Chapter 16, Problem 23P

To determine

The plan which should be chosen.

Answer to Problem 23P

The benefit to cost ratio using $PW$ method is $0.66$ which is less than $1$ then plan $A$ is chosen.

Explanation of Solution

Given:

The table for data given is shown below.

Year	Cash flow $A$	Cash flows $B$
0	$\$300$	$\$450$
$1-15\left(A_1\right)$	$\$75$	$\$100$
$15\left(F_1\right)$	$\$250$	$\$50$
$16-30\left(A_2\right)$	$\$125$	$\$125$
$30\left(F_2\right)$	$\$300$	$\$0$
$31-40\left(A_3\right)$	$\$250$	$\$125$
$40\left(F_3\right)$	$\$0$	$\$150$

Table (1)

All the amounts are in thousands.

Concept used:

Write the expression for present worth of benefits.

$P{W}_B=F_3\left[\frac{1}{{\left(1+i\right)}^n}\right]$ ...... (I)

Here, the present worth is PW, final cash flow at stage $3$ is $F_3$, interest rate is $i$ and time period is $n$.

Write the expression for present worth of costs.

$P{W}_C=\left[\begin{array}{l}P+A_1\left[\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right]-F_1\left[\frac{1}{{\left(1+i\right)}^n}\right]\\ -F_2\left[\frac{1}{{\left(1+i\right)}^n}\right]-A_3\left[\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right]\left[\frac{1}{{\left(1+i\right)}^n}\right]\end{array}\right]$ ...... (II)

Here, first cost is $P$, annual maintenance at stage $1$ is $A_1$,, annual maintenance at stage $3$ is $A_3$, final cost at stage $1$ is $F_1$ and final cost at stage $3$ is $F_3$.

Write the expression for incremental benefit to cost ratio using $PW$ method.

$\text{Incremental B-C ratio}=\left(\frac{P{W}_B}{P{W}_C}\right)$ ...... (III)

If the value for incremental benefit to cost ratio is greater than $1$ then plan $B$ is chosen otherwise the plan $A$ is chosen.

Calculations:

Calculate the difference between cash flows $A$ and cash flow $B$.

Year	Cash flow $A$	Cash flows $B$	Incremental analysis $\left(B-A\right)$
0	$\$300$	$\$450$	$\$150$
$1-15\left(A_1\right)$	$\$75$	$\$100$	$\$25$
$15\left(F_1\right)$	$\$250$	$\$50$	$-\$200$
$16-30\left(A_2\right)$	$\$125$	$\$125$	$\$0$
$30\left(F_2\right)$	$\$300$	$\$0$	$-\$300$
$31-40\left(A_3\right)$	$\$250$	$\$125$	$-\$125$
$40\left(F_3\right)$	$\$0$	$\$150$	$\$150$

Table (2)

Calculate the present worth of benefits.

Substitute $\$150000$ for $F_3$, $7\%$ for $i$ and $40$ for $n$ in Equation (I).

$\begin{array}{c}P{W}_B=\$150000\left[\frac{1}{{\left(1+0.07\right)}^{40}}\right]\\ =\$150000\left[\frac{1}{{\left(1.07\right)}^{40}}\right]\\ =\$150000\left(0.0668\right)\\ =\$10020\end{array}$

Calculate the present worth of costs.

Substitute $\$150000$ for $P$, $\$25000$ for $A_1$, $\$200000$ for $F_1$, $\$300000$ for $F_2$, $\$125000$ for $A_3$, $7\%$ for $i$ in Equation (II).

$\begin{array}{c}P{W}_C=\left[\begin{array}{c}\$150000+\$25000\left[\frac{{\left(1+0.07\right)}^{15}-1}{0.07{\left(1+0.07\right)}^{15}}\right]\\ -\$200000\left[\frac{1}{{\left(1+0.07\right)}^{15}}\right]-\$300000\left[\frac{1}{{\left(1+0.07\right)}^{30}}\right]-\\ \$125000\left[\frac{{\left(1+0.07\right)}^{10}-1}{0.07{\left(1+0.07\right)}^{10}}\right]\left[\frac{1}{{\left(1+0.07\right)}^{30}}\right]\end{array}\right]\\ =\left[\begin{array}{l}\$150000+\$25000\left(9.108\right)\\ -\$200000\left(0.3624\right)-\$300000\left(0.1314\right)\\ -\$125000\left(7.024\right)\left(0.1314\right)\end{array}\right]\\ =\left(\begin{array}{l}\$150000+\$227700\\ -\$72480-\$39420\\ -\$115369.2\end{array}\right)\\ =\$150430.8\end{array}$

Calculate the incrementalbenefit to cost ratio using $PW$ method.

Substitute $\$10020$ for $P{W}_B$, $\$150430.8$ for $P{W}_C$ in Equation (III).

$\begin{array}{c}\text{Incremental B-C ratio}=\left(\frac{\$10020}{\$150430.8}\right)\\ =0.66\end{array}$

The benefit to cost ratio using $PW$ method is $0.66$ which is less than $1$ then plan $A$ is chosen.

Conclusion:

Thus, the benefit to cost ratio using $PW$ method is $0.66$ which is less than $1$ ; hence plan $A$ is chosen.