USF IMSE EGN 3615 MID-TERM I(U18) Problem 3. Which process line (A or B) should
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USF IMSE EGN 3615 MID-TERM I(U18) Problem 3. Which process line (A or B) should be built for a new chemical? The expected market for the chemical is 20 years. A 9% annual rate is used to evaluate the new process facilities, which are c better choice save in terms of present value? For your calculations in parts (a), (b) and (c), use the LCM method ompared with present worth. How much does the First cost Annual O&M; cost Salvage value Life $18M $6M $5M 10 years S26M SSM S6M 20 years (a) (8 pts) Calculate the present worth of process line A (b) (8 pts) Calculate the present worth of process line B (c) (4 pts) Which process line should be selected and how much can be saved?Explanation / Answer
Alternative A
Alternative B
First Cost
18,000,000
26,000,000
Annual O & M Cost
6,000,000
5,000,000
Salvage Value
5,000,000
6,000,000
Life (in year)
10
20
Rate of interest is 9%.
Lives of the alternatives are not equal. Use the common multiple method and convert the unequal life into equal life. The common multiple of 10 and 20 is 20.
a. Present Worth of A
NPW = 18,000,000 + 18,000,000 (P/F, 9%, 10) + 6,000,000 (P/A, 9%, 20) –
5,000,000 (P/F, 9%, 10) - 5,000,000 (P/F, 9%, 20)
NPW = 18,000,000 + 18,000,000 (0.4224) + 6,000,000 (9.1285) –
5,000,000 (0.4224) - 5,000,000 (0.1784)
NPW = 77,370,200
b. Present Worth of B
NPW = 26,000,000 + 5,000,000 (P/A, 9%, 20) – 6,000,000 (P/F, 9%, 20)
NPW = 26,000,000 + 5,000,000 (9.1285) – 6,000,000 (0.1784)
NPW = 70,572,100
c. Process B should be selected because if less cost. If process B is selected then the cost savings will be (77,370,200 - 70,572,100) 6,798,100.
Alternative A
Alternative B
First Cost
18,000,000
26,000,000
Annual O & M Cost
6,000,000
5,000,000
Salvage Value
5,000,000
6,000,000
Life (in year)
10
20
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