In: Economics
1. A piece of equipment was purchased new at $35,000. The salvage value is $1500 after its 8 year service life. Determine the depreciation and blue book value for the life of the asset, assuming:
a) straight line depreciation
b) MACRS (use 5 yr table)
c) SOYD method
d) DDB method
(a) SL method
Annual depreciation ($) = (Cost - Salvage value) / Life = (35,000 - 1,500) / 8 = 33,500 / 8 = 4,187.5
SLM | |||
Year | Asset Cost ($) | Annual Depreciation ($) | End-of-Year Book Value ($) |
1 | 35,000.0 | 4,187.5 | 30,812.5 |
2 | 30,812.5 | 4,187.5 | 26,625.0 |
3 | 26,625.0 | 4,187.5 | 22,437.5 |
4 | 22,437.5 | 4,187.5 | 18,250.0 |
5 | 18,250.0 | 4,187.5 | 14,062.5 |
6 | 14,062.5 | 4,187.5 | 9,875.0 |
7 | 9,875.0 | 4,187.5 | 5,687.5 |
8 | 5,687.5 | 4,187.5 | 1,500.0 |
(b) MACRS
This method ignores salvage value.
MACRS | ||||
Year | Beginning-of-year Book Value ($) | Depreciation Rate (%) | Annual Depreciation ($) | End-of-Year Book Value ($) |
1 | 35,000.00 | 20 | 7,000.00 | 28,000.00 |
2 | 28,000.00 | 32 | 8,960.00 | 19,040.00 |
3 | 19,040.00 | 19.2 | 3,655.68 | 15,384.32 |
4 | 15,384.32 | 11.52 | 1,772.27 | 13,612.05 |
5 | 13,612.05 | 11.52 | 1,568.11 | 12,043.94 |
6 | 12,043.94 | 5.76 | 693.73 | 11,350.21 |
(c) SOYD method
Sum-of-years-digit (SOYD) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36
Annual depreciation in year N = (Cost - Salvage value) x (Number of years remaining at beginning of year N / SOYD)
= $(35,000 - 1,500) / (Number of years remaining at beginning of year N / SOYD)
= $33,500 / (Number of years remaining at beginning of year N / SOYD)
SOYD | ||||
Year | Cost ($) | Depreciation Rate | Annual Depreciation ($) | End-of-Year Book Value ($) |
1 | 33,500 | 8/36 | 7,444.44 | 26,055.56 |
2 | 33,500 | 7/36 | 6,513.89 | 19,541.67 |
3 | 33,500 | 6/36 | 5,583.33 | 13,958.33 |
4 | 33,500 | 5/36 | 4,652.78 | 9,305.56 |
5 | 33,500 | 4/36 | 3,722.22 | 5,583.33 |
6 | 33,500 | 3/36 | 2,791.67 | 2,791.67 |
7 | 33,500 | 2/36 | 1,861.11 | 930.56 |
8 | 33,500 | 1/36 | 930.56 | 0.00 |
(d) DDB Method
SLM depreciation rate = 1/Useful life = 1/8 = 0.125
DDB Depreciation rate = 2 x SLM rate = 2 x 0.125 = 0.25
Depreciation schedule is as follows. This method ignores salvage value.
DDB | ||||
Year | Beginning-of-year Book Value ($) | Depreciation Rate | Annual Depreciation ($) | End-of-Year Book Value ($) |
1 | 35,000.00 | 0.25 | 8,750.00 | 26,250.00 |
2 | 26,250.00 | 0.25 | 6,562.50 | 19,687.50 |
3 | 19,687.50 | 0.25 | 4,921.88 | 14,765.63 |
4 | 14,765.63 | 0.25 | 3,691.41 | 11,074.22 |
5 | 11,074.22 | 0.25 | 2,768.55 | 8,305.66 |
6 | 8,305.66 | 0.25 | 2,076.42 | 6,229.25 |
7 | 6,229.25 | 0.25 | 1,557.31 | 4,671.94 |
8 | 4,671.94 | 0.25 | 1,167.98 | 3,503.95 |