1. Benefit/Cost Ratios, Internal rate of Return, valuation, and depreciation

2. Calculating an unknown interest rate Cash flow must have costs and receipts Interest rate offer by a cash flow results in the present worth of receipts and costs being equal PW of Costs (i) = PW of Receipts (i) solve for I – Non linear equations – No direct method of solution – Multiple Roots – Provides unrealistic solutions when cash flow has wide swings in magnitude

3. Two Opportunities What is the Rate of Return

4. Although they both offer the same arithmetic sum they offer very different rates of return Opportunity 1 ROR = 15.1% Opportunity 2 ROR = 41.1% Internal rate of return assumes that funds can be reinvested at the same rate of return. What is the ROR if the payment of -$130,000 is made in year 10 rather than year 3 Some times there is more than one interest rates

5. Benefit to Cost Ratios Benefit to cost is preferred analysis convention although most times the same answer would results. – Conventions Reduction in travel time, crash costs, operating costs are generally considered benefits of public investment – E.G., delay reduction Costs are the first costs and operating costs need to provide and support the investment – E.G., annual operating and maintenance costs

6. What is the B/C of the Following operations Use a five percent per year interest rate

7. What do you think about the B/C Ratio What happens if the discount rate increase or lowers Benefit to cost ratios – Non-mutually exclusive alternatives – Mutually exclusive alternatives

8. Non-mutually exclusive project Example: A state department of transportation has several major interstate improvements targeted within one urban are and has to choose which ones to fund from this year’s budget. (network level analysis) The total budget available is $8,000,000. The current social discount rate is 8%. All project are to be evaluated over 20 years

9. Benefit to Cost Ratio Which projects should you select if you have only $8,000,000

10. Order by increasing costs Run out Of money

11. Non-mutually exclusive project example with Internal rate of return Internal rate of return is the interest rate that meets the follow condition

12. Ordering of Projects Rate of Return

13. Continuous alternatives Examples: Thickness of pavement Diameter of pipe Number of new employees Example problem: You are asked to specify the number of trains to provide on a subways system to minimize the total of the costs of equipment (trains) and the delay experienced by passengers. Givens:Line is 20 miles long Trains average 30 mpg User delay = $5.00/hr Costs per train First cost = $5,800,000 Energy cost per year = $80,000 Labor and maint. Cost per year = $200,000 Life = 40 years Salvage value = 0

14. Rail transit example The increase in the number trains decrease the headway. Short headways = lower user costs Long headway = lower train costs It takes 80 minutes to make a round trip Headway = 80/x X = the number of trains Expected delay = Headway/2 Weekdays per year = 255 Total yearly delay cost = (80/x)200,000*255(5/60) = $3.4 x10 8 /x Annual cost per train = $200,000 + $80,000 + $5.8x10 6 (A/P,10%,40yr) = $873,108

15. Rail example continued Total annual costs = $873,108(x) + $3.4x10 8 /x TC = Total cost A = $873,108 B = $ 3.4x10 8 TC = Ax + B/x dTC/dx = A + -b/x 2 x = B/A x = 3.4x10 8 /873,108 = 19.7 or 20 trains

16. Three project – three evaluation methods Social Discount Rate is 10%

17. Benefit to Cost Ratio

18. Maximum present worth

19. Internal rate of return

20. Summary

21. Discrete, Mutually Exclusive alternatives Steps – Order with increasing first cost of investment – Do-nothing alternative is the defender, all other projects are challengers. – Compare B/C for all alternatives: Greater than one, retain challenger Less than one, eliminate challenger All less than one, the defender is the most cost effective and stop – Of those challengers with B/C greater than one, arrange in order of first and return to step 2.

22. Example Roadway alignment alternatives Plan HPlan JPlan K First cost$110,000$700,000$1,300,000 Annual O&M$35,000$21,000$17,000 Salvage$0$300,000$550,000 A Plan H Plan K Plan J B

23. Example continued Traffic demand is expected to increase through the next 10 years and then stay at a constant level from year 10 to year 20. User cost of H = $210,000 in year one with a $10,000 per year increase till year 10 User cost of j = $157,500 in year one with a $7,500 per year increase till year 10 User cost of K = $136,500 in year one with a $6,500 per year increase till year 10 Use a 7 percent per year discount rate

24. Example continued

25. First Round

26. Second Round

27. Financial Sciences Economics and Accounting

28. Assigning value or a loss in value to an asset Important for annual tax consequences Important when making replacement disposal decisions Important when making rehab decisions Important concepts – Sunk costs are only important when they impact future costs

29. Depreciation Accounting methodology for assisting loss in value and value over time Book Value Time Depreciation Life Straight Line Accelerated

30. Uses of depreciation Accounting analysis Sinking fund Replacement decisions making Rehabilitation/maintenance decision making Important concept – Sunk costs should never be considered in future decisions

31. Asset lives Depreciation life – Life for accounting purposes Economic life – Life which minimized average life time costs Physical life – When asset can no long perform function Function life – When technology make asset obsolete

32. Example Life Costs of Asset Minimum

33. Replacement Decision- Making

34. Valuation methods Historical value Fair market value Potential future earnings The benefits accrued by users or to the economy Replacement cost – Wear and tear (condition- based valuation) Inflated historical costs – Wear and tear (condition-base valuation)

35. Problems with historical value Brooklyn bridge – built in 1883 1.14 Miles Long Carries 144,000 vpd, 4,000 ped/day Original cost $15,100,000 NYDOT spends as much $100,000,000/yr on maintenance Replacement value – 1 to 2 billion dollars

36. Problems with condition based valuation Value is measured against static standard Value is unrelated to contribution to agency objective Value is unrelated to its contribution to user benefits

37. Condition based valuation Time - Wear Performance Actual S-SHAPED DETERIORATION CURVES

38. Concept - Elements Standardized taxonomy for maintenance Taken from Paul Thompson and Richard Shepard Deck Girder Diaphragm Abutment Cap Column Footing Caltrans bridge valuation example

39. Serviceability is suspect due to loss of material Active attack by physical or chemical processes, no damage Protective materials or systems have partially or completely failed Protective materials and systems sound and functioning as intended Concept – Condition States 1. Protected 2. Exposed 3. Attacked 4. Damaged 5. Failed Element no longer serves its intended function

40. Severity weighting factor Number of Possible Condition States State 1 WF State 2 WF State 3 WF State 4 WF State 5 WF 3 Condition States 1.000.500.00 4 Condition States 1.000.670.330.00 5 Condition States 1.000.750.500.250.00

41. Current value of each element Current element value = (quantity in Condition state x WF x FC) WF = Weight factor for the severity of the deterioration as determine in table. FC = Failure costs of the element (cost to rehabilitate or replace an element if it fails.

42. Core Element Condition & Extent Data ElementTotal Quantity UnitsState 1State 2State 3State 4State 5Unit Failure Cost Concrete Deck 300Sq Meters 0030000$600 Steel Girder100Meters6134500$3,500 Reinforced Concrete Abutment 24Meters240000$7,700 Reinforced Concrete Column 4Each40000$9,000 Joint Seal24Meters0024na $556

43. Bridge valuation calculation ElementCalculation Current Element Value Concrete Deck 300 x 0.5 x 600$90,000 Steel Girder((61 X 1.0)+(34x0.75)+(5x0.5))x3,500$311,500 RC Abutment 24 x 1.0 x 7,700$184,800 RC Column4 x 1.9 x 9,000$36,000 Joint Seal24 x 0.0 x 556$0 Total Current Value of Bridge$622,300

44. Benefit based approached Takes into account benefit stream from asset Differentiates between economic analysis and financial analysis Difficult to estimate