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Homework 5 Solution

Amount of work host is a 26 with a token it machine

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Unformatted text preview: bytes, ever. the highest one seen so far ever arrives, i It is not possible to send anywsizea sequen than = 51. lower than greater routers. In all casespacket ith ARP query+ number the 802.3 MAC-level the the mobile host, it broadcasts an is asking for rejected as being obsolete be sent. e router has more recent data. address of has to 5 approximately address. amount of work host is a 26. With a token it machine with promiscuous mode, since th mobile queueing48 24. The router the might µsec, 200,000 cells/sec can reading all frames holds not 13. Conceivably everydo go into that IP the same When the Each cell dropped data bytes or 384 bits. big very data rate is thendoubt Mbps.is normally done is onto the LAN, but this is net There is little 76.8 that packet, no matter how Theit is. inefficient. Instead, what processing 10 pack- This algorithm has a few problems, but they are manageable. First, if the sequence n that the home agent tricks token bucket schemeill sec to drain an 8 on 1000 ets Anaive bytes eachrapmuch more onfusion wprocessingTshaping. iofnewby is o use a the 27. 4Theof 100 answer saysisthat the6router into than4/3 reit is. the e solutAhost etoken tis put into32-bit sequence numb . n ATM network uses around, c work thinking traffic h packet megabit a at Mbps it takes for ign 1 mobile w h re responding to ARP one link token packet for one an which contains 48137 requests. When good pe second, that interval, for bytes. However,µsec. Each statewrong, because cell, IP it would take more years to What is ound, so this pos bucket. every 5 this answer is isthe routerrgetsduring packet destined bytes of data. wrap ar bucket the mobile host, it broadcastsanswer query obtained the 802.3 tokens arrive. Thecan any packets can be asking forby using MAC-level correct an ARP 25. It isthe maximumto send be ignored.greater thanWhen bytes, ever.the formula not possible sustainable data rate? 1024 the mobile host is not address M !the machine with we get S address.! 1) or 1.6 sec. of #). Substituting, that IP = 8 / (6 S = C/ ( 26. With a token every Second, if a router ever crashes, itEach lose holds 48 its sequence number. If it starts 5 µsec, 200,000 cells/sec can be sent. cell track of 28. Call the length of the maximum burst interval $t. In the will extreme case, the data bytes or 384 bits. the next packetis then 76.8 Mbps. as a duplicate. The net data rate will be rejected , bucket is full at the0start of the interval (1 Mbyte) and another 10$t Mbytes come in during the interval.at The output takes 4/3 sec to drain an burst con27. The naive answer says that 6 Mbps it during the transmission 8 megabit hird, if a is wrong, number we get 1 in...
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