Question

For th below two machines and based on CC analysis which machine we should select? MARR=10%

	Machine A	Machine B
First cost, $	29,601	129,434
Annual cost, $/year	13,018	8,452
Salvage value, $	5,328	-
Life, years	3	infinite

Answer the below question:

B- the CC for machine B=

Answer 1

Ans. Capatalized cost of machine A is the infinite stream of equivalent annual present value of all the annual costs and salvage of the machine,

CC = Equivalent annual worth [-29601 - 13018/(1+0.10) -13018/(1+0.10)^2 - 13018/(1+0.10)^3 + 5328/(1+0.10)^3]/0.10

=> CC = -23311.331/0.10 = -233113.31

Thus, capatalized cost of machine A is $233113.31

Capatalized cost of machine B is the present value of all the annual costs machine,

CC = -129434 - 8452/(1+0.10) -8452/(1+0.10)^2 - 8452/(1+0.10)^3 -.......infinity

=> CC = -129434 - 8452/0.10 [Using the formula for infinite sum of geometric progression]

=> CC = -213954

Thus, capatalized cost of machine B is $213954

Thus, with lower CC, machine B should be selected.

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