Homework 5 Solution

1t sec path pick a packet thus the words an occasional

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Unformatted text preview: o the wrong destination and acceptedsource router can send the 2(n ! 1)T sec, before the as genuine. Put delivered the subnet,shortest path. Now remove all the arcs used in in other 8. secondroute using the the throughput is one packet every 2(n ! 1)T sec. path Pick a packet. Thus, the words, an occasional noise burst could change a perfectly legal packet for one 26 just found, and run the shortest path algorithm again. The second path will be PROBLEM SOLUTIONS FOR CHAPTER 5 destination into the failure of any host makes either 1, 2, or vice versa. It is 22. Each to survive a perfectly legal packet in the first path, and 3 hops. The proable packet emitted by the source line for another destination. bability sent it all theone hop is p.host takes the packet it makes two hops is conceivable, though, destination The probability ABCF ABEF, ABEG 7. only be follow makes that this heuristic may fail even, though ,two line-disjoint It will that when of the following routes: ABCDthat from the destination, 2 p (1 ! exist. To,probability that it a begins3propagating !shouldThe used. path hops is (1 used 24. paths . The acknowledgement The router.p,)AGHF solve it and AGEF makesnumber of hops p ) . Itbe mean full AGHD Now theAGEB, correctly, . max-flow algorithmback. is takes two length packet can expect to travel then the source sum can send the transits aof the subnet, 2(n ! 1)T sec,isbefore the weightedrouter of these three 9. probabilities,using2(11,shortestNotice Now for p = 0all the arcsis 3 hopsthe path Going via B or p ! 8. Pick a route gives the3p 6, 14, 18, 12,that remove the mean used in and for path. 8). second packet. Thus, the+ 3. throughput is one packet every 2(n ! 1)T sec. Going via D gives thehop. With 0 < p < just 1 the meanrun(19, shortest path9, 10). 1, multipleThe second pathmay be found, and is 1 15, 9, 3, algorithm again. transmissions will be p= Going The mean failure of any line in e either 1, 2, by hops. The the able packet gives (12, 11, source host 9). thepack found or realizing that part 22. Each tovia E emittedBuilding ttransmissions can beets is and3vicehversa. pro- is determining when to build the by the 8, 14, 5, tat needed. survive the number of he link smakes first path, easy. T e hardIt is conceivable, it a successfulhop heuristicuild them periodica) ,,which s, will bability thatofthough, that thisyissptoThemaythe way is though 2two hat iweat isegular intervals. Another possibility makes one transmission all fail even (1 ! p lly t line-disjoint possibilit i . b probability gives (11, 6, 0, hops r probability minimum for each destination e...
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