Engineering Economic Analysis

276 OTHER ANALYSISTECHNIQUES

SOLUTION

Assuming a 12-year analysis period, the cash flow table is:

Year

o

1-5

Machine X

-$200

+95

{

+95

-200

+50

+95

{

+95

+50

6

7-11

12

We will solve the problem using

Machine Y

-$700

+120

+120

+120

+120

+150

B EUAB

C EUAC

and considering the salvage value of the machines to be reductions in cost, rather than increases

in benefits. This choice affects the ratio value, but not the decision.

Machine X

EUAC=200(Aj P,10%, 6)- 50(Aj F, 10%, 6)

=200(0.2296) - 50(0.1296) =46 - 6=$40

EUAB=$95

Note that this assumes the replacement for the last 6 years has identical costs. Under these

circumstances, the EUAC for the fi~st6 years equals the EUAC for all 12 years.

Machine Y

EUAC=700(Aj P,10%, 12) - 150(Aj F,10%, 12)

- 700(0.1468) - 150(0.0468)=103 -7 =$96

EUAB=$120

Machine Y- Machine X

dB =120- 95 _ 25=0.45

= ;; LSC= 96 - 40 56 =

Since the incremental benefit-costratio isless than 1, it represents an undesirable increment of

investmenf! We-=thereforechoosethe lower-cost alternative-Machine Xi' If we had computed

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