1. Economic Evaluation of Pavement Alternatives CEE 320 Steve Muench 9/12/2015 1 The majority roads are public facilities built with tax money to serve the public. Economic analysis can form the basis upon which the pavement engineer can make his/her final judgment. User cost must be consider along with agency economic limitations and availability of resources.

2. Pavement Cost Agency Cost : Initial or Capital Cost Maintenance Cost Routine Major Rehabilitation Cost Liability Cost Road User Cost : Taxes and Fees Running Cost Vehicle Fuel Added User Cost Delay (Time is money) Fuel Accidents

3. Pavement Benefits VS Cost A.Positive (Benefits) In-coming 1. Salvage values of capital investment 2. User time 3. Saved vehicle costs 4. Saved accident costs? a. monetary value of property damage is traditionally estimated b. monetary value of lives lost and sometimes injuries is problematic

4. Pavement Benefits VS Cost B.Negative (Costs) Out-Going 1. Capital 2. Maintenance and Rehabilitation 3. Liability

5. Economic Analysis A. Compound Amount Factor (CAF): How much will X be worth after n at i per period CAF = (1 + i) n example: How much 300 million Dollars will equal after 10 years at 2%/yr compounded annually x = current value, and y = future value x = $300 M n = 10 years Solution: Y = 300 (1 +.02) 10 = 300 (1.22) = $365.7M

6. Economic Analysis B. Present Worth Factor (PWF): How much will X at i per period after n periods be worth today? PWF = example: salvage value of the pavement will be $100,000 at end of 12 year service life, interest or discount rate = 6% Present value = Y = = (0.49696936) * 100000 = $49696.94

7. Economic Analysis C.Series Present Worth Factor (SPWF): What is X per period for n periods at i rate per period worth today? Example: The Ministry of Public Works & Housing is planning an annual maintenance strategy for Amman-Zarka Expressway equal JD1 million per year each of the next ten years. Assuming a 6% interest rate, how much will these payments equal as of today ? Answer: SPWF(6%,10 periods) = 7.3601 Present worth = Y = 7.3601*JD1 million = JD7.36 million

8. Economic Analysis D.Capital Recovery Factor (CRF): What Y/period for n periods at i rate per period will equal X today? Example: Suppose i =6% compounded annually The Greater Amman Municipality is borrowing JD7.36 million in 2013 for new pavement overlays. How much money must it pay annually over the next 10 years to retire the debt? Answer: CRF (6%, 10 periods) = 0.1359 Annual payment = Y = (0.1359) * $7.36 million = $1 million

9. Economic Analysis Example: As a Pavement Engineer, you are hired by the Greater Amman Municipality to choose the most economical pavement alternatives shown in the table below. Suppose i =5% compounded annually, and the Salvage value at the end of pavement life is 30% of the first capital (initial) cost. Additionally, an annual routine maintenance will be conducted through the first 10 years of the pavement life for both alternatives at JD0.1/ m 2

10. Economic Analysis Alternative No.1 (20 years life) Alternative No. 2 (20 years life) Surface: 15 cm @ JD20/m 2 Surface: 10 cm @ JD14/m 2 Base: 25 cm @ JD5/m 2 Subbase: 40 cm @ JD3/m 2 Seal @ 10 years with JD2/m 2 Overlay, 5 cm @ 10 years with JD20/m 2

11. Economic Analysis (Solution:) Computing the present value of the alternatives: JD6.6 JD22 JD0.1/yr JD20 n=20yr n=10yr n=0yr n=0 JD28 JD0.1/yr JD8.4 JD2 n=20yr n=10 Alt. No.1 Alt. No.2 36.4 22.6

12. Economic Analysis (Solution:) Alternative 1 Initial Cost = (20+5+3) = JD28/ m 2 Present Value = JD28 + JD0.1(SPWF,5%, 10 yr) + JD2(PWF,5%,10yr) – JD(28*0.3) (PWF,5%,20 yr) Present Value = JD28 + JD0.1(7.722) + JD2(0.614) –JD8.4(0.377) = JD 26.833/m 2 BETTER Alt Alternative 2 Initial Cost = (14+5+3) = JD22/ m 2 Present Value = JD22 + JD0.1(SPWF,5%, 10 yr) + JD20(PWF,5%,10yr) – JD(22*0.3) (PWF,5%,20 yr) Present Value = JD22 + JD0.1(7.722) + JD20(0.614) –JD6.6(0.377) = JD 32.564/m 2