Chapter 8, Problem 24P

(a):

To determine

Calculate the incremental rate of return.

Explanation of Solution

Table-1 shows the cash flow.

Table -1

Alternative	1	2
First cost (F)	-15,000	-25,000
AOC (AC) per year	-1,600	-400
Salvage value (SV)	3,000	4,000
Time period (n)	2	4

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.

$\begin{array}{c}-\left(\text{F2}-\text{F1}\right)=\left(\begin{array}{l}\text{AC2}\\ -\text{AC1}\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n2}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n2}}}\right)-\frac{\left(\text{SV1}+\text{F1}\right)}{{\left(1+\text{i}\right)}^{\text{n1}}}+\frac{\text{SV2}-\text{SV1}}{{\left(1+\text{i}\right)}^{\text{n2}}}\\ -\left(\begin{array}{l}-25,000\\ -\left(-15,000\right)\end{array}\right)=\left(\begin{array}{l}-\text{400}\\ -\left(-1,600\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{4}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{4}}}\right)-\frac{\left(\begin{array}{l}\text{3,000}\\ +\left(-15,000\right)\end{array}\right)}{{\left(1+\text{i}\right)}^{\text{2}}}+\frac{\left(\begin{array}{l}\text{4,000}\\ -\text{3,000}\end{array}\right)}{{\left(1+\text{i}\right)}^{\text{n2}}}\\ 10,000=1,200\left(\frac{{\left(1+\text{i}\right)}^{\text{4}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{4}}}\right)+\frac{12,000}{{\left(1+\text{i}\right)}^{\text{2}}}+\frac{1,000}{{\left(1+\text{i}\right)}^{\text{4}}}\end{array}$

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

$\begin{array}{l}10,000=1,200\left(\frac{{\left(1+\text{0}\text{.3}\right)}^{\text{4}}-1}{\text{0}\text{.3}{\left(1+\text{0}\text{.3}\right)}^{\text{4}}}\right)+\frac{12,000}{{\left(1+\text{0}\text{.3}\right)}^{\text{2}}}+\frac{1,000}{{\left(1+\text{0}\text{.3}\right)}^{\text{4}}}\\ 10,000=1,200\left(\frac{2.8561-1}{\text{0}\text{.3}\left(2.8561\right)}\right)+\frac{12,000}{1.69}+\frac{1,000}{2.8561}\\ 10,000=1,200\left(\frac{1.8561}{0.85683}\right)+7,100.59+350.13\\ 10,000=1,200\left(2.1662\right)+7,100.59+350.13\\ 10,000=2,599.44+7,100.59+350.13\\ 10,000<10,050.16\end{array}$

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

$\begin{array}{l}10,000=1,200\left(\frac{{\left(1+\text{0}\text{.3031}\right)}^{\text{4}}-1}{\text{0}\text{.3031}{\left(1+\text{0}\text{.3031}\right)}^{\text{4}}}\right)+\frac{12,000}{{\left(1+\text{0}\text{.3031}\right)}^{\text{2}}}+\frac{1,000}{{\left(1+\text{0}\text{.3031}\right)}^{\text{4}}}\\ 10,000=1,200\left(\frac{2.8834-1}{\text{0}\text{.3031}\left(2.8834\right)}\right)+\frac{12,000}{1.6981}+\frac{1,000}{2.8834}\\ 10,000=1,200\left(2.1549\right)+7,066.72+346.81\\ 10,000=2,585.88+7,066.72+346.81\\ 10,000=9,999.41\end{array}$

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.

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:

$\begin{array}{c}-\left(\text{F2}-\text{F1}\right)=\left(\begin{array}{l}\text{AC2}\\ -\text{AC1}\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\text{SV2}-\text{SV1}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ -\left(\begin{array}{l}-25,000\\ -\left(-15,000\right)\end{array}\right)=\left(\begin{array}{l}-\text{400}\\ -\left(-1,600\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{4}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{4}}}\right)+\frac{\left(\text{4,000}-\text{3,000}\right)}{{\left(1+\text{i}\right)}^{\text{n2}}}\\ 10,000=1,200\left(\frac{{\left(1+\text{i}\right)}^{\text{4}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{4}}}\right)+\frac{1,000}{{\left(1+\text{i}\right)}^{\text{4}}}\end{array}$

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

$\begin{array}{l}10,000=1,200\left(\frac{{\left(1-0.17\right)}^{\text{4}}-1}{-\text{0}\text{.17}{\left(1-0.17\right)}^{\text{4}}}\right)+\frac{1,000}{{\left(1-0.17\right)}^{\text{4}}}\\ 10,000=1,200\left(\frac{0.474583-1}{-\text{0}\text{.17}\left(0.474583\right)}\right)+\frac{1,000}{0.474583}\\ 10,000=1,200\left(\frac{-0.525417}{-0.080679}\right)+2,107.11\\ 10,000=1,200\left(6.5124\right)+2,107.11\\ 10,000=7,814.88+2,107.11\\ 10,000>9,921.99\end{array}$

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

$\begin{array}{l}10,000=1,200\left(\frac{{\left(1-0.1722\right)}^{\text{4}}-1}{-\text{0}\text{.1722}{\left(1-0.1722\right)}^{\text{4}}}\right)+\frac{1,000}{{\left(1-0.1722\right)}^{\text{4}}}\\ 10,000=1,200\left(\frac{0.469571-1}{-\text{0}\text{.1722}\left(0.469571\right)}\right)+\frac{1,000}{0.469571}\\ 10,000=1,200\left(\frac{-0.530429}{-0.08086}\right)+\frac{1,000}{0.469571}\\ 10,000=1,200\left(6.5598\right)+2,129.6\\ 10,000=7,871.66+2,107.11\\ 10,000\approx 9,978.87\end{array}$

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.