A small hydraulic machine is being analyzed for acquisition. The engineering economist is sure only of its cost, which is $5,000. For the values of the other three data, he conducts some research. Based on input from an insurance company about the machine’s useful life and salvage value, and from

Question

A small hydraulic machine is being analyzed for acquisition. The engineering economist is sure only of its cost, which is $5,000. For the values of the other three data, he conducts some research. Based on input from an insurance company about the machine’s useful life and salvage value, and from within the company about net annual benefit, he has determined the following most likely values, along with their optimistic and pessimistic values.

	Optimistic	Most Likely	Pessimistic
Net Annual Benefit	$3,000	$2,200	$1,000
Useful Life (yrs)	7	6	5
Salvage Value	$500	$300	$0

Should the machine be funded for investment if the payback period for approval is not to exceed 2 years, and MARR is 10%? Assume that the data follow β-distribution.

Accepted Answer

Under β-distribution, as just discussed, the most likely value is given fourfold weighting than the optimistic or pessimistic value. In other words, the most likely value is counted four times, giving a total of six data values. The mean value of the data is therefore obtained by adding these six data values and dividing the total by six, as follows:

Mean [latex]= \frac{Optimistic Value + 4 (Most likely Value) + Pessimistic Value }{6}[/latex]

For the net annual benefit, the mean value is:

[latex]Benefit_{ mean} = \frac{\$3,000+4(\$2,200)+\$1,000}{6}[/latex]

[latex]=\$2,133[/latex]

The mean value of useful life is:

[latex]Life_{ mean} = \frac{5+4 imes 6+7}{7}[/latex]

[latex]=6[/latex] years

The mean salvage value is:

[latex]Salvage_{mean}=\frac{\$500+4 imes \$300+0}{6}[/latex]

[latex]=\$283[/latex]

The problem is now solved the usual way, as in other chapters, using the mean values of the parameters. So, A = $2,133, n = 6 years, and S = $283. Assuming the salvage value to be an element of the $5,000 investment which is reduced by recovery through salvaging,

Investment [latex]=\$5,000-283({P}/{F},10\%,6)[/latex]

[latex]=\$5,000-283({P}/{F},10\%,6)[/latex]

[latex]=\$5,000-283 imes 0.5645[/latex]

[latex]=\$5,000-\$160[/latex]

[latex]=\$4,840[/latex]

Payback period [latex]={Investment}/{Annual benefit:}[/latex]

[latex]={\$4,840}/{\$2,133}[/latex]

[latex]=2.27[/latex] years

Since the payback period is greater than the desirable 2 years, the machine should not be funded for investment.

Question 15.5: A small hydraulic machine is being analyzed for acquisition....

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