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 26P
To determine

Calculate incremental rate of return.

Expert Solution & Answer
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Explanation of Solution

Table-1 shows the cash flow (Values are in 1,000 units).

Table -1

AlternativeTI
First cost (F)-5,000-6,500
AOC (AC) per year-1,000-650
Expected revenue (R) per year2,5002,500
Salvage value (SV)100200
Time period (n)55

MARR is 25%.

Incremental rate of return of alternatives T and Do nothing can be calculated as follows:

F=(R+AC)((1+i)n1i(1+i)n)+SV(1+i)n5,000=(2,500+(1,000))((1+i)51i(1+i)5)+100(1+i)55,000=1,500((1+i)51i(1+i)5)+100(1+i)5

Substitute the incremental rate of return as 15% by trial-and-error method in the above equation.

5,000=1,500((1+0.15)510.15(1+0.15)5)+100(1+0.15)55,000=1,500(2.01135710.15(2.011357))+1002.0113575,000=1,500(1.0113570.301704)+1002.0113575,000=1,500(3.35215)+49.725,000=5,028.23+49.725,000=5,077.95

The calculated value is greater than the present value of the incremental first cost. Thus, increase the incremental rate of return to 15.65%.

5,000=1,500((1+0.1565)510.1565(1+0.1565)5)+100(1+0.1565)55,000=1,500(2.06884610.1565(2.068846))+1002.0688465,000=1,500(1.0688460.323774)+1002.0688465,000=1,500(3.30121)+48.345,000=4,951.82+49.725,0005,001.54

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is ~15.65%. Since the incremental rate of return is less than MARR, reject the alternative T and select the do nothing.

Incremental rate of return of alternatives I and Do nothing can be calculated as follows:

F=(R+AC)((1+i)n1i(1+i)n)+SV(1+i)n6,500=(2,500+(650))((1+i)51i(1+i)5)+200(1+i)56,500=1,850((1+i)51i(1+i)5)+200(1+i)5

Substitute the incremental rate of return as 13% by trial-and-error method in the above equation.

6,500=1,850((1+0.13)510.13(1+0.13)5)+200(1+0.13)56,500=1,850(1.84243510.13(1.842435))+2001.8424356,500=1,850(0.8424350.239517)+2001.8424356,500=1,850(3.517224)+108.556,500=6,506.86+108.556,500<6,615.41

The calculated value is greater than the present value of the incremental first cost. Thus, increase the incremental rate of return to 13.72%.

6,500=1,850((1+0.1372)510.1372(1+0.1372)5)+200(1+0.1372)56,500=1,850(1.90188510.1372(1.901885))+2001.9018856,500=1,850(0.9018850.260939)+2001.9018856,500=1,850(3.456306)+105.166,500=6,394.17+108.556,5006,502.72

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is 13.72%. Since the incremental rate of return is less than MARR, reject the alternative I and select do nothing.

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