Chapter 12, Problem 29P

To determine

Calculate the internal rate of return.

Explanation of Solution

The first cost (FC) is $400,000. Annual operating cost (AC) is $75,000. Annual savings (AS) is $192,000. Salvage value (SV) is 20% of the first cost. Profitability index (i) is 12%. Time period (n) is 5 years.

The internal rate of return (ir) can be calculated using the following formula:

$\begin{array}{c}\text{FC}=\left(\text{AS}-\text{AC}\right)\left(\frac{{\left(1+\text{ir}\right)}^{\text{n}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{n}}}\right)+\frac{\text{FC}\times \text{SV}}{{\left(1+\text{ir}\right)}^{\text{n}}}\\ 400,000=\left(192,000-75,000\right)\left(\frac{{\left(1+\text{ir}\right)}^{\text{5}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{5}}}\right)+\frac{\text{400,000}\times \text{0}\text{.2}}{{\left(1+\text{ir}\right)}^{\text{5}}}\\ 400,000=\left(117,000\right)\left(\frac{{\left(1+\text{ir}\right)}^{\text{5}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{5}}}\right)+\frac{80,000}{{\left(1+\text{ir}\right)}^{\text{5}}}\end{array}$

Substitute ir as 0.12 by trial and error method in the above calculation.   

$\begin{array}{l}400,000=\left(117,000\right)\left(\frac{{\left(1+\text{0}\text{.12}\right)}^{\text{5}}-1}{\text{0}\text{.12}{\left(1+0.12\right)}^{\text{5}}}\right)+\frac{80,000}{{\left(1+\text{0}\text{.12}\right)}^{\text{5}}}\\ 400,000=\left(117,000\right)\left(3.6048\right)+\frac{80,000}{1.7623}\\ 400,000=421,761.6+45,395.22\\ 400,000<467,156.82\end{array}$

The calculated value is less than the first cost. Thus, increase the ir value to 0.1809.

$\begin{array}{l}400,000=\left(117,000\right)\left(\frac{{\left(1+\text{0}\text{.1809}\right)}^{\text{5}}-1}{\text{0}\text{.1809}{\left(1+0.1809\right)}^{\text{5}}}\right)+\frac{80,000}{{\left(1+\text{0}\text{.1809}\right)}^{\text{5}}}\\ 400,000=\left(117,000\right)\left(3.1211\right)+\frac{80,000}{2.2965}\\ 400,000=365,168.7+34,835.62\\ 400,000\approx 400,00.32\end{array}$

Since the calculated value is equal to the first cost, the value of ir is confirmed as 18.09%.

The value PI can be calculated as follows:

$\begin{array}{c}\text{PI}=\frac{\left(\text{AS}-\text{AC}\right)\left(\frac{{\left(1+\text{ir}\right)}^{\text{n}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{n}}}\right)+\frac{\text{FC}\times \text{SV}}{{\left(1+\text{ir}\right)}^{\text{n}}}}{\text{FC}}\\ =\frac{\left(117,000\right)\left(\frac{{\left(1+\text{0}\text{.12}\right)}^{\text{5}}-1}{\text{0}\text{.12}{\left(1+0.12\right)}^{\text{5}}}\right)+\frac{80,000}{{\left(1+\text{0}\text{.12}\right)}^{\text{5}}}}{400,000}\\ \frac{=\left(117,000\right)\left(3.6048\right)+\frac{80,000}{1.7623}}{400,000}\\ =\frac{421,761.6+45,395.22}{400,000}\\ =\frac{467,156.82}{400,000}\\ =1.17\end{array}$

The value of PI is 1.17.

The value PW can be calculated as follows:

$\begin{array}{c}\text{PW}=-\text{FC}+\left(\text{AS}-\text{AC}\right)\left(\frac{{\left(1+\text{ir}\right)}^{\text{n}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{n}}}\right)+\frac{\text{FC}\times \text{SV}}{{\left(1+\text{ir}\right)}^{\text{n}}}\\ =-400,000+\left(117,000\right)\left(\frac{{\left(1+\text{0}\text{.12}\right)}^{\text{5}}-1}{\text{0}\text{.12}{\left(1+0.12\right)}^{\text{5}}}\right)+\frac{80,000}{{\left(1+\text{0}\text{.12}\right)}^{\text{5}}}\\ =-400,000+\left(117,000\right)\left(3.6048\right)+\frac{80,000}{1.7623}\\ =-400,000+421,761.6+45,395.22\\ =67,156.82\end{array}$

The present wroth (PW) is $67,156.82.