Chapter 13, Problem 28P

To determine

(a)

The minimum cost life and the minimum $EUAC$ if the new car is bought.

Answer to Problem 28P

The minimum cost of life is $\$2300$.

The minimum equivalent uniform annual cost is $\$19700$.

Explanation of Solution

Given:

The annual cost of new car is $\$22000$.

The depreciation rate for new as well as old car is $10\%$.

The maintenance cost is $\$100$ per year during $2$ years warranty period.

The maintenance cost after the warranty period is $\$600$ per year with an annual increase of $\$400$ per year.

The rate of interest is $10\%$.

Concept used:

Write the expression for minimum cost of life.

$C_L=D+M$ ...... (I)

Here, depreciation is $D$ and maintenance cost is $M$.

Write the expression for minimum equivalent uniform annual cost.

$EUA{C}_{\min }=A-C_L$ ...... (II)

Here, annual cost of car is $A$ and minimum cost of life is $C_L$.

Calculations:

Calculate the depreciation for the new car.

$\begin{array}{c}D=10\%\times \$22000\\ =\frac{10}{100}\times \$22000\\ =\$2200\end{array}$

Calculate the minimum cost of life.

Substitute $\$2200$ for $D$ and $\$100$ for $M$ in Equation (I).

$\begin{array}{c}{C}_L=\$2200+\$100\\ =\$2300\end{array}$

The minimum cost of life is $\$2300$.

Calculate the minimum equivalent uniform annual cost.

Substitute $\$22000$ for $A$ and $\$2300$ for $C_L$. in Equation (II).

$\begin{array}{c}EUA{C}_{\min }=\$22000-\$2300\\ =\$19700\end{array}$

The minimum equivalent uniform annual cost is $\$19700$.

Conclusion:

Thus, the minimum cost of life is $\$2300$.

The minimum equivalent uniform annual cost is $\$19700$.

To determine

(b)

The minimum cost life and the minimum $EUAC$ when 2 years old car is bought.

Answer to Problem 28P

The minimum cost of life is $\$2382$.

The minimum equivalent uniform annual cost is $\$15438$.

Explanation of Solution

Given:

He buys the car after it is $2$ years old.

Calculations:

Calculate the cost of car after 2 years.

$\begin{array}{c}A=\$22000-\left(10\%\times \$22000\right)-\left(10\%\times 90\%\times \$22000\right)\\ =\$22000-\left(\frac{10}{100}\times \$22000\right)-\left(\frac{10}{100}\times \frac{90}{100}\times \$22000\right)\\ =\$22000-\$2200-\$1980\\ =\$17820\end{array}$

Calculate the depreciation for the car.

$\begin{array}{c}D=\left(10\%\right)\times \$17820\\ =\frac{10}{100}\times \$17820\\ =\$1782\end{array}$

Calculate the minimum cost of life.

Substitute $\$1782$ for $D$ and $\$600$ for $M$ in Equation (I).

$\begin{array}{c}{C}_L=\$1782+\$600\\ =\$2382\end{array}$

The minimum cost of life is $\$2382$.

Calculate the minimum equivalent uniform annual cost.

Substitute $\$17820$ for $A$ and $\$2382$ for $C_L$. in Equation (II).

$\begin{array}{c}EUA{C}_{\min }=\$17820-\$2382\\ =\$15438\end{array}$

The minimum equivalent uniform annual cost is $\$15438$.

Conclusion:

Thus, the minimum cost of life is $\$2382$.

The minimum equivalent uniform annual cost is $\$15438$.

To determine

(c)

The minimum cost life and the minimum $EUAC$ when 4 years old car is bought.

Answer to Problem 28P

The minimum cost of life is $\$3043.42$.

The minimum equivalent uniform annual cost is $\$11390.78$.

Explanation of Solution

Given:

He buys the car after it is $4$ years old.

Calculations:

Calculate the cost of car after 4 years.

$\begin{array}{c}A=\$17820-\left(10\%\times \$17820\right)-\left(10\%\times 90\%\times \$17820\right)\\ =\$17820-\left(\frac{10}{100}\times \$17820\right)-\left(\frac{10}{100}\times \frac{90}{100}\times \$17820\right)\\ =\$17820-\$1782-\$1603.8\\ =\$14434.2\end{array}$

Calculate the depreciation for the car.

$\begin{array}{c}D=\left(10\%\right)\times \$14434.2\\ =\frac{10}{100}\times \$14434.2\\ =\$1443.42\end{array}$

Calculate the minimum cost of life.

Substitute $\$1443.42$ for $D$ and $\left(\$600+\$1000\right)$ for $M$ in Equation (I).

$\begin{array}{c}{C}_L=\$1443.42+\left(\$600+\$1000\right)\\ =\$3043.42\end{array}$

The minimum cost of life is $\$3043.42$.

Calculate the minimum equivalent uniform annual cost.

Substitute $\$14434.2$ for $A$ and $\$3043.42$ for $C_L$. in Equation (II).

$\begin{array}{c}EUA{C}_{\min }=\$14434.2-\$3043.42\\ =\$11390.78\end{array}$

The minimum equivalent uniform annual cost is $\$11390.78$.

Conclusion:

Thus, the minimum cost of life is $\$3043.42$.

The minimum equivalent uniform annual cost is $\$11390.78$.

To determine

(d)

The minimum cost life and minimum $EUAC$ when 6 years old car is bought.

Answer to Problem 28P

The minimum cost of life is $\$5969.1$.

The minimum equivalent uniform annual cost is $\$5721.9$.

Explanation of Solution

Given:

He buys the car after it is $6$ years old.

Calculations:

Calculate the cost of car after 6 years.

$\begin{array}{c}A=\$14434.2-\left(10\%\times \$14434.2\right)-\left(10\%\times 90\%\times \$14434.2\right)\\ =\$14434.2-\left(\frac{10}{100}\times \$14434.2\right)-\left(\frac{10}{100}\times \frac{90}{100}\times \$14434.2\right)\\ =\$14434.2-\$1443.42-\$1299.08\\ =\$11691\end{array}$

Calculate the depreciation for new car.

$\begin{array}{c}D=\left(10\%\right)\times \$11691\\ =\frac{10}{100}\times \$11691\\ =\$1169.1\end{array}$

Calculate the minimum cost of life.

Substitute $\$1169.1$ for $D$ and $\left(\$600+\$1000+\$1400+\$1800\right)$ for $M$ in Equation (I).

$\begin{array}{c}{C}_L=\$1169.1+\left(\$600+\$1000+\$1400+\$1800\right)\\ =\$5969.1\end{array}$

The minimum cost of life is $\$5969.1$.

Calculate the minimum equivalent uniform annual cost.

Substitute $\$11691$ for $A$ and $\$5969.1$ for $C_L$. in Equation (II).

$\begin{array}{c}EUA{C}_{\min }=\$11691-\$5969.1\\ =\$5721.9\end{array}$

The minimum equivalent uniform annual cost is $\$5721.9$.

Conclusion:

Thus, the minimum cost of life is $\$5969.1$.

The minimum equivalent uniform annual cost is $\$5721.9$.

To determine

(e)

The strategy that should be recommended.

Answer to Problem 28P

The best strategy recommended is to buy the car after it is $6\text{years}$ old.

Explanation of Solution

The strategy which gives minimum equivalent uniform annual cost is considered the best one.

The minimum cost of life when 6 years old car is bought is $\$5969.1$ and the minimum equivalent uniform annual cost is $\$5721.9$.

Thus, it is considered as the best strategy as it yields the minimum equivalent uniform annual cost.