So the probability there exist undetected bit errors

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Unformatted text preview: y of s − t flow network G. We will find the pmf of Y. If none of the links fail, the network capacity is 30, because flow can be carried from s to t at rate 30 by sending flow over links 1, 2, 3, and 4 from left to right at the link capacities, and sending flow up the middle link at rate 10. Clearly 30 is the largest possible value of Y , and if any of the links fail then Y is less than 30. Thus, pY (30) = q1 q2 q3 q4 q5 . At the other extreme, Y = 0 if and only if every s − t path has at least one failed link. Thus, pY (0) is the same as the outage probability of s − t network E considered previously. The other two possible values of Y are 10 or 20. A little thought shows that Y = 20 requires that both links 2 and 3 work. If links 2 and 3 work, then Y = 20 if link 5 fails and links 1 and 4 do not fail, or if link 5 works and at least one of links 1 or 4 fails. Therefore, pY (20) = q2 q3 (p5 q1 q4 + q5 (p1 + p4 − p1 p4 )). Finally, pY (10) = 1 − pY (0) − pY (20) − pY (30). A more systematic way to calculate the pmf of Y is to enumerate all the...
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