Chapter 7, Problem 12P

(a)

To determine

Calculate the rate of return.

Explanation of Solution

Investment is (I) is $650,000. Equivalent annual revenue (A) is $105,000. Salvage value (SV) is 50,000. Time period (n) is 5.

Rate of return (i) can be calculated as follows:

$\begin{array}{c}\text{I}=\text{A}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\text{SV}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ 650,000=105,000\left(\frac{{\left(1+\text{i}\right)}^{\text{5}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{5}}}\right)+\frac{\text{50,000}}{{\left(1+\text{i}\right)}^{\text{5}}}\end{array}$

The normal revenue is $575,000 $\left(\left(105,000\times 5\right)+50,000\right)$ which is less than the investment. Thus, rate of return is negative. Substitute the rate of return as -3% by trial and error method in the above Equation.

$\begin{array}{c}650,000=105,000\left(\frac{{\left(1+\left(-0.03\right)\right)}^{\text{5}}-1}{\left(-0.03\right){\left(1+\left(-0.03\right)\right)}^{\text{5}}}\right)+\frac{\text{50,000}}{{\left(1+\left(-0.03\right)\right)}^{\text{5}}}\\ 650,000=105,000\left(\frac{{\left(0.97\right)}^{\text{5}}-1}{\left(-0.03\right){\left(0.97\right)}^{\text{5}}}\right)+\frac{\text{50,000}}{{\left(0.97\right)}^{\text{5}}}\\ 650,000=105,000\left(\frac{0.858734-1}{\left(-0.03\right)\left(0.858734\right)}\right)+\frac{\text{50,000}}{0.858734}\\ 650,000=105,000\left(\frac{-0.141266}{-0.025762}\right)+58,225.25\\ 650,000=105,000\left(5.483503\right)+58,225.25\\ 650,000=575,767.82+58,225.25\\ 650,000>633,993.07\end{array}$

When the rate of return as -3% is substituted, the calculated value is less than the investment and thus, the rate of return is reduced to 3.742%.

$\begin{array}{c}650,000=105,000\left(\frac{{\left(1+\left(-0.03742\right)\right)}^{\text{5}}-1}{\left(-0.03742\right){\left(1+\left(-0.03742\right)\right)}^{\text{5}}}\right)+\frac{\text{50,000}}{{\left(1+\left(-0.03742\right)\right)}^{\text{5}}}\\ 650,000=105,000\left(\frac{{\left(0.96258\right)}^{\text{5}}-1}{\left(-0.03742\right){\left(0.96258\right)}^{\text{5}}}\right)+\frac{\text{50,000}}{{\left(0.96258\right)}^{\text{5}}}\\ 650,000=105,000\left(\frac{0.826388-1}{\left(-0.03742\right)\left(0.826388\right)}\right)+\frac{\text{50,000}}{0.826388}\\ 650,000=105,000\left(\frac{-0.173612}{-0.030923}\right)+60,504.27\\ 650,000=105,000\left(5.614332\right)+60,504.27\\ 650,000=589,504.86+60,504.27\\ 650,000\approx 650,009.13\end{array}$

The calculated value is nearly equal to investment with rate of return 3.742%. Thus, it is confirmed that the rate of return is 3.742%.

(b)

To determine

Calculate the equivalent annual value.

Explanation of Solution

Investment is (I) is $650,000. Salvage value (SV) is 50,000. Time period (n) is 5. Rate of return (i) is 15%.

Equivalent annual value (A) can be calculated as follows:

$\begin{array}{c}\text{I}=\text{A}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\text{SV}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ 650,000=\text{A}\left(\frac{{\left(1+\text{0}\text{.15}\right)}^{\text{5}}-1}{\text{0}\text{.15}{\left(1+\text{0}\text{.15}\right)}^{\text{5}}}\right)+\frac{\text{50,000}}{{\left(1+\text{0}\text{.15}\right)}^{\text{5}}}\\ 650,000=\text{A}\left(\frac{2.011357-1}{\text{0}\text{.15}\left(2.011357\right)}\right)+\frac{\text{50,000}}{2.011357}\\ 650,000=\text{A}\left(\frac{1.011357}{0.301704}\right)+24,858.84\\ 650,000=\text{A}\left(3.35215\right)+24,858.84\\ \text{A}\left(3.35215\right)=650,000-24,858.84\\ \text{A}=\frac{625,141.6}{3.35215}\\ =186,489.61\end{array}$

The equivalent annual value is $186,489.61.