Engineering Economy
Engineering Economy
8th Edition
ISBN: 9780073523439
Author: Leland T Blank Professor Emeritus, Anthony Tarquin
Publisher: McGraw-Hill Education
Question
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Chapter 8, Problem 25P

(a):

To determine

Calculate the incremental rate of return.

(a):

Expert Solution
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Explanation of Solution

Table-1 shows the cash flow.

Table -1

AlternativePQ
First cost (F)-18,000-35,000
AOC (AC) per year-4,000-3,600
Salvage value (SV)1,0002,700
Time period (n)36

MARR is 10%.

Incremental rate of return of alternatives Y and X can be calculated as follows: Since alternative 1 has 2-year life time, which is less than alternative 2, its cash flow has to be repeated for two years.

(FQFP)=(ACQACP)((1+i)nQ1i(1+i)nQ)(SVP+FP)(1+i)nP+SVQSVP(1+i)nQ(35,000(18,000))=(3,600(4,000))((1+i)61i(1+i)6)(1,000+(18,000))(1+i)3+(2,7001,000)(1+i)617,000=400((1+i)61i(1+i)6)+17,000(1+i)3+1,700(1+i)6

Substitute the incremental rate of return as 6% by trial-and-error method in the above equation.

17,000=400((1+0.06)610.06(1+0.06)6)+17,000(1+0.06)3+1,700(1+0.06)617,000=400(1.418510.06(1.4185))+17,0001.191+1,700(1.4185)17,000=400(0.41850.0851)+17,0001.191+1,700(1.4185)17,000=400(4.9177)+14,273.72+1,198.4517,000=1,967.08+14,273.72+1,198.4517,000<17,439.25

The calculated value is greater than the present value of the incremental first cost. Thus, increase the incremental rate of return to 6.84%.

17,000=400((1+0.0684)610.0684(1+0.0684)6)+17,000(1+0.0684)3+1,700(1+0.0684)617,000=400(1.487310.0684(1.4873))+17,0001.2196+1,700(1.4873)17,000=400(0.48730.1017)+17,0001.2196+1,700(1.4873)17,000=400(4.7915)+13,938.99+1,143.0117,000=1,916.6+13,938.99+1,143.0117,00016,998.6

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is 0.0684%. Since the incremental rate of return is less than MARR, select the alternative P (Initial alternative).

(b):

To determine

Calculate the annual worth.

(b):

Expert Solution
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Explanation of Solution

Annual worth of project P (AW) can be calculated as follows:

AW=F(i(1+i)n(1+i)n1)+AC+SV(i(1+i)n1)=18,000(0.1(1+0.1)3(1+0.1)31)4,000+1,000(0.1(1+0.1)31)=18,000(0.1(1.331)1.3311)4,000+1,000(0.11.3311)=18,000(0.13310.331)4,000+1,000(0.10.331)=18,000(0.402115)4,000+1,000(0.302115)=7,238.074,000+302.12=10,935.96

Annual worth of alternative P is -$10,935.96.

Annual worth of project Q (AW) can be calculated as follows:

AW=F(i(1+i)n(1+i)n1)+AC+SV(i(1+i)n1)=35,000(0.1(1+0.1)6(1+0.1)61)3,600+2,700(0.1(1+0.1)61)=35,000(0.1(1.771561)1.7715611)3,600+2,700(0.11.7715611)=35,000(0.1771560.771561)3,600+2,700(0.10.771561)=35,000(0.229607)3,600+2,700(0.129607)=8,036.253,600+349.94=11,286.31

Annual worth of alternative Q is -$11,286.31. Since the annual worth of alternative P is greater than alternate Q, select alternate P.

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