Chapter 11, Problem 10P

To determine

(a)

The depreciation schedule using MACRS method.

Answer to Problem 10P

The depreciation schedule using MACRS is shown below.

Year $\left(t\right)$	Depreciation $d_{\text{t}}=r_{\text{t}}\times B$
$1$	$\$7,000$
$2$	$\$11,200$
$3$	$\$6,720$
$4$	$\$4,032$
$5$	$\$4,032$
$6$	$\$2,016$
Total	$\$35,000$

Explanation of Solution

Given:

Cost of the small motorboat is $\$35,000$.

Time period is $5\text{years}$.

Concept used:

Write the expression to calculate the depreciation value for the motorboat.

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

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

Calculation:

The given asset comes under MACRS “automobiles and trucks” designation. Thus it is considered as a 5 year property with ADR of 5 years. Using MACRS GDS 5 year property table calculate the depreciation for the motorboat.

Calculate the depreciation for motorboat.

Substitute, $\$35,000$ for $B$ and $20\%$ for $r_{\text{t}}$ in Equation (I)

$\begin{array}{c}{d}_{\text{t}}=\$35,000\times 20\%\\ =\$7,000\end{array}$

Calculate the depreciation for 5 years and enter them in a table below.

Year $\left(t\right)$	MACRS $\left(r_{\text{t}}\right)$	Cost $\left(B\right)$	Depreciation $d_{\text{t}}=r_{\text{t}}\times B$
$1$	$20\%$	$\$35,000$	$\$7,000$
$2$	$32\%$	$\$35,000$	$\$11,200$
$3$	$19.20\%$	$\$35,000$	$\$6,720$
$4$	$11.52\%$	$\$35,000$	$\$4,032$
$5$	$11.52\%$	$\$35,000$	$\$4,032$
$6$	$5.76\%$	$\$35,000$	$\$2,016$
Total	$100\%$		$\$35,000$

Conclusion:

The depreciation schedule using MACRS is shown below.

Year $\left(t\right)$	Depreciation $d_{\text{t}}=r_{\text{t}}\times B$
$1$	$\$7,000$
$2$	$\$11,200$
$3$	$\$6,720$
$4$	$\$4,032$
$5$	$\$4,032$
$6$	$\$2,016$
Total	$\$35,000$

To determine

(b)

The depreciation schedule using SOYD method.

Answer to Problem 10P

The depreciation schedule using SOYD is shown below.

Year $\left(t\right)$	Depreciation $\left(d_{\text{t}}\right)$
$1$	$\$9,167$
$2$	$\$7,333$
$3$	$5,500$
$4$	$\$3,667$
$5$	$\$1,833$
Total	$\$27,500$

Explanation of Solution

Given:

Cost of the small motorboat is $\$35,000$.

Time period is $5\text{years}$.

Salvage value is $\$7,500$.

Concept used:

SOYD is known as sum of year’s digits depreciation.

Write the expression to calculate the depreciation of the motorboat.

$\begin{array}{c}{d}_{\text{t}}=\frac{n-t+1}{SOYD}\left(B-S\right)\\ d_{\text{t}}=\frac{n-t+1}{\left[\frac{n\left(n+1\right)}{2}\right]}\left(B-S\right)\end{array}$ ...... (II)

Here, the depreciation value is $d_{\text{t}}$, total number of years is $n$, cost of the motorboat is $B$, particular year is $t$ and salvage value after depreciable life is $S$.

Calculation:

Calculate the depreciation using SOYD for the motorboat.

Substitute, $5$ for $n$, $\$35,000$ for $B$ and $\$7,500$ for $S$ in Equation (II).

$\begin{array}{c}{d}_{\text{t}}=\frac{5-t+1}{\left[\frac{5\left(5+1\right)}{2}\right]}\left(\$35,000-\$7500\right)\\ =\frac{\left(6-t\right)}{15}\times \left(\$27,500\right)\\ =\$1,833.33\left(6-t\right)\end{array}$ ...... (III)

Calculate the depreciation value for 5 years.

Year $\left(t\right)$	Depreciation $\left(d_{\text{t}}\right)$
$1$	$\$9,167$
$2$	$\$7,333$
$3$	$5,500$
$4$	$\$3,667$
$5$	$\$1,833$
Total	$\$27,500$

Here, calculate the depreciation from Equation (III).

Conclusion:

The depreciation schedule using SOYD is shown below.

Year $\left(t\right)$	Depreciation $\left(d_{\text{t}}\right)$
$1$	$\$9,167$
$2$	$\$7,333$
$3$	$5,500$
$4$	$\$3,667$
$5$	$\$1,833$
Total	$\$27,500$