1 / 23

IENG 217 Cost Estimating for Engineers

IENG 217 Cost Estimating for Engineers. Project Estimating. Hoover Dam. U.S. Reclamation Service opened debate 1926 Six State Colorado River Act, 1928 Plans released 1931, RFP Bureau completed its estimates 3 bids, 2 disqualified Winning bid by 6 companies with bid price at $48,890,955

niabi
Download Presentation

IENG 217 Cost Estimating for Engineers

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. IENG 217Cost Estimating for Engineers Project Estimating

  2. Hoover Dam • U.S. Reclamation Service opened debate 1926 • Six State Colorado River Act, 1928 • Plans released 1931, RFP • Bureau completed its estimates • 3 bids, 2 disqualified • Winning bid by 6 companies with bid price at $48,890,955 • Winning bid $24,000 above Bureau estimates

  3. Project Methods • Power Law and sizing CERs • Cost estimating relationships • Factor

  4. Power Law and Sizing • In general, costs do not rise in strict proportion to size, and it is this principle that is the basis for the CER

  5. Power Law and Sizing

  6. Power Law and Sizing • Ten years ago BHPL built a 100 MW coal generation plant for $100 million. BHPL is considering a 150 MW plant of the same general design. The value of m is 0.6. The price index 10 years ago was 180 and is now 194. A substation and distribution line, separate from the design, is $23 million. Estimate the cost for the project under consideration.

  7. Class Problem

  8. CER

  9. Factor Method Uses a ratio or percentage approach; useful for plant and industrial construction applications

  10. Factor Basic Item Cost Factor Method

  11. Adjustment for Inflation

  12. Example; Plant Project

  13. Example; Plant Project

  14. Example; Plant Project 4.1 1.7 1.1

  15. Example; Plant Project

  16. Other Project Methods • Expected Value • Range • Percentile • Simulation

  17. A A A A A MARR = 15% 1 2 3 4 5 10,000 Expected Value Suppose we have the following cash flow diagram. NPW = -10,000 + A(P/A, 15, 5)

  18. 2 , 000 p  1 / 6   A  3 , 000 p  2 / 3   4 , 000 p  1 / 6  Expected Value Now suppose that the annual return A is a random variable governed by the discrete distribution:

  19. 2 , 000 p  1 / 6   A  3 , 000 p  2 / 3   4 , 000 p  1 / 6  Expected Value For A = 2,000, we have NPW = -10,000 + 2,000(P/A, 15, 5) = -3,296

  20. 2 , 000 p  1 / 6   A  3 , 000 p  2 / 3   4 , 000 p  1 / 6  Expected Value For A = 3,000, we have NPW = -10,000 + 3,000(P/A, 15, 5) = 56

  21. 2 , 000 p  1 / 6   A  3 , 000 p  2 / 3   4 , 000 p  1 / 6  Expected Value For A = 4,000, we have NPW = -10,000 + 4,000(P/A, 15, 5) = 3,409

  22. Expected Value There is a one-for-one mapping for each value of A, a random variable, to each value of NPW, also a random variable. A 2,000 3,000 4,000 p(A) 1/6 2/3 1/6 NPW -3,296 56 3,409 p(NPW) 1/6 2/3 1/6

  23. Expected Value E[Return] = (1/6)-3,296 + (2/3)56 + (1/6)3,409 = $56 A 2,000 3,000 4,000 p(A) 1/6 2/3 1/6 NPW -3,296 56 3,409 p(NPW) 1/6 2/3 1/6

More Related