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,
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,
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
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
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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- Fall '13
- Networking, Shortest path problem, bability thatofthough