Engineering Economic Analysis

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148 PRESENTWORTH ANALYSIS

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A firm is considering which of two mechanical devices to install to reduce costs in a particular

situation. Both devices cost $1000 and have useful lives of 5 years and no salvage value. Device

Acan be expected to result in $300 savings annually.DeviceBwill provide cost savings of $400

the first year but will decline $50 annually, making the second-year savings $350, the third-y~ar

savings $300, and so forth. With interest at 7%, which device should the firm purchase?

SOLUTION

The analysis period can convenientlybe selected as the useful life of the devices, or 5 years. Since

both devices cost $1000, there is a fixed input (cost) of $1000 regardless of whetherAorBis

chosen. The appropriate decision criterion is to choose the alternativethat maximizes the present

worth of benefits.

PW of Benefits PW of Benefits

PW of benefitsA=300(PjA,7%, 5) = 300(4.100)= $1230

PW of benefitsB=400(PjA, 7%, 5) - 50(PjG, 7%, 5)

=400(4.100) - 50(7.647) =$1257.65

DeviceBhas the larger present worth of benefits and is, therefore, the preferred alternative.It is

worth noting that, if we ignore the time value of money,both alternativesprovide $1500 worth of

benefits over the 5-year period. DeviceBprovides greater benefits in the first 2 years and smaller

benefits in the last 2 years. This more rapid flow of benefits fromB,although the total magnitude

equals that ofA,results in a greater present worth of benefits.

Wayne County will build an aqueduct to bring water in from the upper part of the state. I~can

be built at a reduced size now for $300 million and be enlarged 25 years hence for an additional

$350 million. An alternativeis to construct the full-sized aqueduct now for $400 million.

Both alternativeswould provide the needed capacity for the 50-year analysis period. Mainte-

nanCecosts are small and~may.be ignored. At 6% interest, which alternatiye should be selected?

DeviceA DeviceB

A= 300^400350

i

300

r t t t t r t

(^250200)
t t
0-1-2-3-4-5 01-2-3-4-5
!
n= 5 years
!
n= 5 years