Question

# Q8

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

 Machine A Machine B First cost, \$ 21,439 124,845 Annual cost, \$/year 13,160 7,237 Salvage value, \$ 5,419 - Life, years 3 infinite

B- the CC for machine B=

SOLUTION :

r = MARR = 10% = 0.1

=> (1 + r) = 1.1

PV of costs of Machine A :

= 21439 + 13160/ 1.1 + 13160/1.1^2 + 13160/1.1^3 - 5419/1.1^3

= 50094.60 (\$)

PV of costs of Machine B :

= 124845 + 7237 / 0.1

= 197215 (\$)

Therefore, as per CC, Machine A has lower cost. So Machine A should be selected.

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