asset is purchased and installed in 1972 at a cost of 36000

asset is purchased and installed in 1972 at a cost of $36,000. The asset has an 8- iation life and an estimated salvage value of zero. Determine the de book tion amount D, for each of the first 3 years and the book balance pre value) BB, at the end of each of those years, using the following depre SL, SOYD, 150% DB, and DDB. Ignore half-y ear conventions, ITC, a nd Sec. 179

Solution

SL Method: Depreciation under SL=(Cost-Salvage value)/depreciation life=(36000-0)/8=4500 $4500 for each of the 3 years Book value Year 1 36000-4500 31500 Year 2 31500-4500 27000 Year 3 27000-4500 22500 SOYD Method: Sum of years of depreciation life=1+2+3+4+5+6+7+8=36 Depreciation for year 1=Cost*8/36 Depreciation for year 2=Cost*7/36 Depreciation for year 3=Cost*6/36 Year Depreciation Book value 1 8000 28000 (36000-8000) 2 7000 21800 (28800-7000) 3 6000 14800 (21800-7000) 150% DB Method: Depreciation rate=150% of straight line depreciation rate=150%*(1/8) =18.75% Depreciation for year 1=Cost*18.75% Depreciation for year 2=Book value*18.75% Depreciation for year 3=Book value*18.75% Year Depreciation Book value 1 6750 29250 (36000*18.75%) (36000-6750) 2 5484 23766 (29250*18.75%) (29250-5484) 3 4456 19310 (23766*18.75%) (23766-4456) DDB Method: Depreciation rate=200% of straight line depreciation rate=200%*(1/8) =25% Depreciation for year 1=Cost*25% Depreciation for year 2=Book value*25% Depreciation for year 3=Book value*25% Year Depreciation Book value 1 9000 27000 (36000*25%) (36000-9000) 2 6750 20250 (27000*25%) (27000-6750) 3 5063 15188 (20250*25%) (20250-5063)