how we got the table

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, $7.500. It kept, its originally cost $19.500, and has a present realizable MV of $7,500 MVs and annual expenses are expected to be as follows: End of Year, MV, End of Yeark Annual Expenses, E $6,000 $8.250 4.500 3.000 11.700 1,500 13.200 9.900 Determine the most economical period to keep the defender before replan it (if at all) with the present challenger of Example 9-4. The before-tax cost capital (MARR) is 10% per year. Solution 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) (5) Annual Expenses (E) (6) (= (3) + (4) + (5) Total (Marginal) Cost for Year, EUAC through Yeark (TC) (3) Cost of (1) (2) Loss in Market Capital = 10% End of MV, End Value (MV) of Beginning Year, of Year, k during Yeark of Year MV 0 $7,500 1 6 .000 $1,500 5750 4,500 1,500 600 3 3,000 1,500 4 1.500 1,500 300 N =1 $8,250 9.900 11,700 13,200 $10,500 12,000 13,650 15.000 $10,500 11,214 11,950 12,607 * EUAC = 1 E.TC/CP/F 10% ] A/P,108, k)