计算机网络原理习题讲解

/ChapterIWhatisthedifferencebetweenahostandanendsystem?Listthetypesofendsystems.IsaWebserveranendsystem?Whatisaclientprogram?Whatisaserverprogram?Doesaserverprogramrequestandreceiveservicesfromaclientprogram?Listsixaccesstechnologies.Classifyeachoneasresidentialaccess,companyaccess,ormobileaccess.Dial-upmodems,HFC,andDSLareallusedforresidentialaccess.Foreachoftheseaccesstechnologies,providearangeoftransmissionratesandcommentonwhetherthetransmissionrateissharedordedicated.DescribethemostpopularwirelessInternetaccesstechnologiestoday.Compareandcontrastthem.Whatadvantagedoesacircuit-switchednetworkhaveoverapacket-switchednetwork?WhatadvantagesdoesTDMhaveoverFDMinacircuit-switchednetwork?Considersendingapacketfromasourcehosttoadestinationhostoverafixedroute.Listthedelaycomponentsintheend-to-enddelay.Whichofthesedelaysareconstantandwhicharevariable?Howlongdoesittakeapacketoflength2,000bytestopropagateoveralinkofdistance2,000km,propagationspeedm/s,andtransmissionrate2Mbps?Moregenerally,howlongdoesittakeapacketoflengthLtopropagateoveralinkofdistanced,propagationspeeds,andtransmissionrateRbps?Doesthisdelaydependonpacketlength?Doesthisdelaydependontransmissionrate?WhatarethefivelayersintheInternetprotocolstack?Whataretheprincipalresponsibilitiesofeachoftheselayers?WhichlayersintheInternetprotocolstackdoesarouterprocess?Whichlayersdoesalink-layerswitchprocess?Whichlayersdoesahostprocess?Whatisanapplication-layermessage?Atransport-layersegment?Anetwork-layerdatagram?Alink-layerframe?Thiselementaryproblembeginstoexplorepropagationdelayandtransmissiondelay,twocentralconceptsindatanetworking.Considertwohosts,AandB,connectedbyasinglelinkofrateRbps.Supposethatthetwohostsareseparatedbymmeters,andsupposethepropagationspeedalongthelinkissmeters/sec.HostAistosendapacketofsizeLbitstoHostB.a.Expressthepropagationdelay,,intermsofmands.b.Determinethetransmissiontimeofthepacket,,intermsofLandR.c.Ignoringprocessingandqueuingdelays,obtainanexpressionfortheend-to-enddelay.d.SupposeHostAbeginstotransmitthepacketattimet=0.Attime,whereisthelastbitofthepacket?e.Supposeisgreaterthan.Attimet=,whereisthefirstbitofthepacket?f.Supposeislessthan.Attimet=,whereisthefirstbitofthepacket?g.Suppose,L=100bits,andR=28kbps.Findthedistancemsothatequals.Inmodernpacket-switchednetworks,thesourcehostsegmentslong,application-layermessages<forexample,animageoramusicfile>intosmallerpacketsandsendsthepacketsintothenetwork.Thereceiverthenreassemblesthepacketsbackintotheoriginalmessage.Werefertothisprocessasmessagesegmentation.Figure1.24illustratestheend-to-endtransportofamessagewithandwithoutmessagesegmentation.ConsideramessagethatisbitslongthatistobesentfromsourcetodestinationinFigure1.24.Supposeeachlinkinthefigureis2Mbps.Ignorepropagation,queuing,andprocessingdelays.a.Considersendingthemessagefromsourcetodestinationwithoutmessagesegmentation.Howlongdoesittaketomovethemessagefromthesourcehosttothefirstpacketswitch?Keepinginmindthateachswitchusesstore-and-forwardpacketswitching,whatisthetotaltimetomovethemessagefromsourcehosttodestinationhost?b.Nowsupposethatthemessageissegmentedinto4,000packets,witheachpacketbeing2,000bitslong.Howlongdoesittaketomovethefirstpacketfromsourcehosttothefirstswitch?Whenthefirstpacketisbeingsentfromthefirstswitchtothesecondswitch,thesecondpacketisbeingsentfromthesourcehosttothefirstswitch.Atwhattimewillthesecondpacketbefullyreceivedatthefirstswitch?c.Howlongdoesittaketomovethefilefromsourcehosttodestinationhostwhenmessagesegmentationisused?Comparethisresultwithyouranswerinpart<a>andcomment.d.Discussthedrawbacksofmessagesegmentation.下列说法中,正确的是<>。A.在较小范围内布置的一定是局域网,币在较大范围内布置的一定是广域网B.城域网是连接广域网而覆盖园区的网络C.城域网是为淘汰局域网和广域网而提出的一种网络技术D.局域网是基于广播技术发展起来的网络,广域网是基于交换技术发展起来的向络解答:D。通常而言,局域网的覆盖范围较小,而广域网的覆盖范围较大,但这并不绝对。有时候在一个不大的范围内采用广域网,这取决于应用的需要和是否采用单一网络等多种因素。特别是局域网技术的进步,使得其覆盖范围越来越大,达到几十千米的范围。城域网是利用广域网技术、满足一定区域需求的一种网络,事实上,城域网的范围弹性非常大。最初的局域网采用广播技术,这种技术一直被沿用,而广域网最初使用的是交换技术,也一直被沿用。・相对于o⒏的7层参考模型的低4层,TCP/IP协议集内对应的层次有<>。A.传输层、互联网层、网络接口层和物理层B.传输层、互联网层、网络接口层C.传输层、互联网层、ATM层和物理层D.传输层、网络层、数据链路蜃和物理层解答:B。根据TCP/P分层模型可知,其对应OSI低4层的分别是传输层、互联网层、网络接口层。在C/S模式的网络中,最恰当的是<>。A.客户机提出请求,服务器响应请求、进行处理并返回结果B.服务器有时可以同时为多个客户机服务C.客户机可以将服务器的资源各份在本地,以避免向服务器请求服务D.服务器永远是网络的瓶颈解答:A。根据C/S模式的定义,选项A描述了C/S模式的基本工作流程。服务器必须总能而不是有时可以同时为多个客户机服务,否则网络就没有了存在的价值。由于服务器的资源太庞大,而且很多资源因为知识产权、保密、管理复杂等一系列的原因,使客户机不可能都把服务器的资源备份到本地。从表面上看,服务器可能是网络的瓶颈,但事实上:在多数情况下网络的主要瓶颈不在服务器.而在通信线路。比较分组交换与报文交换,说明分组交换优越的原因。解答:报文交换网络与分组交换的原理都是:将用户数据加上源地址、自的地址、长度、校验码等辅助信息封装成PDU,发送给下一个节点。下一个节点收到后先暂存报文,待输出线路空闲时再转发给下一个节点,重复这一过程直到到达目的节点。每个PDU可单独选择到达目的节点的路径。这种方式也称为存储―转发方式。两者的不同之处是:分组交换所生成的PDU的长度较短,而且是固定的;而报文交换的PDU的长度不是固定的。正是这一差别,使得分组交换具有独特的优点:缓冲区易于管理;分组的平均延迟更小,网络中占用的平均缓冲区更少;更易标准化;更适合应用。所以现在的主流网络基本上都可以看成是分组交换网络。单顶选择题[1]第一个分组交换网是<A>。A.ARPAnetB.X.25C.以太网D.Internet[2]在大多数网络中,数据链路层都是用请求重发已损坏了帧的办法来解决发送出错问题。如果一个帧被损坏的概率是p,而且确认信息不会丢失,则发送一帧的平均发送次数是<d>A.1+pB.I-pC.1/<1+p>D.1/<1-p>[3]物理层的电气特性规定的特性包括〔bA.接插件的形式B.信号的电压值C.电缆的长度D.各引脚的功能[4]网卡是完成<b>的功能。A.物理层B.数据链路层C.物理层和数据链路层D.数据链路层和网络层[5]通信子网不包括<d>。A.物理层B.数据链路层C.网络层D.传输层[6]当数据由端系统A传至端系统B时,不参与数据封装工作的是<a>。A.物理层B.数据链路层C.应用层D.表示层[7]RFC是〔。〔XX大学20XX试题A.因特网标准的形式B.一种网络协议C.一种网络文件格式D.一种网络技术解析：所有的因特网标准都是以RFC的形式在因特网上发表。RFC<reguestforcomments>的意思就是"请求评论"。所有的RFC文档都可以从因特网上免费下载。但应注意,并非所有的RFC文档都是因特网标准,只有一小部分RFC文档最后才能变成因特网标准。RFC接收到时间的先后从小到大编上序号<即RFCxxxx,这里xxxx是阿拉伯数字>。一个RFC文档更新后就使用一个新的编号,并在文档中指出原来老编号的RFC文档已成为陈旧的。简言之,RFC是因特网标准的形式。所以选项A为正确答案。答案:A[8]在OSI的七层模型中.工作在第三层以上的网间连接设备是<>。<华中科技大学20XX试题>A.集线器B.网关C.网桥D.中继器解析:集线器属于LAN与大型机以及LAN与WAN的互连。网挢工作于数据链数据通信系统中的基础设备,应用于OSI参考模型第一层。集线器的设计目标主要是优化网络布线结构.简化网络管理.主要功能是对接收到的信号进行再生整形放大.以扩大网络的传输距离.同时把所有节点集中在以它为中心的结点上。网关亦称网间协议转换器,工作于OSI/RM的传输层、会话层、表示层和应用层。网关不仅具有路由器的全部功能,同时还可以完成因操作系统差异引起的通信协议之间的转换。网关可用于LAN-LAN、路层。它要求两个互连网络在数据链路层以上采用相同或兼容的网络协议。中继器是最简单的网络互连设备,主要完成物理层的功能,负责在两个结点的物理层上按位传递信息,完成信号的复制、调整和放大功能,以此来延长网络的长度。它位于0SI参考模型中物理层。由此可知,网关工作于OSI/RM的传输层、会话层、表示层和应用层。所以选项B为正确答案。答案：B[9]在OSI七层结构模型中.处于数据链路层于传输层之间的是〔<华中科技大学20XX试题>物理层B.网络层C.会话层D.表示层解析：OSI/RM网络结构模型将计算机网络体系结构的通信协议规定为物理层、数据链路层、网络层、传输层、会话层、表示层、应用层.共七层。因此,网络层处于数据链路层与传输层之间。所以选项B为正确答案。答案:B[10]完成路径选择功能是在OSI参考模型的〔。〔华中科技大学20XX试题A.物理层B.数据链路层C.网络层D.传输层解析:物理层:主要是利用物理传输介质为数据链路层提供物理连接,以便透明地传递比特流。数据链路层:分为MAC和LLC,传送以帧为单位的数据,采用差错控制,流量控制方法。网络层:实现路由选择、拥塞控制和网络互连功能,使用TCP和UDP协议。传输层:是向用户提供可靠的端到端服务,透明地传送报文,使用TCP协议。由此可知,网络层具有路径选择的功能。所以选项C为正确答案。答案:C[11]在TCP/IP协议簇的层次中.解决计算机之间通信问题是在〔。〔华中科技大学20XX试题A.网络接口层B.网际层C.传输层D.应用层解析:TCP/IP协议族把整个协议分成四个层次:<1>网络接口层:负责接收P数据报,并把该数据报发送到相应的网络上。从理论上讲,该层不是TCP/IP协议的组成部分,但它是TCP/IP协议的基础,是各种网络与TCP/IP协议的接口。.<2>网络层<也叫网际层>:网络层解决了计算机到计算机通信的问题。因特网在该层的协议主要有网络互联协议IP、网间控制报文协议ICMP、地址解析协议ARP等。<3>传输层:传输层提供一个应用程序到另一个应用程序之间端到端的通信。因特网在该层的协议主要有传输控制协议TCP、用户数据报协议UDP等。<4>应用层:是TCP/IP协议的最高层,与0sI参考模型的上三层的功能类似。因特网在该层的协议主要有文件传输协议FTP、远程终端访问协议Telnet、简单邮件传输协议sMTP和域名服务协议DNS等。由此可知,网际层解决了计算机到计算机通信的问题。所以选项B为正确答案。答案:B[12]TCP/IP参考模型的网际层用于实现地址转换的协议有<>。A.ARPB.ICMPC.UDPD.TCP解析:ARP地址解析协议就是主机在发送帧前,将目标IP地址转换成目标MAC地址的过程。ARP协议的基本功能就是通过目标设备的IP地址,查询目标设备的MAC地址,以保证通信的顺利进行。所以选项A正确。ICMP协议是TCP/IP协议集中的一个子协议,属于网络层协议,主要用于在主机与路由器之间传递控制信息,包括报告错误、交换受限控制和状态信息等。因此,ICMP协议不具备地址转换功能,排除选项B。用户数据报协议<UDP>是ISO参考模型中一种无连接的传输层协议,提供面向事务的简单不可靠信息传送服务。UDP协议基本上是IP协议与上层协议的接口。由此可知,UDP不具备地址转换功能,排除选项C。TCP是面向连接的传输层协议,.它提供的是一种虚电路方式的运输服务。因此.排除选项D。简述面向连接服务于面向非连接服务的特点。解析:面向连接服务是电话系统服务模式的抽象。每一次完整的数据传输都必须经过建立连接、数据传输和终止连接三个过程。无连接服务是邮政系统服务模式的抽象。在无连接服务的情况下,两个实体之间的通信不需要先建立好一个连接,因此其下层的有关资源不需要事先进行预定保留。这些资源将在数据传输时动态地进行分配。答案:面向连接服务的特点是,在数据交换之前,必须先建立连接。当数据交换结束后,则应终止这个连接。在数据传输过程中,各数据报地址不需要携带目的地址,而是使用连接号。接收到的数据与发送方的数据在内容和顺序上是一致的。~无连接服务的特点,每个报文带有完整的目的地址,每个报文在系统中独立传送。无连接服务不能保证报文到达的先后顺序,先发送煦报文不一定先到。无连接服务不保证报文传输的可靠性。Chapter2Foracommunicationsessionbetweenapairofprocesses,whichprocessistheclientandwhichistheserver?Whatisthedifferencebetweennetworkarchitectureandapplicationarchitecture?Whatinformationisusedbyaprocessrunningononehosttoidentifyaprocessrunningonanotherhost?Supposeyouwantedtodoatransactionfromaremoteclienttoaserverasfastaspossible.WouldyouuseUDPorTCP?Why?Listthefourbroadclassesofservicesthatatransportprotocolcanprovide.Foreachoftheserviceclasses,indicateifeitherUDPorTCP<orboth>providessuchaservice.WhydoHTTP,FTP,SMTP,andPOP3runontopofTCPratherthanonUDP?Whatismeantbyahandshakingprotocol?SupposeAlice,withaWeb-basede-mailaccount<suchasHotmailorgmail>,sendsamessagetoBob,whoaccesseshismailfromhismailserverusingPOP3.DiscusshowthemessagegetsfromAlice'shosttoBob'shost.Besuretolisttheseriesofapplication-layerprotocolsthatareusedtomovethemessagebetweenthetwohosts.Fromauser'sperspective,whatisthedifferencebetweenthedownload-and-deletemodeandthedownload-and-keepmodeinPOP3?Isitpossibleforanorganization'sWebserverandmailservertohaveexactlythesamealiasforahostname<forexample,>?WhatwouldbethetypefortheRRthatcontainsthehostnameofthemailserver?WhyisitsaidthatFTPsendscontrolinformation"out-of-band"?Trueorfalse?a.AuserrequestsaWebpagethatconsistsofsometextandthreeimages.Forthispage,theclientwillsendonerequestmessageandreceivefourresponsemessages.b.TwodistinctWebpages<forexample,/students.html>canbesentoverthesamepersistentconnection.c.Withnonpersistentconnectionsbetweenbrowserandoriginserver,itispossibleforasingleTCPsegmenttocarrytwodistinctHTTPrequestmessages.d.TheDate:headerintheHTTPresponsemessageindicateswhentheobjectintheresponsewaslastmodified.SupposewithinyourWebbrowseryouclickonalinktoobtainaWebpage.TheIPaddressfortheassociatedURLisnotcachedinyourlocalhost,soaDNSlookupisnecessarytoobtaintheIPaddress.SupposethatnDNSserversarevisitedbeforeyourhostreceivestheIPaddressfromDNS;thesuccessivevisitsincuranRTTof.FurthersupposethattheWebpageassociatedwiththelinkcontainsexactlyoneobject,consistingofasmallamountofHTMLtext.LetdenotetheRTTbetweenthelocalhostandtheservercontainingtheobject.Assumingzerotransmissiontimeoftheobject,howmuchtimeelapsesfromwhentheclientclicksonthelinkuntiltheclientreceivestheobject?ReferringtoProblemP7,supposetheHTMLfilereferencesthreeverysmallobjectsonthesameserver.Neglectingtransmissiontimes,howmuchtimeelapseswitha.Non-persistentHTTPwithnoparallelTCPconnections?b.Non-persistentHTTPwithparallelconnections?c.PersistentHTIP?通知第二次试验时间：11月8日.上午10:00~12:00地点：九教北401试验内容：《实验指南》实验四、实验五、实验六；参考GBN源代码资料存储在：邮箱：密码：123456abcdChapter3〔教材R3DescribewhyanapplicationdevelopermightchoosetorunanapplicationoverUDPratherthanTCP.A:AnapplicationdevelopermaynotwantitsapplicationtouseTCP’scongestioncontrol,whichcanthrottletheapplication’ssendingrateattimesofcongestion.Often,designersofIPtelephonyandIPvideoconferenceapplicationschoosetoruntheirapplicationsoverUDPbecausetheywanttoavoidTCP’scongestioncontrol.Also,someapplicationsdonotneedthereliabledatatransferprovidedbyTCP.〔教材R4WhyisitthatvoiceandvideotrafficisoftensentoverTCPratherthanUDPintoday'sInternet.<Hint:TheanswerwearelookingforhasnothingtodowithTCP'scongestion-controlmechanism.>A:SincemostfirewallsareconfiguredtoblockUDPtraffic,usingTCPforvideoandvoicetrafficletsthetrafficthoughthefirewalls.〔教材R5IsitpossibleforanapplicationtoenjoyreliabledatatransferevenwhentheapplicationrunsoverUDP?Ifso,how?A:Yes.Theapplicationdevelopercanputreliabledatatransferintotheapplicationlayerprotocol.Thiswouldrequireasignificantamountofworkanddebugging,however.〔教材R6ConsideraTCPconnectionbetweenHostAandHostB.SupposethattheTCPsegmentstravelingfromHostAtoHostBhavesourceportnumberXanddestinationportnumbery.WhatarethesourceanddestinationportnumbersforthesegmentstravelingfromHostBtoHostA?A:Sourceportnumberyanddestinationportnumberx.〔教材R7SupposeaprocessinHostChasaUDPsocketwithportnumber6789.SupposebothHostAandHostBeachsendaUDPsegmenttoHostCwithdestinationportnumber6789.WillbothofthesesegmentsbedirectedtothesamesocketatHostC?Ifso,howwilltheprocessatHostCknowthatthesetwosegmentsoriginatedfromtwodifferenthosts?A:Yes,bothsegmentswillbedirectedtothesamesocket.Foreachreceivedsegment,atthesocketinterface,theoperatingsystemwillprovidetheprocesswiththeIPaddressestodeterminetheoriginsoftheindividualsegments.〔教材R9Inourrdtprotocols,whydidweneedtointroducesequencenumbers?A:Sequencenumbersarerequiredforareceivertofindoutwhetheranarrivingpacketcontainsnewdataorisaretransmission.〔教材10Inourrdtprotocols,whydidweneedtointroducetimers?A:Tohandlelossesinthechannel.IftheACKforatransmittedpacketisnotreceivedwithinthedurationofthetimerforthepacket,thepacket<oritsACKorNACK>isassumedtohavebeenlost.Hence,thepacketisretransmitted.〔教材R14SupposeHostAsendstwoTCPsegmentsbacktobacktoHostBoveraTCPconnection.Thefirstsegmenthassequencenumber90;thesecondhassequencenumber110.a.Howmuchdataisinthefirstsegment?b.SupposethatthefirstsegmentislostbutthesecondsegmentarrivesatB.IntheacknowledgmentthatHostBsendstoHostA,whatwillbetheacknowledgmentnumber?A:a>20bytesb>acknumber=90〔教材R15Trueorfalse?a.ThesizeoftheTCPRcvWindowneverchangesthroughoutthedurationoftheconnection.b.SupposeHostAissendingHostBalargefileoveraTCPconnection.ThenumberofunacknowledgedbytesthatAsendscannotexceedthesizeofthereceivebuffer.c.HostAissendingHostBalargefileoveraTCPconnection.AssumeHostBhasnodatatosendHostA.HostBwillnotsendacknowledgmentstoHostAbecauseHostBcannotpiggybacktheacknowledgmensondata.d.TheTCPsegmenthasafieldinitsheaderforRcvWindow.e.SupposeHostAissendingalargefiletoHostBoveraTCPconnection.Ifthesequencenumberforasegmentofthisconnectionism,thenthesequencenumberforthesubsequentsegmentwillnecessarilybem+1.f.SupposethatthelastSampleRTTinaTCPconnectionisequalto1sec.ThecurrentvalueofTimeoutrntervalfortheconnectionwillnecessarilybe>1sec.g.SupposeHostAsendsonesegmentwithsequencenumber38and4bytesofdataoveraTCPconnectiontoHostB.Inthissamesegmenttheacknowledgmentnumberisnecessarily42.A:a>false;b>false;needconsiderretransmitpacketc>false;d>true;e>false;f>false;g>false;〔教材P1SupposeclientAinitiatesaTelnetsessionwithServerS.Ataboutthesametime,clientBalsoinitiatesaTelnetsessionwithServerS.Providepossiblesourceanddestinationportnumbersfora.ThesegmentssentfromAtoS.b.ThesegmentssentfromBtoS.c.ThesegmentssentfromStoA.d.ThesegmentssentfromStoB.e.IfAandBaredifferenthosts,isitpossiblethatthesourceportnumberinthesegmentsfromAtoSisthesameasthatfromBtoS?f.Howaboutiftheyarethesamehost?A:〔教材P2ConsiderFigure3.5.Whatarethesourceanddestinationportvaluesinthesegmentsflowingfromtheserverbacktotheclients'processes?WhataretheIPaddressesinthenetwork-layerdatagramcarryingthetransport-layersegments?A:SupposetheIPaddressesofthehostsA,B,andCarea,b,c,respectively.<Notethata,b,caredistinct.>TohostA:Sourceport=80,sourceIPaddress=b,destport=26145,destIPaddress=aTohostC,leftprocess:Sourceport=80,sourceIPaddress=b,destport=7532,destIPaddress=cTohostC,rightprocess:Sourceport=80,sourceIPaddress=b,destport=26145,destIPaddress=c〔教材P19Answertrueorfalsetothefollowingquestionsandbrieflyjustifyyouranswer:a.WiththeSRprotocol,itispossibleforthesendertoreceiveanACKforapacketthatfallsoutsideofitscurrentwindowb.WithGBN,itispossibleforthesendertoreceiveanACKforapacketthatfallsoutsideofitscurrentwindow.c.Thealternating-bitprotocolisthesameastheSRprotocolwithasenderandreceiverwindowsizeof1.d.Thealternating-bitprotocolisthesameastheGBNprotocolwithasenderandreceiverwindowsizeof1.Answer:True.Supposethesenderhasawindowsizeof3andsendspackets1,2,3att0.Att1<t1>t0>thereceiverACKS1,2,3.Att2<t2>t1>thesendertimesoutandresends1,2,3.Att3thereceiverreceivestheduplicatesandre-acknowledges1,2,3.Att4thesenderreceivestheACKsthatthereceiversentatt1andadvancesitswindowto4,5,6.Att5thesenderreceivestheACKs1,2,3thereceiversentatt2.TheseACKsareoutsideitswindow.True.Byessentiallythesamescenarioasin<a>.True.True.Notethatwithawindowsizeof1,SR,GBN,andthealternatingbitprotocolarefunctionallyequivalent.Thewindowsizeof1precludesthepossibilityofout-of-orderpackets<withinthewindow>.AcumulativeACKisjustanordinaryACKinthissituation,sinceitcanonlyrefertothesinglepacketwithinthewindow.〔教材P24HostAandBarecommunicatingoveraTCPconnection,andHostBhasalreadyreceivedfromAallbytesupthroughbyte358.SupposeHostAthensendstwosegmentstoHostBback-to-back.Thefirstandsecondsegmentscontain50and80bytesofdata,respectively.Inthefirstsegment,thesequencenumberis359,thesourceportnumberis1028,andthedestinationportnumberis80.HostBsendsanacknowledgementwheneveritreceivesasegmentfromHostA.a.InthesecondsegmentsentfromHostAtoB,whatarethesequencenumber,sourceportnumber,anddestinationportnumber?b.Ifthefirstsegmentarrivesbeforethesecondsegment,intheacknowledgementofthefirstarrivingsegment,whatistheacknowledgmentnumber,thesourceportnumber,andthedestinationportnumber?c.Ifthesecondsegmentarrivesbeforethefirstsegment,intheacknowledgementofthefirstarrivingsegment,whatistheacknowledgmentnumber?d.SupposethetwosegmentssentbyAarriveinorderatB.Thefirstacknowledgementislostandthesecondacknowledgementarrivesafterthefirsttimeoutinterval.Drawatimingdiagram,showingthesesegmentsandallothersegmentsandacknowledgementssent.<Assumethereisnoadditionalpacketloss.>Foreachsegmentinyourfigure,providethesequencenumberandthenumberofbytesofdata;foreachacknowledgementthatyouadd,providetheacknowledgementnumber.Answer:InthesecondsegmentfromHostAtoB,thesequencenumberis409,sourceportnumberis1028anddestinationportnumberis80.Ifthefirstsegmentarrivesbeforethesecond,intheacknowledgementofthefirstarrivingsegment,theacknowledgementnumberis409,thesourceportnumberis80andthedestinationportnumberis1028.Ifthesecondsegmentarrivesbeforethefirstsegment,intheacknowledgementofthefirstarrivingsegment,theacknowledgementnumberis359,indicatingthatitisstillwaitingforbytes359andonwards.〔教材P25HostAandBaredirectlyconnectedwitha200Mbpslink.ThereisoneTCPconnectionbetweenthetwohosts,andHostAissendingtoHostBanenormousfileoverthisconnection.HostAcansendapplicationdataintothelinkat100MbpsbutHostBcanreadoutofitsTCPreceivebufferatamaximumrateof50Mbps.DescribetheeffectofTCPflowcontrol.Answer:HostAsendsdataintothereceivebufferfasterthanHostBcanremovedatafromthebuffer.Thereceivebufferfillsupatarateofroughly50Mbps.Whenthebufferisfull,HostBsignalstoHostAtostopsendingdatabysettingRcvWindow=0.HostAthenstopssendinguntilitreceivesaTCPsegmentwithRcvWindow>0.HostAwillthusrepeatedlystopandstartsendingasafunctionoftheRcvWindowvaluesitreceivesfromHostB.Onaverage,thelong-termrateatwhichHostAsendsdatatoHostBaspartofthisconnectionisnomorethan50Mbps.〔教材P30InSection,wesawthatTCPwaitsuntilithasreceivedthreeduplicateACKsbeforeperformingafastretransmit.WhydoyouthinktheTCPdesignerschosenottoperformafastretransmitafterthefirstduplicateACKforasegmentisreceived?Answer:Supposepacketsn,n+1,andn+2aresent,andthatpacketnisreceivedandACKed.Ifpacketsn+1andn+2arereorderedalongtheend-to-end-path<i.e.,arereceivedintheordern+2,n+1>thenthereceiptofpacketn+2willgenerateaduplicateackfornandwouldtriggeraretransmissionunderapolicyofwaitingonlyforsecondduplicateACKforretransmission.BywaitingforatripleduplicateACK,itmustbethecasethattwo???threepacketsafterpacketnarecorrectlyreceived,whilen+1wasnotreceived.ThedesignersofthetripleduplicateACKschemeprobablyfeltthatwaitingfortwopackets<ratherthan1>wastherighttradeoffbetweentriggeringaquickretransmissionwhenneeded,butnotretransmittingprematurelyinthefaceofpacketreordering.〔教材P34ConsiderthefollowingplotofTCPwindowsizeasafunctionoftime.AssumingTCPRenoistheprotocolexperiencingthebehaviorshownabove,answerthefollowingquestions.Inallcases,youshouldprovideashortdiscussionjustifyingyouranswer.a.IdentifytheintervalsoftimewhenTCPslowstartisoperating.b.IdentifytheintervalsoftimewhenTCPcongestionavoidanceisoperating.c.Afterthe16thtransmissionround,issegmentlossdetectedbyatripleduplicateACKorbyatimeout?d.Afterthe22ndtransmissionround,issegmentlossdetectedbyatripleduplicateACKorbyatimeout?e.WhatistheinitialvalueofThresholdatthefirsttransmissionround?f.WhatisthevalueofThresholdatthe18thtransmissionround?g.WhatisthevalueofThresholdatthe24thtransmissionround?h.Duringwhattransmissionroundisthe70thsegmentsent?i.Assumingapacketlossisdetectedafterthe26throundbythereceiptofatripleduplicateACK,whatwillbethevaluesofthecongestionwindowsizeandofThreshold?Answer:TCPslowstartisoperatingintheintervals[1,6]and[23,26]TCPcongestionadvoidanceisoperatingintheintervals[6,16]and[17,22]Afterthe16thtransmissionround,packetlossisrecognizedbyatripleduplicateACK.Iftherewasatimeout,thecongestionwindowsizewouldhavedroppedto1.Afterthe22ndtransmissionround,segmentlossisdetectedduetotimeout,andhencethecongestionwindowsizeissetto1.Thethresholdisinitially32,sinceitisatthiswindowsizethatslowtartstopsandcongestionavoidancebegins.Thethresholdissettohalfthevalueofthecongestionwindowwhenpacketlossisdetected.Whenlossisdetectedduringtransmissionround16,thecongestionwindowssizeis42.Hencethethresholdis21duringthe18thtransmissionround.Thethresholdissettohalfthevalueofthecongestionwindowwhenpacketlossisdetected.Whentimerouteventoccurredduringtransmissionround23,thecongestionwindowssizeis26.Hencethethresholdis13duringthe24thtransmissionround.Duringthe1sttransmissionround,packet1issent;packet2-3aresentinthe2ndtransmissionround;packets4-7aresentinthe3rdtransmissionround;packets8-15aresentinthe4thtransmissionround;packets15-31aresentinthe5thtransmissionround;packets32-63aresentinthe6thtransmissionround;packets64–96aresentinthe7thtransmissionround.Thuspacket70issentinthe7thtransmissionround.Thecongestionwindowandthresholdwillbesettohalfthecurrentvalueofthecongestionwindow<8>whenthelossoccurred.Thusthenewvaluesofthethresholdandwindowwillbe4.〔教材P38HostAissendinganenormousfiletoHostBoveraTCPconnection.Overthisconnectionthereisneveranypacketlossandthetimersneverexpire.DenotethetransmissionrateofthelinkconnectingHostAtotheInternetbyRbps.SupposethattheprocessinHostAiscapableofsendingdataintoitsTCPsocketatarateSbps,whereS=10*R.FurthersupposethattheTCPreceivebufferislargeenoughtoholdtheentirefile,andthesendbuffercanholdonlyonepercentofthefile.WhatwouldpreventtheprocessinHostAfromcontinuouslypassingdatatoitsTCPsocketatrateSbps?TCPflowcontrol?TCPcongestioncontrol?Orsomethingelse?Elaborate.Answer:Inthisproblem,thereisnodangerinoverflowingthereceiversincethereceiver’sreceivebuffercanholdtheentirefile.Also,becausethereisnolossandacknowledgementsarereturnedbeforetimersexpire,TCPcongestioncontroldoesnotthrottlethesender.However,theprocessinhostAwillnotcontinuouslypassdatatothesocketbecausethesendbufferwillquicklyfillup.Oncethesendbufferbecomesfull,theprocesswillpassdataatanaveragerateorR<<S.在TCP协议中,为了使通信不致发生混乱,引入了所谓套接字的概念,这里,套接字由<>和IP地址两部分组成。A。端口号B.域名C。接口D。物理地址解析:端口号和IP地址合起来,称为套接字,套接字可以在全网范围内唯一标识一个端口。在TCP协议中,一条连接两端的套接字就可以唯一标识该连接了。所以选项A为正确答案。答案:A面向连接的传输有三个过程:连接建立、<>和连接释放。A．连接请求B。连接应答C。数据传输D。数据共享解析:面向连接服务具有连接建立、数据传输和连接释放这三个阶段。所以选项C为正确答案。答案:C试述UDP和TCP协议的主要特点及它们的使用场合。<华中科技大学20XX试题>解析:用户数据报协议是对IP协议组的扩充,它增加了一种机制,发送方使用这种机制可以区分一台计算机上的多个接收者。每个UDP报文除了包含某用户进程发送数据外,还有报文目的端口的编号和报文源端口的编号,从而使UDP的这种扩充,在两个用户进程之间的递送数据报成为可能。TCP提供的是一种可靠的数据流服务。当传送受差错干扰的数据,或基础网络故障,或网络负荷太重而使网络基本传输系统<无连接报文递交系统>不能正常工作时,就需要通过TCP这样的协议来保证通信的可靠。答案:UDP是一个简单的面向数据报的传输层协议。应用进程的每个输出操作都产生一个UDP数据报,并组装成一份待发送的IP数据报中发送。UDP提供不可靠、无连接的数据报服务,它把应用程序传给IP层的数据发送出去,但是并不保证它们能到达目的地。因此,uDP通常用于不要求可靠传输的场合,另外也常用于客户机/服务器模式中。TCP协议被用来在一个不可靠的互联网中为应用程序提供可靠的端点间的字节流服务。所有TCP连接都是全双工和点对点的,因而TCP不支持广播和组播的功能。TCP实体间以"段"为单位进行数据交换。为实现可靠的数据传输服务,TCP提供了对段的检错、应答、重传和排序的功能,提供了可靠地建立连接和拆除连接的方法,还提供了流量控制和阻塞控制的机制。TCP适用于传输大量重要数据的场合。在使用TCP协议传送数据时,如果有一个确认报文段丢失了,也不一定会引起对方数据的重传。试说明为什么?解析:本题考查的是TCP重传机制。TCP连接的一个重要的特性就是为上层服务提供了一个可靠的数据流。由于TCP是建立在不可靠的IP层的基础之上的,因此就必然涉及报文丢失的问题,这样,报文的重传就成了保证数据可靠到达的一个重要机制。这方面TCP采取了超时重传的策略,对每个TCP连接都维护一个计时器,每发送一个报文:就设置一次计时器,只要计时器设置的重传时间已到但仍然没有收到相应的确认信息,就重传这一报文。答案:对方还未来得及重传,就收到了对更高序号的确认,相当于对连同被丢失确认的报文段一并确认。在连续ARQ协议中,设编号用3位,而发送窗口=8,试找出一种情况,使得在此