《計(jì)算機(jī)網(wǎng)絡(luò)(第四版)》習(xí)題答案

計(jì)算機(jī)網(wǎng)絡(luò)(第四版)習(xí)題答案第1頁(yè)共59頁(yè)1Theperformanceofaclient-serversystemisinfluencedbytwonetworkfactors:thebandwidthofthenetwork(howmanybits/secitcantransport)andthelatency(howmanysecondsittakesforthebittogetfromtheclienttotheserver).Giveanexampleofanetworkthatexhibitshighbandwidthandhighlatency.Thengiveanexampleofonewithlowbandwidthandlowlatency.客戶(hù)-服務(wù)器系統(tǒng)的性能會(huì)受到兩個(gè)網(wǎng)絡(luò)因素的影響：網(wǎng)絡(luò)的帶寬（每秒可以傳輸多少位數(shù)據(jù)）和延遲（將第一個(gè)數(shù)據(jù)位從客戶(hù)端傳送到服務(wù)器端需要多少秒時(shí)間）。請(qǐng)給出一個(gè)網(wǎng)絡(luò)的例子，它具有高帶寬和高延遲。然后再給出另一個(gè)網(wǎng)絡(luò)的例子，它具有低帶寬和低延遲。答：橫貫大陸的光纖連接可以有很多千兆56kbpsBesidesbandwidthandlatency,whatotherparameterisneededtogiveagoodcharacterizationthequalityofserviceofferedbyanetworkusedfordigitizedvoicetraffic?除了帶寬和延遲以外數(shù)？聲音的傳輸需要相應(yīng)的固定時(shí)間1-6 Aclient-serversystemusesasatellitenetwork,withthesatelliteataheightof40,000km.Whatthebest-casedelayinresponsetoarequest?最佳情形下的延遲是什么？60,000光速為秒，約533。1-9 Agroupof2n-1routersareinterconnectedinacentralizedbinarytree,witharouterateachnode.Routericommunicateswithrouterjbysendingamessagetotherootofthetree.Therootthensendsthemessagebackdowntoj.Deriveanapproximateexpressionforthemeannumberofhopspermessageforlargen,assumingthatallrouterpairsareequallylikely.在一個(gè)集中式的二叉樹(shù)上，個(gè)路由器相互連接起來(lái)；每個(gè)樹(shù)節(jié)點(diǎn)上都有一個(gè)路由器。路由第2頁(yè)共59頁(yè)的路由器都是等概率出現(xiàn)的，請(qǐng)推導(dǎo)出很大時(shí)每條消息的平均跳數(shù)的一個(gè)近似表達(dá)式。，深度為n層需要n-1。從根到n-1層的路徑的路由器和n --2跳步。因此，路徑長(zhǎng)l為或表達(dá)式歸約為l＝n－2，平均的路由器到路由器路徑為2n-4。1-11Whataretworeasonsforusinglayeredprotocols?請(qǐng)說(shuō)出使用分層協(xié)議的兩個(gè)理由？不會(huì)影響高層或低層的協(xié)議。Whatistheprincipaldifferencebetweenconnectionlesscommunicationandconnection-orientedcommunication?在無(wú)連接通信和面向連接的通信二者之間，最主要的區(qū)別是什么？答：主要的區(qū)別有兩條。其一：面向連接通信分為三個(gè)階段，第一是建立連接，在此階段，發(fā)出一個(gè)建立連接的請(qǐng)求。只有在連接成功建立之后，才能開(kāi)始數(shù)據(jù)傳輸，這是第二階段。接著，當(dāng)數(shù)據(jù)傳輸完畢，必須釋放連接。而無(wú)連接通信沒(méi)有這么多階段，它直接進(jìn)行數(shù)據(jù)傳輸。據(jù)的順序一致。Twonetworkseachprovidereliableconnection-orientedservice.Oneofthemoffersareliablestreamandtheotheroffersareliablemessagestream.Aretheseidentical?Ifso,whyisthedistinctionmade?Ifnot,giveanexampleofhowtheydiffer.兩個(gè)網(wǎng)絡(luò)都可以提供可靠的面向連接的服務(wù)其。中一個(gè)提供可靠的字節(jié)流，另一個(gè)提供可靠的報(bào)第3頁(yè)共59頁(yè)個(gè)例子說(shuō)明它們?nèi)绾尾煌?02410242048字節(jié)。對(duì)于報(bào)文流，接受方將得到兩個(gè)報(bào)文。每個(gè)報(bào)102420481-17Insomenetworks,thedatalinklayerhandlestransmissionerrorsbyrequestingdamagedframestoberetransmitted.Iftheprobabilityofaframe'sbeingdamagedisp,whatisthemeannumberoftransmissionsrequiredtosendaframe?Assumethatacknowledgementsareneverlost.幀請(qǐng)求正好是k次的概率Pk，就是起初的k-1次嘗試都失敗的概率。pk-1,乘以第k次傳輸成功的概率。平均傳輸次數(shù)就是1-22WhatisthemaindifferencebetweenTCPandUDP?TCP和UDP之間最主要的區(qū)別是什么？TCPUDP是一種數(shù)據(jù)報(bào)服務(wù)。1-25Whenafileistransferredbetweentwocomputers,twoacknowledgementstrategiesarepossible.Inthefirstone,thefileischoppedupintopackets,whichareindividuallyacknowledgedbythereceiver,butthefiletransferasawholeisnotacknowledged.Inthesecondone,thepacketsarenotacknowledgedindividually,buttheentirefileisacknowledgedwhenitarrives.Discussthesetwoapproaches.當(dāng)一個(gè)文件在兩臺(tái)計(jì)算機(jī)之間傳輸?shù)臅r(shí)候被確認(rèn)。在第二種策略中，這些分組并沒(méi)有被單獨(dú)地確認(rèn)，但是當(dāng)整個(gè)文件到達(dá)的時(shí)候，它會(huì)被確認(rèn)。請(qǐng)討論這兩種方案。第4頁(yè)共59頁(yè)如果網(wǎng)絡(luò)容易丟失分組從而減少了確認(rèn)的次數(shù)，節(jié)省了帶寬；不過(guò)，即使有單個(gè)分組丟失，也需要重傳整個(gè)文件。Howlongwasabitontheoriginal802.3standardinmeters?Useatransmissionspeedof10andassumethepropagationspeedincoaxis2/3thespeedoflightinvacuum.標(biāo)準(zhǔn)中，一位是多長(zhǎng)（按米來(lái)計(jì)算）？請(qǐng)使電纜的傳播速度是真空中光速2/3.200,000km/sec,200m/μsec10Mbps,傳輸一位需要,0.1μsec,20米。Animageis1024x768pixelswith3bytes/pixel.Assumetheimageisuncompressed.Howdoesittaketotransmititovera56-kbpsmodemchannel?Overa1-Mbpscablemodem?Overa10-MbpsEthernet?Over100-MbpsEthernet?的電纜調(diào)制解調(diào)器（cablemodem）呢？通過(guò)10Mbps的因特網(wǎng)呢？通過(guò)100Mbps的因特網(wǎng)呢？圖像是×3bytes2,359,296bytes.就是18,874,368bits.56,000,337.042sec.1,000,000,要大約18.874sec.10,000,000bits/sec,需要大約1.887sec.100,000,000bits/sec,需要大約0.189sec.Wirelessnetworksareeasytoinstall,whichmakestheminexpensivesinceinstallationcostsusuallyfarovershadowequipmentcosts.Nevertheless,theyalsohavesomedisadvantages.Nametwothem.無(wú)線(xiàn)網(wǎng)絡(luò)很容易安裝比例。然而，它們也有一些缺點(diǎn)。請(qǐng)說(shuō)出兩個(gè)缺點(diǎn)。第5頁(yè)共59頁(yè)一個(gè)缺點(diǎn)是安全性L(fǎng)isttwoadvantagesandtwodisadvantagesofhavinginternationalstandardsfornetworkprotocols.請(qǐng)列舉出網(wǎng)絡(luò)協(xié)議國(guó)際標(biāo)準(zhǔn)化的兩個(gè)優(yōu)點(diǎn)和缺點(diǎn)。優(yōu)點(diǎn)了新的更好的技術(shù)或方法，也難以替換。第6頁(yè)共59頁(yè)2Anoiseless4-kHzchannelissampledevery1msec.Whatisthemaximumdatarate?一條無(wú)噪聲4kHz信道按照每1ms一次進(jìn)行采樣，請(qǐng)問(wèn)最大數(shù)據(jù)傳輸率是多少？2Hlog2V。因此最大數(shù)據(jù)傳輸率決定于每次采樣所產(chǎn)生的比特?cái)?shù)，如果每次采樣產(chǎn)的，其最大數(shù)據(jù)傳輸率由香農(nóng)定律給出。Televisionchannelsare6MHzwide.Howmanybits/seccanbesentiffour-leveldigitalareused?Assumeanoiselesschannel.噪聲信道。答：采樣頻率12MHz，每次采樣2bit，總的數(shù)據(jù)率為24Mbps。Ifabinarysignalissentovera3-kHzchannelwhosesignal-to-noiseratiois20dB,whatismaximumachievabledatarate?信道上發(fā)送一個(gè)二進(jìn)制信號(hào)，該信道的信噪比輸率為多少？答：信噪比20dB即S/N= 由于由香農(nóng)定理，該信道的信道容量3log2(1 +100) 。又根據(jù)乃奎斯特定理，發(fā)送二進(jìn)制信號(hào)3kHz信道的最大數(shù)據(jù)傳輸速率為2*3log22=6 。所以可以取得的最大數(shù)據(jù)傳輸速率為6kbps。Whatsignal-to-noiseratioisneededtoputaT1carrierona50-kHzline?答：為發(fā)送T1信號(hào)，我們需要第7頁(yè)共59頁(yè)所以，在50kHz線(xiàn)路上使用T1載波需要93dB的信噪比。Howmuchbandwidthistherein0.1micronofspectrumatawavelengthof1micron?答：因此，在0.1的頻段中可以有30THz。Itisdesiredtosendasequenceofcomputerscreenimagesoveranopticalfiber.Thescreenisx640pixels,eachpixelbeing24bits.Thereare60screenimagespersecond.Howmuchbandwidthisneeded,andhowmanymicronsofwavelengthareneededforthisbandat1.30microns?246幅屏幕圖像。請(qǐng)問(wèn)，需要多少帶寬？1.3波長(zhǎng)上，這段帶寬需要多少?答：數(shù)據(jù)速率480× 即。需要442Mbps的帶寬，對(duì)應(yīng)的波長(zhǎng)范圍是 。2-18Asimpletelephonesystemconsistsoftwoendofficesandasingletollofficetowhicheachendofficeisconnectedbya1-MHzfull-duplextrunk.Theaveragetelephoneisusedtomakefourcallsper8-hourworkday.Themeancalldurationis6min.Tenpercentofthecallsarelong-distance(i.e.,pass第8頁(yè)共59頁(yè)throughthetolloffice).Whatisthemaximumnumberoftelephonesanendofficecansupport?(Assume4kHzpercircuit.)小時(shí)的工作日中，平均每部電話(huà)的呼叫是長(zhǎng)途（即通過(guò)長(zhǎng)途局）。請(qǐng)問(wèn)一個(gè)端局能夠支持最多多少部電話(huà)？（假設(shè)每條線(xiàn)0.56分鐘。因此一部電話(huà)每小時(shí)占用一條電3分，即20200部電話(huà)占用一條完全時(shí)間的長(zhǎng)途線(xiàn)路。局間干線(xiàn)復(fù)用1000000/4000=250200電話(huà)，因此，一個(gè)端局可以支持的電話(huà)部數(shù)。2-22AmodemconstellationdiagramsimilartoFig.2-25hasdatapointsatthefollowingcoordinates:(1,1),(1,-1),(-1,1),and(-1,-1).Howmanybpscanamodemwiththeseparametersachieveat1200baud?2.2的調(diào)制解調(diào)器星座圖有以下幾個(gè)坐標(biāo)點(diǎn)）--）-，。請(qǐng)問(wèn)一個(gè)具備這些參數(shù)的調(diào)制解調(diào)器每個(gè)波特有41200。Tensignals,eachrequiring4000Hz,aremultiplexedontoasinglechannelusingFDM.Howmuchminimumbandwidthisrequiredforthemultiplexedchannel?AssumethattheguardbandsareHzwide.最小要求多少帶寬？假設(shè)防護(hù)頻段寬。第9頁(yè)共59頁(yè)104000Hz9個(gè)防護(hù)頻段來(lái)避免干擾。最小帶寬需求=43,600Hz.WhyhasthePCMsamplingtimebeensetat12sc?答的采樣時(shí)間對(duì)應(yīng)于每8000次采樣。一個(gè)典型的電話(huà)通道。根據(jù)奈奎斯特理，為獲取一4kHz的通道中的全部信息需要每8000次的采樣頻率。實(shí)際上額定帶寬稍有些少，截止點(diǎn)并不清)WhatisthepercentoverheadonaT1carrier;thatis,whatpercentofthe1.544Mbpsaredeliveredtotheenduser?193位中的等于。Whatisthedifference,ifany,betweenthedemodulatorpartofamodemandthecoderpartofcodec?(Afterall,bothconvertanalogsignalstodigitalones.)（余弦）波，產(chǎn)生數(shù)字信號(hào)。第10頁(yè)共59頁(yè)Asignalistransmitteddigitallyovera4-kHznoiselesschannelwithonesampleeverys.Howmanybitspersecondareactuallysentforeachoftheseencodingmethods?(a)CCITT2.048Mbpsstandard.(b)DPCMwitha4-bitrelativesignalvalue.(c)Deltamodulation.答2.048Mbps標(biāo)準(zhǔn)用32個(gè)8位數(shù)據(jù)樣本組成一125 的基本幀個(gè)信道用傳信息個(gè)信道用于傳控制信號(hào)。在每一4kHz信道上發(fā)送的數(shù)據(jù)率就是。44kHz信道實(shí)際。st可以選擇一個(gè)合適的量化??1bit時(shí)對(duì)應(yīng)每個(gè)4kHz。2-39Whatistheessentialdifferencebetweenmessageswitchingandpacketswitching?將被拆分成多個(gè)報(bào)文。Threepacket-switchingnetworkseachcontainnnodes.Thefirstnetworkhasastartopologywith第11頁(yè)共59頁(yè)acentralswitch,thesecondisa(bidirectional)ring,andthethirdisfullyinterconnected,withawirefromeverynodetoeveryothernode.Whatarethebest-,average-,and-worstcasetransmissionpathsinhops?三個(gè)分組交換網(wǎng)絡(luò)每個(gè)包二個(gè)網(wǎng)絡(luò)是一個(gè)雙向環(huán)答：Thethreenetworkshavethefollowingproperties:，最差為，平均為，最差為，平均為如果考慮n為奇偶數(shù)，則n為奇數(shù)時(shí)，最壞為n-/，平均為n為偶數(shù)時(shí)，最壞為平均為n2/4(n 1)全連接：最好為1，最差為1，平均為1。Comparethedelayinsendinganx-bitmessageoverak-hoppathinacircuit-switchednetworkandina(lightlyloaded)packet-switchednetwork.Thecircuitsetuptimeisssec,thepropagationisdsecperhop,thepacketsizeispbits,andthedatarateisbbps.Underwhatconditionsdoesthepacketnetworkhavealowerdelay?請(qǐng)比較一下在一個(gè)電路交換網(wǎng)絡(luò)中和在一個(gè)負(fù)載較輕的分組交換網(wǎng)絡(luò)中，x位消息的延遲情況。電路建立的時(shí)間b。在什么條件下分組網(wǎng)絡(luò)的延遲比較短？對(duì)于電路交換，t=s時(shí)電路建立起來(lái)t s+ x/d 時(shí)報(bào)文的最后一位發(fā)送完畢t =s+ x/b+kd時(shí)報(bào)文到達(dá)目的地。而對(duì)于分組交換，最后一位t=x/b 時(shí)發(fā)送完畢。為到達(dá)最終目的地，最后一個(gè)分組必須被中間的路由器重k 1次，每次重發(fā)花時(shí)p/以總的延遲為為了使分組交換比電路交換快，必須：第12頁(yè)共59頁(yè)所以：Supposethatxbitsofuserdataaretobetransmittedoverak-hoppathinapacket-switchednetworkasaseriesofpackets,eachcontainingpdatabitsandhheaderbits,withxp+h.Thebitratethelinesisbbpsandthepropagationdelayisnegligible.Whatvalueofpminimizesthetotaldelay?跳到路徑向前傳b值使總延遲最小？x/p，因此總的數(shù)據(jù)加上頭信息交通量(p+h)x/p位。源端發(fā)送這些位需要時(shí)間(p+h (k-1)(p +h )/b因此我們得到的總的延遲為對(duì)該函數(shù)求p的導(dǎo)數(shù)，得到令得到因?yàn)楣蕰r(shí)能使總的延遲最小。

/p第13頁(yè)共59頁(yè)Inatypicalmobilephonesystemwithhexagonalcells,itisforbiddentoreuseafrequencybandanadjacentcell.If840frequenciesareavailable,howmanycanbeusedinagivencell?在一個(gè)典型的移動(dòng)電話(huà)系統(tǒng)中個(gè)頻率可以使用的話(huà)，則任何一個(gè)給定的單元內(nèi)可以使用多少個(gè)頻率？每個(gè)單元有6個(gè)鄰居。如果中間的單元使用頻段組合A，它的六個(gè)鄰居可以分別使用的頻段組合B,C,B,C,B,3個(gè)單一的單元。因此，每個(gè)單元可以使280個(gè)頻率。2-50SupposethatA,B,andCaresimultaneouslytransmitting0bits,usingaCDMAsystemwiththechipsequencesofFig.2-45(b).Whatistheresultingchipsequence?FIG2-45(b)結(jié)果是通過(guò)C求反再將這三個(gè)碼片序列相加得到的結(jié)果(+3+1+1 1 3 1 1+1).2-53ACDMAreceivergetsthefollowingchips:(-1+1-3+1-1-3+1+1).AssumingthechipsequencesdefinedinFig.2-45(b),whichstationstransmitted,andwhichbitsdideachonesend?CDM接收器得到了下面的時(shí)間片-1+1-3+1-1-3+1+）。假設(shè)時(shí)間片序列如2.45義，請(qǐng)問(wèn)那些移動(dòng)站傳輸了數(shù)據(jù)？每個(gè)站發(fā)送了什么位？Justcomputethefournormalizedinner號(hào)之誤？(1+13+113+1+1)d(1 1 1+1+1 1+1+1)/8=1(1+13+113+1+1)d(1 1+1 1+1+1+1 1)/8= 1( 1+13+113+1+1)d(1+1 1+1+1+1 1 1)/8=0( 1+1

3+1 1

3+1+1)d(

1+1 1 1 1 1+1 1)/8=1AD發(fā)送了1,B發(fā)送了0,C沒(méi)有發(fā)送。第14頁(yè)共59頁(yè)3Anupper-layerpacketissplitinto10frames,eachofwhichhasan80percentchanceofarrivingundamaged.Ifnoerrorcontrolisdonebythedatalinkprotocol,howmanytimesmustthemessagesentonaveragetogettheentirethingthrough?供錯(cuò)誤控制的話(huà)，請(qǐng)問(wèn)，該報(bào)文平均需要發(fā)送多少次才能完整地到達(dá)接收方？0.8。為使信息完整的到達(dá)接收方，發(fā)送一次成功的概率p，二次成功的概率三次成功概率(1-p )2次成功的概率(1-p)i-1 因此平均的發(fā)送次數(shù)等于：Thefollowingcharacterencodingisusedinadatalinkprotocol:A:01000111;B:11100011;FLAG:01111110;ESC:11100000Showthebitsequencetransmitted(inbinary)forthefour-characterframe:BESCFLAGwheneachofthefollowingframingmethodsareused:(a)Charactercount.(b)Flagbyteswithbytestuffing.(c)Startingandendingflagbytes,withbitstuffing.結(jié)果是(a)0000010001000111111000111110000001111110(b)0111111001000111111000111110000011100000111000000111111001111110(c)011111100100011111010001111100000001111101001111110Abitstring,0111101111101111110,needstobetransmittedatthedatalinklayer.Whatisstringactuallytransmittedafterbitstuffing?011110111110111111的是什么？第15頁(yè)共59頁(yè)輸出是1110111110011111010.Whenbitstuffingisused,isitpossiblefortheloss,insertion,ormodificationofasinglebittocauseanerrornotdetectedbythechecksum?Ifnot,whynot?Ifso,how?Doesthechecksumplayarolehere?假設(shè)使用了位填充成幀方法，請(qǐng)問(wèn)，因?yàn)閬G失一位，插入一位，或者篡改一位而引起的錯(cuò)誤是否有可能通過(guò)校驗(yàn)和檢測(cè)出來(lái)？如果不能的話(huà)請(qǐng)，問(wèn)為什么？如果能的話(huà)，請(qǐng)問(wèn)校驗(yàn)和長(zhǎng)度在這里是如何起作用的？答：可能。假定原來(lái)的正文包含位序01111110作為數(shù)據(jù)。位填充之后，這個(gè)序列將變成0616位的內(nèi)容碰巧經(jīng)驗(yàn)證后仍然正確的概率。如果這種概率的條件成立了，就會(huì)導(dǎo)致不正確的幀被接收。顯然，檢驗(yàn)和段越長(zhǎng)，傳輸錯(cuò)誤不被發(fā)現(xiàn)的概率會(huì)越低，但該概率永遠(yuǎn)不等于零。DatalinkprotocolsalmostalwaysputtheCRCinatrailerratherthaninaheader.Why?數(shù)據(jù)鏈路協(xié)議幾乎總是將CRC放在尾部，而不是頭部，為什么？是在發(fā)送期間進(jìn)行計(jì)算的CRC編碼附CRC放在幀的頭部，那么就要在發(fā)送之前把整個(gè)幀先檢查一遍來(lái)計(jì)算CRC放在尾部就可以把處理時(shí)間減半。Achannelhasabitrateof4kbpsandapropagationdelayof20msec.Forwhatrangeofframe第16頁(yè)共59頁(yè)sizesdoesstop-and-waitgiveanefficiencyofatleast50percent?。請(qǐng)問(wèn)幀的大小在什么范圍內(nèi)，答：當(dāng)發(fā)送一幀的時(shí)間等于信道的傳播延遲2?；蛘哒f(shuō)，當(dāng)發(fā)送?，F(xiàn)在發(fā)送速率為4Mb/s，發(fā)送一位需要0.25。只有在幀長(zhǎng)不小于160kb時(shí)，停等協(xié)議的效率才會(huì)至少達(dá)到50%。A3000-km-longT1trunkisusedtotransmit64-byteframesusingprotocol5.Ifthepropagationspeedis6ec/km,howmanybitsshouldthesequencenumbersbe?字節(jié)的幀，兩端使用了協(xié)公里，則序列號(hào)應(yīng)該有多少位？答；為了有效運(yùn)行，序列空間（實(shí)際上就是發(fā)送窗口大?。┍仨氉銐虻拇螅栽试S發(fā)送方在收到第一個(gè)確認(rèn)應(yīng)答之前可以不斷發(fā)送。信號(hào)在線(xiàn)路上的傳播時(shí)間為18000 ，即。在T1速率，發(fā)64字節(jié)的數(shù)據(jù)幀需花的時(shí)間648(1.536106) =0.33。所以，發(fā)送的第一幀從開(kāi)始發(fā)送起18.33ms間（忽略不計(jì)）。發(fā)送方應(yīng)該有足夠大的窗口，從而能夠連續(xù)發(fā)送36.33ms。36.33/0.33=110也就是說(shuō)，為充滿(mǎn)線(xiàn)路管道，需要至1107位。第17頁(yè)共59頁(yè)Inprotocol3,isitpossiblethatthesenderstartsthetimerwhenitisalreadyrunning?Ifso,mightthisoccur?Ifnot,whyisitimpossible?一個(gè)幀將在定時(shí)器仍在運(yùn)行的情況下被發(fā)送。Imagineaslidingwindowprotocolusingsomanybitsforsequencenumbersthatwraparoundneveroccurs.Whatrelationsmustholdamongthefourwindowedgesandthewindowsize,whichconstantandthesameforboththesenderandthereceiver.想象這樣一個(gè)滑動(dòng)窗口協(xié)議，它的序列號(hào)有非常多的位，所以序列號(hào)幾乎永遠(yuǎn)不會(huì)回轉(zhuǎn)。4令發(fā)送方窗口為(Sl,Su)接收方窗口為(Rl,Ru)，令窗口大小為W。二者必須保持的關(guān)系是：0≤Su?Sl+1≤W1Ru?Rl+1=WSl≤Rl≤Su+1Iftheprocedurebetweeninprotocol5checkedfortheconditionabcinsteadoftheconditionabc,wouldthathaveanyeffectontheprotocol'scorrectnessorefficiency?Explainyouranswer.有影響嗎？解釋你的答案。答：改變檢查條件后，協(xié)議將變得不正確。假定使3位序列號(hào)，考慮下列協(xié)議運(yùn)行過(guò)程：A站剛發(fā)出7A站收到，并發(fā)送0~6號(hào)B。考察A站在r.rack=7next_frame_to_send_=。修改后的檢查條件將被置成“真”，不會(huì)報(bào)告已發(fā)現(xiàn)的丟失幀錯(cuò)誤，而第18頁(yè)共59頁(yè)所以結(jié)論是：為保證協(xié)議的正確性，已接收的確認(rèn)應(yīng)答號(hào)應(yīng)該小于下一個(gè)要發(fā)送的序列號(hào)。Inprotocol6,whenadataframearrives,acheckismadetoseeifthesequencenumberfromtheoneexpectedandno_nakistrue.Ifbothconditionshold,aNAKissent.Otherwise,theauxiliarytimerisstarted.Supposethattheelseclausewereomitted.Wouldthischangeaffecttheprotocol'scorrectness?字句被省略掉，這種改變會(huì)影響協(xié)議的正確性嗎？答：可能導(dǎo)致死鎖。假定有一組幀正確到達(dá)，并被接收。然后，接收方會(huì)向前移動(dòng)窗口。現(xiàn)在假定所有的確認(rèn)幀都丟失了發(fā)送一個(gè)。然后NONAKNAKONAK為偽，所以不會(huì)再發(fā)送NA，從而產(chǎn)生死鎖。如果設(shè)置輔助計(jì)數(shù)器（ls”子句），NA，終究會(huì)使雙方重新獲得同步。Supposethatthethree-statementwhileloopneartheendofprotocol6wereremovedfromthecode.Wouldthisaffectthecorrectnessoftheprotocolorjusttheperformance?Explainyour中接近尾部的內(nèi)含三條語(yǔ)句還是僅僅影響協(xié)議的性能？請(qǐng)解釋答案。沒(méi)有這一段程序，發(fā)送方會(huì)一直保持超時(shí)條件，從而使得協(xié)議的運(yùn)行不能向前進(jìn)展。SupposethatthecaseforchecksumerrorswereremovedfromtheswitchstatementofprotocolHowwouldthischangeaffecttheoperationoftheprotocol?這樣將使得NAK的作用失效，于是我們將退回到超時(shí)。盡管效率會(huì)降低，正確性卻不會(huì)受到影響。NAK不是必不可少的。Inprotocol6thecodeforframe_arrivalhasasectionusedforNAKs.Thissectionisinvokedifthe第19頁(yè)共59頁(yè)incomingframeisaNAKandanotherconditionismet.Giveascenariowherethepresenceofthisotherconditionisessential.在協(xié)中，針frame_arrival 代碼中有一部分被用。如果收到的幀是一并且另一個(gè)條件也滿(mǎn)足的話(huà)是非常關(guān)鍵的。

考慮下列操作細(xì)節(jié)：A站發(fā)送0號(hào)幀給BBACK幀，但ACKA站發(fā)生超時(shí)，重發(fā)0號(hào)幀。但B10A0不成立，所以用不著選擇性重01號(hào)幀。這個(gè)例子就說(shuō)明了這段程序中的另一個(gè)條件，即也是重要的。Imaginethatyouarewritingthedatalinklayersoftwareforalineusedtosenddatatoyou,butfromyou.TheotherendusesHDLC,witha3-bitsequencenumberandawindowsizeofsevenframes.Youwouldliketobufferasmanyout-of-sequenceframesaspossibletoenhanceefficiency,butyouarenotallowedtomodifythesoftwareonthesendingside.Isitpossibletohaveareceiverwindowgreaterthan1,andstillguaranteethattheprotocolwillneverfail?Ifso,whatisthelargestwindowthatcanbesafelyused?想象你正在編寫(xiě)一個(gè)數(shù)據(jù)鏈路層軟件幀的窗口。你希望將亂序的幀盡可0至670號(hào)0號(hào)幀到達(dá)接收方時(shí)，它將會(huì)被緩存保留，接收方確6號(hào)幀。當(dāng)7號(hào)幀到來(lái)的時(shí)70。Inprotocol6,MAX_SEQ=2n-1.Whilethisconditionisobviouslydesirabletomakeefficient第20頁(yè)共59頁(yè)useofheaderbits,wehavenotdemonstratedthatitisessential.DoestheprotocolworkcorrectlyforMAX_SEQ=4,forexample?這個(gè)條件確實(shí)很關(guān)鍵。例如，協(xié)議的時(shí)候也能夠正確地工作嗎？。因此在該協(xié)議中，偶數(shù)序號(hào)使用緩沖。這種映射意味著幀40034000中，變量MaxSeq是奇數(shù)才能正確的工作性質(zhì)。Framesof1000bitsaresentovera1-Mbpschannelusingageostationarysatellitewhosepropagationtimefromtheearthis270msec.Acknowledgementsarealwayspiggybackedontoframes.Theheadersareveryshort.Three-bitsequencenumbersareused.Whatisthemaximumachievablechannelutilizationfor(a)Stop-and-wait.(b)Protocol5.(c)Protocol6.(a(bc6答：對(duì)應(yīng)三種協(xié)議的窗口大小值分別7。使用衛(wèi)星信道端到端的典型傳輸延遲，以1Mb/s長(zhǎng)的幀的發(fā)送時(shí)間為1m。我們用t=0t=1mst=271ms時(shí)，第一t=272mt=542m。如果在542ms內(nèi)可以發(fā)送k間為（a）k=1，最大信道利用率=1/542=0.18%第21頁(yè)共59頁(yè)=7/542=1.29%=4/542=0.74%Computethefractionofthebandwidththatiswastedonoverhead(headersandretransmissions)forprotocol6onaheavily-loaded50-kbpssatellitechannelwithdataframesconsistingof40headerand3960databits.Assumethatthesignalpropagationtimefromtheearthtothesatelliteis270msec.ACKframesneveroccur.NAKframesare40bits.Theerrorratefordataframesis1percent,andtheerrorforNAKframesisnegligible.Thesequencenumbersare8bits.答：使用選擇性重傳滑動(dòng)窗口協(xié)議，序列號(hào)長(zhǎng)度8。衛(wèi)星信道端到端的傳輸延遲是270m50kb/s4000bt3960+40.02*4000=80m。我們用t=0表示傳輸開(kāi)始時(shí)間，那么t=80m，第一幀發(fā)送完畢；t=270+80=350m，第一幀完全到達(dá)接收方t=350+80=430m，對(duì)第一幀作捎帶確認(rèn)的反向數(shù)據(jù)，發(fā)送128幀時(shí)間，對(duì)于3960404000*0.01=40NAK40*1/100=0.40位，所以每396080.4位。80.4/(3960+80.4)=1.99。3-32A100-km-longcablerunsattheT1datarate.Thepropagationspeedinthecableis2/3thespeedoflightinvacuum.Howmanybitsfitinthecable?答：在該電纜中的傳播速度是每秒鐘200000km，即每毫秒200km，因此100km的電纜將會(huì)在0.5ms內(nèi)填滿(mǎn)T1速率125 傳送一193位的幀0.5ms可以傳4個(gè)T1幀即193*4=772bt第22頁(yè)共59頁(yè)第23頁(yè)共59頁(yè)4Forthisproblem,useaformulafromthischapter,butfirststatetheformula.Framesarriveata100-Mbpschannelfortransmission.Ifthechannelisbusywhenaframearrives,itwaitsitsturninaqueue.Framelengthisexponentiallydistributedwithameanof10,000bits/frame.Foreachofthefollowingframearrivalrates,givethedelayexperiencedbytheaverageframe,includingbothqueueingtimeandtransmissiontime.(a)90frames/sec.(b)900frames/sec.(c)9000frames/sec.(timedelay,T// achannelofcapacityCbps// withanarrivalrateof frames/sec)這個(gè)公式是4.1.1段落給出的Markov排隊(duì)問(wèn)題的標(biāo)準(zhǔn)公式也就是， 這里C=108、， (a)0.1msec,(b)0.11msec,(c)1msec.對(duì)于c的情況，我們操作一個(gè)帶來(lái)10倍延遲的排隊(duì)系統(tǒng)AgroupofNstationssharea56-kbpspureALOHAchannel.Eachstationoutputsa1000-bitframeonanaverageofonceevery100sec,evenifthepreviousonehasnotyetbeensent(e.g.,stationscanbufferoutgoingframes).WhatisthemaximumvalueofN?幀還沒(méi)有被送出，它也這樣進(jìn)行（比如這些站可以將送出的幀緩存起來(lái)）答：對(duì)于純可用的帶寬Kb/s =10.304 Kb/。每個(gè)站需要的帶寬為而所以最多可以1030個(gè)站即N的最大值ConsiderthedelayofpureALOHAversusslottedALOHAatlowload.Whichoneisless?youranswer.第24頁(yè)共59頁(yè)樣，平均會(huì)引入半個(gè)時(shí)隙的延遲。因此，ALOHA的延遲比較小。TenthousandairlinereservationstationsarecompetingfortheuseofasingleslottedALOHAchannel.Theaveragestationmakes18requests/hour.Aslotis1e.Whatistheapproximatechannelload?每個(gè)終端每10000200秒做10000508000。AlargepopulationofALOHAusersmanagestogenerate50requests/sec,includingbothoriginalsandretransmissions.Timeisslottedinunitsof40msec.(a)Whatisthechanceofsuccessonthefirstattempt?(b)Whatistheprobabilityofexactlykcollisionsandthenasuccess?(c)Whatistheexpectednumberoftransmissionattemptsneeded?(a)首次發(fā)送成功的幾率是多少(b次沖突之后成功的概率是多?(c是多少？k幀的概率服從泊松分布生成0幀的概率為e-G對(duì)于純的G=e-2G對(duì)于分隙的ALOHA，由于沖突危險(xiǎn)區(qū)減少為原來(lái)的一半，任一幀時(shí)內(nèi)無(wú)其他幀發(fā)送的概率是e-G。40m2550G=因此，首次嘗試的成功率是e-2=1/e2第25頁(yè)共59頁(yè)（b）嘗試k次才能發(fā)送成功的概率（k-1次沖突，第k次才成功）為：那么每幀傳送次數(shù)的數(shù)學(xué)期望為MeasurementsofaslottedALOHAchannelwithaninfinitenumberofusersshowthat10oftheslotsareidle.(a)Whatisthechannelload,G?(b)Whatisthethroughput?(c)Isthechannelunderloadedoroverloaded?ALOH10(a該信道是載荷不足，還是過(guò)載?答：）peG，因此＝-lnp0=-ln0.2.3e-G，G 因?yàn)槊慨?dāng)G>1時(shí)，信道總是過(guò)載的，因此在這里信道是過(guò)載的。4-8 Howlongdoesastation,s,havetowaitintheworstcasebeforeitcanstarttransmittingitsframeoveraLANthatuses(a)thebasicbit-mapprotocol?(b)MokandWard'sprotocolwithpermutingstationnumbers?使用了下列協(xié)議，請(qǐng)問(wèn)在最差情況下，一個(gè)時(shí)間？）基本的位圖協(xié)議）Mok-War協(xié)議s是編號(hào)最小的站點(diǎn)N+(N-1)N+(N-1)d位時(shí)間；(b)s具有其中最小的虛擬站編號(hào)s在其它個(gè)站各發(fā)送了一個(gè)幀之后將獲得傳輸機(jī)會(huì)，以及每個(gè)大小og2N的N個(gè)爭(zhēng)用周期。等待時(shí)間是(N+1)×d+N×log2Nbits.第26頁(yè)共59頁(yè)4-10Sixteenstations,numbered1through16,arecontendingfortheuseofasharedchannelbyusingtheadaptivetreewalkprotocol.Ifallthestationswhoseaddressesareprimenumberssuddenlybecomereadyatonce,howmanybitslotsareneededtoresolvethecontention?號(hào)為素?cái)?shù)的所有站突然間全部要發(fā)送幀，請(qǐng)問(wèn)需要多少位時(shí)槽才能解決競(jìng)爭(zhēng)？（見(jiàn)圖BCBD節(jié)點(diǎn)以下各站來(lái)競(jìng)爭(zhēng)信道。本題中，站111313如下：733第27頁(yè)共59頁(yè)第十時(shí)隙：74-14Sixstations,AthroughF,communicateusingtheMACAprotocol.Isitpossiblethattwotransmissionstakeplacesimultaneously?Explainyouranswer.F發(fā)送時(shí)A也能向B送。ConsiderbuildingaCSMA/CDnetworkrunningat1Gbpsovera1-kmcablewithnorepeaters.Thesignalspeedinthecableis200,000km/sec.Whatistheminimumframesize?（無(wú)中繼器。請(qǐng)問(wèn)最小的幀長(zhǎng)度為多少？答對(duì)于1km電纜單程傳播時(shí)間1/200000 即5 來(lái)回路程傳播時(shí)間2t=10為了能夠按CSMA/CD工作，最小幀的發(fā)射時(shí)間不能小10 。以1Gb/s速率工作可以發(fā)送的比特?cái)?shù)等于：10000bit1250字節(jié)長(zhǎng)。AnIPpackettobetransmittedbyEthernetis60byteslong,includingallitsheaders.IfLLCisinuse,ispaddingneededintheEthernetframe,andifso,howmanybytes?字節(jié)長(zhǎng)，其中包括所有的頭部。如果沒(méi)有使太網(wǎng)幀中需要填補(bǔ)字節(jié)碼？如果需要的話(huà)，請(qǐng)問(wèn)需要填補(bǔ)多少字節(jié)？18bytes報(bào)文是6078bytes,已經(jīng)超過(guò)了64-byte的最小限制。因此，不需要填補(bǔ)。第28頁(yè)共59頁(yè)Ethernetframesmustbeatleast64byteslongtoensurethatthetransmitterisstillgoingintheofacollisionatthefarendofthecable.FastEthernethasthesame64-byteminimumframesizebutcangetthebitsouttentimesfaster.Howisitpossibletomaintainthesameminimumframesize?快速以太網(wǎng)的最大線(xiàn)路長(zhǎng)度是以太網(wǎng)1/10。SomebooksquotethemaximumsizeofanEthernetframeas1518bytesinsteadof1500bytes.theywrong?Explainyouranswer.有效載荷是1500bytes,但將目的地址、源地址、類(lèi)1518.ConsidertheinterconnectedLANsshownsinFig.4-44.AssumethathostsaandbareonLAN1,isonLAN2,anddisonLAN8.Initially,hashtablesinallbridgesareemptyandthespanningtreeshowninFig4-44(b)isused.Showhowthehashtablesofdifferentbridgeschangeaftereachofthefollowingeventshappeninsequence,first(a)then(b)andsoon.(a)asendstod.(b)csendstoa.(c)dsendstoc.(d)dmovestoLAN6.(e)dsendstoa.上。所示的生成樹(shù)。在下面給出的每個(gè)事件依次發(fā)生以后發(fā)送幀第一個(gè)幀會(huì)被每個(gè)網(wǎng)橋轉(zhuǎn)發(fā)為aDLAN2a第29頁(yè)共59頁(yè)被網(wǎng)橋B,D和A看到。這些網(wǎng)橋會(huì)在它們的散列表中添加一個(gè)目的地c的新項(xiàng)目。例如，網(wǎng)橋DLAN2c轉(zhuǎn)發(fā)幀的項(xiàng)目。第三個(gè)信息會(huì)被網(wǎng)橋H,D,A和B看到。這些網(wǎng)橋會(huì)在它們的散列表中添加一個(gè)目的地d被網(wǎng)橋E,C,B,和AE和CdD,B和Ad的散列表項(xiàng)目。OneconsequenceofusingaspanningtreetoforwardframesinanextendedLANisthatsomebridgesmaynotparticipateatallinforwardingframes.IdentifythreesuchbridgesinFig.4-44.Isanyreasonforkeepingthesebridges,eventhoughtheyarenotusedforforwarding?中使用生成樹(shù)來(lái)轉(zhuǎn)發(fā)幀的一個(gè)結(jié)果是，有的網(wǎng)橋可能根本不參與幀的轉(zhuǎn)發(fā)過(guò)留這些網(wǎng)橋呢？G,IJ沒(méi)有被用來(lái)轉(zhuǎn)發(fā)任何幀。在一個(gè)擴(kuò)展LAN中具有回路的主要原因是增加可靠）能包括一個(gè)或更多不屬于先前生成樹(shù)部分的網(wǎng)橋。Brieflydescribethedifferencebetweenstore-and-forwardandcut-throughswitches.存儲(chǔ)-轉(zhuǎn)發(fā)型交換機(jī)完整存儲(chǔ)輸入的每個(gè)幀，然后檢查并轉(zhuǎn)發(fā)。直通型交換機(jī)在輸入幀沒(méi)有全部到達(dá)之前就開(kāi)始轉(zhuǎn)發(fā)。一得到目的地址，轉(zhuǎn)發(fā)就開(kāi)始了。Store-and-forwardswitcheshaveanadvantageovercut-throughswitcheswithrespecttoframes.Explainwhatitis.Store-and-forwardswitchesstoreentireframesbeforeforwardingthem.Afteraframecomesin,thechecksumcanbeverified.Iftheframeisdamaged,itisdiscardedimmediately.Withcut=through,damagedframescannotbediscardedbytheswitchbecausebythetimetheerrorisdetected,theframeisalreadygone.Tryingtodealwiththeproblemislikelockingthebarndoorafterthehorsehasescaped.存儲(chǔ)-轉(zhuǎn)發(fā)型交換機(jī)在轉(zhuǎn)發(fā)幀之前存儲(chǔ)整個(gè)幀。當(dāng)一個(gè)幀到達(dá)時(shí)，校驗(yàn)和將被驗(yàn)證。如果幀已被損壞，它將被立即丟棄。在直通型交換機(jī)，損壞的幀不能被交換機(jī)丟棄因。為當(dāng)錯(cuò)誤被檢測(cè)到時(shí)，幀已經(jīng)過(guò)去了。想要處第30頁(yè)共59頁(yè)理這個(gè)問(wèn)題就像是在馬已經(jīng)逃逸之后再鎖上牲口棚。第31頁(yè)共59頁(yè)5Givetwoexamplecomputerapplicationsforwhichconnection-orientedserviceisNowgivetwoexamplesforwhichconnectionlessserviceisbest.Arethereanycircumstanceswhenconnection-orientedservicewill(oratleastshould)packetsoutoforder?Explain.排在前面等待程序處理的任何數(shù)據(jù)（即已經(jīng)鍵入但尚未被程序讀取的數(shù)據(jù)）。Datagramsubnetsrouteeachpacketasaseparateunit,independentofallothers.Virtual-circuitsubnetsdonothavetodothis,sinceeachdatapacketfollowsapredeterminedroute.Doesthisobservationmeanthatvirtual-circuitsubnetsdonotneedthecapabilitytorouteisolatedpacketsfromarbitrarysourcetoanarbitrarydestination?Explainyouranswer.能力。Considerthefollowingdesignproblemconcerningimplementationofvirtual-circuitservice.Ifvirtualcircuitsareusedinternaltothesubnet,eachdatapacketmusthavea3-byteheaderandeachmusttieup8bytesofstorageforcircuitidentification.Ifdatagramsareusedinternally,15-byteheadersareneededbutnoroutertablespaceisrequired.Transmissioncapacitycosts1centper106bytes,perhop.Veryfastroutermemorycanbepurchasedfor1centperbyteandisdepreciatedovertwoyears,assuminga40-hourbusinessweek.Thestatisticallyaveragesessionrunsfor1000sec,inwhichtime200packetsaretransmitted.Themeanpacketrequiresfourhops.Whichimplementationischeaper,andbyhowmuch?第32頁(yè)共59頁(yè)答：虛電路實(shí)現(xiàn)需要1000秒內(nèi)固定分5*8=40字節(jié)的存儲(chǔ)器。數(shù)據(jù)報(bào)實(shí)現(xiàn)需要比虛電路現(xiàn)多傳送的頭信息的容量等(15-3) 字跳段現(xiàn)在的問(wèn)題就變成40000字秒的存儲(chǔ)器對(duì)9600字跳段的電路容量。如果存儲(chǔ)器的使用期為兩年，即秒一個(gè)字秒的代價(jià)為1/(=分那么40000字秒的代價(jià)為2.7毫分另一方面?zhèn)€字跳段代價(jià)是10-6分個(gè)字跳段的代為10-6× 分，即9.6毫分，即在1000秒內(nèi)的時(shí)間內(nèi)便宜大6.9毫分。Assumingthatallroutersandhostsareworkingproperlyandthatallsoftwareinbothisfreeoferrors,isthereanychance,howeversmall,thatapacketwillbedeliveredtothewrongdestination?答有可能大的突發(fā)噪聲可能破壞分組使用k位的檢驗(yàn)和差錯(cuò)仍然有2 k的概率被漏檢。如果分組的目的地段或虛電路號(hào)碼被改組將會(huì)被投遞到錯(cuò)誤的目的地并可能被接收為正確的分組換句話(huà)說(shuō)偶然的突發(fā)噪聲可能把送往一個(gè)目的地的完全合法的分組改變成送往另個(gè)目的地的也是完全合法的分組。ConsiderthenetworkofFig.5-7,butignoretheweightsonthelines.Supposethatitusesastheroutingalgorithm.IfapacketsentbyAtoDhasamaximumhopcountof3,listalltheroutesitwilltake.Alsotellhowmanyhopsworthofbandwidthitconsumes.第33頁(yè)共59頁(yè)路徑將依次為下面的路:ABCD,ABCF,ABEF,ABEG,AGHD,AGHF,, AGEF.用到的跳數(shù)是24。ConsiderthesubnetofFig.5-13(a).Distancevectorroutingisused,andthefollowingvectorshavejustcomeintorouterC:fromB:(5,0,8,12,6,2);fromD:(16,12,6,0,9,10);andfromE:(7,6,3,9,4).ThemeasureddelaystoB,D,andE,are6,3,and5,respectively.WhatisC'snewroutingtable?Giveboththeoutgoinglinetouseandtheexpecteddelay.第34頁(yè)共59頁(yè)答：通過(guò)B給出通過(guò)D給出通過(guò)E給出取到達(dá)每一目的地的最小值C除外）Ifdelaysarerecordedas8-bitnumbersina50-routernetwork,anddelayvectorsareexchangedtwiceasecond,howmuchbandwidthper(full-duplex)lineischewedupbythedistributedroutingalgorithm?Assumethateachrouterhasthreelinestootherrouters.2次，因此即在每條線(xiàn)路的每個(gè)方向上消耗的帶寬都800。5-12Forhierarchicalroutingwith4800routers,whatregionandclustersizesshouldbechosentominimizethesizeoftheroutingtableforathree-layerhierarchy?Agoodstartingplaceisthehypothesisthatasolutionwithkclustersofkregionsofkroutersisclosetooptimal,whichmeansthatkisaboutthecuberootof4800(around16).Usetrialanderrortocheckoutcombinationswhereallthreeparametersareinthegeneralvicinityof16.所謂分級(jí)路由，就是將路由器按區(qū)REGIO）何為分組選擇路由到達(dá)目的地的細(xì)節(jié)第35頁(yè)共59頁(yè)結(jié)構(gòu)是不夠的，還可以把區(qū)組合成簇CLUSTE），把簇再組合成域ZON），對(duì)于等級(jí)式路由，在路由表中對(duì)應(yīng)