Chapter 16, Problem 30P

To determine

The best design.

Answer to Problem 30P

The best design is alternative $A$.

Explanation of Solution

Given:

Interest rate is $8\%$.

Project life is $30$ years.

Concept used:

Write the expression to calculate the $B/C$ ratio using annual worth analysis.

$\begin{array}{c}B/C=\frac{A{W}_{NetBenefits}}{A{W}_{Cost}}\\ =\frac{A+F\frac{\left(i\right)}{{\left(1+i\right)}^n-1}}{P\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}}\end{array}$ ...... (I)

Here, the net benefit is $A$, the salvage value is $F$, the interest rate is $i$ and the number of year is $n$.

Calculation:

Alternative $A$.

Calculate the net benefit for alternative $A$.

$\text{Net}\text{Benefit}=\left(\text{Annual}\text{Benefit}\right)-\left(\text{O\&M}\text{Cost}\right)-\left(\text{Annual}\text{Disbenifit}\right)$ ...... (II)

Substitute $\$2200$ for $\text{Annual}\text{Benefit}$, $\$550$ for $\text{O\&M}\text{Cost}$ and $\$350$ for $\text{Annual}\text{Disbenefit}$ in Equation (II).

$\begin{array}{c}\text{Net}\text{Benefit}=\$2200-\$550-\$350\\ =\$1300\end{array}$

Calculate the $B/C$ ratio.

Substitute $\$1300$ for $A$, $\$1000$ for $F$, $\$9500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$\begin{array}{c}B/C=\frac{\$1300+\left(\$1000\frac{\left(0.08\right)}{{\left(1+0.08\right)}^{30}-1}\right)}{\left(\$9500\frac{0.08{\left(1+0.08\right)}^{30}}{{\left(1+0.08\right)}^{30}-1}\right)}\\ =\frac{\$1300+\$1000\times 0.0088}{\$9500\times 0.0888}\\ =\$1.55\end{array}$

Thus, the $B/C$ for alternative $A$ is $1.55$.

Alternative $B$.

Calculate the net benefit for alternative $B$.

Substitute $\$1500$ for $\text{Annual}\text{Benefit}$, $\$175$ for $\text{O\&M}\text{Cost}$ and $\$150$ for $\text{Annual}\text{Disbenefit}$ in Equation (II).

$\begin{array}{c}\text{Net}\text{Benefit}=\$1500-\$175-\$150\\ =\$1175\end{array}$

Calculate the $B/C$ ratio.

Substitute $\$1175$ for $A$, $\$6000$ for $F$, $\$12500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$\begin{array}{c}B/C=\frac{\$1175+\left(\$6000\frac{\left(0.08\right)}{{\left(1+0.08\right)}^{30}-1}\right)}{\left(\$12500\frac{0.08{\left(1+0.08\right)}^{30}}{{\left(1+0.08\right)}^{30}-1}\right)}\\ =\frac{\$1175+\$6000\times 0.0088}{\$12500\times 0.0888}\\ =1.106\end{array}$

Thus, the $B/C$ for alternative $B$ is $1.55$.

Alternative $C$.

Calculate the net benefit for alternative $C$.

Substitute $\$1000$ for $\text{Annual}\text{Benefit}$, $\$325$ for $\text{O\&M}\text{Cost}$ and $\$75$ for $\text{Annual}\text{Disbenefit}$ in Equation (II).

$\begin{array}{c}\text{Net}\text{Benefit}=\$1000-\$325-\$75\\ =\$600\end{array}$

Calculate the $B/C$ ratio.

Substitute $\$600$ for $A$, $\$3500$ for $F$, $\$14000$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$\begin{array}{c}B/C=\frac{\$600+\left(\$3500\frac{\left(0.08\right)}{{\left(1+0.08\right)}^{30}-1}\right)}{\left(\$14000\frac{0.08{\left(1+0.08\right)}^{30}}{{\left(1+0.08\right)}^{30}-1}\right)}\\ =\frac{\$600+\$3500\times 0.0088}{\$14000\times 0.0888}\\ =0.507\end{array}$

Thus, the $B/C$ for alternative $C$ is $0.507$.

Alternative $D$.

Calculate the net benefit for alternative $D$.

Substitute $\$2500$ for $\text{Annual}\text{Benefit}$, $\$145$ for $\text{O\&M}\text{Cost}$ and $\$700$ for $\text{Annual}\text{Disbenefit}$ in Equation (II).

$\begin{array}{c}\text{Net}\text{Benefit}=\$2500-\$145-\$700\\ =\$1655\end{array}$

Calculate the $B/C$ ratio.

Substitute $\$1655$ for $A$, $\$7500$ for $F$, $\$15750$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$\begin{array}{c}B/C=\frac{\$1655+\left(\$7500\frac{\left(0.08\right)}{{\left(1+0.08\right)}^{30}-1}\right)}{\left(\$15750\frac{0.08{\left(1+0.08\right)}^{30}}{{\left(1+0.08\right)}^{30}-1}\right)}\\ =\frac{\$1655+\$7500\times 0.0088}{\$15750\times 0.0888}\\ =1.18\end{array}$

Thus, the $B/C$ for alternative $B$ is $1.18$.

Conclusion:

The $B/C$ ratio for the alternative $A$ is the highest.

Thus, the best design is alternative $A$.