Question

QUESTION 3

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

	Machine A	Machine B	Machine C
First cost, $	26,538	30000	10000
Annual cost, $/year	8,060	6,000	4,000
Salvage value, $	4,000	5,000	1,000
Life, years	3	6	2

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 A	Machine B	Machine C
First cost, $	15000	21,667	10000
Annual cost, $/year	8,870	6,000	4,000
Salvage value, $	4,000	5,000	1,000
Life, years	3	6	2

Answer the below questions:

B- AW for machine B=

Answer 1

3)

Annual worth = Net present worth $\times$ Capital recovery factor

8060 8060 8060 4000 Net present worth machine A = 26538 + (1 + 0.10)1' (1 +0.10)' (1 +0.10) 3 (1 + 0.103 + +

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) (1 +0.10) 5 (1 +0.10) 5 (1 + 0.106

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

Net present worth machine C 4000 4000 1000 10000+ + (1 +0.10)1' (1 +0.10) 2 (1 +0.10)

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

8870 8870 8870 Net present worth machine A = 15000 + 4000 + + (1 + 0.101 '(1 + 0.102' (1 + 0.10) 3 (1 + 0.103

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

6000 6000 6000 6000 6000 6000 5000 Net present worth machine B = 21667 + + + + + + (1+0.101'(1 + 0.10)'(1 + 0.10) 3 (1+0.10)' (1 + 0.10) 5 (1 + 0.10)6 (1 + 0.10)

Net present worth of machine B = $ 44976.2

Annual worth machine B = $ 44976.2 $\times$ 0.2296

Annual worth machine B = $ 10326.53

Net present worth machine C 4000 4000 1000 10000+ + (1 +0.10)1' (1 +0.10) 2 (1 +0.10)

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.

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)