Chapter 11, Problem 1P

To determine

The depreciation and book value over the asset’s $10$ year life of use in a table.

Answer to Problem 1P

The depreciation and book value over the asset’s $10$ year life of use in a table is shown below.

Year $\left(t\right)$	MACRS $\left(r_{\text{t}}\right)$	Cost basis $\left(B\right)$	Depreciation $d_{\text{t}}=r_{\text{t}}\times B$	CumulativeDepreciation $\left(Cu{m}_{d_{\text{t}}}\right)$	Book value at the end of the year $B{V}_{\text{t}}=B-Cu{m}_{d_{\text{t}}}$
$1$	$14.29\%$	$\$20,000$	$\$2,858$	$\$2,858$	$\$17,142$
$2$	$24.49\%$	$\$20,000$	$\$4,898$	$\$7,756$	$\$12,244$
$3$	$17.49\%$	$\$20,000$	$\$3,498$	$\$11,254$	$\$8,746$
$4$	$12.49\%$	$\$20,000$	$\$2,498$	$\$13,752$	$\$6,248$
$5$	$8.93\%$	$\$20,000$	$\$1,786$	$\$15,538$	$\$4,462$
$6$	$8.92\%$	$\$20,000$	$\$1,784$	$\$17,322$	$\$2,678$
$7$	$8.93\%$	$\$20,000$	$\$1,786$	$\$19,108$	$\$892$
$8$	$4.46\%$	$\$20,000$	$\$892$	$\$20,000$	$\$0$
Total	$100\%$		$\$20,000$

Explanation of Solution

Given:

The cost basis of the asset is $\$20,000$.

Concept used:

MACRS is known as modified accelerated cost recovery system and is a modified method to calculate depreciation.

Write the expression to calculate the depreciation value.

$d_{\text{t}}=B\times r_{\text{t}}$ ...... (I)

Here, the depreciation value is $d_{\text{t}}$, the cost is $B$ and the MACRS percentage is $r_{\text{t}}$.

Write the expression to calculate the book value at the end of the year.

$BV=B-Cu{m}_{d_{\text{t}}}$ ...... (II)

Here, book value at the end of the year is $BV$, cost is $B$, and the cumulative depreciation is $Cu{m}_{d_{\text{t}}}$.

Calculation:

The asset fits into MACRS “all property not assigned to another class” designation. Thus it is considered as 7 year property. Use MACRS GDS 7 year property percentages.

Calculate the depreciation.

Substitute $\$20,000$ for $B$ and $14.29\%$ for $r_{\text{t}}$ in Equation (I).

$\begin{array}{c}{d}_{\text{t}}=\$20,000\times 14.29\%\\ =\$2,858\end{array}$

Calculate the book value at the end of the year.

Substitute $\$20,000$ for $B$ and $\$2,858$ for $Cu{m}_{d_{\text{t}}}$ in Equation (II).

$\begin{array}{c}BV=\$20,000-\$2,858\\ =\$17,142\end{array}$

Calculate the depreciation and book value for 7 years and enter them in a table below.

Year $\left(t\right)$	MACRS $\left(r_{\text{t}}\right)$	Cost basis $\left(B\right)$	Depreciation $d_{\text{t}}=r_{\text{t}}\times B$	CumulativeDepreciation $\left(Cu{m}_{d_{\text{t}}}\right)$	Book value at the end of the year $B{V}_{\text{t}}=B-Cu{m}_{d_{\text{t}}}$
$1$	$14.29\%$	$\$20,000$	$\$2,858$	$\$2,858$	$\$17,142$
$2$	$24.49\%$	$\$20,000$	$\$4,898$	$\$7,756$	$\$12,244$
$3$	$17.49\%$	$\$20,000$	$\$3,498$	$\$11,254$	$\$8,746$
$4$	$12.49\%$	$\$20,000$	$\$2,498$	$\$13,752$	$\$6,248$
$5$	$8.93\%$	$\$20,000$	$\$1,786$	$\$15,538$	$\$4,462$
$6$	$8.92\%$	$\$20,000$	$\$1,784$	$\$17,322$	$\$2,678$
$7$	$8.93\%$	$\$20,000$	$\$1,786$	$\$19,108$	$\$892$
$8$	$4.46\%$	$\$20,000$	$\$892$	$\$20,000$	$\$0$
Total	$100\%$		$\$20,000$

Conclusion:

The depreciation and book value over the asset’s $10$ year life of use in a table is shown below.

Year $\left(t\right)$	MACRS $\left(r_{\text{t}}\right)$	Cost basis $\left(B\right)$	Depreciation $d_{\text{t}}=r_{\text{t}}\times B$	CumulativeDepreciation $\left(Cu{m}_{d_{\text{t}}}\right)$	Book value at the end of the year $B{V}_{\text{t}}=B-Cu{m}_{d_{\text{t}}}$
$1$	$14.29\%$	$\$20,000$	$\$2,858$	$\$2,858$	$\$17,142$
$2$	$24.49\%$	$\$20,000$	$\$4,898$	$\$7,756$	$\$12,244$
$3$	$17.49\%$	$\$20,000$	$\$3,498$	$\$11,254$	$\$8,746$
$4$	$12.49\%$	$\$20,000$	$\$2,498$	$\$13,752$	$\$6,248$
$5$	$8.93\%$	$\$20,000$	$\$1,786$	$\$15,538$	$\$4,462$
$6$	$8.92\%$	$\$20,000$	$\$1,784$	$\$17,322$	$\$2,678$
$7$	$8.93\%$	$\$20,000$	$\$1,786$	$\$19,108$	$\$892$
$8$	$4.46\%$	$\$20,000$	$\$892$	$\$20,000$	$\$0$
Total	$100\%$		$\$20,000$