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ENGM 661 Engineering Economics for Managers

ENGM 661 Engineering Economics for Managers

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ENGM 661 Engineering Economics for Managers

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  1. ENGM 661 Engineering Economics for Managers InvestmentWorth

  2. Investment Worth MARR Suppose a company can earn 12% / annum in U. S. Treasury bills No way would they ever invest in a project earning < 12% Def:The Investment Worth of all projects are measured at the Minimum Attractive Rate of Return (MARR) of a company.

  3. MARR MARRis company specific • utilities - MARR = 10 - 15% • mutuals - MARR = 12 - 18% • new venture - MARR = 20 - 30% MARR based on • firms cost of capital • Price Index • Treasury bills

  4. Investment Worth Alternatives • NPW(MARR) > 0 Good Investment

  5. Investment Worth Alternatives • NPW(MARR) > 0 Good Investment • EUAW(MARR) > 0 Good Investment

  6. Investment Worth Alternatives • NPW(MARR) > 0 Good Investment • EUAW(MARR) > 0 Good Investment • IRR > MARR Good Investment

  7. Present Worth Example: Suppose you buy and sell a piece of equipment. Purchase Price $16,000 Sell Price (5 years) $ 4,000 Annual Maintenance $ 3,000 Net Profit Contribution $ 6,000 MARR 12% Is it worth it to the company to buy the machine?

  8. 4,000 4,000 6,000 3,000 0 0 5 5 3,000 16,000 Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5) 16,000

  9. 4,000 4,000 6,000 3,000 0 0 5 5 3,000 16,000 Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5) = -16 +3(3.6048) + 4(.5674) 16,000

  10. 4,000 4,000 6,000 3,000 0 0 5 5 3,000 16,000 Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5) = -16 +3(3.6048) + 4(.5674) = -2.916 = -$2,916 16,000

  11. Annual Worth Annual Worth (AW or EUAW) AW(i) = PW(i) (A/P, i%, n) = [ At (P/F, i%, t)](A/P, i%, n) AW(i) = Annual Worth of Investment AW(i) > 0 **OK Investment** 

  12. 4,000 3,000 0 5 16,000 Annual Worth; Example Repeating our PW example, we have AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)

  13. 4,000 3,000 0 5 16,000 Annual Worth; Example Repeating our PW example, we have AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5) = -16(.2774) + 3 + 4(.1574)

  14. 4,000 3,000 0 5 16,000 Annual Worth; Example Repeating our PW example, we have AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5) = -16(.2774) + 3 + 4(.1574) = -.808 = -$808

  15. 4,000 3,000 0 5 16,000 Alternately AW(12) = PW(12) (A/P, 12%, 5) = -2.92 (.2774) = - $810 < 0 NO GOOD

  16. Internal Rate of Return Internal Rate-of-Return IRR- internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment**

  17. Internal Rate of Return Internal Rate-of-Return IRR- internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = At(1 + i*)n - t 

  18. Internal Rate of Return Internal Rate-of-Return IRR- internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = At(1 + i*)n - t PWrevenue(i*) = PWcosts(i*) 

  19. 4,000 3,000 0 5 16,000 Internal Rate of Return Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5)

  20. 4,000 3,000 0 5 16,000 Internal Rate of Return Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5)

  21. 4,000 3,000 0 5 16,000 Internal Rate of Return Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5) i* = 5 1/4 % i* < MARR

  22. Public School Funding

  23. Public School Funding 216% 16 yrs

  24. 216 100 1 2 3 16 F = P(F/P,i*,16) (F/P,i*,16) = F/P = 2.16 (1+i*)16 = 2.16 School Funding

  25. 216 100 1 2 3 16 School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701

  26. 216 100 1 2 3 16 School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 ln(1+i*) = .0481

  27. 216 100 1 2 3 16 School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 ln(1+i*) = .0481 (1+i*) = e.0481 = 1.0493

  28. 216 100 1 2 3 16 School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 ln(1+i*) = .0481 (1+i*) = e.0481 = 1.0493 i* = .0493 = 4.93%

  29. 216 100 1 2 3 16 School Funding We know i = 4.93%, is that significant growth?

  30. 216 100 1 2 3 16 School Funding We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period.

  31. 216 100 1 2 3 16 i  j . 0493  . 0350 d   1  j 1 . 0350 School Funding We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period. d= 1.4%

  32. 216 100 1 2 3 16 School Funding We know that d, the real increase in school funding after we discount for the effects of inflation is 1.4%. So schools have experienced a real increase in funding?

  33. 216 100 1 2 3 16 School Funding We know that d, the real increase in school funding after we discount for the effects of inflation is 1.4%. So schools have experienced a real increase in funding? Rapid City growth rate 3% / yr. 

  34. Summary • NPW > 0 Good Investment

  35. Summary • NPW > 0 Good Investment • EUAW > 0 Good Investment

  36. Summary • NPW > 0 Good Investment • EUAW > 0 Good Investment • IRR > MARR Good Investment

  37. Summary • NPW > 0 Good Investment • EUAW > 0 Good Investment • IRR > MARR Good Investment Note: If NPW > 0 EUAW > 0 IRR > MARR

  38. Internal Rate of Return Internal Rate-of-Return IRR- internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = At(1 + i*)n - t PWrevenue(i*) = PWcosts(i*) 

  39. 4,000 3,000 0 5 16,000 Internal Rate of Return Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5)

  40. 4,000 3,000 0 5 16,000 Internal Rate of Return Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5) i* = 5 1/4 % i* < MARR

  41. Spreadsheet Example

  42. 4,100 2,520 0 1 2 3 n 1,000 5,580 IRR Problems Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR).

  43. 4,100 2,520 0 1 2 3 n 1,000 5,580 IRR Problems Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR). PWR(i*) = PWC(i*) 4,100(1+i*)-1 + 2,520(1+i*)-3 = 1,000 + 5,580(1+i*)-2

  44. NPV vs. Interest $25 $20 $15 $10 Net Present Value $5 $0 0% 10% 20% 30% 40% 50% 60% ($5) Interest Rate IRR Problems

  45. External Rate of Return Purpose: to get around a problem of multiple roots in IRR method Notation: At = net cash flow of investment in period t At , At > 0 0 , else -At , At < 0 0 , else rt = reinvestment rate (+) cash flows (MARR) i’ = rate return (-) cash flows  Rt =  Ct =

  46. External Rate of Return Method find i = ERR such that Rt (1 + rt) n - t = Ct (1 + i’) n - t Evaluation If i’ = ERR > MARR Investment is Good  

  47. 4,100 2,520 0 2 3 1 1,000 5,580 External Rate of Return ExampleMARR = 15% Rt (1 + .15) n - t = Ct (1 + i’) n - t 4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1 i’ = .1505  

  48. 4,100 2,520 0 2 3 1 1,000 5,580 External Rate of Return ExampleMARR = 15% Rt (1 + .15) n - t = Ct (1 + i’) n - t 4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1 i’ = .1505 ERR > MARR  

  49. 4,100 2,520 0 2 3 1 1,000 5,580 External Rate of Return ExampleMARR = 15% Rt (1 + .15) n - t = Ct (1 + i’) n - t 4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1 i’ = .1505 ERR > MARR Good Investment  

  50. Critical Thinking IRR < MARR a. IRR < MARR < ERR b. IRR < ERR < MARR c. ERR < IRR < MARR