Chapter 10, Problem 29P

To determine

(a)

The expected value of the present worth.

Answer to Problem 29P

The expected value of the present worth is $\$10,158.5$.

Explanation of Solution

Given:

The present worth of optimistic projection is $\$94,125$.

The present worth of most-likely projection is $\$25,300$.

The present worth of pessimistic projection is $-\$71,055$.

The probability of optimistic projection is $20\%$.

The probability of most-likely projection is $50\%$.

The probability of pessimistic projection are $30\%$.

Concept used:

Write the expression to calculate the probable present worth.

$PW=W\times P$ ...... (I).

Here, the probable present worth is $PW$, the present worth is $W$ and the probability is $P$.

Write the expression to calculate the expected value of the present worth.

$EW=\left(P{W}_O+P{W}_M+P{W}_P\right)$ ...... (II).

Here, the expected value of the present worth is $EW$, the probable optimistic present worth is $P{W}_O$, the probable most-likely present worth is $P{W}_M$ and the probable pessimistic present worth is $P{W}_P$.

Calculation:

Calculate the probable optimistic present worth.

Substitute $\$94,125$ for $W_O$ and $0.2$ for $P_O$ in the Equation-(I).

$\begin{array}{c}P{W}_O=W_O\times P_O\\ =\$94,125\times 0.2\\ =\$18,825\end{array}$

Calculate the probable most-likely present worth.

Substitute $\$25,300$ for $W_M$ and $0.5$ for $P_M$ in the Equation-(I).

$\begin{array}{c}P{W}_M=W_M\times P_M\\ =\$25,300\times 0.5\\ =\$12,650\end{array}$

Calculate the probable pessimistic present worth.

Substitute $-\$71,055$ for $W_P$ and $0.3$ for $P_P$ in the Equation-(I).

$\begin{array}{c}P{W}_P=W_P\times P_P\\ =-\$71,055\times 0.3\\ =-\$21,316.5\end{array}$

Calculate the expected value of the present worth.

Substitute $\$18,825$ for $P{W}_O$, $\$12,650$ for $P{W}_M$ and $-\$21,316.5$ for $P{W}_P$ in Equation-(II).

$\begin{array}{c}EW=\left(P{W}_O+P{W}_M+P{W}_P\right)\\ =\left(\$18,825+\$12,650-\$21,316.5\right)\\ =\$10,158.5\end{array}$.

Conclusion:

Thus, the expected value of the present worth is $\$10,158.5$.

To determine

(b)

The expected value of annual savings and the corresponding present worth.

Answer to Problem 29P

The expected value of annual savings is $\$18,900$.

The present worth is $\$10,158.5$.

Explanation of Solution

Given information:

The saving of optimistic projection is $\$25,000$.

The saving of most-likely projection is $\$20,000$.

The saving of pessimistic projection is $\$13,000$.

The probability of optimistic projection is $20\%$.

The probability of most-likely projection is $50\%$.

The probability of pessimistic projection is $30\%$.

The interest rate is $6\%$.

The transmission line have a life of $30$ years.

The outflow is $\$250,000$.

Concept used:

Write the expression to calculate the probable saving.

$PS=S\times P$ ...... (III).

Here, the probable saving is $PS$, the saving is $S$ and probability is $P$.

Write the expression to calculate the expected value of saving & annual inflow.

$ES=P{S}_0+P{S}_M+P{S}_P$ ...... (IV).

Here, the expected value of saving & annual inflow is $ES$, the probable optimistic saving is $P{S}_O$, the probable most-likely saving is $P{S}_M$ and the probable pessimistic saving is $P{S}_P$.

Write the expression to calculate $\left(\frac{P}{A},6\%,30\text{years}\right)$.

$\left(\frac{P}{A},i,N\right)=\frac{{\left(1+i\right)}^N-1}{i\times {\left(i+1\right)}^N}$ ...... (V).

Here, the interest rate is $i$ and the time period is $N$.

Write the expression to calculate the present worth.

$PW=-\left(\text{outflow}\right)+\left[ES\times \left(\frac{P}{A},6\%,30\text{years}\right)\right]$ ...... (VI).

Here, the expected value of saving & annual inflow is $ES$.

Calculation:

Calculate the probable optimistic saving.

Substitute $\$25,000$ for $S_O$ and $0.2$ for $P_O$ in Equation-(III).

$\begin{array}{c}P{S}_O=S_O\times P_O\\ =\$25,000\times 0.2\\ =\$5,000\end{array}$

Calculate the probable most-likely saving.

Substitute $\$20,000$ for $S_M$ and $0.5$ for $P_M$ in Equation-(III).

$\begin{array}{c}P{S}_M=S_M\times P_M\\ =\$20,000\times 0.5\\ =\$10,000\end{array}$

Calculate the probable pessimistic saving.

Substitute $\$13,000$ for $S_P$ and $0.3$ for $P_P$ in Equation-(III).

$\begin{array}{c}P{S}_P=S_P\times P_P\\ =\$13,000\times 0.3\\ =\$3,900\end{array}$

Calculate the expected value of saving & annual inflow.

Substitute $\$5,000$ for $P{S}_O$, $\$10,000$ for $P{S}_M$ and $\$3,900$ for $P{S}_P$ in Equation-(IV).

$\begin{array}{c}ES=P{S}_0+P{S}_M+P{S}_P\\ =\$5,000+\$10,000+\$3,900\\ =\$18,900\end{array}$

The annual inflow is $\$18,900$.

Calculate the value of $\left(\frac{P}{A},6\%,30\text{years}\right)$.

Substitute $6\%$ for $i$ and $30\text{year}$ for $N$ Equation-(V).

$\begin{array}{c}\left(\frac{P}{A},i,N\right)=\frac{{\left(1+i\right)}^N-1}{i\times {\left(i+1\right)}^N}\\ =\frac{{\left(1+0.06\right)}^{30}-1}{0.06\times {\left(0.06+1\right)}^{30}}\\ =13.765\end{array}$

Calculate the present worth.

Substitute $\$250,000$ for $\text{outflow}$, $\$18,900$ for $ES$ and $13.765$ for $\frac{P}{A},6\%,30\text{years}$ value in Equation-(VI).

$\begin{array}{c}PW=-\left(\text{outflow}\right)+\left[ES\times \left(\frac{P}{A},6\%,30\text{years}\right)\right]\\ =-\left(\$250,000\right)+\left(\$18,900\times 13.765\right)\\ =-\$250,000+\$260,158.5\\ =\$10,158.5\end{array}$

The present worth is $\$10,158.5$.

Conclusion:

Thus, the expected value of annual savings is $\$18,900$.

The present worth is $\$10,158.5$.

To determine

(c)

If the answers (a) and (b) match.

Answer to Problem 29P

Yes, the answer in both (a) and (b) part are same.

Explanation of Solution

In part (a) the different probable present worth has been calculated for each saving and expected present worth has been calculated.

In part (b) the different probable saving has been calculated for each probability and annual saving has been calculated. Then by annual saving and outflow the expected present worth has been calculated.

Both the parts come to the same equation. This is why the net expected value is same.

Conclusion:

Yes, the answer in both (a) and (b) part are same.