Life Cost Analysis (LCA) 1. Premature Collapses 2. Product Life vs. Design Life 3. Life Cycle Cost Analysis (LCA) of Culvert Systems

Buried Pipe Must Perform Two Critical Functions ! Conduit Buried Pipe WHAT IS THE FUNCTION OF OUR BURIED INFRASTRUCTURE? AND WHY CAN THESE SYSTEMS CAUSE US PROBLEMS Structure

Report Card for America’s Infrastructure - 2003 AMERICA’S SANITARY AND STORM SEWER’S

INCLUDES SANITARY AND STORM SEWER Report Card for America’s Infrastructure - 2005 WASTEWATER = D- INCLUDES SANITARY AND STORM SEWER IT IS ACCORDING TO THE ASCE OUR NATIONS WASTE WATER SYSTEMS - DOWN FROM PREVIOUS REPORT CARD D TO D-. MOST OF OUR NATIONS SYSTEMS ARE OVER 100 YEARS OLD. AN EXAMPLE OF THIS PROBLEM IS OUR INTERSTATES WHICH ARE APPROXIMATELY 45 YEARS OLD AND THE DESIGN LIFE FOR MANY OF THE CULVERTS USED IN THIS SYSTEM ARE APPROX 50 YEARS. THINK ABOUT IT

What do we do???

Tangible Life Cost Factors Planning Specifications Hydrology Hydraulics Structures Installation Durability Maintenance Economics

How much do they ACTUALLY cost? Sudden Culvert Failures…. How much do they ACTUALLY cost? The Intangible

Intangible Life Cost Factors Tremendous delay to road users Economic loss of business Political implications due to News Media Coverage Owner Owner liability WE CAN REALLY RECORD THESE COSTS BUT FACTORS CAN BE ASSIGNED IS RELATION TO THE ROAD USAGE LEVEL AND TIME. I think this is me

Road Closed but they’re Open for Business 72” Aluminum Culvert Failure Cost = Up to $500,000 Major Highway CLOSED Business Profit Decrease Detour Longer Employee Commute Public Driving Confusion Officials Manning Roadblocks THIS IS AN EXAMPLE OF THE TANGLABLE AND INTANGLABLE AFFECTS OF A 72” Aluminum CULVERT FAILURE IN HILTON HEAD, SC The sinkhole was caused when a pipe which lets tides flow under the road began leaking. The hole caused it to suck sand inside, away from the surrounding area. The hollow eventually caused the land above to collapse.

LIFE ?

FABRICATION INSTALLATION DURABILITY

Fabrication Hand shake---team work for correct installation

Installation Teamwork between producer, owner, contractor, and the engineer.

“…dumping a burning river of gasoline into city storm sewers.” Durability “…dumping a burning river of gasoline into city storm sewers.”

Does this really happen? Smoldering charcoal from tailgate party grill dumped into a storm sewer inlet are suspected as the cause of a plastic pipe failure.

Material Service Life (Years) US Army Corps of Engineers – Conduits, Culverts, and Pipe Pipe Material Material Service Life (Years) RCP 70 –100 Aluminized CMP 50 HDPE

U.S. Army Corps of Engineers EM 1110-2-2902 Conduits, Culverts and Pipes (1) Concrete Most studies estimated product service life for concrete pipe to be between 70 and 100 years. Of nine state highway departments, three listed the life as 100 years, five states stated between 70 and 100 years, and one state gave 50 years. (2) Steel Corrugated steel pipe usually fails due to corrosion of the invert or the exterior of the pipe. Properly applied coatings can extend the product life to at least 50 years for most environments. (3) Aluminum. Aluminum pipe is usually affected more by soil-side corrosion than by corrosion of the invert. Long-term performance is difficult to predict because of a relatively short history of use, but the designer should not expect a product service life of greater than 50 years. (4) Plastic. Many different materials fall under the general category of plastic. Each of these materials may have some unique applications where it is suitable or unsuitable. Performance history of plastic pipe is limited. A designer should not expect a product service life of Excerpt from Paragraph 1-4, Life Cycle Design

Project Design Life FACILITY PROJECT DESIGN LIFE Storm Sewer System 100 years or greater Sanitary Sewer System 100 years or greater Expressway Culverts 100 years or greater Arterial Culverts 50 to 75 years Collector Culverts 50 to 75 years Local/Rural Culverts 25 to 50 years

ASTM Life Cost Analysis Planning Specifications Hydrology Hydraulics Structures Installation Durability Maintenance Economics

Life Cycle Cost Project Design Life Material Service Life Initial Cost Interest (Discount) Rate Inflation Rate Maintenance Cost Rehabilitation Cost Replacement Cost Replacement Cost (Direct and Indirect) Residual Value

Interest Inflation Factor Cost Effective Least Cost Sinking Fund Service Life Present Value Life Cycle Compound Interest Simple Interest Interest Rate Constant Dollars Capital Recovery Current Dollars Inflation Rate Nominal Rate Discount Rate Real Rate Amortization Interest Inflation Factor

Product Life = Design Life

Effective Cost Case 1: Product Life = Design Life EC = P Where: EC = Effective Cost, dollars P = Bid Price, dollars

Effective Cost Case 2: Product Life < Design Life EC = P + (P x IF x PVF) Where: IF = Inflation Factor PVF = Present Value Factor

INFLATION FACTOR & PRESENT VALUE FACTOR Where: I = Inflation Rate n = material life, years i = interest rate IF= The inflation factor converts the current bid price to future replacement costs. The present value factor converts future replacement costs to today’s dollars PVF= PERCENT

Least Cost (Life Cycle) Analysis $5.00 P(1 + I) n n years $1.00X(1+0.04) 36 $1.00X4.10 = $4.10 = 36 yrs. PVF = 1 (1 + i) n PRESENT VALUE $1.00 PVF = 1 = 0.123 (1+0.06) 36 PVF x $4.10 = $0.50 $0.50 PRESENT VALUE $1.50 Interest (Discount) Rate, i = 6% Inflation Rate, I = 4% - ( i- I ) = 2% $0 PRESENT n = 36 yrs. FUTURE TIME

Project Design Life, n = 40 yrs. Least Cost (Life Cycle) Analysis $5.00 n years = 40 yrs. $1.00(1+0.04)20 = $2.19 PRESENT VALUE ( 1 + I ) n $1.00 =0.312 n 1 (1+0.06)20 1 + I 1 + i PRESENT VALUE Material B $1.69 Material A $1.00 1 (1 + i) n $0.688 $1.00(0.981)20 = $0.683 n = 20 yrs. $0 PRESENT FUTURE TIME Interest (Discount) Rate, i = 6% Inflation Rate, I = 4% ( i-I ) = 2% (1+I/1+i) = 0.981 Project Design Life, n = 40 yrs. Material A, n = 100 years Material B, n = 20 years

User Delay Costs (Indirect Costs) Joseph Perrin’s Research: D = AADT * t * d *(cv * vv * vof + cf * vf) Where: AADT = Annual Average Daily Traffic of the roadway which the culvert is being installed t = the average increase in delay or congestion the installation is causing to each vehicle per day, in hours d = the number of days the project will take cv = the average rate of person-delay, in dollars per hour vv = the percentage of passenger vehicles traffic vof = the vehicle occupancy factor cf = the average rate of freight-delay, in dollars per hour vf = the percentage of truck traffic

User Delay Costs (Indirect Costs) Average Established Delay Costs as of 2005, in Dollars: cv = $18.62 per person / hour of delay cf = $52.86 per freight /hour of delay Typical Traffic Assumptions: vv = 97% vehicle passenger traffic vf = 3% truck traffic vof =1.2 persons per vehicle A one-hour delay on a roadway carrying an Annual Average Daily Traffic of 20,000 vehicles costs the public over $450,000 every day. Indirect costs include traffic user costs, economic loss of business, and political implications and liability for the owner. While these costs are not directly absorbed by the owner, they should be considered in a Least Cost Analysis. National Compensation Survey: Occupational Wages in the United States, Department of Labor Statistics, United States Department of Labor, June 2005. The Economic Costs of Culvert Failures, Joseph Perrin, Jr., November 2003.

Effective Cost Case 3: Product Life > Design Life EC = P - S Where: S = Residual Value S = P (F)np (ns/n)

S = P (F)np (ns/n) Effective Cost Where: S = Residual Value P = Present Cost F = Inflation/Interest Factor = Product Life np = Design Life ns = Number of years the product life exceeds the design life 1 + I 1 + i

Least Cost (Life Cycle) Analysis $5.00 n years = 40 yrs. (1+0.04)20 = $2.19 ( 1 + I ) n PRESENT VALUE $1.00 =0.312 n 1 (1+0.06)20 1 + I 1 + i PRESENT VALUE Material B $1.69 Material A $1.00 1 (1 = i) n (0.981)20 =0.683 - $0.28 $0.72 n = 20 yrs. $0 PRESENT FUTURE TIME Residual Value S = P (F)np (ns/n) = $1.00(0.981) 40 X (60/100) = $0.28

Maintenance & Rehabilitation Costs Where: M = Present value of expected maintenance costs N = Present value of expected rehabilitation costs Cm = Annual Maintenance / Rehabilitation Cost

Example Problem : Find: A 75-year design life has been assigned to a storm sewer project to be constructed for a private subdivision. Two alternative pipe with different wall thicknesses are included in the bid documents. Material A with a project bid price of $300,000 has been assigned a 60- year service life with an annual maintenance cost of $6,000/year. To meet the project design life, a $75,000 rehabilitation cost will have to be incurred at the end of the 60-year service life. Material B has an “in ground” cost of $345,000 with a 100-year projected service life. The annual maintenance cost has been estimated at $5,000/year. Planning and design costs applicable to all alternatives are $150,000. Based on historical data, a 5% inflation rate and 7.15% interest (discount) rate is appropriate for this project. Find: The most cost effective material with the lowest LCA.

Example Problem np

Least Cost (Life Cycle) Analysis $500,000 (1 + 0.04) 50 = $3.2M $450,000 PRESENT VALUE 1 =0.0543 (1 + 0.06)50 PRESENT VALUE HDPE $623,700 Concrete Pipe $500,000 $173,700 (0.981)50 = 0.386 n=50yrs. n=100yrs $0 FUTURE PRESENT TIME Interest (Discount) Rate, i = 6% - Inflation Rate, I = 4% - ( i-I ) = 2% & (1+I/1+I) = 0.981 HDPE Service Life, nalt= 50 yrs. Concrete Pipe Service Life, nconc= 100 yrs.

Example: Given: A culvert is to be installed under a primary road: Design life = 100 years for the project Bid price: Concrete pipe = $750,000 (100-year service life) HDPE pipe was $600,000 (50-year service life) Traffic AADT of 10,000 vehicles 60 days of delays for replacement 30 minute average dealy Average rate over time: interest = 9% inflation = 7%

Find: The effective cost of the two alternates by least cost analysis method, and select the most economical pipe material.

Direct Effective Cost 1+I ECHDPE = PHDPE 1+ 1+i 1+ 0.07 However, the HDPE pipe will need to be replaced at the end of nHDPE, years to have a total service life equal to 100 years. n 1+I ECHDPE = PHDPE 1+ 1+i 50 1+ 0.07 ECHDPE = 600,000 1+ 1+ 0.09 ECHDPE = $837,692

Solution: The service life of the pipe is based on the U.S. Army Corps of Engineers guidelines. Effective Concrete Cost = Bid Price ECConc = PConc = $750,000

User Delay Costs (Indirect Costs) D = AADT * t * d *(cv * vv * vof + cf * vf) AADT = 10,000 vehicle t = 0.5 hours d = 60 days cv = $18.62 per person / hour of delay cf = $52.86 per freight /hour of delay vv = 97% vehicle passenger traffic vf = 3% truck traffic vof =1.2 persons per vehicle D = 10,000*0.5*60*(18.62*0.97*1.2+52.86* 0.03) D = $6,977,840 when the replacement is incurred

User Delay Costs (Indirect Costs) 50 1+ 0.07 IECHDPE = 6,977,840 1+ 0.09 IECHDPE = $2,812,070 today

Life Cycle Cost Project Design Life Material Service Life Initial Cost Interest (Discount) Rate Inflation Rate Maintenance Cost Rehabilitation Cost Replacement Cost Replacement Cost (Direct and Indirect) Residual Value

It is unwise to pay too much, but it is worse to pay too little. When you pay too much, you lose a little money. When you pay too little, you sometimes lose everything, because the thing you bought was incapable of doing the thing it was bought to do. The common law of business balance prohibits paying a little and getting a lot-it can’t be done. If you deal with the lowest bidder, it is well to add something for the risk you run. And, if you do that, you will have enough to pay for something better. JOHN RUSKIN 1819-1900, renowned English critic, social commentator, and economist of the Victorian Age

“The bitterness of poor quality remains long after the sweetness of low prices is gone.” (Unknown)

Questions www.concrete-pipe.org