Methods for Finding the
Rate of Return
Lecture No. 25
Chapter 7
Contemporary Engineering Economics
Copyright © 2006
Contemporary Engineering Economics, 4th
edition, © 2007
Simple versus Nonsimple Investments

Simple Investment: The project with only one
sign change in the net cash flow

Nonsimple investment: an investment in
which more than one sign change occurs in
the net cash flow series
Contemporary Engineering Economics, 4th
edition, © 2007
Example 7.1 Investment
Classification
Net Cash Flow
Period Project Project Project
n
A
B
C
0
1
2
3
4
-1,000
-500
800
1,500
2,000
-1,000
3,900
-5,030
2,145
1,000
-450
-450
-450
Project A: a simple investment
Project B: a nonsimple investment
Project C: a simple borrowing
Contemporary Engineering Economics, 4th
edition, © 2007
Predicting Multiple i*s

Net Cash Flow Rule of Signs
The number of real i*s that are greater than -100% for a
project with N periods is never greater than the number
of sign changes in the sequence of the cash flows. A
zero cash flow is ignored.

Accumulated Cash Flow Sign Test
If the sequence of accumulated cash flow series starts
negatively and changes sign only once, then a unique
positive i* exists
Contemporary Engineering Economics, 4th
edition, © 2007
Net Cash Flow Rule of Signs
- 100% < i *< infinity
• Net Cash Flow Rule of Signs
No. of real RORs (i*s)
<
No. of sign changes in the project
cash flows
Contemporary Engineering Economics, 4th
edition, © 2007
Example
n
0
1
2
3
4
5
6
Net Cash flow
Sign Change
-$100
-$20
$50
0
$60
-$30
$100
1
1
1
• No. of real i*s <= 3
• This implies that the project could have
(0, 1, 2, or 3) i*s but NOT more than 3.
Contemporary Engineering Economics, 4th
edition, © 2007
Accumulated Cash Flow Sign Test
Step 1: Find the accounting sum of net cash
flows at the end of each period over the life
of the project
Period
(n)
Cash Flow
(An )
Sum
Sn
0
A0
S0  A0
1
A1
S1  S0  A1
2
A2
S2  S1  A2


N
AN

SN  SN 1  AN
Step 2: If the series Sn starts negatively and
changes sign ONLY ONCE, there exists a
unique positive i*.
Contemporary Engineering Economics, 4th
edition, © 2007
Example
n
0
1
2
3
4
5
6
An
-$100
-$20
$50
0
$60
-$30
$100
Sn
-$100
-$120
-$70
-$70
-$10
-$40
$60
Sign change
1
• No of sign change = 1, indicating a unique i*.
• i* = 10.46%
Contemporary Engineering Economics, 4th
edition, © 2007
Practice Problem
$3,900
0
$2,145
2
1
3
$1,000
$5,030
• Is this a simple investment?
• How many RORs (i*s) can you expect from
examining the cash flows?
• Can you tell whether or not this investment has a
unique rate of return?
Contemporary Engineering Economics, 4th
edition, © 2007
Computational Methods




Using Excel’s Financial Command
Direct Solution Method
Trial-and-Error Method (works only for simple
investment)
Cash Flow Analyzer – Online Financial
Calculator
Excel command to find the rate of return:
=IRR(cell range, guess)
e.g., =IRR(C0:C7, 10%)
Contemporary Engineering Economics, 4th
edition, © 2007
Finding Rate of Return on Excel
Period
(N)
Cash
Flow
0
-$1,000
1
-500
2
800
3
1,500
4
2,000
A
1 Period
2
3
4
5
6
7
8
9 IRR =
10
11
B
Cash Flow
0
1
2
3
4
=IRR(B3:B7,10%)
Contemporary Engineering Economics, 4th
edition, © 2007
-1000
-500
800
1500
2000
44%
C
Other Computational Methods
n
Direct
Solution
Direct
Solution
Log
Quadratic
Project A
Project B
Trial &
Error
Method
Computer
Solution
Method
Project C
Project D
0
-$1,000
-$2,000
-$75,000
-$10,000
1
0
1,300
24,400
20,000
2
0
1,500
27,340
20,000
3
0
55,760
25,000
4
1,500
Contemporary Engineering Economics, 4th
edition, © 2007
Direct Solution Methods
• Project A
• Project B
$1,000  $1,500( P / F , i ,4)
$1,000  $1,500(1  i ) 4
0.6667  (1  i ) 4
ln 0.6667
 ln(1  i )
4
0.101365  ln(1  i )
e
0.101365
 1 i
i  e 0.101365  1
 10.67%
PW (i )  $2,000 
$1,300 $1,500

0
2
(1  i ) (1  i )
1
, then
1 i
PW (i )  2,000  1,300 x  1,500 x 2
Let x 
Solve for x:
x  0.8 or -1.667
Solving for i yields
1
1
 i  25%,  1667
.

 i  160%
1 i
1 i
Since  100%  i  , the project's i *  25%.
0.8 
Contemporary Engineering Economics, 4th
edition, © 2007
Trial and Error Method – Project C


Step 1: Guess an interest
rate, say, i = 15%
Step 2: Compute PW(i)
at the guessed i value.
PW (15%) = $3,553

Step 3: If PW(i) > 0, then
increase i. If PW(i) < 0,
then decrease i.

Step 4: If you bracket the
solution, you use a linear
interpolation to approximate
the solution
3,553
0
-749
15%
i
18%
PW(18%) = -$749
Note: This method works only
for finding i* for simple investments.
 3,553 
i  15%  3%

 3,553  749 
Contemporary Engineering Economics, 4th
edition, © 2007
 17.45%
Graphical Method
Step 1: Create the NPW
profile.
Step 2: Find the point at
which the curve crosses
the horizontal axis closely
approximates i*
Contemporary Engineering Economics, 4th
edition, © 2007
Using Cash Flow Analyzer – Project D
Output
196%
Input data
Contemporary Engineering Economics, 4th
edition, © 2007