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, $	15,567	30000	10000
Annual cost, $/year	9,535	6,000	4,000
Salvage value, $	4,000	5,000	1,000
Life, years	3	6	2

Answer the below questions :

A- AW for machine A=

Answer 1

Answer:
		Machine A	Machine B	Machine C
	First cost (a)	15,567.00	30,000.00	10,000.00
	Life	3.00	6.00	2.00
	Annual cost	9,535.00	6,000.00	4,000.00
	PV of annual cost (b)	9,535 * PVAF(10%,3)	6,000 * PVAF(10%,6)	4,000 * PVAF(10%,2)
		9,535 * 2.487	6,000 * 4.355	4,000 * 1.7355
		23,713.55	26,130.00	6,942.00
	Salvage value	4,000.00	5,000.00	1,000.00
	Pv of salvage ©	4,000 * PVIF(10%,3)	5,000 * PVIF(10%,6)	1,000 * PVIF(10%,2)
		4,000 * 0.7513	5,000 * 0.5645	1,000 * 0.8264
		3,005.20	2,822.50	826.40
	NPV (-a-b+c)	(36,275.35)	(53,307.50)	(16,115.60)
	AW	(14,585.99)	(12,240.53)	(9,285.85)
	(NPV/ PVAF)
	So, based on AW analysis, Machine C should be selected as its annual cost is the lowest.
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