Chapter 13, Problem 37P

(a):

To determine

Calculate the payback period with interest rate (i) 0%.

Explanation of Solution

SA: First cost (FC) is $40,000. Net annual income (NI) is $10,000.

AU: First cost (FC) is $90,000. Net annual income (NI) is $15,000.

Payback period (n) for SA can be calculated as follows:

$\begin{array}{c}\text{Payback period}=\frac{\text{FC}}{\text{AI}}\\ =\frac{40,000}{10,000}\\ =4\end{array}$

Payback period is 4 years.

Payback period (n) for AU can be calculated as follows:

$\begin{array}{c}\text{Payback period}=\frac{\text{FC}}{\text{AI}}\\ =\frac{90,000}{15,000}\\ =6\end{array}$

Payback period is 6 years. Since the payback period for the alternative SA is less than 5 years, select the alternate SA.

(b):

To determine

Calculate the payback period with interest rate (i) 10%.

Explanation of Solution

Payback period (n) for alternate SA can be calculated as follows:

$\begin{array}{c}\text{AI}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)=\text{I}\\ 10,000\left(\frac{{\left(1+0.1\right)}^{\text{n}}-1}{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{n}}}\right)=\text{40,000}\\ \left(\frac{{\left(1+0.1\right)}^{\text{n}}-1}{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{n}}}\right)\text{=}\frac{40,000}{10,000}\\ \left(\frac{{\left(1+0.1\right)}^{\text{n}}-1}{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{n}}}\right)=4\end{array}$

Substitute n as 5 by trial and error method to verify the time period in the above calculation.

$\begin{array}{c}\frac{{\left(1+0.1\right)}^{\text{5}}-1}{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{5}}}=4\\ \left(\frac{1.61051-1}{\text{0}\text{.1}\left(1.61051\right)}\right)=4\\ \frac{0.61051}{0.161051}=4\\ 3.7907<4\end{array}$

Since the calculated present value factor to the annual value is less than the given value, the payback period is more than 5 years.

Payback period (n) for alternate AU can be calculated as follows:

$\begin{array}{c}\text{AI}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)=\text{I}\\ 15,000\left(\frac{{\left(1+0.1\right)}^{\text{n}}-1}{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{n}}}\right)=\text{90,000}\\ \left(\frac{{\left(1+0.1\right)}^{\text{n}}-1}{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{n}}}\right)\text{=}\frac{90,000}{15,000}\\ \left(\frac{{\left(1+0.1\right)}^{\text{n}}-1}{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{n}}}\right)=6\end{array}$

Substitute n as 5 by trial and error method to verify the time period in the above calculation.

$\begin{array}{c}\frac{{\left(1+0.1\right)}^{\text{5}}-1}{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{5}}}=6\\ \left(\frac{1.61051-1}{\text{0}\text{.1}\left(1.61051\right)}\right)=6\\ \frac{0.61051}{0.161051}=6\\ 3.7907<6\end{array}$

Since the calculated present value factor to the annual value is less than the given value, the payback period is more than 5 years. Since both the alternate’s payback periods are greater than 5 years, both the alternates should not be selected.