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CHAPTER1

15.With16bits,wecanrepresentupto216differentcolors.

17.

a.Meshtopology'.Ifoneconnectionfails,theotherconnectionswillstillbeworking.

b.Startopology:Theotherdeviceswillstillbeabletosenddatathroughthehub;

therewillbenoaccesstothedevicewhichhasthefailedconnectiontothehub.

c.BusTopology:Alltransmissionstopsifthefailureisinthebus.Ifthedrop-line

fails,onlythecorrespondingdevicecannotoperate.

2

d.RingTopology,.Thefailedconnectionmaydisablethewholenetworkunlessit

isadualringorthereisaby-passmechanism.

19.Theoretically,inaringtopology,unpluggingonestation,interruptsthering.However,

mostringnetworksuseamechanismthatbypassesthestation;theringcan

continueitsoperation.

21.SeeFigure1.1

23.

a.E-mailisnotaninteractiveapplication.Evenifitisdeliveredimmediately,it

maystayinthemail-boxofthereceiverforawhile.Itisnotsensitivetodelay.

b.Wenormallydonotexpectafiletobecopiedimmediately.Itisnotverysensitive

todelay.

c.SurfingtheInternetistheanapplicationverysensitivetodelay.Weexceptto

getaccesstothesitewearesearching.

25.Thetelephonenetworkwasoriginallydesignedforvoicecommunication;the

Internetwasoriginallydesignedfordatacommunication.Thetwonetworksare

similarinthefactthatbotharemadeofinterconnectionsofsmallnetworks.The

telephonenetwork,aswewillseeinfuturechapters,ismostlyacircuit-switched

network;theInternetismostlyapacket-switchednetwork.

Figure1.1SolutiontoExercise21

Station

StationStation

Repeater

Station

StationStation

Repeater

Station

StationStation

Repeater

Hub

1

CHAPTER2

Exercises

15.TheInternationalStandardsOrganization,ortheInternationalOrganizationof

Standards.(ISO)isamultinationalbodydedicatedtoworldwideagreementon

internationalstandards.AnISOstandardthatcoversallaspectsofnetworkcommunications

istheOpenSystemsInterconnection(OSI)model.

2

17.

a.Reliableprocess-to-processdelivery:transportlayer

b.Routeselection:networklayer

c.Definingframes:datalinklayer

d.Providinguserservices:applicationlayer

e.Transmissionofbitsacrossthemedium:physicallayer

19.

a.Formatandcodeconversionservices:presentationlayer

b.Establishing,managing,andterminatingsessions:sessionlayer

c.Ensuringreliabletransmissionofdata:datalinkandtransportlayers

d.Log-inandlog-outprocedures:sessionlayer

e.Providingindependencefromdifferentdatarepresentation:presentationlayer

21.SeeFigure2.1.

23.Beforeusingthedestinationaddressinanintermediateorthedestinationnode,the

packetgoesthrougherrorcheckingthatmayhelpthenodefindthecorruption

(withahighprobability)anddiscardthepacket.Normallytheupperlayerprotocol

willinformthesourcetoresendthepacket.

25.Theerrorsbetweenthenodescanbedetectedbythedatalinklayercontrol,butthe

erroratthenode(betweeninputportandoutputport)ofthenodecannotbe

detectedbythedatalinklayer.

Figure2.1SolutiontoExercise21

B/42C/82

A/40

Sender

Sender

LANILAN2

RI

D/80

4240ADijDataT28082ADijDataT2

CHAPTER3

17.

a.f=l/T=l/(5s)=0.2Hz

b.f=l/T=1/(12闖=83333Hz=83.333x103Hz=83.333KHz

c.f=l/T=1/(220ns)=4550000Hz=4.55x106Hz=4.55MHz

19.SeeFigure3.1

21.Eachsignalisasimplesignalinthiscase.Thebandwidthofasimplesignalis

zero.Sothebandwidthofbothsignalsarethesame.

23.

a.(10/1000)s=0.01s

b.(8/1000)s=0.008s=8ms

2

c.((100,000x8)/1000)s=800s

25.Thesignalmakes8cyclesin4ms.Thefrequencyis8/(4ms)=2KHz

27.Thesignalisperiodic,sothefrequencydomainismadeofdiscretefrequencies,as

showninFigure3.2.

29.

Usingthefirstharmonic,datarate=2x6MHz=12Mbps

Usingthreeharmonics,datarate=(2x6MHz)/3=4Mbps

Usingfiveharmonics,datarate=(2x6MHz)/5=2.4Mbps

31.-10=101ogio(P2/5)-logio(P2/5)=-1-(P2/5)=10-1->P2=0.5W

33.100,000bits/5Kbps=20s

35.1pmx1000=1000pm=1mm

37.Wehave

4,000log2(l+10/0.005)=43,866bps

39.Torepresent1024colors,weneedlog21024=10(seeAppendixC)bits.Thetotal

numberofbitsare,therefore,

1200x10()0x1()=12,000,000bits

41.Wehave

SNR=(signalpower)/(noisepower).

However,powerisproportionaltothesquareofvoltage.Thismeanswehave

Figure3.1SolutiontoExercise19

Figure3.2SolutiontoExercise27

02()5()100200

Frequencydomain

Bandwidth=2(X)-0=200

Amplitude

1()volts

Frequency

3()

KHz

1()

KHz

3

SNR=[(signalvoltage)2]/[(noisevoltage"]=

[(signalvoltage)/(noisevoltage)]!=20z=400

Wethenhave

SNRdB=10logioSNR=26.02

43.

a.Thedatarateisdoubled(C2=2xCi).

b.WhentheSNRisdoubled,thedatarateincreasesslightly.Wecansaythat,

approximately,(C2=Ci+1).

45.Wehave

transmissiontime=(packetlength)/(bandwidth)=

(8,000,000bits)/(200,000bps)=40s

47.

a.Numberofbits=bandwidthxdelay=1Mbpsx2ms=2000bits

b.Numberofbits=bandwidthxdelay=10Mbpsx2ms=20,000bits

c.Numberofbits=bandwidthxdelay=100Mbpsx2ms=200,000bits

1

CHAPTER4

13.Weusetheformulas=cxNx(1/r)foreachcase.Weletc=1/2.

a.r=1-?s=(1/2)x(1Mbps)x1/1=500kbaud

b.r=1/2fs=(1/2)x(1Mbps)xl/(l/2)=1Mbaud

c.r=2-s=(l⑵x(1Mbps)x1/2=250Kbaud

d.r=4/3-s=(l/2)x(1Mbps)/l/(4/3)=375Kbaud

15.SeeFigure4.1Bandwidthisproportionalto(3/8)Nwhichiswithintherangein

Table4.1(B=0toN)fortheNRZ-Lscheme.

17.SeeFigure4.2.Bandwidthisproportionalto(12.5/8)Nwhichiswithintherange

inTable4.1(B=NtoB=2N)fortheManchesterscheme.

2

19.SeeFigure4.3.Bisproportionalto(5.25/16)NwhichisinsiderangeinTable4.1

(B=0toN/2)for2B/1Q.

21.Thedatastreamcanbefoundas

a.NRZ-I:10011001.

b.DifferentialManchester:11000100.

c.AMI:01110001.

23.Thedatarateis100Kbps.Foreachcase,wefirstneedtocalculatethevaluef7N.

WethenuseFigure4.8inthetexttofindP(energyperHz).Allcalculationsare

approximations.

a.f/N=0/100=0-P=0.0

b.f/N=50/100=1/2-P=0.3

c.f/N=100/100=1-P=0.4

d.f/N=150/100=1.5-P=0.0

Figure4.1SolutiontoExercise15

Figure4.2SolutiontoExercise17

00000000

1111111100110011

01010101

Casea

Caseb

Casec

Cased

AverageNumberofChanges=(0+0+8+4)/4=3forN=8

B(3/8)N

00000000

1111111100110011

01010101

Casea

Caseb

Casec

Cased

AverageNumberofChanges=(15+15+8+12)/4=12.5forN=8

B(12.5/8)N

3

25.In5B/6B,wehave25=32datasequencesand26=64codesequences.Thenumber

ofunusedcodesequencesis64-32=32.In3B/4B,wehave23=8data

sequencesand24=16codesequences.Thenumberofunusedcodesequencesis

16-8=8.

27

a.Inalow-passsignal,theminimumfrequency0.Therefore,wehave

fmax=0+200=200KHz.Tfs=2x200,000=400,000samples/s

b.Inabandpasssignal,themaximumfrequencyisequaltotheminimumfrequency

plusthebandwidth.Therefore,wehave

fmax=100+200=300KHz.-?fs=2x300,000=600,000samples/s

29.Themaximumdataratecanbecalculatedas

Nmax=2xBxnb=2x200KHzxlog24=800kbps

31.Wecancalculatethedatarateforeachscheme:

Figure4.3SolutiontoExercise19

a.NRZN=2xB=2x1MHz=2Mbps

b.Manchester-?N=1xB=1x1MHz=1Mbps

c.MLT-3N=3xB=3x1MHz=3Mbps

d.2B1QN=4xB=4x1MHz=4Mbps

1111111111111111

00000000000000000110011001100110

+3

+1

-3

-1

+3

+1

-3

-1

+3

+1

-3

-1

0011001100110011

+3

+1

-3

-1

Casea

Caseb

Casec

Cased

AverageNumberofChanges=(0+7+7+7)/4=5.25forN=16

B(5.25/8)N

1

CHAPTER5

11.WeusetheformulaS=(1/r)xN,butfirstweneedtocalculatethe

valueofrfor

eachcase.

a.r=log22=1S=(1/1)x(2000bps)=2000baud

b.r=log22=1-S=(1/1)x(4000bps)=4000baud

c.r=log24=2->S=(1/2)x(6000bps)=3000baud

d.r=log264=6-S=(1/6)x(36,000bps)=6000baud

2

13.Weusetheformular=logzLtocalculatethevalueofrforeachcase.

15.SeeFigure5.1

a.ThisisASK.Therearetwopeakamplitudesbothwiththesamephase(0

degrees).ThevaluesofthepeakamplitudesareAi=2(thedistancebetween

thefirstdotandtheorigin)andA2=3(thedistancebetweentheseconddotand

theorigin).

b.ThisisBPSK,Thereisonlyonepeakamplitude(3).Thedistancebetweeneach

dotandtheoriginis3.However,wehavetwophases,0and180degrees.

c.ThiscanbeeitherQPSK(oneamplitude,fourphases)or4-QAM(oneamplitude

andfourphases).Theamplitudeisthedistancebetweenapointandthe

origin,whichis(22+22)1/2=2.83.

d.ThisisalsoBPSK.Thepeakamplitudeis2,butthistimethephasesare90and

270degrees.

17.WeusetheformulaB=(1+d)x(1/r)xN,butfirstweneedtocalculatethe

valueofrforeachcase.

a.Iog24=2

b.Iog28=3

c.Iog24=2

d.Iog2128=7

Figure5.1SolutiontoExercise/5

a.r=1B=(l+1)x(1/1)x(4000bps)=8000Hz

b.r=1fB=(l+1)x(1/1)x(4000bps)+4KHz=8000Hz

c.r=2->B=(l+1)x(1/2)x(4000bps)=2000Hz

d.r=4-B=(l+1)x(1/4)x(4000bps)=1000Hz

23-33

-2

-2

2

2

-2

2

a.b.

I

II

Q

Q

Q

c.d.

I

Q

3

19.

First,wecalculatethebandwidthforeachchannel=(1MHz)/10=100KHz.We

thenfindthevalueofrforeachchannel:

B=(l+d)x(1/r)x(N)-r=N/B-r=(lMbps/100KHz)=10

Wecanthencalculatethenumberoflevels:L=2r=2io=1024.Thismeansthat

thatweneeda1024-QAMtechniquetoachievethisdatarate.

21.

a.BAM=2xB=2x5=10KHz

b.BFM=2x(1+p)xB=2x(1+5)x5=60KHz

C.BPM=2x(1+p)xB=2x(1+1)x5=20KHz

1

CHAPTER6

13.Tomultiplex10voicechannels,weneednineguardbands.Therequiredbandwidth

isthenB=(4KHz)x10+(500Hz)x9=44.5KHz

15.

a.Grouplevel:overhead=48KHz-(12x4KHz)=0Hz.

b.Supergrouplevel:overhead=240KHz-(5x48KHz)=0Hz.

2

c.Mastergroup:overhead=2520KHz-(10x240KHz)=120KHz.

d.JumboGroup:overhead=16.984MHz-(6x2.52MHz)=1.864MHz.

17.

a.Eachoutputframecarries2bitsfromeachsourceplusoneextrabitforsynchronization.

Framesize=20x2+1=41bits.

b.Eachframecarries2bitfromeachsource.Framerate=100,000/2=50,000

frames/s.

c.Frameduration=1/(framerate)=1/50,000=20ps.

d.Datarate=(50,000frames/s)x(41bits/frame)=2.05Mbps.Theoutputdata

ratehereisslightlylessthantheoneinExercise16.

e.Ineachframe40bitsoutof41areuseful.Efficiency=40/41=97.5%.Efficiency

isbetterthantheoneinExercise16.

19.Wecombinesix200-kbpssourcesintothree400-kbps.Nowwehaveseven400-

kbpschannel.

a.Eachoutputframecarries1bitfromeachoftheseven400-kbpsline.Frame

size=7x1=7bits.

b.Eachframecarries1bitfromeach400-kbpssource.Framerate=400,000

frames/s.

c.Frameduration=1/(framerate)=1MOO,000=2.5ps.

d.Outputdatarate=(400,000frames/s)x(7bits/frame)=2.8Mbps.Wecanalso

calculatetheoutputdatarateasthesumofinputdataratebecausethereisno

synchronizingbits.Outputdatarate=6x200+4x400=2.8Mbps.

21.Weneedtoaddextrabitstothesecondsourcetomakebothrates=190kbps.Now

wehavetwosources,eachof190Kbps.

a.Theframecarries1bitfromeachsource.Framesize=1+1=2bits.

b.Eachframecarries1bitfromeach190-kbpssource.Framerate=190,000

frames/s.

c.Frameduration=1/(framerate)=1/190,000=5.3ps.

d.Outputdatarate=(190,000frames/s)x(2bits/frame)=380kbps.Herethe

outputbitrateisgreaterthanthesumoftheinputrates(370kbps)becauseof

extrabitsaddedtothesecondsource.

23.SeeFigure6.1.

25.SeeFigure6.2.

Figure6.1SolutiontoExercise23

OLELYIEBHH

TDM

3

27.Thenumberofhops=100KHz/4KHz=25.SoweneedIog225=4.64=5bits

29.Randomnumbersare11,13,10,6,12,3,8,9ascalculatedbelow:

Figure6.2SolutiontoExercise25

Ni=ll

N2=(5+7x11)mod17-1=13

N3=(5+7x13)mod17-1=10

N4=(5+7x10)mod17-1=6

N5=(5+7x6)mod17-1=12

N6=(5+7x12)mod17-1=3

N7=(5+7x3)mod17-1=8

N8=(5+7x8)mod17-1=9

000000011000

101010100111

10100000

10100111

TDM

4

1

CHAPTER7

11.SeeTable7.1(thevaluesareapproximate).

13.WecanuseTable7.1tofindthepowerfordifferentfrequencies:

Table7.1SolutiontoExercise11

DistancedBat1KHzdBat10KHzdBat100KHz

1Km-3-5-7

10Km-30-50-70

15Km-45-75T05

20Km-60-100-140

1KHzdB=-3P2=Pixi0—3/]()=100.23mw

10KHzdB=-5P2=Pix10-5/10=63.25mvv

2

Thetableshowsthatthepowerfor100KHzisreducedalmost5times,whichmay

notbeacceptableforsomeapplications.

15.WefirstmakeTable7.2fromFigure7.9(inthetextbook).

Ifweconsiderthebandwidthtostartfromzero,wecansaythatthebandwidth

decreaseswithdistance.Forexample,ifwecantolerateamaximumattenuationof

-50dB(loss),thenwecangivethefollowinglistingofdistanceversusbandwidth.

17.Wecanusetheformulaf=c/Atofindthecorrespondingfrequencyforeachwave

lengthasshownbelow(cisthespeedofpropagation):

a.B=[(2x108)/1000x10-9]-[(2x10s)/1200x10-9]=33THz

b.B=[(2x108)/1000x10-9]-[(2x10s)/1400x10-9]=57THz

19.SeeTable7.3(Thevaluesareapproximate).

21.SeeFigure7.1.

a.Theincidentangle(40degrees)issmallerthanthecriticalangle(60degrees).

Wehaverefraction.Thelightrayentersintothelessdensemedium.

b.Theincidentangle(60degrees)isthesameasthecriticalangle(60degrees).

Wehaverefraction.Thelightraytravelsalongtheinterface.

100KHzdB=-7P2=PIX]O-7/IO=39.90mw

Table7.2SolutiontoExercise15

DistancedBat1KHzdBat10KHzdBat100KHz

1Km-3-7-20

10Km-30-70-200

15Km-45-105-300

20Km-60-140-400

DistanceBandwidth

1Km100KHz

10Km1KHz

15Km1KHz

20Km0KHz

Table7.3SolutiontoExercise19

DistancedBat800nmdBat1000nmdBat1200nm

1Km-3-1.1-0.5

10Km-30-11-5

15Km-45-16.5-7.5

20Km-60-22-10

3

c.Theincidentangle(80degrees)isgreaterthanthecriticalangle(60degrees).

Wehavereflection.Thelightrayreturnsbacktothemoredensemedium.

Figure7.1SolutiontoExercise21

Criticalangle=60

Criticalangle=60

Criticalangle=60

Refraction

b.60degrees

Reflection

c.80degrees

Criticalangle

Criticalangle

a.40degrees

Refraction

Criticalangle

1

CHAPTER8

11.Weassumethatthesetupphaseisatwo-waycommunicationandtheteardown

phaseisaone-waycommunication.Thesetwophasesarecommonforallthree

cases.Thedelayforthesetwophasescanbecalculatedasthreepropagationdelays

andthreetransmissiondelaysor

3[(5000km)/(2xiOsm/s)]+3[(1000bits/1Mbps)]=75ms+3ms=78ms

Weassumethatthedatatransferisinonedirection;thetotaldelayisthen

delayforsetupandteardown+propagationdelay+transmissiondelay

a.78+25+1=704

b.78+25+100=203ms

2

c.78+25+1000="03心

d.Incasea,wehave104ms.Incasebwehave203/100=2.03ms.Incasec,we

have1103/1000=1.101ms.Theratioforcasecisthesmallestbecauseweuse

onesetupandteardownphasetosendmoredata.

13.

a.Inacircuit-switchednetwork,end-to-endaddressingisneededduringthesetup

andteardownphasetocreateaconnectionforthewholedatatransferphase.

Aftertheconnectionismade,thedataflowtravelsthroughthealready-reserved

resources.Theswitchesremainconnectedfortheentiredurationofthedata

transfer;thereisnoneedforfurtheraddressing.

b.Inadatagramnetwork,eachpacketisindependent.Theroutingofapacketis

doneforeachindividualpacket.Eachpacket,therefore,needstocarryanendto-

endaddress.Thereisnosetupandteardownphasesinadatagramnetwork

(connectionlesstransmission).Theentriesintheroutingtablearesomehow

permanentandmadebyotherprocessessuchasroutingprotocols.

c.Inavirtual-circuitnetwork,thereisaneedforend-to-endaddressingduring

thesetupandteardownphasestomakethecorrespondingentryintheswitching

table.Theentryismadeforeachrequestforconnection.Duringthedatatransfer

phase,eachpacketneedstocarryavirtual-circuitidentifiertoshowwhich

virtual-circuitthatparticularpacketfollows.

15.Incircuit-switchedandvirtual-circuitnetworks,wearedealingwithconnections.

Aconnectionneedstobemadebeforethedatatransfercantakeplace.Inthecase

ofacircuit-switchednetwork,aphysicalconnectionisestablishedduringthesetup

phaseandtheisbrokenduringtheteardownphase.Inthecaseofavirtual-circuit

network,avirtualconnectionismadeduringsetupandisbrokenduringtheteardown

phase;theconnectionisvirtual,becauseitisanentryinthetable.Thesetwo

typesofnetworksareconsideredconnection-oriented.Inthecaseofadatagram

networknoconnectionismade.Anytimeaswitchinthistypeofnetworkreceives

apacket,itconsultsitstableforroutinginformation.Thistypeofnetworkisconsidered

aconnectionlessnetwork.

17.

Packet1:2

Packet2:3

Packet3:3

Packet4:2

19.

a.Inadatagramnetwork,thedestinationaddressesareunique.Theycannotbe

duplicatedintheroutingtable.

b.Inavirtual-circuitnetwork,theVCIsarelocal.AVCIisuniqueonlyinrelationship

toaport.Inotherwords,the(port,VCI)combinationisunique.This

meansthatwecanhavetwoentrieswiththesameinputoroutputports.Wecan

havetwoentrieswiththesameVCIs.However,wecannothavetwoentries

withthesame(port,VCI)pair.

3

21.

a.If〃>k,annxkcrossbarislikeamultiplexerthatcombinesninputsintokoutputs.

However,weneedtoknowthataregularmultiplexerdiscussedinChapter

6isnx1.

b.If〃v￡an〃xkcrossbarislikeademultiplexerthatdividesninputsintokoutputs.

However,weneedtoknowthataregulardemultiplexerdiscussedin

Chapter6is1xn.

23.

a.SeeFigure8.1.

b.Thetotalnumberofcrosspointsare

Numberofcrosspoints=10(10x6)+6(10x10)+10(6x10)=1800

c.Onlysixsimultaneousconnectionsarepossibleforeachcrossbaratthefirst

stage.Thismeansthatthetotalnumberofsimultaneousconnectionsis60.

d.Ifweuseonecrossbar(100x100),allinputlinescanhaveaconnectionatthe

sametime,whichmeans100simultaneousconnections.

e.Theblockingfactoris60/100or60percent.

25.

a.Totalcrosspoints=N2=1OOO2=l9000,000

b.TotalcrosspointsN4N[(2N)I/2-1]N174,886.Withlessthan200,000crosspoints

wecandesignathree-stageswitch.Wecanusen=(N/2)I/2=23and

choosek=45.Thetotalnumberofcrosspointsis178,200.

Figure8.1SolutiontoExercise23Parta

Stage1Stage2

10x6

10x10

10x10

10

Crossbars

10

Crossbars

6

Crossbars

Stage3

n=10

n=10

n=10

N二100

n=10

n=10

n=10

10x6N=100

10x6

6x10

6^10

6x10

CHAPTER9

11.Packet-switchednetworksarewellsuitedforcarryingdatainpackets.Theend-toend

addressingorlocaladdressing(VCI)occupiesafieldineachpacket.Telephone

networksweredesignedtocarryvoice,whichwasnotpacketized.Acircuit-

switchednetwork,whichdedicatesresourcesforthewholedurationofthe

conversation,ismoresuitableforthistypeofcommunication.

2

13.Inatelephonenetwork,thetelephonenumbersofthecallerandcalleeareserving

assourceanddestinationaddresses.Theseareusedonlyduringthesetup(dialing)

andteardown(hangingup)phases.

15.SeeFigure9.1.

17.

19.Wecancalculatetimebasedontheassumptionof1()Mbpsdatarate:

Time=(1,000,000x8)/10,000,000=0.8seconds

21.Thecablemodemtechnologyisbasedonthebus(orrathertree)topology.The

cableisdistributedintheareaandcustomershavetosharetheavailablebandwidth.

Thismeansifallneighborstrytotransferdata,theeffectivedataratewillbe

decreased.

Figure9.1SolutiontoExercise15

a.V.32-Time=(1,000,000x8)/9600=834s

b.V.32bistTime=(1,000,000x8)/14400之556s

c.V.90-Time=(1,000,000x8)/56000=143s

V.32V.32bisV.90

10kbps9600bps

14.4kbps

56kbps

如kbps

30kbps

40kbps

50kbps

60kbps

1

CHAPTER10

11.Wecansaythat(vulnerablebits)=(datarate)x(burstduration)

Comment:Thelastexampleshowshowanoiseofsmalldurationcanaffectso

manybitsifthedatarateishigh.

13.Thecodewordfordataword10is101.Thiscodewordwillbechangedto010ifa

3-bitbursterroroccurs.Thispatternisnotoneofthevalidcodewords,sothe

receiverdetectstheerroranddiscardsthereceivedpattern.

15.

a.d(10000,00000)=1

b.d(10101,10000)=2

c.d(llll,1111)=0

d.d(000,000)=0

Comment:Partcanddshowthatthedistancebetweenacodewordanditselfis0.

17.

a.01

b.error

c.00

d.error

19.Wecheckfiverandomcases.Allareinthecode.

21.Weshowthedataword,codeword,thecorruptedcodeword,thesyndrome,andthe

interpretationofeachcase:

a.Dataword:0100-Codeword:0100011—Corrupted:1100011-s2siso=110

Changeb3(Table10.5)一Correctedcodeword:0100011-dataword:0100

Thedatawordiscorrectlyfound.

b.Dataword:0111-?Codeword:0111001—Corrupted:0011001->s2siso=011

Changeb2(Table10.5)-Correctedcodeword:0111001-dataword:0111

Thedatawordiscorrectlyfound.

c.Dataword:1111-Codeword:1111111-?Corrupted:()111110-S2siso=111

Changebi(Table10.5)->Correctedcodeword:0101110^dataword:0101

Thedatawordisfound,butitisincorrect.C(7,4)cannotcorrecttwoerrors.

a.vulnerablebits=(1,500)x(2x10-3)=3bits

b.vulnerablebits=(12x103)x(2x10-3)=24bits

c.vulnerablebits=(100x103)x(2x10-3)=200bits

d.vulnerablebits=(100x106)x(2x10-3)=200,000bits

I.(1st)?(2nd)=(2nd)

IL(2nd)?(3th)=(4th)

III.(3rd)?(4th)=(2nd)

IV.(4th)?(5th)=(8th)

V.(5th)?(6th)=(2nd)

d.Dataword:()000-Codeword:0000000-Corrupted:1100001->s2slso=100

Changeq2(Table10.5)一Correctedcodeword:1100101->dataword:11(X)

Thedatawordisfound,butitisincorrect.C(7,4)cannotcorrectthreeerrors.

23.Weneedtofindk=2m-1-mN11.Weusetrialanderrortofindtheright

answer:

a.Letm=1k=2m-1-m=2i-1-1二0(notacceptable)

b.Letm=2k=2m-1-m=22-1-2二1(notacceptable)

c.Letm=3k=2m-1-m=23-1-3=4(notacceptable)

d.Letm=4k=2m-1-m=24-1-4=11(acceptable)

Comment:ThecodeisC(15,11)withdmin=3.

25.

a.101110-X5+X3+X2+X

b.101110f101110000(Three0sareaddedtotheright)

C.X3X(X5+X3+X2+X)=X8+X6+X5+X4

d.101110->10(Thefourrightmostbitsaredeleted)

e.x-4x(x5+X3+X2+x)=x(Notethatnegativepowersaredeleted)

27.CRC-8generatorisxs+x2+x+i.

a.Ithasmorethanonetermandthecoefficientofxois1.Itcandetectasingle-bit

error.

b.Thepolynomialisofdegree8,whichmeansthatthenumberofcheckbits

(remainder)r=8.Itwilldetectallbursterrorsofsize8orless.

c.Bursterrorsofsize9aredetectedmostofthetime,buttheyslipbywithprobability

(l/2)r-ior(1/2)8-1*0.008.Thismeans8outof1000bursterrorsofsize9

areleftundetected.

d.Bursterrorsofsize15aredetectedmostofthetime,buttheyslipbywithprobability

(l/2)ror(1/2)8(0.004.Thismeans4outof1000bursterrorsofsize15

areleftundetected.

29.Weneedtoaddallbitsmodulo-2(XORing).However,itissimplertocountthe

numberof1sandmakethemevenbyaddinga0ora1.Wehaveshowntheparity

bitinthecodewordincolorandseparateforemphasis.

31.Figure10.1showsthegenerationofthecodewordatthesenderandthechecking

ofthereceivedcodewordatthereceiverusingpolynomialdivision.

DatawordNumberofIsParityCodeword

a.1001011-4(even)->001001011

b.(X)01100-2(even)-000001100

c.100(X)00-1(odd)-111000000

d.1110111-6(even)->001110111

4

33.Figure10.2showsthechecksumtosend(0x0000).Thisexampleshowsthatthe

checksumcanbeall0s.ItcanbeallIsonlyifalldataitemsareall0,which

meansnodataatall.

Figure10.1SolutiontoExercise31

Figure10.2SolutiontoExercise33

Codeword

X7+X5+X2+X+1

X7+X4+X3+X+1

X4+X2+X+IX11+X9+X6+X5+X4

X"+X<)+X6+X5+X4+

XU+XJ+XX+X?

X8+X7+X6+X5+X4

X8+X6+X5+X4

X7

X7+X5+X4+X3

X5+X4+X3

X5+X3+X2+X

X4+X2+X

X4+X2+X+1

1

1

Dataword

Sender

Quotient

Divisor

Remainder

Codeword

X7+X5+X2+X+I

X4X3X1

1

x7++++

X4+X2+X+1Xll+X9+X6+X5+X4+

Xn+Xy+Xft+Xs+Xj+

Xll+Xo+Xs+X?

XS+X7+X6+X5+X4

X8+X6+X5+X4

X7

X7+X5+X4+X3

X5+X4+X3

X5+X3+X2+X

X4+X2+X

X4X2X1

0

1

+++

+

1

Dataword

Quotient

Divisor

Remainder

Receiver

Checksum(initial)

Sum

4567

BA98

FFFF

0000Checksum(tosend)

0000

1

CHAPTER11

13.Wegiveaverysimplesolution.EverytimeweencounteranESCorflagcharacter,

weinsertanextraESCcharacterinthedatapartoftheframe(seeFigure11.1).

15.Wewritetwoverysimplealgorithms.Weassumethataframeismadeofaonebyte

beginningflag,variable-lengthdata(possiblybyte-stuffed),andaone-byte

endingflag;weignoretheheaderandtrailer.Wealsoassumethatthereisnoerror

duringthetransmission.

a.Algorithm11.1canbeusedatthesendersite.ItinsertsoneESCcharacter

wheneveraflagorESCcharacterisencountered.

b.Algorithm11.2canbeusedatthereceiversite.

17.Afive-bitsequencenumbercancreatesequencenumbersfrom0to31.The

sequencenumberintheNthpacketis(Nmod32).Thismeansthatthe101th

packethasthesequencenumber(101mod32)or5.

Figure11.1SolutiontoExercise13

Algorithm11.1Sender'ssitesolutiontoExercise15

InsertFrame(one-byteflag);//Insertbeginningflag

while(morecharactersindatabuffer)

(

ExtractBuffer(character);

if(characterisflagorESC)InsertFrame(ESC);//Bytestuff

InsertFrame(character);