1. IENG 217 Cost 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. 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

9. CER

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

11. Factor Method Basic Item Cost Factor

12. Adjustment for Inflation

13. Example; Plant Project

15. 4.1 1.7 1.1

16. Example; Plant Project

17. Other Project Methods Expected Value Range Percentile Simulation

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

19. Expected Value Now suppose that the annual return A is a random variable governed by the discrete distribution: A p p p          200016 3 23 4 16,/,/,/

20. Expected Value A p p p          200016 3 23 4 16,/,/,/ For A = 2,000, we have NPW = -10,000 + 2,000(P/A, 15, 5) = -3,296

21. Expected Value A p p p          200016 3 23 4 16,/,/,/ For A = 3,000, we have NPW = -10,000 + 3,000(P/A, 15, 5) = 56

22. Expected Value A p p p          200016 3 23 4 16,/,/,/ For A = 4,000, we have NPW = -10,000 + 4,000(P/A, 15, 5) = 3,409

23. 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

24. 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