Homework Answers
i = 10%
Economic service life analysis
| Year | Discount factor | O&M cost | PV (O&M) | Cumulative (O&M) | Cumulative (O&M) + Initial Cost | Salvage value | PV (Salvage value) | NPV | (A/P,10%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+15000 | G | H=G*B | I=F-H | J | K = I*J |
| 1 | 0.90909 | 2500.00 | 2272.73 | 2272.73 | 17272.73 | 12800.00 | 11636.36 | 5636.36 | 1.10000 | 6200.00 |
| 2 | 0.82645 | 3200.00 | 2644.63 | 4917.36 | 19917.36 | 8100.00 | 6694.21 | 13223.14 | 0.57619 | 7619.05 |
| 3 | 0.75131 | 5300.00 | 3981.97 | 8899.32 | 23899.32 | 5300.00 | 3981.97 | 19917.36 | 0.40211 | 8009.06 |
| 4 | 0.68301 | 6500.00 | 4439.59 | 13338.91 | 28338.91 | 3500.00 | 2390.55 | 25948.36 | 0.31547 | 8185.95 |
| 5 | 0.62092 | 7800.00 | 4843.19 | 18182.10 | 33182.10 | 0.00 | 0.00 | 33182.10 | 0.26380 | 8753.35 |
| Discount factor | 1/(1+0.1)^n | |||||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | |||||||||
As EUAC is lowest in yr 1, minimum EUAC = 6200
Economic service life = 1 yr
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