computer network课后答案1.docx

computer network课后答案1.docx

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computernetwork课后答案1

Chapter1ReviewQuestions

1.Thereisnodifference.Throughoutthistext,thewords“host”and“endsystem”areusedinterchangeably.EndsystemsincludePCs,workstations,Webservers,mailservers,Internet-connectedPDAs,WebTVs,etc.

3.Anetworkingprogramusuallyhastwoprograms,eachrunningonadifferenthost,communicatingwitheachother.Theprogramthatinitiatesthecommunicationistheclient.Typically,theclientprogramrequestsandreceivesservicesfromtheserverprogram.

22.application-layermessage:

datawhichanapplicationwantstosendandpassedontothetransportlayer;transport-layersegment:

generatedbythetransportlayerandencapsulatesapplication-layermessagewithtransportlayerheader;network-layerdatagram:

encapsulatestransport-layersegmentwithanetwork-layerheader;linklayerframe:

encapsulatesnetwork-layerdatagramwithalink-layerheader.

23.Routersprocesslayers1through3.（Thisisalittlebitofawhitelie,asmodernrouterssometimesactasfirewallsorcachingcomponents,andprocesslayerfouraswell.Linklayerswitchesprocesslayers1through2.Hostsprocessallfivelayers.

Chapter1Problems

Problem6.

admsprop=/seconds.

bdLRtrans=/seconds.

cd（m/sL/Rendtoend=+?

?

seconds.

dThebitisjustleavingHostA.

eThefirstbitisinthelinkandhasnotreachedHostB.

fThefirstbithasreachedHostB.

gWant

（2.510893

2810

1008

3≠=

≠

=S=

R

mLkm.

Problem10.

IttakesLN/RsecondstotransmittheNpackets.Thus,thebufferisemptywhenabatchofNpacketsarrive.

ThefirstoftheNpacketshasnoqueueingdelay.The2ndpackethasaqueueingdelay

ofL/Rseconds.Thenthpackethasadelayof（n?

1L/Rseconds.Theaveragedelayis

2

（1

2

1（1/111（1

10

?

=

?

┣?

=┣=?

==

N

R

NNL

RN

nL

RN

nLRL

N

N

n

N

n

.

Problem11.

aThetransmissiondelayisL/R.Thetotaldelayis

I

LR

R

L

RI

IL

?

+=

?

1

/

（1

bLetx=L/R.

Totaldelay=

ax

x

1?

Problem14.

a40,000bits

b40,000bits

cthebandwidth-delayproductofalinkisthemaximumnumberofbitsthatcanbein

thelink

d1bitis250meterslong,whichislongerthanafootballfield

es/R

Problem18.

a150msec

b1,500,000bits

c600,000,000bits

Chapter2ReviewQuestions

1.TheWeb:

HTTP;filetransfer:

FTP;remotelogin:

Telnet;NetworkNews:

NNTP;email:

SMTP.

5.No.Asstatedinthetext,allcommunicationsessionshaveaclientsideandaserverside.InaP2Pfile-sharingapplication,thepeerthatisreceivingafileistypicallytheclientandthepeerthatissendingthefileistypicallytheserver.

6.TheIPaddressofthedestinationhostandtheportnumberofthedestinationsocket.

10.Theapplicationsassociatedwiththoseprotocolsrequirethatallapplicationdatabereceivedinthecorrectorderandwithoutgaps.TCPprovidesthisservicewhereasUDPdoesnot.

22.WiththeUDPserver,thereisnowelcomingsocket,andalldatafromdifferentclientsenterstheserverthroughthisonesocket.WiththeTCPserver,thereisawelcomingsocket,andeachtimeaclientinitiatesaconnectiontotheserver,anewsocketiscreated.Thus,tosupportnsimultaneousconnections,theserverwouldneedn+1sockets.Chapter2Problems

Problem1.

aF

bT

cF

dF

Problem4.

Applicationlayerprotocols:

DNSandHTTP

Transportlayerprotocols:

UDPforDNS;TCPforHTTP

Problem9.

aThetimetotransmitanobjectofsizeLoveralinkorrateRisL/R.TheaveragetimeistheaveragesizeoftheobjectdividedbyR:

┐=（900,000bits/（1,500,000bits/sec=.6sec

Thetrafficintensityonthelinkis（1.5requests/sec（.6msec/request=.9.Thus,the

averageaccessdelayis（.6sec/（1-.9=6seconds.Thetotalaverageresponsetimeistherefore6sec+2sec=8sec.

bThetrafficintensityontheaccesslinkisreducedby40%sincethe40%oftherequestsaresatisfiedwithintheinstitutionalnetwork.Thustheaverageaccessdelay

is（.6sec/[1쭯（.6（.9]=1.2seconds.Theresponsetimeisapproximatelyzeroiftherequestissatisfiedbythecache（whichhappenswithprobability.4;theaverageresponsetimeis1.2sec+2sec=3.2secforcachemisses（whichhappens60%ofthetime.Sotheaverageresponsetimeis（.4（0sec+（.6（3.2sec=1.92seconds.Thustheaverageresponsetimeisreducedfrom8secto1.92sec.Chapter3ReviewQuestions

1.Sourceportnumberyanddestinationportnumberx.

2.AnapplicationdevelopermaynotwantitsapplicationtouseTCP’scongestioncontrol,whichcanthrottletheapplication’ssendingrateattimesofcongestion.Often,designersofIPtelephonyandIP

videoconferenceapplicationschoosetoruntheirapplicationsoverUDPbecausetheywanttoavoidTCP’scongestioncontrol.Also,someapplicationsdonotneedthereliabledatatransferprovidedbyTCP.

4.afalsebfalsectruedfalseetrueffalsegfalse

5.a20bytesbacknumber=90

Chapter3Problems

Problem1.

sourceportnumbers

destinationportnumbers

aA∪S46723

bB∪S51323

cS∪A23467

dS∪B23513

eYes.

fNo.

Problem3.

11000101

01110000

01010101

+

00010001

01001100

11000101

+

One'scomplement=11101110.

Todetecterrors,thereceiveraddsthefourwords（thethreeoriginalwordsandthechecksum.Ifthesumcontainsazero,thereceiverknowstherehasbeenanerror.Allone-biterrorswillbedetected,buttwo-biterrorscanbeundetected（e.g.,ifthelastdigitofthefirstwordisconvertedtoa0andthelastdigitofthesecondwordisconvertedtoa1.

Problem19.

aTrue.Supposethesenderhasawindowsizeof3andsendspackets1,2,3att0.Att1（t1>t0thereceiverACKS1,2,3.Att2（t2>t1thesendertimesoutandresends1,2,3.Att3thereceiverreceivestheduplicatesandre-acknowledges1,2,3.Att4thesenderreceivesthe

ACKsthatthereceiversentatt1andadvancesitsnwindowto4,5,6.Att5thesenderreceivestheACKs1,2,3thereceiversentatt2.TheseACKsareoutsideitswindow.bTrue.Byessentiallythesamescenarioasin（a.cTrue.dTrue.Notethatwithawindowsizeof1,SR,GBN,andthealternatingbitprotocolarefunctionallyequivalent.Thewindowsizeof1precludesthepossibilityofout-of-orderpackets（withinthewindow.AcumulativeACKisjustanordinaryACKinthissituation,sinceitcanonlyrefertothesinglepacketwithinthewindow.Chapter4ReviewQuestions3.Forwardingisaboutmovingapacketfromarouter’sinputlinktotheappropriateoutputlink.Routingisaboutdeterminingtheend-to-routesbetweensourcesanddestinations.8.switchingviamemory;switchingviaabus;switchingviaaninterconnectionnetwork12.Yes.Theyhaveoneaddressforeachinterface.16.50%overhead34.FalseChapter4ProblemsProblem7.aPrefixMatchLinkInterface11100000011100001000000001111000012otherwise3bPrefixmatchforfirstaddressis4thentry:

linkinterface3Prefixmatchforsecondaddressis2ndentry:

linkinterface1Prefixmatchforfirstaddressis3rdentry:

linkinterface2Problem8.DestinationAddressRangeLinkInterface00000000through00011111101000000

through10111111110000000through21011111111000000through311111111numberofaddressesineachrange=26=64Problem13.AnyIPaddressinrange101.101.101.65to101.101.101.127Fourequalsizesubnets:

101.101.101.64/28,101.101.101.80/28,101.101.101.96/28,101.101.101.112/28Problem16.MP3filesize=4millionbytes.AssumethedataiscarriedinTCPsegments,witheachTCPsegmentalsohaving20bytesofheader.Theneachdatagramcancarry1500-40=1460bytesoftheMP3fileNumberoffragmentsrequired=274014604106=?

?

?

?

?

?

≠=.Allbutthelastfragmentwillbe1,500bytes;thelastfragmentwillbe1060+40=1100bytes.Theoffsetswillbemultiplesof185（asinexampleinSection4.4.1.Chapter5ReviewQuestions1.AlthougheachlinkguaranteesthatanIPdatagramsentoverthelinkwillbereceivedattheotherendofthelinkwithouterrors,itisnotguaranteedthatIPdatagramswillarriveattheultimatedestinationintheproperorder.WithIP,datagramsinthesameTCPconnectioncantakedifferentroutesinthenetwork,andthereforearriveoutoforder.TCPisstillneededtoprovidethereceivingendoftheapplicationthebytestreaminthecorrectorder.Also,IPcanlosepacketsduetorouting

loopsorequipmentfailures.7.248MACaddresses;232IPv4addresses;2128IPv6addresses.Chapter5ProblemsProblem4.Ifwedivide1001into10101010000weget10010111,witharemainderofR=001.Problem11.Att=0Atransmits.Att=576,Awouldfinishtransmitting.Intheworstcase,Bbeginstransmittingattimet=224.Attimet=224+225=449B'sfirstbitarrivesatA.Because449<576,Aabortsbeforecompletingthetransmissionofthepacket,asitissupposedtodo.ThusAcannotfinishtransmittingbeforeitdetectsthatBtransmitted.ThisimpliesthatifAdoesnotdetectthepresenceofahost,thennootherhostbeginstransmittingwhileAistransmitting.