Question 9.6: Economic Life of a Defender (Existing Asset) It is desired t...
Economic Life of a Defender (Existing Asset)
It is desired to determine how much longer a forklift truck should remain in service before it is replaced by the new truck (challenger) for which data were given in Example 9-4 and Table 9-2. The defender in this case is two years old, originally cost $19,500, and has a present realizable MV of $7,500. If kept, its MV^{s} and annual expenses are expected to be as follows:
| End of Year, k | MV, End of Year k | Annual Expenses, E_{k} |
| 1 | $6,000 | $8,250 |
| 2 | 4,500 | 9,900 |
| 3 | 3,000 | 11,700 |
| 4 | 1,500 | 13,200 |
Determine the most economical period to keep the defender before replacing it (if at all) with the present challenger of Example 9-4. The before-tax cost of capital (MARR) is 10% per year .
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 | $1,500 | $750 | $8,250 | $10,500 | 10500 N^{\ast }_{C}=3 |
| 2 | 16,875 | 1,500 | 600 | 9,900 | 12,000 | 11,214 |
| 3 | 12,750 | 1,500 | 450 | 11,700 | 13,650 | 11,950 |
| 4 | 9,750 | 1,500 | 300 | 13,200 | 15,000 | 12,607 |
| 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 ) |
Table 9-3 shows the calculation of total cost for each year (marginal cost) and the EUAC at the end of each year for the defender based on the format used in Table 9-2. Note that the minimum EUAC of $10,500 corresponds to keeping the defender for one more year. The marginal cost of keeping the truck for the second year is, however, $12,000, which is still less than the minimum EUAC for the challenger (i.e., $12,918, from Example 9-4). The marginal cost for keeping the defender the third year and beyond is greater than the $12,918 minimum EUAC for the challenger. Based on the available data shown, it would be most economical to keep the defender for two more years and then to replace it with the challenger.
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 ) | ||||||