Www.izmirekonomi.edu.tr L2: Time Value of Money ECON 320 Engineering Economics Mahmut Ali GOKCE Industrial Systems Engineering Computer Sciences.

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L2: Time Value of Money

ECON 320 Engineering EconomicsMahmut Ali GOKCEIndustrial Systems EngineeringComputer Sciences

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Chapter 2Time Value of Money

Interest: The Cost of Money

Economic Equivalence Interest Formulas –

Single Cash Flows Equal-Payment Series Dealing with Gradient

Series Composite Cash Flows.

Power-Ball Lottery

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Decision Dilemma—Take a Lump Sum or Annual Installments

A suburban Chicago couple won the Power-ball.

They had to choose between a single lump sum $104 million, or $198 million paid out over 25 years (or $7.92 million per year).

The winning couple opted for the lump sum.

Did they make the right choice? What basis do we make such an economic comparison?

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Option A

(Lump Sum)

Option B

(Installment Plan)

0

1

2

3

25

$104 M

$7.92 M

$7.92 M

$7.92 M

$7.92 M

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What Do We Need to Know?

To make such comparisons (the lottery decision problem), we must be able to compare the value of money at different point in time.

To do this, we need to develop a method for reducing a sequence of benefits and costs to a single point in time. Then, we will make our comparisons on that basis.

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Time Value of Money Money has a time value

because it can earn more money over time (earning power).

Money has a time value because its purchasing power changes over time (inflation).

Time value of money is measured in terms of interest rate.

Interest is the cost of money—a cost to the borrower and an earning to the lender

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Delaying Consumption

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N = 0 $100

N = 1 $104(inflation rate = 4%)

N = 0 $100

N = 1 $106(earning rate =6%)

Case 2: Earning power exceeds inflation

N = 0 $100

N = 1 $108(inflation rate = 8%)

N = 0 $100

N = 1 $106(earning rate =6%)

Case 1: Inflation exceeds earning power

Cost of RefrigeratorAccount Value

N = 0 $100

N = 1 $104(inflation rate = 4%)

N = 0 $100

N = 1 $106(earning rate =6%)

Case 2: Earning power exceeds inflation

N = 0 $100

N = 1 $108(inflation rate = 8%)

N = 0 $100

N = 1 $106(earning rate =6%)

Case 1: Inflation exceeds earning power

Cost of RefrigeratorAccount Value

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$$

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Which Repayment Plan?End of Year Receipts Payments

Plan 1 Plan 2

Year 0 $20,000.00 $200.00 $200.00

Year 1 5,141.85 0

Year 2 5,141.85 0

Year 3 5,141.85 0

Year 4 5,141.85 0

Year 5 5,141.85 30,772.48The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate)

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Cash Flow Diagram

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End-of-Period Convention

Beginning ofInterest period

End of interest period

0 1

10

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Methods of Calculating Interest Simple interest: the practice of charging an

interest rate only to an initial sum (principal amount).

Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

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Simple Interest

P = Principal amount i = Interest rate N = Number of

interest periods Example:

P = $1,000 i = 8% N = 3 years

End of Year

Beginning

Balance

Interest earned

Ending Balance

0 $1,000

1 $1,000 $80 $1,080

2 $1,080 $80 $1,160

3 $1,160 $80 $1,240

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Simple Interest Formula

( )

where

= Principal amount

= simple interest rate

= number of interest periods

= total amount accumulated at the end of period

F P iP N

P

i

N

F N

$1,000 (0.08)($1,000)(3)

$1,240

F

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Compound Interest

Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

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Compound Interest

P = Principal amount i = Interest rate N = Number of

interest periods Example:

P = $1,000 i = 8% N = 3 years

End of Year

Beginning Balance

Interest earned

Ending Balance

0 $1,000

1 $1,000 $80 $1,080

2 $1,080 $86.40 $1,166.40

3 $1,166.40 $93.31 $1,259.71

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Compounding Process

$1,000

$1,080

$1,080

$1,166.40

$1,166.40

$1,259.710

1

2

3

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0

$1,000

$1,259.71

1 2

3

3$1,000(1 0.08)

$1,259.71

F

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Compound Interest Formula

1

22 1

0 :

1: (1 )

2 : (1 ) (1 )

: (1 )N

n P

n F P i

n F F i P i

n N F P i

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Some Fundamental Laws

2

F m a

V i R

E m c

The Fundamental Law of Engineering Economy

(1 )NF P i

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Compound Interest

“The greatest mathematical discovery of all time,”

Albert Einstein

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Practice Problem: Warren Buffett’s Berkshire Hathaway Went public in 1965: $18

per share Worth today (August 22,

2003): $76,200 Annual compound growth:

24.58% Current market value:

$100.36 Billion If he lives till 100 (current

age: 73 years as of 2003), his company’s total market value will be ?

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Market Value

Assume that the company’s stock will continue to appreciate at an annual rate of 24.58% for the next 27 years.

27$100.36(1 0.2458)

$37.902 trillions

F

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EXCEL Template

In 1626 the Indians sold Manhattan Island to Peter Minuit Of the Dutch West Company for $24.

• If they saved just $1 from the proceeds in a bank account that paid 8% interest, how much would their descendents have now?

• As of Year 2003, the total US population would be close to 275 millions. If the total sum would be distributed equally among the population, how much would each person receive?

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Excel Solution

$1

8%

377 years

P

i

N

FV(8%,377,0,1)= $3,988,006,142,690

$3,988,006,142,690

275,000,000

$14,502

A

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Excel Worksheet

A B C

1 P 1

2 i 8%

3 N 377

4 FV

5 FV(8%,377,0,1)= $3,988,006,142,690

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Practice Problem

Problem Statement

If you deposit $100 now (n = 0) and $200 two years from now (n = 2) in a savings account that pays 10% interest, how much would you have at the end of year 10?

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Solution

0 1 2 3 4 5 6 7 8 9 10

$100$200

F

10

8

$100(1 0.10) $100(2.59) $259

$200(1 0.10) $200(2.14) $429

$259 $429 $688F

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Practice problem

Problem Statement

Consider the following sequence of deposits and withdrawals over a period of 4 years. If you earn 10% interest, what would be the balance at the end of 4 years?

$1,000 $1,500

$1,210

0 1

2 3

4?

$1,000

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$1,000$1,500

$1,210

0 1

2

3

4

?

$1,000

$1,100

$2,100 $2,310

-$1,210

$1,100

$1,210

+ $1,500

$2,710

$2,981

$1,000

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SolutionEnd of Period

Beginning

balance

Deposit made

Withdraw Ending

balance

n = 0 0 $1,000 0 $1,000

n = 1 $1,000(1 + 0.10) =$1,100

$1,000 0 $2,100

n = 2 $2,100(1 + 0.10) =$2,310

0 $1,210 $1,100

n = 3 $1,100(1 + 0.10) =$1,210

$1,500 0 $2,710

n = 4 $2,710(1 + 0.10) =$2,981

0 0 $2,981