Machine A costs $350,000 to purchase, result in electricity bills of $100,000 per year, and last...

Machine A costs $350,000 to purchase, result in electricity bills of $100,000 per year, and last for 10 years. Machine B costs $550,000 to purchase, result in electricity bills of $80,000 per year, and last for 15 years. The discount rate is 12%. What are the equivalent annual costs for two models? Which model is more cost-effective?

Expert Solution

Machine A
Year	Annual Cost (C)	PV Factor Calculation	PV Factor @ 12 % (F)	PV (= C x F)
0	$350,000	1/(1+0.12)^0	1	$350,000
1	$100,000	1/(1+0.12)^1	0.892857143	$89,286
2	$100,000	1/(1+0.12)^2	0.797193878	$79,719
3	$100,000	1/(1+0.12)^3	0.711780248	$71,178
4	$100,000	1/(1+0.12)^4	0.635518078	$63,552
5	$100,000	1/(1+0.12)^5	0.567426856	$56,743
6	$100,000	1/(1+0.12)^6	0.506631121	$50,663
7	$100,000	1/(1+0.12)^7	0.452349215	$45,235
8	$100,000	1/(1+0.12)^8	0.403883228	$40,388
9	$100,000	1/(1+0.12)^9	0.360610025	$36,061
10	$100,000	1/(1+0.12)^10	0.321973237	$32,197
			NPV	$915,022

Machine B
Year	Annual Cost (C)	PV Factor Calculation	PV Factor @ 12 % (F)	PV (= C x F)
0	$550,000	1/(1+0.12)^0	1	$550,000
1	$80,000	1/(1+0.12)^1	0.892857143	$71,429
2	$80,000	1/(1+0.12)^2	0.797193878	$63,776
3	$80,000	1/(1+0.12)^3	0.711780248	$56,942
4	$80,000	1/(1+0.12)^4	0.635518078	$50,841
5	$80,000	1/(1+0.12)^5	0.567426856	$45,394
6	$80,000	1/(1+0.12)^6	0.506631121	$40,530
7	$80,000	1/(1+0.12)^7	0.452349215	$36,188
8	$80,000	1/(1+0.12)^8	0.403883228	$32,311
9	$80,000	1/(1+0.12)^9	0.360610025	$28,849
10	$80,000	1/(1+0.12)^10	0.321973237	$25,758
11	$80,000	1/(1+0.12)^11	0.287476104	$22,998
12	$80,000	1/(1+0.12)^12	0.256675093	$20,534
13	$80,000	1/(1+0.12)^13	0.22917419	$18,334
14	$80,000	1/(1+0.12)^14	0.204619813	$16,370
15	$80,000	1/(1+0.12)^15	0.182696261	$14,616
			NPV	$1,094,869

Equivalent Annual Cost, EAC = NPV x Discount rate/1 - (1 + Discount rate)^{-Number of periods}

EAC for Machine A = $ 915,022 x 0.12/[1-(1+0.12)^-10]

= $ 915,022 x 0.12/[1-0.321973237]

= $ 915,022 x 0.12/ 0.678026763

= $ 109,802.68/0.678026763

= $ 161,944.46

EAC for Machine B = $ 1,094,869 x 0.12/[1-(1+0.12)^-15]

= $ 1,094,869 x 0.12/[1-0.182696261]

= $ 1,094,869 x 0.12/0.817303739

= $ $131,384.30/0.817303739

= $ 160,753.33

Machine B is more cost effective as EAC of Machine B is less than that of Machine A.

jeff jeffy answered 2 years ago

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