Solution Manual For Data Networks, 2nd Edition

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SOLUTIONSMANUAL|SecondEditionDIMITRIBERTSEKASMassachusertsInstituteofTechnologyROBERTGALLAGERMassachusettsInstituteofTechnologyPRENTICEHALL,UpperSaddleRiver,NewJersey07458

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DownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com

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SolutionsManual--DataNetworks,2/EbyDimitriBertsekasandRobertGallagerTABLEOFCONTENTSCHAPTER3toouueereereereseesssesseessessssssssssssssasssssesesssesssssssssssssassnsssssssossessesesssmsensnsessossessssseenses23CHAPTER4cooovoneereeneneeseesesssssssssasasassnssssssssssesssessessssssssssasnnsnsassessosessessassnsessesssssseneess|CHAPTER©coveveaeerneesseeecemmaaensssssessesssesssesesasmssmsenesssessssssssssssssssessansssessassssesesssesmsnesssssssessseees148+StudyXY

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CHAPTER1SOLUTIONS1.1Thereare250,000pixelspersquareinch,andmultiplyingbythenumberofsquareinchesandthenumberofbitsperpixelgives5.61x108bits.1.2a)Thereare16x109bitsgoingintothenetworkperhour.Thusthereare48x109bitsperhourtravelingthroughthenetwork,or13.33millionbitspersecond.Thisrequires209linksof64kbit/sec.each.b)Sinceatelephoneconversationrequirestwopeople,and10%ofthepeoplearebusyontheaverage,wehave50,000simultaneouscallsontheaverage,whichrequires150,000linksontheaverage.Boththeanswerina)andb)mustbemultipliedbysomefactortoprovideenoughlinkstoavoidcongestion(andtoprovidelocalaccessloopstoeachtelephone),butthepointoftheproblemistoillustratehowlittledata,bothinabsoluteandcomparativeterms,isrequiredforordinarydatatransactionsbypeople.1.3Therearetwopossibleinterpretationsoftheproblem.Inthefirst,packetscanbearbitrarilydelayedorlostandcanalsogetoutoforderinthenetwork.Inthisinterpretation,ifapacketfromAtoBissentattimeTandnotreceivedbysomelatertimet,thereisnowaytotellwhetherthatpacketwilleverarrivelater.ThusifanydatapacketorprotocolpacketfromAtoBislost,nodeBcanneverterminatewiththeassurancethatitwillneverreceiveanotherpacket.Inthesecondinterpretation,packetscanbearbitrarilydelayedorlost,butcannotgetoutoforder.Assumethateachnodeisinitiallyinacommunicationstate,exchangingdatapackets.Theneachnode,perhapsatdifferenttimes,goesintoastateorsetofstatesinwhichitsendsprotocolpacketsinanattempttoterminate.Assumethatanodecanenterthefinalterminationstateonlyonthereceiptofoneoftheseprotocolpackets(sincetiminginformationcannothelp,sincethereisnosideinformation,andsinceanydatapacketcouldbefollowedbyanotherdatapacket).Asinthethreearmyproblem,assumeanyparticularorderinginwhichthetwonodesreceiveprotocolpackets.Thefirstnodetoreceiveaprotocolpacketcannotgotothefinalterminationstatesinceithasnoassurancethatanyprotocolpacketwilleverbereceivedbytheothernode,andthusnoassurancethattheothernodewilleverterminate.Thenextprotocolpackettobereceivedthenfindsneithernodeinthefinalterminationstate.Thusagainthereceivingnodecannotterminatewithoutthepossibilitythattheothernodewillreceivenomoreprotocolpacketsandthusneverterminate.Thesamesituationoccursoneachreceivedprotocolpacket,andthusitisimpossibletoguaranteethatbothnodescaneventuallyterminate.Thisisessentiallythesameargumentasusedforthethreearmyproblem.—StudyXY

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CHAPTER2SOLUTIONS2.1Letx(t)betheoutputforthesinglepulseshowninFig.2.3(a)andlety(t)betheoutputforthesequenceofpulsesinFig.2.3(b).Theinputfor2.3(b)isthesumofsixinputpulsesofthetypein2.3(a);thefirstsuchpulseisidenticaltothatof2.3(a),thesecondisdelayedbyTtimeunits,thethirdisinvertedanddelayedby2Ttimeunits,etc.Fromthetimeinvarianceproperty,theresponsetothesecondpulseaboveisx(t-T)(i.e.x(t)delayedbyT);fromthetimeinvarianceandlinearity,theresponsetothethirdpulseis-x(t-2T).Usinglinearitytoaddtheresponsestothesixpulses,theoveralloutputisy(©)=x(®)+x(t-T)-x(t-2T)+x(t-3T)-x(t-4T)-x(t-5T)Toputtheresultinmoreexplicitform,notethat0Ht<0XO=J1_g;0<t<TEnet>TThustheresponsefromtheithpulse(1<i<6)iszerouptotime(i-1)T.Fort<0,then,y©=0;from0<t<TyO=x@®=1-¢2T;0<t<TFromT<1t<2T,y(©)=x(t)+x(t-T)=(e2-1)eT4[1-£20DM=1-eTSimilarly,for2T£t<3T,y(®)=x(t)+x(t-T)-x(t-2T)=(€2-1)eVT+(2-1)e2T/T_[]_e202DIT)=-1+Qet1)e?T;2T<t<3TAsimilaranalysisforeachsubsequentintervalleadstoy(t)=1-(2e6-2e4+1)e2/T;3T<t<4T=-1+(2e8-2e6+2e4-eT;4T<t<6T=-(e12-2e8+2e6-2et+eT|126T+StudyXY

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Thesolutioniscontinuousovertwithslopediscontinuitiesat0,2T,3T,4T,and6T;thevalueofy(t)atthesepointsisy(0)=0;y2T)=.982;y(3T)=-.732;y(4T)=.766;y(6T)=-968.Anotherapproachtotheproblemthatgetsthesolutionwithlessworkistousex(t)tofirstfintheresponsetoaunitstepandthenviewy(t)astheresponsetoasumofdisplacedunitsteps.2.2Fromtheconvolutionequation,Eq.(2.1),theoutput(t)is-Trt)=Is(Dh(t-t)dt=Ih(t-t)dt~=0Notethath(t-T)=oe*t-0fort-1>0,(i.e.fort<t),andh(t-t)=0fort>t.Thusfort<0,h(t-)=0throughouttheintegrationintervalabove.For0<t<T,wethenhavetTw=ears|0dr=1-¢™;0<t<T=0=Fort2T,h(t-T)=0e"*(t-®overtheentireintegrationintervalandT0)=]aeVgr=Dg>T=0Thustheresponseincreasestowards1for0<t<Twiththeexponentialdecayfactora,andthen,fort>T,decaystoward0.2.3FromEg.(2.1),=|7hydeUsingT'=t-Tasthevariableofintegrationforanygivent,=|STp(tdr_omI(pyar=2gwhereH(f)isasgiveninEq.(2.3).2.4+ooh(t)=[H(fel2nfdf+StudyXY

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SinceH(f)is1fromfgtofoand0elsewhere,wecanintegrateexp(j2rft)from-fjtofo,obtainingbn1ofrf]=sin(2nfor)(1)=Zrlexplnfl)—exp(—i2nfyt)]=—Notethatthisimpulseresponseisunrealizableinthesensethattheresponsestartsbeforetheimpulse(and,evenworse,startsaninfinitetimebeforetheimpulse).Nonetheless,suchidealfiltersareusefulabstractionsinpractice.2.5Thefunctions1(t)iscompressedbyafactorofBonthetimeaxisasshownbelow-AA-ARARS,(f)=i5;(e2dr=I(Bye2d=idt1(f=|seo5s(%)[soBBBS(f)ThusS(f)isattenuatedbyafactorofBinamplitudeandexpandedbyafactorof§onthefrequencyscale;compressingafunctionintimeexpandsitinfrequencyandviceversa.2.6a)Weusethefactthatcos(x)=[exp(jx)+exp(-jx)]/2.ThustheFouriertransformofs(t)cos(2nfor)is=exp(j2rfyt)+exp(—j2nfytIspTREE7PCR)exp(—j2nft)dt=Ruexp[—2n(f-fo)]dt+0expl[—j2n(f+fo)dtS(f-fy) S(f+fy)STtz+StudyXY

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b)Hereweusetheidentitycos2(x)=[1+cos(2x)}/2.ThustheFouriertransformofs(t)cos2(2pfot)istheFouriertransformofs(t)/2plustheFouriertransformofs(t)cos[2p(2fp)t]/2.Usingtheresultinparta,thisisS()/2+S(f-2fo)/4+S(f+2fp)/4.2.7a)E{frametimeon9600bpslink}=1000bits/9600bps=0.104sec.E{frametimeon50,000bpslink}=0.02sec.b)E{timefor106frameson9600bpslink}=1.04-105sec.E{timefor106frameson50,000bpslink}=2-104sec.Sincetheframelengthsarestatisticallyindependent,thevarianceofthetotalnumberofbitsin106framesis106timesthevarianceforoneframe.Thusthestandarddeviationofthetotalnumberofbitsin106framesis103timesthestandarddeviationofthebitsinoneframeor5:10bits.ThestandarddeviationofthetransmissiontimeisthenS.D.{timefor106frameson9600bpslink}=5-105/9600=52sec.S.D.{timefor106frameson50,000bpslink}=5:105/50,000=10sec.¢)Thepointofalltheabovecalculationsistoseethat,foralargenumberofframes,theexpectedtimetotransmittheframesisverymuchlargerthanthestandarddeviationofthetransmissiontime;thatis,thetimeperframe,averagedoveraverylongsequenceofframes,isclosetotheexpectedframetimewithhighprobability.One'sintuitionwouldthensuggestthatthenumberofframesperunitlime,averagedoveraverylongtimeperiod,isclosetothereciprocaloftheexpectedframetimewithhighprobability.Thisintuitioniscorrectandfollowseitherfromrenewaltheoryorfromdirectanalysis.Thusthereciprocaloftheexpectedframetimeistherateofframetransmissionsintheusualsenseoftheword"rate".2.8Letxijbethebitinrowi,columnj.Thentheithhorizontalparitycheckishi=Zjxjwherethesummationissummationmodulo2.Summingbothsidesofthisequation(modulo2)overtherowsi,wehaveZihi=ZijxijThisshowsthatthemodulo2sumofallthehorizontalparitychecksisthesameasthemodulo2sumofallthedatabits.Thecorrespondingargumentoncolumnsshowsthatthemodulo2sumoftheverticalparitychecksisthesame.+StudyXY

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2.9a)Anypatternoftheform--110--—--011--—-101--willfailtobedetectedbyhorizontalandverticalparitychecks.Moreformally,foranythreerowsiy,i2,andi3,andanythreecolumnsjj,j2,and3,apatternofsixerrorsinpositions(i130),(3172)»(2J2)»(i2j3)»(3j1),and(i3j3)willfailtobedetected.b)Thefourerrorsmustbeconfinedtotworows,twoerrorsineach,andtotwocolumns,twoerrorsineach;thatis,geometrically,theymustoccurattheverticesofarectanglewithinthearray.AssumingthatthedatapartofthearrayisJbyK,thenthearrayincludingtheparitycheckbitsisJ+1byK+1.Thereare(J+1)J/2differentpossiblepairsofrows(countingtherowofverticalparitychecks),and(K+1)K/2possiblepairsofcolumns(countingthecolumnofhorizontalchecks).Thusthereare(J+1)(K+1)JK/4undetectablepatternsoffourerrors.2.10Letx=(x1,X2,--XN)andx’=(x'},X'2,---X'N)beanytwodistinctcodewordsinaparitycheckcode.HereN=K+Listhelengthofthecodewords(KdatabitsplusLcheckbits).Lety=(y1,-YN)beanygivenbinarystringoflengthN.LetD(x,y)bethedistancebetweenxandy(i.e.thenumberofpositionsiforwhichx;#y;).SimilarlyletD(x',y)andD(x,x’)bethedistancesbetweenx'andyandbetweenxandx'.WenowshowthatD(x.x")<D(x,y)+D(x"y)Toseethis,visualizechangingD(x,y)bitsinxtoobtainy,andthenchangingD(x',y)bitsinytoobtainx".Ifnobithasbeenchangedtwiceingoingfromxtoyandthentox',thenitwasnecessarytochangeD(x,y)+D(x,y)bitstochangextox’andtheaboveinequalityissatisfiedwithequality.Ifsomebitshavebeenchangedtwice(i.e.x;=Xj#yjforsomei)thenstrictinequalityholdsabove.Bydefinitionoftheminimumdistancedofacode,D(x,x)2d.Thus,usingtheaboveinequality,ifD(x,y)<d/2,thenD(x"y)>d/2.Nowsupposethatcodewordxissentandfewerthand/2errorsoccur.ThenthereceivedstringysatisfiesD(x,y)<d/2andforeveryothercodewordx',D(x',y)>d/2.Thusadecoderthatmapsyintotheclosestcodewordmustselectx,showingthatnodecodingerrorcanbemadeiffewerthand/2channelerrorsoccur.Notethatthisargumentappliestoanybinarycoderatherthanjustparitycheckcodes.2.11Thefirstcodewordgiven,1001011hasonlythefirstdatabitequalto1andhasthefirst,third,andfourthparitychecksequalto1.Thusthoseparitychecksmustcheckonthefirstdatabit.Similarly,fromthesecondcodeword,weseethatthefirst,second,andfourthparitychecksmustcheckonthesecondbit.Fromthethirdcodeword,thefirst,second,andthirdparitycheckeachcheckonthethirddatabit.Thus+StudyXY

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C1=S81+82+583cp=572+83c3=5]+83C4=81+82Thesetofallcodewordsisgivenby00000000011110100101110101010101101011001111001101111000Theminimumdistanceofthecodeis4,ascanbeseenbycomparingallpairsofcodewords.Aneasierwaytofindtheminimumdistanceofaparitycheckcodeistoobservethatifxandx'areeachcodewords,thenx+x'(usingmodulo2componentwiseaddition)isalsoacodeword.Ontheotherhand,x+x'hasa1inapositionifandonlyifxandx’differinthatposition.Thusthedistancebetweenxandx'isthenumberofonesinx+x".Itfollowsthattheminimumdistanceofaparitycheckcodeistheminimum,overallnon-zerocodewords,ofthenumberofonesineachcodeword.2.12Dp*+p%D+1)D'+D+D?p’+Dp%p*+D’p?=Remainder2.13Letz(D)=Di+zj.Di"!+..+Diandassumei<j.MultiplyingG(D)timesZ(D)thenyieldsg(D)z(D)=DL+(zj.1+gL.)DLA+(72+g17j-1+gL)2++(g1+zi)DI*1+DiClearlythecoefficientofDL+jandthecoefficientofDiareeach1,yieldingthedesiredtwonon-zeroterms.Theabovecasei<jariseswheneverz(D)hasmorethanonenon-zeroterm.Forthecaseinwhichz(D)hasonlyonenon-zeroterm,i.e.z(D)=Djforsomej,wehaveg(D)z(D)=DL#j+gi|DL#i-1++Diwhichagainhasatleasttwonon-zeroterms.2.14Supposeg(D)contains(1+D)asafactor;thusg(D)=(1+D)h(D)forsomepolynomialh(D).Substituting1forDandevaluatingwithmodulo2arithmetic,wegetg(1)=0becauseoftheterm(1+D)=(1+1)=0.Lete(D)bethepolynomialforsomearbitraryundetectableerrorsequence.Thene(D)=g(D)z(D)forsomez(D),andhencee(1)=g(1)z(1)=0.Nowe(D)=Z;eiD},soe(1)=Ziej.Thuse(1)=0impliesthatanevennumberofelementsej+StudyXY

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are1;i.e.thate(D)correspondstoanevennumberoferrors.Thusallundetectableerrorsequencescontainanevennumberoferrors;anyerrorsequencewithanoddnumberoferrorsisdetected.2.15a)LetDi*L,dividedbyg(D),havethequotientz®(D)andremainderc®(D)sothatDi+L=gD)z()(D)+ci}(D)Muitiplyingbys;andsummingoveri,s(D)DL=5;5iz®(D)+Z;sic(D)SinceZ;sic®(D)hasdegreelessthanL,thismustbetheremainder(andZ;siz@(D)thequotient)ondividings(D)DLbyg(D).Thus¢(D)=%;sic)(D).b)Twopolynomialsareequalifandonlyiftheircoefficientsareequal,sotheabovepolynomialequalityimpliescj=Zisici®2.16a)Considerthetwoscenariosbelowandnotethatthesescenariosareindistinguishabletothereceiver.ofx]olx|K\fApacket19accepted+StudyXY

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orpacket19acceptedIfthereceiverreleasesthepacketasx3inthequestionedreception,thenanerroroccursonscenario2.Ifthereceiverreturnsanackbutdoesn'treleaseapacket(i.e.theappropriateactionforscenario2),thenunderscenario1,thetransmittererroneouslygoesontopacket3.Finally,ifthereceiverreturnsanak,theproblemisonlypostponedsincethetransmitterwouldthentransmit(2,x7)inscenario1and(2,x1)inscenario2.Asexplainedonpage66,packetsxandx;mightbeidenticalbitstrings,sothereceivercannotresolveitsambiguitybythebitvaluesofthepackets.b)Thescenariosbelowdemonstrateincorrectoperationforthemodifiedconditions.olx|][ofx»|[1]xpacket19accepted)[ox]packet19?accepted+StudyXxy

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2.17a)T=Ti+Ts+2T4b)q=(1-po(1-pp)Apacketistransmittedoncewithprobabilityq,twicewithprobability(1-q)q,threetimeswithprobability(1-q)2g,etc.Thustheexpectednumberoftransmissionsofagivenpacketisx1E{transmissionsperpacket}=yiq1—q)"™=3i=lToverifytheabovesummation,notethatforanyx,0<x<1,CeVddNTER12x“2%=mxTax\Tx)Tq?i=1i=liol(1-x)Usingxfor(1-q)abovegivesthedesiredresult.©)E({tmeperpacket}=(T;+Tr+2T@)/q=(1.3)/0.998=1.303Notethatp;andpghaveverylittleeffectonE[timeperpacket]instopandwaitsystemsunlesstheyareunusuallylarge.2.180Packets12346delivered4ICNCANENodeA\A/\/\#t—pAsn|©1234|s|eWindow[0,3][1,4]2,51|[3.61|(4.71+StudyXxy

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Assumethatthetransmitteralwayssendsthenextpacketinorderuntilreachingtheendofthewindow,andthengoesbacktothebeginningofthewindow.0Packets12deliveredRNoTolofvJu[1|NodeA\A\f\f\/\/\/t—Window[0.3][1,4][2,5]2.19Thesimplestsuchdeadlockoccursifthereissufficientpropagationdelayinthesystemthateachsidecansendn-1frames(containingpacketsnumbered0ton-2)beforefinishingreceiptofthefirstframefromtheotherside.Inthiscase,thenthframefromeachsidewillcarrythepacketnumberedn-1withoutackinganypacketsfromtheotherside.Thuseachsidewillgobacktopacket0,butintheabsenceoferrors,eachsidewillbelookingforpacketnbytimetherepeatofpacket0occurs.Eachsidewillthencyclefrom0ton-1,andneithersidewilleverreceiveanyacks.Thediagrambelowillustratesthisforn=3.ThefirstnumberineachframepositionisSNandthesecondisRN.012(0,0)2,0)(0,0)(1,0)NNPOPaPaVaVaN0,0)(1,0)(2,0)0,0)(1,0)0122.20ThesimplestexampleisfornodeAtosendpackets0throughn-1inthefirstnframes.Incaseofdelayedacknowledgements(i.e.noreturnpacketsintheinterim),nodeAgoesbackandretransmitspacket0.Iftheothernodehasreceivedallthepackets,itiswaitingforpacketn,andifthemodulusmequalsn,thisrepeatofpacket0isinterpretedaspacketn.+StudyXxy

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TherighthandsideofEq.(2.24)issatisfiedwithequalityifSN=SNpn(t;)+n-1.ThisoccursifnodeAsendspackets0throughn-1inthefirstnframeswithnoreturnpacketsfromnodeB.ThelastsuchframehasSN=n-1,whereasSNp;,atthattime(sayt;)is0.Continuingthisscenario,wefindanexamplewheretherighthandsideofEq.(2.25)issatisfiedwithequality.Ifalltheframesabovearecorrectlyreceived,thenafterthelastframe,RNbecomesequalton.IfanotherframeissentfromA(nowcallthistimet;)andifSNisstillO,thenwhenitisreceivedatB(sayattp),wehaveRN(t2)=SNppn(t1)+n.2.21LetRN(t)bethevalueofRNatnodeBatanarbitrarytimeT;SNp;nisthesmallestpacketnumbernotyetacknowledgedatAattimet(whichisregardedasfixed)andSNmax-1isthelargestpacketnumbersentfromAattimet.SinceRN(1)isnondecreasingin7,itissufficienttoshowthatRN(t+Tp+Tg)€SNpaxandtoshowthatRN(t-Try-Tq)2SNpjn.Forthefirstinequality,notethatthepacketnumberedSNyax(bydefinitionofSNpax)hasnotenteredtheDLCunitatnodeAbytimet,andthuscannothavestartedtransmissionbytimet.SincethereisadelayofatleastTm+T4fromthetimeapackettransmissionstartsuntilthecompletionofitsreception,packetSNpaxcannothavebeenreceivedbytimet+Tp+T4.Becauseofthecorrecmessoftheprotocol,RN(t+Ty+Tg)canbenogreaterthanthenumberofapacketnotyetreceived,i.e.SNmax.Forthesecondinequality,notethatforthetransmittertohaveagivenvalueofSNppattimet,thatvaluemusthavebeentransmittedearlierastherequestnumberinaframecomingbackfromnodeB.Thelatesttimethatsuchaframecouldhavebeenformedist-Tm-T4,sobythistimeRNmusthavebeenatleastSNmin.2.22a)Ifthetransmitterneverhastogobackorwaitintheabsenceoferrors,thenitcansendacontinuousstreamofnewpacketsintheabsenceoferrors.Inorderforsuchacontinuousstreamtobesent,eachpacketmustbeacknowledged(i.e.SNp;nmustadvancebeyondthepacket'snumber)beforethenextn-1framescompletetransmission.Thusthesen-1frametransmissiontimesareinaracewiththetime,first,forthegivenpackettopropagateoverthechanneland,second,fortheacknowledgementtowaitforthefeedbackframeinprogress,thenwaittobetransmittedinthenextfeedbackframeandpropagatedbacktotheoriginaltransmitter.Inorderforthefeedbacktoalwayswintherace,theminimumtimeforthen-1framestobetransmittedmustbegreaterthanthemaximumtimeforthefeedback,i.e.,(0-1)Tin>2Td+2TmaxTmax<[(0-1)/2]Tin-Tqb)Ifanisolatederroroccursinthefeedbackdirection,thefeedbackcouldbeheldupforoneadditonalframe,leadingto(0-1)Tmin>2Td+3TmaxTmax<[(0-1)/3]Tmin-(2/3)Tq+StudyXY

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2.23AfteragivenpacketistransmittedfromnodeA,thesecondsubsequentframetransmissionterminationfromBcarriestheacknowledgement(recallthattheframetransmissioninprogressfromBwhenAfinishesitstransmissioncannotcarrytheackforthattransmission;recallalsothatpropagationandprocessingdelaysarenegligible.Thusqistheprobabilityofn-1frameterminationsfromAbeforethesecondframeterminationfromB.Thiscanberephrasedastheprobabilitythatoutofthenextnframeterminationsfromeithernode,eithern-1orncomefromnodeA.SincesuccessiveframeterminationsareequallylikelyandindependentlyfromAorB,thisprobabilityisn_nlon_-na=pTan2-@*2i=n-12.24Ifanisolatederrorinthefeedbackdirectionoccurs,thentheackforagivenpacketisheldupbyoneframeinthefeedbackdirection(i.e.,thenumberRNinthefeedbackframefollowingthefeedbackframeinerrorreacknowledgestheoldpacketaswellasanynewpacketthatmighthavebeenreceivedintheinterim).Thusqisnowtheprobabilityofn-1frameterminationsfromAbefore3frameterminationsfromB(onefortheframeinprogress,onefortheframeinerror,andonefortheframeactuallycarryingtheack;seethesolutiontoproblem2.23).Thisistheprobabilitythatn-1ormoreofthenextn+1frameterminationscomefromA;sinceeachterminationisfromAorBindependentlyandwithequalprobability,[a+]_n:-n-1-n-1CED);a2[n+2+(n+1)n/2]2i=n-12.25Asinthesolutiontoproblem2.23,qistheprobabilityofn-1frameterminationscomingfromnodeAbeforetwoframeterminationscomefromnodeB.FrameterminationsfromA(andsimilarlyfromB)canberegardedasalternatepointsinaPoissonpointprocessfromA(orfromB).Therearetwocasestoconsider.Inthefirst,theinitalframeisreceivedfromAafteranevennumberedpointinthePoissonprocessatB,andinthesecond,theinitialframeisreceivedafteranoddnumberedpoint.Inthefirstcase,qistheprobabilitythat2n-2PoissoneventsfromAoccurbefore4PoissoneventsoccurfromB.Thisistheprobability,inacombinedPoissonpointprocessofPoissoneventsfromAandB,that2n-2ormorePoissoneventscomefromAoutofthenext2n+1eventsinthecombinedprocess.Inthesecondcase,qistheprobabilitythat2n-2PoissoneventsfromAoccurbefore3eventsoccurfromB.Sincethesecasesareequallylikely,2n+1@n+1)!2nQn)!1n+)!\oon1|gnm)!|,-2na=135Lo)=p;SE:2.40i1(2n+1-1)!2.4,il(2n-i)!2.26Weviewthesystemfromthereceiverandaskfortheexpectednumberofframes,7,arrivingatthereceiverstartingimmediatelyafteraframecontainingapacketthatisaccepted+StudyXY

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