plugin-HW_03_Solutions

# plugin-HW_03_Solutions - HW#3 Solutions(Fall 2011 Assigned...

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1 HW #3 Solutions (Fall 2011) Assigned problems Ang & Tang: 2.41, 2.42, 2.58, 3.8, 3.9, 3.12 2.41 Let L, N, H denote the event of low, normal and high demand respectively; also O and G denote oil and gas supply is low respectively. (a) Given normal energy demand, probability of energy crisis = P(E N) = P(O G) = P(O)+P(G)-P(OG) = P(O)+P(G)-P(G O)P(O) = 0.2 + 0.4 – 0.5x0.2 = 0.5 (b) P(E) = P(E L) P(L) + P(E N) P(N) + P(E H) P(H) = P(OG)x0.3 + 0.5x0.6 +1x0.1 = P(G O)P(O)x0.3 + 0.3 + 0.1 = 0.43

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2.42 Given: P(H=1) = 0.2, P(H=2) = 0.05, P(H=0) = 0.75 Where H = number of hurricanes in a year P(J=H=1) = P(D H=1) = 0.99 P(J=H=2) = P(D H=2) = 0.80 Where J and D denote survival of jacket and deck substructure respectively Assume J and D are statistically independent (a) For the case of one hurricane, i.e. H=1 P(damage) = 1-P(JD H=1) = 1- P(J H=1) P(D H=1) = 1 - 0.99x0.99 = 0.0199 (b) For next year where the number of hurricanes is not known. P(no damage) = P(JD H=0)P(H=0)+ P(JD H=1)P(H=1)+ P(JD H=2)P(H=2) = 1x0.75 + 0.99x0.99x0.2 + 0.8x0.8x0.05 = 0.75 + 0.196 + 0.032 = 0.978 (c) P(H=0 JD) = (0 ) ( 0 ) 10 . 7 5 0.767 ( ) 0.978 PJDH PH PJH == ×
3 2.58 Given: P(L) = 15/(15+4+1) = 15/20 = 0.75 P(M) = 4/20 = 0.20 P(H) = 1/20 = 0.05 P(P) = 0.2 where P denotes poorly constructed

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## This note was uploaded on 01/02/2012 for the course CIVIL ENG 408 at USC.

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plugin-HW_03_Solutions - HW#3 Solutions(Fall 2011 Assigned...

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