Chapter 16, Problem 32P

To determine

If the road should be repaired and when.

Answer to Problem 32P

Yes, the road should be repaired $4$ years from now..

Explanation of Solution

Given:

The MARR is $15\%$.

Concept used:

Net present worth is the difference between the net present cash flow and the net present cash out flow.

Calculation:

Write the expression to calculate the net present worth for year $0$.

$NP{W}_{YEAR0}=\left[\begin{array}{l}-P+\frac{F}{{\left(1+i\right)}^n}+G\left(\frac{{\left(1+i\right)}^n-in-1}{i^2{\left(1+i\right)}^n}\right)\\ +A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)\left(\frac{1}{{\left(1+i\right)}^n}\right)\end{array}\right]$ ...... (I)

Here, the initial cost is $P$, the annual benefit is $A$, the future worth is $F$, the gradient is $G$, the interest rate is $i$ and the number of period is $n$.

Substitute $\$150000$ for $P$, $\$5000$ for $F$, $\$10000$ for $G$, $\$50000$ for $A$, $0.15$ for $i$ and $5$ for $n$ in Equation (I).

$\begin{array}{c}NP{W}_{YEAR0}=\left[\begin{array}{l}-\$150000+\frac{\$5000}{{\left(1+0.15\right)}^1}+\$10000\left(\frac{{\left(1+0.15\right)}^5-\left(0.15\times 5\right)-1}{{0.15}^2{\left(1+0.15\right)}^5}\right)\\ +\$50000\left(\frac{{\left(1+0.15\right)}^5-1}{0.15{\left(1+0.15\right)}^5}\right)\left(\frac{1}{{\left(1+0.15\right)}^5}\right)\end{array}\right]\\ =-\$150000+\frac{\$5000}{\left(1.15\right)}+\$10000\left(5.775\right)+\$50000\left(1.6666\right)\\ =-\$150000+\$4347.8+\$57750+\$83330\\ =-\$4572\end{array}$

Write the expression to calculate the net present worth for year $2$.

$NP{W}_{YEAR2}=\left[\begin{array}{l}-P+\frac{A_1}{{\left(1+i\right)}^n}+G\left(\frac{{\left(1+i\right)}^n-in-1}{i^2{\left(1+i\right)}^n}\right)\\ +A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)\left(\frac{1}{{\left(1+i\right)}^n}\right)\end{array}\right]$ ...... (II)

Substitute $\$150000$ for $P$, $\$20000$ for $A_1$, $\$10000$ for $G$, $\$50000$ for $A$ and $0.15$ for $i$ in Equation (II).

$\begin{array}{c}NP{W}_{YEAR2}=\left[\begin{array}{l}-150000+\frac{20000\left({\left(1.15\right)}^3-1\right)}{0.15{\left(1+0.15\right)}^3}\\ +\$10000\left(\frac{{\left(1+0.15\right)}^3-\left(0.15\times 3\right)-1}{{0.15}^2{\left(1+0.15\right)}^3}\right)\\ +\$50000\left(\frac{{\left(1+0.15\right)}^7-1}{0.15{\left(1+0.15\right)}^7}\right)\left(\frac{1}{{\left(1+0.15\right)}^3}\right)\end{array}\right]\\ =-150000+\$20000\times 2.283+\$10000\left(2.0712\right)+\$50000\left(2.73554\right)\\ =-\$150000+\$45660+\$20712+\$136777\\ =\$53149\end{array}$

Convert the net present worth for year $2$ to the net present worth for year $0$.

$NP{W}_0=\frac{NP{W}_{YEAR2}}{{\left(1+i\right)}^n}$ ...... (III)

Substitute $\$53149$ for $NP{W}_{YEAR2}$ and $0.15$ for $i$ and $2$ for $n$ in Equation (III).

$\begin{array}{c}NP{W}_0=\frac{\$53149}{{\left(1+0.15\right)}^2}\\ =\$40188\end{array}$

Write the expression to calculate the net present worth for year $4$.

$NP{W}_{YEAR2}=\left[-P+\frac{A_1}{{\left(1+i\right)}^n}+A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)\left(\frac{1}{{\left(1+i\right)}^n}\right)\right]$ ...... (IV)

Substitute $\$150000$ for $P$, $\$40000$ for $A_1$, $\$50000$ for $A$ and $0.15$ for $i$ in Equation (IV).

$\begin{array}{c}NP{W}_{YEAR4}=-\$150000+\frac{40000}{{\left(1+0.15\right)}^1}+\$50000\left(\frac{{\left(1+0.15\right)}^9-1}{0.15{\left(1+0.15\right)}^9}\right)\left(\frac{1}{{\left(1+0.15\right)}^1}\right)\\ =-\$150000+\frac{40000}{{\left(1.15\right)}^1}+\$50000\left(4.1492\right)\\ =-\$150000+\$34783+\$207460\\ =\$92243\end{array}$

Convert the net present worth for year $4$ to the net present worth for year $0$.

Substitute $\$92243$ for $NP{W}_{YEAR\text{}4}$, $0.15$ for $i$ and $4$ for $n$ in Equation (III).

$\begin{array}{c}NP{W}_0=\frac{\$92243}{{\left(1+0.15\right)}^4}\\ =\$52740\end{array}$

Write the expression to calculate the net present worth for year $5$.

$NP{W}_{YEAR5}=\left[-P+A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)\left(\frac{1}{{\left(1+i\right)}^n}\right)\right]$ ...... (V)

Substitute $\$150000$ for $P$, $\$50000$ for $A$ and $0.15$ for $i$ in Equation (V).

$\begin{array}{c}NP{W}_{YEAR5}=-\$150000+\$50000\left(\frac{{\left(1+0.15\right)}^{10}-1}{0.15{\left(1+0.15\right)}^{10}}\right)\\ =-\$150000+\$50000\left(5.01876\right)\\ =-\$150000+\$250938\\ =\$100938\end{array}$

Convert the net present worth for year $5$ to the net present worth for year $0$.

Substitute $\$100938$ for $NP{W}_{YEAR\text{}n5}$, $0.15$ for $i$ and $5$ for $n$ in Equation (III).

$\begin{array}{c}NP{W}_0=\frac{100938}{{\left(1+0.15\right)}^5}\\ =\$50184\end{array}$.

Conclusion:

The net present worth for now is in negative hence, the road should be repaired, Since year $4$ has the highest net present worth at year $0$, the road should be repaired $4$ years from now..