3232ex7-dep-methods - 986.67 1466.67 3,467 5 3,467 693.33...

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Depreciation Examples Depreciation Factors P = $10,000 S = $2000 N = 5 years Depreciation method: Straight Line (SL) D(N)=(P-S)/N= 1600 Year B(N-1) D(N) B(N) 0 10,000 1 10,000 1600 8,400 2 8,400 1600 6,800 3 6,800 1600 5,200 4 5,200 1600 3,600 5 3,600 1600 2,000 Depreciation method: Declining balance (DB) d=1/5 Year B(N-1) D(N) B(N) 0 10,000 1 10,000 2000 8,000 2 8,000 1600 6,400 3 6,400 1280 5,120 4 5,120 1024 4,096 5 4,096 819.2 3,277 Notes: (Asset/Balance sheet perspective) 1. If B(N) would go less than expected salvage, then stop B(N) at the salvage value. 2. If B(N) doesn't reach the expected salvage in last year of expected life, then convert to SL method when annual depreciation from DB is less than running annual depreciation from SL (next ex)
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Depreciation method: DB with SL adjustment d=1/5 Year B(N-1) D(N); DB D(N); SL B(N) 0 10,000 1 10,000 2000.00 1600.00 8,000 2 8,000 1600.00 1500.00 6,400 3 6,400 1280.00 1466.67 4,933 4 4,933
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Unformatted text preview: 986.67 1466.67 3,467 5 3,467 693.33 1466.67 2,000 Depreciation method: Double Declining balance (DDB) d=1/5*2=2/5 Year B(N-1) D(N) B(N) 10,000 1 10,000 4000 6,000 2 6,000 2400 3,600 3 3,600 1440 2,160 4 2,160 160 2,000 5 2,000 2,000 Depreciation method: Sum of Years Digits (SOYD) MD = 1+2+3+4+5 = (5)(5+1)/2 = 15 Year P-S Factor; Num Factor; Den Factor D(N) B(N) 10000.00 1 8000 5 15 0.33 2666.67 7333.33 2 8000 4 15 0.27 2133.33 5200.00 3 8000 3 15 0.2 1600.00 3600.00 4 8000 2 15 0.13 1066.67 2533.33 5 8000 1 15 0.07 533.33 2000.00 Note: Partial D(4) to reach salvage value. Then no depreciation in year 5. Note: Other depreciation methods Method of depreciation: Units of production D(N) = (Units consumed in year N)/(Lifetime units available) * (P-S)...
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3232ex7-dep-methods - 986.67 1466.67 3,467 5 3,467 693.33...

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