Chapter 7, Problem 21P

To determine

Calculate the rate of return.

Explanation of Solution

Initial cost (C) is -$210,000. First year cost (C1) is -$150,000. Second year saving (S) is $100 and savings increases (SI) by 60,000 thereafter. Time period (n) is 4.

Rate of return (i) for infinitive time period can be calculated as follows:

$\begin{array}{c}-\text{C}=\frac{\text{C1}}{{\left(1+\text{i}\right)}^1}+\left(\text{S}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\text{SI}\frac{1}{\text{i}}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(1+\text{i}\right)}^{\text{n}}}\right)\right)\left(\frac{1}{{\left(1+\text{i}\right)}^{\text{n}}}\right)\\ \text{210,000}=\frac{-\text{150,000}}{{\left(1+\text{i}\right)}^1}+\left(\begin{array}{l}\text{100,000}\left(\frac{{\left(1+\text{i}\right)}^{\text{4}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{4}}}\right)\\ +\text{60,000}\times \frac{1}{\text{i}}\left(\frac{{\left(1+\text{i}\right)}^{\text{4}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{4}}}-\frac{\text{1}}{{\left(1+\text{i}\right)}^{\text{4}}}\right)\end{array}\right)\left(\frac{1}{{\left(1+\text{i}\right)}^{\text{1}}}\right)\end{array}$

Substitute the rate of return as 24% by trial and error method in the above Equation.

$\begin{array}{l}210,000=\frac{-\text{150,000}}{{\left(1+\text{0}\text{.24}\right)}^1}+\left(\begin{array}{l}\text{100,000}\left(\frac{{\left(1+\text{0}\text{.24}\right)}^{\text{4}}-1}{\text{0}\text{.24}{\left(1+\text{0}\text{.24}\right)}^{\text{4}}}\right)+\text{60,000}\\ \times \frac{1}{\text{0}\text{.24}}\left(\frac{{\left(1+\text{0}\text{.24}\right)}^{\text{4}}-1}{\text{0}\text{.24}{\left(1+\text{0}\text{.24}\right)}^{\text{4}}}-\frac{\text{4}}{{\left(1+\text{0}\text{.24}\right)}^{\text{4}}}\right)\end{array}\right)\left(\frac{\text{1}}{{\left(1+\text{0}\text{.24}\right)}^1}\right)\\ \text{210,000}=\frac{-\text{150,000}}{1.24}+\left(\begin{array}{l}\text{100,000}\left(\frac{2.364214-1}{\text{0}\text{.24}\left(2.364214\right)}\right)+\text{60,000}\\ \times \frac{1}{\text{0}\text{.24}}\left(\frac{2.364214-1}{\text{0}\text{.24}\left(2.364214\right)}-\frac{\text{4}}{2.364214}\right)\end{array}\right)\left(\frac{\text{1}}{1.24}\right)\end{array}$

$\begin{array}{l}\text{210,000}=-120,967.74+\left(\begin{array}{l}\text{100,000}\left(\frac{1.364214}{0.567144}\right)+\text{60,000}\\ \times 4.1667\left(\frac{1.364214}{0.567144}-\frac{\text{4}}{2.364214}\right)\end{array}\right)\left(0.806452\right)\\ \text{210,000}=-120,967.74+\left(\begin{array}{l}\text{100,000}\left(2.40541\right)+\text{60,000}\\ \times 4.1667\left(2.40541-1.69187\right)\end{array}\right)\left(0.806452\right)\\ \text{210,000}=-120,967.74+\left(240,541+\text{60,000}\times 4.1667\left(0.71354\right)\right)\left(0.806452\right)\\ \text{210,000}=-120,967.74+\left(240,541+178,386.43\right)\left(0.806452\right)\\ \text{210,000}=-120,967.74+337,877.86\\ \text{210,000}<216,877.12\end{array}$

When the rate of return is substituted as 24%, the calculated value is greater than the initial cost. Thus, the rate of return increases to 24.7%.

$\begin{array}{l}210,000=\frac{-\text{150,000}}{{\left(1+\text{0}\text{.247}\right)}^1}+\left(\begin{array}{l}\text{100,000}\left(\frac{{\left(1+\text{0}\text{.247}\right)}^{\text{4}}-1}{\text{0}\text{.247}{\left(1+\text{0}\text{.247}\right)}^{\text{4}}}\right)+\text{60,000}\\ \times \frac{1}{\text{0}\text{.247}}\left(\frac{{\left(1+\text{0}\text{.247}\right)}^{\text{4}}-1}{\text{0}\text{.247}{\left(1+\text{0}\text{.247}\right)}^{\text{4}}}-\frac{\text{4}}{{\left(1+\text{0}\text{.247}\right)}^{\text{4}}}\right)\end{array}\right)\left(\frac{\text{1}}{{\left(1+\text{0}\text{.247}\right)}^{\text{1}}}\right)\\ \text{210,000}=-120,288.69+\left(\begin{array}{l}\text{100,000}\left(\frac{2.418053-1}{\text{0}\text{.247}\left(2.418053\right)}\right)+\text{60,000}\\ \times \frac{1}{\text{0}\text{.247}}\left(\frac{2.418053-1}{\text{0}\text{.247}\left(2.418053\right)}-\frac{\text{4}}{2.418053}\right)\end{array}\right)\left(\frac{\text{1}}{1.247}\right)\\ \text{210,000}=-120,288.69+\left(\begin{array}{l}\text{100,000}\left(\frac{1.418053}{0.597259}\right)+\text{60,000}\\ \times 4.048583\left(\frac{1.418053}{0.597259}-\frac{\text{4}}{2.418053}\right)\end{array}\right)\left(0.801925\right)\end{array}$

$\begin{array}{l}210,000=-120,288.69+\left(\begin{array}{l}\text{100,000}\left(2.374268\right)+\text{60,000}\\ \times 4.048583\left(2.374268-1.654223\right)\end{array}\right)\left(0.801925\right)\\ 210,000=-120,288.69+\left(237,426.8+\text{60,000}\times 4.048583\left(0.720045\right)\right)\left(0.801925\right)\\ 210,000=-120,288.69+\left(237,426.8+174,909.72\right)\left(0.801925\right)\\ 210,000=-120,288.69+330,662.96\\ 210,000=210,374.27\end{array}$

The calculated value is nearly equal to the initial cost with rate of return 24.7%. Thus, it is confirmed that the rate of return is 24.7%.