Question

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

	Machine A	Machine B	Machine C
First cost, $	15000	30000	11,937
Annual cost, $/year	13,944	6,000	4,000
Salvage value, $	4,000	5,000	1,000
Life, years	3	6	2

Answer the below questions :

C- AW for machine C =

Answer 1

Machine A -

Calculate the annual worth of Machine A -

AW = -First cost(A/P, i, n) - Annual cost + Salvage value(A/F, i, n)

AW = -15,000(A/P, 10%, 3) - 13,944 + 4,000(A/F, 10%, 3)

AW = [-15,000 * 0.40211] - 13,944 + [4,000 * 0.30211]

AW = -6,031.65 - 13,944 + 1,208.44

AW = -18,767.21

The AW for machine A is $ -18,767.21

Machine B -

Calculate the annual worth of Machine B -

AW = -First cost(A/P, i, n) - Annual cost + Salvage value(A/F, i, n)

AW = -30,000(A/P, 10%, 6) - 6,000 + 5,000(A/F, 10%, 6)

AW = [-30,000 * 0.22961] - 6,000 + [5,000 * 0.12961]

AW = -6,888.3 - 6,000 + 648.05

AW = -12,240.25

The AW for machine B is $ -12,240.25

Machine C -

Calculate the annual worth of Machine C -

AW = -First cost(A/P, i, n) - Annual cost + Salvage value(A/F, i, n)

AW = -11,937(A/P, 10%, 2) - 4,000 + 1,000(A/F, 10%, 2)

AW = [-11,937 * 0.57619] - 4,000 + [1,000 * 0.47619]

AW = -6,877.98 - 4,000 + 476.19

AW = -10,401.79

The AW for machine C is $ -10,401.79

The annual worth of Machine C is numerically highest.

Thus,

Machine C should be selected based in AW analysis.