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Decision Analysis

Decision Analysis. Ultimate objective of all engineering analysis Uncertainty always exist, hence satisfactory performance not guaranteed More conservative design reduces risk Same design SF for all? Component vs. System Risk Proper tradeoff between risk and investment.

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Decision Analysis

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  1. Decision Analysis • Ultimate objective of all engineering analysis • Uncertainty always exist, hence satisfactory performance not guaranteed • More conservative design reduces risk • Same design SF for all? • Component vs. System Risk • Proper tradeoff between risk and investment

  2. Solution by Calculus • Set up objective function where ’s are decision variables • From solution to yields optimal values of decision variables

  3. Contractor – submit a bid(Example 2.1) Bid Ratio, R =

  4. To determine optimal bid and optimal R • Establish the objective function

  5. Case 2: Include Idling Cost

  6. h? Cofferdam for construction of Bridge Pier (2 yrs)(Example 2.2)

  7. h? Information • Floods occur according to Poisson process with mean rate of 1.5/yr • Elevation of each flood – exponential with mean 5 feet • Each overtopping  loss due to pumping + delay = $25,000 • Construction cost,

  8. E (loss of flood) Expected damage cost, C

  9. Total Cost  = 8.05 ft

  10. Cost as Functions of cofferdam elevation above normal water level 8.05

  11. Limitation of this Approach • Objective function may not be continuous function of decision variables • Alternatives may be discrete e.g. dam for flood control (height, location, other schemes) • Consequences may be more than monetary costs • Alternative may include acquiring new information before final decision • Should we acquire or not?

  12. Pump System (100/120 gal/min) Q = 95 or 120 gal/min? Bentonite Seal Seepage under the Embankment (Example 2.4) Cooling Lake Embankment

  13. 95 120 95 120 Decision tree for seepage problem Q1(0.9) Pump System A (100) Add Pump System C Q2(0.1) Pump System B (120) Q1(0.9) Q2(0.1) Seal

  14. Alternatives Uncertainties Consequences Decision Tree Model Decision Node Chance Node

  15. Example 2.17 Click to enlarge

  16. Example 2.17 Click to enlarge

  17. Decision Criteria • Pessimistic  Minimize max loss  Install • Optimistic  Maximize max gain  Not Install

  18. 3. Maximum EMV (Expected Monetary Value) E(I) = 0.1x(-2000)+0.9x(-2000) = -2000 E(II) = 0.1x(-10000)+0.9x(0) =-1000

  19. Ex. 2.9 Decision tree for construction project

  20. Spillway Decisions Alternatives Capital Cost Annual OMR Cost • No Change 0 0 • Lengthening spillway 1.04M 0 • Plus lowering crest, installing 1.30M 0 flashboard • Plus considerable crest lowering, 3.90M 0 installing radial gates • 50years service; Discount rate 6%

  21. Spillway Decisions Summary of Annual Costs (in Dollars) Total Annual Cost =Capital Cost x crf (i,n) +Annual DMR Cost +Expected Risk Cost (annual)

  22. Discount factors Given A to find P: Given P to find A: Where i = int. rate per period n= no. of periods

  23. E2.11 Spillway Design

  24. E2.11 Spillway Design Risk Cost

  25. EH 0.7 A (small) 0 -100 -50 -20 EL 0.3 EH 0.7 B (large) EL 0.3 Hence, A is the optimal alternative. Ex. 2.6 Prior Analysis E(A) = 0.7 x 0 + 0.3 x (-100) = -30 E(B) = 0.7 x (-50) + 0.3 x (-20) = -41

  26. Lab. Model test on Efficiency (Cost $10,000) will indicate: HR (high rating) MR (medium rating) LR (Low rating) HR 0.8 HR 0.1 If EH MR 0.15 If EL MR 0.2 LR 0.05 LR 0.7 e.g. If the process is actually high efficiency (EH), then the probability that the test will indicate HR is 0.8.

  27. EH 0.95 A (small) -10 -110 -60 -30 EL 0.05 EH 0.95 B (large) EL 0.05 Suppose the test indicate HR Test HR

  28. > 30 good news Suppose the test indicate HR • Similarly, P(EL|HR) = 0.05 • E(A|HR) =0.95x(-10)+0.05x(-110) = -15 • E(B|HR) =0.95x(-10)+0.05x(-110) = -58.5

  29. EH 0.637 A (small) -10 -110 -60 -30 EL 0.363 EH 0.637 B (large) EL 0.363 < -30 Suppose the test indicate MR • E(A|MR) = -46.3 • E(B|MR) = -49.1 Test MR

  30. < -30 Suppose the test indicate LR • E(A|LR) = -95.7 • E(B|LR) = -34.3 Only if the test showed HR,  saved money; otherwise, more money with test

  31. Should test be performed? Preposterior analysis E(Test) =0.59x(-15)+ 0.165 x(-46.3)+0.245 x(-34.3) = -24.86 Better than -30 (without test)

  32. Procedure for Preposterior Analysis • Determine updated probabilities using Bayes Theorem; • Sub-tree analysis –Identify optimal alternative and maximum utility; • Determine the best alternative in the next decision node (to the left); • If Experimental alternative is optimal, wait for experimental outcome and select corresponding optimal alternative.

  33. Subtree B Subtree C Procedure for Preposterior Analysis B C

  34. EMV of test alternative excluding test cost EMV of optimal alternative without Test Value of Information (VI) • VI = E(T) – E( ) VI = (-24.86 + 10) – (- 30) = 15.14 (max. paid for that specific Test)

  35. Suppose someone comes up with a better test, say cost 25,000, but doesn’t know that exact reliability, should the test be performed?

  36. VPI = E(PT) - E( ) P(EH0) = P(EH0|EH) P(EH) + P(EH0|EL) P(EL) = 1 x 0.7 + 0 x 0.3 = 0.7 E(PT) = 0.7 x 0 + 0.3 x (-20) = -6 VPI = -6 – (-30) = 24 Max. that should be paid for any information

  37. Sensitivity Analysis • If the probability estimates are off by +10%, would the alternative previously chosen be still optimal? Method 1: Repeat analysis with several values of p Method 2: Determine value of probability p that decision is switched

  38. EH p A 0 -100 -50 -20 EL 1-p EH p B EL 1-p E(A) = p x 0 + (1-p) x (-100) E(B) = p x (-50) + (1-p) x (-20)

  39. VI VPI Sensitivity of Decision to Probability p<0.62 • E(B) >E(A) P>0.62 • E(B) <E(A) E(PT) =px0+(1-p)(-20) = -20(1-p) E(T)

  40. Levee Elevation Decision • Annual max. Flood Level: median 10, c.o.v. 20% • Cost of construction:a1: $ 2 million a2: $ 2.5 million • Service Life: 20 years • Average annual damage cost due to inadequate protection: $ 2 million

  41. E(C)=10.594 2x10.594 pwf (20yrs, 7%) 2.731 2.641 Levee Elevation Decision • Annual max. Flood Level: median 10, c.o.v. 20% H=10’ H=14’ H=16’

  42. Value of Perfect Information Max. Amount to be paid for verifying type of distribution of annual flood level • E(CPI) = 0.5x2.699 +0.5x2.482 = 2.59 • VPI = 2.614–2.59 = $ 0.024 M

  43. Consider a Game 0 10 ¢ 5 ¢ 0 $1 $0.5 0 $100 $ 50 0 $100M $ 50M 0.5 • E(A) = 0.5 x 0 + 0.5 x 10¢ = 5 ¢ • E(B) = 1.0 x 5 ¢ = 5 ¢ A 0.5 1.0 B EMV criteria may not be applicable We need something else!

  44. EUV criteria • Expected Utility Value • Definition: EUV is the true value to a decision maker with which he/she can make a proper decision based on the relative utility value.

  45. Risk aversive Utility function of monetary value u(d) Risk Indifferent dollars

  46. Maximum Expected Utility Criterion (EUV) If all consequences expressed in monetary terms:

  47. Example u(d) dollars • E( )= 0.1u (-2000)+0.9 u(-2000) = = -1.49 • E( )= 0.1u (-10000)+0.9 u(0) =0.1x( ) +0.9x( ) = -1.64 -2000 -2000 -10000 0 0 -1

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