Chapter 11, Problem 23QP

a)

Summary Introduction

To determine: The number of miles needed to drive per year for making decision to buy Vehicle LRX 450.

Introduction:

Break-even point is a point at which there is no profit or loss. Under this concept, each sale of product from the point of break-even will generate adequate profit. In case, a dip in sale at this point can incur huge loss for the company.

Answer to Problem 23QP

The number of miles needed to drive per year is 28,747.07 miles per year.

Explanation of Solution

Given information:

LRX 450 Vehicle Company has an annual ownership cost of $300. The list price of Vehicle LRX 450 including taxes is $5,565 and the number of years is 6. The number of miles per gallon is 21 and the cost per gallon on gasoline is $3.25.

Formulae:

The formula to calculate the total additional cost:

$\text{Total additional cost}=\text{List price}+\left(\text{Annual ownership costs}\times \text{Number of years}\right)$

The formula to calculate cost per mile:

$\text{Cost per mile}=\frac{\text{Cost per gallon of gasoline}}{\text{Number of miles per gallon}}$

The formula to calculate the savings per mile driven for Vehicle LRX 450:

$\text{Saving per mile driven for Vehicle LRX 450}=\left(\begin{array}{l}\text{Cost per mile for Vehicle LRX}350\\ -\text{ Cost per mile for Vehicle LRX}450\end{array}\right)$

The formula to calculate the total break-even point:

$\text{Break-even miles}=\frac{\text{Total additional}\text{costs}}{\text{Savings per mile}}$

The formula to calculate the number of miles per year:

$\text{Number of miles per year}=\frac{\text{Total break-even miles}}{\text{Number of years of ownership}}$

Steps to determine the number of miles needed to drive per year for making decision to buy Vehicle LRX 450:

• First, determine the total additional cost of Vehicle $\text{LRX}$ 450.
• Secondly, find out the cost per mile for both vehicles. It is determined by dividing the cost for a gallon of gasoline with the miles per gallon. Even, determine the savings per mile driven for Vehicle $\text{LRX}$ 450 taking into account the difference between the costs per mile for both vehicles.
• Then, estimate the total break-even miles. It is computed by dividing the total additional cost with the saving per miles.
• Finally, the miles required to drive per year is computed by dividing the total break-even miles with the number of years of ownership.

Compute the total additional cost:

$\begin{array}{c}\text{Total additional cost}=\text{List price}+\left(\text{Annual ownership costs}\times \text{Number of years}\right)\\ =\$5,565+\left(\$300\times 6\right)\\ =\$5,565+\$1,800\\ =\$7,365\end{array}$

Hence, the total additional cost is $7,365.

Compute the cost per mile for Vehicle LRX 350:

$\begin{array}{c}\text{Cost per mile for Vehicle LRX 350}=\frac{\text{Cost per gallon of gasoline}}{\text{Number of miles per gallon}}\\ =\frac{\$3.25}{21}\\ =\$0.1547\end{array}$

Hence, the cost per mile for Vehicle LRX 350 is $0.1547.

Compute the cost per mile for Vehicle LRX 450:

$\begin{array}{c}\text{Cost per mile for Vehicle LRX 450}=\frac{\text{Cost per gallon of gasoline}}{\text{Number of miles per gallon}}\\ =\frac{\$3.25}{29}\\ =\$0.1120\end{array}$

Hence, the cost per mile for Vehicle LRX 450 is $0.1120.

Compute the savings per mile driven for Vehicle LRX 450:

$\begin{array}{c}\text{Saving per mile driven for Vehicle LRX }450=\left(\begin{array}{l}\text{Cost per mile for Vehicle LRX }350\\ -\text{Cost per mile for Vehicle LRX }450\end{array}\right)\\ =\$0.1547-\$0.1120\\ =\$0.0427\end{array}$

Hence, the savings per mile driven for Vehicle LRX 450 is $0.0427.

Compute the total break-even point:

$\begin{array}{c}\text{Break-even miles}=\frac{\text{Total additional}\text{costs}}{\text{Savings per mile}}\\ =\frac{\$7,365}{\$0.0427}\\ =172,482.43\text{ miles}\end{array}$

Hence, the total break-even point is 172,482.43 miles.

Compute the miles needed to drive per year:

$\begin{array}{c}\text{Number of miles per year}=\frac{\text{Total breakeven miles}}{\text{Number of years of ownership}}\\ =\frac{172,482.43}{6}\\ =28,747.07\text{miles per year}\end{array}$

Hence, the miles needed to drive per year are 28,747.07miles per year.

b)

Summary Introduction

To determine: The price per gallon on making decision for buying Vehicle LRX 450.

Answer to Problem 23QP

The price per gallon on making decision for buying Vehicle LRX 450 is $6.196.

Explanation of Solution

Given information:

The total number of miles per year is 15,000 and the number of years is 6. The number of miles per gallon is 21.

The formula to calculate the total miles driven:

$\text{Total miles driven}=\left(\begin{array}{c}\text{Total number of miles per year}\\ \times \text{ Number of years of miles driven}\end{array}\right)$

The formula to calculate costing savings needed per mile:

$\text{Cost savings needed per mile}=\frac{\text{Total additional cost}}{\text{Total miles driven}}$

The formula to calculate price per gallon for the miles driven:

$\text{Price per gallon}=\left(\begin{array}{l}\frac{\text{Price per gallon}}{\text{Estimated mileage of Vehicle LRX 450}}\\ -\frac{\text{Price per gallon}}{\text{Estimated mileage of Vehicle LRX 350}}\end{array}\right)$

Compute the total miles driven:

$\begin{array}{c}\text{Total miles driven}=\left(\begin{array}{c}\text{Total number of miles per year}\\ \times \text{ Number of years of miles driven}\end{array}\right)\\ =15,000\times 6\\ =90,000\text{ miles}\end{array}$

Hence, the total miles driven are 90,000 miles.

Compute the costing savings needed per mile:

$\begin{array}{c}\text{Cost savings needed per mile}=\frac{\text{Total additional cost}}{\text{Total miles driven}}\\ =\frac{\$7,365}{90,000}\\ =\$0.0818\end{array}$

Hence, the costing savings needed per mile is $0.0818.

Compute the price per gallon for the miles driven:

Note: Assume the price per gallon has P.

$\begin{array}{c}\text{Price per gallon}=\left(\begin{array}{l}\frac{\text{Price per gallon of Vehicle LRX 350}}{\text{Estimated mileage of Vehicle LRX 350}}\\ -\frac{\text{Price per gallon of Vehicle LRX 450}}{\text{Estimated mileage of Vehicle LRX 4}50}\end{array}\right)\\ \$0.0818=\left(\frac{\text{P}}{21}-\frac{\text{P}}{29}\right)\\ \$0.0818=\text{P}\left(\frac{1}{21}–\frac{1}{29}\right)\\ \$0.0818=\text{P}\left(0.0476-0.0344\right)\\ \$0.0818=\left(\text{P}\times \text{0}\text{.0132}\right)\\ \frac{\$0.0818}{0.0132}=\text{P}\\ \$6.196=\text{P}\end{array}$

Hence, the price per gallon for the miles driven is $6.196.

c)

Summary Introduction

To determine: The number of miles a person drives per year and the price per gallon for the miles driven.

Answer to Problem 23QP

The number of miles a person drives per year is 36,949.96 mile per year.

The price per gallon for the miles driven is $18.909.

Explanation of Solution

Given information:

The interest rate is 10% and the number of years is 6. The annual ownership cost is $300, the vehicle that was driven 15,000 miles and the list price of Vehicle LRS 450 including taxes is $5,565.

The formula to calculate the miles driven by a person peryear using break-even equation:

$\text{Cost}=\left[\begin{array}{l}-\text{ Total list price of Vehicle LRX 450}-\text{Annual ownership costs}\\ \times \left(\begin{array}{l}\text{Present value of an <x-custom-btb-me data-me-id="2167" class="microExplainerHighlight">annuity</x-custom-btb-me>}\\ \text{ of \$1 period for R\% of N period}\end{array}\right)\\ +\left(\begin{array}{l}\text{Savings per mile driven for Vehicle LRX 450}\\ \times \text{ Miles driven per year}\left(\begin{array}{l}\text{Present value of an annuity}\\ \text{ of \$1 period for R\% of N period}\end{array}\right)\end{array}\right)\end{array}\right]$

The formula to calculate the cost savings per gallon of gasoline:

$\text{Cost}=\left[\begin{array}{l}-\text{ Total list price of Vehicle LRX 450}-\text{Annual ownership costs}\\ \times \left(\begin{array}{l}\text{Present value of an annuity}\\ \text{ of \$1 period for R\% of N period}\end{array}\right)\\ +\left(\begin{array}{l}\text{Savings per mile driven for Vehicle LRX 450}\\ \times \text{ Miles driven per year}\left(\begin{array}{l}\text{Present value of an annuity}\\ \text{ of \$1 period for R\% of N period}\end{array}\right)\end{array}\right)\end{array}\right]$

Steps to determine the number of miles a person drive per year and the price per gallon:

• Initially, for determining the number of miles driven, it is required to know the present value of costs and savings. Both present value and savings must be equal to zero. In case, if the miles driven by a person per year are equivalent to the miles driven per year, use break-even formula to determine the miles driven by a person per year.
• Next, determine the cost savings per gallon of gasoline.
• Finally, find the price per gallon for the miles driven.

Compute the miles driven per year using break-even equation:

Note: To determine the present value of annuity of $1 period for 6 period at a discount rate of 10%, refer the PV of an annuity of $1 table. Then find out 10% discount rate and period of 6 years value from the given table. Here, the value for the rate 10% and 6 years period, the value is 4.35526.

$\begin{array}{c}\text{Cost}=\left[\begin{array}{l}-\text{ Total list price of Vehicle LRX 450}-\text{Annual ownership costs}\\ \times \left(\begin{array}{l}\text{Present value of an annuity}\\ \text{ of \$1 period for R\% of N period}\end{array}\right)\\ +\left(\begin{array}{l}\text{Savings per mile driven for Vehicle LRX }450\\ \times \text{ Miles driven per year}\left(\begin{array}{l}\text{Present value of an annuity}\\ \text{ of \$1 period for R\% of N period}\end{array}\right)\end{array}\right)\end{array}\right]\\ =\left[\begin{array}{c}-\$5,565-\$300\left(\begin{array}{l}\text{Present value of an annuity }\\ \text{of }\$1\text{ period for }10\%\text{ of 6 period}\end{array}\right)\\ +\$0.0427\times \text{ Miles driven per year}\times \left(\begin{array}{l}\text{Present value of an annuity }\\ \text{of }\$1\text{ period for }10\%\text{ of 6 period}\end{array}\right)\end{array}\right]\\ =-\$5,565-\left(\$300\times 4.35526\right)+\left(\$0.0427\times \text{Miles driven per year}\times \text{4}.\text{35526}\right)\\ =-\$5,565-\$1,306.57+\left(\$0.0427\times \text{Miles driven per year}\times \text{4}.\text{35526}\right)\\ =-\$6,871.57+\left(\$0.0427\times \text{ Miles driven per year}\times \text{4}.\text{35526}\right)\\ \$6,871.57=\left(\$0.0427\times \text{Miles driven per year}\times \text{4}.\text{35526}\right)\\ \frac{\$6,871.57}{\$0.0427}=\text{Miles driven per year}\times \text{4}.\text{35526}\\ \$160,926.69=\text{Miles driven per year}\times \text{4}.\text{35526}\\ \frac{\$160,926.69}{\text{4}.\text{35526}}=\text{Miles driven per year}\\ 36,949.96=\text{Miles driven per year}\end{array}$

Hence, the miles driven per year are 36,949.96 mile per year.

Compute the cost savings per gallon of gasoline:

$\begin{array}{c}\text{Cost}=\left[\begin{array}{l}-\text{ Total number of miles per year driven}-\text{Annual ownership costs}\\ \times \left(\begin{array}{l}\text{Present value of an annuity}\\ \text{ of \$1 period for R\% of N period}\end{array}\right)\\ +\left(\begin{array}{l}\text{Cost of savings per mile driven}\\ \times \text{ Total number of miles per year driven}\\ \times \left(\begin{array}{l}\text{Present value of an annuity}\\ \text{ of \$1 period for R\% of N period}\end{array}\right)\end{array}\right)\end{array}\right]\\ =\left[\begin{array}{c}-\$15,000-\$300\left(\begin{array}{l}\text{Present value of an annuity }\\ \text{of }\$1\text{ period for }10\%\text{ of 6 period}\end{array}\right)\\ +\text{ Cost of savings per mile driven}\times 15,000\times \left(\begin{array}{l}\text{Present value of an annuity }\\ \text{of }\$1\text{ period for }10\%\text{ of 6 period}\end{array}\right)\end{array}\right]\\ =-\$15,000-\left(\$300\times 4.35526\right)+\left(\text{Cost of savings per mile driven}\times 15,000\times \text{4}.\text{35526}\right)\\ =-\$15,000-\$1,306.57+\left(\text{Cost of savings per mile driven}\times \$\text{65,329}\right)\\ =-\$\text{16,306}\text{.57}+\left(\text{Cost of savings per mile driven}\times \$\text{65,329}\right)\\ \frac{\$\text{16,306}\text{.57}}{\$\text{65,329}}=\text{Cost of savings per mile driven}\\ \$0.24960=\text{Cost of savings per mile driven}\end{array}$

Hence, the cost of saving per miles driven is $0.24960.

Compute the price per gallon for the miles driven:

Note: Assume the price per gallon asP.

$\begin{array}{c}\text{Price per gallon}=\left(\begin{array}{l}\frac{\text{Price per gallon of Vehicle LRX 350}}{\text{Estimated mileage of Vehicle LRX }350}\\ -\frac{\text{Price per gallon of Vehicle LRX }450}{\text{Estimated mileage of Vehicle LRX }450}\end{array}\right)\\ \$0.24960=\left(\frac{\text{P}}{21}-\frac{\text{P}}{29}\right)\\ \$0.24960=\text{P}\left(\frac{1}{21}–\frac{1}{29}\right)\\ \$0.24960=\text{P}\left(0.0476-0.0344\right)\\ \$0.24960=\left(\text{P}\times \text{0}\text{.0132}\right)\\ \frac{\$0.24960}{0.0132}=\text{P}\\ \$18.909=\text{P}\end{array}$

Hence, the price per gallon for the miles driven is $18.909.

d)

Summary Introduction

To discuss: The implication of resale value of each car

Explanation of Solution

As per the computation, both vehicles depreciate by the similar dollar amount. Therefore, the resale value will be the same for both the vehicles.