Question 9.7: Replacement Analysis Using the Coterminated Assumption Suppo...
Replacement Analysis Using the Coterminated Assumption
Suppose that we are faced with the same replacement problem as in Example 9-6, except that the period of needed service is (a) three years or (b) four years. That is, a finite analysis period under the coterminated assumption is being used. In each case, which alternative should be selected?
(a) For a planning horizon of three years, we might intuitively think that either the defender should be kept for three years or it should be replaced immediately by the challenger to serve for the next three years. From Table 9-3, the EUAC for the defender for three years is $11,950; from Table 9-2, the EUAC for the challenger for three years is $12,918. Thus, following this reasoning, the defender would be kept for three years. This, however, is not quite right. Focusing on the Total (Marginal) Cost for Each Year columns, we can see that the defender has the lowest cost in the first two years, but in the third year its cost is $13,650; the EUAC of one year of service for the challenger is only $13,500. Hence, it would be more economical to replace the defender after the second year. This conclusion can be confirmed by enumerating all replacement possibilities and their respective costs and then computing the EUAC for each, as will be done for the four-year planning horizon in Part (b).
TABLE 9-4 Determination of When to Replace the Defender for a Planning Horizon of Four Years [Example 9-7, Part (b)]
| Total (Marginal) Costs for Each Year | ||||||
| Keep Defender for | Keep Challenger for | 1 | 2 | 3 | 4 | EUAC at 10% for 4 Years |
| 0 years | 4 years | $13,500^{a} | $12,375^{a} | $12,813^{a} | $14,275^{a} | $13,211 |
| 1 | 3 | 10,500 | 13,500 | 12,375 | 12,813 | 12,225 |
| 2 | 2 | 10,500 | 12,000 | 13,500 | 12,375 | 12006 ← Least cost alternative |
| 3 | 1 | 10,500 | 12,000 | 13,650 | 13,500 | 12,284 |
| 4 | 0 | 10500^{b} | 12000^{b} | 13,650^{b} | 15,000^{b} | 12,607 |
| ^{a} Column 6 of Table 9-2 | ||||||
| ^{b}Column 6 of Table 9-3. | ||||||
(b) For a planning horizon of four years, the alternatives and their respective costs for each year and the EUAC of each are given in Table 9-4. Thus, the most economical alternative is to keep the defender for two years and then
replace it with the challenger, to be kept for the next two years. The decision to keep the defender for two years happens to be the same as when the repeatability assumption was used, which, of course, would not be true in general.
TABLE 9-3 Determination of the Economic Life N∗ of an Existing Asset (Example 9-6)
| (1) End of Year, k | (2) MV, End of Year, k | (3) Loss in Market Value (MV) during Year k | (4) Cost of Capital = 10% of Beginning of Year MV | (5) Annual Expenses \underline{(E_{k})} | (6) [= (3) + (4) + (5)] Total (Marginal) Cost for Year, \underline{(TC_{k})} | (7) EUAC^{a} through Year k |
| 0 | $7,500 | |||||
| 1 | 6,000 | $1,500 | $750 | $8,250 | $10,500 | 10500 N^{\ast }_{D}=1 |
| 2 | 4,500 | 1,500 | 600 | 9,900 | 12,000 | 11,214 |
| 3 | 3,000 | 1,500 | 450 | 11,700 | 13,650 | 11,950 |
| 4 | 1,500 | 1,500 | 300 | 13,200 | 15,000 | 12,607 |
| ^{a}EUAC_{k}= \left[\sum^{k}_{j}{TC_{j}(P/F, 10\% , j )}\right] (A/p, 10\% , k ) | ||||||
TABLE 9-2 Determination of the Economic Life, N∗, of a New Asset (Example 9-4)
| (1) End of Year, k | (2) MV, End of Year, k | (3) Loss in Market Value (MV) during Year k | (4) Cost of Capital = 10% of Beginning of Year MV | (5) Annual Expenses \underline{(E_{k})} | (6) [= (3) + (4) + (5)] Total (Marginal) Cost for Year, \underline{(TC_{k})} | (7) EUAC^{a} through Year k |
| 0 | $30,000 | |||||
| 1 | 22,500 | $7,500 | $3.000 | $3.000 | $13,500 | 13.500 N^{\ast }_{C}=3 |
| 2 | 16,875 | 5.625 | 2.250 | 4.500 | 12,375 | 12.964 |
| 3 | 12,750 | 4.125 | 1.688 | 7.000 | 12.813 | 12.918 |
| 4 | 9,750 | 3.000 | 1.275 | 10.000 | 14.275 | 13.211 |
| 5 | 7,125 | 2,625 | 975 | 13,000 | 16,600 | 13,766 |
| ^{a}EUAC_{k}= \left[\sum^{k}_{j}{TC_{j}(P/F, 10\% , j )}\right] (A/p, 10\% , k ) |