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, $ | 21,302 | 30000 | 10000 |
Annual cost, $/year | 9,993 | 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=
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 | 26,087 | 10000 |
Annual cost, $/year | 9,164 | 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=
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 | 30000 | 13,830 |
Annual cost, $/year | 18,135 | 6,000 | 4,000 |
Salvage value, $ | 4,000 | 5,000 | 1,000 |
Life, years | 3 | 6 | 2 |
Answer the below questions :
C- AW for machine C =
D- Based on the AW value you got in the previous 3 questions, which machine we should select? type you explanation below
A.
Machine A
Annual worth = 21302(A/P,10,3) + 9993 – 4000(A/F,10,3)
Using DCIF Tables
Annual worth = 21302(0.4021) + 9993 – 4000(0.3021)
Annual worth = $17350.13B.
Machine B
Annual worth = 26087(A/P,10,6) + 6000 – 5000(A/F,10,6)
Using DCIF Tables
Annual worth = 26087(0.2296) + 6000 – 5000(0.1296)
Annual worth = $11341.6
C.
Machine C
Annual worth = 13830(A/P,10,2) + 4000 – 1000(A/F,10,2)
Using DCIF Tables
Annual worth = 13830(0.5762) + 4000 – 1000(0.4762)
Annual worth = $11492.7
D.Based on AW worth Machine B is to be selected as it has minimum Annual worth
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