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Cost Analysis and Estimating for Engineering and Management

Cost Analysis and Estimating for Engineering and Management. Chapter 10 Engineering Economy. Overview. Methods to Determine Project Returns Impact of Time Value of Money (Interest) Simple and Compound Interest Calculations Methods to Evaluate Projects Various Rates of Return

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Cost Analysis and Estimating for Engineering and Management

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  1. Cost Analysis and Estimatingfor Engineering and Management Chapter 10 Engineering Economy

  2. Overview • Methods to Determine Project Returns • Impact of Time Value of Money (Interest) • Simple and Compound Interest Calculations • Methods to Evaluate Projects • Various Rates of Return • Evaluation of Replacement Alternatives • Tax Effects on Decisions

  3. Money • All Projects Require Money • All Should Return Money (Payback) • When Large Sums of Money or Longer Times Are Involved – Interest Becomes Important • Decisions Are Based on Money

  4. Capital • Monetary Assets • Should Be Employed to Earn a Return • Invested • In Ordinary Financial Transactions • In Some Endeavor (Production/Service) • Need to Evaluate Investments

  5. Monitoring Return • Ways to Express Return • Total Dollars • Ratio with Sales • Return on Investment • Calculation of Return • Average Annual Rate of Return • Payback Period • Engineering-Economy Rate of Return

  6. Average Annual Rate of Return • Yearly After Tax Profit from Activity • Total Investment in the Activity • Alternately Eq 10.1 Eq 10.2

  7. Payback Period • How Many Years to Earn Back the Investment • Capital Liquidity Importance Eq 10.3

  8. Payback Caveats • Assumes Equal Annual Earnings • Does Not Account for • Interest • Depreciation / Obsolescence • Earnings After Payback

  9. Interest • Choice of Investments • Traditional (Bank, Stock Market, etc.) • Capital into Endeavor (Equipment, etc.) • Time Value of Money • Interest Calculations • Simple • Compound

  10. Simple Interest • Interest I = Pni • Total Including Principal F = P + I = P(1+ni) • Payments Are at the End of the Periods

  11. Compound Interest • Interest Adds to Principal • Interest Earns Interest • In Subsequent Periods • Single Payment Compound Amount Factor (1+i)n • Used In All Variations

  12. Various Interest Calculations • Net Present Worth • Net Future Worth • Net Equivalent Annual Worth • Rate of Return • All Generally Yield Consistent Recommendations

  13. Year Cost Revenue 0 $1025 $0 1 0 450 2 0 425 3 0 400 Set Up for Discussions • Example Situation • Investment $1025, for 3 Years • Factors Can Be Calculated or Found In Tables (10% and 20% in Appendices)

  14. Period n PW Factor at 10% Cash Flow Amount Year 1 to zero 1 0.9091 $450 $409 Year 2 to zero 2 0.8264 425 351 Year 3 to zero 3 0.7513 400 301 Total $1061 Less proposed investment 0 1.0000 1025 1025 Net present worth $36 Net Present Worth • Convert All Amounts to Present Value

  15. Period n Compound Amount Factor 10% Cash Flow Amount Year 1 to 3 2 1.2100 $450 $545 Year 2 to 3 1 1.1000 425 468 Year 3 0 1.0000 400 400 $1413 Compounded to terminal year at 10% 3 1.3310 1025 1364 Net future worth $ 49 Net Future Worth • Convert All Amounts to Future Value

  16. Net Equivalent Annual Worth • Find Present Value of Receipts • Calculate Annual Equivalent • Calculate Net Annual Equivalent Worth

  17. Amount Anticipated annual receipts $427 Less equal annual equivalent worth 14 Equal annual receipts $413 1025  1.10 = $1128 Less payment 413 715 715  1.10 = $787 Less payment 413 374 374  1.10 = $413 Less payment 413 0 Net Equivalent Annual Worth Example

  18. Rate of Return • Other Methods Assume an Interest Rate • Find the Interest Rate to Yield Same Return as Investment • Found By Trial-and-Error or Interpolation

  19. Example at 5%

  20. Example at 15%

  21. Interpolate

  22. Method Amount Net present worth at 10% $36 Net future worth at 10% $49 Net annual equivalent worth at 10% $14 Rate of return 12.3% Compare the Methods • Compare Present and Future Worths • Rate of Return Finds Actual Interest • Other Methods Assume an Interest

  23. Engineering Economy • P, F, A, i, and n • Functional Notation F=P(P/F, i%, n) P/F means, Find P Given F • 6 Methods (Equations) • 2 Single Payment • 4 Uniform Payment Series

  24. Factor Name Functional Notation Equation Equation Number Compound-amount factor (F/P, i%, n) (10.7) Present-worth factor (P/F, i%, n) (10.8) Single Payment Methods

  25. Single Payment Diagrams

  26. Factor Name Functional Notation Equation Equation Number Sinking-fund factor (A/F, i%, n) (10.9) Capital-recovery factor (A/P, i%, n) (10.10) Compound-amount factor (F/A, i%, n) (10.11) Present-worth factor (P/A, i%, n) (10.12) Uniform Payment Series

  27. Uniform Series Diagrams

  28. Uniform Series Diagrams

  29. Assumptions • Cash Flow at Year End (Lump Sum) • Ignore Inflation • Special Handling of Taxes • Constant Interest Rate • Predetermined MARR • Save Non-Quantifiable Effects for End

  30. Interest • Cost of Using Capital • Internal Rate of Return (IRR) • Minimum Acceptable Rate of Return (MARR) • External Rate of Return (ERR) • Percentage Showing Yield on Different Uses of Capital

  31. IRR • Based Solely on Project’s Cash Flow • Before Taxes • Find IRR for Present Worth of $0 • Required IRR < the Projected IRR • Approve Project • Compare Alternatives

  32. MARR • Rate Set By Management • Sets Lower Limit on Acceptable Return • Rations Capital to Avoid Unproductive Investments

  33. Comparison of Alternatives • Feasible Alternatives • Mutually Exclusive • “Do Nothing” Is an Alternative • May Omit Costs/Revenues If the Same • Find Alternative with Highest Return or Lowest Cost • Equal/Unequal Lives Handled Differently

  34. Automatic Semiautomatic Investment cost $1,250,000 $800,000 Annual Utilities 570,000 480,000 Labor, annual 50,000 700,000 Floor space 80,000 50,000 Maintenance 140,000 90,000 Equipment life 5 years 5 years Equal Life Example (i=10%)

  35. Cash Flows

  36. Automatic Semiautomatic Annual capital recovery cost, P(A/P, 10%,5) $329,750 $211,040 Annual cost 1,002,500 1,478,000 Comparative annual cost $1,332,250 $1,689,040 Comparison for Example

  37. Unequal Life • Comparison Must Be for Equal Output • Extra Cost Involved with Extra Output • Different Useful Lives for Alternatives • Analysis Period Common Multiple of Useful Lives • Example I1 – 3 Years, I2 – 4 Years • Common Multiple = 12 Years

  38. Investments I1 I2 Initial cost $230,000 $320,000 Annual operating cost 25,000 0 Salvage value 0 40,000 Economic life, year 3 4 Multiple of lives 12 12 Repeated investments 3 2 MARR before taxes 10% 10% Unequal Life Example

  39. 12 Year Cash Flow

  40. Comparison • Present Worth for 12 Years PW (I1) = – $800,520 PW (I2) = – $629,116 • Select Investment 2, Lower Cost • Assuming Revenues Are Equal • Study Period May Be Excessive • Situation May Vary from Assumptions

  41. Restricted Life • Project Limited to 2 Years • Assume Salvage Value = $0 PW(I1) = – $273,388 PW(I2) = – $320,000 • Compare to Include Salvage Value (320,000-273,338)/(P/F,10%,2)=$56,400 • Decision Depends on Salvage Cost of I2

  42. Replacement • Existing Alternative In Operation • Economic Facts Are Different for Challenger • Equipment Life – Period of Lower Cost • Need Value for Existing Equipment • Don’t Use Book or Trade In Value • Need the Market Value

  43. Equivalent Annual Cost for Replacement • Includes Resale Value (Fs) Or Eq 10.13

  44. Year Defender (D) Challenger (C) Operating cost Salvage value Operating cost Salvage value 0 $120,000 $350,000 1 $34,000 70,000 $3000 310,000 2 39,000 40,000 10,000 270,000 3 46,000 25,000 12,000 240,000 4 56,000 10,000 15,000 210,000 5 20,000 170,000 6 31,000 120,000 Example • i = 20%, Life = 4 Years

  45. Results • Existing (Defender) EAC(D) = – $86,618 • Challenger EAC(C) = – $105,208 • Challenger Has Higher Cost Don’t Replace

  46. Effects of Taxes • Taxable Income • Before Tax Cash Flow Less Depreciation and Other Charges • Rate of Return After Tax MARR  (Before Tax MARR)(1-t) Before Tax MARR = 20%, t = 38% After Tax MARR = About 12.4% • Timing of Cash Flow Has Impacts

  47. Considerations for After Tax Analysis • Costs, Savings, Revenues • Depreciation • Taxable Income • Cash Flow Effects • Engineering Economy Analysis • Decision Process • Consider Non-Economic Factors

  48. Yr End Before Tax Dep. Deductible Charges 35% Tax Savings After Tax Cash Flow 0 -$3 -$3 1 -25 $3.3 -$28.3 -$9.9 -15.1 2 -14 3.3 -17.3 -6.1 -7.9 3 -14 3.4 -17.4 -6.1 -7.9 3 2 2 0 -$28 -$28 1 -7 9.2 -$16.2 -$5.7 -1.3 2 -7 12.6 -19.6 -6.9 -0.1 3 -7 6.2 -13.2 -4.6 -2.4 After Tax Example Defender Challenger

  49. Outcome EAC(D) = – $11.1 Million EAC(C) = – $12.5 Million • Different Depreciation Methods Used • Favor Challenger • Costs Represent Disbursements • Thus Positive Costs • Select Challenger

  50. Summary • Methods of Project Alternative Evaluation • How to Use and Apply Interest Calculations • How to Evaluate Replacements • What Is the Impact of Taxes for Economic Decisions

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