Engineering Economic Analysis

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478 SELECTION OF A MINIMUM ATTRACTIVERATEOF RETURN

10 I 12

o 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400

CumulativeInvestment($ X 103)

FIGURE 15-2 Cumulative investment required for all projects at or above a given rate of return.

and many must be rejected. Obviously, we want to ensure thatall the selected projects are

better than the best rejected project.To do this, we must know something about the rate of

return on the best rejected project. The best rejected project is the best opportunity forgone,

and this in turn is called theopportunity cost.

Opportunity cost=Cost of the best opportunity forgone

=Rate of return on the best rejected project

If one could predict the opportunity cost for some future period (like the next 12 months),

this rate of return could be one way tojudge whether to accept or reject any proposed capital

expenditure.

Consider the situation represented by Figures 15-1 and 15-2. For a capital expenditure budget of

$1.2 million ($1.2 x 106),what is the opportunity cost?

.SOLUTION\.1-' '. ' =

FroqlFigtiry 15-2 we see tbat the eight projects with a rate of return of 20% Orqlore require a
cUqlp.lativewvestqlent6f$1.2 (x 106). We would ta,keOPthese projects andreject the other fourC""~'~":;:;;:,'..": ,,,.,,,,:,,:",,,.:,4>:,,:,,, , "'." ' '/~:;.-;"-.~ "'' =." .~:.' -i:i..::. ,," ,,;',cM' ,__
(7, 'trI, 10, and 12)witli rates Qfr;etuI'J:lof 18%~o{less.]he bestreject~d projectis 7, and irhas-ah
I 18% rate ofTettitn.1'huStheopPorWnity co,stis 18%,

l ...

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45

40

35

30

Be^25

<U

'S<U 20

15

1: t 5 [ 6

Project Number

8 I 9 I 7 I 11