Engineering Economic Analysis

AnnualCash Flow Analysis 183

A finn is considering which of two devices to install to reduce costs in a particular situation.

Both devices cost $1000 and have useful lives of 5 years with no salvage value. DeviceAcan 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 year savings

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

DeviceA

EUAB=$300

DeviceB

EUAB= 400 - 50(AjG, 7%, 5)

== 400 - 50(1.865)

=$306.75 ,

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To maximize EUAB, select DeviceB.

Example 6-5 was presented earlier, as Example 5-1, where we found:

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

=300(4.100) =$1230

This is converted to EUAB by multiplying by the capital recovery factor:

EUABA = 1230(Aj P,7%, 5) = 1230(0.2439) = $300

PW of benefitsB=400(P j A,7%, 5) - 50(P jG, 7%, 5)

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

and, hence,

EUABB=1257.65(Aj P,7%, 5) = 1257.65(0.2439)

=$306.75

We see, therefore, that it is easy to convert the present worth analysis results into the annual

cash flow analysis results. We could go from annual cash flow to present worth just as

easily, by using the series present worth factor. And, of course, both methods indicate that

DeviceBis the preferred alternative.

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