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ExercisesreviewforFirst3Chapters11.13Whatistheprincipaldifferencebetweenconnectionlesscommunicationandconnection-orientedcommunication?Connection-orientedcommunicationhasthreephases.Intheestablishmentphase,arequestismadetosetupaconnection.Onlyafterthisphasehasbeensuccessfullycompletedcanthedatatransferphasebestartedanddatatransported.Thencomesthereleasephase.Connectionlesscommunicationdoesnothavethesephases.Itjustsendsthedata.51.24(e4:1.27)Howlongwasabitontheoriginal802.3standardinmeters?Useatransmissionspeedof10Mbpsandassumethepropagationspeedincoaxis2/3thespeedoflightinvacuum.在原先的802.3标准中,一比特有多少米长?传输速度为10Mbps,传播速度为真空中光速的2/3。Answer:Thespeedoflightincoaxisabout200,000km/sec,whichis200meters/sec.At10Mbps,ittakes0.1sectotransmitabit.Thus,thebitlasts0.1secintime,duringwhichitpropagates20meters.Thus,abitis20meterslonghere.在同轴电缆中光速为2*106km/s,即200m/S。10Mbps的传输速度,1比特要持续0.1S,在这段时间内它传播了20米。因此这里1bit有20m长。6信号的传播速度为v=2/3光速=2/3*3*108m/s,10Mbps即10Mbits用时1秒,所以1bit用时t=1/(10*106)s。1bit时间内信号延伸了v*t=……2.2Anoiseless4-kHzchannelissampledevery1msec.Whatisthemaximumdatarate?Ifanarbitrarysignalhasbeenrunthroughalow-passfilterofbandwidthB,thefilteredsignalcanbecompletelyreconstructedbymakingonly2B(exact)samplespersecond.Anoiselesschannelcancarryanarbitrarilylargeamountofinformation,nomatterhowoftenitissampled.Justsendalotofdatapersample.Forthe4kHzchannel,make8000samples/sec.Ifeachsampleis16bits,thechannelcansend128kbps.Ifeachsampleis1024bits,thechannelcansend8.192Mbps.Thekeywordhereis‘‘noiseless.’’Withanormal4kHzchannel,theShannonlimitwouldnotallowthis.7Max.datarate=2Blog2Vb/s2.3(e4:2.4)Ifabinarysignalissentovera3-kHzchannelwhosesignal-to-noiseratiois20dB,whatisthemaximumachievabledatarate?Asignal-to-noiseratioof20dBmeansS/N=100.Sincelog2101isabout6.658,theShannonlimitisabout19.975kbps.TheNyquistlimitis6kbps.ThebottleneckisthereforetheNyquistlimit,givingamaximumchannelcapacityof6kbps.8Max.datarate=Blog2(1+S/N)b/s2.36(e4:2.42)Comparethedelayinsendinganx-bitmessageoverak-hoppathinacircuit-switchednetworkandina(lightlyloaded)packet-switchednetwork.Thecircuitsetuptimeisssec,thepropagationdelayisdsecperhop,thepacketsizeispbits,andthedatarateisbbps.Underwhatconditionsdoesthepacketnetworkhavealowerdelay?比较电路交换网和负载轻的分组交换网中,在k跳路径上发送一条长x比特的消息的延时。其中电路建立时间为s秒,传播延迟为每跳d秒,分组大小为p比特,数据率为bbps。在什么条件下分组网的延时低些?Hop:“跳”.从一个设备转发到另一个设备,称为“一跳”11Switching(2)Circuitswitchingrequirescallsetup(connection)beforedataflowssmoothly–Alsoteardownatend(notshown)Packetswitchingtreatsmessagesindependently–Nosetup,butvariablequeuingdelayatroutersCircuit-switchednetworkPackets-switchednetwork建立时间Propagationdelay传播延时PacketsswitchPropagationdelay传播延时dTransmissiontime=p/b传输延时x/bPropagationdelay=d传播延时Transmissiontime=p/b传输延时Withcircuitswitching,att=sthecircuitissetup;att=s+x/bthelastbitissent;att=s+x/b+kdthemessagearrives.Withpacketswitching,thelastbitissentatt=x/b.Togettothefinaldestination,thelastpacketmustberetransmittedk−1timesbyintermediaterouters,eachretransmissiontakingp/bsec,sothetotaldelayisx/b+(k−1)p/b+kd.Packetswitchingisfasterifs(k−1)p/b.用电路交换,t=s秒时电路建立完毕;t=s+x/b时最后一比特传输完毕;t=s+x/b+kd时消息到达。用分组交换,t=x/b时最后一比特传输完毕(离开发送方),其中共[x/p]个分组。每经过一次中间交换需等待p/b的转发时间,共经过k-1个中间交换。k跳,传播延时为kd。故总耗时为x/b+(k−1)p/b+kd。欲x/b+(k−1)p/b+kds+x/b+kd,需s(k−1)p/b.14x/b+(k−1)p/b+kds+x/b+kd•Inadditiontothefastertransmissionundertheseconditions,packetswitchingispreferablewhenfault-tolerant能容忍错误的transmissioninthepresenceofswitchfailuresisdesired.2.37(e4:2.43)Supposethatxbitsofuserdataaretobetransmittedoverak-hoppathinapacket-switchednetworkasaseriesofpackets,eachcontainingpdatabitsandhheaderbits,withxp+h.Thebitrateofthelinesisbbpsandthepropagationdelayisnegligible.Whatvalueofpminimizesthetotaldelay?分组交换,xbits数据,需封装为[x/p]个packet,即待发bit数为s=(p+h)*x/pbits。在t=s/b时最后一比特传输完毕(离开发送方)。每经过一次中间交换需要的传输时间为ti=(p+h)/b,k跳的路径,共经过k-1个中间交换。传播延时忽略不计。故总耗时为s/b+(k-1)*ti=(p+h)*x/p/b+(k−1)(p+h)/b。16在每个中间设备处,一个分组完全到达后缓存起来,排队等候转发。本题说“lightlyloaded”意为无需排队,立即转发。在它转发期间,后续分组陆续到达。即一边接收第2分组,一边转发第1分组。依此类推。每经过一跳(一次中间交换)需要的传输时间为ti=(p+h)/b,k跳的路径,共经过k-1个中间交换。传播延时忽略不计,故总耗时为s/b+(k-1)*ti=(p+h)*x/p/b+(k−1)(p+h)/b。对p求导得:-hx/(bp2)+(k-1)/b,因为导数=0时原函数达到极值,显然是极小值。解方程-hx/(bp2)+(k-1)/b=0得:p=sqrt(hx/(k-1))17(p+h)*x/p/b(p+h)/b(p+h)/b(p+h)/b(p+h)/b到达(忽略传播延时,即路上不耗时间)离开183.8Toprovidemorereliabilitythanasingleparitybitcangive,anerror-detectingcodingschemeusesoneparitybitforcheckingalltheodd-numberedbitsandasecondparitybitforalltheeven-numberedbits.WhatistheHammingdistanceofthiscode?Makingonechangetoanyvalidcharactercannotgenerateanothervalidcharacterduetothenatureofparitybits.Makingtwochangestoevenbitsortwochangestooddbitswillgiveanothervalidcharacter,sothedistanceis2.193.11A12-bitHammingcodewhosehexadecimalvalueis0xE4Farrivesatareceiver.Whatwastheoriginalvalueinhexadecimal?Assumethatnotmorethan1bitisinerror.a1a3a5a7a9a110xE4F=111001001111a1=a3+a5+a7+a9+a111=1+0+0+1+1a2=a3+a6+a7+a10+a111=1+1+0+1+1a4=a5+a6+a7+a120=0+1+0+1a8=a9+a10+a11+a120=1+1+1+1Bit2(aparitybit)isincorrect.The12-bitvaluetransmittedwas0xA4F.Theoriginal8-bitdatavaluewas0xAF(101
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