Engineering Economic Analysis

GraphicalSolutions 249

Initial cost

Uniform annual benefit

End-of-useful-life-salvage value

Usefullife, in years

MachineX

$200

95

50

6

MachineY

$700

120

150

12

,- .-. .'.

.SOLUTION

Noting that for both Machine X and MachineY,the PW of cost reflectsthe conventionof treating

salvage value as a reduction in cost rather than a benefit, and using a 12-yearanalysis period, we

have

MachineX

PW of cost:;::: 200 + (200 --50)(PI F,10%,6) - 50(P/F,10%,12)

= 200 + 150(0.5645)-50(0.3186) = 269

PW of benefit:;:::95(PIA,1O%,12) = 95(6.814):;::: 647

Machine Y

fW of cost:;::: 700- 150(P IF,10%,12)=700 - 150(0.3186)== 652

PW of benefit = 120(PIA,10%,12)-120(6.814) :;:::818

Wben the two alternativesplotted (Figure 8-4), we see that the incrementY-X has a slope much
less than the 10% rate of return line. The rate of return on the increment of investmentis less than
10%; hence, the increment is undesirable. This means that Machine X should be selected rather
than MachineY.

EIGURE8-4 Benefit'-{;ost graph. $900

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