TABLE 2
Salvage value Annual operating
Year at end, $ cost, $
0 12,000
1 8,000 4,700
2 5,000 5,200
3 4,000 5,800
4 3,500 6,600
5 3,000 7,500
6 2,700 8,500
7 2,600 9,700
8 2,500 10,900ment at end ofRth year. Then F^ = F^ 1 (I + O + CR. By retaining the existing machine, the
firm forfeits an income of $12,000, and this is equivalent to making a payment of that
amount now. Then F 0 = $12,000; F 1 = $12,000(1.10) + $4700 = $17,900; F 2 =
$17,900(1.10) + $5200 - $24,890; F 3 = $24,890(1.10) + $5800 - $33,179; etc. The re-
sults are shown in Table 3.
- Compute the annual cost for every prospective remaining life
Let AR = annual cost for a remaining life of R years, and LR — salvage value at end of Rth
year. Then AR = (FR - LjO(SFP, n = R). Thus, with i = 10 percent, A 1 = ($17,900 -
$8000)1 = $9900; A 2 = ($24,890 - $5000)(0.47619) = $9471; A 3 = ($33,179 -
$4000)(0.30211) - $8815; etc. The results are shown in Table 3. - Determine whether the existing machine should be retired now
When an asset is purchased and installed, its resale value drops sharply during the early
years of its life, and thus the firm incurs a rapid loss of capital during those years. The re-
sult is that the annual cost for the remaining life of an existing asset is considerably less
than the annual cost when the asset was first purchased.
Table 3 shows that the optimal remaining life of the existing machine is 5 years and
TABLE 3. Calculation of Annual Cost
Remaining Annual
life, years F, $ Z, $ SFP cost, $
0 12,000
1 17,90 0 8,00 0 1.0000 0 9,90 0
2 24,89 0 5,00 0 0.4761 9 9,47 1
3 33,17 9 4,00 0 0.3021 1 8,81 5
4 43,09 7 3,50 0 0.2154 7 8,53 2
5 54,90 7 3,00 0 0.1638 0 8,50 2
6 68,89 7 2,70 0 0.1296 1 8,58 0
7 85,48 7 2,60 0 0.1054 1 8,73 7
8 104,93 6 2,50 0 0.0874 4 8,95 7