Engineering Economic Analysis

152 PRESENTWORTH ANALYSIS

Assuming the replacement Speedy equipment 5 years hence will also cost $1500,

r

Original Speedy Investment---1

200 200

t t

0-1-2-3-4-5-6-7-8-9-10

1 1.

J

1500 1500

I---ReplacementSpeedy Investment

PW of cost=1500+ (1500- 200)(P/F,7%, 5) - 200(P/F,7%, 10)

= 1500 + 1300(0.7130)- 200(0.5083)

= 1500 + 927 - 102=$2325

For the Allied equipment, on the other hand, we have the following results:.

325

t

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

...

1600

PW of cost= 1600 - 325(P/F,7%, 10)=1600 - 325(0.5083)=$1435

For the fixed output of 10 years of service in the mailroom, the Allied equipment, with it:

smaller present worth of cost, is preferred..

We have seen that setting the analysis period equal to the least common multiple 0

the lives of the two alternativesseems reasonable in the revised Example 5-3. What woul<

one do, however, if the alternatives had useful lives of 7 and 13 years, respectively? Here

the least common multiple of lives is 91 years. An analysis period of 91 years hardly seem:

realistic. Instead, a suitable analysis period should be based on how long the equipment i:

likely to be needed. This may require that terminal values be estimated for the alternative:

at some point prior to the end of their useful lives.

Figure 5-1 graphically represents this concept. As Figure 5-1 indicates, it is not neces.

sary for the analysis period to equal the useful life of an alternative or some multiple of tht

useful life. To properly reflect the situation at the end of the analysis period, an estimate i1

required of the market value of the equipment at that time. The calculations might be easie]

if everything came out even, but it is not essential.

T

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