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, $ | 15000 | 30000 | 11,265 |
Annual cost, $/year | 14,021 | 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 =
The present worth of total outflows for Machine C is the least ($ 17,379) of the three machines @ MARR of 10%, therefore machine C should be selected:
Year | PW factor @ MARR= 10% | Machine A | PW @ MARR | Machine B | PW @ MARR | Machine C | PW @ MARR |
0 | 1 | -15000 | -15000 | -30000 | -30000 | -11265 | -11265 |
1 | 0.909 | -14021 | -12745 | -6000 | -5454 | -4000 | -3636 |
2 | 0.826 | -14021 | -11581 | -6000 | -4956 | -3000 | -2478 |
3 | 0.751 | 10021 | 7526 | -6000 | -4506 | ||
4 | 0.683 | -6000 | -4098 | ||||
5 | 0.621 | -6000 | -3726 | ||||
6 | 0.564 | -1000 | -564 | ||||
Total PW(A) | -31801 | Total PW (B) | -53304 | Total PW(C) | -17379 |
(Note that the outflows in terminal years for each machines is reduced by scarp sale inflows).