Engineering Economy problem. Please, when answering this problem, it will be mor
ID: 1168717 • Letter: E
Question
Engineering Economy problem.
Please, when answering this problem, it will be more than appreciated if you can present clearly the analysis, in order to learn and understand it. Example: Present Value = 20,000(P/F, i, n)+45,000(A/F, i, n)(P/F, i, n)... Thanks in advance!
There are two manufacturing processes that can be used by a company to reduce energy losses in the equipment’s. The following cost information for these two processes is known:
By doing the following analysis, select which manufacturing process must be selected (A or B):
a) Find the Present Value for Process A and for Process B, with an interest rate of 12% annually computed quarterly
b) Find the Quarterly equivalent uniform value for Process A and for Process B with an interest rate of 8% annually computed quarterly
c) Find the Capitalized Cost for Process A and for Process B with an interest rate of 14% annually computed quarterly
Process A Process B Initial Cost $170,000 $220,000 Operation and Maintenance Cost $7,000 quarterly $5,000 quarterly Useful Life (Life Cycle) 2 years 4 years Residual Value $30,000 $42,000Explanation / Answer
(a) 14% annual rate = 12/4 = 3% quarterly rate
PROCESS - A
Present Value, PV = $170,000 + $7,000 x P/A (3%, 8) - $30,000 x P/F (3%, 8)
= $170,000 + ($7,000 x 7.02) - ($30,000 x 0.79)
= $(170,000 + 49,140 - 23,700) = $195,440
PROCESS - B
Present Value, PV = $220,000 + $5,000 x P/A (3%, 16) - $42,000 x P/F (3%, 16)
= $170,000 + ($5,000 x 12.6) - ($42,000 x 0.62)
= $(170,000 + 63,000 - 26,040) = $206,960
(b) 8% annual interest rate = 2% quarterly rate
Process A = $170,000 x A/P (2%, 8) + $7,000 - $30,000 x A/F (2%, 8)
= $170,000 x 0.14 + $7,000 - $30,000 x 0.12
= $(23,800 + 7,000 - 3,600) = $27,200
Process B = $220,000 x A/P (2%, 16) + $5,000 - $42,000 x A/F (2%, 16)
= $220,000 x 0.0737 + $5,000 - $42,000 x 0.0537
= $(16,214 + 5,000 - 2,254) = $18,960
NOTE: Out of 3 questions, the first 2 are answered.
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