Question 10.P.8: To keep an existing machine going for a number of years, an ......
To keep an existing machine going for a number of years, an extensive (and expensive: $4000) overhaul is needed. Maintenance is expected to be $2000 annually for the next 2 years and to increase by $1000 per year after that. The machine has no present or future salvage value. An alternative machine costs $8000 and, owing to its specialized nature, it has no salvage value after it is installed. Maintenance expenses are expected to be $1000 the first year, increasing by $500 per year in subsequent years. If the MARR is 15%, determine the best course of action.
For the defender, writing the first year’s maintenance as $1000 + $1000,
EUAC_1(j) = $4000(A/P, 15%, j)+ $1000(P/F, 15%, 1) (A/P, 15%,j)+ $1000 + $1000(A/G, 15%, j)
and for the challenger,
EUAC_2(j) = $8000(A/P, 15%, j)+ $1000+ $500(A/G, 15%, j)
From the evaluations in Table 10-7, we see that the economic lives are j_1^* = 4 years and j_2^* = 7 years. It follows that the best course of action is to overhaul the current equipment and keep it for four years. Then (provided an analysis indicates that things remain as expected), buy the new equipment and keep it for seven years.
| Table 10-7 | ||
| j | EUAC_1(j) | EUAC_2(j) |
| 1 | $6600 | $10 200 |
| 2 | 4460 | 6 154 |
| 3 | 4040 | 4 957 |
| 4 | 4032 | 4 465 |
| 5 | 4175 | 4 248 |
| 6 | 4 163 | |
| 7 | 4 148 | |
| 8 | 4 173 | |