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For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%

QUESTION 3

For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%



Machine AMachine BMachine C
First cost, $26,5383000010000
Annual cost, $/year8,0606,0004,000
Salvage value, $4,0005,0001,000
Life, years362

Answer the below questions:

A- AW for machine A=


QUESTION 4

For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%



Machine AMachine BMachine C

First cost, $1500021,66710000

Annual cost, $/year8,8706,0004,000

Salvage value, $4,0005,0001,000

Life, years362

Answer the below questions:

B- AW for machine B=

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Answer #1

3)

Annual worth = Net present worth \times Capital recovery factor

1592306880393_image.png

Net present worth of machine A = $ 43576.76784

Annual worth of machine A = $ 43576.76784 \times ( A/P, 10%, 3 years)

( A/P, 10%, 3 years) = Capital recovery factor

Annual worth of machine A = $ 43576.76784 \times 0.4021

Annual worth of machine A = $ 17522.22

+ 6000 6000 6000 6000 6000 6000 5000 Net present worth machine B = 30000 + + + + (1 +0.10)1 (1 +0.102 (1 +0.10) 3 (1 +0.10)

Net present worth machine B = $ 53309.2

Annual worth of machine B = $ 53309.2 \times ( A/P, 10%, 6 years)

Annual worth of machine B = $ 53309.2 \times 0.2296

Annual worth of machine B = $ 12239.79

1592307716177_image.png

Net present worth of machine C = $ 16115.7

Annual worth of machine C =   $ 16115.7 \times ( A/P, 10%, 2 years)

Annual worth of machine C =   $ 16115.7 \times 0.5762

Annual worth of machine C = $ 9285.87

Based on the annual worth analysis, machine C must be selected because it has the lowest annual worth of costs compared to machine A and machine B.

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4)

Annual worth = Net present worth \times Capital recovery factor

1592308246665_image.png

Net present worth of machine A = $ 34053.12

Annual worth = $ 34053.12  \times ( A/P, 10%, 3 years)

Annual worth machine A =   $ 34053.12  \times 0.4021 = $ 13692.76

1592308535887_image.png

Net present worth of machine B = $ 44976.2

Annual worth machine B = $ 44976.2 \times 0.2296

Annual worth machine B = $ 10326.53

1592307716177_image.png

Annual worth of machine C =   $ 16115.7 \times ( A/P, 10%, 2 years)

Annual worth of machine C =   $ 16115.7 \times 0.5762

Annual worth of machine C = $ 9285.87

Based on the annual worth analysis, machine C must be selected because it has the lowest annual worth of costs compared to machine A and machine B.

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Answer #2

SOLUTION :


Q4 :


r = MARR = 10% = 0.1

=> (1 + r) = 1.1


PV of costs of Machine A :


= 26538 + 8060/ 1.1 + 8060/1.1^2 + 8060/1.1^3 - 4000/1.1^3 

= 43576.77 ($) 


PV of costs of Machine B :


= 30000 + 6000 (1.1^6 - 1)/(0.1*1.1^6) - 5000/1.1^6

= 53309.19 ($) 


PV of costs of Machine C :


= 10000 + 4000/1.1 + 4000/1.1^2 - 1000/1.1^2

= 16115.70 ($)


Therefore, as per PV of costs , Machine C has lowest cost. So Machine C should be selected.

(ANSWER)



Q3 :


r = MARR = 10% = 0.1

=> (1 + r) = 1.1


PV of costs of Machine A :


= 15000 + 8870/ 1.1 + 8870/1.1^2 + 8870/1.1^3 - 4000/1.1^3 

= 34053.12 ($) 


PV of costs of Machine B :


= 21667 + 6000 (1.1^6 - 1)/(0.1*1.1^6) - 5000/1.1^6

= 44976.19 ($) 


PV of costs of Machine C :


= 10000 + 4000/1.1 + 4000/1.1^2 - 1000/1.1^2

= 16115.70 ($)


Therefore, as per PV of costs , Machine C has lowest cost. So Machine C should be selected.

(ANSWER)

answered by: Tulsiram Garg
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