 52 weeks).
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646 Chapter 14 Simulation
The simulation in Table 14.10 results in a total annual repair cost of $84,000. However, this is for only 1 year, and thus it is probably not very accurate.
The next step in our simulation analysis is to simulate the machine breakdown and repair system with the maintenance program installed. We will use the revised continuous probability distribution for time between breakdowns and the revised discrete probability distribution for repair time shown in Table 14.9. Table 14.11 illustrates the manual simu- lation of machine breakdowns and repair for 1 year.
Table 14.11 shows that the annual repair cost with the maintenance program totals $42,000. Recall that in the manual simulation shown in Table 14.10, the annual repair cost was $84,000 for the system without the maintenance program. The difference between the two annual repair costs is This figure represents the savings in average annual repair cost with the maintenance program. Because the maintenance pro- gram will cost $20,000 per year, it would seem that the recommended decision would be to implement the maintenance program and generate an expected annual savings of $22,000 per year (i.e., ).
However, let us now concern ourselves with the potential difficulties caused by the fact that we simulated each system (the existing one and the system with the maintenance pro- gram) only once. Because the time between breakdowns and the repair times are proba- bilistic, the simulation results could exhibit significant variation. The only way to be sure of the accuracy of our results is to simulate each system many times and compute an aver- age result. Performing these many simulations manually would obviously require a great deal of time and effort. However, Excel can be used to accomplish the required simulation analysis.
Computer Simulation of the Machine Breakdown Example Using Excel Exhibit 14.7 shows the Excel spreadsheet model of the simulation of our original machine breakdown example simulated manually in Table 14.10. The Excel simulation is for 100 breakdowns. The random numbers in C14:C113 are generated using the RAND() function, which was used in our previous Excel examples. The “Time Between Breakdowns” values in column D are developed using the formula for the continuous cumulative probability func- tion, typed in cell D14 and copied in cells D15:D113.= 4*SQRT(C14),
$42,000 - 20,000 = $22,000
$84,000 - 42,000 = $42,000.
Table 14.11 Simulation of machine
breakdowns and repair with the maintenance program
Time Between Breakdowns, Repair Time, Cost, Cumulative Time,
Breakdowns r1 x (weeks) r2 y (days) $2,000y Σx (weeks)
1 .45 4.03 .19 1 $ 2,000 4.03 2 .90 5.69 .65 2 4,000 9.72 3 .84 5.50 .51 2 4,000 15.22 4 .17 2.47 .17 1 2,000 17.69 5 .74 5.16 .63 2 4,000 22.85 6 .94 5.82 .85 2 4,000 28.67 7 .07 1.59 .37 1 2,000 30.29 8 .15 2.32 .89 2 4,000 32.58 9 .04 1.20 .76 2 4,000 33.78
10 .31 3.34 .71 2 4,000 37.12 11 .07 1.59 .34 1 2,000 38.71 12 .99 5.97 .11 1 2,000 44.68 13 .97 5.91 .27 1 2,000 50.59 14 .73 5.12 .10 1 2,000 55.71
$42,000
Manual simulation is limited because of the amount of real time
required to simulate even one trial.
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Continuous Probability Distributions 647
The “Cumulative Time” in column E is computed by copying the formula in cells E15:E113. The second stream of random numbers in column F is generated using the RAND() function. The “Repair Time” values in column G are generated from the cumu- lative probability distribution in the array of cells B6:C8. As in our previous examples, we
= E14 + D15
Copy = E14 + D15 to E15 : E113.
Copy = VLOOKUP(F14, Lookup,2) to G14 : G113.
From Table 14.8
Spreadsheet frozen at row 24 to show first 10 breakdowns and last 6.
Click on “View“ then on “Freeze Panes“ to freeze panes at row 24.
Exhibit 14.7
Revised formula for time between breakdowns
Probability distribution or repair time from Table 14.9
Exhibit 14.8
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648 Chapter 14 Simulation
name this array “Lookup” and copy the formula in cells G14:G113. The cost values in column H are computed by entering the formula in cell H14 and copying it to cells H15:H113.
The “Average Annual Cost” in cell H7 is computed with the formula =SUM(H14:H113)/ (E113/52). For this, the original problem, the annual cost is $82,397.35, which is not too different from the manual simulation in Table 14.10.
Exhibit 14.8 shows the Excel spreadsheet simulation for the modified breakdown system with the new maintenance program, which was simulated manually in Table 14.11. The two differences in this simulation model are the cumulative probability distribution formulas for the time between breakdowns and the reduced repair time distributions from Table 14.9 in cells A6:C8.
The average annual cost for this model, shown in cell H8, is $44,504.74. This annual cost is only slightly higher than the $42,000 obtained from the manual simulation in Table 14.11. Thus, as before, the decision should be to implement the new maintenance system.
= 2000*G14 = VLOOKUP(F14,Lookup,2)
Statistical Analysis of Simulation Results
In general, the outcomes of a simulation model are statistical measures such as averages, as in the examples presented in this chapter. In our ComputerWorld example, we generated average revenue as a measure of the system we simulated; in the Burlingham Mills queu- ing example, we generated the average waiting time for batches to be dyed; and in the Bigelow Manufacturing Company machine breakdown example, the system was measured in terms of average repair costs. However, we also discussed the care that must be taken in accepting the accuracy of these statistical results because they were frequently based on rel- atively few observations (i.e., simulation replications). Thus, as part of the simulation process, these statistical results are typically subjected to additional statistical analysis to determine their degree of accuracy.
One of the most frequently used tools for the analysis of the statistical validity of simu- lations results is confidence limits. Confidence limits can be developed within Excel for the averages resulting from simulation models in several different ways. Recall that the statis- tical formulas for 95% confidence limits are
where is the mean and s is the sample standard deviation from a sample of size n from any population. Although we cannot be sure that the sample mean will exactly equal the population mean, we can be 95% confident that the true population mean will be between the upper confidence limit (UCL) and lower confidence limit (LCL) computed using these formulas.
Exhibit 14.9 shows the Excel spreadsheet for our machine breakdown example (from Exhibit 14.8), with the upper and lower confidence limits for average repair cost in cells L13 and L14. Cell L11 contains the average repair cost (for each incidence of a breakdown), computed by using the formula Cell L12 contains the sample standard deviation, computed by using the formula The upper con- fidence limit is computed in cell L13 by using the formula shown on the formula bar at the top of the spreadsheet, and the lower control limit is computed similarly. Thus, we can be 95% confident that the true average repair cost for the population is between $3,248.50 and $3,751.50.
= STDEV(H14:H113). = AVERAGE(H14:H113).
x
lower confidence limit = x - (1.96)(s>2n) upper confidence limit = x + (1.96)(s>2n)
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Statistical Analysis of Simulation Results 649
Confidence limits
Exhibit 14.9
Confidence limits plus several additional statistics can also be obtained by using the “Data Analysis” option from the “Data” menu. Select the “Data Analysis” option from the “Data” menu at the top of the spreadsheet, and then from the resulting menu select “Descriptive Statistics.” This will result in a dialog box like the one shown in Exhibit 14.10. This box, completed as shown, results in the summary statistics for repair costs shown in cells J8:K23 in Exhibit 14.11. These summary statistics include the mean, standard devia- tion, and confidence limits we computed in Exhibit 14.9, plus several other statistics.
Cost ($) values in column H
Specifies the location of the statistical summary report on the spreadsheet.
This copies the label in H13 onto the statistical summary report; if the first row is a data point, don’t check this.
Exhibit 14.10
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650 Chapter 14 Simulation
Crystal Ball
So far in this chapter we have used simulation examples that included mostly discrete prob- ability distributions that we set up on an Excel spreadsheet. These are the easiest types of probability distributions to work with in spreadsheets. However, many realistic problems contain more complex probability distributions, like the normal distribution, which are not discrete but are continuous, or they include discrete probability distributions that are more difficult to work with than the simple ones we have used. However, there are several simulation add-ins for Excel that provide the user with the capability to perform simula- tion analysis, using a variety of different probability distributions in a spreadsheet envi- ronment. One of these add-ins is Crystal Ball, published by Oracle; it is available for download from the Crystal Ball Web site using the licensing instructions that accompany this text. Crystal Ball is a risk analysis and forecasting program that uses Monte Carlo sim- ulation to forecast a statistical range of results possible for a given situation. In this section we will provide an overview of how to apply Crystal Ball to a simple example for profit (break-even) analysis that we first introduced in Chapter 1.
Simulation of a Profit Analysis Model In Chapter 1 we used a simple example for the Western Clothing Company to demonstrate break-even and profit analysis. In that example, Western Clothing Company produced denim jeans. The price (p) for jeans was $23, the variable cost was $8 per pair of jeans, and the fixed cost was $10,000. Given these parameters, we formulated a profit (Z) function as follows:
Our objective in that analysis was to determine the break-even volume, v, that would result in no profit or loss. This was accomplished by setting and solving the profit function for v, as follows:
v = cf
p - cv
Z = 0
Z = vp - cf - vcv
(cf) (cv)
Statistical summary report
Exhibit 14.11
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Crystal Ball 651
1This simulation example is located on the CD accompanying this text in a file titled “Exhibit 14.12,” located in the Crystal Ball folder.
= C4*C5�C6�(C4*C7)
1. Click on cell C4 to define normal distribution parameters.
2. Click on “Define Assumption.”
Exhibit 14.12
Substituting the values for and into this formula resulted in the break-even volume:
To demonstrate the use of Crystal Ball, we will modify that example. First, we will assume that volume is actually volume demanded and that it is a random variable defined by a normal probability distribution, with a mean value of 1,050 pairs of jeans and a stan- dard deviation of 410.
Furthermore, we will assume that the price is not fixed but is also uncertain and defined by a uniform probability distribution (from $20 to $26) and that variable cost is not a constant value but defined by a triangular probability distribution. Instead of seeking to determine the break-even volume, we will simulate the profit model, given probabilistic demand, price, and variable costs to determine average profit and the probability that Western Clothing will break even.
The first thing we need to do is access Crystal Ball, which you can download according to the instructions provided with this text.
Exhibit 14.12 shows the Excel spreadsheet for our example.1 We have described the pa- rameters of each probability distribution in our profit model next to its corresponding cell. For example, cell C4 contains the probability distribution for demand. We want to gener- ate demand values in this cell according to the probability distribution for demand (i.e., Monte Carlo simulation). We also want to do this in cell C5 for price and in cell C7 for vari- able cost. This is the same process we used in our earlier ComputerWorld example to gen- erate demand values from a discrete probability distribution, using random numbers. Notice that cell C9 contains our formula for profit, This is the only cell formula in our spreadsheet.
= C4*C5 - C6 - (C4*C7).
= 666.7 pairs of jeans
v = 10,000
23 - 8
cvp, cf,
To set up the normal probability distribution for demand, we first enter the mean value 1,050 in cell C4. Cells require some initial value to start with. Next we click on “Define Assumption” from the top of the spreadsheet, as shown in Exhibit 14.12, which will result in the menu of distributions shown in Exhibit 14.13.
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652 Chapter 14 Simulation
The Distribution drop-down menu window includes several different probability distribu- tions we can use. Because we have indicated that demand is defined by a normal distri- bution, we click it. This will result in the window for the normal distribution shown in Exhibit 14.14.
The “Name” value in the box at the top of the window in Exhibit 14.14 was automati- cally pulled from the spreadsheet, where it is the heading “volume (v) ”; however, a new or different name could be typed in. Next, we click on “Mean” or use the Tab key to toggle down to the “Mean” display in the lower-left-hand corner of this window. Because we entered the mean value of 1,050 in cell C4 on our spreadsheet, this value will already be shown in this window. Next, we click on “Std. Dev.” or use the Tab key to move to the Std. Dev. window and enter the standard deviation of 410. Then we click on the “Enter” button, which will configure the normal distribution figure in the window, and then we click on “OK.”
=
Click on "Normal" distribution.
Exhibit 14.13
Name pulled from original spreadsheet
1. Click on Windows Tab key to enter Mean and standard deviation.
2. Click “Enter” to configure distribution in window.
3. Click on “OK” to return to the spreadsheet.
Exhibit 14.14
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Crystal Ball 653
We will repeat this same process to enter the parameters for the uniform distribution for price in cell C5. First, we enter the value for price, 23, in cell C5. Next (with cell C5 activated), we click on “Define Assumption” at the top of the spreadsheet, as shown earlier in Exhibit 14.13. The Distribution menu window will again appear, and this time we click on “Uniform Distribution.” This results in the Uniform Distribution window shown in Exhibit 14.15.
As before, the “Name” value, “price (p),” was pulled from the original spreadsheet in Exhibit 14.12. Next, we click on “Minimum” or use the Tab key to move to the “Minimum” display at the bottom of the window and enter 20, the lower limit of the uniform distribu- tion specified in the problem statement. Next, we activate the “Maximum” display window and enter 26. Then we click on the “Enter” button to configure the distribution graph in the window. Finally, we click on “OK” to exit this window.
We repeat the same process to enter the triangular distribution parameters in cell C7. A triangular probability distribution is defined by three estimated values—a minimum, a most likely, and a maximum. It is a very useful approximation when enough data points do not exist to allow for the construction of a distribution, but the user can estimate what the endpoints and the midpoint of the distribution might be. Clicking on “Define Assumption” and then selecting the triangular distribution from the Distribution menu results in the window shown in Exhibit 14.16.
We enter the “Minimum” value of 6.75, the “Likeliest” value of 8.00, and the “Maximum” value of 9.10. Clicking on “Enter” will configure the graph of the triangular distribution shown in the window. We click on “OK” to exit this window and return to the spreadsheet.
Next, we click on cell C9 on our original spreadsheet. Recall that this is the cell in which we entered our profit formula in Exhibit 14.12. The profit value of 5,750, computed from the other cell values entered on the original spreadsheet, will be shown in cell C9. We click on “Define Forecast” at the top of the spreadsheet as shown in Exhibit 14.17. This will result in the “Define Forecast” window also shown in Exhibit 14.17. The heading “Profit(Z) ” will already be entered from the spreadsheet. We click on the “Units” display and enter “dollars.” We then click on “OK” to exit this window. This completes the process of enter- ing our simulation parameters and data. Exhibit 14.18 shows the spreadsheet with changes resulting from the parameter inputs. The next step is to run the simulation.
=
Enter minimum and maximum values for the distribution.
Exhibit 14.15
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654 Chapter 14 Simulation
Enter three estimates.
Exhibit 14.16
1. Click on “Define Forecast.”
2. Type in units.
Exhibit 14.17
2. Click on “Start” to begin the simulation.
1. Click on “Run Preferences” to indicate the number of trials and seed number.
Exhibit 14.18
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Crystal Ball 655
The mechanics of the simulation are similar to those of our previous Excel spreadsheet models. Using random numbers, we want to generate a value for demand in cell C4, then a value for price in C5, and then a value for variable cost in C7. These three values are then substituted into the profit formula in cell C9 to compute a profit value. This represents one repetition, or trial, of the simulation. The simulation is run for many trials in order to develop a distribution for profit.
To run the simulation, we click on “Run Preferences” at the top of the spreadsheet in Exhibit 14.18. This activates the window shown in Exhibit 14.19. We then enter the num- ber of simulations for the simulation run. For this example we will run the simulation for 5,000 trials. Next, we click on “Sampling” at the top of this window to activate the window shown in Exhibit 14.20. In this window we must enter the seed value for a sequence of random numbers for the simulation, which is always 999. We click on “OK” and then go back to the spreadsheet. From the spreadsheet menu (Exhibit 14.21), we click on “Start,” which will run the simulation. Exhibit 14.21 shows the simulation window with the simu- lation completed for 5,000 trials and the frequency distribution for this simulation.
1. Enter the number of trials.
2. Go to the “Sampling” screen to enter the seed number.
Exhibit 14.19
Click here to repeat the same simulation.
Exhibit 14.20
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656 Chapter 14 Simulation
A statistical summary report for this simulation can be obtained by clicking on “View” at the top of the “Forecast” window and then selecting “Statistics” from the drop-down menu. This results in the window shown in Exhibit 14.22. You can return to the Forecast window by selecting “Frequency” from the “View” menu at the top of the Statistics window.
Click here to reset the simulation and run it again.
Set new run preferences.
Click on “Statistics” to go to the statistical summary screen.
2. Click here to start the simulation.
1. Click here to establish the number of trials and the seed number.
Exhibit 14.21
Click on “View” to return to the Frequency window.
Exhibit 14.22
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Crystal Ball 657
In our original example formulated in Chapter 1, we wanted to determine the break- even volume. In this revised example, Western Clothing Company wants to know the aver- age profit and the probability that it will break even from this simulation analysis. The mean profit (from the “Statistics” window in Exhibit 14.22) is $5,918.89. The probability of breaking even is determined by clicking on the arrow on the left side of the horizontal axis of the window shown in Exhibit 14.23 and “grabbing” it and moving it to “0.00,” or by clicking on the lower limit, currently set at “Infinity”; we change this to 0 and press the Enter key. This will shift the lower limit to zero, the break-even point. The frequency chart that shows the location of the new lower limit and the “Certainty” of zero profit is shown as 81.36% at the bottom of the window as shown in Exhibit 14.23. Thus, there is a .8136 probability that the company will break even.
Probability (.8136) that the company will break even
Move arrow to “0.00” or set lower limit equal to 0.00.
Exhibit 14.23
We have demonstrated using Crystal Ball with a straightforward example that was not very complex or detailed. Crystal Ball has the capability to perform much more sophisti- cated simulation analyses than what we have shown in this section. However, the demon- stration of these capabilities and other features of Crystal Ball would require more space and in-depth coverage than is possible here. However, although using Crystal Ball to simulate more complex situations requires a greater degree of knowledge than we have provided, this basic introduction to and demonstration of Crystal Ball provide a good starting point to understanding the basic features of Crystal Ball and its use for simula- tion analysis.
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658 Chapter 14 Simulation
Verification of the Simulation Model
Even though we may be able to verify the statistical results of a simulation model, we still may not know whether the model actually replicates what is going on in the real world. The user of simulation generally wants to be certain that the model is internally correct and that all the operations performed in the simulation are logical and mathematically correct. An old adage often associated with computer simulation is “garbage in, garbage out.” To gain some assurances about the validity of simulation results, there are several testing proce- dures that the user of a simulation model can apply.
First, the simulation model can be run for short periods of time or for only a few sim- ulation trials. This allows the user to compare the results with manually derived solutions (as we did in the examples in this chapter) to check for discrepancies. Another means of testing is to divide the model into parts and simulate each part separately. This reduces the complexity of seeking out errors in the model. Similarly, the mathematical relationships in the simulation model can be simplified so that they can more easily be tested to see if the model is operating correctly.
To determine whether the model reliably represents the system being simulated, the sim- ulation results can sometimes be compared with actual real-world data. Several statistical tests are available for performing this type of analysis. However, when a model is developed to simulate a new or unique system, there is no realistic way to ensure that the results are valid.
An additional problem in determining whether a simulation model is a valid represen- tation of the system under analysis relates to starting conditions. Should the simulation be started with the system empty (e.g., should we start by simulating a queuing system with no customers in line), or should the simulation be started as close as possible to normal operating conditions? Another problem, as we have already seen, is the determination of how long the simulation should be run to reach true steady-state conditions, if indeed a steady state exists.
In general, a standard, foolproof procedure for validation is simply not possible. In many cases, the user of a simulation model must rely on the expertise and experience of whoever develops the model.
Simulation models must be validated to make sure they are
accurately replicating the system being simulated.
Sometimes manual simulation of several trials is a good way to
validate a simulation.
Areas of Simulation Application
Simulation is one of the most useful of all management science techniques. The reason for this popularity is that simulation can be applied to a number of problems that are too dif- ficult to model and solve analytically. Some analysts feel that complex systems should be studied via simulation whether or not they can be analyzed analytically because simulation provides such an easy vehicle for experimenting on the system. As a result, simulation has been applied to a wide range of problems. Surveys conducted during the 1990s indicated that a large majority of major corporations use simulation in such functional areas as pro- duction, corporate planning, engineering, financial analysis, research and development, marketing, information systems, and personnel. Following are descriptions of some of the most common applications of simulation.
Queuing A major application of simulation has been in the analysis of queuing systems. As indicated in Chapter 13, the assumptions required to solve the operating characteristic formulas are relatively restrictive. For the more complex queuing systems (which result from a
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Areas of Simulation Application 659
relaxation of these assumptions), it is not possible to develop analytical formulas, and sim- ulation is often the only available means of analysis.
Inventory Control Most people are aware that product demand is an essential component in determining the amount of inventory a commercial enterprise should keep. Most of the mathematical formulas used to analyze inventory systems make the assumption that this demand is certain (i.e., not a random variable). In practice, however, demand is rarely known with certainty. Simulation is one of the few means for analyzing inventory systems in which demand is a random variable, reflecting demand uncertainty. Inventory control is dis- cussed in Chapter 16.
Production and Manufacturing Simulation is often applied to production problems, such as production scheduling, pro- duction sequencing, assembly line balancing (of work-in-process inventory), plant layout, and plant location analysis. It is surprising how often various production processes can be viewed as queuing systems that can be analyzed only by using simulation. Because machine breakdowns typically occur according to some probability distributions, maintenance problems are also frequently analyzed using simulation.
Finance Capital budgeting problems require estimates of cash flows, which are often a result of many random variables. Simulation has been used to generate values of the various con- tributing factors to derive estimates of cash flows. Simulation has also been used to deter- mine the inputs into rate of return calculations in which the inputs are random variables, such as market size, selling price, growth rate, and market share.
Marketing Marketing problems typically include numerous random variables, such as market size and type, and consumer preferences. Simulation can be used to ascertain how a particular mar- ket might react to the introduction of a product or to an advertising campaign for an exist- ing product. Another area in marketing where simulation is applied is the analysis of distribution channels to determine the most efficient distribution system.
Public Service Operations The operations of police departments, fire departments, post offices, hospitals, court sys- tems, airports, and other public systems have all been analyzed by using simulation. Typically, such operations are so complex and contain so many random variables that no technique except simulation can be employed for analysis.
Environmental and Resource Analysis Some of the more recent innovative applications of simulation have been directed at prob- lems in the environment. Simulation models have been developed to ascertain the impact on the environment of projects such as nuclear power plants, reservoirs, highways, and dams. In many cases, these models include measures to analyze the financial feasibility of
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Summary
Simulation has become an increasingly important management science technique inrecent years. Various surveys have shown simulation to be one of the techniques most widely applied to real-world problems. Evidence of this popularity is the number of spe- cialized simulation languages that have been developed by the computer industry and academia to deal with complex problem areas.
The popularity of simulation is due in large part to the flexibility it allows in analyzing systems, compared to more confining analytical techniques. In other words, the problem does not have to fit the model (or technique)—the simulation model can be constructed to fit the problem. A primary benefit of simulation analysis is that it enables us to experiment
660 Chapter 14 Simulation
such projects. Other models have been developed to simulate pollution conditions. In the area of resource analysis, numerous models have been developed in recent years to simu- late energy systems and the feasibility of alternative energy sources.
M a n a g e m e n t S c i e n c e A p p l i c a t i o n Simulating a 10-km Race in Boulder, Colorado
The Bolder Boulder, a popular 10-kilometer race held eachMemorial Day in Colorado, attracts many of the world’s best runners among its 20,000 participants. The race starts at the Bank of Boulder at the northeastern corner of the city, winds through the city streets, and ends at the University of Colorado’s football stadium in the center of the city. As the race grew in size (from 2,200 participants in 1979 to 20,000 in 1985), its quality suffered from overcrowding problems, espe- cially at the finish line, where runners are individually tagged as they finish. Large waiting lines built up at the finish line, caus- ing many complaints from the participants.
To correct this problem, race management implemented an interval-start system in 1986, wherein 24 groups of up to 1,000 runners each were started at 1-minute intervals. Although this solution alleviated the problem of street crowding, it did not solve the queuing problem at the finish line.
A simulation model of the race was then developed to evalu- ate several possible solutions—specifically, increasing the number of finish line chutes from the 8 used previously to either 12 or 15. The model was also used to identify a set of block-start intervals that would eliminate finish line queuing problems with either chute scenario. Recommendations based on the simulation model were for a 12-chute finish line configuration and specific block-start intervals. The race conducted using the recommenda- tions from the simulation model was flawless. The actual race behavior was almost identical to the simulation results. No over- crowding or queuing problems occurred at the finish line. The simulation model was used to fine-tune the 1986 and 1987 races, which were also conducted with virtually no problems.
Source: R. Farina et al., “The Computer Runs the Bolder Boulder: A Simulation of a Major Running Race,” Interfaces 19, no. 2 (March–April 1989): 48–55.
Simulation is one of the most important and widely used
management science techniques.
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Example Problem Solution 661
with the model. For example, in our queuing example we could expand the model to rep- resent more service facilities, more queues, and different arrival and service times; and we could observe their effects on the results. In many analytical cases, such experimentation is limited by the availability of an applicable formula. That is, by changing various parts of the problem, we may create a problem for which we have no specific analytical formula. Simulation, however, is not subject to such limitations. Simulation is limited only by one’s ability to develop a computer program.
Simulation is a management science technique that does not usually result in an optimal solution. Generally, a simulation model reflects the operation of a system, and the results of the model are in the form of operating statistics, such as averages. However, optimal solutions can sometimes be obtained for simulation models by employing search techniques.
However, in spite of its versatility, simulation has limitations and must be used with cau- tion. One limitation is that simulation models are typically unstructured and must be developed for a system or problem that is also unstructured. Unlike some of the structured techniques presented in this text, they cannot simply be applied to a specific type of prob- lem. As a result, developing simulation models often requires imagination and intuitiveness that are not required by some of the more straightforward solution techniques we have pre- sented. In addition, the validation of simulation models is an area of serious concern. It is often impossible realistically to validate simulation results, to know if they accurately reflect the system under analysis. This problem has become an area of such concern that “output analysis” of simulation results is developing into a new field of study. Another lim- iting factor in simulation is the cost in money and time of model building. Because simu- lation models are developed for unstructured systems, they often take large amounts of staff time, computer time, and money to develop and run. For many business companies, these costs can be prohibitive.
Simulation provides a laboratory for experimentation on a real
system.
Simulation does not usually provide a recommended decision as does an optimization model; it
provides operating characteristics.
Simulation has certain limitations.
References
Hammersly, J. M., and Handscomb, D. C. Monte Carlo Methods. New
York: John Wiley & Sons, 1964.
Meier, R. C., Newell, W. T., and Pazer, H. L. Simulation in Business
and Economics. Upper Saddle River, NJ: Prentice Hall, 1969.
Mize, J., and Cox, G. Essentials of Simulation. Upper Saddle River, NJ:
Prentice Hall, 1968.
Naylor, T. H., Balintfy, J. L., Burdinck, D. S., and Chu, K. Computer
Simulation Techniques. New York: John Wiley & Sons, 1966.
Tocher, K. D. “Review of Computer Simulation.” Operational
Research Quarterly 16 (June 1965): 189–217.
Van Horne, R. L. “Validation of Simulation Results.” Management
Science 17 (January 1971): 247–57.
The following example problem demonstrates a manual simulation using discrete prob- ability distributions.
Problem Statement
Members of the Willow Creek Emergency Rescue Squad know from past experience that they will receive between zero and six emergency calls each night, according to the fol- lowing discrete probability distribution:
Example Problem Solution
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662 Chapter 14 Simulation
Calls Probability
0 .05 1 .12 2 .15 3 .25 4 .22 5 .15 6 .06
1.00
The rescue squad classifies each emergency call into one of three categories: minor, regular, or major emergency. The probability that a particular call will be each type of emergency is as follows:
Emergency Type Probability
Minor .30 Regular .56 Major .14
1.00
The type of emergency call determines the size of the crew sent in response. A minor emergency requires a two-person crew, a regular call requires a three-person crew, and a major emergency requires a five-person crew.
Simulate the emergency calls received by the rescue squad for 10 nights, compute the average number of each type of emergency call each night, and determine the maximum number of crew members that might be needed on any given night.
Solution
Step 1: Develop Random Number Ranges for the Probability Distributions
Cumulative Random Number Calls Probability Probability Range, r1
0 .05 .05 1–5 1 .12 .17 6–17 2 .15 .32 18–32 3 .25 .57 33–57 4 .22 .79 58–79 5 .15 .94 80–94 6 .06 1.00 95–99, 00
1.00
Emergency Cumulative Random Number Type Probability Probability Range, r2
Minor .30 .30 1–30 Regular .56 .86 31–86 Major .14 1.00 87–99, 00
1.00
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Example Problem Solution 663
Step 2: Set Up a Tabular Simulation
Use the second column of random numbers in Table 14.3:
Number of Emergency Crew Total per Night r1 Calls r2 Type Size Night
1 65 4 71 Regular 3 18 Minor 2 12 Minor 2 17 Minor 2 9
2 48 3 89 Major 5 18 Minor 2 83 Regular 3 10
3 08 1 90 Major 5 5 4 05 0 — — — — 5 89 5 18 Minor 2
08 Minor 2 26 Minor 2 47 Regular 3 94 Major 5 14
6 06 1 72 Regular 3 3 7 62 4 47 Regular 3
68 Regular 3 60 Regular 3 88 Major 5 14
8 17 1 36 Regular 3 3 9 77 4 43 Regular 3
28 Minor 2 31 Regular 3 06 Minor 2 10
10 68 4 39 Regular 3 71 Regular 3 22 Minor 2 76 Regular 3 11
Step 3: Compute Results
If all the calls came in at the same time, the maximum number of squad members required during any 1 night would be 14.
average number of major emergency calls per night = 4
10 = 0.40
average number of regular emergency calls per night = 13
10 = 1.3
average number of minor emergency calls per night = 10
10 = 1.0
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664 Chapter 14 Simulation
Problems
1. The Hoylake Rescue Squad receives an emergency call every 1, 2, 3, 4, 5, or 6 hours, according to the following probability distribution. The squad is on duty 24 hours per day, 7 days per week:
Time Between Emergency Calls (hr.) Probability
1 .05 2 .10 3 .30 4 .30 5 .20 6 .05
1.00
a. Simulate the emergency calls for 3 days (note that this will require a “running,” or cumulative, hourly clock), using the random number table.
b. Compute the average time between calls and compare this value with the expected value of the time between calls from the probability distribution. Why are the results different?
c. How many calls were made during the 3-day period? Can you logically assume that this is an average number of calls per 3-day period? If not, how could you simulate to determine such an average?
2. The time between arrivals of cars at the Petroco Service Station is defined by the following proba- bility distribution:
Time Between Arrivals (min.) Probability
1 .15 2 .30 3 .40 4 .15
1.00
a. Simulate the arrival of cars at the service station for 20 arrivals and compute the average time between arrivals.
b. Simulate the arrival of cars at the service station for 1 hour, using a different stream of random numbers from those used in (a) and compute the average time between arrivals.
c. Compare the results obtained in (a) and (b).
3. The Dynaco Manufacturing Company produces a product in a process consisting of operations of five machines. The probability distribution of the number of machines that will break down in a week follows:
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 665
Machine Breakdowns per Week Probability
0 .10 1 .10 2 .20 3 .25 4 .30 5 .05
1.00
a. Simulate the machine breakdowns per week for 20 weeks. b. Compute the average number of machines that will break down per week.
4. Solve Problem 19 at the end of Chapter 12 by using simulation.
5. Simulate the decision situation described in Problem 16(a) at the end of Chapter 12 for 20 weeks, and recommend the best decision.
6. Every time a machine breaks down at the Dynaco Manufacturing Company (Problem 3), either 1, 2, or 3 hours are required to fix it, according to the following probability distribution:
Repair Time (hr.) Probability
1 .30 2 .50 3 .20
1.00
a. Simulate the repair time for 20 weeks and then compute the average weekly repair time. b. If the random numbers that are used to simulate breakdowns per week are also used to
simulate repair time per breakdown, will the results be affected in any way? Explain. c. If it costs $50 per hour to repair a machine when it breaks down (including lost
productivity), determine the average weekly breakdown cost. d. The Dynaco Company is considering a preventive maintenance program that would alter
the probabilities of machine breakdowns per week as shown in the following table:
Machine Breakdowns per Week Probability
0 .20 1 .30 2 .20 3 .15 4 .10 5 .05
1.00
The weekly cost of the preventive maintenance program is $150. Using simulation, determine whether the company should institute the preventive maintenance program.
7. Sound Warehouse in Georgetown sells CD players (with speakers), which it orders from Fuji Elec- tronics in Japan. Because of shipping and handling costs, each order must be for five CD players.
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666 Chapter 14 Simulation
Because of the time it takes to receive an order, the warehouse outlet places an order every time the present stock drops to five CD players. It costs $100 to place an order. It costs the warehouse $400 in lost sales when a customer asks for a CD player and the warehouse is out of stock. It costs $40 to keep each CD player stored in the warehouse. If a customer cannot purchase a CD player when it is requested, the customer will not wait until one comes in but will go to a competitor. The fol- lowing probability distribution for demand for CD players has been determined:
Demand per Month Probability
0 .04 1 .08 2 .28 3 .40 4 .16 5 .02 6 .02
1.00
The time required to receive an order once it is placed has the following probability distribution:
Time to Receive an Order (mo.) Probability
1 .60 2 .30 3 .10
1.00
The warehouse has five CD players in stock. Orders are always received at the beginning of the week. Simulate Sound Warehouse’s ordering and sales policy for 20 months, using the first column of random numbers in Table 14.3. Compute the average monthly cost.
8. First American Bank is trying to determine whether it should install one or two drive-through teller windows. The following probability distributions for arrival intervals and service times have been developed from historical data:
Time Between Automobile Arrivals (min.) Probability
1 .20 2 .60 3 .10 4 .10
1.00
Service Time (min.) Probability
2 .10 3 .40 4 .20 5 .20 6 .10
1.00
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 667
Assume that in the two-server system, an arriving car will join the shorter queue. When the queues are of equal length, there is a 50–50 chance the driver will enter the queue for either window.
a. Simulate both the one- and two-teller systems. Compute the average queue length, waiting time, and percentage utilization for each system.
b. Discuss your results in (a) and suggest the degree to which they could be used to make a decision about which system to employ.
9. The time between arrivals of oil tankers at a loading dock at Prudhoe Bay is given by the following probability distribution:
Time between Ship Arrivals (days) Probability
1 .05 2 .10 3 .20 4 .30 5 .20 6 .10 7 .05
1.00
The time required to fill a tanker with oil and prepare it for sea is given by the following prob- ability distribution:
Time to Fill and Prepare (days) Probability
3 .10 4 .20 5 .40 6 .30
1.00
a. Simulate the movement of tankers to and from the single loading dock for the first 20 arrivals. Compute the average time between arrivals, average waiting time to load, and average number of tankers waiting to be loaded.
b. Discuss any hesitation you might have about using your results for decision making.
10. The Saki automobile dealer in the Minneapolis–St. Paul area orders the Saki sport compact, which gets 50 miles per gallon of gasoline, from the manufacturer in Japan. However, the dealer never knows for sure how many months it will take to receive an order once it is placed. It can take 1, 2, or 3 months, with the following probabilities:
Months to Receive an Order Probability
1 .50 2 .30 3 .20
1.00
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668 Chapter 14 Simulation
The demand per month is given by the following distribution:
Demand per Month (cars) Probability
1 .10 2 .30 3 .40 4 .20
1.00
The dealer orders when the number of cars on the lot gets down to a certain level. To determine the appropriate level of cars to use as an indicator of when to order, the dealer needs to know how many cars will be demanded during the time required to receive an order. Simulate the demand for 30 orders and compute the average number of cars demanded during the time required to receive an order. At what level of cars in stock should the dealer place an order?
11. State University is playing Tech in their annual football game on Saturday. A sportswriter has scouted each team all season and accumulated the following data: State runs four basic plays—a sweep, a pass, a draw, and an off tackle; Tech uses three basic defenses—a wide tackle, an Oklahoma, and a blitz. The number of yards State will gain for each play against each defense is shown in the following table:
Tech Defense
State Play Wide Tackle Oklahoma Blitz
Sweep –3 5 12 Pass 12 4 �10 Draw 2 1 20 Off tackle 7 3 �3
The probability that State will run each of its four plays is shown in the following table:
Play Probability
Sweep .10 Pass .20 Draw .20 Off tackle .50
The probability that Tech will use each of its defenses follows:
Defense Probability
Wide tackle .30 Oklahoma .50 Blitz .20
The sportswriter estimates that State will run 40 plays during the game. The sportswriter believes that if State gains 300 or more yards, it will win; however, if Tech holds State to fewer than 300 yards, it will win. Use simulation to determine which team the sportswriter will predict to win the game.
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 669
12. Each semester, the students in the college of business at State University must have their course schedules approved by the college adviser. The students line up in the hallway outside the adviser’s office. The students arrive at the office according to the following probability distribution:
Time Between Arrivals (min.) Probability
4 .20 5 .30 6 .40 7 .10
1.00
The time required by the adviser to examine and approve a schedule corresponds to the follow- ing probability distribution:
Schedule Approval (min.) Probability
6 .30 7 .50 8 .20
1.00
Simulate this course approval system for 90 minutes. Compute the average queue length and the average time a student must wait. Discuss these results.
13. A city is served by two newspapers—the Tribune and the Daily News. Each Sunday readers purchase one of the newspapers at a stand. The following matrix contains the probabilities of a customer’s buying a particular newspaper in a week, given the newspaper purchased the previous Sunday:
This Sunday Next Sunday
Simulate a customer’s purchase of newspapers for 20 weeks to determine the steady-state proba- bilities that a customer will buy each newspaper in the long run.
14. Loebuck Grocery orders milk from a dairy on a weekly basis. The manager of the store has devel- oped the following probability distribution for demand per week (in cases):
Demand (cases) Probability
15 .20 16 .25 17 .40 18 .15
1.00
The milk costs the grocery $10 per case and sells for $16 per case. The carrying cost is $0.50 per case per week, and the shortage cost is $1 per case per week. Simulate the ordering system for Loebuck
Tribune Daily News
Tribune
Daily News
.
.
.
.
65
45
35
55
⎤
⎦ ⎥
⎡
⎣ ⎢ ⎢
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670 Chapter 14 Simulation
Grocery for 20 weeks. Use a weekly order size of 16 cases of milk and compute the average weekly profit for this order size. Explain how the complete simulation for determining order size would be developed for this problem.
15. The Paymore Rental Car Agency rents cars in a small town. It wants to determine how many rental cars it should maintain. Based on market projections and historical data, the manager has deter- mined probability distributions for the number of rentals per day and rental duration (in days only) as shown in the following tables:
Number of Customers/Day Probability
0 .20 1 .20 2 .50 3 .10
1.00
Rental Duration (days) Probability
1 .10 2 .30 3 .40 4 .10 5 .10
1.00
Design a simulation experiment for the car agency and simulate using a fleet of four rental cars for 10 days. Compute the probability that the agency will not have a car available upon demand. Should the agency expand its fleet? Explain how a simulation experiment could be designed to determine the optimal fleet size for the Paymore Agency.
16. A CPM/PERT project network has probabilistic activity times (x) as shown on each branch of the network; for example, activity 1–3 has a .40 probability that it will be completed in 6 weeks and a .60 probability it will be completed in 10 weeks:
1
2 4
3 x
6 10
.4
.6
x
5 9
.6
.4 x
2 6 8
.2
.7
.1
x
3 5
.5
.5
x
1 3 6
.3
.3
.4
5 6
x
1 2
.4
.6
x
3 5 7
.2
.5
.3 P(x)
P(x)
P(x)
P(x)
P(x)
P(x)
P(x)
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 671
Simulate the project network 10 times and determine the critical path each time. Compute the average critical path time and the frequency at which each path is critical. How does this simula- tion analysis of the critical path method compare with regular CPM/PERT analysis?
17. A robbery has just been committed at the Corner Market in the downtown area of the city. The market owner was able to activate the alarm, and the robber fled on foot. Police officers arrived a few minutes later and asked the owner, “How long ago did the robber leave?” “He left only a few minutes ago,” the store owner responded. “He’s probably 10 blocks away by now,” one of the officers said to the other. “Not likely,” said the store owner. “He was so stoned on drugs that I bet even if he has run 10 blocks, he’s still only within a few blocks of here! He’s probably just running in circles!”
Perform a simulation experiment that will test the store owner’s hypothesis. Assume that at each corner of a city block there is an equal chance that the robber will go in any one of the four possi- ble directions: north, south, east, or west. Simulate for five trials and then indicate in how many of the trials the robber is within 2 blocks of the store.
18. Compcomm, Inc., is an international communications and information technology company that has seen the value of its common stock appreciate substantially in recent years. A stock analyst would like to use simulation to predict the stock prices of Compcomm for an extended period. Based on historical data, the analyst has developed the following probability distribution for the movement of Compcomm stock prices per day:
Stock Price Movement Probability
Increase .45 Same .30 Decrease .25
1.00
The analyst has also developed the following probability distributions for the amount of the increases or decreases in the stock price per day:
Probability
Stock Price Change Increase Decrease
1/8 .40 .12 1/4 .17 .15 3/8 .12 .18 1/2 .10 .21 5/8 .08 .14 3/4 .07 .10 7/8 .04 .05 1 .02 .05
1.00 1.00
The price of the stock is currently 62.
Develop a Monte Carlo simulation model to track the stock price of Compcomm stock and simu- late for 30 days. Indicate the new stock price at the end of the 30 days. How would this model be expanded to conduct a complete simulation of 1 year’s stock price movement?
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672 Chapter 14 Simulation
19. The emergency room of the community hospital in Farmburg has one receptionist, one doctor, and one nurse. The emergency room opens at time zero, and patients begin to arrive some time later. Patients arrive at the emergency room according to the following probability distribution:
Time Between Arrivals (min.) Probability
5 .06 10 .10 15 .23 20 .29 25 .18 30 .14
1.00
The attention needed by a patient who comes to the emergency room is defined by the following probability distribution:
Patient Needs to See Probability
Doctor alone .50 Nurse alone .20 Both .30
1.00
If a patient needs to see both the doctor and the nurse, he or she cannot see one before the other—that is, the patient must wait to see both together.
The length of the patient’s visit (in minutes) is defined by the following probability distributions:
Doctor Probability Nurse Probability Both Probability
10 .22 5 .08 15 .07 15 .31 10 .24 20 .16 20 .25 15 .51 25 .21 25 .12 20 .17 30 .28 30 .10 1.00 35 .17
1.00 40 .11 1.00
Simulate the arrival of 20 patients to the emergency room and compute the probability that a patient must wait and the average waiting time. Based on this one simulation, does it appear that this system provides adequate patient care?
20. The Western Outfitters Store specializes in denim jeans. The variable cost of the jeans varies accord- ing to several factors, including the cost of the jeans from the distributor, labor costs, handling, packaging, and so on. Price also is a random variable that varies according to competitors’ prices.
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 673
Sales volume also varies each month. The probability distributions for volume, price, and variable costs each month are as follows:
Sales Volume Probability
300 .12 400 .18 500 .20 600 .23 700 .17 800 .10
1.00
Price Probability
$22 .07 23 .16 24 .24 25 .25 26 .18 27 .10
1.00
Variable Cost Probability
$ 8 .17 9 .32
10 .29 11 .14 12 .08
1.00
Fixed costs are $9,000 per month for the store.
Simulate 20 months of store sales and compute the probability that the store will at least break even and the average profit (or loss).
21. Randolph College and Salem College are within 20 miles of each other, and the students at each college frequently date each other. The students at Randolph College are debating how good their dates are at Salem College. The Randolph students have sampled several hundred of their fellow students and asked them to rate their dates from 1 to 5 (in which 1 is excellent and 5 is poor) according to physical attractiveness, intelligence, and personality. Following are the resulting prob- ability distributions for these three traits for students at Salem College:
Physical Attractiveness Probability
1 .27 2 .35 3 .14 4 .09 5 .15
1.00
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674 Chapter 14 Simulation
Intelligence Probability
1 .10 2 .16 3 .45 4 .17 5 .12
1.00
Personality Probability
1 .15 2 .30 3 .33 4 .07 5 .15
1.00
Simulate 20 dates and compute an average overall rating of the Salem students.
22. In Problem 21 discuss how you might assess the accuracy of the average rating for Salem College students based on only 20 simulated dates.
23. Burlingham Mills produces denim cloth that it sells to jeans manufacturers. It is negotiating a con- tract with Troy Clothing Company to provide denim cloth on a weekly basis. Burlingham has established its monthly available production capacity for this contract to be between 0 and 600 yards, according to the following probability distribution:
Troy Clothing’s weekly demand for denim cloth varies according to the following probability distribution:
Demand (yd.) Probability
0 .03 100 .12 200 .20 300 .35 400 .20 500 .10
1.00
Simulate Troy Clothing’s cloth orders for 20 weeks and determine the average weekly capacity and demand. Also determine the probability that Burlingham will have sufficient capacity to meet demand.
f (x) = x
180,000 , 0 … x … 600 yd.
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Problems 675
24. A baseball game consists of plays that can be described as follows:
Play Description
No advance An out where no runners can advance. This includes strikeouts, pop-ups, short flies, and the like.
Groundout Each runner can advance one base. Possible double play Double play if there is a runner on first base and fewer
than two outs. The lead runner who can be forced is out; runners not out advance one base. If there is no runner on first or there are two outs, this play is treated as a “no advance.”
Long fly A runner on third base can score. Very long fly Runners on second and third base advance one base. Walk Includes a hit batter. Infield single All runners advance one base. Outfield single A runner on first base advances one base, but a
runner on second or third base scores. Long single All runners can advance a maximum of two bases. Double Runners can advance a maximum of two bases. Long double All runners score. Triple Home run
Note: Singles also include a factor for errors, allowing the batter to reach first base.
Distributions for these plays for two teams, the White Sox (visitors) and the Yankees (home), are as follows:
2This problem was adapted from R. E. Trueman, “A Computer Simulation Model of Baseball: With Particular Application to Strategy Analysis,” in R. E. Machol, S. P. Ladany, and D. G. Morrison, eds., Management Science in Sports (New York: North Holland Publishing Co., 1976), 1–14.
Team: White Sox
Play Probability
No advance .03 Groundout .39 Possible double play .06 Long fly .09 Very long fly .08 Walk .06 Infield single .02 Outfield single .10 Long single .03 Double .04 Long double .05 Triple .02 Home run .03
1.00
Team: Yankees
Play Probability
No advance .04 Groundout .38 Possible double play .04 Long fly .10 Very long fly .06 Walk .07 Infield single .04 Outfield single .10 Long single .04 Double .05 Long double .03 Triple .01 Home run .04
1.00
Simulate a nine-inning baseball game using the preceding information.2
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676 Chapter 14 Simulation
25. For the ComputerWorld example in this chapter, recreate the simulation shown in Exhibit 14.2 using Crystal Ball. Assume that demand is normally distributed with a mean of 1.5 laptops and a standard deviation of 0.8. Using Crystal Ball determine average weekly demand and average weekly revenue.
26. Perform the Crystal Ball simulation in Problem 25, assuming that demand is normally distributed with a mean of 2.5 and a standard deviation of 1.2 laptops.
27. For the Bigelow Manufacturing example in this chapter, re-create the simulation shown in Table 14.10 and Exhibit 14.7 using Crystal Ball. Assume that the repair time is normally distributed with a mean of 2.15 days and a standard deviation of 0.8 day. Assume that the time between breakdowns is defined by the triangular distribution used in the example. Using Crystal Ball, determine the average annual number of breakdowns, average annual repair time, and average annual repair cost.
28. For the Bigelow Manufacturing example in this chapter, re-create the simulation for the improved maintenance program shown in Table 14.11 and Exhibit 14.8 using Crystal Ball. For the improved program assume that the repair time is normally distributed with a mean of 1.70 days and a stan- dard deviation of 0.6 day. Assume that the time between breakdowns is defined by the triangular dis- tribution for the improved maintenance program used in the example. Using Crystal Ball, determine the average annual number of breakdowns, average annual repair time, and average annual repair cost. Compare this improved maintenance system with the current one (Problem 27), and indicate whether it should be adopted given the cost of improving the system (i.e., $20,000).
29. Tracy McCoy is shopping for a new car. She has identified a particular sports utility vehicle she likes but has heard that it has high maintenance costs. Tracy has decided to develop a simulation model to help her estimate maintenance costs for the life of the car. Tracy estimates that the projected life of the car with the first owner (before it is sold) is uniformly distributed with a minimum of 2.0 years and a maximum of 8.0 years. Furthermore, she believes that the miles she will drive the car each year can be defined by a triangular distribution with a minimum value of 3,700 miles, a maximum value of 14,500 miles, and a most likely value of 9,000 miles. She has determined from automobile association data that the maintenance cost per mile driven for the vehicle she is interested in is normally distrib- uted, with a mean of $0.08 per mile and a standard deviation of $0.02 per mile. Using Crystal Ball, develop a simulation model (using 1,000 trials) and determine the average maintenance cost for the life of the car with Tracy and the probability that the cost will be less than $3,000.
30. In Problem 20, assume that the sales volume for Western Outfitters Store is normally distributed, with a mean of 600 pairs of jeans and a standard deviation of 200; the price is uniformly distrib- uted, with a minimum of $22 and a maximum of $28; and the variable cost is defined by a trian- gular distribution with a minimum value of $6, a maximum of $11, and a most likely value of $9. Develop a simulation model by using Crystal Ball (with 1,000 trials) and determine the average profit and the probability that Western Outfitters will break even.
31. In Problem 21, assume that the students at Randolph College have redefined the probability dis- tributions of their dates at Salem College as follows: Physical attractiveness is uniformly distributed from 1 to 5; intelligence is defined by a triangular distribution with a minimum rating of 1, a max- imum of 5, and a most likely of 2; and personality is defined by a triangular distribution with a minimum of 1, a maximum of 5, and a most likely rating of 3. Develop a simulation model by using Crystal Ball and determine the average date rating (for 1,000 trials). Also compute the probability that the rating will be “better” than 3.0.
32. In Problem 23, assume that production capacity at Burlingham Mills for the Troy Clothing Company contract is normally distributed, with a mean of 320 yards per month and a standard deviation of 120 yards, and that Troy Clothing’s demand is uniformly distributed between 0 and 500 yards. Develop a simulation model by using Crystal Ball and determine the average monthly short- age or surplus for denim cloth (for 1,000 trials). Also determine the probability that Burlingham will always have sufficient production capacity.
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Problems 677
33. Erin Jones has $100,000 and, to diversify, she wants to invest equal amounts of $50,000 each in two mutual funds selected from a list of four possible mutual funds. She wants to invest for a 3-year period. She has used historical data from the four funds plus data from the market to determine the mean and standard deviation (normally distributed) of the annual return for each fund, as follows:
Return (r)
Fund μ σ
1. Internet .20 .09 2. Index .12 .04 3. Entertainment .16 .10 4. Growth .14 .06
The possible combinations of two investment funds are (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4). a. Use Crystal Ball to simulate each of the investment combinations to determine the expected
return in 3 years. (Note that the formula for the future value, FV, of a current investment, P, with return r for n years in the future is ) Indicate which investment combination has the highest expected return.
b. Erin wants to reduce her risk as much as possible. She knows that if she invests her $100,000 in a CD at the bank, she is guaranteed a return of $20,000 after 3 years. Using the frequency charts for the simulation runs in Crystal Ball, determine which combination of investments will result in the greatest probability of receiving a return of $120,000 or greater.
34. In Chapter 16, the formula for the optimal order quantity of an item, Q, given its demand, D, order cost, and the cost of holding, or carrying, an item in inventory, is as follows:
The total inventory cost formula is
Ordering cost, and carrying cost, are generally values that the company is often able to deter- mine with certainty because they are internal costs, whereas demand, D, is usually not known with cer- tainty because it is external to the company. However, in the order quantity formula given here, demand is treated as if it were certain. To consider the uncertainty of demand, it must be simulated.
Using Crystal Ball, simulate the preceding formulas for Q and TC to determine their average val- ues for an item, with and demand, D, that is normally distributed with a mean of 10,000 and a standard deviation of 4,000.
35. The Management Science Association (MSA) has arranged to hold its annual conference at the Riverside Hotel in Orlando next year. Based on historical data, the MSA believes the number of rooms it will need for its members attending the conference is normally distributed, with a mean of 800 and a standard deviation of 270. The MSA can reserve rooms now (1 year prior to the confer- ence) for $80; however, for any rooms not reserved now, the cost will be at the hotel’s regular room rate of $120. The MSA guarantees the room rate of $80 to its members. If its members reserve fewer than the number of rooms it reserves, MSA must pay the hotel for the difference, at the $80 room rate. If MSA does not reserve enough rooms, it must pay the extra cost—that is, $40 per room.
a. Using Crystal Ball, determine whether the MSA should reserve 600, 700, 800, 900, or 1,000 rooms in advance to realize the lowest total cost.
b. Can you determine a more exact value for the number of rooms to reserve to minimize cost?
Co = $150, Cc = $0.75,
Cc,Co,
TC = CoD
Q + Cc
Q
2
Q = A 2C0D
Cc
Cc,Co,
FVn = Pr(1 + r)n.
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678 Chapter 14 Simulation
36. In Chapter 8, Figure 8.6 shows a simplified project network for building a house, as follows:
There are four paths through this network:
Path A: 1–2–3–4–6–7 Path B: 1–2–3–4–5–6–7 Path C: 1–2–4–6–7 Path D: 1–2–4–5–6–7
The time parameters (in weeks) defining a triangular probability distribution for each activity are provided as follows:
Time Parameters
Activity Minimum Likeliest Maximum
1–2 1 3 5 2–3 1 2 4 2–4 0.5 1 2 3–4 0 0 0 4–5 1 2 3 4–6 1 3 6 5–6 1 2 4 6–7 1 2 4
a. Using Crystal Ball, simulate each path in the network and identify the longest path (i.e., the critical path).
b. Observing the simulation run frequency chart for path A, determine the probability that this path will exceed the critical path time. What does this tell you about the simulation results for a project network versus an analytical result?
1 2
3
5
4 6 7 3
Lay foundation
Design house and obtain financing
Order materials
Select paint
Select carpet
Build house
Finish work
Dummy
1
2 0
1 1
13
Case Problem
JET Copies
James Banks was standing in line next to Robin Cole at Klecko’sCopy Center, waiting to use one of the copy machines. “Gee, Robin, I hate this,” he said. “We have to drive all the way over here
from Southgate and then wait in line to use these copy machines. I hate wasting time like this.”
“I know what you mean,” said Robin. “And look who’s here. A lot of these students are from Southgate Apartments or one of the other apartments near us. It seems as though it would be more logical if Klecko’s would move its operation over to us, instead of all of us coming over here.”
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
Case Problems 679
James looked around and noticed what Robin was talking about. Robin and he were students at State University, and most of the customers at Klecko’s were also students. As Robin suggested, a lot of the people waiting were State students who lived at Southgate Apartments, where James also lived with Ernie Moore. This gave James an idea, which he shared with Ernie and their friend Terri Jones when he got home later that evening.
“Look, you guys, I’ve got an idea to make some money,” James started. “Let’s open a copy business! All we have to do is buy a copier, put it in Terri’s duplex next door, and sell copies. I know we can get customers because I’ve just seen them all at Klecko’s. If we provide a copy service right here in the Southgate complex, we’ll make a killing.”
Terri and Ernie liked the idea, so the three decided to go into the copying business. They would call it JET Copies, named for James, Ernie, and Terri. Their first step was to purchase a copier. They bought one like the one used in the college of business office at State for $18,000. (Terri’s parents provided a loan.) The com- pany that sold them the copier touted the copier’s reliability, but after they bought it, Ernie talked with someone in the dean’s office at State, who told him that the University’s copier broke down fre- quently and when it did, it often took between 1 and 4 days to get it repaired. When Ernie told this to Terri and James, they became worried. If the copier broke down frequently and was not in use for long periods while they waited for a repair person to come fix it, they could lose a lot of revenue. As a result, James, Ernie, and Terri thought they might need to purchase a smaller backup copier for $8,000 to use when the main copier broke down. However, before they approached Terri’s parents for another loan, they wanted to have an estimate of just how much money they might lose if they did not have a backup copier. To get this esti- mate, they decided to develop a simulation model because they were studying simulation in one of their classes at State.
To develop a simulation model, they first needed to know how frequently the copier might break down—specifically, the time between breakdowns. No one could provide them with an exact probability distribution, but from talking to staff members in the college of business, James estimated that the time between break- downs was probably between 0 and 6 weeks, with the probability increasing the longer the copier went without breaking down. Thus, the probability distribution of breakdowns generally looked like the following:
Next, they needed to know how long it would take to get the copier repaired when it broke down. They had a service contract with the dealer that “guaranteed” prompt repair service. However, Terri gathered some data from the college of business from which she developed the following probability distribution of repair times:
Repair Time (days) Probability
1 .20 2 .45 3 .25 4 .10
1.00
Finally, they needed to estimate how much business they would lose while the copier was waiting for repair. The three of them had only a vague idea of how much business they would do but finally estimated that they would sell between 2,000 and 8,000 copies per day at $0.10 per copy. However, they had no idea about what kind of probability distribution to use for this range of values. Therefore, they decided to use a uniform probability distribution between 2,000 and 8,000 copies to estimate the number of copies they would sell per day.
James, Ernie, and Terri decided that if their loss of revenue due to machine downtime during 1 year was $12,000 or more, they should purchase a backup copier. Thus, they needed to simulate the break- down and repair process for a number of years to obtain an average annual loss of revenue. However, before programming the simulation model, they decided to conduct a manual simulation of this process for 1 year to see if the model was working correctly. Perform this manual simulation for JET Copies and determine the loss of revenue for 1 year.
6 x, weeks
.33
f(x)
0
Case Problem
Benefit–Cost Analysis of the Spradlin Bluff River Project
The U.S. Army Corps of Engineers has historically constructeddams on various rivers in the southeastern United States. Its primary instrument for evaluating and selecting among many projects under consideration is benefit–cost analysis. The Corps estimates both the annual benefits deriving from a project in several different categories and the annual costs and then divides
the total benefits by the total costs to develop a benefit–cost ratio. This ratio is then used by the Corps and Congress to compare numerous projects under consideration and select those for fund- ing. A benefit–cost ratio greater than 1.0 indicates that the bene- fits are greater than the costs; and the higher a project’s benefit–cost ratio, the more likely it is to be selected over projects with lower ratios.
The Corps is evaluating a project to construct a dam over the Spradlin Bluff River in southwest Georgia. The Corps has iden- tified six traditional areas in which benefits will accrue: flood
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.
680 Chapter 14 Simulation
100 years (at a rate of interest specified by the government), and annual operation and maintenance costs. The cost estimates for this project are as follows:
Estimate
Cost Minimum Most Likely Maximum
Annualized capital cost $12,890,750 $14,150,500 $19,075,900 Operation and
maintenance 3,483,500 4,890,000 7,350,800
Using Crystal Ball, determine a simulated mean benefit–cost ratio and standard deviation.
What is the probability that this project will have a benefit–cost ratio greater than 1.0?
control, hydroelectric power, improved navigation, recreation, fish and wildlife, and area commercial redevelopment. The Corps has made three estimates (in dollars) for each benefit—a minimum possible value, a most likely value, and a maximum benefit value. These benefit estimates are as follows:
Estimate
Benefit Minimum Most Likely Maximum
Flood control $ 1,695,200 $ 2,347,800 $ 3,570,600 Hydroelectric power 8,068,250 11,845,000 14,845,000 Navigation 50,400 64,000 109,500 Recreation 6,404,000 9,774,000 14,566,000 Fish and wildlife 104,300 255,000 455,300 Area redevelopment 0 1,630,000 2,385,000
There are two categories of costs associated with a construc- tion project of this type—the total capital cost, annualized over
Case Problem
Disaster Planning at Tech
Concerned about recent weather-related disasters, fires, andother calamities at universities around the country, university administrators at Tech have initiated several planning projects to determine how effectively local emergency facilities can handle such situations. One of these projects has focused on the trans- port of disaster victims from campus to the five major hospitals in the area: Montgomery Regional, Raeford Memorial, County General, Lewis Galt, and HGA Healthcare. The project team would like to determine how many victims each hospital might expect in a disaster and how long it would take to transport vic- tims to the hospitals. However, one of the problems the project team faces is the lack of data on disasters, since they occur so infrequently. The project team has looked at disasters at other schools and has estimated that the minimum number of victims that would qualify an event as a disaster for the purpose of initi- ating a disaster plan is 10. The team has further estimated that the largest number of victims in any disaster would be 200, and based on limited data from other schools, they believe the most likely number of disaster victims is approximately 50. Because of the lack of data, it is assumed that these parameters best define a tri- angular distribution. The emergency facilities and capabilities at the five area hospitals vary. It has been estimated that in the event of a disaster situation, the victims should be dispersed to the hospitals on a percentage basis based on the hospitals’ relative emer- gency capabilities, as follows: 25% should be sent to Montgomery Regional, 30% to Raeford Memorial, 15% to County General, 10% to Lewis Galt, and 20% to HGA Healthcare. The proximity of the hospitals to Tech also varies. It is estimated that transport times to
each of the hospitals is exponentially distributed with an average time of 5 minutes to Montgomery Regional, 10 minutes to Raeford Memorial, 20 minutes to County General, 20 minutes to Lewis Galt, and 15 minutes to HGA Healthcare. (It is assumed that each hospital has two emergency vehicles, so that one leaves Tech when the other leaves the hospital, and consequently, one arrives at Tech when the other arrives at the hospital. Thus, the total transport time will be the sum of transporting each victim to a specific hospital.)
a. Perform a simulation analysis using Crystal Ball to deter- mine the average number of victims that can be expected at each hospital and the average total time required to trans- port victims to each hospital.
b. Suppose that the project team believes they cannot confi- dently assume that the number of victims will follow a triangular distribution using the parameters they have estimated. Instead, they believe that the number of vic- tims is best estimated using a normal distribution with the following parameters for each hospital: a mean of 6 minutes and a standard deviation of 4 minutes for Montgomery Regional; a mean of 11 minutes and a standard deviation of 4 minutes for Raeford Memorial; a mean of 22 minutes and a standard deviation of 8 min- utes for County General; a mean of 22 minutes and a standard deviation of 9 minutes for Lewis Galt; and a mean of 15 minutes and a standard deviation of 5 minutes for HGA Healthcare. Perform a simulation analysis using this revised information.
c. Discuss how this information might be used for planning purposes. How might the simulation be altered or changed to provide additional useful information?
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Introduction to Management Science, Tenth Edition, by Bernard W. Taylor III. Published by Prentice Hall. Copyright © 2010 by Pearson Education, Inc.