Engineering Economic Analysis

248 INCREMENTALANALYSIS

$30

$27

$24

CI)

tE $21G.>

c::

~$18

'5

'€ $15o

~c::$12

G.>

p..,~ $9

Alternative 1

$6

$3

$3 $6 $9 $12. $15 $18 $21 .$24 $27 $30

PresentWorthof Cost

FIGURE 8-2 Benefit-cost gra.ph for Example 8-1

with a one-year analysis period..

Since

$30

$27.

$24

Rate of Return

Alternative 2'8~o ~o~~~o~o/ ~\:J

11)/\~

Difference. /./ ;v~4-

Betweenthe ~J' 0 // / .h.b~o'.;;

Alternatives' I,i / /\:J~o

'"

~c::$21

<U

Eo$18

'E $15o

~$12

5

p..,~ $9

Alternative 1

$6

$3

$3 $6 $9 $12 $15 $18 $21 $24 $27 $30

PresentWorthof Cost

FIGURE 8-3 Benefit-cost graph for Alternatives 1

and 2 with a one-year analysis period.

[Higher-cost Alt. 2]=[Lower-costAlt. 1] + [Differencebetween them]

the difference between the alternatives can be represented by a line shown in Figure 8-3.

Viewed in this manner, we clearly see that Alt. 2 may be considered to be two' separate

increments of investment. The first increment is Alt. 1 and the second one is the difference

between the alternatives.Thus we will select Alt. 2 if the difference between the alternativesis a

desirable increment of investment.

Since the slope of the line represents a rate of return, we see that the increment is desirable if

the slope of the increment is greater than the slope of the 6% line that corresponds to NPW=O.

We can see that the slope is greater; hence, the increment of investment is, attractive. In fact, a

careful examination shows that the "Difference between the alternatives"line has the same slope

as. the 30% rate of .return line. We can say, therefore, that the incremental rate of return from

selecting Alt. 2 rather than Alt. 1 is 30%. This is the same as was computed in Example 7-6. We

conclude that Alt. 2 is the preferred alternative.

Solve Example 7-10 by means of a benefit-cost graph. Two machines are being considered for

purchase. If the minimum attractive rate of return (MARR) is 10%, which machine should be

.pought?

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