Engineering Economic Analysis

Rateof Return Analysis 215

The choic~ between the two alternatives reduces to an examination of the differences between

them. We can compute the rate of return on the differences between the alternatives.Writing the

Continuous Alternatives

Year

o

1

Alt. 1

-$10

'+15

Alt. 2

-$20

+28

Alt. 2-Alt. 1

-$20 - (-$10) =-$10

+28 - (+15) =+13

PW of cost of differences (Alt. 2 - Alt. 1)=PW of benefit of differences (Alt. 2- Alt. 1)

10 =13(PjF, i, 1)

Thus,

10

(PjF,i, 1)=- 13 =0.7692

One can see that if $10 increases to $13 in one year, the interest rate must be 30%. The compound
interest tables confirm this conclusion. The 30% rate of return on the difference between the
alternatives is far higher than the 6% MARR. The additional $10 investment to obtain Alt. 2 is
superior to investing the $10 elsewhere at 6%. To obtain this desirable increment of investment,
with its 30% rate of return, Alt. 2 is selected.

To understand more about Example 7-6, compute the rate of return for each alternative.

Alternative 7

PW of cost of Alt. 1=PW of benefit of Alt. 1

$10=$15(P j F, i, 1)

Thus,

10

(PjF, i,1)=- =0.6667

15

From the compound interest tables: rate of return=50%.

Alternative 2

PW of cost of Alt. 2=PW of benefit of Alt.2

$20 = $28(P j F, i, 1) I~

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