# Week 10 assignment

10/31/19

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EE 200: Engineering Economics 2 Time Value of Money

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136 ENGINEERING ECONOMICS

ENGINEERING ECONOMICS

Factor Name Converts Symbol Formula Single Payment Compound Amount to F given P (F/P, i%, n) (1 + i)

n

Single Payment Present Worth to P given F (P/F, i%, n) (1 + i)

–n

Uniform Series Sinking Fund to A given F (A/F, i%, n) ( ) 11+ ni

i

Capital Recovery to A given P (A/P, i%, n) ( )

( ) 11 1 +

+ n

n

i ii

Uniform Series Compound Amount to F given A (F/A, i%, n)

( ) i i n 11+

Uniform Series Present Worth to P given A (P/A, i%, n)

( ) ( )n

n

ii i +

+ 1

11

Uniform Gradient Present Worth to P given G (P/G, i%, n)

( ) ( ) ( )nn

n

ii n

ii i

++ +

11 11

2

Uniform Gradient † Future Worth to F given G (F/G, i%, n)

( ) i n

i i n+

2 11

Uniform Gradient Uniform Series to A given G (A/G, i%, n) ( ) 11

1 + ni n

i

NOMENCLATURE AND DEFINITIONS A �������������Uniform amount per interest period B �������������Benefit BV �����������Book value C �������������Cost d��������������In ation ad usted interest rate per interest period Dj ������������Depreciation in year j EV �����������Expected value F �������������Future worth, value, or amount f �������������� eneral in ation rate per interest period G �������������Uniform gradient amount per interest period i ��������������Interest rate per interest period ie �������������Annual effective interest rate MARR ����Minimum acceptable/attractive rate of return m �������������Number of compounding periods per year n��������������Number of compounding periods; or the expected

life of an asset P �������������Present worth, value, or amount r ��������������Nominal annual interest rate Sn ������������Expected salvage value in year n

Subscripts j ����������� at time j n����������� at time n †����������� F/G = (F/A – n)/i = (F/A) × (A/G)

Risk Risk is the chance of an outcome other than what is planned to occur or expected in the analysis�

NON-ANNUAL COMPOUNDING

i m r1 1e m

= + -b l

BREAK-EVEN ANALYSIS By altering the value of any one of the variables in a situation, holding all of the other values constant, it is possible to find a value for that variable that makes the two alternatives equally economical� This value is the break-even point� Break-even analysis is used to describe the percentage of capacity of operation for a manufacturing plant at which income will ust cover expenses. The payback period is the period of time required for the profit or other benefits of an investment to equal the cost of the investment�

Time Value of Money Relating Future and Present Worth Shows Unreasonable Costs

3 𝐹 is Future Worth 𝑃 is Present Worth 𝑛 is the number of compounding periods 𝐴 is Uniform Amount per interest period (Payment) 𝑖 is the interest rate per period

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10/31/19

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College Costs Have Historically Increased Faster than Inflation: Compound Interest Model

• Umich: Total Annual Cost o 1980: \$3,500 o 2020: \$30,000 o At 3% annual inflation o Should be: \$11,417 o From 𝐹 = 𝑃 1 + 𝑖 )

§ 𝑃 = \$3,500 § 𝑖 = 0.03 = 3% annual

inflation

§ 𝑛 = 40 years

• Effective Umich Inflation rate • Given 𝐹, 𝑃, 𝑛; Find 𝑖 • Rearranging: 𝐹 = 𝑃 1 + 𝑖 )

o 𝑖 = 2 3

4 5 − 1

o 𝑖 = 0.0055 = 5.5%

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Probably a bit high ee200.m Part A

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Fred Wilma Barney Betty Dino Description

Age 35 30 25 20 20 Age starting retirement savings

n yrs 20 20 15 10 45 Years making deposit of A k\$

N yrs 30 35 40 45 45 Years of investing (65 – Age)

A k\$ 10 7.5 5 5 2 Annual lump sum deposit

• 5 Cases shown below • Assumptions

o Stock market average annual rate of return is 10%

o Contribute same annual amount for 𝑛 years starting at 𝐴𝑔𝑒

• Assumptions (cont.) o Investments compound after that

for N − 𝑛 years to Age 65 o Determine projected retirement

account value at Age 65

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136 ENGINEERING ECONOMICS

ENGINEERING ECONOMICS

Factor Name Converts Symbol Formula Single Payment Compound Amount to F given P (F/P, i%, n) (1 + i)

n

Single Payment Present Worth to P given F (P/F, i%, n) (1 + i)

–n

Uniform Series Sinking Fund to A given F (A/F, i%, n) ( ) 11+ ni

i

Capital Recovery to A given P (A/P, i%, n) ( )

( ) 11 1 +

+ n

n

i ii

Uniform Series Compound Amount to F given A (F/A, i%, n)

( ) i i n 11+

Uniform Series Present Worth to P given A (P/A, i%, n)

( ) ( )n

n

ii i +

+ 1

11

Uniform Gradient Present Worth to P given G (P/G, i%, n)

( ) ( ) ( )nn

n

ii n

ii i

++ +

11 11

2

Uniform Gradient † Future Worth to F given G (F/G, i%, n)

( ) i n

i i n+

2 11

Uniform Gradient Uniform Series to A given G (A/G, i%, n) ( ) 11

1 + ni n

i

NOMENCLATURE AND DEFINITIONS A �������������Uniform amount per interest period B �������������Benefit BV �����������Book value C �������������Cost d��������������In ation ad usted interest rate per interest period Dj ������������Depreciation in year j EV �����������Expected value F �������������Future worth, value, or amount f �������������� eneral in ation rate per interest period G �������������Uniform gradient amount per interest period i ��������������Interest rate per interest period ie �������������Annual effective interest rate MARR ����Minimum acceptable/attractive rate of return m �������������Number of compounding periods per year n��������������Number of compounding periods; or the expected

life of an asset P �������������Present worth, value, or amount r ��������������Nominal annual interest rate Sn ������������Expected salvage value in year n

Subscripts j ����������� at time j n����������� at time n †����������� F/G = (F/A – n)/i = (F/A) × (A/G)

Risk Risk is the chance of an outcome other than what is planned to occur or expected in the analysis�

NON-ANNUAL COMPOUNDING

i m r1 1e m

= + -b l

BREAK-EVEN ANALYSIS By altering the value of any one of the variables in a situation, holding all of the other values constant, it is possible to find a value for that variable that makes the two alternatives equally economical� This value is the break-even point� Break-even analysis is used to describe the percentage of capacity of operation for a manufacturing plant at which income will ust cover expenses. The payback period is the period of time required for the profit or other benefits of an investment to equal the cost of the investment�

Time Value of Money To Get Future Worth from Payments (A) and the 2 Equations

8 𝐹 is Future Worth 𝑃 is Present Worth 𝑛 is the number of compounding periods 𝐴 is Uniform Amount per interest period (Payment) 𝑖 is the interest rate per period

1. 𝐹: from portion of time making investment deposits

2. 𝐹; from portion of time investments compound

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Analysis Solution Derived from Time Value of Money Equations

𝐹: = 𝐴 1 + 𝑖 ) − 1

𝑖

𝐹; = 𝐹: 1 + 𝑖 <=)

= 𝐴 1 + 𝑖 ) − 1

𝑖 1 + 𝑖 <=)

• Future Value from o Annual deposits of 𝐴 o For 𝑛 years

• Investment compounding o Of present value 𝐹: o For N− 𝑛 years

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10/31/19

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Lesson: Save Money When You’re Young!

• Corollary: o Don’t borrow o Except for a house

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136 ENGINEERING ECONOMICS

ENGINEERING ECONOMICS

Factor Name Converts Symbol Formula Single Payment Compound Amount to F given P (F/P, i%, n) (1 + i)

n

Single Payment Present Worth to P given F (P/F, i%, n) (1 + i)

–n

Uniform Series Sinking Fund to A given F (A/F, i%, n) ( ) 11+ ni

i

Capital Recovery to A given P (A/P, i%, n) ( )

( ) 11 1 +

+ n

n

i ii

Uniform Series Compound Amount to F given A (F/A, i%, n)

( ) i i n 11+

Uniform Series Present Worth to P given A (P/A, i%, n)

( ) ( )n

n

ii i +

+ 1

11

Uniform Gradient Present Worth to P given G (P/G, i%, n)

( ) ( ) ( )nn

n

ii n

ii i

++ +

11 11

2

Uniform Gradient † Future Worth to F given G (F/G, i%, n)

( ) i n

i i n+

2 11

Uniform Gradient Uniform Series to A given G (A/G, i%, n) ( ) 11

1 + ni n

i

NOMENCLATURE AND DEFINITIONS A �������������Uniform amount per interest period B �������������Benefit BV �����������Book value C �������������Cost d��������������In ation ad usted interest rate per interest period Dj ������������Depreciation in year j EV �����������Expected value F �������������Future worth, value, or amount f �������������� eneral in ation rate per interest period G �������������Uniform gradient amount per interest period i ��������������Interest rate per interest period ie �������������Annual effective interest rate MARR ����Minimum acceptable/attractive rate of return m �������������Number of compounding periods per year n��������������Number of compounding periods; or the expected

life of an asset P �������������Present worth, value, or amount r ��������������Nominal annual interest rate Sn ������������Expected salvage value in year n

Subscripts j ����������� at time j n����������� at time n †����������� F/G = (F/A – n)/i = (F/A) × (A/G)

Risk Risk is the chance of an outcome other than what is planned to occur or expected in the analysis�

NON-ANNUAL COMPOUNDING

i m r1 1e m

= + -b l

BREAK-EVEN ANALYSIS By altering the value of any one of the variables in a situation, holding all of the other values constant, it is possible to find a value for that variable that makes the two alternatives equally economical� This value is the break-even point� Break-even analysis is used to describe the percentage of capacity of operation for a manufacturing plant at which income will ust cover expenses. The payback period is the period of time required for the profit or other benefits of an investment to equal the cost of the investment�

Loan Payments and Interest Paid Analysis

13 𝐹 is Future Worth 𝑃 is Present Worth 𝑛 is the number of compounding periods 𝐴 is Uniform Amount per interest period (Payment) 𝑖 is the interest rate per period

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10/31/19

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Loans Can be Very Costly

• CSULB: Total Annual Cost o 2020: \$25,000 o Loan Payment

o 𝐴 = 𝑃 > :?> 5

:?> 5=: = \$290.30

§ 𝑃 = \$25,000

§ 𝑖 = B.BC :;

for 7% annual interest

§ 𝑛 = 10 ⋅ 12 = 120 months o Cumulative Interest Paid o 𝐶> = 𝑛 ⋅ 𝐴 − 𝑃 = \$9,833

• Calculations in Matlab o ee200.m Part C o Also in Excel: LoanAmort.xlsx

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Credit Cards Are Very Bad Trap: Min Payment is Negative Amortization

• Causes principal owed to grow o Assumed Principal: \$20,000

o Interest: 𝑖 = B.:H :;

for 18% annual interest

o 𝐴 = 𝑃 > :?> 5

:?> 5=: = \$360.38

o Min Pmt of 𝐴 = \$50 o Debt grows rapidly

• Calculations in Excel o LoanAmort.xlsx

:; for 5% annual

interest

o 𝐴 = 𝑃 > :?> 5

:?> 5=: = \$2,684.12

o 30-year term

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10/31/19

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136 ENGINEERING ECONOMICS

ENGINEERING ECONOMICS

Factor Name Converts Symbol Formula Single Payment Compound Amount to F given P (F/P, i%, n) (1 + i)

n

Single Payment Present Worth to P given F (P/F, i%, n) (1 + i)

–n

Uniform Series Sinking Fund to A given F (A/F, i%, n) ( ) 11+ ni

i

Capital Recovery to A given P (A/P, i%, n) ( )

( ) 11 1 +

+ n

n

i ii

Uniform Series Compound Amount to F given A (F/A, i%, n)

( ) i i n 11+

Uniform Series Present Worth to P given A (P/A, i%, n)

( ) ( )n

n

ii i +

+ 1

11

Uniform Gradient Present Worth to P given G (P/G, i%, n)

( ) ( ) ( )nn

n

ii n

ii i

++ +

11 11

2

Uniform Gradient † Future Worth to F given G (F/G, i%, n)

( ) i n

i i n+

2 11

Uniform Gradient Uniform Series to A given G (A/G, i%, n) ( ) 11

1 + ni n

i

NOMENCLATURE AND DEFINITIONS A �������������Uniform amount per interest period B �������������Benefit BV �����������Book value C �������������Cost d��������������In ation ad usted interest rate per interest period Dj ������������Depreciation in year j EV �����������Expected value F �������������Future worth, value, or amount f �������������� eneral in ation rate per interest period G �������������Uniform gradient amount per interest period i ��������������Interest rate per interest period ie �������������Annual effective interest rate MARR ����Minimum acceptable/attractive rate of return m �������������Number of compounding periods per year n��������������Number of compounding periods; or the expected

life of an asset P �������������Present worth, value, or amount r ��������������Nominal annual interest rate Sn ������������Expected salvage value in year n

Subscripts j ����������� at time j n����������� at time n †����������� F/G = (F/A – n)/i = (F/A) × (A/G)

Risk Risk is the chance of an outcome other than what is planned to occur or expected in the analysis�

NON-ANNUAL COMPOUNDING

i m r1 1e m

= + -b l

BREAK-EVEN ANALYSIS By altering the value of any one of the variables in a situation, holding all of the other values constant, it is possible to find a value for that variable that makes the two alternatives equally economical� This value is the break-even point� Break-even analysis is used to describe the percentage of capacity of operation for a manufacturing plant at which income will ust cover expenses. The payback period is the period of time required for the profit or other benefits of an investment to equal the cost of the investment�

These Formulas Form the Basis of Engineering Economic Trade-offs and Decision Making

18 𝐹 is Future Worth 𝑃 is Present Worth 𝑛 is the number of compounding periods 𝐴 is Uniform Amount per interest period (Payment) 𝑖 is the interest rate per period

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