Chapter 18, Problem 1P

To determine

Calculate annual worth.

Explanation of Solution

Investment (INV) is $850,000. Lowest range of revenue (LR) is $290,000 per year. Highest range of revenue (HR) is $$325,000 per year. Time period (n) is 5. Rate of return (i) is 2%.

Annual worth (AW) for the lowest range of earning can be calculated as follows.

$\begin{array}{c}\text{AW}=-\text{INV}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)+\text{LR}\\ =-\text{850,000}\left(\frac{\text{0}\text{.2}{\left(1+\text{0}\text{.2}\right)}^{\text{5}}}{{\left(1+\text{0}\text{.2}\right)}^{\text{5}}-1}\right)+\text{290,000}\\ =-\text{850,000}\left(0.33438\right)+\text{290,000}\\ =5,777\end{array}$

Annual worth of the project with the lowest expected earnings per year is $5,777.

Annual worth (AW) for the highest range of earning can be calculated as follows.

$\begin{array}{c}\text{AW}=-\text{INV}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)+\text{HR}\\ =-\text{850,000}\left(\frac{\text{0}\text{.2}{\left(1+\text{0}\text{.2}\right)}^{\text{5}}}{{\left(1+\text{0}\text{.2}\right)}^{\text{5}}-1}\right)+\text{325,000}\\ =-\text{850,000}\left(0.33438\right)+\text{325,000}\\ =40,777\end{array}$

Annual worth of the project with highest expected earnings per year is $40,777.