Chapter 12, Problem 27P

(a):

To determine

Calculate the internal rate of return.

Explanation of Solution

The initial investment (IN) is $750,000. Net income (NY) is $135,000 per year. Time period (n) is 10. MARR (i) is 12%.

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

$\begin{array}{c}\text{IN}=\text{NY}\left(\frac{{\left(1+\text{ir}\right)}^{\text{n}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{n}}}\right)\\ 750,000=135,000\left(\frac{{\left(1+\text{ir}\right)}^{\text{n}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{n}}}\right)\\ \left(\frac{{\left(1+\text{ir}\right)}^{\text{10}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{10}}}\right)=\frac{750,000}{135,000}\\ \left(\frac{{\left(1+\text{ir}\right)}^{\text{10}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{10}}}\right)=5.5556\end{array}$

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

$\begin{array}{c}\left(\frac{{\left(1+0.12\right)}^{10}-1}{0.12{\left(1+0.12\right)}^{10}}\right)=5.5556\\ \left(\frac{3.1058-1}{0.12\left(3.1058\right)}\right)=5.5556\\ \left(\frac{2.1058}{0.3727}\right)=5.5556\\ 5.6501>5.5556\end{array}$

The calculated value is less than the present value factor to the annual value. Thus, increase the ir value to 0.124.

$\begin{array}{c}\left(\frac{{\left(1+0.124\right)}^{10}-1}{0.124{\left(1+0.124\right)}^{10}}\right)=5.5556\\ \left(\frac{3.2189-1}{0.124\left(3.2189\right)}\right)=5.5556\\ \left(\frac{2.2189}{0.3991}\right)=5.5556\\ 5.5598\approx 5.5556\end{array}$

Since the calculated value is equal to the present value factor to the annual value, the value of ir is confirmed as 12.4%.

The value PI can be calculated as follows:

$\begin{array}{c}\text{PI}=\frac{\text{NY}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)}{\text{IN}}\\ =\frac{135,000\left(\frac{{\left(1+0.12\right)}^{10}-1}{0.12{\left(1+0.12\right)}^{10}}\right)}{750,000}\\ =\frac{135,000\left(\frac{3.1058-1}{0.12\left(3.1058\right)}\right)}{750,000}\\ =\frac{135,000\left(\frac{2.1058}{0.3727}\right)}{750,000}\\ =\frac{135,000\left(5.6501\right)}{750,000}\\ =1.02\end{array}$

The value of PI is 1.02.

The value PW can be calculated as follows:

$\begin{array}{c}\text{PW}=-\text{IN}+\text{NY}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)\\ =-750,000+135,000\left(\frac{{\left(1+0.12\right)}^{10}-1}{0.12{\left(1+0.12\right)}^{10}}\right)\\ =-750,000+135,000\left(\frac{3.1058-1}{0.12\left(3.1058\right)}\right)\\ =-750,000+135,000\left(\frac{2.1058}{0.3727}\right)\\ =-750,000+135,000\left(5.6501\right)\\ =-750,000+762,763.5\\ =12,763.5\end{array}$

The present wroth (PW) is $12,763.5.

(b):

To determine

The measures for which the project is economically justified.

Explanation of Solution

The internal rate of return is greater than MARR. PI is greater than 1 and the present worth is greater than 0. Thus, the project is economically justified by these three measures.

(c):

To determine

Calculate the breakeven rate of interest.

Explanation of Solution

The breakeven interest rate can be calculated using the following spreadsheet function:

= RATE(10,135000,-750000)

The above function gives the value of 12.4148%. Thus, the breakeven interest rate is 12.4148%.