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

2. 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

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

4. INCLUDES SANITARY AND STORM SEWER
Report Card for America’s Infrastructure
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

5. What do we do???

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

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

8. 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

9. 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.

10. LIFE ?

11. FABRICATION
INSTALLATION
DURABILITY

12. Fabrication
Hand shake---team work for correct installation

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

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

15. 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.

16. 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

17. 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

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

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

20. 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

21. 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

24. Product Life = Design Life

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

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

27. 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

30. 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 + i) n
PRESENT VALUE
$1.00
PVF = = 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

31. 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

32. 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

33. 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.

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

35. 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

36. 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

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

38. 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 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.

39. Example Problem np

42. Least Cost (Life Cycle) Analysis
$500,000
( ) 50 = $3.2M
$450,000
PRESENT VALUE 1
=0.0543
( )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.

44. 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%

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

46. 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
ECHDPE = 600,000 1+
ECHDPE = $837,692

47. 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

48. 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* * 0.03)
D = $6,977,840 when the replacement is incurred

49. User Delay Costs (Indirect Costs)50
IECHDPE = 6,977,840
IECHDPE = $2,812,070 today

50. 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

51. 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 , renowned English critic, social commentator, and economist of the Victorian Age

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

53. Questions