Economic Evaluation of Economic Evaluation of Pavement AlternativesPavement Alternatives

CEE 320Steve Muench

04/20/23 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.

Pavement Cost

Agency Cost:Initial or Capital CostMaintenance Cost

RoutineMajor Rehabilitation CostLiability Cost

Road User Cost:Taxes and FeesRunning Cost

VehicleFuel

Added User CostDelay (Time is money)FuelAccidents

Pavement Benefits VS Cost

A. Positive (Benefits) In-coming1. Salvage values of capital investment

2. User time

3. Saved vehicle costs

4. Saved accident costs?a. monetary value of property damage is

traditionally estimatedb. monetary value of lives lost and

sometimes injuries is problematic

Pavement Benefits VS Cost

B. Negative (Costs) Out-Going1. Capital

2. Maintenance and Rehabilitation

3. Liability

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 yearsat 2%/yr compounded annually

x = current value, and y = future valuex = $300 M n = 10 years

Solution:

Y = 300 (1 + .02)10 = 300 (1.22) = $365.7M

X

Y?

n periods

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

i n11

X

Y?

100000*1

06.0112

Economic AnalysisC. 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) = 7.3601Present worth = Y = 7.3601*JD1 million = JD7.36 million

ii

in

n

SPWF1

)1( 1 Y?

knownperiod

X

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) = 0.1359Annual payment = Y = (0.1359) * $7.36 million = $1 million

1)1(

)1(

i

in

iCRF

n

unknown?period

Y

X

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/m2

Economic Analysis

Alternative No.1

(20 years life)

Alternative No. 2

(20 years life)Surface: 15 cm @ JD20/m2 Surface: 10 cm @ JD14/m2

Base: 25 cm @ JD5/m2 Base: 25 cm @ JD5/m2

Subbase: 40 cm @ JD3/m2 Subbase: 40 cm @ JD3/m2

Seal @ 10 years with JD2/m2 Overlay, 5 cm @ 10 years with JD20/m2

Economic Analysis (Solution:)

Computing the present value of the alternatives:

JD6.6

JD22

JD0.1/yr

JD20

n=20yr

n=10yrn=0yr

n=0

JD28

JD0.1/yr

JD8.4

JD2

n=20yr

n=10

Alt. No.1

Alt. No.2

36.4

22.6

Economic Analysis (Solution:)

Alternative 1Initial Cost = (20+5+3) = JD28/m2

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/m2 BETTER Alt

Alternative 2Initial Cost = (14+5+3) = JD22/m2

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/m2