A new machine with expected useful life of 5 years is being planned for acquisition next year when it will cost $6,000. This cost includes maintenance for the first 2 years of the purchase. The maintenance for the subsequent 3 years of the useful life is projected to cost $300, $500, and $750

Question

A new machine with expected useful life of 5 years is being planned for acquisition next year when it will cost $6,000. This cost includes maintenance for the first 2 years of the purchase. The maintenance for the subsequent 3 years of the useful life is projected to cost $300, $500, and $750 respectively. How much should be budgeted now for the purchase and maintenance of this machine if the yearly interest rate is 6%, compounded annually?

Accepted Answer

Let us sketch a cashflow diagram to help comprehend the problem. Such a diagram is shown in Fig. 4.2. If you have difficulty sketching or understanding this diagram, then reread Chapter 2.

The diagram shows that the machine will be purchased next year for $6,000 (vector down at period 1 since it is a cost). There are no cashflows at periods 2 and 3 since the purchase cost includes the maintenance for the first 2 years. The maintenance costs of the subsequent 3 years are displayed as cashflows at periods 4, 5, and 6. Note that the $300 cost during year 3–4 is posted at period 4 following the period-end assumption. So are the other two maintenance costs.

Altogether there are four costs, which will be paid for from the budget this year, i.e., from P. Thus, P must be [latex]equivalent[/latex] to these costs. Note that vector P is [latex]shown^{4}[/latex] dotted to highlight that in itself it is [latex]not a cost[/latex], but a budgeted sum which should be enough to pay for the four future costs.

The equivalency between P and the four costs suggests that P can be determined simply by adding the P-value of each future cost. Which equation is appropriate for finding P for a given F? It is Equation (3.7a). Thus, we get

[latex]P=6000({P}/{F},6\%,1)+300({P}/{F},6\%,4) +500({P}/{F},6\%,5)+750({P}/{F},6\%,6)[/latex]

Substituting the values of the functional factors from the interest table for 6%, we have

[latex]P=6,000 imes 0.9434+300 imes 0.7921+500 imes 0.7473+750 imes 0.7050[/latex]

[latex]=5,660.40+237.63+373.65+528.75[/latex]

[latex]=\$6,800[/latex]

Thus, the machine should be budgeted this year for $6,800. At the end of the first year, the balance will be $6,800(1 + 0.06) = $7,208, of which $6,000 will be used for the purchase. The remaining $1,208 earning interest on the balance with annual compounding will fully pay for the three maintenance costs.

Question 4.1: A new machine with expected useful life of 5 years is being ...

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