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 Winning bid $24,000 above Bureau estimates

Project Methods Power Law and sizing CERs Cost estimating relationships Factor

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

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.

Class Problem

CER

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

Factor Method Basic Item Cost Factor

Adjustment for Inflation

Example; Plant Project

Example; Plant Project

Other Project Methods Expected Value Range Percentile Simulation

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

Expected Value Now suppose that the annual return A is a random variable governed by the discrete distribution: A p p p          ,/,/,/

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

Expected Value A p p p          ,/,/,/ For A = 3,000, we have NPW = -10, ,000(P/A, 15, 5) = 56

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

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, ,409 p( NPW ) 1/6 2/3 1/6

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, ,409 p( NPW ) 1/6 2/3 1/6