计网第五次作业.docx
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计网第五次作业.docx
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计网第五次作业
ReviewQuestions:
1.Whataresomeofthepossibleservicesthatalink-layerprotocolcanoffertothenetworklayer?
Whichoftheselink-layerserviceshavecorrespondingservicesinIP?
InTCP?
链路层协议提供给网络层的服务有哪些?
哪些是给IP的?
哪些是给TCP的?
Linkaccess,framing,reliabledeliverybetweenadjacentnodes,flowcontrol,errordetection,errorcorrection,half-duplexandfull-duplex.
InIP:
framing,errordetection.
InTCP:
framing,reliabledeliverybetweenadjacentnodes,errordetection,half-duplexandfull-duplex.
2.IfallthelinksintheInternetweretoprovidereliabledeliveryservice,wouldtheTCPreliabledeliveryserviceberedundant?
Whyorwhynot?
不会多余,因为TCP保证的是传输层的数据传送,而link提供稳定可靠传输保证链路层的传输稳定,二者不完全重叠,所以TCP可靠传输也不会多余。
3.InSection5.3,welistedfourdesirablecharacteristicsofabroadcastchannel.WhichofthesecharacteristicsdoesslottedALOHAhave?
Whichofthesecharacteristicsdoestokenpassinghave?
4.SupposetwonodesstarttotransmitatthesametimeapacketoflengthLoverabroadcastchannelofrateR.Denotethepropagationdelaybetweenthetwonodesasdprop.Willtherebeacollisionifdprop Whyorwhynot? 因为L/R=包传递的时间,如果dprop 7.SupposenodesA,B,andCeachattachtothesamebroadcastLAN(throughtheiradapters).IfAsendsthousandsofIPdatagramstoBwitheachencapsulatingframeaddressedtotheMACaddressofB,willC’sadapterprocesstheseframes? Ifso,willC’sadapterpasstheIPdatagramsintheseframestothenetworklayerC? HowwouldyouranswerschangeifAsendsframeswiththeMACbroadcastaddress? 不会,C会拆封帧从而读取报头的MAC,因为每一个host的MAC都唯一,C读取到数据报中的MAC和自己的不一样就不会继续拆封数据报,不会投递给C。 当使用LAN口广播地址的时候,C的适配器就会拆封帧,向C传递数据。 8.HowbigistheMACaddressspace? TheIPv4addressspace? TheIPv6addressspace? 2^48,2^32,2^128. 9.WhyisanARPquerysentwithinabroadcastframe? WhyisanARPresponsesentwithinaframewithaspecificdestinationMACaddress? 因为新加入网络的主机是不知道路由器的IP的,自己也没有IP,所以只能广播才能得到IP。 因为每个主机的MAC地址都是唯一的,而ARP建立转发表的时候会带上MAC地址。 12.InCSMA/CD,afterthefifthcollision,whatistheprobabilitythatanodechoosesK=4? TheresultK=4correspondstoadelayofhowmanysecondsona10MbpsEthernet? 可能,因为第五次K的取值范围是0-(2^5-1)即0-31。 Bittime=1bit/R=1bit/10Mbps=1msec,K=4,waittime=4*512*1msec=2048msec。 Problems: 1.Supposetheinformationcontentofapacketisthebitpattern1110110010001010andanevenparityschemeisbeingused.Whatwouldthevalueofthefieldcontainingtheparitybitsbeforthecaseofatwo-dimensionalparityscheme? Youranswershouldbesuchthataminimum-lengthchecksumfieldisused.假设一个数据包的信息含量是XXX,使用偶校验方案。 采用二位奇偶校验方案的字段包含的奇偶校验位的字段的值是多少? 答案要使用最小长度校验。 10100 10100 10100 10111 00011 2.Supposetheinformationportionofapacket(DinFigure5.4)contains10bytesconsistingofthe8-bitunsignedbinaryASCIIrepresentationoftheintegers0through9.ComputetheInternetchecksumforthisdata.假设一个包的信息的一部分包括10bytes组成的8-bit无符号二进制码表示的整数0-9,计算该数据的网络校验。 算校验码先把0-9加起来 0000000000000001 0000001000000011 0000010000000101 0000011000000111 0000100000001001 0001010000011001 →→→→取反可得 校验码为1110101111100110 3.Considerthepreviousproblem,butinsteadofcontainingthebinaryofthenumbers0through9supposethese10bytescontain.ComputetheInternetchecksumforthisdata. a.thebinaryrepresentationofthenumbers1through10. 0000000100000010 0000001100000100 0000010100000110 0000011100001000 0000100100010100 0001101100011110 →→→→取反可得 1110010011100001 b.theASCIIrepresentationofthelettersAthroughJ(uppercase). 0100000101000010 0100001101000100 0100010101000110 0100011101001000 0100100101001010 0101100001011111 →→→→取反可得 1010011110100000 c.theASCIIrepresentationofthelettersathroughj(lowercase).小写字母表示 0110000101100010 0110001101100100 0110010101100110 0110011101101000 0110100101101010 1111100111111101 →→→→取反可得 0000011000000010 6.Considerthepreviousproblem,butsupposethatDhasthevalue a.10010001. b.10100011. c.01010101. 前一题题目: Considerthe7-bitgenerator,G=10011,andsupposethatDhasthevalue1010101010.WhatisthevalueofR? 求余而已,记住不要做减法而是做与运算就好。 a.R=001 b.R=101 c.R=101 12.ConsiderthreeLANsinterconnectedbytworouters,asshowninFigure5.38. a.Redrawthediagramtoincludeadapters.重新画图 b.AssignIPaddressestoalloftheinterfaces.ForSubnet1useaddressesoftheform111.111.111.xxx;forSubnet2usesaddressesoftheform122.122.122.xxx;andforSubnet3useaddressesoftheform133.133.133.xxx.所有的接口分配IP地址。 c.AssignMACaddressestoalloftheadapters. a.b.c如图 d.ConsidersendinganIPdatagramfromHostAtoHostF.SupposealloftheARPtablesareuptodate.Enumerateallthesteps,asdoneforthesingle-routerexampleinSection5.4.2. 1.hostA发送一个数据报,通过转发表查询F的IP,向路由器1发送,其中destinationIP为133.133.133.12,MAC未知,sourceIP为111.111.111.12,sourceMAC为aa-aa-aa-aa-aa-aa。 2.适配器更改destination的IP为111.111.111.12,MAC地址变为gg-gg-gg-gg-gg-gg 3.路由器1发现目标IP和MAC不属于子网1中任何host,属于子网3(图中忘了画了,意会一下)。 于是根据转发表向路由器2进行转发。 Destination的IP为122.122.122.20,MAC为ii-ii-ii-ii-ii-ii。 4.路由器2收到了数据报,发现hostF在自己的子网内,于是修改destination的IP为133.133.133.12,MAC地址为ff-ff-ff-ff-ff-ff。 修改sourceIP为133.133.133.20,MAC为jj-jj-jj-jj-jj-jj,然后向F发送数据报。 5.F收到来自A的数据报。 e.Repeat(d),nowassumingthattheARPtableinthesendinghostisempty(andtheothertablesareuptodate). 发送方的ARP表为空,首先需要建立ARP表 1.hostA发送一个广播,destinationIP是255.255.255.255,MAC为空。 SourceIP为111.111.111.12,MAC为aa-aa-aa-aa-aa-aa 2.适配器收到了来自hostA的数据报,更新自己的ARP表,同时发送一个ACK给hostA,告诉hostA自己的IP、MAC。 3.hostA建立ARP表 4.如d小问所答,开始进行数据发送。 14.RecallthatwiththeCSMA/CDprotocol,theadapterwaitsK·512bittimesafteracollision,whereKisdrawnrandomly.ForK=100,howlongdoestheadapterwaituntilreturningtoStep2fora10MbpsEthernet? Fora100MbpsEthernet? 当网速=10Mbps时,bittime=1bit/10Mbps t=100*512*1/(10^6)=5.12msec 当网速=100Mbps时,bittime=1bit/100Mbps t=100*512*1/(10^7)=0.512msec 15.SupposenodesAandBareonthesame10MbpsEthernetbus,andthepropagationdelaybetweenthetwonodesis225bittimes.SupposeAandBsendframesatthesametime,theframescollide,andthenAandBchoosedifferentvaluesofKintheCSMA/CDalgorithm.Assumingnoothernodesareactive,cantheretransmissionsfromAandBcollide? Forourpurposes,itsufficestoworkoutthefollowingexample.SupposeAandBbegintransmissionatt=0bittimes.Theybothdetectcollisionsatt=225bittimes.Theyfinishtransmittingajamsignalatt=225+48=273bittimes.SupposeKA=0andKB=1.AtwhattimedoesBscheduleitsretransmission? AtwhattimedoesAbegintransmission? (Note: ThenodesmustwaitforanidlechannelafterreturningtoStep2—seeprotocol.)AtwhattimedoesA’ssignalreachB? DoesBrefrainfromtransmittingatitsscheduledtime? 1.因为A的K值=0,所以A从273bittime开始检测是否冲突 2.由于传输延迟的问题,B的最后一个bit要等到273+225=498bittime才能传到A,也就是说此时A检测到没有冲突。 3.A传输前先等待96个bittime,即t=498+96=594bittime的时候A开始传输数据 4.因为propagationdelay=225bittime,所以t=594+225=819bittime的时候A传输完毕 5.因为KB=1,当B等待1*512*1=512bittime的时候,B可以重传,此时B开始检测信道是否空闲,此时t=273+512=785bittime 6.此时B检测到信道忙,因此不能重传,等待信道空闲 7.当t=819bittime的时候信道空闲,此时B等待96bittime,即t=96+819=915bittime的时候B可以开始重传 16.SupposenodesAandBareonthesame10MbpsEthernetbus,andthepropagationdelaybetweenthetwonodesis225bittimes.SupposenodeAbeginstransmittingaframeand,beforeitfinishes,nodeBbeginstransmittingaframe.CanAfinishtransmittingbeforeitdetectsthatBhastransmitted? Whyorwhynot? Iftheanswerisyes,thenAincorrectlybelievesthatitsframewassuccessfullytransmittedwithoutacollision.Hint: Supposeattimet=0bittimes,Abeginstransmittingaframe.Intheworstcase,Atransmitsaminimum-sizedframeof512+64bittimes.SoAwouldfinishtransmittingtheframeatt=512+64bittimes.Thus,theanswerisno,ifB’ssignalreachesAbeforebittimet=512+64bits.Intheworstcase,whendoesB’ssignalreachA? 因为A最快传输时间t=512+64bittime,看了答案,微积分神马的可以让我shi了…… LetYbearandomvariabledenotingthenumberofslotsuntilasuccess: P(Y=m)=β(1-β)^(m-1) Whereβistheprobabilityofasuccess. Thisisageometricdistribution,whichhasmean1/β.ThenumberofconsecutivewastedslotsisX=Y–1that x=E[X]=E[Y]-1=(1-β)/β β=Np(1-p)^(N-1) x={1–Np*(1-p)^(N-1)}/{Np*(1-p)^(N-1)} efficiency=k/(k+x)=k/{k+[1–Np*(1-p)^(N-1)]/[Np(1-p)^(N-1)]} 19.Supposetwonodes,AandB,areattachedtooppositeendsofa900mcable,andthattheyeachhaveoneframeof1,000bits(includingallheadersandpreambles)tosendtoeachother.Bothnodesattempttotransmitattimet=0.SupposetherearefourrepeatersbetweenAandB,eachinsertinga20-bitdelay.Assumethetransmissionrateis10Mbps,andCSMA/CDwithbackoffintervalsofmultiplesof512bitsisused.Afterthefirstcollision,AdrawsK=0andBdrawsK=1intheexponentialbackoffprotocol.Ignorethejamsignalandthe96-bittimedelay. a.Whatistheone-waypropagationdelay(includingrepeaterdelays)betweenAandBinseconds? Assumethatthesignalpropagationspeedis2·10^8m/sec. 传输时间t0=900m/[2*(10^8)m/sec]=4.5μsec Repeater传输延迟t1=4*20bit/10Mbps=8μsec T=t0+t1=12.5μsec b.Atwhattime(inseconds)isA’spacketcompletelydeliveredatB? (为什么这里答案没有96个bittime的等待? ? ? ) 1bittime=1/10Mbps=0.1μsec 当t=12.5μsec时,A、B检测到冲突,停止传输。 A的K值=0,于是A立即开始检测信道是否空闲,当B传输延迟结束后,信道
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