ECE313.Lecture20

ECE313.Lecture20 - ECE 313 Probability with Engineering...

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System Reliability II Professor Dilip V. Sarwate Department of Electrical and Computer Engineering © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign. All Rights Reserved ECE 313 Probability with Engineering Applications
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ECE 313 - Lecture 20 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 2 of 36 Communication Networks Communication networks consist of links over which messages can pass between nodes (or terminals or hosts) Messages to distant nodes have to pass over multiple links and multiple nodes The intermediate links can fail The intermediate nodes can fail What is the probability that two nodes can communicate?
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ECE 313 - Lecture 20 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 3 of 36 Links only; we don’t do nodes A node that fails makes all the links connected to it inoperable All paths between other nodes that pass through the failed node are not available In the simple analyses presented in this course, we assume that nodes do not fail The only failures that we are concerned with are the link failures
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ECE 313 - Lecture 20 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 4 of 36 Communication over parallel links The diagram illustrates two parallel links connecting a transmitter and a receiver R T The transmitter can send messages as long as at least one link is viable Both links must fail in order for communication to be impossible
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ECE 313 - Lecture 20 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 5 of 36 Some notation Let V i denote the event that link #i is viable that is, link #i has not failed The complementary event is F i V (F = V c ) denotes the event that there is (is not) a communication path from T to R R T #1 #2
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ECE 313 - Lecture 20 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 6 of 36 As easy as rolling off a log… V = V 1 V 2 F = F 1 F 2 P(F) = P(F 1 F 2 ) = P(F 1 )P(F 2 ) = p 1 p 2 if the events F 1 and F 2 can be assumed to be independent Similarly, P(V) = P(V 1 ) + P(V 2 ) – P(V 1 V 2 ) = P(V 1 )+P(V 2 ) – P(V 1 )P(V 2 ) R T #1 #2
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ECE 313 - Lecture 20 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 7 of 36 Let’s be sure we understand… V = V 1 V 2 F = F 1 F 2 P(F) = P(F 1 F 2 ) = P(F 1 )P(F 2 ) = p 1 p 2 if the events F 1 and F 2 can be assumed to be independent P(V) = P(V 1 ) + P(V 2 ) – P(V 1 V 2 ) = P(V 1 ) + P(V 2 ) – P(V 1 )P(V 2 ) = q 1 + q 2 – q 1 q 2
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ECE 313 - Lecture 20 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 8 of 36 Did matters improve?
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ECE313.Lecture20 - ECE 313 Probability with Engineering...

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