Chapter 6, Problem 70P

To determine

The EUAW of the alternatives and see if there is a cross over point in the decision.

Answer to Problem 70P

Therefore, the most economical choice is, Diesel.

Explanation of Solution

The cars average distance covered by car is $80000\text{km}$ a year and interest is $6\%$.

Calculation:

Calculate the annual cost of each fuel for each case.

$\text{Annual}\text{cost}=\frac{\text{Distance}\text{covered}}{\text{Mileage}\left(\text{km}/\text{litre}\right)}\times \text{Fuel}\text{cost}\text{per}\text{litre}$         ...... (I)

Calculate the annual cost for diesel.

Substitute $80000\text{km}$ for distance covered and $16\text{km}/\text{liter}$ for mileage and $88\%$ for fuel cost per litre in Equation (I).

$\begin{array}{c}\text{Annual}\text{cost}=\frac{80000\text{km}}{\text{16}\left(\text{km}/\text{litre}\right)}\times \text{0}\text{.88}\\ =\$4400\end{array}$

Calculate the annual cost for Gasoline.

Substitute $80000\text{km}$ for distance covered and $11\text{km}/\text{liter}$ for mileage and $92\%$ for fuel cost per litre in Equation (I).

$\begin{array}{c}\text{Annual}\text{cost}=\frac{80000\text{km}}{\text{11}\left(\text{km}/\text{litre}\right)}\times \text{0}\text{.92}\\ =\$6691\end{array}$

Alternative1: Diesel

Given:

Vehicle cost is $\$24000$

Annual repairs $\$900$

Annual insurance premium is $\$1000$

Salvage value is $\$4000$

Interest rate $6\%$

Time period is $5\text{years}$

$\begin{array}{l}EUAC=\left(P-S\right)\left(\frac{A}{P},i,n\right)+\text{other}\text{annual}\text{cost}+\text{tax}\\ EUAC=\left(P-S\right)\left(\frac{A}{P},i,n\right)+\$4000\times 0.06+\$4400+\$900+\$1000\end{array}$         ...... (II)

Here, vehicle cost is P, salvage value is S.

Calculate the factor $\left(\frac{A}{P},i,n\right)$.

$\left(\frac{A}{P},i,n\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$         ...... (III)

Substitute $6\%$ for interest and $5\text{years}$ in Equation (III).

$\begin{array}{c}\left(\frac{A}{P},i,n\right)=\left[\frac{0.06{\left(1+0.06\right)}^5}{{\left(1+0.06\right)}^5-1}\right]\\ =0.2374\end{array}$

Substitute $0.2374$ for $\left(\frac{A}{P},i,n\right)$, $\$24000$ for P, and $\$4000$ for S in Equation (II).

$\begin{array}{c}\text{EUAC}=\left(\$240000-\$4000-S\right)\times 0.2374+\$4000\times 0.06+\$4400+\$900+\$1000\\ =\$11288\end{array}$

Here, EUAB is zero, thus, $\text{EUAC}=\text{EUAW}$ in this case.

Alternative 2: Gasoline

Given:

Vehicle cost is $\$19000$

Annual repairs $\$700$

Annual insurance premium is $\$1000$

Salvage value is $\$6000$

Interest rate $6\%$

Time period is $4\text{years}$

$\begin{array}{l}\text{EUAC}=\left(P-S\right)\left(\frac{A}{P},i,n\right)+\text{other}\text{annual}\text{cost}+\text{tax}\\ \text{EUAC}=\left(P-S\right)\left(\frac{A}{P},i,n\right)+\$6000\times 0.06+\$6691+\$700+\$1000\end{array}$         ...... (IV)

Here, vehicle cost is P, salvage value is S.

Calculate the factor $\left(\frac{A}{P},i,n\right)$.

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

Substitute $6\%$ for interest and $4\text{years}$ in Equation (V).

$\begin{array}{c}\left(\frac{A}{P},i,n\right)=\left[\frac{0.06{\left(1+0.06\right)}^4}{{\left(1+0.06\right)}^4-1}\right]\\ =0.2886\end{array}$

Substitute $0.2886$ for $\left(\frac{A}{P},i,n\right)$, $\$19000$ for P, and $\$6000$ for S in Equation (IV).

$\begin{array}{l}\text{EUAC}=\left(\$19000-\$6000\right)\times 0.2886+\$6000\times 0.06+\$6691+\$700\\ =\$12503\end{array}$

Here, EUAB is zero, thus, $\text{EUAC}=\text{EUAW}$ in this case.

Tabulate the EUAC of alternatives A and B.

		Outflows
Year	Diesel	Gasoline	Interest rate $2\%$	EUAW A	EUAW B
$0$	$-\$11288$	$-\$12503$	$0$	$\$1612.57$	$\$2083.83$
$1$	$-\$2258$	$-\$100$	$0.2$	$\$3131.56$	$\$3759.72$
$2$	$-\$2258$	$-\$600$	$0.4$	$\$4988.42$	$\$5767.13$
$3$	$-\$2258$	$-\$1100$	$0.6$	$\$7034.87$	$\$7977.28$
$4$	$-\$2258$	$-\$1600$	$0.8$	$\$9180.35$	$\$10305.39$
$5$	$-\$2258$	$-\$2100$	$2$	$\$9725.94$	$\$10899.43$

    

    Figure (1)

Cross over point is a point on the chart when an indicator and security intersect and is used to predict any movements in the market. Thus, in this case the lines of both alternatives are parallel so there is no cross over point.

Conclusion:

Therefore, the most economical choice is, Diesel.