Chapter 9, Problem 1P

To determine

Calculate the present worth.

Explanation of Solution

Time period is denoted by n and the interest is denoted by i. The present worth (PW) of defender (Old boat) can be calculated as follows.

$\begin{array}{c}{\text{PW}}_{\text{Defender}}=\text{MV}+\text{Expense}\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)+\text{Total mile}\times \text{Mileage}\times {\text{Cost}}_{\text{Fuel}}-\frac{{\text{MV}}_{\text{End of the life}}}{{\left(\text{1}+\text{i}\right)}^{\text{n}}}\\ =3,000+\text{400}\left(\frac{{\left(\text{1}+\text{0}\text{.15}\right)}^{\text{2}}-1}{\text{0}\text{.15}{\left(\text{1}+\text{0}\text{.15}\right)}^{\text{2}}}\right)+\text{200}\times \text{2}\times \text{5}-\frac{\text{500}}{{\left(\text{1}+\text{0}\text{.15}\right)}^{\text{2}}}\\ =4,085\end{array}$   

Annual worth of defender is $4,085.

The present worth (PW) of challenger (New boat) can be calculated as follows.

$\begin{array}{c}{\text{PW}}_{\text{Defender}}=\text{Investment}+\text{Expense}\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)+\text{Total mile}\times \text{Mileage}\times {\text{Cost}}_{\text{Fuel}}-\frac{{\text{MV}}_{\text{End of the life}}}{{\left(\text{1}+\text{i}\right)}^{\text{n}}}\\ =10,000+\text{200}\times \text{10}\times \text{5}-\frac{\text{7,000}}{{\left(\text{1}+\text{0}\text{.15}\right)}^{\text{2}}}\\ =4,870\end{array}$   

Present worth is $4,870. Since the present worth of the cost is less for the defender, retain the defender one more year.