Problem 6-39 (algorithmic) Question Help The cash flows in the table below repre
ID: 1142151 • Letter: P
Question
Problem 6-39 (algorithmic) Question Help The cash flows in the table below represent the potential annual savings associated with two different types of production processes, each of which requires an investment of $31,000. Assume an interest rate of 9% Click the icon to view the data for cash flows Click the icon to view the interest factors for discrete compounding when i-9% per year (a) Determine the equivalent annua More Info The equivalent annual savings for Process A $31,000 $18,490 $16,070 $13,650 $11,230 ProcessB $31,000 $15,600 $15,600 $15,600 $15,600 2 4 Print Done Enter your answer in the answer parts remaining Clear All Check AnswerExplanation / Answer
a)
The equivalent annual saving for process A are $5,549
The equivalent annual saving for process B are $6,030
Explanation:
Process A:
Annual savings decreases by 18490 - 16,070 = 2,420 each subsequent year
The equivalent annual saving = -31,000(A/P, 9%, 4) + [18,490 - 2,420(A/G, 9%, 4)]
= -31,000(0.3087) + [18,490 - 2,420(1.393)]
= -9,569.7 + 15,118.94
= $5,549.24
We can round it as $5,549.
Process B:
The equivalent annual saving = -31,000(A/P, 9%, 4) + 15,600
= -31,000(0.3087) + 15,600
= -9,569.7 + 15,600
= $6,030.3
We can round it as $6,030.
b) The hourly saving for process A are $5,549 / 3,000 = $1.85
The hourly saving for process B are $6,030 / 3,000 = $2.01
c) Process B should be selected. Because annual savings as well as hourly savings of process B is greater than process B.
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