contemporary engineering economics, 4 th edition ©2007 time value of money lecture no.4 chapter 3...

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Contemporary Engineering Economics, 4 th edition ©2007 Time Value of Money Lecture No.4 Chapter 3 Contemporary Engineering Economics Copyright © 2006

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Contemporary Engineering

Economics, 4th edition ©2007

Time Value of Money

Lecture No.4Chapter 3Contemporary Engineering EconomicsCopyright © 2006

Contemporary Engineering

Economics, 4th edition © 2007

Chapter Opening Story —Take a Lump Sum or Annual Installments

Mrs. Louise Outing won a lottery worth $5.6 million.

Before playing the lottery, she was offered to choose between a single lump sum $2.912 million, or $5.6 million paid out over 20 years (or $280,000 per year).

She ended up taking the annual installment option, as she forgot to mark the “Cash Value box”, by default.

What basis do we compare these two options?

Contemporary Engineering

Economics, 4th edition © 2007

Year Option A

(Lump Sum)

Option B

(Installment Plan)

0

1

2

3

19

$2.912M $283,770

$280,000

$280,000

$280,000

$280,000

Contemporary Engineering

Economics, 4th edition © 2007

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.

Contemporary Engineering

Economics, 4th edition © 2007

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 This a two-edged sword whereby earning

grows, but purchasing power decreases (due to inflation), as time goes by.

Contemporary Engineering

Economics, 4th edition © 2007

The Interest Rate

Contemporary Engineering

Economics, 4th edition © 2007

Cash Flow Transactions for Two Types of Loan RepaymentEnd 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)

Contemporary Engineering

Economics, 4th edition © 2007

Cash Flow Diagram for Plan 2

Contemporary Engineering

Economics, 4th edition © 2007

End-of-Period Convention

Contemporary Engineering

Economics, 4th edition © 2007

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.

Contemporary Engineering

Economics, 4th edition © 2007

Simple Interest

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

interest periods Example:

P = $1,000 i = 10% N = 3 years

End of Year

Beginning

Balance

Interest earned

Ending Balance

0 $1,000

1 $1,000 $100 $1,100

2 $1,100 $100 $1,200

3 $1,200 $100 $1,300

Contemporary Engineering

Economics, 4th edition © 2007

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 (0.10)($1,000)(3)

$1,300

F

Contemporary Engineering

Economics, 4th edition © 2007

Compound Interest

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

interest periods Example:

P = $1,000 i = 10% N = 3 years

End of Year

Beginning Balance

Interest earned

Ending Balance

0 $1,000

1 $1,000 $100 $1,100

2 $1,100 $110 $1,210

3 $1,210 $121 $1,331

Contemporary Engineering

Economics, 4th edition © 2007

Compounding Process

$1,000

$1,100

$1,100

$1,210

$1,210

$1,3310

1

2

3

Contemporary Engineering

Economics, 4th edition © 2007

0

$1,000

$1,331

1 2

3

3$1,000(1 0.10)

$1,331

F

Cash Flow Diagram

Contemporary Engineering

Economics, 4th edition © 2007

Relationship Between Simple Interest and Compound Interest

Contemporary Engineering

Economics, 4th edition © 2007

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

Contemporary Engineering

Economics, 4th edition © 2007

Some Fundamental Laws

2

F m a

V i R

E m c

The Fundamental Law of Engineering Economy

(1 )NF P i

Contemporary Engineering

Economics, 4th edition ©2007

Compound Interest

“The greatest mathematical discovery of all time,”

Albert Einstein

Contemporary Engineering

Economics, 4th edition © 2007

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

per share Worth today (June 22,

2006): $91,980 Annual compound growth:

23.15% Current market value:

$115.802 Billion If his company continues to

grow at the current pace, what will be his company’s total market value when reaches 100? ( lives till 100 (76 years as of 2006)

Contemporary Engineering

Economics, 4th edition © 2007

Market Value

Assume that the company’s stock will continue to appreciate at an annual rate of 23.15% for the next 24 years.

24$115.802 (1 0.2315)

$17.145 trillions

F M

Contemporary Engineering

Economics, 4th edition © 2007

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 2006, the total US population would be close to 300 millions. If the total sum would be distributed equally among the population, how much would each person receive?

Contemporary Engineering

Economics, 4th edition © 2007

Excel Solution

380

$1

8%

380 years

$1(1 0.08) $5,023,739,194,020

P

i

N

F

=FV(8%,380,0,1)= $5,023,739,194,020

$5,023,739,194,020Amount per person

300,000,000

$16,746

Contemporary Engineering

Economics, 4th edition © 2007

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?

Contemporary Engineering

Economics, 4th edition © 2007

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

Contemporary Engineering

Economics, 4th edition © 2007

Practice problem

Problem Statement

Consider the following sequence of deposits and withdrawals over a period of 4 years. If you earn a 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

Contemporary Engineering

Economics, 4th edition © 2007

$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

Contemporary Engineering

Economics, 4th edition © 2007

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