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

2. 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

3. Contemporary Engineering Economics, 4th edition © 2007
The Interest Rate
Contemporary Engineering Economics, 4th edition © 2007

4. Contemporary Engineering Economics, 4th edition © 2007
Cash Flow Transactions for Two Types of Loan Repayment
End of Year
Receipts
Payments
Plan 1
Plan 2
Year 0
$20,000.00
$200.00
Year 1
5,141.85
1,800
Year 2
Year 3
Year 4
Year 5
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

5. Cash Flow Diagram for Plan 1
Contemporary Engineering Economics, 4th edition © 2007

6. 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-of-period 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

7. End-of-Period Convention
Contemporary Engineering Economics, 4th edition © 2007

8. 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practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.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

9. 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
$1,000 1
$100
$1,100 2
$1,200 3
$1,300
Contemporary Engineering Economics, 4th edition © 2007

10. Simple Interest Formula
Contemporary Engineering Economics, 4th edition © 2007

11. 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
$1,000 1
$100
$1,100 2
$110
$1,210 3
$121
$1,331
Contemporary Engineering Economics, 4th edition © 2007

12. Contemporary Engineering Economics, 4th edition © 2007
Compounding Process
$1,100
$1,210
$1,331 1
$1,000 2 3
$1,100
$1,210
Contemporary Engineering Economics, 4th edition © 2007

13. Contemporary Engineering Economics, 4th edition © 2007
Cash Flow Diagram
$1,331 1 2 3
$1,000
Contemporary Engineering Economics, 4th edition © 2007

14. Relationship Between Simple Interest and Compound Interest
Contemporary Engineering Economics, 4th edition © 2007

15. 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

16. Contemporary Engineering Economics, 4th edition © 2007
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 ( )20 = $6,433.29B
Contemporary Engineering Economics, 4th edition © 2007

17. 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

18. Contemporary Engineering Economics, 4th edition © 2007
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

19. 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

20. 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 1 4 2 3
$1,500
$1,000
$1,000
Contemporary Engineering Economics, 4th edition © 2007

21. Contemporary Engineering Economics, 4th edition © 2007
Solution
End of Period
Beginning
balance
Deposit made
Withdraw
Ending
n = 0
$1,000
n = 1
$1,000( ) =$1,100
$2,100
n = 2
$2,100( ) =$2,310
$1,210
$1,100
n = 3
$1,100( ) =$1,210
$1,500
$2,710
n = 4
$2,710( ) =$2,981
$2,981
Contemporary Engineering Economics, 4th edition © 2007