Chapter 10, Problem 27P

To determine

(a)

The expected value of equivalent uniform annual cost.

Answer to Problem 27P

The expected value of equivalent uniform annual cost is $\$7021.35$.

Explanation of Solution

Given:

Optimistic probability is $20\%$, most likely probability is $50\%$ and pessimistic probability is $30\%$.

From question number $2$ obtain equivalent uniform annual cost $\left(EUAC\right)$ for optimistic, most likely and pessimistic case.

The values are:

Cases	EUAC
Pessimistic $\left(5000\text{miles}/\text{year}\right)$	$\$3783.50$
Most likely $\left(10000\text{miles}/\text{year}\right)$	$\$6727$
Optimistic $\left(15000\text{miles}/\text{year}\right)$	$\$9670.50$

Concept used:

The expected value of equivalent uniform annual cost can be calculated by the summation of the product of equivalent uniform cost and their probabilities in optimistic, most likely and in pessimistic case.

Calculation:

Let expected value in optimistic case be ${\left(E\right)}_O$.

Calculate the expected value in optimistic case ${\left(E\right)}_O$.

${\left(E\right)}_O=\left(EUA{C}_o\times P_o\right)$

Here, probability is $P_o$ and $EUA{C}_o$ in optimistic case.

Substitute $\$3783.50$ for $EUA{C}_o$ and $0.20$ for $P_o$.

$\begin{array}{c}{\left(E\right)}_O=\left(\$3783.50\times 0.20\right)\\ =\$756.70\end{array}$

Let expected value in most likely case be ${\left(E\right)}_M$.

Calculate the expected value in most likely case ${\left(E\right)}_M$.

${\left(E\right)}_M=\left(EUA{C}_m\times P_m\right)$

Here, probability is $P_m$ and $EUA{C}_m$ in most likely case.

Substitute, $\$6727$ for $EUA{C}_m$ and $0.50$ for $P_m$.

$\begin{array}{c}{\left(E\right)}_M=\left(\$6727\times 0.50\right)\\ =\$3363.50\end{array}$

Let expected value in pessimistic case be ${\left(E\right)}_P$

Calculate the expected value in pessimistic case ${\left(E\right)}_P$.

${\left(E\right)}_P=\left(EUA{C}_p\times P_p\right)$

Here, probability is $P_p$ and $EUA{C}_p$ in pessimistic case.

Substitute $\$9670.50$ for $EUA{C}_p$ and $0.30$ for $P_p$.

$\begin{array}{c}{\left(E\right)}_P=\left(\$9670.50\times 0.30\right)\\ =\$2901.15\end{array}$

Calculate the expected value of equivalent uniform of annual cost.

Let the expected value of equivalent uniform of annual cost denoted by ${\left(E\right)}_{EUAC}$.

${\left(E\right)}_{EUAC}={\left(E\right)}_O+{\left(E\right)}_M+{\left(E\right)}_P$

Substitute $\$756.70$ for ${\left(E\right)}_O$, $\$3363.50$ for ${\left(E\right)}_M$ and $\$2901.15$ for ${\left(E\right)}_P$.

$\begin{array}{c}{\left(E\right)}_{EUAC}=\$756.70+\$3363.50+\$2901.15\\ =\$7020.35\end{array}$.

Conclusion:

Thus, the expected value of equivalent uniform of annual cost is $\$7021.35$.

To determine

(b)

The expected value for number of miles and their corresponding annual cost.

Answer to Problem 27P

The expected value for number of miles are $10500$.

The corresponding annual cost is $\$7373.54$.

Explanation of Solution

Given:

Fuel cost is $\$1500$.

Cost for oil, repair, tires is $\$1500$ per year.

Initial investment is $\$14000$.

Drop in cent per mile is $10\text{cent}$.

Interest rate is $6\%$.

Number of years is $5$.

Concept used:

The expected of number of miles is determined by the summation of the product of number of miles in different situations and their respective probabilities.

Calculation:

Calculate the expected number of miles in optimistic case ${\left(E_M\right)}_O$.

${\left(E_M\right)}_O={\left(N_M\times P\right)}_O$

Here, probability is $P$ and number of miles in optimistic case is $N_M$.

Substitute, $5000$ for $N_M$ and $0.20$ for $P$.

$\begin{array}{c}{\left(E_M\right)}_O=\left(5000\times 0.20\right)\\ =1000\end{array}$

Let expected number of miles in most likely case be ${\left(E_M\right)}_M$.

Calculate the expected number of miles in most likely case.

${\left(E_M\right)}_M={\left(N_M\times P\right)}_M$

Substitute, $10000$ for $N_M$ and $0.50$ for $P$.

$\begin{array}{c}{\left(E_M\right)}_M=\left(10000\times 0.50\right)\\ =5000\end{array}$

Let expected number of years in pessimistic case be ${\left(E_M\right)}_P$.

Calculate the expected number of miles in pessimistic case.

${\left(E_M\right)}_P={\left(N_M\times P\right)}_P$

Substitute, $15000$ for $N_M$ and $0.30$ for $P$.

$\begin{array}{c}{\left(E_M\right)}_P=\left(15000\times 0.30\right)\\ =4500\end{array}$

Calculate the expected number of miles ${\left(E\right)}_r$.

${\left(E\right)}_r={\left(E_M\right)}_O+{\left(E_M\right)}_M+{\left(E_M\right)}_P$

Substitute, $1000$ for ${\left(E_M\right)}_O$, $5000$ for ${\left(E_M\right)}_M$, $4500$ for ${\left(E_M\right)}_P$.

$\begin{array}{c}{\left(E\right)}_r=1000+5000+4500\\ =10500\text{miles}\end{array}$

The annual cost is the sum of fuel cost and oil, repair and tires cost per year.

Calculate the annual cost $\left(a\right)$.

$a=f+t$

Here, fuel cost is $f$ and oil, repair and tires cost is $t$.

Substitute, $\$1500$ for $f$ and $\$1500$ for $t$.

$\begin{array}{c}a=\$1500+\$1500\\ =\$3000\end{array}$

Calculate the salvage value per year $\left(S\right)$.

$S=\left(E_M\right)\times \frac{d}{100}$

Here, expected number of miles per year is $\left(E_M\right)$ and drop in cent per mile is $d$.

Substitute, $10500$ for $\left(E_M\right)$ and $10$ for $d$.

$\begin{array}{c}S=10500\times \frac{10}{100}\\ =\$1050\end{array}$

The total annual cost is the sum of salvage cost and annual cost.

Calculate the total annual cost $\left(A\right)$.

$A=a+S$

Substitute, $\$3000$ for $a$ and $\$1050$ for $S$.

$\begin{array}{c}A=\$3000+\$1050\\ =\$4050\end{array}$

Calculate the equivalent annual cost. ${\left(A\right)}_e$.

${\left(A\right)}_e=\text{Annual}\text{cost}+\left(\frac{r\times \text{initial}\text{investemnt}}{1-\frac{1}{{\left(1+r\right)}^n}}\right)$

Here, interest rate is $r$ and number of years is $n$.

Substitute, $\$4050$ for $\text{Annual}\text{cost}$, $0.06$ for $r$, $\$14000$ for $\text{Initial}\text{investemnt}$ and $5$ for $n$.

$\begin{array}{c}{\left(A\right)}_e=\$4050+\left(\frac{0.06\times \$14000}{1-\frac{1}{{\left(1+.06\right)}^5}}\right)\\ =\$7373.54\end{array}$.

Conclusion:

Thus, the expected value for number of miles are $10500$.

The corresponding annual cost is $\$7373.54$.

To determine

(c)

If the answers in part $\left(a\right)$ and part $\left(b\right)$ match.

Answer to Problem 27P

Answer in part $\left(a\right)$ and part $\left(b\right)$ are approximately the same.

Explanation of Solution

Answer in part $\left(a\right)$ and part $\left(b\right)$ are approximately the same.

This is because in part $\left(a\right)$, the expected value of equivalent uniform annual cost is calculated with each possible mileage which is adjusted in accordance to their probability.

In part $\left(b\right)$, the annual cost is calculated by first adjusting with their probability and then is calculated for possible mileage.

The same equation is obtained in both the parts. Therefore, answers are approximately same.