Composite Method

The first cost of a battery-operated machine is $2,500, and its salvage value at the end of 5-year useful life is estimated to be $100. Determine its depreciation schedule based on the DDB-SL $^{21}$ composite method.

Question

Composite Method

The first cost of a battery-operated machine is $2,500, and its salvage value at the end of 5-year useful life is estimated to be $100. Determine its depreciation schedule based on the DDB-SL[latex]^{21}[/latex] composite method.

Accepted Answer

From the discussions in Sections 10.4 and 10.5 as well as in the earlier sections, and also from Fig. 10.5, depreciation will be most rapid under the DDB-SL composite method.
It is given that the total depreciation = P-S = $2,500 – $100 = $2,400. With the SL-method, the depreciation rate for a 5-year asset is 1/5, and therefore the yearly SL-charge[latex]^{22}[/latex] will be (1/5) × $2,400 = $480.

The DDB depreciation rate is twice that of the SL-rate, i.e., 2 × (1/5) = 2/5, and therefore the charges for the 5 years will be:

[latex]1st year depreciation =({2}/{N}) imes Book Value[/latex]

[latex]=({2}/{5})(2,500)[/latex]

[latex]=\$1,000[/latex]

[latex]2nd year depreciation =({2}/{N}) imes Book Value[/latex]

[latex]=({2}/{5})(2,500-1,000)[/latex]

[latex]=\$600[/latex]

[latex]3rd year depreciation =({2}/{N}) imes Book Value[/latex]

[latex]=({2}/{5})(2,500-1,600)[/latex]

[latex]=\$360[/latex]

[latex]4th year depreciation =({2}/{N}) imes Book Value[/latex]

[latex]=({2}/{5})(2,500-1,960)[/latex]

[latex]=\$216[/latex]

[latex]5th year depreciation =({2}/{N}) imes Book Value[/latex]

[latex]=({2}/{5})(2,500-2,176)[/latex]

[latex]=\$130[/latex]

The DDB depreciation schedule is thus:

Total:	5	4	3	2	1	Year
$2,306	$130	$216	$360	$600	$1,000	DDB Depr.

Since the total DDB depreciation of $2,306 is lower than the maximum allowed ($2,400), a switch-over to the SL-method is called for[latex]^{23}[/latex] . This will enable the total depreciation charged to exactly equal the maximum allowable.

To do this we compare for each year the DDB-charge with the SL-charge and choose the greater of the two. For year 1, the DDB charge at $1,000 is more than the $480 SL-charge. So, we opt for DDB depreciation of $1,000 for year 1. For year 2, the DDB charge is $600, while the SL-charge[latex]^{24}[/latex] will be:

[latex]=({1}/{4})[/latex] (total depreciation remaining to be charged)

[latex]=0.25 imes (2,400-1,00)[/latex]

[latex]=\$350[/latex]

So, for year 2 too, we choose DDB since its $600 charge is greater than the $350 SL-charge.

We now check the third year to see whether it makes sense to switch. From the previous table, the DDB-charge for year 3 is $360. The cumulative depreciation charged so far is $1,000 + $600 = $1,600. So, the SL-method yields a depreciation charge of (1/3) × (2,400 – 1,600) = $267, which is less than $360. Thus, switching to the SL-method is not yet desirable. We thus “stick with” the DDB for the third year too.

Next check for year 4. The SL-charge for the fourth year is (½)(2,400 – 1,960) = $220, which is greater than the DDB charge of $216 for year 4. Hence, switching to the SL-method at year 4 makes sense. The switching means charging $220 for year 4, and also[latex]^{25}[/latex] for year 5. The composite DDB-SL depreciation schedule will thus be:

Year	1	2	3	4	5	Total:
Composite Depr.	$1,000	$600	$360	$220	$220	$2,400

The total of $2,400 being equal to P-S confirms the accuracy of the results.

[latex]^{21}[/latex]Begins with the DDB-method and switches to the SL-method.

[latex]^{22}[/latex]This annual charge will change depending on when we switch over to the SL-method, as discussed later.

[latex]^{23}[/latex]Had this total been greater than $2,400, we would have reduced the charges for the latter years, since tax laws do not allow total depreciation to exceed P-S. For example, if S is $500 in Example 10.10, the cumulative DDB depreciation charge of $2,306 is greater than P-S (= $2,500 – $500 = $2,000). In such a case, no composite depreciation is warranted for. With S as $500, the results of Example 10.10 would have been:

Year	1	2	3	4	5	Total:
DDB Depr.	$1,000	$600	$360	$40	$0	$2,000

[latex]^{24}[/latex]If we switch to the SL-method now at the end of year 1, then N is 4, and therefore depreciation rate is 1/4. Since $1,000 has already been charged for the first year by the DDB-method, depreciation that remains to be charged is $2,400 (maximum allowed) minus $1,000.

[latex]^{25}[/latex]Once we switch to the SL-method, the depreciation for the subsequent years remains the same due to the linear nature of the method.

Question 10.10: Composite Method The first cost of a battery-operated machin...

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