3第三章课后习题及答案Word格式文档下载.docx
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3第三章课后习题及答案Word格式文档下载.docx
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a.CallthisprotocolSimpleTransportProtocol(STP).Atthesenderside,STPacceptsfromthesendingprocessachunkofdatanotexceeding1196bytes,adestinationhostaddress,andadestinationportnumber.STPaddsafour-byteheadertoeachchunkandputstheportnumberofthedestinationprocessinthisheader.STPthengivesthedestinationhostaddressandtheresultingsegmenttothenetworklayer.ThenetworklayerdeliversthesegmenttoSTPatthedestinationhost.STPthenexaminestheportnumberinthesegment,extractsthedatafromthesegment,andpassesthedatatotheprocessidentifiedbytheportnumber.
b.Thesegmentnowhastwoheaderfields:
asourceportfieldanddestinationportfield.Atthesenderside,STPacceptsachunkofdatanotexceeding1192bytes,adestinationhostaddress,asourceportnumber,andadestinationportnumber.STPcreatesasegmentwhichcontainstheapplicationdata,sourceportnumber,anddestinationportnumber.Itthengivesthesegmentandthedestinationhostaddresstothenetworklayer.Afterreceivingthesegment,STPatthereceivinghostgivestheapplicationprocesstheapplicationdataandthesourceportnumber.
c.No,thetransportlayerdoesnothavetodoanythinginthecore;
thetransportlayer“lives”intheendsystems.
2.(Q2)Consideraplanetwhereeveryonebelongstoafamilyofsix,everyfamilylivesinitsownhouse,eachhousehasauniqueaddress,andeachpersoninagivenhousehasauniquename.Supposethisplanethasamailservicethatdeliverslettersformsourcehousetodestinationhouse.Themailservicerequiresthat(i)theletterbeinanenvelopeandthat(ii)theaddressofthedestinationhouse(andnothingmore)beclearlywrittenontheenvelope.Supposeeachfamilyhasadelegatefamilymemberwhocollectsanddistributeslettersfortheotherfamilymembers.Thelettersdonotnecessarilyprovideanyindicationoftherecipientsoftheletters.
a.UsingthesolutiontoProblemQ1aboveasinspiration,describeaprotocolthatthedelegatescanusetodeliverlettersfromasendingfamilymembertoareceivingfamilymember.
b.Inyourprotocol,doesthemailserviceeverhavetoopentheenvelopeandexaminetheletterinordertoprovideitsservice.
a.Forsendingaletter,thefamilymemberisrequiredtogivethedelegatetheletteritself,theaddressofthedestinationhouse,andthenameoftherecipient.Thedelegateclearlywritestherecipient’snameonthetopoftheletter.Thedelegatethenputstheletterinanenvelopeandwritestheaddressofthedestinationhouseontheenvelope.Thedelegatethengivesthelettertotheplanet’smailservice.Atthereceivingside,thedelegatereceivestheletterfromthemailservice,takestheletteroutoftheenvelope,andtakesnoteoftherecipientnamewrittenatthetopoftheletter.Thedelegatethangivesthelettertothefamilymemberwiththisname.
b.No,themailservicedoesnothavetoopentheenvelope;
itonlyexaminestheaddressontheenvelope.
3.(Q3)DescribewhyanapplicationdevelopermightchoosetorunanapplicationoverUDPratherthanTCP.
AnapplicationdevelopermaynotwantitsapplicationtouseTCP’scongestioncontrol,whichcanthrottletheapplication’ssendingrateattimesofcongestion.Often,designersofIPtelephonyandIPvideoconferenceapplicationschoosetoruntheirapplicationsoverUDPbecausetheywanttoavoidTCP’scongestioncontrol.Also,someapplicationsdonotneedthereliabledatatransferprovidedbyTCP.
4.(P1)SupposeClientAinitiatesaTelnetsessionwithServerS.Ataboutthesametime,ClientBalsoinitiatesaTelnetsessionwithServerS.Providepossiblesourceanddestinationportnumbersfor
a.ThesegmentsentfromAtoB.
b.ThesegmentsentfromBtoS.
c.ThesegmentsentfromStoA.
d.ThesegmentsentfromStoB.
e.IfAandBaredifferenthosts,isitpossiblethatthesourceportnumberinthesegmentfromAtoSisthesameasthatfromBtoS
f.Howaboutiftheyarethesamehost
sourceportnumbers
destinationportnumbers
a
A→S
467
23
b
B→S
513
c
S→A
d
S→B
eYes.
fNo.
5.(P2)ConsiderFigureWhatarethesourceanddestinationportvaluesinthesegmentsflowingformtheserverbacktotheclients’processesWhataretheIPaddressesinthenetwork-layerdatagramscarryingthetransport-layersegments
SupposetheIPaddressesofthehostsA,B,andCarea,b,c,respectively.(Notethata,b,caredistinct.)
TohostA:
Sourceport=80,sourceIPaddress=b,destport=26145,destIPaddress=a
TohostC,leftprocess:
Sourceport=80,sourceIPaddress=b,destport=7532,destIPaddress=c
TohostC,rightprocess:
Sourceport=80,sourceIPaddress=b,destport=26145,destIPaddress=c
6.(P3)UDPandTCPuse1scomplementfortheirchecksums.Supposeyouhavethefollowingthree8-bitbytes:
01101010,01001111,01110011.Whatisthe1scomplementofthesumofthese8-bitbytes(NotethatalthoughUDPandTCPuse16-bitwordsincomputingthechecksum,forthisproblemyouarebeingaskedtoconsider8-bitsums.)Showallwork.WhyisitthatUDPtakesthe1scomplementofthesum;
thatis,whynotjustsuethesumWiththe1scomplementscheme,howdoesthereceiverdetecterrorsIsitpossiblethata1-biterrorwillgoundetectedHowabouta2-biterror
01010101
+
01110000
11000101
01001100
00010001
One'
scomplement=11101110.
Todetecterrors,thereceiveraddsthefourwords(thethreeoriginalwordsandthechecksum).Ifthesumcontainsazero,thereceiverknowstherehasbeenanerror.Allone-biterrorswillbedetected,buttwo-biterrorscanbeundetected.,ifthelastdigitofthefirstwordisconvertedtoa0andthelastdigitofthesecondwordisconvertedtoa1).
7.(P4)SupposethattheUDPreceivercomputestheInternetchecksumforthereceivedUDPsegmentandfindsthatitmatchesthevaluecarriedinthechecksumfield.CanthereceiverbeabsolutelycertainthatnobiterrorshaveoccurredExplain.
No,thereceivercannotbeabsolutelycertainthatnobiterrorshaveoccurred.Thisisbecauseofthemannerinwhichthechecksumforthepacketiscalculated.Ifthecorrespondingbits(thatwouldbeaddedtogether)oftwo16-bitwordsinthepacketwere0and1thenevenifthesegetflippedto1and0respectively,thesumstillremainsthesame.Hence,the1scomplementthereceivercalculateswillalsobethesame.Thismeansthechecksumwillverifyeveniftherewastransmissionerror.
8.(P5)a.Supposeyouhavethefollowing2bytes:
01001010and01111001.Whatisthe1scomplementofsumofthese2bytes
b.Supposeyouhavethefollowing2bytes:
and01101110.Whatisthe1scomplementofsumofthese2bytes
c.Forthebytesinpart(a),giveanexamplewhereonebitisflippedineachofthe2bytesandyetthe1scomplementdoesn’tchange.
a.Addingthetwobytesgives.Takingtheone’scomplementgives01100010
b.Addingthetwobytesgives00011110;
theone’scomplementgives.
c.firstbyte=00110101;
secondbyte=01101000.
9.(P6)Considerourmotivationforcorrectingprotocol.Showthatthereceiver,shownin
thefigureonthefollowingpage,whenoperatingwiththesendershowinFigure,canleadthesenderandreceivertoenterintoadeadlockstate,whereeachiswaitingforaneventthatwillneveroccur.
Supposethesenderisinstate“Waitforcall1fromabove”andthereceiver(thereceivershowninthehomeworkproblem)isinstate“Waitfor1frombelow.”Thesendersendsapacketwithsequencenumber1,andtransitionsto“WaitforACKorNAK1,”waitingforanACKorNAK.Supposenowthereceiverreceivesthepacketwithsequencenumber1correctly,sendsanACK,andtransitionstostate“Waitfor0frombelow,”waitingforadatapacketwithsequencenumber0.However,theACKiscorrupted.WhenthesendergetsthecorruptedACK,itresendsthepacketwithsequencenumber1.However,thereceiveriswaitingforapacketwithsequencenumber0and(asshowninthehomeworkproblem)alwayssendsaNAKwhenitdoesn'
tgetapacketwithsequencenumber0.Hencethesenderwillalwaysbesendingapacketwithsequencenumber1,andthereceiverwillalwaysbeNAKingthatpacket.Neitherwillprogressforwardfromthatstate.
10.(P7)DrawtheFSMforthereceiversideofprotocol
Thesendersideofprotocoldiffersfromthesendersideofprotocolinthat
timeoutshavebeenadded.Wehaveseenthattheintroductionoftimeoutsaddsthepossibilityofduplicatepacketsintothesender-to-receiverdatastream.However,thereceiverinprotocolcanalreadyhandleduplicatepackets.(Receiver-sideduplicatesinrdtwouldariseifthereceiv
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