plugin-HW_08_Solutions

Plugin-HW_08_Solutio - HW#8Solutions AngandTang:4.3,4.8,4.13,4.33,4.38 4.3(a T = total power supply = N T T T = 100 200 400 = 700 Where T = 152 402

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HW #8 Solutions Ang and Tang: 4.3, 4.8, 4.13, 4.33, 4.38
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4.3 (a) T = total power supply = ) , ( T T N σ μ Where 100 200 400 700 T = + + = 2 2 2 15 40 100 108.74 T = + + = (b) P(Normal weather) = P(W) = 2/3 P(Extreme weather) = P(E) = 1/3 P(Power shortage) = P(S) = P(S W)P(W)+P(S E)P(E) = P(T<400)x2/3 + P(T<600)x1/3 = 400 700 2 600 700 1 [ ( )] [ ( )] 108.74 3 108.74 3 Φ × + Φ × = 2 1 [ ( 2.7589)] [ ( 0.9196)] 3 3 Φ − × + Φ − × = 0.0029x0.667 + 0.1789x0.333 = 0.06 (c) P(W S) = ( ) ( ) 0.0029 0.667 0.03 ( ) 0.0616 P S W P W P S × = = (d) P(all individual power source can meet respective demand) = P(N>0.15x400)P(F>0.3x400)P(H>0.55x400) = 60 100 120 200 220 400 [1 ( )][1 ( ( )] 15 40 100 −Φ = Φ (2.67)x Φ (2)x Φ (1.8) = 0.9962x0.9772x0.9641 = 0.9385 P(at least one source not able to supply respective allocation) = 0.06
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4.8 P(T<30)=P(T<30|N=0)P(N=0)+P(T<30|N=1)P(N=1)+P(T<30|N=2)P(N=2) 3 . 0 ) 3 2 25 40 30 ( 5 . 0 ) 3 25 35 30 ( 2 . 0 ) 5 30 30 ( 2 2 × × + Φ + × + Φ + × Φ = 3 . 0 ) 525 . 1 ( 5 . 0 ) 857 . 0 ( 2 . 0 5 . 0 × Φ + × Φ + × = =0.217
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4.13 (a) P(X 2)=1-P(X=0)-P(X=1) 0 5 1 4 5 4 5 5 1 0.6 0.4 0.6 0.4 0 1 1 0.4 5 0.6 0.4 0.913     = −         = − − × × = (b) N H , N B are the number of highway and building jobs won P(N H =1 N B =0)=P(N H =1)P(N B =0) = 2 0 2 1 4 . 0 6 . 0 0
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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_08_Solutio - HW#8Solutions AngandTang:4.3,4.8,4.13,4.33,4.38 4.3(a T = total power supply = N T T T = 100 200 400 = 700 Where T = 152 402

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