L03 (a) Any one (20 Marks) A Dubai Sewage Company is trying to decide among thre

Question

L03 (a) Any one (20 Marks) A Dubai Sewage Company is trying to decide among three alternatives of waste dewatering processes. The costs associated with these alternatives are shown below. Alternative Y will need an upgrade of $9700 at the end of year 2. At the end of year 2, alternative Z would be replaced with another alternative Z having the same installed and operating costs. If the MARR is 10% per year, which alternative should be chosen? Use Annual worth Analysis. (See table below) Z. Alternative Installed costs $68.500 $48,500 $33,500 Annual operating costs $6000 4000$5000 Overhaul cost in year 2 $9700 Salvage value 33,250 $28,250 $15,750 Useful life, years Perform the annual worth analysis for each alternative. The MARR is 10% per year and select the best one.

Explanation / Answer

Calculate Annual Worth of Alternative X -

AW = -$68,500(A/P, 10%, 8) -$6,000 + $33,250(A/F, 10%, 8)

AW = (-$68,500 * 0.1874) - $6,000 + ($33,250 * 0.0874)

AW = -$12,836.9 - $6,000 + $2,906.05

AW = -$15,930.85

The Annual Worth of Alternative X is -$15,930.85

Calculate the annual worth of Alternative Y -

AW = -$48,500(A/P, 10%, 4) - $4,000 - $9,700(A/F, 10%, 2) + $28,250(A/F, 10%, 4)

AW = (-$48,500 * 0.3155) - $4,000 - ($9,700 * 0.4762) + ($28,250 * 0.2155)

AW = -$15,301.75 - $4,000 - $4,619.14 + $6,087.87

AW = -$17,833.02

The Annual Worth of Alternative Y is -$17,833.02

Calculate the annual worth of alternative Z -

AW = -$33,500(A/P, 10%, 2) - $5,000 + $15,750(A/F, 10%, 2)

AW = (-$33,500 * 0.5762) - $5,000 + ($15,750 * 0.4762)

AW = -$19,302.7 - $5,000 + $7,500.15

AW = -$16,802.55

The Annual Worth of Alternative Z is -$16,802.55

As per the annual worth method, the alternative with numerically highest annual worth should be selected.

Since, AWX > AWZ > AWY

So, Alternative X is best and should be selected.

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