Engineering Economic Analysis

158 PRESENTWORTH ANALYSIS

Multiple Alternatives

So far the discussion has been based on examples with only two alternatives.But multipll

alternative problems may be solved by exactly the same methods employed for problen

with two alternatives. (The only reason for avoiding multiple alternatives was to simpli1

the examples.) Examples 5-8 and 5-9 have multiple alternatives.

A contractor has been awarded the contract to construct a 6-mile-Iong tunnel in the mountains.

During the 5-year construction period, the contractor will need water from a nearby stream. He

will construct a pipeline to convey the water to the main construction yard. An analysis of costs

for various pipe sizes is as follows:

PipeSizes (in.)

2 3 4 6

Installed cost of pipeline and pump

Cost per hour for pumping

$22,000

$1.20

$23,000

$0.65

$25,000

$0.50

$30,000

$0.40

The pipe and pump will have a salvage value at the end of 5 years, equal to the cost to remove

them. The pump will operate 2000 hours per year. The lowest interest rate at which the contractor

is willing to invest money is 7%. (The minimum required interest rate for investedmoney is called

theminimum attractive rate of return,or MARR.) Select the alternative with the least present

worth of cost.

SdllJTI ON-,. -.'.. .-.'.' ....

We can comptite the present worth of cost for each alternative. For each pipe size, the present

worth of cost is equal to the installed cost of the pipeline and pump plus the present worth of

5 years of pumping costs.

PipeSize (in.)

2 6

II~I

Installed cost of pipeline and pump

1.20 x 2000 hr x(P/A,7%, 5)

0.65 x 2000 hr x4.100

0.50 x 2000 hr x 4.100

0.40 x 2000 hr x 4.100

Present worth of cost

$22,000

9,840

$30,000

3,280;

$31,840 $33,280

Select the 3 in. pipe size.

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