Chapter 12, Problem 74P

To determine

The machine to be selected.

Answer to Problem 74P

Machine (III) should be selected.

Explanation of Solution

Given:

The taxable income is $\$150,000$.

After tax MARR is $15\%$.

The selling price is $25\%$ of the purchase price.

Project life is $10\text{years}$.

Concept used:

Calculations:

Make the table for alternatives and their data.

Alternative	Initial Cost	Before tax cost	Salvage Value $\left(25\%\text{of}\text{initial}\text{cost}\right)$
(I)	$\$9000$	$\$25$	$\$2250$
(II)	$\$8000$	$\$200$	$\$2000$
(III)	$\$7500$	$\$300$	$\$1875$
(IV)	$\$6200$	$\$600$	$\$1550$

Table-(1)

Assume the tax rate applicable here is $40\%$.

From the observations, the first alternative having cost of just $\$25$ per year over the time frame thus this alternative is rejected by observation only.

As compared to alternative (III) and (IV), the alternative (II) also has a low cost per year over the time frame.

Hence, alternative (II) is also rejected.

Now the alternatives (III) and (IV) are left.

Consider alternative (III).

Calculate the depreciation under MACRS rate as shown below.

Year	Cost Basis $\left(a\right)$	MACRS rate $\left(b\right)$	Depreciation $c=\left(a\times b\right)$
$1$	$\$7500$	$20\%$	$\$1500$
$2$	$\$7500$	$32\%$	$\$2400$
$3$	$\$7500$	$19.2\%$	$\$1440$
$4$	$\$7500$	$11.52\%$	$\$864$
$5$	$\$7500$	$11.52\%$	$\$864$
$6$	$\$7500$	$5.76\%$	$\$432$

Table-(2)

Make the before and after tax cash flow table.

Year	Before taxCash flow $\left(a\right)$	Depreciation $\left(b\right)$	Taxable income $c=\left(a-b\right)$	Income taxes $d=-\left(c\times 0.4\right)$	After taxCash flow $e=\left(a+d\right)$
$0$	$-\$7500$				$-\$7500$
$1$	$\$300$	$\$1500$	$-\$1200$	$\$480$	$\$780$
$2$	$\$300$	$\$2400$	$-\$2100$	$\$840$	$\$1140$
$3$	$\$300$	$\$1440$	$-\$1140$	$\$456$	$\$756$
$4$	$\$300$	$\$864$	$-\$564$	$\$225.6$	$\$525.6$
$5$	$\$300$	$\$864$	$-\$564$	$\$225.6$	$\$525.6$
$6$	$\$300$	$\$432$	$-\$132$	$52.8	$352.8$
$7$	$\$300$	$0$	$\$300$	$-\$120$	$\$180$
$8$	$\$300$	$0$	$\$300$	$-\$120$	$\$180$
$9$	$\$300$	$0$	$\$300$	$-\$120$	$\$180$
$10$	$\$300$ $\$1875$	$0$	$\$300$	$-\$120$	$\$180$ $\$1875$

Table-(3)

Write the expression to calculate the present value with non uniform cash flow.

$\begin{array}{c}P={\displaystyle \sum \frac{A}{{\left(1+i\right)}^n}}\\ P=\left(\begin{array}{l}\frac{A_1}{{\left(1+i\right)}^1}+\frac{A_2}{{\left(1+i\right)}^2}+\frac{A_3}{{\left(1+i\right)}^3}+\frac{A_4}{{\left(1+i\right)}^4}+\frac{\left(A_5\right)}{{\left(1+i\right)}^5}+\frac{\left(A_6\right)}{{\left(1+i\right)}^6}+\frac{\left(A_7\right)}{{\left(1+i\right)}^7}\\ +\frac{\left(A_8\right)}{{\left(1+i\right)}^8}+\frac{\left(A_9\right)}{{\left(1+i\right)}^9}+\frac{\left(A_{10}\right)}{{\left(1+i\right)}^{10}}+\frac{\left(SV\right)}{{\left(1+i\right)}^{10}}\end{array}\right)\end{array}$ ....... (I)

Here, the present value is P, the annual benefit is A, the salvage value is SV and the rate is i.

Substitute the value of $A$, $P$ and SV from Table-(3) in Equation (I).

$\$7500=\left(\begin{array}{l}\frac{\$780}{{\left(1+i\right)}^1}+\frac{\$1140}{{\left(1+i\right)}^2}+\frac{\$756}{{\left(1+i\right)}^3}+\frac{\$525.6}{{\left(1+i\right)}^4}+\frac{\left(\$525.6\right)}{{\left(1+i\right)}^5}+\frac{\left(352.8\right)}{{\left(1+i\right)}^6}\\ +\frac{\left(\$180\right)}{{\left(1+i\right)}^7}+\frac{\left(\$180\right)}{{\left(1+i\right)}^8}+\frac{\left(\$180\right)}{{\left(1+i\right)}^9}+\frac{\left(\$180\right)}{{\left(1+i\right)}^{10}}+\frac{\left(\$1875\right)}{{\left(1+i\right)}^{10}}\end{array}\right)$ ....... (II)

Solve Equation (II) for $i$.

$\begin{array}{l}i=0.203\\ i=\left(0.203\times 100\right)\%\\ i=20.3\%\end{array}$

Consider alternative (IV).

Calculate the depreciation under MACRS rate as shown below.

Year	Cost Basis $\left(a\right)$	MACRS rate $\left(b\right)$	Depreciation $c=\left(a\times b\right)$
$1$	$\$6200$	$20\%$	$\$1240$
$2$	$\$6200$	$32\%$	$\$1984$
$3$	$\$6200$	$19.2\%$	$\$1190.4$
$4$	$\$6200$	$11.52\%$	$\$714.24$
$5$	$\$6200$	$11.52\%$	$\$714.24$
$6$	$\$6200$	$5.76\%$	$\$357.12$

Table-(4)

Make the before and after tax cash flow table.

Year	Before taxCash flow $\left(a\right)$	Depreciation $\left(b\right)$	Taxable income $c=\left(a-b\right)$	Income taxes $d=-\left(c\times 0.4\right)$	After taxCash flow $e=\left(a+d\right)$
$0$	$-\$6200$				$-\$6200$
$1$	$\$600$	$\$1240$	$-\$640$	$\$256$	$\$856$
$2$	$\$600$	$\$1984$	$-\$1384$	$\$553.6$	$\$1153.6$
$3$	$\$600$	$\$1190.4$	$-\$590.4$	$\$236.16$	$\$836.16$
$4$	$\$600$	$\$714.24$	$-\$114.24$	$\$45.696$	$\$645.696$
$5$	$\$600$	$\$714.24$	$-\$114.24$	$\$45.696$	$\$645.696$
$6$	$\$600$	$\$357.12$	$-\$242.88$	$\$97.152$	$\$697.152$
$7$	$\$600$	$0$	$\$600$	$-\$240$	$\$360$
$8$	$\$600$	$0$	$\$600$	$-\$240$	$\$360$
$9$	$\$600$	$0$	$\$600$	$-\$240$	$\$360$
$10$	$\$600$ $\$1550$	$0$	$\$600$	$-\$240$	$\$360$ $\$1550$

Table-(5)

Substitute the value of $A$, $P$ and SV from Table-(5) in Equation (I).

$\$6200=\left(\begin{array}{l}\frac{\$856}{{\left(1+i\right)}^1}+\frac{\$1153.6}{{\left(1+i\right)}^2}+\frac{\$836.16}{{\left(1+i\right)}^3}+\frac{\$645.696}{{\left(1+i\right)}^4}\\ +\frac{\left(\$645.696\right)}{{\left(1+i\right)}^5}+\frac{\left(\$697.152\right)}{{\left(1+i\right)}^6}+\frac{\left(\$360\right)}{{\left(1+i\right)}^7}+\frac{\left(\$360\right)}{{\left(1+i\right)}^8}\\ +\frac{\left(\$360\right)}{{\left(1+i\right)}^9}+\frac{\left(\$360\right)}{{\left(1+i\right)}^{10}}+\frac{\left(\$1550\right)}{{\left(1+i\right)}^{10}}\end{array}\right)$ ....... (III)

Solve Equation (III) for $i$.

$\begin{array}{l}i=0.044\\ i=\left(0.044\times 100\right)\%\\ i=4.4\%\end{array}$.

Conclusion:

The after tax rate of return of alternative (III) is greater than alternative (II) and also greater than the MARR.

Hence, machine (III) should be selected.