In: Economics
For the below ME alternatives , which machine should be selected based on the PW analysis. MARR=10%
Machine A | Machine B | Machine C | |
First cost, $ | 22,421 | 30000 | 10000 |
Annual cost, $/year | 8,497 | 6,000 | 4,000 |
Salvage value, $ | 4,000 | 5,000 | 1,000 |
Life, years | 3 | 6 | 2 |
Answer the below questions :
A- PW for machine A=
Machine A:
First Cost = 22,421
Annual Cost = 8,497
Salvage Value = 4,000
Life = 3
Year | Annual Cost | Present worth of annual cost | Rough work to calculate present worth |
1 | 8,497 | 7,724.55 | [8,497 / 1.1^1] |
2 | 8,497 | 7,022.31 | [8,497 / 1.1^2] |
3 | 8,497 | 6,383.92 | [8,497 / 1.1^3] |
21,130.78 |
Present worth of salvage at the end of 3 years = [4,000 / 1.1^3] = 3,005.26
Present worth of machine A = - First cost - Present worth of annual maintenance cost + Present worth of salvage value = - 22,421 - 21,130.78 + 3,005.26 = - 40,546.52