Question 2 Compare the machines shown below on the basis of their capitalized cost. Use 10% per year Machine 2 Machine 1 20,000 9000 4000 First cost,S Annual cost,S/year Salvage value, S Life, years...

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Question 2 Compare the machines shown below on the basis of their capitalized cost. Use 10% per year Machine 2 Machine 1 20,0

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Answer 1

Answer #1

Case of Machine 1

Initial Cost=Co=-20000

Annual Cost=R=-9000

Salvage =S=$4000

Useful life=n=3 year

AW=-Co*(A/P,10%,3)-R+S*(A/F,10%,3)

AW=-20000*(A/P,10%,3)-9000+4000*(A/F,10%,3)

Let us calculate interest factors

(A/P, i, n) = 1-

0.10 (A/P. 0.10.3) = 0.402 1+0.10

(A/F,i,n)-+in -1

0.10 (A/F, 0.10.3) = (1 +0.10)3-1 0.302 (1+0.10)3

AW=-20000*402-9000+4000*0.302=-$15832

Capitalized cost=AW/i=15832/10%=$158320

Now let us consider the case of machine 2

Initial Cost=Co=-100000

Annual Cost=R=-7000

Useful life=n=infinite

PW of annual Costs=PWC=R/i=-7000/10%=-$70000

Capitalized Cost=-Co+PWC=-100000-70000=-$170000

Absolute value of capitalized cost is lower in case of Machine 1

So, Machine 1 should be selected.

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Question 2 Compare the machines shown below on the basis of their capitalized cost. Use 10% per year Machine 2 Machine 1 20,000 9000 4000 First cost,S Annual cost,S/year Salvage value, S Life, years...