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2013/6/21SolutionChapter114.Thepropagationdelayisthetimethatisrequiredfortheenergyofasignaltopropagatefromonepointtoanother.a.Findthepropagationdelayforasignaltraversingthefollowingnetworksatthespeedoflightincable(2.3x108meters/second):iibd10yacircuitboard10cmyaroom10myabuilding100myametropolitanarea100kmyacontinent5000kmyupanddowntoageostationarysatellite2x36000kmyupanddowntoageostationarysatellite2x36000km2013/6/22Solution:yTofindthepropagationdelay,dividedistancebythespeedoflightincable.Thuswehave:yacircuitboardtprop=4.347x10-10seconds8yaroomtprop=4.3478x10-8secondsyabuildingtprop=4.3478x10-7secondsyametropolitanareatprop=4.3478x10-4secondsyacontinenttprop=0.02174secondsyupanddowntoageostationarysatellitetprop=0.31304secondsb.Howmanybitsareintransitduringthepropagationdelayintheabovecases,ifbitsareenteringtheabovenetworksatthefollowingtransmissionspeeds:10,000bits/second;1megabit/second;100megabits/second;10gigabits/second.Solution:fyThenumberofbitsintransitisobtainedbymultiplyingthetransmissionrateRbythepropagationdelay:Distance(m)10Kbps1Mbps100Mbps10Gbps0.14.347x10−64.347x10−40.043474.3478104.3478x10−40.0434784.3478434.7801004.3478x10−30.4347843.4784347.8001000004.3478434.78434784.3478x1065000000217.42174021740002.174x108720000003130.4313040313040003.1304x1092013/6/2315.Inproblem14,howlongdoesittaketosendanL-bytefileandtoreceivea1-byteacknowledgmentback?LetL=109bytes.SliSolution:yThetotaltimerequiredtosendafileandreceiveanacknowledgmentofitsreceiptisgivenby:yttotal=Lmessage/R+Lack/R+2*tprop=Lmessage/R+Lack/R+2*d/cywhereListhemessagelengthinbitsListheywhereLmessageisthemessagelengthinbits,Lackistheacknowledgmentlengthinbits,Risthetransmissionbitrate,disthedistancetraversed,andcisthespeedoflight.Theaboveequationshowsthattherearetwomainfactorsthatdeterminetotaldelay:1.MessageandACKtransmissiontime,whichdependsonthemessagelengthandthetransmissionbitrate;2.Propagationdelay,whichdependssolelyondistance.yWhenthepropagationdelayissmall,messageandACKtransmissiontimesdeterminethetotaldelayOntheothertransmissiontimesdeterminethetotaldelay.Ontheotherhand,whenthebitratebecomesverylarge,thepropagationdelayprovidesadelaycomponentthatcannotbereducednomatterhowfastthetransmissionratebecomes.yThetablesbelowshowthetwomaincomponentsofthetotaldelayinmicrosecondsThemessagetransmissiontimeisdelayinmicroseconds.Themessagetransmissiontimeisshowninredandthepropagationdelayisshowninblue.Theentriesinthetotaldelaytablesarecoloredaccordingtowhichdelaycomponentisdominant.2013/6/24Message+ACKdelay@10kbpsmessage+ACKdelay@1Mbpsmessage+ACKdelay@100Mbpsmessage+ACKdelay@10Gbps8.00E+118.00E+098.00E+078.00E+05Table1:Messagelength=109bytesDistance(meters)2*prop.delay(microseconds)Totaldelay@10kbpstotaldelay@1Mbpstotaldelay@100Mbpstotaldelay@10Gbps0.10.000878.00E+118.00E+098.00E+078.00E+05100.0869578.00E+118.00E+098.00E+078.00E+051000.8695658.00E+118.00E+098.00E+078.00E+05100000869.56528.00E+118.00E+098.00E+078.01E+05500000043478.268.00E+118.00E+098.00E+078.43E+05720000003130448.00E+118.00E+098.06E+071.13E+06Message+ACKdelay@10kbpsmessage+ACKdelay@1Mbpsmessage+ACKdelay@100Mbpsmessage+ACKdelay@10Gbps8.00E+118.00E+098.00E+078.00E+05Table1:Messagelength=1000bytesDistance(meters)2*prop.delay(microseconds)Totaldelay@10kbpstotaldelay@1Mbpstotaldelay@100Mbpstotaldelay@10Gbps0.10.00087800800.0018008.0008780.080870.80167100.086957800800.0878008.0869680.166960.8877571000.869565800800.8698008.8695780.949571.670365100000869.5652801669.5658877.56522949.6452870.366500000043478.26844278.26051486.260943558.3443479.06720000003130441113044321044313124313044.82013/6/21SolutionChapter219.SupposeanapplicationlayerentitywantstosendanL-bytemessagetoitspeerprocess,usinganexistingTCPconnection.TheTCPsegmentconsistsofthemessageplus20bytesofheader.ThesegmentisencapsulatedintoanIPpacketthathasanadditional20bytesofheaderTheIPpacketinturngoesanadditional20bytesofheader.TheIPpacketinturngoesinsideanEthernetframethathas18bytesofheaderandtrailer.Whatpercentageofthetransmittedbitsinthephysicallayercorrespondtomessageinformation,ifL=100bytes,500bytes,1000bytes?2013/6/22Solution:yTCP/IPoverEthernetallowsdataframeswithapayloadsizeupto1460bytes.Therefore,L=100,500and1000bytesarewithinthislimit.yThhdildyThemessageoverheadincludes:yTCP:20bytesofheaderyIP:20bytesofheaderyEthernet:total18bytesofheaderandtrailer.yThereforeyL=100bytes100/158=633%efficiencyyL=100bytes,100/158=63.3%efficiency.yL=500bytes,500/558=89.6%efficiency.yL=1000bytes,1000/1058=94.5%efficiency.20.SupposethattheTCPentityreceivesa1.5megabytefilefromtheapplicationlayerandthattheIPlayeriswillingtocarryblocksofmaximumsize1500bytes.Calculatetheamountofoverheadincurredfromsegmentingthefileintopacket-sizedunits.units.Solution:y1500-20-20=1460bytesy1.5Mbyte/1460byte=1027.4≈1028,therefore1028blocksareneededtotransferthefile.yOverhead=(20+20)×1028=41120bytesOverhead(20+20)×102841120bytesyOverhead=41120/(41120+1.5M)=2.668%≈2.7%2013/6/2321.SupposeaTCPentityreceivesadigitalvoicestreamfromtheapplicationlayer.Thevoicestreamarrives