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The Zed Company is considering the purchase of a new machine. The new machine will have zero salvage value and a useful life

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Answer #1

1)

In case of machine A

AW of savings =AW(B)=$700

AW of O&M Costs=AW(OM)=$300

Annualized initial Cost=AW(Co)=2000*(A/P,0.03,7)

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

(A/P, 0.03,7) 03.70.03 0.160506

Annualized initial Cost=2000*0.160506=$321.01

AW (B) 700 B - CRatio 1.1272 1.13 AW (Co)+AW (OM) 321.01300

2)

In case of machine B

AW of savings =AW(B)=$700

AW of O&M Costs=AW(OM)=$400

Annualized initial Cost=AW(Co)=3000*(A/P,0.03,7)=3000*0.160506=481.52

AW (B) 700 -0.7941 ะ 0.79 AW (Co AW (OM) 481.52400

3)

In case of machine C

AW of savings =AW(B)=$3000

AW of O&M Costs=AW(OM)=$700

Annualized initial Cost=AW(Co)=8000*(A/P,0.03,7)=8000*0.160506=1284.05

AW (B) 30001.51211.51 B-CRatio AW (CoAW (OM) 1284.05700

4)

B-C ratio is less than 1 for machine B. So, Machine B should be disregarded from further analysis.

5)

Machine A and Machine C are left for further analysis.

Machine A has a lower initial cost, we would take machine A as base project.

Incremental Initial Cost=Co=8000-2000=$6000

AW of incremental savings =AW(B)=3000-700=$2300

AW of incremental O&M Costs=AW(OM)=700-300=$400

Annualized incremental initial Cost=AW(Co)=6000*(A/P,0.03,7)=6000*0.160506=963.04

AW (B) 2300 B - C Ratio forA(C- A) AW (Co)+AW (OM) 963.04 400

B-C Ratio for▲(C-A) 1.6874 ~ 1.69

6)

Incremental B-C is higher than 1 for incremental analysis (C-A). Machine C should be selected.

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