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 24P

(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

Alternative12
First cost (F)-15,000-25,000
AOC (AC) per year-1,600-400
Salvage value (SV)3,0004,000
Time period (n)24

MARR is 20%.

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.

(F2F1)=(AC2AC1)((1+i)n21i(1+i)n2)(SV1+F1)(1+i)n1+SV2SV1(1+i)n2(25,000(15,000))=(400(1,600))((1+i)41i(1+i)4)(3,000+(15,000))(1+i)2+(4,0003,000)(1+i)n210,000=1,200((1+i)41i(1+i)4)+12,000(1+i)2+1,000(1+i)4

Substitute the incremental rate of return as 30% by trial-and-error method in the above equation.

10,000=1,200((1+0.3)410.3(1+0.3)4)+12,000(1+0.3)2+1,000(1+0.3)410,000=1,200(2.856110.3(2.8561))+12,0001.69+1,0002.856110,000=1,200(1.85610.85683)+7,100.59+350.1310,000=1,200(2.1662)+7,100.59+350.1310,000=2,599.44+7,100.59+350.1310,000<10,050.16

The calculated value is greater than the present value of the incremental first cost. Thus, increase the incremental rate of return to 30.31%.

10,000=1,200((1+0.3031)410.3031(1+0.3031)4)+12,000(1+0.3031)2+1,000(1+0.3031)410,000=1,200(2.883410.3031(2.8834))+12,0001.6981+1,0002.883410,000=1,200(2.1549)+7,066.72+346.8110,000=2,585.88+7,066.72+346.8110,000=9,999.41

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is 30.31%. Since the incremental rate of return is greater than MARR, select the alternative 2.

(b):

To determine

Calculate incremental rate of return.

(b):

Expert Solution
Check Mark

Explanation of Solution

Alternative 2 has 4 years. Thus, time period (n) is 4. Incremental rate of return of alternatives Y and X can be calculated as follows:

(F2F1)=(AC2AC1)((1+i)n1i(1+i)n)+SV2SV1(1+i)n(25,000(15,000))=(400(1,600))((1+i)41i(1+i)4)+(4,0003,000)(1+i)n210,000=1,200((1+i)41i(1+i)4)+1,000(1+i)4

Substitute the incremental rate of return as -17% by trial-and-error method in the above equation.

10,000=1,200((10.17)410.17(10.17)4)+1,000(10.17)410,000=1,200(0.47458310.17(0.474583))+1,0000.47458310,000=1,200(0.5254170.080679)+2,107.1110,000=1,200(6.5124)+2,107.1110,000=7,814.88+2,107.1110,000>9,921.99

The calculated value is less than the present value of the incremental first cost. Thus, decrease the incremental rate of return to -17.22%.

10,000=1,200((10.1722)410.1722(10.1722)4)+1,000(10.1722)410,000=1,200(0.46957110.1722(0.469571))+1,0000.46957110,000=1,200(0.5304290.08086)+1,0000.46957110,000=1,200(6.5598)+2,129.610,000=7,871.66+2,107.1110,0009,978.87

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is -17.22%. Since the incremental rate of return is negative, select the initial alternative 1.

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