计算机网络自顶向下(第七版)课后答案-英文

ComputerNetworking:ATop-DownApproach,

7thEdition

SolutionstoReviewQuestionsandProblems

VersionDate:December2016

Thisdocumentcontainsthesolutionstoreviewquestionsandproblemsforthe7theditionofComputerNetworking:ATop-DownApproachbyJimKuroseandKeithRoss.ThesesolutionsarebeingmadeavailabletoinstructorsONLY.PleasedoNOTcopyordistributethisdocumenttoothers(evenotherinstructors).Pleasedonotpostanysolutionsonapublicly-availableWebsite.We’llbehappytoprovideacopy(up-to-date)ofthissolutionmanualourselvestoanyonewhoasks.

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

Allmaterial©copyright1996-2016byJ.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

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

traceroute

andintheWindowscommandpromptyoucanuse

tracert

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

Networkarchitecturereferstotheorganizationofthecommunicationprocessintolayers(e.g.,thefive-layerInternetarchitecture).Applicationarchitecture,ontheotherhand,isdesignedbyanapplicationdeveloperanddictatesthebroadstructureoftheapplication(e.g.,client-serverorP2P).

Theprocesswhichinit