In: Finance
A company is considering the purchase of new equipment for its production area. The equipment has an initial cost of $ 3,000 with operation and maintenance costs, as well as the market liquidation value as shown in the following table:
Year | Costs of operation | Rescue value |
1 | $1,000 | $1,500 |
2 | $1,700 | $1,000 |
3 | $2,400 | $500 |
4 | $3,100 | $0 |
Determine the Optimal Economic Life of this investment, if the MARR of the company is 12%
- | Particulars | Cahflow | PVF @ 12% | 1 Year | 2 Years | 3 Years | 4 Years |
A | Initial Cost (Year 0) | $ 3,000 | 1 | $ 3,000 | $ 3,000 | $ 3,000 | $ 3,000 |
B | PV of Cost of Operation | ||||||
. | Year 1 | $ 1,000 | 0.8929 | $ 892.86 | $ 892.86 | $ 892.86 | $ 892.86 |
. | Year 2 | $ 1,700 | 0.7972 | $ 1,355.23 | $ 1,355.23 | $ 1,355.23 | |
. | Year 3 | $ 2,400 | 0.7118 | $ 1,708.27 | $ 1,708.27 | ||
. | Year 4 | $ 3,100 | 0.6355 | $ 1,970.11 | |||
. | PV of Cost of Operation | $ 892.86 | $ 2,248.09 | $ 3,956.36 | $ 5,926.47 | ||
C | PV of Salvage Value | ||||||
. | Year 1 | $ 1,500 | 0.8929 | $ 1,339.29 | |||
. | Year 2 | $ 1,000 | 0.7972 | $ 797.19 | |||
. | Year 3 | $ 500 | 0.7118 | $ 355.89 | |||
. | Year 4 | $ - | 0.6355 | $ - | |||
. | PV of Salvage Value | $ 1,339.29 | $ 797.19 | $ 355.89 | $ - | ||
D | PV of Cost for the given Life | A+B-C | $ 2,553.57 | $ 4,450.89 | $ 6,600.47 | $ 8,926.47 | |
E | PVAF @ 12% | 1.6900 | 2.4017 | 3.0375 | |||
F | Equivalent Annual Cost | $ 2,553.57 | $ 2,633.66 | $ 2,748.29 | $ 2,938.75 |
As Equivalent annual Cost is Less in case of Machine Used for 1 Year, Optimal Economic Life of the Investment is 1 Year.
Computation of PVAF:
Year(n) | r | 1+r | (1+r)^-n | 1- [(1+r)^-n] | [1- [(1+r)^-n]] /r |
2 | 12.00% | 1.1200 | 0.7972 | 0.2028 | 1.6900 |
3 | 12.00% | 1.1200 | 0.7118 | 0.2882 | 2.4017 |
4 | 12.00% | 1.1200 | 0.6355 | 0.3645 | 3.0375 |