Download 2) Time Value of Money

Time Value of Money
Lecture No.4
Chapter 3
Contemporary Engineering Economics
Copyright © 2006
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
1,800
1,800
1,800
1,800
21,800.00
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 1
Contemporary Engineering Economics, 4th
edition © 2007
End-of-Period Convention



In practice, cash flows can occur at the
beginning or in the middle of an interest period,
or indeed, at practically any point in time.
One of the simplifying assumptions we make in
engineering economic analysis is the end-ofperiod convention.
End-of-period convention:
Unless otherwise mentioned, all cash flow
transactions occur at the end of an interest
period.
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that has not been withdrawn.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
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  (0.10)($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)3
$1,000
 $1,331
Contemporary Engineering Economics, 4th
edition © 2007
Relationship Between Simple Interest and
Compound Interest
Contemporary Engineering Economics, 4th
edition © 2007
Warren Buffett’s
Berkshire Hathaway
Went
public in 1965:
$18 per share
Worth today (April 05,
2010): $121,700 per
share
Annual compound
growth: 21.65%
Current market value:
$127.7 Billion
If his company continues to grow at the current pace,
what will be his company’s total market value when he
reaches 100? (80 years as of 2010)

Contemporary Engineering Economics, 5th edition ©
2010
Market Value

Assume that the company’s stock will continue to
appreciate at an annual rate of 21.65% for the
next 20 years.
F = $127.7B (1 + 0.2165)20 = $6,433.29B
Contemporary Engineering Economics, 4th
edition © 2007
Example: Comparing Simple with
Compound Interest
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 2010, the total US population would be close to 308
millions. If the total sum would be distributed equally
among the population, how much would each person
receive?

Contemporary Engineering Economics, 5th edition ©
2010
Excel Solution
P = $1
i = 8%
N = 384 years
F = $1 (1+0.08)384 = $ 6,834,741,711,384.36
~ $ 6.8 Trillion
Excel Formula:
F = FV(8%,384,0,1)
= $6,834,741,711,384.36
Amount per person = F/308 Million = $22,190.72
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
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
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