Engineering Economic Analysis

Analysis Period 187

PumpBin Example 6-7 is now believed to have a 9-year useful life. Assuming the same initial

cost and salvage value, compare it with PumpAusing the same 7% interest rate.

If we assume that the need forAorBwill exist for some continuing period?the comparison of

annual costs for the unequal lives is an acceptable technique. For 12 years of PumpA:

EUAC=(7000 - 1500)(AI P,7%, 12) + 1500(0.07) $797

For 9 years of PumpB:

EUAC- (5000-1000)(AI p, 7%, 9) + 1000(0.07). $684

For minimum EUAC, select PumpB.

Infinite Analysis Period

At times we have an alternativewith a limited (finite)useful life in an infinite analysisperiod

situation. The equivalent uniform annual cost may be computed for the limited life. The

assumption of identical replacement (replacementshave identical costs, performance, etc.)

is often appropriate. Based on this assumption,the same EUACoccurs for each replacement

of the limited-life alternative. The EUAC for the infinite analysis period is therefore equal

to the EUAC computed for the limited life. With identical replacement,

EUACinfinite analysisperiod=EUACfor limited lifen

A somewhat different situation occurs when there is an alternative with an infinite life in a

problem with an infinite analysis period:

EUACnfinite analysisperiod=P(AI P, i,(0) + Any other annual costs

Whenn=00, we haveA=Pi and, hence, (AI P, i,(0) equalsi.

EUACinfinite analysisperiod=Pi+ Any other annual costs

In the construction of an aqueduct to expand the water supply of a city, ther~are t~o aJ-ternatives
for a particular portion of the aqueduct. Either a tunnel can be constructed tlu:ougha mountain,
or a pipeline can be laid to go around the mountain. If there is a permanent need for the aqueduct, ,
J

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