Chapter 6, Problem 1ST

To determine

Calculate the annual worth.

Explanation of Solution

Time period is denoted by n and interest rate is denoted by i. Annual worth (AW) can be calculated as follows.

$\begin{array}{c}\text{AW}=\left(\begin{array}{l}\text{Revenue}+{\text{Revenue}}_{\text{Increase}}\left(\frac{\text{1}}{\text{i}}-\frac{\text{n}}{{\left(\text{1}+\text{i}\right)}^{\text{n}}-\text{1}}\right)\\ -\left(\begin{array}{l}\text{Investment}\left(\frac{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}{{\left(\text{1}+\text{i}\right)}^{\text{n}}-\text{1}}\right)\\ +\text{Cost}+\text{Increasing cost}\left(\frac{\text{1}}{\text{i}}-\frac{\text{n}}{{\left(\text{1}+\text{i}\right)}^{\text{n}}-\text{1}}\right)\end{array}\right)\end{array}\right)\\ =\left(\begin{array}{l}\text{50,000}+\text{30,000}\left(\frac{1}{0.12}-\frac{5}{{\left(1+0.12\right)}^5-1}\right)\\ -\left(\begin{array}{l}\text{80,000}\left(\frac{0.12{\left(1+0.12\right)}^5}{{\left(1+0.12\right)}^5-1}\right)+\\ \text{20,000}+\text{10,000}\left(\frac{1}{0.12}-\frac{5}{{\left(1+0.12\right)}^5-1}\right)\end{array}\right)\end{array}\right)\\ =43,299.11\end{array}$

 Annual worth is $43,299.1.  Thus, option ‘d’ is correct.