Chapter 10, Problem 36P

To determine

i.

Present Worth of first Cost and revenue

Expected Worth of Present worth.

Answer to Problem 36P

Present worth of first cost is $95500.

Expected Worth of Present worth is -$19295.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Calculation:

First Costs	Probability	Net Revenues	Probability
$300,000	0.2	$70,000	0.3
$400,000	0.5	$90,000	0.5
$600,000	0.3	$100,000	0.2

$\begin{array}{l}PW=-C+A(P/A,i,n)\\ =-\$300000+\$70000(P/A,0.12,10)\\ =-\$300000+\$70000\times 5.650\\ =-\$300000+\$395500\\ =\$95500\end{array}$

Where,

PW is the present value.

C is the initial cost of the option examined.

A is the annual amount that is consistent in the cash flow series.

P is the present worth value of the time series.

i is the interest rate.

n is the number of terms that the money is for.

Present worth for pessimistic is $95500.

First Costs	Probability	Net Revenues	Probability	Present Worth
$300,000	0.2	$70,000	0.3	$95,500
$400,000	0.5	$90,000	0.5	($108,500)
$600,000	0.3	$100,000	0.2	$35,000

Expected worth of present worth

Useful Life = 10 year

Interest Rate = 12%

$\begin{array}{l}\text{EV}\text{of}\text{Present}\text{worth}\\ \text{=}{\text{PW}}_{\text{P}}\text{×}\text{P}\text{(joint}\text{Probability)}\\ \text{+}{\text{PW}}_{\text{M}}\text{×}\text{P}\text{(joint}\text{Probability)}\\ \text{+}{\text{PW}}_{\text{O}}\text{×}\text{P}\text{(joint}\text{Probability)}\end{array}$

$\begin{array}{l}=\$95500\times 0.06+\$108500\times 0.25+\$35000\times 0.06\\ =-\$19295\end{array}$.

Conclusion:

Expected Worth of Present worth is -$19295

Present worth of first cost is $95500.

To determine

ii.

Expected first costs, net revenues and present worth for the expected values.

Answer to Problem 36P

Present worth of expected value is -$45900.

Explanation of Solution

Calculation:

$\begin{array}{l}PW=-Expected(first\cos t)+Expected(netrevenue)A(P/A,i,n)\\ =-\$440000+\$86000\times 5.650\\ =-\$440000+\$485900\\ =-\$45900\end{array}$

Expected first cost = $44000

Expected net Revenue

$\begin{array}{l}\text{E}\text{(Net}\text{Revenue)}\\ \text{=}\text{0}\text{.3×\$70000+0}\text{.5×\$90000+0}\text{.2×\$100000}\\ \text{=}\text{\$21000+\$45000+\$20000}\\ \text{=}\text{\$86000}\end{array}$

Expected Net Worth = $86000

$\begin{array}{l}PW=-Expected(first\cos t)+Expected(netrevenue)A(P/A,i,n)\\ =-\$440000+\$86000\times 5.650\\ =-\$440000+\$485900\\ =-\$45900\end{array}$.

Conclusion:

Present worth of expected value is -$45900.

To determine

iii.

Answers in part a and b match or not.

Answer to Problem 36P

Both values in both parts don’t match.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Concept used:

Both values in both parts don’t match. As in the first part, joint probability is used to get the expected worth of present worth for the first value and revenue whereas in second part expected first cost and and expected net revenue is calculated separately and after that present worth is calculated.

Conclusion:

Both values in both parts don’t match.