TCPIP协议族第四版课后答案

1CHAPTER1IntroductionExercises1.AsofJanuary2009,theRFCwiththehighestnumberisRFC5459,titledRTPPay-loadFormatUpdate.3.RFC2014:ThisRFCdiscussestheIRTFworkinggroupguidelinesandproce-dures.5.RFC3692andRFC1410aretwoexamplesofexperimentalRFCs.7.ThemainRFCforFTPisRFC959thathasbecomethestandardSTD0009.9.ThemainRFCrelatedtoTCPisRFC793(J.Postel)thathasbecomethestandardSTD0007.21CHAPTER2TheOSIModelandtheTCP/IPProtocolSuiteExercises1.TheInternationalStandardsOrganization(ISO)isamultinationalbodydedicatedtoworldwideagreementoninternationalstandards.AnISOstandardthatcoversallaspectsofnetworkcommunicationsistheOpenSystemsInterconnection(OSI)model.3.a.Transportlayerb.Networklayerc.Datalinklayerd.Applicationlayere.Physicallayer5.a.Presentationlayerb.Sessionlayerc.Datalinkandtransportlayersd.Sessionlayere.Presentationlayer7.Ifwethinkabouttheswitchasapassiveone(notabridge),Figure2.E7showsthesolution.Chapter02.fmPage1Saturday,June13,20098:05PM29.Theheaderatthetransportlayershouldatleastincludethesourceanddestinationportnumber.Thismeansthesizeoftheheaderisatleast2+2=4bytes.11.Theheaderatthedatalinklayershouldatleastincludethephysicalsourceanddestinationaddresses.Thismeansthesizeoftheheaderisatleast6+6=12bytes.13.Atthephysicallayer,thesignalrepresentingthebitstreamisbroadcasttoallsta-tionsinanetwork.Everystationreceivesit;thereisnoneedforaddressesinthislayer.15.Thedestinationaddressisneededtodefinetherecipientofthemessage;thesourceaddressisneededifthereceiverofthemessagehastorespondortheintermediatenodeshastoreportanyerrorthesource.Figure2.E7SolutiontoExercise7APhysicalPhysicalDatalinkDatalinkR1BBNetworkNetworkTransportTransportApplicationApplicationMessageD5D5D5D5MessageAR1Link2Link1Chapter02.fmPage2Saturday,June13,20098:05PM1CHAPTER3UnderlyingTechnologiesExercises1.WeknowthatD=T×V,whereDisthedistance,Tisthetime,andVistheveloc-ityorspeed.Inotherwords,T=D/V.Weinsertthecorrespondingvaluestofindthetimeneededforabittotravelthecable.3.Assumethattheminimumframesizeis65bytesor520bits.WehaveL=T×R,whereListhelengthoftheframe,Tisthetime,andtheRisthedatarate.WecansayT=L/R.Thetimecanbecalculatedas5.Thepaddingneedstomakethesizeofthedatasection46bytes.Ifthedatareceivedfromtheupperlayeris42bytes,weneed46−42=4bytesofpadding.7.a.Similarities:Eachstationhasanequalrighttothemedium.Eachstationsensesthemedium.b.Differences:CSMA/CD:Astationcansendifitsensesnosignalontheline.CSMA/CA:Astationneedstoinformotherstationsthatitneedsthemediumforaspecificamountoftime.CSMA/CD:Acollisioncanoccur.CSMA/CA:Collisionsareavoided.T=D/V=(2500meters)/(200,000,000meters/second)=0.0000125s=12.5μsT=L/R=(520bits)/(10,000,000)bits/second=0.000052s=52μs28.SeeTable3.E8.Table3.E8Exercise8FieldsIEEE802.3IEEE802.11Destinationaddress6Sourceaddress6Address16Address26Address36Address46FC2D/ID2SC2PDUlength2Dataandpadding1500Framebody2312FCS(CRC)441CHAPTER4IntroductiontoNetworkLayerExercises1.Wementiononeadvantageandonedisadvantageforaconnectionlessservice:a.Theconnectionlessservicehasatleastoneadvantage.Aconnectionlessserviceissimple.Thesource,destination,andtheroutersneedtodealwitheachpacketindividuallywithoutconsideringtherelationshipbetweenthem.Thismeanstherearenosetupandteardownphases.Noextrapacketsareexchangedbetweenthesourceandthedestinationforthesetwophases.b.Theconnectionlessservicehasatleastonedisadvantage.Thepacketsmayarriveoutoforder;theupperlayerthatreceivethemneedstoreorderthem.3.Ann-bitlabelcancreate2ndifferentvirtual-circuitidentifier.5.Eachpacketstartedfromthesourceneedstohaveafragmentationidentification,whichisrepeatedineachfragment.Thedestinationcomputerusethisidentifica-tiontoreassembleallfragmentsbelongingtothesamepacket.7.Thedelayintheconnection-orientedserviceisalwaysmorethanthedelayintheconnectionlessservicenomatterthemessageislongorshort.However,theratiooftheoverheaddelay(setupandteardownphases)tothedatatransferdelay(trans-missionandpropagation)issmallerforalongmessagethanashortmessageinaconnection-orientedservice.9.Arouterisnormallyconnectedtodifferentlink(networks),eachwithdifferentMTU.ThelinkfromwhichthepacketisreceivedmayhavealargerMTUthanthelinktowhichthepacketissent,whichmeansthatrouterneedstofragmentthepacket.WewillseeinChapter27thatIPv6doesnotallowfragmentationattherouter,whichmeansthesourceneedstodosomeinvestigationandmakethepacketsmallenoughtobecarriedthroughalllinks.11.Afragmentmayhavebeenlostandneverarrives.Thedestinationhostcannotwaitforever.Thedestinationhoststartsatimeandafterthetime-out,itcansendsanerrormessage(seeChapter9)toinformthesourcehostthatthepacketislostand,ifnecessary,shouldberesent.Thetime-outdurationcanbebasedontheinforma-2tionthedestinationhostmaycollectaboutthestatusoftheInternet.Iftherearemanydelaysinthepreviouspacketarrivals,itmeansthattheInternetiscongested,andthefragmentmayarrivesoon(itisnotnecessarilylost).1CHAPTER5IPv4AddressesExercises1.a.28=256addressesb.216=65536addressesc.264=1.846744737×1019addresses3.310=59,049addresses5.a.0x72220208b.0x810E0608c.0xD022360Cd.0xEE2202017.a.(8bits)/(4bitsperhexdigits)=2hexdigitsb.(16bi