川大林锋计网第三次作业任务.docx

川大林锋计网第三次作业任务.docx

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川大林锋计网第三次作业任务

ReviewQuestions:

3.ConsideraTCPconnectionbetweenHostAandHostB.SupposethattheTCPsegmentstravelingfromHostAtoHostBhavesourceportnumberxanddestinationportnumbery.WhatarethesourceanddestinationportnumbersforthesegmentstravelingfromHostBtoHostA?

Thesourceportnumberisy,thedestinationportnumberisx.

5.WhyisitthatvoiceandvideotrafficisoftensentoverTCPratherthanUDPintoday’sInternet.（Hint:

TheanswerwearelookingforhasnothingtodowithTCP’scongestion-controlmechanism.）

因为UDP虽然传输速度快，不需要建立对等连接，但是容易丢包，而且数据包顺序容易混乱；TCP传输更稳定，文件没有丢失或者损失。

6.IsitpossibleforanapplicationtoenjoyreliabledatatransferevenwhentheapplicationrunsoverUDP?

Ifso,how?

是的，只需要在UDP的协议上添加验证的数据，比如给每个包传输时加个头，并且设置返回值，还有很多其他方法在此不一一列举了。

7.SupposeaprocessinHostChasaUDPsocketwithportnumber6789.SupposebothHostAandHostBeachsendsaUDPsegmenttoHostCwithdestinationportnumber6789.WillbothofthesesegmentsbedirectedtothesamesocketatHostC?

Ifso,howwilltheprocessatHostCknowthatthesetwosegmentsoriginatedfromtwodifferenthosts?

正解，两个部分将针对同一接口。

在套接字接口，对于每个收到的片段，操作系统将提供过程与IP地址确定各段的起源。

8.SupposethataWebserverrunsinHostConport80.SupposethisWebserverusespersistentconnections,andiscurrentlyreceivingrequestsfromtwodifferentHosts,AandB.ArealloftherequestsbeingsentthroughthesamesocketatHostC?

Iftheyarebeingpassedthroughdifferentsockets,dobothofthesocketshaveport80?

Discussandexplain.

对于每个坚持的连接，web服务器会创建一个连接套接字，每个连接套接字由4个数组定义。

当主机C收到一个请求的时候，它会检查数据报和段中的4个区域，以此决定用哪个区域来接收TCP段。

因此，A和B传输的套接字是不一样的。

每一个套接字的标识符都使用80端口作为目的的，但是不同的资源有不同的IP地址。

和UDP不同，当传输层传输一个TCP段至应用层的时候，它并不需要明确指定IP地址，因为套接字标识符会帮它指定。

14.Trueorfalse?

（F）a.HostAissendingHostBalargefileoveraTCPconnection.AssumeHostBhasnodatatosendHostA.HostBwillnotsendacknowledgmentstoHostAbecauseHostBcannotpiggybacktheacknowledgmentsondata.

（F）b.ThesizeoftheTCPrwndneverchangesthroughouttheduration（持续时间）oftheconnection.

（T）c.SupposeHostAissendingHostBalargefileoveraTCPconnection.ThenumberofunacknowledgedbytesthatAsendscannotexceedthesizeofthereceivebuffer.

（F）d.SupposeHostAissendingalargefiletoHostBoveraTCPconnection.Ifthesequencenumberforasegmentofthisconnectionism,thenthesequencenumberforthesubsequentsegmentwillnecessarilybem+1.

（T）e.TheTCPsegmenthasafieldinitsheaderforrwnd.

（F）f.SupposethatthelastSampleRTTinaTCPconnectionisequalto1sec.ThecurrentvalueofTimeoutIntervalfortheconnectionwillnecessarilybe≥1sec.ComputerNetworkingfile:

///C|/Users/liwin/Desktop/新建文件夹/dummy_split_025.html[2012/4/1512:

46:

20]

（）g.SupposeHostAsendsonesegmentwithsequencenumber38and4bytesofdataoveraTCPconnectiontoHostB.Inthissamesegmenttheacknowledgmentnumberisnecessarily42.

15.SupposeHostAsendstwoTCPsegmentsbacktobacktoHostBoveraTCPconnection.Thefirstsegmenthassequencenumber90;thesecondhassequencenumber110.

a.Howmuchdataisinthefirstsegment?

110-90=20

b.SupposethatthefirstsegmentislostbutthesecondsegmentarrivesatB.IntheacknowledgmentthatHostBsendstoHostA,whatwillbetheacknowledgmentnumber?

90

17.SupposetwoTCPconnectionsarepresentoversomebottlenecklinkofrateRbps.Bothconnectionshaveahugefiletosend（inthesamedirectionoverthebottlenecklink）.Thetransmissionsofthefilesstartatthesametime.WhattransmissionratewouldTCPliketogivetoeachoftheconnections?

R/2

Problems:

1.SupposeClientAinitiatesaTelnetsessionwithServerS.Ataboutthesametime,ClientBalsoinitiatesaTelnetsessionwithServerS.Providepossiblesourceanddestinationportnumbersfor

a.ThesegmentssentfromAtoS.

Thesourceportnumberis467,destinationnumberis23.

b.ThesegmentssentfromBtoS.

Thesourceportis513,thedestinationnumberis23.

c.ThesegmentssentfromStoA.

Thesourceportis23,thedestinationnumberis467.

d.ThesegmentssentfromStoB.

Thesourceportis23,thedestinationnumberis513.

e.IfAandBaredifferenthosts,isitpossiblethatthesourceportnumberinthesegmentsfromAtoSisthesameasthatfromBtoS?

Yes.

f.Howaboutiftheyarethesamehost?

No.

2.ConsiderFigure3.5.Whatarethesourcesanddestinationportvaluesinthesegmentsflowingfromtheserverbacktotheclients’processes?

在段中，资源和目的端口从服务器返回用户进程的价值是什么？

WhataretheIPaddressesinthenetwork-layerdatagramscarryingthetransport-layersegments?

网络层数据报中携带的传输层段里的IP地址是什么？

<1>返回值包含了客户端和服务器的IP地址以及port码，还有用户索求的内容。

<2>网络层数据报中携带的IP地址包含目的端口的IP地址、port码、传输层添加的头（message）、网络层添加的段（segment）。

5.a.Supposeyouhavethefollowing2bytes:

01011100and01010110.Whatisthe1scomplementofthesumofthese2bytes?

01011100+01010110=10110010

取反=0110010

b.Supposeyouhavethefollowing2bytes:

11011010and00110110.Whatisthe1scomplementofthesumofthese2bytes?

11011010+00110110=100010000

去首位补至末位=00010001

取反=11101110

c.Forthebytesinpart（a）,giveanexamplewhereonebitisflippedineachofthe2bytesandyetthe1scomplementdoesn’tchange.

18.ConsidertheGBNprotocolwithasenderwindowsizeof3andasequencenumberrangeof1,024.Supposethatattimet,thenextin-orderpacketthatthereceiverisexpectinghasasequencenumberofk.Assumethatthemediumdoesnotreordermessages.Answerthefollowingquestions:

a.Whatarethepossiblesetsofsequencenumbersinsidethesender’swindowattimet?

Justifyyouranswer.

因为窗口大小为3，接收方期待的下一列序号为k，所以t时刻发送窗口中序列号应为k,k+1,k+2。

b.WhatareallpossiblevaluesoftheACKfieldinallpossiblemessagescurrentlypropagatingbacktothesenderattimet?

Justifyyouranswer.

由于序列号为0-1024，所以在t时刻，所有可能的ack序号应为0≤ackNumber≤1024

19.Answertrueorfalsetothefollowingquestionsandbrieflyjustifyyouranswer:

（T）a.WiththeSR（selectiverepeat）protocol,itispossibleforthesendertoreceiveanACKforapacketthatfallsoutsideofitscurrentwindow.

（T）b.WithGBN,itispossibleforthesendertoreceiveanACKforapacketthatfallsoutsideofitscurrentwindow.

（T）c.Thealternating-bitprotocolisthesameastheSRprotocolwithasenderandreceiverwindowsizeof1.

（T）d.Thealternating-bitprotocol（rdt3.0）isthesameastheGBNprotocolwithasenderandreceiverwindowsizeComputerNetworkingfile:

///C|/Users/liwin/Desktop/新建文件夹/dummy_split_025.html[2012/4/1512:

46:

20]of1.

21.ConsidertheGBNandSRprotocols.Supposethesequencenumberspaceisofsizek.WhatisthelargestallowablesenderwindowthatwillavoidtheoccurrenceofproblemssuchasthatinFigure3.27foreachoftheseprotocols?

首先我们必须保证发送窗口和接受窗口中序列号都不重复。

假设窗口的序列号为a，那么序列号k必须保持k≥2a，才能使窗口中的序列号不重复。

22.WehavesaidthatanapplicationmaychooseUDPforatransportprotocolbecauseUDPoffersfinerapplicationcontrol（thanTCP）ofwhatdataissentinasegmentandwhen.

a.Whydoesanapplicationhavemorecontrolofwhatdataissentinasegment?

看了答案，不理解为什么答案只说明TCP和UDP传输的特点。

我的理解是应该说明传输层对TCP和UDP的控制和转换问题，但是我不会，坐等老师上课解，另附标准答案，希望助教大大不要怪我照抄。

Considersending anapplication messageoveratransportprotocol.WithTCP,the application writes data totheconnection‘ssendbufferandTCPwillgrabbyteswithoutnecessarilyputtingasinglemessageintheTCPsegment;TCPmayput more orlessthanasinglemessageinasegment.UDP,ontheotherhand,encapsulatesinasegmentwhateverthe application givesit;sothat,ifthe application givesUDP anapplication message,thismessagewillbethepayload of theUDPsegment.Thus,withtheUDP, anapplication has more control of what data issentinasegment. 

b.Whydoesanapplicationhavemorecontrolonwhenthesegmentissent?

WithTCP,duetoflow control andcongestion control,theremaybesignificantdelayfromthetimewhen an application writes data toitssendbufferuntilwhenthe data isgiventothenetworklayer.UDP does not have delaysduetoflowcontrol andcongestion control.

23.ConsidertransferringanenormousfileofLbytesfromHostAtoHostB.AssumeanMSS（managementsupportsystem）of1460bytes.

a.WhatisthemaximumvalueofLsuchthatTCPsequencenumbersarenotexhausted?

RecallthattheTCPsequencenumberfieldhas4bytes.

因为TCP序号范围有4bytes，则L最大为2^32bytes

b.FortheLyouobtainin（a）,findhowlongittakestotransmitthefile.Assumethatatotalof66bytesoftransport,network,anddata-linkheaderareaddedtoeachsegmentbeforetheresultingpacketissentoutovera155Mbpslink.IgnoreflowcontrolandcongestioncontrolsoAcanpumpoutthesegmentsbacktobackandcontinuously.

由于不需要考虑流控和拥塞控制，传输速度为155Mbps，每段加66bytes大小的头，首先计算一共分多少段：

2^32bytes/1460bytes=2941758段

每段加一个头，则头大小的和为：

2941758X66bytes=194156028bytes，加起来得出总共需传输194156028bytes+2^32bytes=4489123324bytes=35912986592bits的数据。

用10Mbps的速度传输则时间为35912986592bits/Mbps=3591s=59.85min。

24.HostAandBarecommunicatingoveraTCPconnection,andHostBhasalreadyreceivedfromAallbytesupthroughbyte126.SupposeHostAthensendstwosegmentstoHostBback-to-back.Thefirstandsecondsegmentscontain70and50bytesofdata,respectively.Inthefirstsegment,thesequencenumberis127,thesourceportnumberis302,andthedestinationportnumberis80.HostBsendsanacknowledgementwheneveritreceivesasegmentfromHostA.

a.InthesecondsegmentsentfromHostAtoB,whatarethesequencenumber,sourceportnumber,anddestinationportnumber?

Thesequencenumberis247,thesourcenumberis302,thedestinationnumberis80.

b.Ifthefirstsegmentarrivesbeforethesecon