Chapter D, Problem 73P

To determine

The closest equivalent annual cost of ownership.

Answer to Problem 73P

Therefore the correct option is $\left(c\right)\$21844$.

Explanation of Solution

Given:

Cost of the product is $\$120,000$.

Salvage value is $\$10,000$.

Time period is $10$ years.

Annual repair cost is $\$5000$.

Increase in the annual cost is $\$500$.

Interest is $10\%$.

Concept used:

Write the expression to calculate the net present worth.

$NPV=A\left(\frac{1}{{\left(1+i\right)}^n}\right)$

Here, the amount is $A$, interest is $i$ and time period is $n$.

Calculation:

Calculate the cash flow for the last year.

$\begin{array}{c}{A}_{10}=\$10000-\$7500\\ =\$2500\end{array}$

Write the expression to calculate the net present value.

$NPV=\left[\begin{array}{l}\$120,000+5,000\left(\frac{1}{{\left(1+i\right)}^5}\right)+5,500\left(\frac{1}{{\left(1+i\right)}^6}\right)+6,000\left(\frac{1}{{\left(1+i\right)}^7}\right)\\ +6,500\left(\frac{1}{{\left(1+i\right)}^8}\right)+7,000\left(\frac{1}{{\left(1+i\right)}^9}\right)-2500\left(\frac{1}{{\left(1+i\right)}^{10}}\right)\end{array}\right]$ ...... (I)

Substitute, $10\%$ for $i$ in Equation (I).

$\begin{array}{c}NPV=\left[\begin{array}{l}\$120,000+5,000\left(\frac{1}{{\left(1+0.10\right)}^5}\right)+5,500\left(\frac{1}{{\left(1+0.10\right)}^6}\right)+6,000\left(\frac{1}{{\left(1+0.10\right)}^7}\right)\\ +6,500\left(\frac{1}{{\left(1+0.10\right)}^8}\right)+7,000\left(\frac{1}{{\left(1+0.10\right)}^9}\right)-2500\left(\frac{1}{{\left(1+0.10\right)}^{10}}\right)\end{array}\right]\\ =\left[\begin{array}{l}120,000+5,000\left(0.62092\right)+5,500\left(0.56447\right)+6,000\left(0.51316\right)\\ +6,500\left(0.46651\right)+7,000\left(0.42409\right)-2500\left(0.38554\right)\end{array}\right]\\ =\left[\begin{array}{l}120,000+3104.6+3104.585+3078.96\\ +3032.32+2968.63-963.85\end{array}\right]\\ =\$134,325.94\end{array}$

Write the expression to calculate equivalent uniform annual cost by using $PMT$ function.

$EUAC=PMT\left(r,n,NPV\right)$ ...... (II)

Here, rate of interest is $r$, number of years is $n$ and net present value is $NPV$.

Substitute, $10\%$ for $r$, $10$ for $n$ and $\$134,325.94$ for $NPV$ in Equation (II)

$\begin{array}{c}EUAC=PMT\left(0.10,10,\$134325.94\right)\\ =\frac{\$134,325.94}{\left(\frac{1-\frac{1}{{\left(1+0.10\right)}^{10}}}{0.10}\right)}\\ =\frac{\$134,325.94}{6.145}\\ =\$21,859.38\end{array}$.

Conclusion:

Therefore, the correct option is $\left(c\right)\$21844$.