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Unformatted text preview: Chapter 12 12.1 (a) BV 3 = 100,000 (40,000+24,000+14,000) = $22,000 (b) Market value today and S = $20,000, while BV 3 = $22,000. (c) d 1 = 0.40; d 2 = 0.24; d 3 = 0.14 Percent written off is the sum = 0.78 or 78%. 12.2 (a) Depreciation = Rate(495,000) Year Depr BV 1 $164,98 4 $330,016 2 220,02 8 109,988 3 73,30 9 36,679 4 36,67 8 (b) BV 3 = $36,679 while market value is much higher at $150,000. The rates reduce the asset to zero salvage, not recognizing the $150,000 salvage. 12.3 (a, b) Tax depreciation: D t = Rate(BV t1 ) Book depreciation: D t = Rate(40,000) Tax Book Year Depr BV Depr BV 40,00 40,000 1 16,00 24,00 10,00 30,000 2 9,60 14,40 10,00 20,000 3 5,76 8,64 10,00 10,000 4 3,45 6 5,18 4 10,00 A spreadsheet solution with graphs follows. 12 1 12.4 Productive life Time the asset is actually expected to be in service. Tax recovery period Time allow by tax laws to depreciate the assets value to salvage (or zero). Book recovery period Time used on company accounting books for depreciation to salvage (or zero) Straight Line (SL) Depreciation 12.5 (a) d = 1/10 = 0.1 or 10% (b) S = 0.15(475,000) = $71,250 D t = (550,000 71,250)/10 = $47,875 per year (c) BV 5 = 550,000 5(47,875) = $310,625 (d) BV 10 = 550,000 10(47,875) = $71,250 = S 12.6 Spreadsheet solution uses SL depreciation. D t = (350,000 50,000)/5 = $60,000 per year 12 2 12.7 Book: D t = (150,000 25,000)/9.5 = $13,158 per year Tax: D t = (150,000 0)5 = $ 30,000 per year Spreadsheet solution uses SL depreciation for book and tax purposes to plot BV. 12.8 D t = 18,900 = (P 0.25P)/8 0.75P = 8(18,900) P = $201,600 12.9 DDB: D 3 = (2/5)(500,000)(3/5) 2 = $72,000 150%DB: d = 1.5(1/5) = 0.3 D 3 = (0.3)(500,000)(0.7) 2 = $73,500 SL: D 3 = (500,00050,000)/5 = $90,000 12.10 The DDB salvage amount is calculated using d = 0.4. BV 5 = B(1d) t = 500,000(.6) 5 = $38,880 Considerably less than the estimated S = $50,000. For 150% DB, implied salvage is calculated using d = 0.3. BV 5 = 500,000(.7) 5 = $84,035 Considerably more than the estimated S = $50,000. 12.11 Use Equation [12.7] with d = 0.4. d 1 = 0.4(0.6) = 0.40 d 2 = 0.4(0.6) 1 = 0.24 d 3 = 0.4(0.6) 2 = 0.144 d 4 = 0.4(0.6) 3 = 0.0864 d 5 = 0.4(0.6) 4 = 0.05184 12 3 12.12 Spreadsheet solution determines BV values and plots. Note the DDB depreciation in year 5 is $14,800, so BV 5 = S = $50,000 is maintained. 12.13 The d is the fixed percentage (in decimal form) by which the BV is reduced each year, while d max is a specific value of d for DDB, that is, double the SL rate of d = 1/n. Finally, the annual depreciation rate, d t is the fraction of P written off each year. In DB depreciation, d t and d are not equal, whereas in SL depreciation, they are equal....
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 Spring '08
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