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For the below ME alternatives , which machine should be selected based on the PW analysis....

For the below ME alternatives , which machine should be selected based on the PW analysis. MARR is 10%.

Machine A Machine B Machine C
First cost, $ 15000 30000 13,435
Annual cost, $/year   16,207                6,000             4,000
Salvage value, $                  4,000                5,000             1,000
Life, years 3 6 2

Answer the below questions :

C- PW for machine C =

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Homework Answers

Answer #1

Machine C:

First Cost = 13,435

Annual cost = 4,000

Salvage = 1,000

Life = 2 years

MARR = 10%

Present value of annual cost of 1st year = [4,000 / (1 + 0.1)^1] = 3,636.364

Present value of annual cost of 2nd year = [4,000 / (1 + 0.1)^2] = 3,305.785

Present value of annual cost of both year = 3,636.364 + 3,305.785 = 6,942.149

Present value of salvage value after 2 years = [Salvage Value / (1 + MARR)^Life] = [1,000 / (1 + 0.1)^2] = 826.44

Present value of machine = Annual Cost - Present value of annual Cost + Salvage Value = 13,435 - 6,942.149 + 826.44 = 7,319.291

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