3-3 Machine \"A\" has an initial cost of $50,000, an estimated service period of
ID: 1132421 • Letter: 3
Question
3-3 Machine "A" has an initial cost of $50,000, an estimated service period of 10 years, and an estimated salvage value of S10,000 at the end of the 10 years. Estimated end-of-year annual disbursements for operation and maintenance are $5,000. A major overhaul costing $10,000 will be required at the end of 5 years. An alternate Machine "B" has an initial cost of $40,000 and an estimated zero salvage value at the end of the 10-year service period with estimated end-of-year disbursements for operation and maintenance of $8,000 for the first year, $8,500 for the second year, and increasing $500 each year thereafter. Using a minimum ROR of 10%, compare the present worth costs of 10-year service from Machines "A" and "B."Explanation / Answer
Machine A
Machine B
First Cost = 50,000
Annual Operating Cost = 5,000
Salvage Value = 10,000
Major overhaul cost at 5th year = 10,000
Life = 10 years
ROR = 10%
NPW = 50,000 + 5000 (P/A, 10%, 10) +
10,000 (P/F, 10%, 5) –
10,000 (P/F, 10%, 10)
NPW = 50,000 + 5000 (6.145) +
10,000 (.6209) –
10,000 (.3855)
NPW = 83,079
First Cost = 40,000
Annual Operating Cost = 8,000 in the 1st year that increases by 500 each year
Life = 10 years
ROR = 10%
First compensating the gradients and calculating uniform cash flow
A1 = 8,000
G = 500
A = A1 + G (A/G, 10%, 10)
A = 8,000 + 500 (3.725) = 9,862.5
NPW = 40,000 + 9,862.5 (P/A, 10%, 10)
NPW = 40,000 + 9,862.5 (6.145)
NPW = 1,00,605
As both the machines are cost dominated, select the lowest cost machine, i.e. Machine A should be selected because of lowest cost.
Machine A
Machine B
First Cost = 50,000
Annual Operating Cost = 5,000
Salvage Value = 10,000
Major overhaul cost at 5th year = 10,000
Life = 10 years
ROR = 10%
NPW = 50,000 + 5000 (P/A, 10%, 10) +
10,000 (P/F, 10%, 5) –
10,000 (P/F, 10%, 10)
NPW = 50,000 + 5000 (6.145) +
10,000 (.6209) –
10,000 (.3855)
NPW = 83,079
First Cost = 40,000
Annual Operating Cost = 8,000 in the 1st year that increases by 500 each year
Life = 10 years
ROR = 10%
First compensating the gradients and calculating uniform cash flow
A1 = 8,000
G = 500
A = A1 + G (A/G, 10%, 10)
A = 8,000 + 500 (3.725) = 9,862.5
NPW = 40,000 + 9,862.5 (P/A, 10%, 10)
NPW = 40,000 + 9,862.5 (6.145)
NPW = 1,00,605
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