Engineering Economy
Engineering Economy
8th Edition
ISBN: 9780073523439
Author: Leland T Blank Professor Emeritus, Anthony Tarquin
Publisher: McGraw-Hill Education
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Chapter 13, Problem 41P

(a):

To determine

Selection of the alternative using the payback period analysis.

(a):

Expert Solution
Check Mark

Explanation of Solution

Alternate A: First cost (FC) is $300,000. Revenue (RE) is $60,000 per year.

Alternate B: First cost (FC) is $300,000. Revenue is $10,000 first year and increase (AI) by $15,000 per year.

MARR (i) is 8%.

Payback period (n) for Alternate A can be calculated as follows:

RE((1+i)n1i(1+i)n)=FC60,000((1+0.08)n10.08(1+0.08)n)=300,000((1+0.08)n10.08(1+0.08)n)=300,00060,000((1+0.08)n10.08(1+0.08)n)=5

From 8% interest table, it is known that the payback period is between the years 6 and 7 years.

Payback period (n) for Alternate B can be calculated as follows:

RE((1+i)n1i(1+i)n)+AI×1i((1+i)n1i(1+i)nn(1+i)n)=FC10,000((1+0.08)n10.08(1+0.08)n)+15,000×10.08((1+0.08)n10.08(1+0.08)nn(1+0.08)n)=300,000

Substitute n as 7 years by trial and error method to verify the payback period.

10,000((1+0.08)710.08(1+0.08)7)+15,000×10.08((1+0.08)710.08(1+0.08)77(1+0.08)7)=300,00010,000(5.20637)+15,000×12.5(5.206374.08443)=300,000262,427.45<300,000

Since the calculated value is less than the first cost, payback period should increase. Thus, the payback period for Alternate B is greater than 7 years. The payback period for alternate A is less than 7 years. Since the payback period for alternate A is less, select Alternate A.

(b):

To determine

Selection of the alternative using the present worth analysis.

(b):

Expert Solution
Check Mark

Explanation of Solution

Time period (n) is 10.

Present worth (PW) for Alternate A can be calculated as follows:

PW=FC+RE((1+i)n1i(1+i)n)=300,000+60,000((1+0.08)1010.08(1+0.08)n)=300,000+60,000(6.7101)=300,000+402,606=102,606

Present worth for Alternate A is $102,606.

Present worth (PW) for Alternate B can be calculated as follows:

PW=RE((1+i)n1i(1+i)n)+AI×1i((1+i)n1i(1+i)nn(1+i)n)FC=10,000((1+0.08)1010.08(1+0.08)10)+15,000×10.08((1+0.08)1010.08(1+0.08)1010(1+0.08)10)300,000=10,000(6.7101)+15,000×12.5(6.71014.511)300,000=67,101+389,652300,000=156,753

Present worth for Alternate B is $156,753. Since the present worth of Alternate B is greater, select Alternate B.

(c):

To determine

Selection of the alternative using the spreadsheet.

(c):

Expert Solution
Check Mark

Explanation of Solution

Payback period and present worth of the alternate can be calculated using the spreadsheet as follows:

Engineering Economy, Chapter 13, Problem 41P

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