Engineering Economic Analysis

Analysis Period

Two pumps are being considered for purchase. If interest is 7%, which pump should be bought?

Cost Depletion

End-of-useful-life salvage value

Useful life, in years

PumpA

$7000

1500

12

PumpB

$5000

1000

6

The annual cost for 12 years of PumpAcan be Jound by using Equation 6-4:

EUAC=(P - S)(Aj P, i, n)+Si

=(7000- 1500)(Aj P,7%,12) + 1500(0.07)

=5500(0.1259) + 105=$797

Now compute the annual cost for 6 years of PumpB:

EUAC=(5000 - 1000)(Aj P,7%, 6) + 1000(0.07)

=4000(0.2098) + 70= $909

For a common analysis period of 12 years, we need to replace PumpBat the end of its 6-year

useful life. If we assume that another pumpB' can be obtained, having the same $5000 initial

cost, $1000 salvage value, and 6-year life, the cash flow wilt be as follows:

1000 1000

t .t

0-1-2-3-4"---5-6~7-8--9 10 "11 "12

1 1

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5000 5000

1 Pump6 yearsB I, ReplacementPump6 years B'

For the 12-year analysis period, the annual cost for PumpBis

EUAC=[5000- 1000(PIF, 7%, 6) + 5000(P j F,7%, 6)

• 1000(Pj F,7%, 12)] x(AlP, 7%, 12)

[5000 -1000(0.6663) + 5000(0.6663) - 1000(0.4449,11x(0.125Q)

= (5000 - 666 + 3331 - 444)(0.1259)

=(7211)(Q.,!.459)= $9£)9 =

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