L2: Time Value of Money
ECON 320 Engineering Economics
Mahmut Ali GOKCE
Industrial Systems Engineering
Computer Sciences
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Chapter 2
Time 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)
0
1
2
3
25
Option B
(Installment
Plan)
$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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Account Value
Case 1:
Inflation
exceeds
earning power
N = 0 $100
N = 1 $106
Case 2:
Earning power
exceeds
inflation
N = 0 $100
N = 1 $106
Cost of Refrigerator
N = 0 $100
N = 1 $108
(earning rate =6%) (inflation rate = 8%)
N = 0 $100
N = 1 $104
(earning rate =6%) (inflation rate = 4%)
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Which Repayment Plan?
End of Year
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
Receipts
$20,000.00
Payments
Plan 1
$200.00
5,141.85
5,141.85
5,141.85
5,141.85
5,141.85
Plan 2
$200.00
0
0
0
0
30,772.48
The 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
0
Beginning of
Interest period
0
1
End of interest
period
1
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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
Beginnin
g
Balance
Interest
earned
0
Ending
Balance
$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
F  P  (iP ) N
where
P = Principal amount
i = simple interest rate
N = number of interest periods
F = total amount accumulated at the end of period N
F  $1, 000  (0.08)($1, 000)(3)
 $1, 240
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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
End
 i = Interest rate
of
 N = Number of
Year
interest periods
0
 Example:
 P = $1,000
 i = 8%
 N = 3 years
Beginning
Balance
Interest
earned
Ending
Balance
$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
$1,080
$1,166.40
0
$1,259.71
1
$1,000
2
3
$1,080
$1,166.40
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$1,259.71
0
1
2
3
F  $1, 000(1  0.08)3
$1,000
 $1, 259.71
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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 )
n  N : F  P(1  i )
2
N
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Some Fundamental Laws
F
V
E

m

i R

m
a
c
2
The Fundamental Law of Engineering Economy
F  P(1  i)
N
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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.
F  $100.36(1  0.2458)
 $37.902 trillions
27
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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
P  $1
i  8%
N  377 years
FV(8%,377,0,1)
= $3,988,006,142,690
$3,988,006,142,690
A
275,000,000
 $14,502
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Excel Worksheet
A
B
1
P
1
2
i
8%
3
N
377
4
FV
5
C
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
F
0
1
2
3
4
5
6
7
8
9
10
$100(1  0.10)10  $100(2.59)  $259
$100
$200
$200(1  0.10)8  $200(2.14)  $429
F  $259  $429  $688
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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?
$1,210
0
1
4
2
$1,000 $1,000
?
3
$1,500
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?
$1,210
0
1
3
2
$1,000
$1,000
4
$1,500
$1,100
$1,000
$2,100
$2,310
$1,210
-$1,210
+ $1,500
$1,100
$2,710
$2,981
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Solution
End 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
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