A certain narrow-aisle forklift truck costs $4,000. When it is replaced, it will be replaced with a truck of the same kind. Operating costs for this truck are $500 for the first year and increase at the rate of $40 per year squared thereafter. Technological improvements reduce operating costs by an estimated $30 per year. The salvage value of the trucks declines linearly over their seven-year life. The desired rate of return is to be 20 percent before taxes.
When should the equipment be replaced?
SOLUTION
Given:
Initial cost of equipment = $4,000
Operating costs 500 + 40(t – 1)2 – 30(t – 1)
Salvage value Sn = I(1 – t/7)
Rate of return on investment = 20%
Replacement is expected to be with equipment of like kind
The best replacement year can be found by comparing the equivalent annual cost of a sequence of similar equipment replaced every n years. The equivalent annual cost is:
ACn = [I + ∑nj=1 (Cj / 1 + i)j) – (Sn/ (1 + i)n) ] [I (1 + i)n / (1 + i)n – 1]
Solving this equation for different years is facilitated if the equation is set up in tabular form, as shown in Table 12-6.
The equipment should be replaced at the end of the third year of service although a 5- year replacement cycle is also attractive.
TABLE 12-6 Equivalent Annual Cost Computations for Problem 8
| (1) | (2) | (3) | (4) | | (5)=(1+2–3)(4) | (5)=(1+2–3)(4) | |
|---|
| Year, | | Operating | Salvage | Factor | Factor | Equivalent annual | Equivalent annual | |
| Year, | Initial | costs, | value, | i (1+ i)n | i (1+ i)n | Equivalent annual | Equivalent annual | |
| n | investment, I | ∑Cj / (1+ i) j | S n / (1+i)n | | | cost, ACn | cost, ACn | |
| n | investment, I | ∑Cj / (1+ i) j | S n / (1+i)n | (1+i)n 1 | | cost, ACn | cost, ACn | |
| 1 | $4,000 | $416 | $2,857 | 1.20 | | $1,871 | | |
| 2 | 4,000 | 770a | 1,984 | 0.65 | | 1,821 | | |
| 3 | 4,000 | 1,117b | 1,323 | 0.47 | | 1,783 | ⇐ | |
| 4 | 4,000 | 1,488 | 827 | 0.39 | | 1,818 | | |
| 5 | 4,000 | 1,898 | 459 | 0.33 | | 1,795 | | |
| 6 | 4,000 | 2,350 | 191 | 0.30 | | 1,848 | | |
| 7 | 4,000 | 2,841 | 0 | 0.28 | | 1,915 | | |
a $416 + [500 + 40(2 – 1)2 – 30(2 – 1)] / (1 + 0.20)2 = $770
b $770 + [500 + 40(3 – 1)2 – 30 (3-1)] / 1 + 0.20)3 = $1,117