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CVE 4073/5073

CVE 4073/5073. Construction Cost Estimating Class #2: Concept Estimating Prof. Ralph V. Locurcio, PE. Estimating defined… .

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CVE 4073/5073

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  1. CVE 4073/5073 Construction Cost Estimating Class #2: Concept Estimating Prof. Ralph V. Locurcio, PE

  2. Estimating defined… • A complexprocess involving the collection of available information on the scopeof a project, the expected resourceconsumption, and future changes in resource costs. • This information is synthesized in a visualizationof the construction process and translated into an approximationof the final cost.

  3. Qualifications of an estimator… • Extensive knowledge of: • Construction processes & methods • Materials & methods • Contracts & work practices • Construction documents & details • Background business factors • Design & code requirements • Verbal & written communications • Good common sense & judgment

  4. Estimating & Bidding Yes No

  5. Key Estimating Activities

  6. Components of an estimate… • Systematic analysis of: • Project components • Equipment needed • Construction methods • Temporary work items • Support activities • Externalities

  7. Two broad types of estimates… • Conceptual • Used for early decisions & feasibility • Rough estimate using minimal data • Relies heavily on past data • Level of accuracy about 70% • Detailed • Used for major decisions; i.e. bids & budgets • Based on final design documents • Level of accuracy up to 95%

  8. Accuracy increases over time… 100 90 60 30 % Design 0 25 50 90 100 % Accuracy

  9. Conceptual types: • Based on prior experience: • Weight check… weight of equipment • Cost-capacity factor- ratio of Q/$ • Comparative cost of structure – $/unit • Feasibility estimates • Appropriations estimates • Time & location adjustments

  10. Comparative cost of structure… • Basis is cost per unit based on past experience with similar units. • Compare structures of like construction, (e.g. school: steel, concrete, quality, local labor, weather) • Develop cost per unit, e.g. A$/pupil = A Cost/A pupils • Apply to new construction: B Cost = B pupils x A$/pupil

  11. Examples: • Schools: $/pupil • Bridges: $/feet of span • Stadium: $/seat • Hospital: $/beds • Offices: $/ft2 • Warehouses: $/ft3

  12. Variations of comparative cost… • Plant cost ratio… process plant ratio differentiates machine ft2 from admin ft2 • Floor area… cost/ft2 … but • Total horizontal area… all floors equal • Finished floor area… $ based on type finish • Cubic foot of volume… floor height varies

  13. Parametric estimates… • Used for capital budgeting • Expect 80-90% accuracy • Based on historical data base • Methods: • Panel… hist data on “typical” panels • Bay… hist data on “typical” bays • Plant component… equip cost + other TPC = ET/(1-PT) Where ET=tot equip cost/(1-% all other cost items) ET = $$$ equip/(1-70%)

  14. Time & Location adjustments… • Cost indexing • Based on similar “group” of key items • Compare “group” over time or location • Final cost based on ratio x calculated cost: Est Cost = T2/T1 x Calc Cost Est Cost = (T2/T1)(L2/L1) x Calc Cost

  15. Cost forecasting… • “The future ain’t what it used to be”… Yogi Berra • Need to project costs to mid-point of the construction year for budgeting • Generally average past 3-5 years cost data & project forward

  16. Detailed estimates… • Based on three primary factors: A. Scope… quantities B. Constructability… methods C. Risk… externalities

  17. A. Scope factors… • Technology in project - complexity • Milestone deadlines for work - timing • Material & equipment - availability • Staffing needed – skill & availability • Contract terms – allocation of risk • Amount of competition - markets

  18. B. Constructability factors… • Construction quality – skill required • Allowable tolerances – productivity • Complexity of design – learning curve • Flexibility of methods – special terms

  19. C. Risk factors… • Unspecified material or workmanship • Differing site conditions • Error in bidding • Changes in cost over time • Subcontractor error or failures • Weather & environmental issues • Strikes & labor issues • Utilities & support issues

  20. Estimating process… 1. Determine project characteristics: Scope, constructability, risk 2. Examine the project design & site 3. Structure the estimate 4. Determine elements of cost 5. Calculate estimate

  21. Examine Project Design… • Technical specifications • Referenced standards • Project drawings • Testing and performance • Special methods or procedures • Physical elements & site characteristics • Errors & omissions

  22. Structure the estimate… • Structure often follows specification template • Ensure nothing left out • Need a plan for completing the project • Logical flow of resources & materials • Productivity factors

  23. Determine elements of cost… • Labor- craft & skill resources; sub-contractors; union labor; wage rates; training; “effective” labor rates; productivity factors (weather, complexity, experience, management) • Material- price, shipping, availability, storage, relation to specifications • Equipment- purchase vs. lease, single use, down time, cost of maintenance • Capital- interest rates, payment flow, retainage • Time- overhead costs, required completion date,

  24. Calculate the estimate… • Material take-off from plans • Quantities of all construction materials • Compare with specifications • Labor cost for construction or installation • Effective labor rates • Team composition & duration • Construction equipment schedule • On hand vs. lease or purchase • Time on job • Follow elements of cost • Ensure all items are covered

  25. Some Example Problems • Calculate the estimated cost • Solution in class

  26. Conceptual Estimates • Based on historical data • Accuracy = +30% • Used for planning purposes when no design exists • Used to check validity of detailed estimate

  27. Cost of equipment method… • Assumes cost of equipment is dominant factor in plant cost • Estimate based on % of cost allocated to equipment relative to total cost of facility • Estimated cost is proportional to known cost • Must adjust for time & location

  28. Example 1… 5 min. • Assume that the equipment portion of a chemical processing plant constructed in 2006 cost $1.2M and the total plant cost was $2.5M. Therefore the equipment portion of the total cost was 46%. Estimate the cost of a new plant with the same processes if the equipment for the new plant has been determined to cost $2.7M. Assume time and location factors are negligible in this case. • Solution:

  29. Example 2… • Assume that the equipment portion of a chemical processing plant constructed in 2006 cost $1.2M and the total plant cost was $2.5M. Therefore the equipment portion of the total cost was 46%. Estimate the cost of a new plant with the same process if the equipment for the new plant has been determined to cost $2.4M. Assume time and location factors are negligible in this case. • Solution: C2/E2 = C1/E1 … solve for C2 C2 = [C1/E1] x E2 C2 = [$2.5M/$1.2M] x $2.4M C2 = [2.08] x $2.7 = $5,625,000M

  30. Volumetric unit measure… • Similar to comparison of $/sf unit measure • Used when facility has differing ceiling heights • Calculate the $/cf of known facility • Determine $/cf of known facility • Calculate total cf of new facility • Using known $/cf calculate cost of new facility • Adjust for time and location

  31. Example 2… 10 min. • Determine the cost of a new warehouse which will have 3 distinct storage areas with differing ceiling heights. Area 1 = 12’; Area 2=14’ and Area 3 = 10’. The floor area of all three areas is 4000sf. A similar warehouse was constructed earlier this year for $2.4M. In that building the floor area was 8000sf and the ceiling height was a uniform 14’. Assume time and location factors are negligible in this case. • Solution:

  32. Example 2… • Determine the cost of a new warehouse which will have 3 distinct storage areas with differing ceiling heights. Area 1 = 12’; Area 2=14’ and Area 3 = 10’. The floor area of each area is 4000sf. A similar warehouse was constructed earlier this year for $2.4M. In that building the floor area was 8000sf and the ceiling height was a uniform 14’. Assume time and location factors are negligible in this case. • Solution: Existing Building A = 8000sf x 14’ = 112,000 cf Cost/cf = $2,400,000 / 112,000 cf = $21.43/cf New Building A1 = 4000sf x 12’ = 48,000 cf A2 = 4000sf x 14’ = 56,000 cf A3 = 4000sf x 10’ = 40,000 cf Total = 144,000 cf Cost = 144,000 cf x $21.43/cf = $3,085,714

  33. Adjusting for time… • Prices “escalate” over time due to inflation • Must adjust estimated cost to “mid-point” of construction. • Determine period of construction • Determine start date of construction • Determine mid-point of construction = average cost • Determine “escalation rate” = given index • Multiply estimated cost by “escalation rate” • Note: escalation is compounded, ie. increase is cumulative over number of years of escalation

  34. Adjusting for location… • Prices differ by region of the country • Must consult a reliable index to determine “relative” cost of construction between regions • RS Means Cost Data or Engineering News Record are examples • Adjustment is a simple ratio multiplied times the estimated construction cost. • So… C2/C1 = L2/L1 X Estimated Construction Cost and C2 = C1 x L2/L1

  35. Example 3… 15 min. • You have been asked to prepare a conceptual cost estimate for the construction of a new university dormitory at Florida Tech to open in 2010. You have 2005 RS Means data that provides a cost of $30,000/bed for similar dormitories in the northeast of the US, and a construction period of 2 years. The Means data states that construction costs escalate at 10% per year in the southeast. If the dorm is to house 200 students at Florida Tech and the Means Cost Comparison Index between the northeast and southeast is 0.80, what is the estimated cost for the new library at Florida Tech at the mid-point of construction? • Estimated Florida Tech cost:______________________

  36. Solution to pop quiz… • $30,000 per bed x 200 beds = $6,000,000… in the northeast in 2005 • In the southeast in 2005… $6,000,000 x 0.80 = $4,800,000 • Construction start is… Occupancy – 2 years = 2010 – 2 = 2008 • Mid-point of construction is = 2008 + 1 = 2009 • Years of escalation… 2009 – 2005 = 4 yrs

  37. Solution to pop quiz… • Escalation: $4,800,000 x 1.10 = $5,280,000 (2005-06) $5,280,000 x 1.10 = $5,808,000 (2006-07) $5,808,000 x 1.10 = $6,388,800 (2007-08) $6,388,800 x 1.10 = $7,027,680 (2008-09)

  38. Solution… Therefore… the estimated cost of the new dormitory for Florida Tech at the mid-point of construction in 2009 is: $7,027,680

  39. Last Slide…

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