Question

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?

4 of 7 (2 complete) HT ic) Defender Challenger Year MV at EOY 0&M Costs 0&M Costs MV at EOY 1 $26,500 $19,500 $38,500 $14000

pter 9 Homewo rk 4 of 7 (2 complete) HW Sco ic) $14000 $38,500 $19,500 $26,500 1 16,500 31,500 22,000 23,000 2 19,000 24,500

23,000 2,000 31,500 T6,500 10 000 24500 19 500 24-600 X More Info Discrete Compounding; i-8% Uniform Series Single Payment Co

Que 23,000 22,000 31,500 24 600 16,500 10 000 10 500 24 600 More Info - X To Find F Given P To Find P To Find F Given A To Fi

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Answer #1

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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