Chapter 6, Problem 49P

To determine

The more economical choice.

Answer to Problem 49P

The most economical choice is, Diesel.

Explanation of Solution

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

The interest is $6\%$.

Calculation:

Calculate the annual cost of each fuel for each case.

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

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

Alternative 1: Diesel

Given:

Vehicle cost is $\$24000$

Annual repairs $\$900$

Annual insurance premium is $\$1000$

Salvage value is $\$4000$

Interest rate $6\%$

Time period is $5\text{years}$.

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

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

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).

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

Conclusion:

EUAC of diesel is less than the Gasoline. Thus Diesel should be selected.