Two s t ow networks are shown in figure 213 we assume

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Unformatted text preview: at is, for network C, P (F ) = p1 p3 + p2 p4 − p1 p2 p3 p4 . If pi = 0.001 for all i, P (F ) ≈ 0.000002, or about half the outage probability for network C. Finally, consider network E. One way to approach the computation of P (F ) is to use the law of total probability, and consider two cases: either link 5 fails or it doesn’t fail. So P (F ) = c c P (F |F5 )P (F5 ) + F (F |F5 )P (F5 ). Now P (F |F5 ) is the network outage probability if link 5 fails. If link 5 fails we can erase it from the picture, and the remaining network is identical to network C . So P (F |F5 ) for Network E is equal to P (F ) for network C. If link 5 does not fail, the network c becomes equivalent to network D. So P (F |F5 ) for Network E is equal to P (F ) for network C. Combining these yields: P (outage in network E ) = p5 P (outage in network C ) + (1 − p5 )P (outage in network D). Another way to calculate P (F ) for network E is by closer comparison to network D. Outage occurs in network E on the event (F1 F3 )...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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