I Need Help In Manufacturing System. I Attached The Assignment

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EXAM_1-Answers.docx

EXAM #1 Grade…………. /15

IE 470- Manufacturing Systems

First Name: ………………………………

Last Name: ……………………………….

University ID: ……………………………

1. Review Questions (0.5-point each)

A. What are the three capabilities that a manufacturing system must possess in order to be flexible?

Answer: As identified in the text, the three capabilities are (1) identification of the different work units, (2) quick changeover of operating instructions, and (3) quick changeover of the physical setup.

B. What is the difference between a machine cluster and machine cell?

Answer: A machine cluster is defined as a collection of two or more machines producing parts or products with identical cycle times and serviced by one worker. By contract, a machine cell consists of one or more machines organized to produce a family of parts or products.

C. What do the terms starving and blocking mean?

Answer: Starving is the situation in which the assembly operator has completed the assigned task on the current work unit, but the next unit has not yet arrived at the station. The worker is thus starved for work. Blocking means that the operator has completed the assigned task on the current work unit but cannot pass the unit to the downstream station because that worker is not yet ready to receive it. The operator is therefore blocked from working.

D. The theoretical minimum number of workers on an assembly line w* is the minimum integer that is greater than the ratio of the work content time Twc divided by the cycle time Tc. Two factors are identified in the text that make it difficult to achieve this minimum value in practice. Name the two factors and briefly explain each.

Answer: The two factors are (1) repositioning losses, in which either the worker or the work unit must be positioned at the start of each work cycle, and (2) the line balancing problem, which refers to the difficulty encountered in allocating all of the work elements in the work content time equally among workers on the line.

2. A plastic injection molding plant will be built to produce 6 million molded parts per year. The plant will run three 8 hour shifts per day, five days per week, and 50 weeks per year. For planning purposes, the average batch size = 6000 moldings, average changeover time between batches = 6 hrs, and average molding cycle time per part = 30 sec. Assume scrap rate = 2 percent, and average uptime proportion (reliability) per molding machine = 97 %, which applies to both run time and changeover time. How many molding machines are required in the new plant? (3 points)

Solution: Production: WL = = 51,020.4 hr/yr

Setup: number batches/yr = = 1000 batches = 1000 setups

WL = 1000(6) = 6000 hr/yr

AT = 3(5)(50)(8)(0.97) = 5820 hr/yr per machine

n = = 8.766 + 1.031 = 9.797 10 molding machines

3. The CNC grinding section has a large number of machines devoted to grinding shafts for the automotive industry. The grinding machine cycle takes 3.6 min. At the end of this cycle an operator must be present to unload and load parts, which takes 40 sec. (a) determine how many grinding machines the worker can service if it takes 20 sec to walk between the machines and no machine idle time is allowed. (b) How many seconds during the work cycle is the worker idle? (c) What is the hourly production rate of this machine cluster? (3 points)

Solution: (a) n = = 256/60 = 4.27 Use n1 = 4 grinding machines

(b) Worker idle time IT = 256 - 4(60) = 256 - 240 = 16 sec

(c) Tc = 256 sec = 4.267 min Rc = 4 = 56.25 pc/hr

4. Production rate for a certain assembled product is 47.5 units per hour. The assembly work content time = 32 min of direct manual labor. The line operates at 95% uptime. Ten workstations have two workers on opposite sides of the line so that both sides of the product can be worked on simultaneously. The remaining stations have one worker. Repositioning time lost by each worker is 0.2 min/cycle. It is known that the number of workers on the line is two more than the number required for perfect balance. Determine (a) number of workers, (b) number of workstations, (c) balance efficiency, and (d) average manning level. (3 points)

Solution: (a) Tc = = 1.2 min, Ts = 1.2 - 0.2 = 1.0 min

For perfect balance, Eb = 1.0. w = Minimum Integer = 32 workers.

With 2 more workers than required for perfect balance, w = 32 + 2 = 34 workers.

(b) n = 20/2 + (34 - 20) = 10 + 14 = 24 stations.

(c) Eb = = 0.941 = 94.1%

(d) M = 34/24 = 1.417

5. A manual assembly line operates with a mechanized conveyor. The conveyor moves at a speed of 5 ft/min, and the spacing between base parts launched onto the line is 4 ft. It has been determined that the line operates best when there is one worker per station and each station is 6 ft long. There are 14 work elements that must be accomplished to complete the assembly, and the element times and precedence requirements are listed in the table below. Determine (a) feed rate and corresponding cycle time, (b) tolerance time for each worker, and (c) ideal minimum number of workers on the line. (d) Draw the precedence diagram for the problem. (e) Determine an efficient line balancing solution. (f) For your solution, determine the balance delay. (4 points)

Element

Te

Preceded by:

Element

Te

Preceded by:

1

0.2 min

-

8

0.2 min

5

2

0.5 min

-

9

0.4 min

5

3

0.2 min

1

10

0.3 min

6, 7

4

0.6 min

1

11

0.1 min

9

5

0.1 min

2

12

0.2 min

8, 10

6

0.2 min

3, 4

13

0.1 min

11

7

0.3 min

4

14

0.3 min

12, 13

Solution: (a) Assume Tr = 0. fp = = 1.25 asbys/min, Tc = 0.8 min/asby

(b) Tt = = 1.2 min/station

(c) Twc = Tek = 0.2 + 0.5 + . . + 0.3 = 3.7 min

w = Minimum Integer = 4.625 5 workers

(d) Precedence diagram

(e) Line balancing solution using the Kilbridge and Wester method

List of elements by precedence columns

Allocation of elements to stations

Element

Te (min)

Column

Station

Element

Te

Te

1

0.2

I

1

1

0.2 min

2

0.5

I

2

0.5 min

3

0.2

II

5

0.1 min

0.8 min

4

0.6

II

2

3

0.2 min

5

0.1

II

4

0.6 min

0.8 min

6

0.2

III

3

6

0.2 min

7

0.3

III

7

0.3 min

8

0.2

III

8

0.2 min

0.7 min

9

0.4

III

4

9

0.4 min

10

0.3

IV

10

0.3 min

0.7 min

11

0.1

IV

5

11

0.1 min

12

0.2

V

12

0.2 min

13

0.1

V

13

0.1 min

14

0.3

VI

14

0.3 min

0.7 min

3.7 min total

(f) Balance delay d = = 0.075 = 7.5%

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