Fragmentation at Network Layer - GeeksforGeeks

Fragment offset (13 bits) – use to identify the sequence of fragments in the frame. It generally indicates a number of data bytes preceding or ... 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26Oct,2021Prerequisite–IPv4DatagramFragmentationandDelays Fragmentationisdonebythenetworklayerwhenthemaximumsizeofdatagramisgreaterthanmaximumsizeofdatathatcanbeheldinaframei.e.,itsMaximumTransmissionUnit(MTU).Thenetworklayerdividesthedatagramreceivedfromthetransportlayerintofragmentssothatdataflowisnotdisrupted.  Sincethereare16bitsfortotallengthinIPheaderso,themaximumsizeofIPdatagram=216–1=65,535bytes.  Itisdonebythenetworklayeratthedestinationsideandisusuallydoneatrouters.Sourcesidedoesnotrequirefragmentationduetowise(good)segmentationbytransportlayeri.e.insteadofdoingsegmentationatthetransportlayerandfragmentationatthenetworklayer,thetransportlayerlooksatdatagramdatalimitandframedatalimitanddoessegmentationinsuchawaythatresultingdatacaneasilyfitinaframewithouttheneedoffragmentation.Receiveridentifiestheframewiththeidentification(16bits)fieldintheIPheader.Eachfragmentofaframehasthesameidentificationnumber.Receiveridentifiesthesequenceofframesusingthefragmentoffset(13bits)fieldintheIPheaderOverheadatthenetworklayerispresentduetotheextraheaderintroducedduetofragmentation.FieldsinIPheaderforfragmentation–Identification(16bits)–usetoidentifyfragmentsofthesameframe.Fragmentoffset(13bits)–usetoidentifythesequenceoffragmentsintheframe.Itgenerallyindicatesanumberofdatabytesprecedingoraheadofthefragment. Maximumfragmentoffsetpossible=(65535–20) =65515 {where65535isthemaximumsizeofdatagramand20istheminimumsizeofIPheader} So,weneedceil(log265515)=16bitsforafragmentoffsetbutthefragmentoffsetfieldhasonly13bits.So,torepresentefficientlyweneedtoscaledownthefragmentoffsetfieldby216/213=8whichactsasascalingfactor.Hence,allfragmentsexceptthelastfragmentshouldhavedatainmultiplesof8sothatfragmentoffset∈N.Morefragments(MF=1bit)–tellsifmorefragmentsareaheadofthisfragmenti.e.ifMF=1,morefragmentsareaheadofthisfragmentandifMF=0,itisthelastfragment.Don’tfragment(DF=1bit)–ifwedon’twantthepackettobefragmentedthenDFisseti.e.DF=1.ReassemblyofFragments–Ittakesplaceonlyatthedestinationandnotatrouterssincepacketstakeanindependentpath(datagrampacketswitching),soallmaynotmeetatarouterandhenceaneedoffragmentationmayariseagain.Thefragmentsmayarriveoutoforderalso.   Algorithm–DestinationshouldidentifythatdatagramisfragmentedfromMF,Fragmentoffsetfield.DestinationshouldidentifyallfragmentsbelongingtosamedatagramfromIdentificationfield.Identifythe1stfragment(offset=0).Identifysubsequentfragmentsusingheaderlength,fragmentoffset.RepeatuntilMF=0.Efficiency–Efficiency(e)=useful/total=(Datawithoutheader)/(Datawithheader)  Throughput=e*B{whereBisbottleneckbandwidth}  Example–AnIProuterwithaMaximumTransmissionUnit(MTU)of200byteshasreceivedanIPpacketofsize520byteswithanIPheaderoflength20bytes.ThevaluesoftherelevantfieldsintheIPheader. Explanation–SinceMTUis200bytesand20bytesisheadersizeso,themaximumlengthofdata=180bytesbutitcan’tberepresentedinfragmentoffsetsinceitisnotdivisibleby8so,themaximumlengthofdatafeasible=176bytes. Numberoffragments=(520/200)=3. Headerlength=5(sincescalingfactoris4therefore,20/4=5) Efficiency,e=(Datawithoutheader)/(Datawithheader)=500/560=89.2%    GATEPracticequestion– GATECS2013|Question65GATE-CS-2014-(Set-3)|Question38GATE-CS-2015(Set2)|Question65GATE-CS-2016(Set1)|Question63GATECS2018|Question37  MyPersonalNotes arrow_drop_upSave LikePreviousIPv4DatagramFragmentationandDelaysNext 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