Chapter D, Problem 74P

To determine

The annual dollar advantage of alternative II over alternative I at an interest rate of $15\%$.

Answer to Problem 74P

The correct option is $\left(b\right)\$3020$.

Explanation of Solution

Given:

The interest rate is $15\%$.

• A company considering two alternatives

	Alternative I	Alternative II
First cost	$\$42,500$	$\$70,000$
Annual maintenance	$6,000$	$4,000$
Annual savings	$18,500$	$20,000$
Salvage value	$12,000$	$25,000$
Useful life of alternative	$3\text{years}$	$6\text{years}$

Calculation:

Write the expression to calculate the net annual savings.

$\text{Net}\text{annual}\text{savings}=\left(\text{Annual}\text{savings}\right)-\left(\text{Annual}\text{maintenance}\right)$ ...... (I)

Calculate the net annual savings for Alternative I.

Substitute, $\$18,500$ for Annual savings and $\$6,000$ for Annual maintenance in Equation (I).

$\begin{array}{c}\text{Net}\text{annual}\text{savings}=\$18,500-\$6,000\\ =\$12,500\end{array}$

Calculate the net annual savings for Alternative II.

Substitute, $\$20,000$ for annual savings and $\$4,000$ for annual maintenance in Equation (I).

$\begin{array}{c}\text{Net}\text{annual}\text{savings}=\$20,000-\$4,000\\ =\$16,000\end{array}$

Write the expression to determine the net cost of replacement.

$\text{Net}\text{cost}\text{of}\text{replacement}=\left(\text{Initial}\text{cost}\right)-\left(\text{Salvage}\text{value}\right)$ ...... (II)

Calculate the net cost of replacement for Alternative I.

At the end of year three the initial cost of $\$42,500$ will be repeated, but the salvage value of $\$12,000$ will also be received.

Substitute, $\$42,500$ for initial cost and $\$12,000$ for salvage value in Equation (II)

$\begin{array}{c}\text{Net}\text{cost}\text{of}\text{replacement}=\$42,500-\$12,000\\ =\$30,500\end{array}$

The annual cash for each project and the cash flow for a period of 6 years is shown below.

Years	Alternative I	Alternative II
$0$	$-\$42,500$	$-\$70,000$
$1$	$\$12,500$	$\$16,000$
$2$	$\$12,500$	$\$16,000$
$3$	$\$12,500+\left(-\$30,500\right)$	$\$16,000$
$4$	$\$12,500$	$\$16,000$
$5$	$\$12,500$	$\$16,000$
$6$	$\$12,500+\$12,000$	$\$16,000+\$25,000$

Write the expression to calculate the net present value of the dollar advantage of alternative II over alternative I.

$\text{Net}\text{present}\text{value}=\left[\begin{array}{l}-\$27,500+\$3,500\left(P/A,i,n_{\text{I}}\right)\\ +\$30,500\left(P/F,i,n_{\text{II}}\right)+13,000\left(P/F,i,n_{\text{I}}\right)\end{array}\right]$ ...... (IV)

Here, the interest rate is i and the time period is n.

Substitute, $15\%$ for $i$ and $6$ years for $n_{\text{I}}$ and $3$ years for $n_{\text{II}}$ in Equation (IV).

$\begin{array}{c}\text{Net}\text{present}\text{value}=\left[\begin{array}{c}-\$27,500+\$3,500\left(\frac{{\left(1+0.15\right)}^6-1}{0.15{\left(1+0.15\right)}^6}\right)\\ +\$30,500\left(\frac{1}{{\left(1+0.15\right)}^3}\right)+13,000\left(\frac{1}{{\left(1+0.15\right)}^6}\right)\end{array}\right]\\ =\left[\begin{array}{l}-\$27,500+\$3,500\left(3.784\right)+\$30,500\left(0.6575\right)\\ +13,000\left(0.4323\right)\end{array}\right]\\ =-\$27,500+\$13,244+\$20,053.75+\$5,619.9\end{array}$

$\text{Net}\text{present}\text{value}=\$11,417.6$

Write the expression to calculate the annual dollar advantage of alternative II over alternative I.

$A=P\left(A/P,i,n\right)$ ...... (V)

Here, the annual dollar advantage is A and the net present value is P.

Substitute, $\$11,417.6$ for net present value, $15\%$ for $i$ and $6$ years for $n$ in Equation (V).

$\begin{array}{c}A=\$11,417.6\left(\frac{0.15{\left(1+0.15\right)}^6}{{\left(1+0.15\right)}^6-1}\right)\\ =\$11,417.6\left(0.2642\right)\\ =\$3,016.52\end{array}$.

Conclusion:

Therefore, the correct option is $\left(b\right)\$3020$.