Example 8D: Depreciation
This example shows double-declining depreciation computed at a five-year useful life with no salvage value. As is often the case with a five-year life, the double-declining method is used for the first three years and the straight-line method is used for the remaining two years. The double-declining method first computes what the straight-line percentage would be. In this case 100% divided by five years equals 20%. The 20% is then doubled. In this case 20% times 2 equals 40%. Then the 40% is multiplied by the remaining balance to be depreciated. Thus 60,000 times 40% for year one equals 24,000 depreciation, with a remaining balance of 36,000. Then 36,000 times 40% for year two equals 14,400 depreciation, and 36,000 minus 14,400 equals 21,600 remaining balance, and so on.
Now note the difference in annual depreciation between Example 8B, using straight-line for all five years, and Example 8D, using the combined double-declining and straight-line methods.
Depreciation Expense Computation: Double-Declining-Balance
Five year useful life; $10,000 salvage value
Year # Annual Depreciation Remaining Balance |
Beginning Balance = 60,000 |
1 24,000* 40,000 |
2 14,000* 40,000 |
3 8,640* 30,000 |
4 6,480** 20,000 |
5 6,480** 10,000 |
*double-declining balance depreciation
**straight-line depreciation for remaining two years (12,960 divided by 2 = 6,480/yr)
Practice Exercise 8–II: Depreciation
Compute the straight-line depreciation for each year for equipment with a cost of $50,000, a five-year useful life, and a $5,000 salvage value.
Assignment Exercise 8–4: Depreciation
Set up a purchase scenario of your own and compute the depreciation with and without salvage value.
Assignment Exercise 8–5: Depreciation Computation: Units-of-Service
Study the “Units of Service” portion of the chapter closely.
Required
- Using the format in Table 8–A-5, compute units of service depreciation using the following assumptions:
Cost to be depreciated = $50,000
Salvage value = zero
Total units of service = 10,000
Units of service per year: Year 1 = 2,200; Year 2 = 2,100; Year 3 = 2,300; Year 4 = 2,200; Year 5 = 200