A city water and waste-water department has a four year old sludge pump that was initially purchased for 65,000. This pump can be kept in service for an additional 4 years, or it can be sold for 36,000 and replaced by a new pump. The purchase price of the replacement pump is 51,500. The projected MVs and operating maintenance cost over the four year planning horizon are shown in the table that follows. Assuming the MARR is 8%, A. determine the economic life of the challenger and B. determine when the defender should be replaced?




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Using Excel to perform Economic life analysis
Economic Life 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,8%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+51500 | G | H=G*B | I=F-H | J | K = I*J |
| 1 | 0.92593 | 14000.00 | 12962.96 | 12962.96 | 64462.96 | 38500.00 | 35648.15 | 28814.81 | 1.08000 | 31120.00 |
| 2 | 0.85734 | 16500.00 | 14146.09 | 27109.05 | 78609.05 | 31500.00 | 27006.17 | 51602.88 | 0.56077 | 28937.31 |
| 3 | 0.79383 | 19000.00 | 15082.81 | 42191.87 | 93691.87 | 24500.00 | 19448.89 | 74242.98 | 0.38803 | 28808.76 |
| 4 | 0.73503 | 21500.00 | 15803.14 | 57995.01 | 109495.01 | 17500.00 | 12863.02 | 96631.99 | 0.30192 | 29175.21 |
| Discount factor | 1/(1+0.08)^n | |||||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | |||||||||
As EUAC is Lowest in yr 3 at 28808.76, therefore Economic life of challenger is 3 yrs
Economic Life 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,8%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+36000 | G | H=G*B | I=F-H | J | K = I*J |
| 1 | 0.92593 | 19500.00 | 18055.56 | 18055.56 | 54055.56 | 26500.00 | 24537.04 | 29518.52 | 1.08000 | 31880.00 |
| 2 | 0.85734 | 22000.00 | 18861.45 | 36917.01 | 72917.01 | 23000.00 | 19718.79 | 53198.22 | 0.56077 | 29831.92 |
| 3 | 0.79383 | 24500.00 | 19448.89 | 56365.90 | 92365.90 | 19500.00 | 15479.73 | 76886.17 | 0.38803 | 29834.41 |
| 4 | 0.73503 | 27000.00 | 19845.81 | 76211.71 | 112211.71 | 16000.00 | 11760.48 | 100451.23 | 0.30192 | 30328.32 |
| Discount factor | 1/(1+0.08)^n | |||||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | |||||||||
As minimum EUAC of Defender is 29831.92 at yr 2, which is more than the minimum EUAC of Challenger of 28808.76, therefore defender should be replaced immediately
Showing formula for economic 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,8%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+36000 | G | H=G*B | I=F-H | J | K = I*J |
| 1 | =1/(1.08)^A44 | 19500 | =C44*B44 | =D44 | =36000+E44 | 26500 | =G44*B44 | =F44-H44 | =0.08*((1 + 0.08)^A44)/((1 + 0.08)^A44-1) | =I44*J44 |
| 2 | =1/(1.08)^A45 | 22000 | =C45*B45 | =E44+D45 | =36000+E45 | 23000 | =G45*B45 | =F45-H45 | =0.08*((1 + 0.08)^A45)/((1 + 0.08)^A45-1) | =I45*J45 |
| 3 | =1/(1.08)^A46 | 24500 | =C46*B46 | =E45+D46 | =36000+E46 | 19500 | =G46*B46 | =F46-H46 | =0.08*((1 + 0.08)^A46)/((1 + 0.08)^A46-1) | =I46*J46 |
| 4 | =1/(1.08)^A47 | 27000 | =C47*B47 | =E46+D47 | =36000+E47 | 16000 | =G47*B47 | =F47-H47 | =0.08*((1 + 0.08)^A47)/((1 + 0.08)^A47-1) | =I47*J47 |
| Discount factor | 1/(1+0.08)^n | |||||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | |||||||||
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