Chapter 8, Problem 22P

To determine

Calculate the incremental rate of return.

Explanation of Solution

Table-1 shows the cash flow.

Table -1

Alternate	V	D
First cost (C)	-250,000	-225,000
Annual cost (O)per year	-231,000	-235,000
Overhaul cost (O3) year 3	0	-26,000
Overhaul cost (O4) year 4	-39,000	0
Salvage value (SV)	50,000	10,000
Life (n)	6	6

MARR is 15%.

Incremental rate of return can be calculated as follows:

$\begin{array}{c}-\left(\text{CV}-\text{CD}\right)=\left(\begin{array}{l}\left(\text{OV}-\text{OD}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\left(\text{O3V}-\text{O3D}\right)}{{\left(1+\text{i}\right)}^{\text{3}}}\\ +\frac{\left(\text{O4V}-\text{O4D}\right)}{{\left(1+\text{i}\right)}^4}+\frac{\left(\text{SVV}-\text{SVD}\right)}{{\left(1+\text{i}\right)}^{\text{n}}}\end{array}\right)\\ -\left(\begin{array}{l}-250,000\\ -\left(-225,000\right)\end{array}\right)=\left(\begin{array}{l}\left(\begin{array}{l}-\text{231,000}\\ -\left(-235,000\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{6}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{6}}}\right)+\frac{\left(\text{0}-\left(-26,000\right)\right)}{{\left(1+\text{i}\right)}^{\text{3}}}\\ +\frac{\left(-\text{39,000}-\text{0}\right)}{{\left(1+\text{i}\right)}^4}+\frac{\left(\text{50,000}-\text{10,000}\right)}{{\left(1+\text{i}\right)}^6}\end{array}\right)\\ 25,000=\left(4,000\left(\frac{{\left(1+\text{i}\right)}^{\text{6}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{6}}}\right)+\frac{26,000}{{\left(1+\text{i}\right)}^{\text{3}}}-\frac{39,000}{{\left(1+\text{i}\right)}^4}+\frac{40,000}{{\left(1+\text{i}\right)}^6}\right)\end{array}$

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

$\begin{array}{l}25,000=\left(4,000\left(\frac{{\left(1+\text{0}\text{.17}\right)}^{\text{6}}-1}{\text{0}\text{.17}{\left(1+\text{0}\text{.17}\right)}^{\text{6}}}\right)+\frac{26,000}{{\left(1+\text{0}\text{.17}\right)}^{\text{3}}}-\frac{39,000}{{\left(1+\text{0}\text{.17}\right)}^4}+\frac{40,000}{{\left(1+\text{0}\text{.17}\right)}^6}\right)\\ 25,000=\left(4,000\left(\frac{2.5652-1}{\text{0}\text{.17}\left(2.5652\right)}\right)+\frac{26,000}{1.6016}-\frac{39,000}{1.8739}+\frac{40,000}{2.5652}\right)\\ 25,000=\left(4,000\left(\frac{1.5652}{0.4361}\right)+\frac{26,000}{1.6016}-\frac{39,000}{1.8739}+\frac{40,000}{2.5652}\right)\\ 25,000=\left(4,000\left(3.5891\right)+16,233.77-20,812.21+15,593.33\right)\\ 25,000=\left(14,356.4+16,233.77-20,812.21+15,593.33\right)\\ 25,000=25,371.29\end{array}$

The calculated value is greater than the present value factor. Thus, increase the incremental rate of return to 17.43%.

$\begin{array}{l}25,000=\left(\begin{array}{l}4,000\left(\frac{{\left(1+\text{0}\text{.1743}\right)}^{\text{6}}-1}{\text{0}\text{.1743}{\left(1+\text{0}\text{.1743}\right)}^{\text{6}}}\right)+\frac{26,000}{{\left(1+\text{0}\text{.1743}\right)}^{\text{3}}}\\ -\frac{39,000}{{\left(1+\text{0}\text{.1743}\right)}^4}+\frac{40,000}{{\left(1+\text{0}\text{.1743}\right)}^6}\end{array}\right)\\ 25,000=\left(4,000\left(\frac{2.6223-1}{\text{0}\text{.1743}\left(2.6223\right)}\right)+\frac{26,000}{1.6193}-\frac{39,000}{1.9016}+\frac{40,000}{2.6223}\right)\\ 25,000=\left(4,000\left(\frac{1.6223}{0.4571}\right)+16,056.32-20,509.05+15,253.78\right)\\ 25,000=\left(4,000\left(3.5491\right)+16,056.32-20,509.05+15,253.78\right)\\ 25,000=\left(14,196.4+16,056.32-20,509.05+15,253.78\right)\\ 25,000=24,997.45\end{array}$

The calculated value is nearly equal to the incremental present value factor. Thus, it is confirmed that the incremental rate of return is 17.43%.

The incremental rate of return from alternative V and alternative D is greater than MARR. Thus, the firm should select the new alternative V.