Homework Answers
Using Excel for economic service analysis
Economic service analysis of Challenger
| 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,12%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+80000 | G | H=G*B | I=F-H | J | K = I*J |
| 1 | 0.89286 | 10000.00 | 8928.57 | 8928.57 | 88928.57 | 70500.00 | 62946.43 | 25982.14 | 1.12000 | 29100.00 |
| 2 | 0.79719 | 11200.00 | 8928.57 | 17857.14 | 97857.14 | 69090.00 | 55078.13 | 42779.02 | 0.59170 | 25312.26 |
| 3 | 0.71178 | 12544.00 | 8928.57 | 26785.71 | 106785.71 | 67708.20 | 48193.36 | 58592.35 | 0.41635 | 24394.87 |
| 4 | 0.63552 | 14049.28 | 8928.57 | 35714.29 | 115714.29 | 66354.04 | 42169.19 | 73545.10 | 0.32923 | 24213.58 |
| 5 | 0.56743 | 15735.19 | 8928.57 | 44642.86 | 124642.86 | 65026.96 | 36898.04 | 87744.82 | 0.27741 | 24341.27 |
| Discount factor | 1/(1+0.12)^n | |||||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | |||||||||
Minimum EUAC is 24213.58 at 4 yrs, so the economic service life is 4 yrs of challenger
Economic service analysis of Defender
| 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,12%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+55000 | G | H=G*B | I=F-H | J | K = I*J |
| 1 | 0.89286 | 9000.00 | 8035.71 | 8035.71 | 63035.71 | 45000.00 | 40178.57 | 22857.14 | 1.12000 | 25600.00 |
| 2 | 0.79719 | 10350.00 | 8250.96 | 16286.67 | 71286.67 | 42300.00 | 33721.30 | 37565.37 | 0.59170 | 22227.36 |
| 3 | 0.71178 | 11902.50 | 8471.96 | 24758.64 | 79758.64 | 39762.00 | 28301.81 | 51456.83 | 0.41635 | 21424.00 |
| 4 | 0.63552 | 13687.88 | 8698.89 | 33457.53 | 88457.53 | 37376.28 | 23753.30 | 64704.23 | 0.32923 | 21302.86 |
| 5 | 0.56743 | 15741.06 | 8931.90 | 42389.43 | 97389.43 | 35133.70 | 19935.81 | 77453.62 | 0.27741 | 21486.39 |
| Discount factor | 1/(1+0.12)^n | |||||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | |||||||||
Minimum EUAC is 21302.86 at 4 yrs, so the economic service life is 4 yrs of defender
b.
As EUAC @ 5 yrs of Defender is 21486 which is less than minimum EUAC of challenger, the defender should be kept for 5 yrs before being replaced by challenger
Showing Formula in Excel for Challenger, defender can be calculated in similar way
| 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,12%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+80000 | G | H=G*B | I=F-H | J | K = I*J |
| 1 | =1/(1.12)^A14 | 10000 | =C14*B14 | =D14 | =80000+E14 | 70500 | =G14*B14 | =F14-H14 | =0.12*((1 + 0.12)^A14)/((1 + 0.12)^A14-1) | =I14*J14 |
| 2 | =1/(1.12)^A15 | =C14*1.12 | =C15*B15 | =E14+D15 | =80000+E15 | =G14*0.98 | =G15*B15 | =F15-H15 | =0.12*((1 + 0.12)^A15)/((1 + 0.12)^A15-1) | =I15*J15 |
| 3 | =1/(1.12)^A16 | =C15*1.12 | =C16*B16 | =E15+D16 | =80000+E16 | =G15*0.98 | =G16*B16 | =F16-H16 | =0.12*((1 + 0.12)^A16)/((1 + 0.12)^A16-1) | =I16*J16 |
| 4 | =1/(1.12)^A17 | =C16*1.12 | =C17*B17 | =E16+D17 | =80000+E17 | =G16*0.98 | =G17*B17 | =F17-H17 | =0.12*((1 + 0.12)^A17)/((1 + 0.12)^A17-1) | =I17*J17 |
| 5 | =1/(1.12)^A18 | =C17*1.12 | =C18*B18 | =E17+D18 | =80000+E18 | =G17*0.98 | =G18*B18 | =F18-H18 | =0.12*((1 + 0.12)^A18)/((1 + 0.12)^A18-1) | =I18*J18 |
| Discount factor | 1/(1+0.12)^n | |||||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | |||||||||
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