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...
View Full Document
- Spring '08
- The Land