Chapter 13, Problem 41P

(a):

To determine

Selection of the alternative using the payback period analysis.

Explanation of Solution

Alternate A: First cost (FC) is $300,000. Revenue (RE) is $60,000 per year.

Alternate B: First cost (FC) is $300,000. Revenue is $10,000 first year and increase (AI) by $15,000 per year.

MARR (i) is 8%.

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

$\begin{array}{c}\text{RE}\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)=\text{FC}\\ \text{60,000}\left(\frac{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}\right)=\text{300,000}\\ \left(\frac{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}\right)=\frac{\text{300,000}}{60,000}\\ \left(\frac{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}\right)=\text{5}\end{array}$

From 8% interest table, it is known that the payback period is between the years 6 and 7 years.

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

$\begin{array}{c}\text{RE}\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)+\text{AI}\times \frac{1}{\text{i}}\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)=\text{FC}\\ \text{10,000}\left(\frac{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}\right)+15,000\times \frac{1}{\text{0}\text{.08}}\left(\frac{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(\text{1}+0.08\right)}^{\text{n}}}\right)=\text{300,000}\end{array}$

Substitute n as 7 years by trial and error method to verify the payback period.

$\begin{array}{c}\text{10,000}\left(\frac{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{7}}-1}{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{7}}}\right)+15,000\times \frac{1}{\text{0}\text{.08}}\left(\frac{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{7}}-1}{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{7}}}-\frac{\text{7}}{{\left(\text{1}+0.08\right)}^{\text{7}}}\right)=\text{300,000}\\ \text{10,000}\left(5.20637\right)+15,000\times 12.5\left(5.20637-4.08443\right)=\text{300,000}\\ 262,427.45<\text{300,000}\end{array}$

Since the calculated value is less than the first cost, payback period should increase. Thus, the payback period for Alternate B is greater than 7 years. The payback period for alternate A is less than 7 years. Since the payback period for alternate A is less, select Alternate A.

(b):

To determine

Selection of the alternative using the present worth analysis.

Explanation of Solution

Time period (n) is 10.

Present worth (PW) for Alternate A can be calculated as follows:

$\begin{array}{c}\text{PW}=-\text{FC}+\text{RE}\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)\\ =-\text{300,000}+\text{60,000}\left(\frac{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{10}}-1}{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}\right)\\ =-\text{300,000}+\text{60,000}\left(6.7101\right)\\ =-\text{300,000}+402,606\\ =102,606\end{array}$

Present worth for Alternate A is $102,606.

Present worth (PW) for Alternate B can be calculated as follows:

$\begin{array}{c}\text{PW}=\text{RE}\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)+\text{AI}\times \frac{1}{\text{i}}\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)-\text{FC}\\ =10,000\left(\frac{{\left(1+0.08\right)}^{10}-1}{0.08{\left(1+0.08\right)}^{10}}\right)+15,000\times \frac{1}{0.08}\left(\frac{{\left(1+0.08\right)}^{10}-1}{0.08{\left(1+0.08\right)}^{10}}-\frac{10}{{\left(1+0.08\right)}^{10}}\right)-300,000\\ =10,000\left(6.7101\right)+15,000\times 12.5\left(6.7101-4.511\right)-300,000\\ =67,101+389,652-300,000\\ =156,753\end{array}$

Present worth for Alternate B is $156,753. Since the present worth of Alternate B is greater, select Alternate B.

(c):

To determine

Selection of the alternative using the spreadsheet.

Explanation of Solution

Payback period and present worth of the alternate can be calculated using the spreadsheet as follows: