A milling machine’s first cost is $30,000. The likely benefits from the machine are $5,000 in the first year, $4,750 in the second year, and so on, decreasing annually by $250. If the machine has a useful life of 10 years and a salvage value of $1,000, what is its ROR?

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Accepted Answer

We are asked to determine the value of ROR. To do this, express the PW of the given cashflows and equate it to zero. That means evaluating [latex]PW_{benefits}[/latex] and [latex]PW_{costs}[/latex] from the machine’s cashflow diagram. The diagram is in Fig. 7.4(a) except for the first cost which is $30,000.

From the tailored cash inflows in Fig. 7.4(b), as explained in the previous example,

[latex]P_{benefits} =\left\{5,000({P}/{A},i,10)-250({P}/{G},i,10)\right\} +1,000({P}/{F},i,10)[/latex]

Thus,

[latex]PW=PW_{benefits} -PW_{costs}[/latex]

[latex]=\left\{5,000({P}/{A},i,10)-250({P}/{G},i,10)\right\} +1,000({P}/{F},i,10) -30,000[/latex]

By imposing the condition PW = 0, [latex]i[/latex] is evaluated from

[latex]5,000({P}/{A},i,10)-250({P}/{G},i,10) +1,000({P}/{F},i,10)-30,000 =0[/latex]

Dividing throughout by 5,000, we get

[latex]({P}/{A},i,10) -0.05 ({P}/{G},i,10) +0.2 ({P}/{F},i,10) -6=0[/latex]

This PW-function can be solved[latex]^{17}[/latex] for [latex]i[/latex] the way explained in Example 7.2, yielding ROR of approximately 6%.

[latex]^{17}[/latex]The trial and error results are tabulated below:

i(%) (P/A,i, 10) – 0.05(P/G,i, 10) + 0.2(P/F,i, 10) – 6

4 8.111 – 0.05 × 33.881 + 0.2 × 0.6756 – 6 = 0.552

5 7.722 – 0.05 × 31.652 + 0.2 × 0.6139 – 6 = 0.262

6 7.360 – 0.05 × 29.602 + 0.2 × 0.5584 – 6 = –0.008

Since the sign changes in between 5% and 6%, interpolate within this range to get ROR = i = 5 + 0.262/(0.008 + 0.262) = 5 + 0.97 = 5.97%

Question 7.4: A milling machine’s first cost is $30,000. The likely benefi...

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