finalexam-universityofwindsor期末考试-温莎大学

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1、FINALEXAMQ.1(15Marks)AnIPdatagramoflength(includingaheaderof20bytes)1800bytesneedstocrossanEthernetfollowedbyaWANtoreachitsdestination.TheMTUforEthernetis1500bytes.FortheWAN,theMTUisgiventobe576bytes.Howmanyfragmentsreachthedestination?Foreachfragment,givethevalue

2、oftheMORE-FRAGMENTS-BIT,theFRAGMENTOFFSETandtheTOTALLENGTHfields.Ans:Thedatagramof1800bytescannotbecarriedinoneunitbyEthernet.Hence2fragmentsarerequired.DataheaderTotalFrag11480201500bytesFrag230020320bytes-------1780bytesofdata.AttheentrytotheWAN,theRouterhastofu

3、rtherfragmentFRAG1.FRAG2goesthroughtheWANasitis.Sinceeverysub-fragmentmusthaveanIPheaderof20bytes,theWANcancarryamaximumdatasizeof556bytes.However556isnotdivisibleby8,asrequiredtheFragmentOffset.Hencewedecidetosend552bytesofdatainthefirstsub-fragmentofFRAG1.DataHe

4、aderTotalFrag1A552+20572Frag1B552+20572Frag1C376+20396-------1480bytesofdata.Henceatdestinationwouldreach4fragmentsnamelyFRAG1A,FRAG1B,FRAG1C,AndFRAG2.MFBFOTLFRAG1A10572FRAG1B169572FRAG1C1138396FRAG20185320Q.2(20Marks)a)DescribetheextraHopsProblem.b)DoesanExterior

5、RouteruseEGPtoexchangereachabilityinformationwithinteriorrouters?WhyistheIPsourcenetworkspecifiedinanEGProutingupdatemessage?c)WhatarecoreRouters?CantheCoreRoutingArchitecturebepartitionedintotwosetsofroutersthatkeeppartialinformationandusedefaultroutes?d)Asetofro

6、utersR1,R2andR3useVectorDistanceRouting.SomeoftheentriesintheroutingtableofR1areasfollows:DestinationDistanceRouteNet155R2Net163R3Net178R2R1doesnothaveanyentryforNet19.R2sendsanupdatemessagetoR1.Themessagecontainsthefollowingitems:DestinationDistanceNet154Net161Ne

7、t175Net192WhatchangeswillR1makeinitsroutingtable?Ans:2(a):ExtraHopsProblemTheviewpointofthecoresystemasaCentralRoutingMechanismtowhichnon-corerouterscansenddatagramsfordeliveryinalargerInternetwheretheNo.ofCOREROUTERS

8、calnetN2localnetBackBoneR2R1R3NoncoreRouterSupposeR3thinksofthecoreasadeliverysystemandchoosesR1todeliverdatagramstoallthosenetworks,whicharenotdirectly