Chapter 11, Problem 11.5E

Depreciation methods; solving for unknowns

• LO11–2

For each of the following depreciable assets, determine the missing amount (?). Abbreviations for depreciation methods are SL for straight line, SYD for sum-of-the-years’-digits, and DDB for double-declining balance.

To determine

Depreciation:

Depreciation refers to the reduction in the monetary value of a fixed asset due to its wear and tear or obsolescence. It is a method of distributing the cost of the fixed assets over its estimated useful life. The following is the formula to calculate the depreciation.

$\text{Depreciation cost = }\frac{\text{Cost of the asset-Salvage value}}{\text{Estimated useful life of the asset}}$

To determine:  The missing amounts.

Explanation of Solution

Asset A:

 In the given case cost of the asset is missing. Given information is residual value $20,000; service life 5 years, depreciation method: double declining balance method and depreciation in year 2 is $24,000.

Double declining balance (DDB) method:

In this method of depreciation, the depreciation is calculated by multiply beginning of year book value, not depreciable base, by an annual rate that is a multiple of the straight line rate.

$\text{Depreciation expense}\text{=}\left(\begin{array}{l}\text{Begining}\text{value}\\ \text{of}\text{the}\text{asset}\end{array}\right)\text{×}\left(\begin{array}{l}\text{Straight}\text{line}\text{rate}\\ \text{of}\text{depreciation }\times \text{ 2}\end{array}\right)$

Determine cost of the asset.

$\begin{array}{c}\$24,000\text{=}\left(\begin{array}{l}\text{Begining}\text{value}\\ \text{of}\text{the}\text{asset A at year 2}\end{array}\right)\text{×}\left(\frac{1}{5}\times \text{ 2}\right)\\ \begin{array}{l}\text{Begining}\text{value}\\ \text{of}\text{the}\text{asset A at year 2}\end{array}\}=\$24,000\times \frac{5}{2}\\ \begin{array}{l}\text{Begining}\text{value}\\ \text{of}\text{the}\text{asset A at year 2}\end{array}\}=\$60,000\end{array}$

Beginning value of the asset A at year 1 is calculated as follows:

$\begin{array}{c}\text{Cost at year 1}-\left(\left(\mathrm{c}\text{ost}\text{at year 1}\right)\times \frac{1}{5}\times 2\right)=\$60,000\\ \text{Cost at year 1}-\text{0}\text{.4}\text{cost}\text{at}\text{year1=}\text{\$60,000}\\ 0.6\text{cost at year 1}\text{=}\text{\$60,000}\\ \text{cost at year 1 = }\frac{\$60,000}{0.6}\\ \text{cost}\text{at}\text{year}\text{1}=\$100,000\end{array}$

Hence, cost at year 1 is $100,000.

Asset B:

In given case residual value is missing. Given information is cost of the asset B is $40,000; service life is 8 years; depreciation method is sum of the year’s digit method; depreciation in year 2 is $7,000.

Sum-of- the-years’ digits (SYD) method:

Sum-of-the years’ digits method determines the depreciation expense by multiplying the depreciable base and declining fraction.

Determine the residual value of asset B

$\text{Depreciation}\text{expense}\text{=}\text{Cost of the asset }\times \left(\begin{array}{l}\text{Number of years of}\\ \text{estimated life remaining }\\ \text{at the beginning of the year}\end{array}\right)\begin{array}{l}\text{÷}\text{SYD}\hfill \end{array}$

$\begin{array}{c}\$7,000=\left(\$40,000-\text{residual value}\right)\times \frac{7}{36}\\ \$7,000\times 36=\left(\$40,000\times 7\right)-\left(\text{residual value}\times 7\right)\\ \$252,000=\$280,000-\left(\text{residual value}\times 7\right)\\ \text{residual value}=\frac{\$280,000-\$252,000}{7}\\ \text{Residual value =}\text{\$4,000}\end{array}$

Working note:

$\begin{array}{c}SYD=\frac{\text{n}\left(n+1\right)}{2}\\ =\frac{8\left(8+1\right)}{2}\\ =36\end{array}$

Conclusion:

Hence, the residual value of the asset B is $4,000.

Asset C:

In the given case for asset C, estimated life is missing. Given information is as follows cost of the asset is $65,000; residual value of the asset is $5,000; depreciation method is straight line method and year 2 depreciation is $6,000.

Straight line method:

Under the straight-line method of depreciation, the same amount of depreciation is allocated every year over the estimated useful life of an asset.

Determine estimated life of the asset C

$\text{Depreciation expense = }\frac{\text{(Cost of the asset-Salvage value)}}{\text{Estimated useful life of the asset}}$

$\begin{array}{c}\$6,000=\frac{\left(\$65,000-\$5,000\right)}{\text{estimated life}}\\ \$6,000\times \text{estimated life =}\text{\$60,000}\\ \text{Estimated life = }\frac{\$60,000}{\$6,000}\\ \text{Estimated life = }10\text{years}\end{array}$

Hence, estimated life or service life of the asset C is 10 Years.

Asset D:

In the given case, method of depreciation is missing. Given information is as follows: cost of the asset D is $230,000; residual value of the asset D is $10,000; service life is 10 years; depreciation for year 2 is $22,000.

Determine the method of depreciation.

$\begin{array}{c}\text{Depreciation expense}\text{=}\frac{\left(\text{cost}\text{of}\text{the}\text{asset}\text{-}\text{residual}\text{value}\right)}{\text{estimated}\text{life}\text{of}\text{the}\text{asset}}\\ \$22,000=\frac{\left(\$230,000-\$10,000\right)}{10\text{years}}\\ \$22,000=\frac{\$220,000}{10\text{years}}\\ \$22,000=\$22,000\end{array}$

Hence, straight line method of depreciation is used to determine the depreciation of asset D.

Asset E:

In the given situation depreciation for year 2 is missing. Given information is as follows: cost of the asset E $200,000; residual value of the asset E is $20,000; service life is 8 years; depreciation method is 150% declining method.

One hundred fifty percent declining balance:

In this method of depreciation, the depreciation is calculated by multiply beginning of year book value, not depreciable base, by an annual rate that is a 150% or 1.5 of the straight line rate.

Determine the depreciation on asset E for year 2.

$\text{Depreciation expense}\text{=}\left(\begin{array}{l}\text{Begining}\text{value}\\ \text{of}\text{the}\text{asset}\end{array}\right)\text{×}\left(\begin{array}{l}\text{Straight}\text{line}\text{rate}\\ \text{of}\text{depreciation }\times \text{ 1}\text{.5}\end{array}\right)$

$\begin{array}{c}\text{Depreciation expense}\text{=}\left(\$200,000\right)\text{×}\left(\frac{1}{8}\times \text{ 1}\text{.5}\right)\\ =\$37,500\end{array}$

Depreciation on asset E for year 1 is $37,500

For year 2 depreciation expense is calculated as follows

$\begin{array}{c}\text{Depreciation expense}\text{=}\left(\$200,000-\$37,500\right)\text{×}\left(\frac{1}{8}\times \text{ 1}\text{.5}\right)\\ =\$162,500\times \frac{1.5}{8}\\ =\$30,469\end{array}$

Hence, depreciation expense for year 2 is $30,469.