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F.Chap10

# 1482 1482110000 1482110000 16302 16302 4 0741

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Unformatted text preview: year 6 = 110,000 – 36,663 – 48,884 – 16,302 – 8,151 = 0 After-tax salvage = 17,000 - . 4(17,000 – 0) = \$10,200 18 18 Example: Seven-Year MACRS Year MACRS MACRS Percent Percent D 1 .1429 .1429(110,000) = 15,719 2 .2449 .2449(110,000) = 26,939 3 .1749 .1749(110,000) = 19,239 4 .1249 .1249(110,000) = 13,739 5 .0893 .0893(110,000) = 9,823 6 .0893 .0893(110,000) = 9,823 BV in year 6 = 110,000 – 15,719 – 26,939 – 19,239 – 13,739 – 9,823 – 9,823 = 14,718 After-tax salvage = 17,000 - . 4(17,000 – 14,718) = 16,087.20 19 19 Example: Replacement Problem Original Machine Initial cost = 100,000 Annual depreciation = 9000 Purchased 5 years ago Book Value = 55,000 Salvage today = 65,000 Salvage in 5 years = Salvage 10,000 10,000 New Machine Initial cost = 150,000 5-year life Salvage in 5 years = 0 Cost savings = 50,000 Cost per year per 3-year MACRS 3-year depreciation depreciation Required return = Required 10% 10% Tax rate = 40% 20 20 Replacement Problem – Replacement Computing Cash Flows Computing Remember that we are interested in Remember incremental cash flows incremental If we buy the new machine, then we will If sell the old machine sell What are the cash flow consequences of What selling the old machine today instead of in 5 years? in 21 21 Replacement Problem – Pro Replacement Forma Income Statements Forma Year 1 2 3 4 5 50,000 50,000 50,000 50,000 50,00...
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