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Time Value of Money
Lecture No. 4
Chapter 3
Contemporary Engineering Economics
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Contemporary Engineering Economics, 6e, GE
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Chapter Opening Story
Take a Lump Sum or Annual Installments
Dearborn couple claimed
Missouri’s largest jackpot:
$293.75 millionin 2012.
 They had two options.
Option 1: Take a lump
sum cash payment of
$192.37 M.
Option 2: Take an annuity
payment of $9.79 M a year
for 30 years.
 Which option would you
recommend?
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What Do We Need to Know?
o Be able to compare the value of money at
different points in time.
o A method for reducing a sequence of
benefits and costs to a single point in time
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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).
Contemporary Engineering Economics, 6e, GE
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The Market Interest Rate
o Interest is the cost of
money, a cost to the
borrower and a profit to
the lender.
o Time value of money is
measured in terms of
market interest rate,
which reflects both
earning and purchasing
power in the financial
market.
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Cash Flow Diagram
(A Graphical Representation of Cash Transactions over Time)
Borrow $20,000 at 9%
interest over 5 years,
requiring $200 loan
origination fee upfront.
The required annual
repayment is $5,141.85
over 5 years.
o n = 0: $20,000
o n = 0: $200
o n = 1 ~ 5: $5,141.85
Contemporary Engineering Economics, 6e, GE
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End-of-Period Convention
Convention: Any cash
flows occurring during the
interest period are
summed to a single
amount and placed at the
end of the interest period.
Logic: This convention
allows financial
institutions to make
interest calculations
easier.
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Methods of Calculating Interest


Simple interest: Charging an interest rate only
to an initial sum (principal amount)
Compound interest: Charging an interest rate to
an initial sum and to any previously
accumulated interest that has not been
withdrawn

Note: Unless otherwise mentioned, all interest rates used in engineering
economic analyses are compound interest rates.
Contemporary Engineering Economics, 6e, GE
Park
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Simple Interest
Formula
F  P  (iP)N
• P = $1,000, i = 10%, N = 3
years
End of
Year
where
P = Principal amount
Beginning
Balance
Interest
Earned
0
i = simple interest rate
Ending
Balance
$1,000
N = number of interest periods
F = total amount accumulated at the end of period N
•
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1
$1,000
$100
$1,100
2
$1,100
$100
$1,200
3
$1,200
$100
$1,300
F = $1,000 + (0.10)($1,000)3
= $1,300
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Compound Interest
Formula
n  0:P
n  1: F1  P(1  i)
n  2: F2  F1 (1  i)  P(1  i)2
n  N : F  P(1  i)N
• P = $1,000, i = 10%, N = 3 years
End of
Year
Beginning
Balance
Interest
Earned
0
Ending
Balance
$1,000
1
$1,000
$100
$1,100
2
$1,100
$110
$1,210
3
$1,210
$121
$1,331
• F = $1,000(1 + 0.10)3 = $1,331
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Compounding Process
$1,100
$1,210
0
$1,331
1
$1,000
2
3
$1,100
$1,210
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The Fundamental Law of Engineering
Economy
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Warren Buffett’s Berkshire
Hathaway
 Went public in 1965: $18 per
share
 Worth today (May 29, 2015):
$214,800 per share
 Annual compound growth:
20.65%
 Current market value:
$179.5 billion
 If his company continues to
grow at the current pace,
what will be his company’s
total market value when he
reaches 100? (He is 85 years
old as of 2015.)
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Assume that the company’s stock
will continue to appreciate at an annual rate
of 20.65% for the next 15 years. The stock
price per share at his 100th birthday would
be
F = 214,800(1 + 0.2065)15 = $3,588,758
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Example 3.2: Comparing Simple with
Compound Interest
In 1626, American Indians sold Manhattan Island to Peter
Minuit of the Dutch West Company for $24.
 Given: If they saved just $1 from the proceeds in a bank
account that paid 8% interest, how much would their
descendents have in 2010?
 Find: As of 2015, the total U.S. population would be close to
308 million. If the total sum would be distributed equally
among the population, how much would each person
receive?
Contemporary Engineering Economics, 6e, GE
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Solution
P  $1
i  8%
N  384 years
F  $1(1  0.08)384  $164,033,801,073,200
$164,033,801,073,200
Amount per person 
308,000,000
 $532,577
Contemporary Engineering Economics, 6e, GE
Park
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