Question

5.30 A company needs to purchase a new machine to maintain its level of production. The company is considering three different machines. The costs, savings and service life related to each machine are listed in the table below.

	Machine A	Machine B	Machine C
First Cost	$37,500	$31,000	$35,000
Annual Savings	$13,500	$12,000	$12,750
Annual Maintenance	$3,000 the first year and increasing by $600 every year thereafter	$2,500	$2,000
Salvage Value	$5,000	$11,000	$13,000
Service Life	6 years	3 years	3 years

Given a MARR of 12% and using the rate of return method, which alternative should be chosen?

Answer 1

We can determine which alternative must be selected by determining the NPW

NPW_A = -37,500 + (13,500-3,000) (P/A,12%,6) -600(P/G,12%,6)+ 5000(P/F,12%,6)

= - 37,500+ 10,500*4.111 -600*8.930 + 5,000*0.5066

= -37,500 + 43,165.50 - 5,358 + 2,518

= $ 2,825.50

Similarly we can determine NPW of Machine B

NPW_B = -31,000 +(12000-2500)*(P/A,12%,6) -20000(P/F,12%,3) + 11,000(P/F,12%,6)

= -31,000+ 9500*4.111 -20000*0.7118 + 11000*0.5036

= - $ 641.90

Now, calculate NPW of Machine C

NPW_C = -35,000 +(12750-2000)(P/A,12%,6) -22000(P/F,12%,3) +13000(P/F,12%,6)

= - 35000 + 10750*4.111 - 22000*0.7118 +13000*0.5036

= $ 80.45

Machine A has highest NPW hence select machine A.

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