Chapter 9, Problem 42P

To determine

(a)

The alternative to be selected using benefit cost ratio and the incremental benefit cost ratio.

Answer to Problem 42P

The alternative to be selected is Alternative $\text{New}\text{School}$

The incremental $\Delta \left(B/C\right)$ ratio is, $1.006$.

Explanation of Solution

Given:

MARR is $8\%$

Useful life is $20$ years.

Calculation:

Write the expression to calculate the present worth of the cost.

$PW=P+A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)$ ....... (I)

Here, the annual operation and maintenance cost is $A$, the initial cost is $P$, the rate of interest is $i$, the present worth is $PW$ and the number of period is $n$.

Calculate the present worth cost of the alternative Expand.

Substitute $\$8.8$ for $P$, $\$0.95$ for $A$, $8\%$ for $i$ and $20$ for $n$ in Equation (I).

$\begin{array}{c}PW=\$8.8+\$0.95\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)\\ =\$8.8+\$0.95\left(9.818\right)\\ =\$18.1271\end{array}$

Calculate the present worth cost of the alternative New school.

Substitute $\$10.4$ for $P$, $\$1.7$ for $A$, $8\%$ for $i$ and $20$ for $n$ in Equation (I).

$\begin{array}{c}PW=\$10.4+\$1.7\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)\\ =\$10.4+\$1.7\left(9.818\right)\\ =\$27.0906\end{array}$

Calculate the present worth benefit of the alternative Expand.

Write the expression to calculate the annual benefit.

$PW=A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)$ ...... (II)

Here, the annual benefit is $A$, the rate of interest is $i$, the present worth is $PW$ and the number of period is $n$.

Calculate the present worth benefit for the Expand.

Substitute $\$2.1$ for $A$, $8\%$ for $i$ and $20$ for $n$ in Equation (II).

$\begin{array}{c}P=\$2.1\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)\\ =\$2.1\left(9.818\right)\\ =\$20.6178\end{array}$

Calculate the present worth benefit for the New school.

Substitute $\$3.1$ for $A$, $8\%$ for $i$ and $20$ for $n$ in Equation (II).

$\begin{array}{c}P=\$3.1\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)\\ =\$3.1\left(9.818\right)\\ =\$30.435\end{array}$ =

Write the expression to calculate the benefit-cost ratio.

$B/C=\frac{B-D}{C}$ ....... (III)

Here, the present worth benefit is $B$, the present worth user dis-benefit is $D$, the present worth cost is $C$.

For Alternative Expand.

Substitute $\$20.6178$ for $B$, $\$18.1271$ for $C$ and $\$0.8$ for $D$ Equation (III).

$\begin{array}{c}B/C=\frac{\$20.6178-\$0.8}{\$18.1271}\\ =1.09\end{array}$

For Alternative New school.

Substitute $\$30.435$ for $B$, $\$27.0906$ for $C$ and $\$0$ for $D$ Equation (III).

$\begin{array}{c}B/C=\frac{\$30.435}{\$27.0906}\\ =1.123\end{array}$

Since the $B/C$ ratio of New school is higher than Expand.

Hence, the alternative New school is desirable.

Calculate the incremental ratio.

${\Delta }_{\text{Cost}}=\left(P{W}_E-P{W}_N\right)$ ...... (IV)

Here the incremental cost between the alternative Expand and New school is ${\Delta }_{\text{Cost}}$, the present worth of cost for alternative Expand is $P{W}_E$ and the present worth of cost for alternative New school is $P{W}_N$.

Substitute $\$18.1271$ for $P{W}_E$ and $\$27.0906$ for $P{W}_N$ in Equation (IV).

$\begin{array}{c}{\Delta }_{\text{Cost}}=\left(\$18.1271-\$27.0906\right)\\ =-\$8.9635\end{array}$

Calculate the ${\Delta }_{Benefit}$ for alternative $\text{Expand}-\text{New}\text{School}$

${\text{Δ}}_{\text{Benefit}}=\left(P_E+P_D-P_S\right)$ ...... (V)

Here the incremental benefit between the alternative Expand and New school is ${\text{Δ}}_{\text{Benefit}}$, the present worth of benefit for alternative Expand is $P_E$ and the present worth of benefit for alternative New school is $P_S$ and present worth of user dis-benefit for Expand is $P_D$.

Substitute $\$20.6178$ for $P_E$ and $\$30.435$ for $P_S$ and $0.8$ for $P_D$ in Equation (V).

$\begin{array}{c}{\Delta }_{Benefit}=\$20.6178+\$0.8-\$30.435\\ =-\$9.0172\end{array}$

Calculate the incremental benefit-cost ratio

$\Delta \left(B/C\right)=\frac{{\text{Δ}}_{\text{Benefit}}}{{\text{Δ}}_{\text{Cost}}}$ ...... (VII)

Substitute $-\$9.0172$ for ${\text{Δ}}_{\text{Benefit}}$ and $-\$8.9635$ for ${\text{Δ}}_{\text{Cost}}$ in Equation (VII).

$\begin{array}{c}\Delta \left(B/C\right)=\frac{-\$9.0172}{-\$8.9635}\\ =1.006\end{array}$.

Conclusion:

The alternative to be selected is Alternative $\text{New}\text{School}$

The incremental $\Delta \left(B/C\right)$ ratio is $1.006$.

To determine

(b)

The alternative to be selected using modified benefit cost ratio and the incremental benefit cost ratio.

Answer to Problem 42P

The alternative to be selected is Alternative $\text{New}\text{School}$

The incremental $\Delta \left(B/C\right)$ ratio is, $2.03$.

Explanation of Solution

Given:

MARR is $8\%$

Useful life is $20$ years.

Calculation:

Calculate the present worth benefit of the alternative $Expand$

Write the expression to calculate the annual benefit.

$PW=A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)-A_{OM}\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)-D$ ...... (VI)

Here, the annual benefit is $A$, the annual operation maintenance cost is $A_{OM}$, the user dis-benefit is $D$ the rate of interest is $i$, the present worth is $PW$ and the number of period is $n$.

Calculate the present worth benefit for the $\text{Expand}$.

Substitute $\$2.1$ for $A$, $8\%$ for $i$ and $20$ for $n$ in Equation (VI).

$\begin{array}{c}P=\$2.1\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-\$0.95\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-0.8\\ =\$2.1\left(9.818\right)-\$0.95\left(9.818\right)-\$0.8\\ =\$10.4907\end{array}$

Calculate the present worth benefit for the $\text{New}\text{School}$.

Substitute $\$3.1$ for $A$, $8\%$ for $i$ and $20$ for $n$ in Equation (VI).

$\begin{array}{c}P=\$3.1\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-\$1.7\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-\$0\\ =\$3.1\left(9.818\right)-\$1.7\left(9.818\right)\\ =\$13.7452\end{array}$

Write the expression to calculate the benefit-cost ratio.

$B/C=\frac{B}{C}$ ....... (VII)

Here, the present worth benefit is $B$, the present worth cost is $C$.

For Alternative $\text{Expand}$.

Substitute $\$10.4907$ for $B$, $\$8.8$ for $C$ in Equation (VII).

$\begin{array}{c}B/C=\frac{\$10.4907}{\$8.8}\\ =1.19\end{array}$

For Alternative $\text{New}\text{School}$.

Substitute $\$13.7452$ for $B$, $\$10.4$ for $C$ Equation (VII).

$\begin{array}{c}B/C=\frac{\$13.7452}{\$10.4}\\ =1.32\end{array}$

Since the $B/C$ ratio of $\text{New}\text{School}$ is higher than $\text{Expand}$.

Hence, the alternative $\text{New}\text{School}$ is desirable.

Calculate the incremental ratio.

${\Delta }_{\text{Cost}}=\left(P{W}_N-P{W}_E\right)$ ...... (VIII)

Here the incremental cost between the alternative $\text{Expand}$ and $\text{New}\text{School}$ is ${\Delta }_{\text{Cost}}$, the present worth of cost for alternative $\text{Expand}$ is $P{W}_E$ and the present worth of cost for alternative $\text{New}\text{School}$ is $P{W}_N$.

Substitute $\$8.8$ for $P{W}_E$ and $\$10.4$ for $P{W}_N$ in Equation (IV).

$\begin{array}{c}{\text{Δ}}_{\text{Cost}}=\left(\$10.4-\$8.8\right)\\ =\$1.6\end{array}$

Calculate the ${\text{Δ}}_{\text{Benefit}}$ for alternative $\text{New}\text{School}-\text{Expand}$.

${\Delta }_{Benefit}=\left(P_N-P_E\right)$ ...... (IX)

Here the incremental benefit between the alternative $\text{Expand}$ and $\text{New}\text{School}$ is ${\Delta }_{Benefit}$, the present worth of benefit for alternative $\text{Expand}$ is $P_E$ and the present worth of benefit for alternative $\text{New}\text{School}$ is $P_N$ and present worth of user dis-benefit for $\text{Expand}$ is $P_D$.

Substitute $\$10.4907$ for $P_E$ and $\$13.4907$ for $P_N$ and $0.8$ for $P_D$ in Equation (V).

$\begin{array}{c}{\text{Δ}}_{\text{Benefit}}=\$13.7452-\$10.4907\\ =\$3.26\end{array}$

Calculate the incremental benefit-cost ratio

$\Delta \left(B/C\right)=\frac{{\text{Δ}}_{\text{Benefit}}}{{\text{Δ}}_{\text{Cost}}}$ ...... (X)

Substitute $\$3.26$ for ${\text{Δ}}_{\text{Benefit}}$ and $\$1.6$ for ${\text{Δ}}_{\text{Cost}}$ in Equation (X).

$\begin{array}{c}\Delta \left(B/C\right)=\frac{\$3.26}{\$1.6}\\ =2.03\end{array}$.

Conclusion:

The alternative to be selected is Alternative $\text{New}\text{School}$

The incremental $\Delta \left(B/C\right)$ ratio is $2.03$.

To determine

(c)

The alternative to be selected using public government version of benefit cost ratio.

Answer to Problem 42P

The alternative to be selected is Alternative $\text{New}\text{School}$.

Explanation of Solution

Given:

MARR is $8\%$.

Useful life is $20$ years.

Calculation:

Write the expression to calculate the present worth of the government cost.

$PW=P+A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)$ ....... (XI)

Here, the annual operation and maintenance cost is $A$, the initial cost is $P$, the rate of interest is $i$, the present worth is $PW$ and the number of period is $n$.

Calculate the present worth cost of the alternative $\text{Expand}$.

Substitute $\$8.8$ for $P$, $\$0.95$ for $A$, $8\%$ for $i$ and $20$ for $n$ in Equation (XI).

$\begin{array}{c}PW=\$8.8+\$0.95\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)\\ =\$8.8+\$0.95\left(9.818\right)\\ =\$18.1271\end{array}$

Calculate the present worth government cost of the alternative $\text{New}\text{School}$.

Substitute $\$10.4$ for $P$, $\$1.7$ for $A$, $8\%$ for $i$ and $20$ for $n$ in Equation (XI).

$\begin{array}{c}PW=\$10.4+\$1.7\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)\\ =\$10.4+\$1.7\left(9.818\right)\\ =\$27.0906\end{array}$

Calculate the present worth benefit of the alternative $\text{Expand}$.

Write the expression to calculate the annual benefit of the public.

$PW=A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)-D$ ...... (XII)

Here, the annual benefit is $A$, the user dis-benefit is $D$ the rate of interest is $i$, the present worth is $PW$ and the number of period is $n$.

Calculate the present worth public benefit for the $\text{expand}$.

Substitute $\$2.1$ for $A$, $8\%$ for $i$, $\$0.8$ for $D$ and $20$ for $n$ in Equation (XII).

$\begin{array}{c}P=\$2.1\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-\$0.8\\ =\$2.1\left(9.818\right)-\$0.8\\ =\$19.8178\end{array}$

Calculate the present worth public benefit for the $\text{New}\text{School}$.

Substitute $\$3.1$ for $A$, $8\%$ for $i$, $0$ for $D$ and $20$ for $n$ in Equation (XII).

$\begin{array}{c}P=\$3.1\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-\$0\\ =\$3.1\left(9.818\right)\\ =\$30.435\end{array}$

Write the expression to calculate the benefit-cost ratio.

$B/C=\frac{B}{C}$ ....... (XIII)

Here, the present worth public benefit is $B$, the present worth of government cost is $C$.

For Alternative $\text{Expand}$.

Substitute $\$19.8178$ for $B$, $\$18.1271$ for $C$ in Equation (XIII).

$\begin{array}{c}B/C=\frac{\$19.8178}{\$18.1271}\\ =1.09\end{array}$

For Alternative $\text{New}\text{School}$.

Substitute $\$30.435$ for $B$, $\$27.0906$ for $C$ Equation (XIII).

$\begin{array}{c}B/C=\frac{\$30.435}{\$27.0906}\\ =1.123\end{array}$

Since the $B/C$ ratio of $\text{New}\text{School}$ is higher than $\text{Expand}$.

Hence, the alternative $\text{New}\text{School}$ is desirable.

To determine

(d)

The alternative to be selected using present worth index.

Answer to Problem 42P

The alternative to be selected is Alternative $\text{New}\text{School}$.

Explanation of Solution

Given:

MARR is $8\%$

Useful life is $20$ years.

Calculation:

Write the expression to calculate the present worth for the alternative $\text{expand}$. $PW=-P+A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)-A_{OM}\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)-D$ ...... (XIV)

Here, the present worth is $PW$, the initial cost is $P$, the annual benefit is $A$, the annual operation and maintenance cost is $A_{OM}$ and the user dis-benefit is $D$.

Substitute $\$8.8$ for $P$, $\$2.1$ for $A$, $\$0.95$ for $A_{OM}$, $0.8$ for $D$, $8\%$ for $i$ and $20$ for $n$ in Equation (XIV).

$\begin{array}{c}PW=-\$8.8+\$2.1\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-\$0.95\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-\$0.8\\ =-\$8.8+\$2.1\left(9.818\right)-\$0.95\left(9.818\right)-\$0.8\\ =-7.11\end{array}$

Calculate the present worth for alternative $\text{New}\text{School}$.

Substitute $\$10.4$ for $P$, $\$3.1$ for $A$, $\$1.7$ for $A_{\text{OM}}$, $0$ for $D$, $8\%$ for $i$ and $20$ for $n$ in Equation (XIV).

$\begin{array}{c}PW=-\$10.4+\$3.1\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-\$1.7\left(\frac{{\left(1+0.08\right)}^{20}-1}{0.08{\left(1+0.08\right)}^{20}}\right)-\$0\\ =-\$10.4+\$3.1\left(9.818\right)-\$1.7\left(9.818\right)\\ =3.3452\end{array}$

The present worth of alternative $\text{New}\text{School}$ is positive.

Hence, the alternative $\text{New}\text{School}$ is desirable.

Conclusion:

The alternative to be selected is Alternative $\text{New}\text{School}$.

To determine

(e)

The consistency of data from $a$ to $d$ and the largest value with reason.

Explanation of Solution

Yes, the result remain constant throughout from $a$ to $d$ that is the alternative $\text{New}\text{School}$ should selected.

The modified benefit cost ratio gives the highest value with an incremental benefit-cost ratio of $2.03$ because in this method all the operation and the maintenance cost and user disbenefit are included as a deduction in Benefits.

To determine

(f)

The ethical issues involved in the overcrowded schools dilemma in terms of stakeholders and impacts.

Explanation of Solution

The ethical issues involved in overcrowded school are as follows:

Physiological Problems: Overcrowded classroom reduces the quality of air for the individual breathe which is not good for the health of any individual.

Temperature also influences social behaviour like aggression, impatience.

Psychological and Sociological Problem: The personal and social development of individual are not improved which is one of the basic principles of sending an individual to school.

The impacts of overcrowded school are as follows:

The academic performances of the individuals become poor.

The health of student is at risk.

The Moral corruption of the students as they are not fed with proper moral teachings.

Moreover, there is a lot of stress on teachers to teach such a large gathering of students.