下載本文檔
版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請進(jìn)行舉報(bào)或認(rèn)領(lǐng)
文檔簡介
第一章(1.21.3節(jié))5.Calculatethetotaltimerequiredtotransfera1,000-KB?leinthefollowingcases,assuminganRTTof100ms,apacketsizeof1-KBdata,andaninitial2×RTTof“handshaking”beforedataissent.(a)Thebandwidthis1.5Mbps,anddatapacketscanbesentcontinuously.(b)Thebandwidthis1.5Mbps,butafterwe?nishsendingeachdatapacketwemustwaitoneRTTbeforesendingthenext.(c)Thebandwidthis“in?nite,”meaningthatwetaketransmittimetobezero,andupto20packetscanbesentperRTT.(d)Thebandwidthisin?nite,andduringthe?rstRTTwecansendonepacket(21?1),duringthesecondRTTwecansendtwopackets(22?1),duringthethirdwecansendfour(23?1),andsoon.(Ajusti?cationforsuchanexponentialincreasewillbegiveninChapter6.)7.Considerapoint-to-pointlink2kminlength.Atwhatbandwidthwouldpropagationdelay(ataspeedof2×108m/sec)equaltransmitdelayfor100-bytepackets?Whatabout512-bytepackets?13.How“wide”isabitona1-Gbpslink?Howlongisabitincopperwire,wherethespeedofpropagationis2.3×108m/s?15.Supposea100-Mbpspoint-to-pointlinkisbeingsetupbetweenEarthandanewlunarcolony.ThedistancefromthemoontoEarthisapproximately385,000km,anddatatravelsoverthelinkatthespeedoflight—3×108m/s.(a)CalculatetheminimumRTTforthelink.(b)UsingtheRTTasthedelay,calculatethedelay×bandwidthproductforthelink.(c)Whatisthesigni?canceofhedelay×bandwidthproductcomputedin(b)?(d)AcameraonthelunarbasetakespicturesofEarthandsavesthemindigitalformattodisk.SupposeMissionControlonEarthwishestodownloadthemostcurrentimage,whichis25MB.Whatistheminimumamountoftimethatwillelapsebetweenwhentherequestforthedatagoesoutandthetransferis?nished?18.Calculatethelatency(from?rstbitsenttolastbitreceived)forthefollowing:(a)A10-MbpsEthernetwithasinglestore-and-forwardswitchinthepath,andapacketsizeof5,000bits.Assumethateachlinkintroducesapropaga-tiondelayof10μs,andthattheswitchbeginsretransmittingimmediatelyafterithas?nishedreceivingthepacket.(b)Sameas(a)butwiththreeswitches.(c)Sameas(a)butassumetheswitchimplements“cut-through”switching:itisabletobeginretransmittingthepacketafterthe?rst200bitshavebeenreceived.第二章(除2.72.9節(jié))1.ShowtheNRZ,Manchester,andNRZIencodingsforthebitpatternshowninFigure2.46.AssumethattheNRZIsignalstartsoutlow.23.ConsideranARQalgorithmrunningovera20-kmpoint-to-point?berlink.(a)Computethepropagationdelayforthislink,assumingthatthespeedoflightis2×108m/sinthe?ber.(b)SuggestasuitabletimeoutvaluefortheARQalgorithmtouse.(c)WhymightitstillbepossiblefortheARQalgorithmtotimeoutandretransmitaframe,giventhistimeoutvalue?26.Thetextsuggeststhattheslidingwindowprotocolcanbeusedtoimplement?owcontrol.WecanimaginedoingthisbyhavingthereceiverdelayACKs,thatis,notsendtheACKuntilthereisfreebufferspacetoholdthenextframe.Indoingso,eachACKwouldsimultaneouslyacknowledgethereceiptofthelastframeandtellthesourcethatthereisnowfreebufferspaceavailabletoholdthenextframe.Explainwhyimplementing?owcontrolinthiswayisnotagoodidea.44.LetAandBbetwostationsattemptingtotransmitonanEthernet.Eachhassteadyqueueofframesreadytosend;A’sframeswillbenumberedA1,A2,andsoon,andB’ssimilarly.LetT=51.2μsbetheexponentialbackoffbaseunit.SupposeAandBsimultaneouslyattempttosendframe1,collide,andhappentochoosebackofftimesof0×Tand1×T,respectively,meaningAwinstheraceandtransmitsA1whileBwaits.Attheendofthistransmission,BwillattempttoretransmitB1whileAwillattempttotransmitA2.These?rstattemptswillcollide,butnowAbacksoffforeither0×Tor1×T,whileBbacksofffortimeequaltooneof0×T,...,3×T.(a)GivetheprobabilitythatAwinsthissecondbackoffraceimmediatelyafterthis?rstcollision,thatis,A’s?rstchoiceofbackofftimek×51.2islessthanB’s.(b)SupposeAwinsthissecondbackoffrace.AtransmitsA3,andwhenitis?nished,AandBcollideagainasAtriestotransmitA4andBtriesoncemoretotransmitB1.GivetheprobabilitythatAwinsthisthirdbackoffraceimmediatelyafterthe?rstcollision.(c)GiveareasonablelowerboundfortheprobabilitythatAwinsallthere-mainingbackoffraces.(d)WhatthenhappenstotheframeB1?ThisscenarioisknownastheEthernetcaptureeffect.48.Repeatthepreviousexercise,nowwiththeassumptionthatEthernetisp-persistentwithp=0.33(thatis,awaitingstationtransmitsimmediatelywithprobabilitypwhenthelinegoesidle,andotherwisedefersone51.2-μsslottimeandrepeatstheprocess).Yourtimelineshouldmeetcriterion(1)ofthepreviousproblem,butinlieuofcriterion(2),youshouldshowatleastonecollisionandatleastonerunoffourdeferralsonanidleline.Again,notethatmanysolutionsarepossible.第三章(3.13.2節(jié))1.UsingtheexamplenetworkgiveninFigure3.30,givethevirtualcircuittablesforalltheswitchesaftereachofthefollowingconnectionsisestablished.Assumethatthesequenceofconnectionsiscumulative,thatis,the?rstconnectionisstillupwhenthesecondconnectionisestablished,andsoon.AlsoassumethattheVCIassignmentalwayspicksthelowestunusedVCIoneachlink,startingwith0.(a)HostAconnectstohostB.(b)HostCconnectstohostG.(c)HostEconnectstohostI.(d)HostDconnectstohostB.(e)HostFconnectstohostJ.(f)HostHconnectstohostA.3.ForthenetworkgiveninFigure3.31,givethedatagramforwardingtableforeachnode.Thelinksarelabeledwithrelativecosts;yourtablesshouldforwardeachpacketviathelowest-costpathtoitsdestination.5.ConsiderthevirtualcircuitswitchesinFigure3.33.Table3.6lists,foreachswitch,whatport,VCI(orVCI,interface)pairsareconnectedtoother.Connectionsarebidirectional.Listallendpoint-to-endpointconnections.13.GiventheextendedLANshowninFigure3.34,indicatewhichportsarenotselectedbythespanningtreealgorithm.15.ConsiderthearrangementoflearningbridgesshowninFigure3.35.Assumingallareinitiallyempty,givetheforwardingtablesforeachofthebridgesB1–B4afterthefollowingtransmissions:■AsendstoC.■CsendstoA.■DsendstoC.Identifyportswiththeuniqueneighborreacheddirectlyfromthatport,thatis,theportsforB1aretobelabeled“A”and“B2.”17.ConsiderhostsX,Y,Z,WandlearningbridgesB1,B2,B3,withinitiallyemptyforwardingtables,asinFigure3.36.(a)SupposeXsendstoZ.WhichbridgeslearnwhereXis?DoesY’snetworkinterfaceseethispacket?(b)SupposeZnowsendstoX.WhichbridgeslearnwhereZis?DoesY’snetworkinterfaceseethispacket?(c)SupposeYnowsendstoX.WhichbridgeslearnwhereYis?DoesZ’snet-workinterfaceseethispacket?(d)Finally,supposeZsendstoY.WhichbridgeslearnwhereZis?DoesW’snetworkinterfaceseethispacket?第四章(4.14.24.3.14.3.54.5節(jié))4.SupposeaTCPmessagethatcontains2,048bytesofdataand20bytesofTCPheaderispassedtoIPfordeliveryacrosstwonetworksoftheInternet(i.e.,fromthesourcehosttoaroutertothedestinationhost).The?rstnetworkuses14-byteheadersandhasanMTUof1,024bytes;theseconduses8-byteheaderswithanMTUof512bytes.Eachnetwork’sMTUgivesthesizeofthelargestIPdatagramthatcanbecarriedinalinklayerframe.Givethesizesandoffsetsofthesequenceoffragmentsdeliveredtothenetworklayeratthedestinationhost.AssumeallIPheadersare20bytes.21.SupposearouterhasbuiltuptheroutingtableshowninTable4.14.Theroutercandeliverpacketsdirectlyoverinterfaces0and1,oritcanforwardpacketstoroutersR2,R3,orR4.Describewhattherouterdoeswithapacketaddressedtoeachofthefollowingdestinations:(a)
溫馨提示
- 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁內(nèi)容里面會有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫網(wǎng)僅提供信息存儲空間,僅對用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對任何下載內(nèi)容負(fù)責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對自己和他人造成任何形式的傷害或損失。
最新文檔
- 新教師崗前培訓(xùn)心得體會
- 2024-2025學(xué)年“逐夢計(jì)劃”環(huán)際大聯(lián)考高二上學(xué)期期中考試生物試題(解析版)
- 2024-2025學(xué)年浙江省9+1高中聯(lián)盟高三上學(xué)期11月期中物理試題(解析版)
- 跳轉(zhuǎn)頁面動態(tài)優(yōu)化-洞察分析
- 2023-2024學(xué)年山東省菏澤市鄄城縣一中高一1月月考生物試題(解析版)
- 云計(jì)算在水文模型中的應(yīng)用-洞察分析
- 旋臂內(nèi)暗物質(zhì)分布-洞察分析
- 體育俱樂部法律與政策研究-洞察分析
- 2023-2024年項(xiàng)目部治理人員安全培訓(xùn)考試題及1套參考答案
- 2023-2024年項(xiàng)目管理人員安全培訓(xùn)考試題及答案(典優(yōu))
- 2025年行政執(zhí)法人員執(zhí)法資格考試必考題庫及答案(共232題)
- 2025年北京探礦工程研究所招聘高校應(yīng)屆畢業(yè)生歷年管理單位筆試遴選500模擬題附帶答案詳解
- 2025-2030年中國新能源汽車行業(yè)市場分析報(bào)告
- 網(wǎng)站建設(shè)合同范本8篇
- 宜賓天原5萬噸氯化法鈦白粉環(huán)評報(bào)告
- GB/T 44888-2024政務(wù)服務(wù)大廳智能化建設(shè)指南
- 2024年工廠股權(quán)轉(zhuǎn)讓盡職調(diào)查報(bào)告3篇
- 2024年行政執(zhí)法考試題庫及答案(題)
- 針灸推拿題庫及參考答案
- 會計(jì)專業(yè)工作簡歷表(中級)
- 頂管施工技術(shù)全面詳解
評論
0/150
提交評論