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ENGM 661 Engineering Economics

ENGM 661 Engineering Economics. Depreciation & Taxes. Taxable Income. + Gross Income - Depreciation Allowance - Interest on Borrowed Money - Other Tax Exemptions = Taxable Income. Corporate Tax Rate. Corporate Tax.

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ENGM 661 Engineering Economics

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  1. ENGM 661 Engineering Economics Depreciation & Taxes

  2. Taxable Income + Gross Income - Depreciation Allowance - Interest on Borrowed Money - Other Tax Exemptions = Taxable Income

  3. Corporate Tax Rate

  4. Corporate Tax Ex: Suppose K-Corp earns $5,000,000 in revenue above manufacturing and operations cost. Suppose further that depreciation costs total $800,000 and interest paid on short and long term debt totals $1,500,000. Compute the tax paid.

  5. Corporate Tax Gross Income $ 5,000,000 Depreciation - 800,000 Interest - 1,500,000 Taxable Income $ 2,700,000

  6. Corporate Tax Gross Income $ 5,000,000 Depreciation - 800,000 Interest - 1,500,000 Taxable Income $ 2,700,000 Tax = $ 113,900 + .35(2,700,000 - 335,000) = $ 941,650

  7. After Tax Cash Flow + Gross Income - Interest = Before Tax Cash Flow

  8. After Tax Cash Flow + Gross Income - Interest = Before Tax Cash Flow - Tax = After Tax Cash Flow

  9. After Tax Cash Flow Ex: Suppose K-Corp earns $5,000,000 in revenue above manufacturing and operations cost. Suppose further that depreciation costs total $800,000 and interest paid on short and long term debt totals $1,500,000. Compute the after tax cash flow.

  10. After Tax Cash Flow Gross Income $ 5,000,000 Depreciation - 800,000 Interest - 1,500,000 Before Tax Cash Flow $ 2,700,000

  11. After Tax Cash Flow Gross Income $ 5,000,000 Interest - 1,500,000 Before Tax Cash Flow $3,500,000 Less Tax 941,650 After Tax Cash Flow $ 2,558,350

  12. Methods of Depreciation • Straight Line (SL) • Sum-of-Years Digits (SYD) • Declining Balance (DB) • Prior to 1981 • Accelerated Cost Recovery System (ACRS) • 1981-86 • Modified Accelerated Cost Recovery (MACRS) • 1986 on

  13. Straight Line (SLD) Let P = Initial Cost n = Useful Life s = Salvage Value year n Dt = Depreciation Allowance in year t Bt = Unrecovered Investment (Book Value) in year t Then Dt = (P - S) / n Bt = P - [ (P - S) / n ]t

  14. Ex: Straight Line Depr. Let P = $100,000 n = 5 years s = $ 20,000 Then Dt = (P - S) / n = $ 16,000 B5 = P - [ (P - S) / n ] 5 = $ 20,000

  15. Declining Balance In declining balance, we write off a constant % , p, of remaining book value D1 = pP , P = initial cost B1 = P - D1 = P - pP = P(1-p) D2 = pB1 = pP(1-p)

  16. Declining Balance In declining balance, we write off a constant % , p, of remaining book value B2 = B1 - D2 = P(1-p) - pB1

  17. Declining Balance In declining balance, we write off a constant % , p, of remaining book value B2 = B1 - D2 = P(1-p) - pB1 = P(1-p) - pP(1-p)

  18. Declining Balance In declining balance, we write off a constant % , p, of remaining book value B2 = B1 - D2 = P(1-p) - pB1 = P(1-p) - pP(1-p) = P(1-p)[1 - p]

  19. Declining Balance In declining balance, we write off a constant % , p, of remaining book value B2 = B1 - D2 = P(1-p) - pB1 = P(1-p) - pP(1-p) = P(1-p)[1 - p] = P(1-p)2

  20. Declining Balance In declining balance, we write off a constant % , p, of remaining book value B2 = B1 - D2 = P(1-p) - pB1 = P(1-p) - pP(1-p) = P(1-p)2 Dt = p [ P (1 - p) t - 1] Bt = P (1 - p) t

  21. Ex: Declining Balance P = $100,000 n = 5 years S = $20,000 p = 2/5 (200% declining balance) Then D1 = (2/5)(100,000) = $40,000 D5 = ? , B5 = ?

  22. Ex: Declining Balance P = $100,000 n = 5 years S = $20,000 p = 2/5 (200% declining balance) Then D1 = (2/5)(100,000) = $40,000 B1 = 100,000 - 40,000 = $ 60,000 D5 = ? , B5 = ?

  23. Ex: Declining Balance P = $100,000 n = 5 years S = $20,000 p = 2/5 (200% declining balance) Then D1 = (2/5)(100,000) = $ 40,000 B1 = 100,000 - 40,000 = $ 60,000 D2 = (2/5)(60,000) = $ 24,000 D5 = ? , B5 = ?

  24. Ex: Declining Balance (cont) • Dt = p [ P (1 - p) t - 1] • D5 = .4(100,000)(.6) 4 • = $ 5,184 • Bt = P (1 - p) t • B5 = 100,000(.6) 5 • = $ 7,776

  25. Ex: Declining Balance (cont) • Dt = p [ P (1 - p) t - 1] • D5 = .4(100,000)(.6) 4 • = $ 5,184 • Bt = P (1 - p) t • B5 = 100,000(.6) 5 • = $ 7,776 Note that Declining Balance will never depreciate book value to $0. It will, however, depreciate past the salvage value

  26. Time Value of Tax Savings(Tax Rate = 40%)

  27. DDB/SL Conversion(Salvage = $0)

  28. DDB/SL Conversion(Salvage = $0)

  29. DDB/SL Conversion(Salvage = $0)

  30. DDB/SL Conversion(Salvage = $0)

  31. DDB/SL Conversion(Salvage = $0)

  32. Class Problem Ex: Suppose K-Corp is interested in purchasing a new conveyor system. The cost of the conveyor is $180,000 and may be depreciated over a 5 year period. K-Corp uses 150% declining balance method with a conversion to straight line. Compute the depreciation schedule over the 5 year period.

  33. Class Problem

  34. Class Problem (p = 1.5/5 = .3)

  35. Class Problem

  36. DDB/SL Conversion(Half-Year Convention)

  37. DDB/SL Conversion(Half-Year Convention)

  38. DDB/SL Conversion(Half-Year Convention)

  39. Class Problem A $180,000 piece of machinery is installed and is to be depreciated over 5 years. You may assume that the salvage value at the end of 5 years is $ 0. The method of depreciation is to be double declining balance with conversion to straight line using the half-year convention (you may only deduct 1/2 year of depreciation in year 1). Establish a table showing the depreciation and the end of year book value for each year.

  40. Class Problem

  41. Solution

  42. MACRS Tables

  43. Modified Accelerated Cost Property Classes 3 yr. - useful life < 4 yrs. autos, tools 5 yr. - 4 yrs. < useful life < 10 yrs. office epuipment, computers, machinery 7 yr. - 10 < UL < 16 office furniture, fixtures, exploration 10 yr. - 16 < UL < 20 vessels, tugs, elevators (grain) 15 yr. - 20 < UL < 25 data communication, sewers, bridges, fencing

  44. MACRS (Cont.) 20 yr. - UL > 25 farm buildings, electric generation 27.5 - residential rental property 31.5 - non-residential real property Depreciation class (3, 5, 7, 10 yr.) uses 200% declining balance switching to straight-line @ optimal year class (15, 20) 150% DB switch to SLD class (27.5, 31.5) use straight-line

  45. After Tax Cash Flow Formulas BTCF = Before Tax Cash Flow = Revenues - Expenses

  46. After Tax Cash Flow Formulas BTCF = Before Tax Cash Flow = Revenues - Expenses TI = Taxable Income = Cash Flow - Interest - Depreciation

  47. After Tax Cash Flow Formulas BTCF = Before Tax Cash Flow = Revenues - Expenses TI = Taxable Income = Cash Flow - Interest - Depreciation Tax = TI * Tax Rate

  48. After Tax Cash Flow Formulas BTCF = Before Tax Cash Flow = Revenues - Expenses TI = Taxable Income = Cash Flow - Interest - Depreciation Tax = TI * Tax Rate ATCF = After Tax Cash Flow = BTCF - Tax

  49. Ex: After Tax Cash Flow

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