1. CVE 4070 Construction Engineering Estimating-2 Prof. Ralph V. Locurcio, PE

2. Example 1…  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 48%. 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:

3. 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 48%. 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: C 2 /E 2 = C 1 /E 1 … solve for C 2 C 2 = [C 1 /E 1 ] x E 2 C 2 = [$2.5m/$1.2m] x $2.4 m C 2 = [2.08] x $2.7 = $5,625,000 m  Solution: TPC = ET/(1-PT) = $2.7/(1-.48) where PT = $1.2m/$2.5m = $5,625,000

4. 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

5. 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:

6. 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

7. 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

8. 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… C 2 /C 1 = L 2 /L 1 X Estimated Construction Cost and C 2 = C 1 x L 2 /L 1

9. Example 3…  You have been asked to prepare a conceptual cost estimate for the construction of a new pedestrian bridge at Florida Tech to open in 2018.  You have 2005 cost data that provides a cost of $175/SF for similar bridges in Florida, and a construction period of 2 years.  Your cost data states that construction costs escalate at 2% per year in the southeast. According to the Cost Comparison Index, the construction cost in the Melbourne is 0.88 compared to the average for Florida.  The new bridge is to be 175’ long and 12’ wide. What is the estimated cost for the new bridge at Florida Tech at the mid-point of construction?  Estimated Florida Tech cost:______________________  Show your work:

10. Handout HW #4 Estimate

12. Solution to pop quiz… 1.$30,000 per bed x 200 beds = $6,000,000… in the northeast in 2007 2.In the southeast in 2007… $6,000,000 x 0.80 = $4,800,000 3.Construction start is… Occupancy – 2 years = 2012 – 2 = 2010 4.Mid-point of construction is = 2010 + 1 = 2011 5.Years of escalation… 2011 – 2007 = 4 yrs Cost Index

13. Solution to pop quiz… 6.Escalation: $4,800,000 x 1.10 = $5,280,000 (2007-08) $5,280,000 x 1.10 = $5,808,000 (2008-09) $5,808,000 x 1.10 = $6,388,800 (2009-10) $6,388,800 x 1.10 = $7,027,680 (2010-11)

14. Solution… Therefore… the estimated cost of the new dormitory for Florida Tech at the mid-point of construction in 2011 is: $7,027,680

15. Last Slide

16. Happy Thanxgiving!!