QUESTION 3
For the below ME alternatives , which machine should be selected based on the PW analysis. MARR=10%
Machine A | Machine B | Machine C | |
First cost, $ | 16,122 | 30000 | 10000 |
Annual cost, $/year | 8,601 | 6,000 | 4,000 |
Salvage value, $ | 4,000 | 5,000 | 1,000 |
Life, years | 3 | 6 | 2 |
Answer the below questions :
A- PW for machine A=
QUESTION 4
For the below ME alternatives , which machine should be selected based on the PW analysis. MARR=10%
Machine A | Machine B | Machine C | |
First cost, $ | 15000 | 25,846 | 10000 |
Annual cost, $/year | 9,857 | 6,000 | 4,000 |
Salvage value, $ | 4,000 | 5,000 | 1,000 |
Life, years | 3 | 6 | 2 |
Answer the below questions :
B- PW for machine B=
QUESTION 5
For the below ME alternatives, which machine should be selected based on the PW analysis. MARR=10%.
Machine A | Machine B | Machine C | |
First cost, $ | 15000 | 30000 | 12,631 |
Annual cost, $/year | 18,241 | 6,000 | 4,000 |
Salvage value, $ | 4,000 | 5,000 | 1,000 |
Life, years | 3 | 6 | 2 |
Answer the below questions :
C- PW for machine C =
Present worth = -P+(AR-Am)(
P/A,i%,n)-+F(P/F,i%,n)
where, P is the initial cost, AR is the annual revenue,
Am is the annual maintenance cost and F is the salvage
value
3) Present worth of machine A =
-16,122+(0-8,601)(P/A,10%,3)+4,000(P/F,10%,3)
= -16,122+(0-8,601)(2.487)+4,000(.7513) = $-34,507.49
4) Present worth of machine B =
-25,846+(0-6,000)(P/A,10%,6)+5,000(P/F,10%,6)
= -25,846+(0-6,000)(4.355)+5,000(.5645)= $-49,153.5
5) Present worth of machine C =
-12,631+(0-4,000)(P/A,10%,2)+1,000(P/F,10%,2)
= -12,631+(0-4,000)(1.736)+1,000(.8264)= $-18,748.6
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