Chapter 10, Problem 15P

To determine

(a)

The joint probability distribution for first cost and net revenue.

Answer to Problem 15P

Net Revenue	First cost	Joint Probability
$\$70,000$	$\$300,000$	$0.06$
$\$90,000$	$\$300,000$	$0.10$
$\$100,000$	$\$300,000$	$0.04$	Optimistic
$\$70,000$	$\$400,000$	$0.15$
$\$90,000$	$\$400,000$	$0.25$	Most likely
$\$100,000$	$\$400,000$	$0.10$
$\$70,000$	$\$600,000$	$0.09$	Pessimistic
$\$90,000$	$\$600,000$	$0.15$
$\$100,000$	$\$600,000$	$0.06$

Explanation of Solution

Calculation:

Calculate the joint probability.

Net Revenue $\left(a\right)$	Probability $\left(b\right)$	First cost $\left(c\right)$	Probability $\left(d\right)$	Joint Probability $\begin{array}{c}\left(e\right)\\ \left(b\times d\right)\end{array}$
$\$70,000$	$0.3$	$\$300,000$	$0.2$	$0.06$
$\$90,000$	$0.5$	$\$300,000$	$0.2$	$0.10$
$\$100,000$	$0.2$	$\$300,000$	$0.2$	$0.04$	Optimistic
$\$70,000$	$0.3$	$\$400,000$	$0.5$	$0.15$
$\$90,000$	$0.5$	$\$400,000$	$0.5$	$0.25$	Most likely
$\$100,000$	$0.2$	$\$400,000$	$0.5$	$0.10$
$\$70,000$	$0.3$	$\$600,000$	$0.3$	$0.09$	Pessimistic
$\$90,000$	$0.5$	$\$600,000$	$0.3$	$0.15$
$\$100,000$	$0.2$	$\$600,000$	$0.3$	$0.06$
Total				$1.00$

Conclusion:

The joint probability is shown in table.

To determine

(b)

The present worth for optimistic, pessimistic and most likely estimates.

Answer to Problem 15P

Present worth for optimistic scenario is $\$265,000$.

Present worth for most likely scenario is $\$108,500$.

Present worth for pessimistic scenario is $\$204,500$.

Explanation of Solution

Calculation:

Write the formula to calculate the present worth for optimistic scenario.

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

Here, Present worth is PW, first cost is P, Annual benefit is A and the rate of interest is i.

Substitute $\$300,000$ for P, $\$100,000$ for $A$, $12\%$ for i and $10$ years for n in Equation (I).

$\begin{array}{c}P{W}_O=-\$300,000+\$100,000\left(\frac{{\left(1+0.12\right)}^{10}-1}{0.12{\left(1+0.12\right)}^{10}}\right)\\ =-\$300,000+\$100,000\left(5.65\right)\\ =\$265,000\end{array}$

Calculate the present worth for most likely scenario.

Substitute $\$400,000$ for P, $\$90,000$ for $A$, $12\%$ for i and $10$ years for n in Equation (I).

$\begin{array}{c}P{W}_M=-\$400,000+\$90,000\left(\frac{{\left(1+0.12\right)}^{10}-1}{0.12{\left(1+0.12\right)}^{10}}\right)\\ =-\$400,000+\$90,000\left(5.65\right)\\ =\$108,500\end{array}$

Calculate the present worth for pessimistic scenario.

Substitute $\$600,000$ for P, $\$70,000$ for $A$, $12\%$ for i and $10$ years for n in Equation (I).

$\begin{array}{c}P{W}_P=-\$600,000+\$70,000\left(\frac{{\left(1+0.12\right)}^{10}-1}{0.12{\left(1+0.12\right)}^{10}}\right)\\ =-\$600,000+\$70,000\left(5.65\right)\\ =\$204,500\end{array}$.

Conclusion:

Thus, the present worth for optimistic scenario is $\$265,000$.

Present worth for most likely scenario is $\$108,500$.

Present worth for pessimistic scenario is $\$204,500$.