Chapter 16, Problem 44P

To determine

(a)

The capital expenditure to build the water project.

Answer to Problem 44P

The capital expenditure to build the water project is, $\$879629.11$.

Explanation of Solution

Given:

					Decades of Focus
	First	Second	Third	Fourth	Fifth
Annual benefits (In thousands of dollars)	$\$60$	$\$64$	$\$66$	$\$66$	$\$70$

Calculation:

Calculate the present worth of project.

$\text{Present}\text{worth}=\left[\begin{array}{l}\$60000\left(\frac{P}{A},0.05,10\right)+\\ \$64000\left(\frac{P}{A},0.05,10\right)\left(\frac{P}{F},0.05,10\right)+\\ \$66000\left(\frac{P}{A},0.05,20\right)\left(\frac{P}{F},0.05,20\right)+\\ \$70000\left(\frac{P}{A},0.05,10\right)\left(\frac{P}{F},0.05,40\right)\end{array}\right]$ ...... (I)

Calculate the factor $\left(\frac{P}{A},0.05,10\right)$.

$\left(\frac{P}{A},0.05,10\right)=\left[\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right]$ ....... (II)

Here, the rate of interest is I and the time period is n.

Substitute $5\%$ for i, and $10$ years for n in Equation (II).

$\begin{array}{c}\left(\frac{P}{A},0.05,10\right)=\left[\frac{{\left(1+0.05\right)}^{10}-1}{0.05{\left(1+0.05\right)}^{10}}\right]\\ =7.722\end{array}$

Calculate the factor $\left(\frac{P}{A},0.05,20\right)$.

Substitute $5\%$ for i, and $20$ years for n in Equation (II).

$\begin{array}{c}\left(\frac{P}{A},0.05,10\right)=\left[\frac{{\left(1+0.05\right)}^{20}-1}{0.05{\left(1+0.05\right)}^{20}}\right]\\ =12.462\end{array}$

Calculate the factor $\left(\frac{P}{F},0.05,10\right)$.

$\left(\frac{P}{F},0.05,10\right)=\frac{1}{{\left(1+i\right)}^n}$ ...... (III)

Substitute $5\%$ for i, and $10$ years for n in Equation (III).

$\begin{array}{c}\left(\frac{P}{F},0.05,10\right)=\frac{1}{{\left(1+0.05\right)}^{10}}\\ =0.6139\end{array}$

Calculate the factor $\left(\frac{P}{F},0.05,20\right)$.

Substitute $5\%$ for i, and $20$ years for n in Equation (III).

$\begin{array}{c}\left(\frac{P}{F},0.05,20\right)=\frac{1}{{\left(1+0.05\right)}^{20}}\\ =0.3769\end{array}$

Calculate the factor $\left(\frac{P}{F},0.05,40\right)$.

Substitute $5\%$ for i, and $40$ years for n in Equation (III).

$\begin{array}{c}\left(\frac{P}{F},0.05,40\right)=\frac{1}{{\left(1+0.05\right)}^{40}}\\ =0.1420\end{array}$

Substitute $7.722$ for $\left(\frac{P}{A},0.05,10\right)$, $12.462$ for $\left(\frac{P}{A},0.05,20\right)$, $0.6139$ for $\left(\frac{P}{F},0.05,10\right)$, $0.3769$ for $\left(\frac{P}{F},0.05,20\right)$, and $0.1420$ for $\left(\frac{P}{F},0.05,40\right)$ in Equation (I)

$\begin{array}{c}\text{Present}\text{worth}=\left[\begin{array}{l}\$60000\times 7.722+\\ \$64000\times 7.722\times 0.6139+\\ \$66000\times 12.462\times 0.3769+\\ \$70000\times 7.722\times 01420\end{array}\right]\\ =\$463200+\$303394.29+\$309997.24+\$76756.68\\ =\$1153468\end{array}$

The annual cost are $\$15000$.

Calculate the justified capital expenditure.

$\text{JCE}={\text{PW}}_{\text{Benefits}}-A\left(\frac{P}{A},i,n\right)$ ...... (IV)

Calculate the value of $A\left(\frac{P}{A},i,n\right)$.

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

Substitute $5\%$ for i, $50$ years for n, and $\$15000$ for A in Equation (V).

$\begin{array}{c}A\left(\frac{P}{A},i,n\right)=\$15000\left[\frac{{\left(1+0.05\right)}^{50}-1}{0.05{\left(1+0.05\right)}^{50}}\right]\\ =\$15000\times 18.256\\ =\$273838.89\end{array}$

Substitute $\$1153468$ for ${\text{PW}}_{\text{Benefits}}$, and $\$273838.89$ for $A\left(\frac{P}{A},i,n\right)$ in equation (IV).

$\begin{array}{c}\text{JCE}=\$1153468-\$273838.89\\ =\$879629.11\end{array}$.

Conclusion:

Therefore, the capital expenditure to build the water project is, $\$879629.11$.

To determine

(b)

The capital expenditure to build the water project.

Answer to Problem 44P

The capital expenditure to build the water project is, $\$558614$.

Explanation of Solution

Calculate the present worth of project.

$\text{Present}\text{worth}=\left[\begin{array}{l}\$60000\left(\frac{P}{A},0.08,10\right)+\\ \$64000\left(\frac{P}{A},0.08,10\right)\left(\frac{P}{F},0.08,10\right)+\\ \$66000\left(\frac{P}{A},0.08,20\right)\left(\frac{P}{F},0.08,20\right)+\\ \$70000\left(\frac{P}{A},0.08,10\right)\left(\frac{P}{F},0.08,40\right)\end{array}\right]$ ...... (VI)

Calculate the factor $\left(\frac{P}{A},0.08,10\right)$.

$\left(\frac{P}{A},0.08,10\right)=\left[\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right]$ ....... (VII)

Substitute $8\%$ for i, and $10$ years for n in Equation (VII).

$\begin{array}{c}\left(\frac{P}{A},0.08,10\right)=\left[\frac{{\left(1+0.08\right)}^{10}-1}{0.08{\left(1+0.08\right)}^{10}}\right]\\ =6.710\end{array}$

Calculate the factor $\left(\frac{P}{A},0.08,20\right)$.

Substitute $8\%$ for i, and $20$ years for n in Equation (VII).

$\begin{array}{c}\left(\frac{P}{A},0.08,10\right)=\left[\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right]\\ =9.818\end{array}$

Calculate the factor $\left(\frac{P}{F},0.08,10\right)$.

$\left(\frac{P}{F},0.08,10\right)=\frac{1}{{\left(1+i\right)}^n}$ ...... (VIII)

Substitute $8\%$ for i, and $10$ years for n in Equation (VIII).

$\begin{array}{c}\left(\frac{P}{F},0.08,10\right)=\frac{1}{{\left(1+0.08\right)}^{10}}\\ =0.4632\end{array}$

Calculate the factor $\left(\frac{P}{F},0.08,20\right)$.

Substitute $8\%$ for i, and $20$ years for n in Equation (VIII).

$\begin{array}{c}\left(\frac{P}{F},0.08,20\right)=\frac{1}{{\left(1+0.08\right)}^{20}}\\ =0.2145\end{array}$

Calculate the factor $\left(\frac{P}{F},0.05,40\right)$.

Substitute $8\%$ for i, and $40$ years for n in Equation (VIII).

$\begin{array}{c}\left(\frac{P}{F},0.08,40\right)=\frac{1}{{\left(1+0.08\right)}^{40}}\\ =0.0460\end{array}$

Substitute $6.710$ for $\left(\frac{P}{A},0.08,10\right)$, $9.818$ for $\left(\frac{P}{A},0.08,20\right)$, $0.4632$ for $\left(\frac{P}{F},0.08,10\right)$, $0.2145$ for $\left(\frac{P}{F},0.08,20\right)$, and $0.0460$ for $\left(\frac{P}{F},0.08,40\right)$ in Equation (VI)

$\begin{array}{c}\text{Present}\text{worth}=\left[\begin{array}{l}\$60000\times 6.710+\\ \$64000\times 6.710\times 0.4632+\\ \$66000\times 9.818\times 0.2145+\\ \$70000\times 6.710\times 0.0460\end{array}\right]\\ =\$742116.23\end{array}$

The annual cost are $\$15000$.

Calculate the justified capital expenditure.

$\text{JCE}={\text{PW}}_{\text{Benefits}}-A\left(\frac{P}{A},i,n\right)$ ...... (IX)

Calculate the value of $A\left(\frac{P}{A},i,n\right)$.

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

Substitute $8\%$ for i, $50$ years for n, and $\$15000$ for A in Equation (V).

$\begin{array}{c}A\left(\frac{P}{A},i,n\right)=\$15000\left[\frac{{\left(1+0.08\right)}^{50}-1}{0.08{\left(1+0.08\right)}^{50}}\right]\\ =\$15000\times 12.233\\ =\$183502.27\end{array}$

Substitute $\$742116.23$ for ${\text{PW}}_{\text{Benefits}}$, and $\$183502.27$ for $A\left(\frac{P}{A},i,n\right)$ in equation (IX).

$\begin{array}{c}\text{JCE}=\$742116.23-\$183502.27\\ =\$558614\end{array}$.

Conclusion:

Therefore, the capital expenditure to build the water project is, $\$558614$.