Time Value of Money
Lecture No.4
Chapter 3
Contemporary Engineering Economics
Copyright © 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
0
1
2
3
Option A
(Lump Sum)
$2.912M
Option B
(Installment Plan)
$280,000
$280,000
$280,000
$280,000
19
$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
Repayment
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)
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
End of
i = Interest rate
Year
N = Number of
0
interest periods
1
Example:



P = $1,000
i = 10%
N = 3 years
Beginning
Balance
Interest
earned
Ending
Balance
$1,000
$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
F  P  ( iP ) N
w h ere
P = P rin cip al am o u n t
i = sim p le in terest rate
N = n u m b er o f in terest p erio d s
F = to tal am o u n t accu m u lated at th e en d o f p erio d N
F  $1, 000  (0.10)($1, 000)(3)
 $1, 300
Contemporary Engineering Economics, 4th
edition © 2007
Compound Interest




P = Principal amount
End
i = Interest rate
of
N = Number of
Year
interest periods
0
Example:



P = $1,000
i = 10%
N = 3 years
Beginning
Balance
Interest
earned
Ending
Balance
$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
$1,100
$1,210
0
$1,331
1
$1,000
2
3
$1,100
$1,210
Contemporary Engineering Economics, 4th
edition © 2007
Cash Flow Diagram
$1,331
0
1
2
3
F  $1, 000(1  0.10)
$1,000
 $1, 331
Contemporary Engineering Economics, 4th
edition © 2007
3
Relationship Between Simple Interest and
Compound Interest
Contemporary Engineering Economics, 4th
edition © 2007
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 )
N
Contemporary Engineering Economics, 4th
edition © 2007
2
Some Fundamental Laws
F

m a
V

iR
E

m c
2
The Fundamental Law of Engineering Economy
F  P (1  i )
Contemporary Engineering Economics, 4th
edition © 2007
N
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.
F  $115.802 M (1  0.2315)
 $17.145 trillions
Contemporary Engineering Economics, 4th
edition © 2007
24
Example
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
P  $1
i  8%
N  3 8 0 years
F  $ 1(1  0 .0 8)
380
 $ 5, 0 2 3, 7 3 9,1 9 4, 0 2 0
=FV(8%,380,0,1)
= $5,023,739,194,020
A m o u n t p er p erso n 
$ 5, 0 2 3, 7 3 9,1 9 4, 0 2 0
3 0 0, 0 0 0, 0 0 0
 $ 1 6, 7 4 6
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
F
0
1
2
3
4
5
6
$ 1 0 0 (1  0 .1 0 )
7
10
8
9
10
 $ 1 0 0 ( 2 .5 9 )  $ 2 5 9
$ 2 0 0 (1  0 .1 0 )  $ 2 0 0 ( 2 .1 4 )  $ 4 2 9
8
$100
$200
F  $259  $429  $688
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?
$1,210
0
1
4
2
$1,000 $1,000
?
3
$1,500
Contemporary Engineering Economics, 4th
edition © 2007
?
$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
Contemporary Engineering Economics, 4th
edition © 2007
$2,981
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
Contemporary Engineering Economics, 4th
edition © 2007