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Economic Analysis in Transportation Systems

Economic Analysis in Transportation Systems. Tapan K. Datta, Ph.D., P.E. CE 7640: Fall 2002. Chapter 5 Interest and Vestcharge. Economic cost of owning or using one product as compared to other alternatives can be assessed correctly if a factor for cost of money or lost opportunity is used.

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Economic Analysis in Transportation Systems

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  1. Economic Analysis in Transportation Systems Tapan K. Datta, Ph.D., P.E. CE 7640: Fall 2002

  2. Chapter 5Interest and Vestcharge Economic cost of owning or using one product as compared to other alternatives can be assessed correctly if a factor for cost of money or lost opportunity is used.

  3. Concept of Interest • Interest: Money paid by borrower for the money loaned to him or her • Rate of Interest: Rate based on amount of loan per unit time Risk and how lenders often vary rates based on credit risk

  4. Return: Monetary return received from operating an enterprise • Net profit received • Rate of Return: Quotient of the net return for one year divided by invested capital. • Dividends: Stock type ventures pay • Dividends to stockholders

  5. Vestcharge • Vestcharge: Charge for the use of money invested in physical assets • it is in lieu of cash returns • Vestcharge is generally less than the interest rate • Prime Rate: The best rate the banks give to their best customers.

  6. Concept of Vestcharge • Interest: may be considered as monetary rent on borrowed money. • Return: Is expected monetary compensation for investing money. • Vestcharge: means an economic charge against the investment.

  7. Interest and Vestcharge In economic analysis the word Vestcharge is used exactly as the terms: 1. Discount Factor 2. Minimum Attractive Rate of Return (MARR) 3. Minimum Attractive Discount Rate

  8. Why use Vestcharge? 1. Money invested in durable and long lasting goods should pay the penalty for failure to be consumed immediately. 2. The investor, the public has alternative uses of the highway tax money for dividend or profit returning investment 3. The risk of all uses of money varies, so there is a need of a factor in the analysis, which provides a measure of alternative risks.

  9. Why use Vestcharge? 4. The disbursement dates vary for a public work project, so a discount factor is needed to bring them all on similar timelines.

  10. Project Estimated Cumulative Probable No. Cost Costs Rate of Return 1 540 K 540 K 22% 2 1,610 K 2,150 K 22% 3 954K 3,104 K 21% : : : : 38 3,610 K 26,936 K 12% 39 200 K 27,136 K 12% 40 805 K 27,941 K 11% 41 512 K 28,453 K 11% : : : : 92 726 K 65,762 K 7% Method of Calculating MARR-Winfrey Budget = $ 28 M Budget allows only up to project No. 40 to be built. The lowest rate of return is 11%.

  11. Winfrey’s Process (MARR) 1. Estimate cost of project 2. Estimate project benefits 3. Estimate Rate of Return for each project 4. Rank order projects from highest to lowest 5. Cut projects as soon as it reaches the budget amount. 6. The lowest Rate of Return is called the MARR and can be used as the vestcharge rate.

  12. Alternative Method 1. Calculate Rate of Return observed in similar projects. 2. Rank order them highest to lowest 3. Use lowest observed Rate of Return (ROR) as vestcharge.

  13. Example Problem Existing Total User Cost per Vehicle Mile = $0.35. Analysis Period = 20 years, Terminal Value = 0. What is the MARR?

  14. Costs and Benefits • Costs: • Initial Investment (construction costs) • Annual Operation and Maintenance (O&M) costs • Benefit: • Reduction in Road User cost • For example, for Project A, the benefit is (0.350-0.335)*(20,800,000) • = $312,000 per year.

  15. Steps to Follow • Set up the EUAC, PW or FW equation for each of the alternatives equal to zero. Annual Benefits Analysis Period = n Annual Costs Initial Investment Project A: EUAC = - 1,000,000*(CR)in=20- 150,000 - 312,000 = 0 Project B: EUAC = - 1,250,000*(CR)in=20- 145,000 + 392,000 = 0 Project C: EUAC = - 1,100,000*(CR)in=20- 140,000 + 214,000 = 0

  16. Steps to Follow • Try with an interest rate, say, i = 15%. For Project A: EUAC = - 1,000,000*(CR)i=15n=20 - 150,000 + 312,000 = - 1,000,000*0.15976- 150,000 + 312,000 = + $2,240 For i = 16%, EUAC = -$6,670  The value of i lies somewhere between 15% and 16%.

  17. Steps to Follow 3. Use linear interpolation to find the exact value of the interest rate. + $2,240 15.25% 16% 15% B A X - $6,670 (0.16 – X)/6670 = (X – 0.15)/2240 By solving the equation, we find X = 0.1525  Rate of Return = 15.25%.

  18. Steps to Follow • Similarly, find the Rate of Return for other projects. In this problem, RORA = 15.25% RORB = 19.16% RORC = 5.68% 5. Rank the projects based on highest to lowest Rate of Return. 6. Conclusion: The Minimum Attractive Rate of Return (MARR) is 5.68%.

  19. Chapter 6: Compound Interest Equations P = a present sum of money; a single payment; the present worth of an annuity or a single sum. F = a sum at a future date, ‘n’ interest payment periods from the present. This is equivalent to P with compound interest rate “i” over ‘n’ periods

  20. Chapter 6: Compound Interest Equations F = P(1 + i)n Compound the amount at the end of ‘n’ periods CA = (1 + i)n F n-1 n 1 2 3 P

  21. Compound Interest Equations P F Present worth of single sum to be withdrawn at the end of “n” periods in the future. 1 1 (1+i)n n-1 n 1 2 3 P = F (1+i)n PW =

  22. Compound Interest Equations Compound amount at the end of “n” periods to which a series of payments of ‘n’ uniform deposits will accumulate. (1+i)n –1 i (1+i)n –1 i F n-1 n 0 1 2 3 A F = A SCA =

  23. Compound Interest Equations F Sinking Fund i (1+i)n - 1 SF = i (1+i)n - 1 A = End of Period annuity amortization payment i = interest rate n = number of interest periods n-1 n 0 1 2 3 A = F A

  24. A n-1 n 0 1 2 3 P SPW = (1 + i)n - 1 i (1 + i)n Compound Interest Equations Present worth of a series of ‘n’ uniform period end withdrawals (1 + i)n - 1 P = A i (1 + i)n

  25. A n-1 n 0 1 2 3 P Compound Interest Equations Capital Recovery with interest amenity which will return ‘n’ period end uniform receipts i(1+i)n (1+i)n-1 i(1+i)n (1+i)n-1 A = P CR =

  26. (CAF)i =6 = 1.5036 n =7 (CAF)i =6 = 2.0121 n =12 Examples • $ 100 kept at 6% compound interest for 7 years, what will be the amount at end of 7 years? F = P(1+i)n = F = 100(CAF) in = 100(CAF)i =6 = 100 (1.5036) F = $ 150.36 • What will be the amount after 12 years, using 6% interest F = 100 (2.0121) = $ 201.21 From Compound Interest Table n =7

  27. *(PWF)i (PWF)i =6 = 0.6651 *(PWF)i =6 n n =7 n =12 Examples • $ 100 due in 7 years. What is its present worth? • P = 100 • P = 100 (.6651) • P = $ 66.51 • What is the present worth of a $100 bond after 12 years? • P = 100 • = 100 (0.4970) = $ 49.70

  28. F = 100 = 100 (8.394) = $ 839.38 (SCA)i =6 n =7 Examples • Annual savings of $ 100 for 7 years at 6%? • What will be annual deposit that will accumulate to $ 100 at 6% after 7 years? A = 100 (SF)67 = 100 (0.11914) A = $ 11.91

  29. Examples • What is the present worth at 6% interest of $ 100 receivable each year for 7 years? P = 100 (SPW)67 = 100 (5.582) P = $ 558.20

  30. Examples • A $ 100 initial fund that earns interest at 6% on its balance could be exactly paid out in 7 years A = 100 (CR)6%7 = 100 (0.17913) A = $ 17.91

  31. Chapter 7- Methods of Economic Analysis Economic analysis is performed for • Project Selection • Project Evaluation Cost Effectiveness Analysis is performed for • $ Cost/Fatality Savings or • $ Cost/Injury Savings

  32. Traditional Analysis Methods 1. Equivalent Uniform Annual Cost Method (EUAC) 2. Present Worth of Costs Method (PWOC) 3. Equivalent Uniform Annual Net Return Method (EUANR) 4. Net Present Value Method (NPV) 5. Benefit to Cost Ratio (B/C) 6. Rate of Return Method (ROR) * Incremental Benefit-Cost Ratio * Cost/Effectiveness Method

  33. Economic Analysis • EUAC = -I(CR)in + T(SF)in - K - U I = Initial cost T = Terminal value K = Total uniform annual costs U = Uniform annual road user costs • PWOC = -I + T(PW)in - K(SPW)in - U(SPW)in • EUANR = -I(CR)in + T(SF)in - K + R R = Uniform annual gross benefit

  34. Economic Analysis B/C = PWOB/PWOC or B/C = EUAB/EUAC

  35. Economic Analysis Problem Highway Industrial Plant Project A1 A2 B3 B4 Base Prop Base Prop I 140K 160K 120K 200K Initial Investment T 40K 50K 10K 18K Terminal Value Total Administ. K 7K 8K 500K 485.5K Oper. & Maint. Uniform Annual U 74K 70K Road user Cost Uniform Road U 15K 14.5K T User Taxes I 8% 8% 8% 8% Vestcharge n 10 10 10 10 Analysis Period These are two mutually exclusive pairs with equal levels of service

  36. EUAC Method EUACA1= -140,000 * CR i=8n=10 + 40,000 * SF i=8n=10 - 7,000 - 74,000 = - 140,000 *(0.149029) + 40,000* (0.0629) - 81,000 EUACA1= - 99,103 EUACA2= -160,000 * CR i=8 n=10 + 50,000 * SF i=8 n=10 -8,000 - 70,000 = -160,000*(0.149029) + 50,000* (0.069029) -78,000 EUACA2= - 98,394

  37. EUAC Method EUACA1 - EUACA2 = +99,103 - 98,394 = 709 A2 Alternative has $709 less annual costs A2 is better than A1.

  38. PWOC Method PWOCA1 = - 140,000 + 40,000 PW810 - 7000 * SPW810 - 74,000 * SPW810 = -140,000 + 40,000*(0.46319) - 7,000 * (6.710081) - 74,000* (6.710081) PWOCA1 = -664,989

  39. PWOC Method PWOCA2 = - 160,000 + 50,000 (PW810) - 8000 * (SPW810) - 70,000 * (SPW810) PWOCA2 = -660,227 PWOCA2 - PWOCA1 = -660,227 - (-664,989) = $4,762 The cost of A1 Alternative is $4762 more than A2  A2 is better.

  40. Equivalent Uniform Annual Net Return Method Reduction in Road user costs is equivalent to cash income. EUANRA2 = -(Ip - IB) (CR810) + (Tp - TB) * (SF810) - (Up - UB) -(Kp - KB) EUANRA2 = - (160K - 140K) (0.149029) + (50K- 40K) (0.069029) – (70 K - 74 K) - (8K-7K) = $709 A2 is better.

  41. Net Present Value Method NPVA2 = -(Ip - IB) + (Tp - TB) * (PW810) – [(Up - UB) + (Kp - KB)] * (SPW810) = -20 K + 10K (0.463193)- (-K+1K)*(6.71008) NPVA2= 4,762 A2 is better.

  42. B/C Ratio B/C = EUAB/EUAC or B/C = PWOB/PWOC Benefit - Reduction in road user costs and reduction in annual expenses together is the benefit - Initial cost and Terminal value should be considered as costs

  43. B/C Ratio (Continued) EUAB = - (UP - UB) - (KP - KB) = - (70,000 - 74,000) - (8,000 - 7,000) EUAB = 3000 EUAC = -(Ip - IB)*(CR810) + (Tp - TB)*(SF810) = -(160K-140K)(0.149029) + (50K- 40K)(.069029) EUAC = - 2291 B/C = + 3000/-2291 = 1.31 1.31 is considered positive since the numerator is positive

  44. Rate of Return Method Assume, i = 10% on the basis of present worth 0 = B – C  0 = -(160,000 – 140,000) + (50,000 – 40,000)*(PW1010) – (70,000 – 74,000) (SPW1010) - (8,000 – 7,000) (SPW1010) = -20,000 + 10,000*(0.385543) + 4,000(6.144567) – 1,000 (6.144567) = $2,288

  45. Rate of Return Method (Continued) Second Iteration: Assume i = 12% 0 = -20,000 + 10,000(0.321973) + 4,000(5.65023) – 1,000 (5.65023) = 171 Rate of Return A2= 10 + 2*[(2288 -171)/2288] Rate of Return = 12.15%

  46. Small Improvement Project

  47. National Safety Council (NSC) For 1996 Injury = $ 34,100 PDO = $ 6,400 Reduction in road user costs related to traffic crashes = 39.27(6,400) + 30.59 (34,100) = 251,328 + 1043119 = $ 1,294,447

  48. Project Cost Location 1 = $ 34,100 Location 2 = $ 35,200 Location 3 = $ 29,400 Total = $ 98,700 Annual Operation and Maintenance cost = $ 1000 per Intersection per year Assume i = 7% and n = 15 years

  49. PWOC and PWOB PWOC = -98,700 – (1000)*(SPW715) = -98,700 – 1000 (9.1079) = -107,807.9 PWOB = 1,294,447*(SPW715) = 11,789,693 B/C = 11 ,789,693/107,807.9 B/C = 109:1

  50. Incremental B/C Ratio Assume that you want to pick 1 alternative amongst 5 solutions. Say, the alternatives were ranked in increasing cost. BCB-C A1 10 2 8 A2 20 14 6 A3 50 25 25 A4 80 30 50 A5 90 45 45

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