Computer-Networking-A-Top-Down-Approach-6th-solutions-计算机网络自顶向下-课后习题答案

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ComputerNetworking:ATop-DownApproach,

6thEdition

SolutionstoReviewQuestionsandProblems

VersionDate:May2012

Thisdocumentcontainsthesolutionstoreviewquestionsandproblemsforthe5theditionofComputerNetworking:ATop-DownApproachbyJimKuroseandKeithRoss.ThesesolutionsarebeingmadeavailabletoinstructorsONLY.PleasedoNOTcopyordistributethisdocumenttoothers<evenotherinstructors>.Pleasedonotpostanysolutionsonapublicly-availableWebsite.We’llbehappytoprovideacopy<up-to-date>ofthissolutionmanualourselvestoanyonewhoasks.

Acknowledgments:Overtheyears,severalstudentsandcolleagueshavehelpeduspreparethissolutionsmanual.SpecialthanksgoestoHongGangZhang,RakeshKumar,PrithulaDhungel,andVijayAnnapureddy.Alsothankstoallthereaderswhohavemadesuggestionsandcorrectederrors.

Allmaterial©copyright1996-2012byJ.F.KuroseandK.W.Ross.Allrightsreserved

Chapter1ReviewQuestions

Thereisnodifference.Throughoutthistext,thewords"host"and"endsystem"areusedinterchangeably.EndsystemsincludePCs,workstations,Webservers,mailservers,PDAs,Internet-connectedgameconsoles,etc.

FromWikipedia:Diplomaticprotocoliscommonlydescribedasasetofinternationalcourtesyrules.Thesewell-establishedandtime-honoredruleshavemadeiteasierfornationsandpeopletoliveandworktogether.Partofprotocolhasalwaysbeentheacknowledgmentofthehierarchicalstandingofallpresent.Protocolrulesarebasedontheprinciplesofcivility.

Standardsareimportantforprotocolssothatpeoplecancreatenetworkingsystemsandproductsthatinteroperate.

1.Dial-upmodemovertelephoneline:home;2.DSLovertelephoneline:homeorsmalloffice;3.CabletoHFC:home;4.100MbpsswitchedEthernet:enterprise;5.Wifi<802.11>:homeandenterprise:6.3Gand4G:wide-areawireless.

HFCbandwidthissharedamongtheusers.Onthedownstreamchannel,allpacketsemanatefromasinglesource,namely,theheadend.Thus,therearenocollisionsinthedownstreamchannel.

InmostAmericancities,thecurrentpossibilitiesinclude:dial-up;DSL;cablemodem;fiber-to-the-home.

7.EthernetLANshavetransmissionratesof10Mbps,100Mbps,1Gbpsand10Gbps.

8.Today,Ethernetmostcommonlyrunsovertwisted-paircopperwire.Italsocanrunoverfibersopticlinks.

9.Dialupmodems:upto56Kbps,bandwidthisdedicated;ADSL:upto24Mbpsdownstreamand2.5Mbpsupstream,bandwidthisdedicated;HFC,ratesupto42.8Mbpsandupstreamratesofupto30.7Mbps,bandwidthisshared.FTTH:2-10Mbpsupload;10-20Mbpsdownload;bandwidthisnotshared.

10.TherearetwopopularwirelessInternetaccesstechnologiestoday:

Wifi<802.11>InawirelessLAN,wirelessuserstransmit/receivepacketsto/fromanbasestation<i.e.,wirelessaccesspoint>withinaradiusoffewtensofmeters.ThebasestationistypicallyconnectedtothewiredInternetandthusservestoconnectwirelessuserstothewirednetwork.

3Gand4Gwide-areawirelessaccessnetworks. Inthesesystems,packetsaretransmittedoverthesamewirelessinfrastructureusedforcellulartelephony,withthebasestationthusbeingmanagedbyatelecommunicationsprovider.Thisprovideswirelessaccesstouserswithina radiusoftensofkilometersofthebasestation.

11.Attimet0thesendinghostbeginstotransmit.Attimet1=L/R1,thesendinghostcompletestransmissionandtheentirepacketisreceivedattherouter<nopropagationdelay>.Becausetherouterhastheentirepacketattimet1,itcanbegintotransmitthepackettothereceivinghostattimet1.Attimet2=t1+L/R2,theroutercompletestransmissionandtheentirepacketisreceivedatthereceivinghost<again,nopropagationdelay>.Thus,theend-to-enddelayisL/R1+L/R2.

12.Acircuit-switchednetworkcanguaranteeacertainamountofend-to-endbandwidthforthedurationofacall.Mostpacket-switchednetworkstoday<includingtheInternet>cannotmakeanyend-to-endguaranteesforbandwidth.FDMrequiressophisticatedanaloghardwaretoshiftsignalintoappropriatefrequencybands.

13.a>2userscanbesupportedbecauseeachuserrequireshalfofthelinkbandwidth.

b>Sinceeachuserrequires1Mbpswhentransmitting,iftwoorfeweruserstransmit simultaneously,amaximumof2Mbpswillberequired.Sincetheavailable bandwidthofthesharedlinkis2Mbps,therewillbenoqueuingdelaybeforethe link.Whereas,ifthreeusers transmitsimultaneously,thebandwidthrequired willbe3Mbpswhichismorethantheavailablebandwidthofthesharedlink.In thiscase,therewillbequeuingdelaybeforethelink.

c>Probabilitythatagivenuseristransmitting=0.2

d>Probabilitythatallthreeusersaretransmittingsimultaneously=

=<0.2>3=0.008.Sincethequeuegrowswhenalltheusersaretransmitting,the fractionoftimeduringwhichthequeuegrows<whichisequaltotheprobability thatallthreeusersaretransmittingsimultaneously>is0.008.

14.IfthetwoISPsdonotpeerwitheachother,thenwhentheysendtraffictoeachothertheyhavetosendthetrafficthroughaproviderISP<intermediary>,towhichtheyhavetopayforcarryingthetraffic.Bypeeringwitheachotherdirectly,thetwoISPscanreducetheirpaymentstotheirproviderISPs.AnInternetExchangePoints<IXP><typicallyinastandalonebuildingwithitsownswitches>isameetingpointwheremultipleISPscanconnectand/orpeertogether.AnISPearnsitsmoneybychargingeachofthetheISPsthatconnecttotheIXParelativelysmallfee,whichmaydependontheamountoftrafficsenttoorreceivedfromtheIXP.

15.Google'sprivatenetworkconnectstogetherallitsdatacenters,bigandsmall.TrafficbetweentheGoogledatacenterspassesoveritsprivatenetworkratherthanoverthepublicInternet.Manyofthesedatacentersarelocatedin,orcloseto,lowertierISPs.Therefore,whenGoogledeliverscontenttoauser,itoftencanbypasshighertierISPs.Whatmotivatescontentproviderstocreatethesenetworks?First,thecontentproviderhasmorecontrolovertheuserexperience,sinceithastousefewintermediaryISPs.Second,itcansavemoneybysendinglesstrafficintoprovidernetworks.Third,ifISPsdecidetochargemoremoneytohighlyprofitablecontentproviders<incountrieswherenetneutralitydoesn'tapply>,thecontentproviderscanavoidtheseextrapayments.

16.Thedelaycomponentsareprocessingdelays,transmissiondelays,propagationdelays,andqueuingdelays.Allofthesedelaysarefixed,exceptforthequeuingdelays,whicharevariable.

17.a>1000km,1Mbps,100bytes

b>100km,1Mbps,100bytes

18.10msec;d/s;no;no

19.a>500kbps

b>64seconds

c>100kbps;320seconds

20.EndsystemAbreaksthelargefileintochunks.Itaddsheadertoeachchunk,therebygeneratingmultiplepacketsfromthefile.TheheaderineachpacketincludestheIPaddressofthedestination<endsystemB>.ThepacketswitchusesthedestinationIPaddressinthepackettodeterminetheoutgoinglink.Askingwhichroadtotakeisanalogoustoapacketaskingwhichoutgoinglinkitshouldbeforwardedon,giventhepacket’sdestinationaddress.

21.Themaximumemissionrateis500packets/secandthemaximumtransmissionrateis

350packets/sec.Thecorrespondingtrafficintensityis500/350=1.43>1.Losswilleventuallyoccurforeachexperiment;butthetimewhenlossfirstoccurswillbedifferentfromoneexperimenttothenextduetotherandomnessintheemissionprocess.

22.Fivegenerictasksareerrorcontrol,flowcontrol,segmentationandreassembly,multiplexing,andconnectionsetup.Yes,thesetaskscanbeduplicatedatdifferentlayers.Forexample,errorcontrolisoftenprovidedatmorethanonelayer.

23.ThefivelayersintheInternetprotocolstackare–fromtoptobottom–theapplicationlayer,thetransportlayer,thenetworklayer,thelinklayer,andthephysicallayer.TheprincipalresponsibilitiesareoutlinedinSection1.5.1.

24.Application-layermessage:datawhichanapplicationwantstosendandpassedontothetransportlayer;transport-layersegment:generatedbythetransportlayerandencapsulatesapplication-layermessagewithtransportlayerheader;network-layerdatagram:encapsulatestransport-layersegmentwithanetwork-layerheader;link-layerframe:encapsulatesnetwork-layerdatagramwithalink-layerheader.

25.Routersprocessnetwork,linkandphysicallayers<layers1through3>.<Thisisalittlebitofawhitelie,asmodernrouterssometimesactasfirewallsorcachingcomponents,andprocessTransportlayeraswell.>Linklayerswitchesprocesslinkandphysicallayers<layers1through2>.Hostsprocessallfivelayers.

26.a>Virus

Requiressomeformofhumaninteractiontospread.Classicexample:E-mailviruses.

b>Worms

Nouserreplicationneeded.WormininfectedhostscansIPaddressesandport numbers,lookingforvulnerableprocessestoinfect.

27.Creationofabotnetrequiresanattackertofindvulnerabilityinsomeapplicationorsystem<e.g.exploitingthebufferoverflowvulnerabilitythatmightexistinanapplication>.Afterfindingthevulnerability,theattackerneedstoscanforhoststhatarevulnerable.Thetargetisbasicallytocompromiseaseriesofsystemsbyexploitingthatparticularvulnerability.Anysystemthatispartofthebotnetcanautomaticallyscanitsenvironmentandpropagatebyexploitingthevulnerability.Animportantpropertyofsuchbotnetsisthattheoriginatorofthebotnetcanremotelycontrolandissuecommandstoallthenodesinthebotnet.Hence,itbecomespossiblefortheattackertoissueacommandtoallthenodes,thattargetasinglenode<forexample,allnodesinthebotnetmightbecommandedbytheattackertosendaTCPSYNmessagetothetarget,whichmightresultinaTCPSYNfloodattackatthetarget>.

28.TrudycanpretendtobeBobtoAlice<andvice-versa>andpartiallyorcompletelymodifythemessage<s>beingsentfromBobtoAlice.Forexample,shecaneasilychangethephrase"Alice,Ioweyou$1000"to"Alice,Ioweyou$10,000".Furthermore,TrudycanevendropthepacketsthatarebeingsentbyBobtoAlice<andvise-versa>,evenifthepacketsfromBobtoAliceareencrypted.

Chapter1Problems

Chapter1-Problem1

Thereisnosinglerightanswertothisquestion.Manyprotocolswoulddothetrick.Here'sasimpleanswerbelow:

MessagesfromATMmachinetoServer

Msgname purpose

HELO<userid> LetserverknowthatthereisacardintheATMmachine

ATMcardtransmitsuserIDtoServer

PASSWD<passwd> UserentersPIN,whichissenttoserver

BALANCE Userrequestsbalance

WITHDRAWL<amount> Useraskstowithdrawmoney

BYE useralldone

MessagesfromServertoATMmachine<display>

Msgname purpose

PASSWD AskuserforPIN<password>

OK lastrequestedoperation<PASSWD,WITHDRAWL>OK

ERR lastrequestedoperation<PASSWD,WITHDRAWL>inERROR

AMOUNT<amt> sentinresponsetoBALANCErequest

BYE userdone,displaywelcomescreenatATM

Correctoperation:

clientserver

HELO<userid> > <checkifvaliduserid>

< PASSWD

PASSWD<passwd> > <checkpassword>

< OK<passwordisOK>

BALANCE >

< AMOUNT<amt>

WITHDRAWL<amt> > checkifenough$tocover withdrawl

< OK

ATMdispenses$

BYE >

< BYE

Insituationwhenthere'snotenoughmoney:

HELO<userid> > <checkifvaliduserid>

< PASSWD

PASSWD<passwd> > <checkpassword>

< OK<passwordisOK>

BALANCE >

< AMOUNT<amt>

WITHDRAWL<amt> > checkifenough$tocoverwithdrawl

< ERR<notenoughfunds>

errormsgdisplayed

no$givenout

BYE >

< BYE

Problem2

AttimeN*<L/R>thefirstpackethasreachedthedestination,thesecondpacketisstoredinthelastrouter,thethirdpacketisstoredinthenext-to-lastrouter,etc.AttimeN*<L/R>+L/R,thesecondpackethasreachedthedestination,thethirdpacketisstoredinthelastrouter,etc.Continuingwiththislogic,weseethatattimeN*<L/R>+<P-1>*<L/R>=<N+P-1>*<L/R>allpacketshavereachedthedestination.

Problem3

a>Acircuit-switchednetworkwouldbewellsuitedtotheapplication,becausetheapplicationinvolveslongsessionswithpredictablesmoothbandwidthrequirements.Sincethetransmissionrateisknownandnotbursty,bandwidthcanbereservedforeachapplicationsessionwithoutsignificantwaste.Inaddition,theoverheadcostsofsettingupandtearingdownconnectionsareamortizedoverthelengthydurationofatypicalapplicationsession.

b>Intheworstcase,alltheapplicationssimultaneouslytransmitoveroneormorenetworklinks.However,sinceeachlinkhassufficientbandwidthtohandlethesumofalloftheapplications'datarates,nocongestion<verylittlequeuing>willoccur.Givensuchgenerouslinkcapacities,thenetworkdoesnotneedcongestioncontrolmechanisms.

Problem4

Betweentheswitchintheupperleftandtheswitchintheupperrightwecanhave4connections.Similarlywecanhavefourconnectionsbetweeneachofthe3otherpairsofadjacentswitches.Thus,thisnetworkcansupportupto16connections.

Wecan4connectionspassingthroughtheswitchintheupper-right-handcornerandanother4connectionspassingthroughtheswitchinthelower-left-handcorner,givingatotalof8connections.

Yes.FortheconnectionsbetweenAandC,weroutetwoconnectionsthroughBandtwoconnectionsthroughD.FortheconnectionsbetweenBandD,weroutetwoconnectionsthroughAandtwoconnectionsthroughC.Inthismanner,thereareatmost4connectionspassingthroughanylink.

Problem5

Tollboothsare75kmapart,andthecarspropagateat100km/hr.Atollboothservicesacaratarateofonecarevery12seconds.

a>Therearetencars.Ittakes120seconds,or2minutes,forthefirsttollboothtoservicethe10cars.Eachofthesecarshasapropagationdelayof45minutes<travel75km>beforearrivingatthesecondtollbooth.Thus,allthecarsarelinedupbeforethesecondtollboothafter47minutes.Thewholeprocessrepeatsitselffortravelingbetweenthesecondandthirdtollbooths.Italsotakes2minutesforthethirdtollboothtoservicethe10cars.Thusthetotaldelayis96minutes.

b>Delaybetweentollboothsis8*12secondsplus45minutes,i.e.,46minutesand36seconds.Thetotaldelayistwicethisamountplus8*12seconds,i.e.,94minutesand48seconds.

Problem6

a>seconds.

b>seconds.

c>seconds.

d>ThebitisjustleavingHostA.

e>ThefirstbitisinthelinkandhasnotreachedHostB.

f>ThefirstbithasreachedHostB.

g>Want

km.

Problem7

Considerthefirstbitinapacket.Beforethisbitcanbetransmitted,allofthebitsinthepacketmustbegenerated.Thisrequires

sec=7msec.

Thetimerequiredtotransmitthepacketis

sec=sec.

Propagationdelay=10msec.

Thedelayuntildecodingis

7msec+sec+10msec=17.224msec

Asimilaranalysisshowsthatallbitsexperienceadelayof17.224msec.

Problem8

a>20userscanbesupported.

b>.

c>.

d>.

Weusethecentrallimittheoremtoapproximatethisprobability.Letbeindependentrandomvariablessuchthat.

"21ormoreusers"

whenisastandardnormalr.v.Thus"21ormoreusers".

Problem9

10,000

Problem10

ThefirstendsystemrequiresL/R1totransmitthepacketontothefirstlink;thepacketpropagatesoverthefirstlinkind1/s1;thepacketswitchaddsaprocessingdelayofdproc;afterreceivingtheentirepacket,thepacketswitchconnectingthefirstandthesecondlinkrequiresL/R2totransmitthepacketontothesecondlink;thepacketpropagatesoverthesecondlinkind2/s2.Similarly,wecanfindthedelaycausedbythesecondswitchandthethirdlink:L/R3,dproc,andd3/s3.

Addingthesefivedelaysgives

dend-end=L/R1+L/R2+L/R3+d1/s1+d2/s2+d3/s3+dproc+dproc

Toanswerthesecondquestion,wesimplyplugthevaluesintotheequationtoget6+6+6+20+16+4+3+3=64msec.

Problem11

Becausebitsareimmediatelytransmitted,thepacketswitchdoesnotintroduceanydelay;inparticular,itdoesnotintroduceatransmissiondelay.Thus,

dend-end=L/R+d1/s1+d2/s2+d3/s3

ForthevaluesinProblem10,weget6+20+16+4=46msec.

Problem12

Thearrivingpacketmustfirstwaitforthelinktotransmit4.5*1,500bytes=6,750bytesor54,000bits.Sincethesebitsaretransmittedat2Mbps,thequeuingdelayis27msec.Generally,thequeuingdelayis<nL+<L-x>>/R.

Problem13

Thequeuingdelayis0forthefirsttransmittedpacket,L/Rforthesecondtransmittedpacket,andgenerally,<n-1>L/Rforthenthtransmittedpacket.Thus,theaveragedelayfortheNpacketsis:

<L/R+2L/R++<N-1>L/R>/N

=L/<RN>*<1+2++<N-1>>

=L/<RN>*N<N-1>/2

=LN<N-1>/<2RN>

=<N-1>L/<2R>

Notethathereweusedthewell-knownfact:

1+2++N=N<N+1>/2

Ittakessecondstotransmitthepackets.Thus,thebufferisemptywhenaeachbatchofpacketsarrive.Thus,theaveragedelayofapacketacrossallbatchesistheaveragedelaywithinonebatch,i.e.,<N-1>L/2R.

Problem14

Thetransmissiondelayis.Thetotaldelayis

Let.

Totaldelay=

Forx=0,thetotaldelay=0;asweincreasex,totaldelayincreases,approachinginfinityasxapproaches1/a.

Problem15

Totaldelay.

Problem16

Thetotalnumberofpacketsinthesystemincludesthoseinthebufferandthepacketthatisbeingtransmitted.So,N=10+1.

Because,so<10+1>=a*<queuingdelay+transmissiondelay>.Thatis,

11=a*<0.01+1/100>=a*<0.01+0.01>.Thus,a=550packets/sec.

Problem17

Therearenodes<thesourcehostandtherouters>.Letdenotetheprocessingdelayatthethnode.Letbethetransmissionrateofthethlinkandlet

.Letbethepropagationdelayacrossthethlink.Then

.

Letdenotetheaveragequeuingdelayatnode.Then

.

Problem18

Onlinuxyoucanusethecommand

andintheWindowscommandpromptyoucanuse

Ineithercase,youwillgetthreedelaymeasurements.Forthosethreemeasurementsyoucancalculatethemeanandstandarddeviation.Repeattheexperimentatdifferenttimesofthedayandcommentonanychanges.

Hereisanexamplesolution:

TraceroutesbetweenSanDiegoSuperComputerCenterand

Theaverage<mean>oftheround-tripdelaysateachofthethreehoursis71.18ms,71.38msand71.55ms,respectively.Thestandarddeviationsare0.075ms,0.21ms,0.05ms,respectively.

Inthisexample,thetracerouteshave12routersinthepathateachofthethreehours.No,thepathsdidn’tchangeduringanyofthehours.

TraceroutepacketspassedthroughfourISPnetworksfromsourcetodestination.Yes,inthisexperimentthelargestdelaysoccurredatpeeringinterfacesbetweenadjacentISPs.

Traceroutesfrom<France>to<USA>.

Theaverageround-tripdelaysateachofthethreehoursare87.09ms,86.35msand86.48ms,respectively.Thestandarddeviationsare0.53ms,0.18ms,0.23ms,respectively.Inthisexample,thereare11routersinthepathateachofthethreehours.No,thepathsdidn’tchangeduringanyofthehours.TraceroutepacketspassedthreeISPnetworksfromsourcetodestination.Yes,inthisexperimentthelargestdelaysoccurredatpeeringinterfacesbetweenadjacentISPs.

Problem19

Anexamplesolution:

TraceroutesfromtwodifferentcitiesinFrancetoNewYorkCityinUnitedStates

InthesetraceroutesfromtwodifferentcitiesinFrancetothesamedestinationhostinUnitedStates,sevenlinksareincommonincludingthetransatlanticlink.

InthisexampleoftraceroutesfromonecityinFranceandfromanothercityinGermanytothesamehostinUnitedStates,threelinksareincommonincludingthetransatlanticlink.

TraceroutestotwodifferentcitiesinChinafromsamehostinUnitedStates

Fivelinksarecommoninthetwotraceroutes.ThetwotraceroutesdivergebeforereachingChina

Problem20

Throughput=min{Rs,Rc,R/M}

Problem21

Ifonlyuseonepath,themaxthroughputisgivenby:

.

Ifuseallpaths,themaxthroughputisgivenby.

Problem22

Probabilityofsuccessfullyreceivingapacketis:ps=<1-p>N.

Thenumberoftransmissionsneededtobeperformeduntilthepacketissuccessfullyreceivedbytheclientisageometricrandomvariablewithsuccessprobabilityps.Thus,theaveragenumberoftransmissionsneededisgivenby:1/ps.Then,theaveragenumberofre-transmissionsneededisgivenby:1/ps-1.

Problem23

Let’scallthefirstpacketAandcallthesecondpacketB.

Ifthebottlenecklinkisthefirstlink,thenpacketBisqueuedatthefirstlinkwaitingforthetransmissionofpacketA.Sothepacketinter-arrivaltimeatthedestinationissimplyL/Rs.

Ifthesecondlinkisthebottlenecklinkandbothpacketsaresentbacktoback,itmustbetruethatthesecondpacketarrivesattheinputqueueofthesecondlinkbeforethesecondlinkfinishesthetransmissionofthefirstpacket.Thatis,

L/Rs+L/Rs+dprop<L/Rs+dprop+L/Rc

Thelefthandsideoftheaboveinequalityrepresentsthetimeneededbythesecondpackettoarriveattheinputqueueofthesecondlink<thesecondlinkhasnotstartedtransmittingthesecondpacketyet>.Therighthandsiderepresentsthetimeneededbythefirstpackettofinishitstransmissionontothesecondlink.

IfwesendthesecondpacketTsecondslater,wewillensurethatthereisnoqueuingdelayforthesecondpacketatthesecondlinkifwehave:

L/Rs+L/Rs+dprop+T>=L/Rs+dprop+L/Rc

Thus,theminimumvalueofTisL/RcL/Rs.

Problem24

40terabytes=40*1012*8bits.So,ifusingthededicatedlink,itwilltake40*1012*8/<100*106>=3200000seconds=37days.ButwithFedExovernightdelivery,youcanguaranteethedataarrivesinoneday,anditshouldcostlessthan$100.

Problem25

160,000bits

160,000bits

Thebandwidth-delayproductofalinkisthemaximumnumberofbitsthatcanbeinthelink.

thewidthofabit=lengthoflink/bandwidth-delayproduct,so1bitis125meterslong,whichislongerthanafootballfield

s/R

Problem26

s/R=20000km,thenR=s/20000km=2.5*108/<2*107>=12.5bps

Problem27

80,000,000bits

800,000bits,thisisbecausethatthemaximumnumberofbitsthatwillbeinthelinkatanygiventime=min<bandwidthdelayproduct,packetsize>=800,000bits.

.25meters

Problem28

ttrans+tprop=400msec+80msec=480msec.

20*<ttrans+2tprop>=20*<20msec+80msec>=2sec.

Breakingupafiletakeslongertotransmitbecauseeachdatapacketanditscorrespondingacknowledgementpacketaddtheirownpropagationdelays.

Problem29

Recallgeostationarysatelliteis36,000kilometersawayfromearthsurface.

150msec

1,500,000bits

600,000,000bits

Problem30

Let’ssupposethepassengerandhis/herbagscorrespondtothedataunitarrivingtothetopoftheprotocolstack.Whenthepassengerchecksin,his/herbagsarechecked,andatagisattachedtothebagsandticket.ThisisadditionalinformationaddedintheBaggagelayerifFigure1.20thatallowstheBaggagelayertoimplementtheserviceorseparatingthepassengersandbaggageonthesendingside,andthenreunitingthem<hopefully!>onthedestinationside.Whenapassengerthenpassesthroughsecurityandadditionalstampisoftenaddedtohis/herticket,indicatingthatthepassengerhaspassedthroughasecuritycheck.Thisinformationisusedtoensure<e.g.,bylaterchecksforthesecurityinformation>securetransferofpeople.

Problem31

Timetosendmessagefromsourcehosttofirstpacketswitch=Withstore-and-forwardswitching,thetotaltimetomovemessagefromsourcehosttodestinationhost=

Timetosend1stpacketfromsourcehosttofirstpacketswitch=..Timeatwhich2ndpacketisreceivedatthefirstswitch=timeatwhich1stpacketisreceivedatthesecondswitch=

Timeatwhich1stpacketisreceivedatthedestinationhost=.Afterthis,every5mseconepacketwillbereceived;thustimeatwhichlast<800th>packetisreceived=.Itcanbeseenthatdelayinusingmessagesegmentationissignificantlyless<almost1/3rd>.

Withoutmessagesegmentation,ifbiterrorsarenottolerated,ifthereisasinglebiterror,thewholemessagehastoberetransmitted<ratherthanasinglepacket>.

Withoutmessagesegmentation,hugepackets<containingHDvideos,forexample>aresentintothenetwork.Routershavetoaccommodatethesehugepackets.Smallerpacketshavetoqueuebehindenormouspacketsandsufferunfairdelays.

Packetshavetobeputinsequenceatthedestination.

Messagesegmentationresultsinmanysmallerpackets.Sinceheadersizeisusuallythesameforallpacketsregardlessoftheirsize,withmessagesegmentationthetotalamountofheaderbytesismore.

Problem32

Yes,thedelaysintheappletcorrespondtothedelaysintheProblem31.Thepropagationdelaysaffecttheoverallend-to-enddelaysbothforpacketswitchingandmessageswitchingequally.

Problem33

ThereareF/Spackets.EachpacketisS=80bits.Timeatwhichthelastpacketisreceivedatthefirstrouterissec.Atthistime,thefirstF/S-2packetsareatthedestination,andtheF/S-1packetisatthesecondrouter.Thelastpacketmustthenbetransmittedbythefirstrouterandthesecondrouter,witheachtransmissiontakingsec.Thusdelayinsendingthewholefileis

TocalculatethevalueofSwhichleadstotheminimumdelay,

Problem34

Thecircuit-switchedtelephonenetworksandtheInternetareconnectedtogetherat"gateways".WhenaSkypeuser<connectedtotheInternet>callsanordinarytelephone,acircuitisestablishedbetweenagatewayandthetelephoneuseroverthecircuitswitchednetwork.Theskypeuser'svoiceissentinpacketsovertheInternettothegateway.Atthegateway,thevoicesignalisreconstructedandthensentoverthecircuit.Intheotherdirection,thevoicesignalissentoverthecircuitswitchednetworktothegateway.ThegatewaypacketizesthevoicesignalandsendsthevoicepacketstotheSkypeuser.

Chapter2ReviewQuestions

TheWeb:;filetransfer:FTP;remotelogin:Telnet;e-mail:SMTP;BitTorrentfilesharing:BitTorrentprotocol

Networkarchitecturereferstotheorganizationofthecommunicationprocessintolayers<e.g.,thefive-layerInternetarchitecture>.Applicationarchitecture,ontheotherhand,isdesignedbyanapplicationdeveloperanddictatesthebroadstructureoftheapplication<e.g.,client-serverorP2P>.

Theprocesswhichinitiatesthecommunicationistheclient;theprocessthatwaitstobecontactedistheserver.

No.InaP2Pfile