1. CIVL 181 tutorial 10 Value of information, expected utility

3. E[L|CN]?

4. Adopted from Question 2.8 In order to protect an $8M dam from overtopping, two spillway systems are proposed. Spillway A ~ 3.5 M, Spillway B ~ 3 M If fail, the spillway and the dam will be destroyed. P (F A ) = 0.15, P (F B ) = 0.3 A test, cost 0.3M, give the following result: If “A is preferred”, P (F A ) = 0.2, P (F B ) = 0.6 If “B is preferred”, P (F A ) = 0.5, P (F B ) = 0.1 The engineer thinks 70% chance will be “B is preferred”

5. Question 2.8 a)Draw a decision tree, include all relevant alternatives Test No test A preffered (0.3) B preffered (0.7) System A System B System A System B System A System B Probability of failure Loss 0.2 0.6 0.5 0.1 0.15 0.3 11.5 11 11.5 11 11.5 11

6. Question 2.8 b) Determine whether the test should be conducted based on EMV No test: E(A) = 0.15 x 11.5 + 0.85 x 3.5 = 4.7, E(B) = 0.3 x 11 + 0.7 x 3 = 5.4A* Test and if A is preferred: E(A) = 5.1, E(B) = 7.8 A* Test and if B is preferred: E(A) = 7.5, E(B) = 3.8B*

7. Question 2.8 Test No test A preferred (0.3) B preferred (0.7) System A* System B System A System B* System A* System B Probability of failure Loss 0.2 0.6 0.5 0.1 0.15 0.3 11.5 11 11.5 11 11.5 11

8. Question 2.8 E (no test) = E (A*) =4.7 E (test) = E(A|A*)P(A*) + E(B|B*)P(B*) = (5.1)(0.3) + (3.8)(0.7) [no test cost] = 4.19 E (test) with test cost = 4.19 + 0.3 = 4.49 So, test c) What is the value of the test? = 4.7 – 4.19 = 0.51

9. Question 2.8 d) Suppose the designer is Risk aversive, such that u(x) = x 1.5, should the test be conducted? Test No test A prefered (0.3) B prefered (0.7) System A* System B System A System B* System A* System B Probability of failure Loss 0.2 0.6 0.5 0.1 0.15 0.3 11.8 11.3 11.8 11.3 11.5 11 utility 40.53 37.98 40.53 37.98 39 36.48

10. Question 2.8 No test: U(A) = 0.15 x 39 + 0.85 x 3.5 2 = 16.2625, U(B) = 0.3 x 36.48 + 0.7 x 3 2 = 17.244A* Test and if A is preferred: U(A) =, U(B) = A* Test and if B is preferred: U(A) =, U(B) = B* …