Chapter 6, Problem 29P

To determine

The equivalent annual cost of owning a car.

Answer to Problem 29P

The equivalent annual cost of owning a car is, $\$7150.485$.

Explanation of Solution

Given:

Vehicle down payment is $\$2200$.

Vehicle payments is $\$5500$ for period $1-4$.

Prepaid insurance is, $\$1500$, which is increased by $8\%$ per year.

Gas, oil, and maintenance cost is, $\$2000$ that increases by $10\%$ per year.

New tires cost is $\$650$ during period $4$ and $800$ during period $8$.

Major maintenance overhaul at period five is, $\$2400$.

Salvage of vehicle at period $9$ is, $\$3750$.

Calculation:

Tabulate the values for calculation of total cost/benefits.

Time	Down payment	Annual payments (a)	Prepaid insurance(b)	Wear and Tear $\left(\text{c}\right)$	Tires(d)	Salvage	Total costs/benefits $\left(\begin{array}{l}a+b\\ +c+d\end{array}\right)$
$0$	$\$2200$						$\$2200$
$1$		$\$5500$	$\$1500$	$\$2000$			$\$9000$
$2$		$\$5500$	$\$1620$	$\$2200$			$\$9320$
$3$		$\$5500$	$\$1749.6$	$\$2420$			$\$9670$
$4$		$\$5500$	$\$1889.57$	$\$2662$	$\$650$		$\$10701.57$
$5$			$\$2040.7$	$\$2928.20$			$\$4969$
$6$			$\$2204$	$\$3221.02$			$\$5425$
$7$			$\$2380.3$	$\$3543.12$			$\$5923.42$
$8$			$\$2570.74$	$\$3897.43$	$\$800$		$\$7268.17$
$9$						$\$3750$	$\$3750$

Write the equation for equivalent annual unit cost.

$\text{EUAC}=\left[\begin{array}{l}\$2200\left(\frac{A}{P},0.05,9\right)+\$9000\left(\frac{P}{F},0.05,1\right)\left(\frac{A}{P},0.05,9\right)\\ +\$9320\left(\frac{P}{F},0.05,2\right)\left(\frac{A}{P},0.05,9\right)\\ +\$9670\left(\frac{P}{F},0.05,3\right)\left(\frac{A}{P},0.05,9\right)\\ +\$10701.57\left(\frac{P}{F},0.05,4\right)\left(\frac{A}{P},0.05,9\right)\\ +\$4969\left(\frac{P}{F},0.05,5\right)\left(\frac{A}{P},0.05,9\right)\\ +\$5425\left(\frac{P}{F},0.05,6\right)\left(\frac{A}{P},0.05,9\right)\\ +\$5923.42\left(\frac{P}{F},0.05,7\right)\left(\frac{A}{P},0.05,9\right)\\ +\$7268.17\left(\frac{P}{F},0.05,8\right)\left(\frac{A}{P},0.05,9\right)\\ -\$3750\left(\frac{A}{F},0.05,9\right)\end{array}\right]$ ...... (I)

Here, the equivalent uniform cost is EUAC, the down payment is D, salvage value is S, present worth is P, future worth is F, annual amount is A, interest rate is i, and number of year is n.

Calculate the factor $\left(\frac{A}{P},i,n\right)$ for $9$ years.

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

Substitute $5\%$ for i, and $9\text{years}$ for n in Equation (II).

$\begin{array}{c}\left(\frac{A}{P},i,n\right)=\left[\frac{0.05{\left(1+0.05\right)}^9}{{\left(1+0.05\right)}^9-1}\right]\\ =0.1407\end{array}$

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

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

Substitute $5\%$ for i, and $9\text{years}$ for n in Equation (III).

$\begin{array}{c}\left(\frac{A}{F},i,n\right)=\left[\frac{0.05}{{\left(1+0.05\right)}^9-1}\right]\\ =0.0907\end{array}$

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

$\left(\frac{P}{F},i,n\right)=\frac{1}{{\left(1+i\right)}^n}$ ...... (IV)

Calculate the factor $\left(\frac{P}{F},i,n\right)$ for $1$ year.

Substitute $5\%$ for i, and $1\text{years}$ for n in Equation (IV).

$\begin{array}{c}\left(\frac{P}{F},i,n\right)=\frac{1}{{\left(1+0.05\right)}^1}\\ =0.9524\end{array}$

Substitute $5\%$ for i, and $2\text{years}$ for n in Equation (IV).

$\begin{array}{c}\left(\frac{P}{F},i,n\right)=\frac{1}{{\left(1+0.05\right)}^2}\\ =0.9070\end{array}$

Substitute $5\%$ for i, and $3\text{years}$ for n in Equation (IV).

$\begin{array}{c}\left(\frac{P}{F},i,n\right)=\frac{1}{{\left(1+0.05\right)}^3}\\ =0.8638\end{array}$

Substitute $5\%$ for i, and $4\text{years}$ for n in Equation (IV).

$\begin{array}{c}\left(\frac{P}{F},i,n\right)=\frac{1}{{\left(1+0.05\right)}^4}\\ =0.8227\end{array}$

Substitute $5\%$ for i, and $5\text{years}$ for n in Equation (IV).

$\begin{array}{c}\left(\frac{P}{F},i,n\right)=\frac{1}{{\left(1+0.05\right)}^5}\\ =0.7835\end{array}$

Substitute $5\%$ for i, and $6\text{years}$ for n in Equation (IV).

$\begin{array}{c}\left(\frac{P}{F},i,n\right)=\frac{1}{{\left(1+0.05\right)}^6}\\ =0.7462\end{array}$

Substitute $5\%$ for i, and $7\text{years}$ for n in Equation (IV).

$\begin{array}{c}\left(\frac{P}{F},i,n\right)=\frac{1}{{\left(1+0.05\right)}^7}\\ =0.7107\end{array}$

Substitute $5\%$ for i, and $8\text{years}$ for n in Equation (IV).

$\begin{array}{c}\left(\frac{P}{F},i,n\right)=\frac{1}{{\left(1+0.05\right)}^8}\\ =0.6768\end{array}$

Substitute all the factors in Equation (I).

$\begin{array}{c}\text{EUAC}=\left[\begin{array}{l}\$2200\times 0.1407+\$9000\times 0.9524\times 0.1407\\ +\$9320\times 0.9070\times 0.1407\\ +\$9670\times 0.8638\times 0.1407\\ +\$10701.57\times 0.8227\times 0.1407\\ +\$4969\times 0.7835\times 0.1407\\ +\$5425\times 0.7462\times 0.1407\\ +\$5923.42\times 0.7107\times 0.1407\\ +\$7268.17\times 0.6768\times 0.1407\\ -\$3750\times 0.0907\end{array}\right]\\ =\left[\begin{array}{l}\$309.54+\$1206.02+\$1189.37\\ +\$1175.25+\$1238.75+\$517.68\\ +\$569.57+\$592.31+\$692.12-\$340.125\end{array}\right]\\ =\$7150.485\end{array}$.

Conclusion:

Therefore, the Equivalent annual cost of owning a car is, $\$7150.485$.