Question

The MARR value is 10%
                                    Machine A         Machine B
Initial Cost ($)    66,000    85,000
Annual MaintenanceCost ($)      12,000               13,000
Annual Income ($) 30,000               34,000
Salvage ($)                               6,000                6,000

Life (years)                           5                       5

What is the Capital Recovery for Machine A?

What is the Annual Worth of Machine A?

What is the Annual Worth of Machine B?

Which machine would you recommend?

Life (years) 5 5

What is the Capital Recovery for Machine A?

What is the Annual Worth of Machine A?

What is the Annual Worth of Machine B?

Answer 1

ANSWER

1)

Capital recovery of Machine A

CR = IC (A/P, r n) +SV (A/F, r n)

CR = -66,000(A/P, 10%, 5) + 6000(A/F,10%,5)

= -66,000 (0.2638) + 6,000 (0.1638)

= -17,410.8 + 982.8

= -$16,428

2)

Annual Worth of machine A

AWA = IC (A/P, r n) +SV (A/F, r n) – Annual maintenance and income

= -66,000(A/P, 10%, 5) + 6000(A/F,10%,5) – 42,000

= -66,000 (0.2638) + 6,000 (0.1638) – 42,000

= -17,410.8 + 982.8 - 42,000

= -$58,428

3)

Annual Worth of machine B

AWB = IC (A/P, r n) +SV (A/F, r n) – Annual maintenance and income

= -85,000(A/P, 10%, 5) + 6000(A/F,10%,5) – 47,000

= -85,000 (0.2638) + 6,000 (0.1638) - 47,000

= -22,423 + 982.8 - 47,000

= -$68,440.2

Which machine would you recommend?

Select machine A since AWB > AWB.

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