Engineering Economic Analysis

Rationing Capital by Present Worth Methods 525

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Locating a value of pin [NPW - p(PWof cost)] by trial and error, we use the following table,

where all amounts are in thousands of dollars.

For a value ofpequal to 0.25, the best selection order is computed to be Projects 1,2,4,5,6,

and 9.(Note:the smallest value forpis 0.239.)

Alternate Formation of Example 17-3

The preceding answer does not agree with the solution obtained in Example 17-2.The difficultyis

that the interest rate used in the present worth calculationsis not equal to the computed cutoffrate

of return. In Example 17-2 the cutoff rate of return was between 14 and 15%, say 14.5%. We wi)l

recompute the present worth solution using MARR = 14.5%, again with amounts in thousands

of dollars.

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Cost

$100

200

100

100

100

50

$650

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Computed

Trialp=0.20 Trialp=0.25

'oject Cost NPW [NPW-p(PWof cost)) Cost [NPW-p(PW of cost)]

(^1) $ 100 $60.04 $40.04 $ 100 $35.04
2 200 67.40 27.40 200 17.40
3 50 11.91 1.91 50 -0.59
4 100 55.48 35.48 100 30.48
5 100 80.52 60.52 (^100) 55.52
6 100 67.10 47.10 100 42.10
7 300 13.46 -46.54 -61.54
8 300 64.38 4.38 300 10.62
(^9) - 50 20.13 10.13 50 7.63

• $1300 $1000

Computed NPW Cost of Projects

Project Cost at 14.5% with Positive NPW

(^1) $100 $22.01 $100
(^2200) 3.87 200
3 50 6.81 50
(^4100) 21.10 100
5 100 28.14 100
(^6100) 17.91 100
7 300 -27.05
8- 300 --,::31.69 =11I '" ::; =
9 50 -1.28

• $650
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