Computer Networks: A Systems Approach Fifth Edition Solutions Manual

122 Pages·2011·1.07 MB·English

Computer Networks: A Systems Approach
Fifth Edition
Solutions Manual
Larry Peterson and Bruce Davie
2011
1
DearInstructor:
ThisInstructors’Manualcontainssolutionstomostoftheexercisesinthefifthedition
ofPetersonandDavie’sComputerNetworks: ASystemsApproach.
Exercises are sorted (roughly) by section, not difficulty. While some exercises are
moredifficultthan others, none are intendedto be fiendishly tricky. A few exercises
(notably,thoughnotexclusively,theonesthatinvolvecalculatingsimpleprobabilities)
requireamodestamountofmathematicalbackground;mostdonot. Thereisasidebar
summarizingmuchoftheapplicablebasicprobabilitytheoryinChapter2.
Anoccasionalexerciseisawkwardlyorambiguouslywordedinthetext. Thismanual
sometimessuggestsbetterversions;alsoseetheerrataatthewebsite.
Whereappropriate,relevantsupplementalfilesforthesesolutions(e.g.programs)have
beenplacedonthetextbookwebsite,http://mkp.com/computer-networks.
Useful other material can also be found there, such as errata, sample programming
assignments,PowerPointlectureslides,andEPSfigures.
If youhaveany questionsaboutthese supportmaterials, please contactyourMorgan
Kaufmann sales representative. If you would like to contribute your own teaching
materialstothissite,pleasecontactourAssociateEditorDavidBevans,
[email protected].
We welcome bug reports and suggestions as to improvementsfor both the exercises
andthesolutions;[email protected].
LarryPeterson
BruceDavie
March,2011
Chapter1 1
Solutions for Chapter 1
3. Wewillcountthetransferascompletedwhenthelastdatabitarrivesatitsdesti-
nation.AnalternativeinterpretationwouldbetocountuntilthelastACKarrives
backatthesender,inwhichcasethetimewouldbehalfanRTT(25ms)longer.
(a) 2initialRTT’s(100ms)+1000KB/1.5Mbps(transmit)+RTT/2(propaga-
tion=25ms)
0.125 + 8Mbit/1.5Mbps= 0.125+5.333 sec = 5.458 sec. If we pay
≈
morecarefulattentiontowhenamegais106versus220,weget
8,192,000bits/1,500,000bps=5.461sec,foratotaldelayof5.586sec.
(b) Totheaboveweaddthetimefor999RTTs(thenumberofRTTsbetween
when packet 1 arrives and packet 1000 arrives), for a total of 5.586 +
49.95=55.536.
(c) Thisis49.5RTTs,plustheinitial2,for2.575seconds.
(d) Rightafterthehandshakingisdonewesendonepacket.OneRTTafterthe
handshakingwesendtwopackets. AtnRTTspasttheinitialhandshaking
wehavesent1+2+4+ +2n =2n+1 1packets. Atn=9wehave
··· −
thusbeenabletosendall1,000packets;thelastbatcharrives0.5RTTlater.
Totaltimeis2+9.5RTTs,or.575sec.
4. Theanswerisinthebook.
5. Propagationdelayis4 103m/(2 108m/s)=2 10−5sec=20µs.100bytes/20µs
× × ×
is5bytes/µs,or5MBps,or40Mbps.For512-bytepackets,thisrisesto204.8Mbps.
6. Theanswerisinthebook.
7. Postaladdressesarestronglyhierarchical(withageographicalhierarchy,which
network addressing may or may not use). Addresses also provide embedded
“routing information”. Unlike typical network addresses, postal addresses are
longandofvariablelengthandcontainacertainamountofredundantinforma-
tion. Thislastattributemakesthemmoretolerantofminorerrorsandinconsis-
tencies. Telephonenumbers,atleastthoseassignedtolandlines,aremoresim-
ilar to network addresses: they are (geographically)hierarchical, fixed-length,
administrativelyassigned, andin more-or-lessone-to-onecorrespondencewith
nodes.
8. One might want addresses to serve as locators, providinghints as to how data
shouldberouted.Oneapproachforthisistomakeaddresseshierarchical.
Another property might be administratively assigned, versus, say, the factory-
assigned addresses used by Ethernet. Other address attributes that might be
relevant are fixed-length v. variable-length, and absolute v. relative (like file
names).
Chapter1 2
Ifyouphoneatoll-freenumberforalargeretailer,anyofdozensofphonesmay
answer. Arguably,then,allthesephoneshavethesamenon-unique“address”.A
moretraditionalapplicationfornon-uniqueaddressesmightbeforreachingany
ofseveralequivalentservers(orrouters). Non-uniqueaddressesarealsouseful
whenglobalreachabilityisnotrequired,suchastoaddressthecomputerswithin
a single corporationwhenthosecomputerscannotbe reachedfromoutside the
corporation.
9. Video oraudioteleconferencetransmissionsamonga reasonablylargenumber
ofwidelyspreadsiteswouldbeanexcellentcandidate: unicastwouldrequirea
separateconnectionbetweeneachpairofsites, whilebroadcastwouldsendfar
toomuchtraffictositesnotinterestedinreceivingit. Deliveryofvideoandaudio
streamsforatelevisionchannelonlytothosehouseholdscurrentlyinterestedin
watchingthatchannelisanotherapplication.
Tryingtoreachanyofseveralequivalentservers,eachofwhichcanprovidethe
answer to some query, would be anotherpossible use, althoughthe receiverof
manyresponsestothequerywouldneedtodealwiththepossiblylargevolume
ofresponses.
10. STDMandFDMbothworkbestforchannelswithconstantanduniformband-
width requirements. For both mechanismsbandwidththat goesunusedby one
channelissimplywasted,notavailabletootherchannels. Computercommuni-
cationsareburstyandhavelongidleperiods;suchusagepatternswouldmagnify
thiswaste.
FDM andSTDMalso requirethatchannelsbeallocated(and,forFDM,beas-
signedbandwidth)wellinadvance.Again,theconnectionrequirementsforcom-
putingtendtobetoodynamicforthis;attheveryleast,thiswouldprettymuch
precludeusingonechannelperconnection.
FDM waspreferredhistoricallyforTV/radiobecauseit is verysimple to build
receivers;italsosupportsdifferentchannelsizes. STDMwaspreferredforvoice
because it makes somewhat more efficient use of the underlyingbandwidth of
the medium, and because channelswith differentcapacities was not originally
anissue.
11. 10Gbps=1010bps,meaningeachbitis10−10 sec(0.1ns)wide. Thelengthin
thewireofsuchabitis.1ns 2.3 108m/sec=0.023mor23mm
× ×
12. xKBis8 1024 xbits. yMbpsisy 106bps;thetransmissiontimewould
× × ×
be8 1024 x/y 106sec=8.192x/yms.
× × ×
13. (a) TheminimumRTTis2 385,000,000m/3 108m/s=2.57seconds.
× ×
(b) Thedelay bandwidthproductis2.57s 1Gbps=2.57Gb=321MB.
× ×
(c) Thisrepresentstheamountofdatathesendercansendbeforeitwouldbe
possibletoreceivearesponse.
Chapter1 3
(d) We require at least one RTT from sending the request before the first bit
ofthepicturecouldbeginarrivingattheground(TCPwouldtakelonger).
25 MB is 200Mb. Assuming bandwidth delay only, it would then take
200Mb/1000Mbps=0.2secondstofinishsending,foratotaltimeof0.2+
2.57=2.77secuntilthelastpicturebitarrivesonearth.
14. Theanswerisinthebook.
15. (a) Delay-sensitive;themessagesexchangedareshort.
(b) Bandwidth-sensitive,particularlyforlargefiles. (Technicallythisdoespre-
sumethattheunderlyingprotocolusesalargemessagesizeorwindowsize;
stop-and-waittransmission(asinSection2.5ofthetext)withasmallmes-
sagesizewouldbedelay-sensitive.)
(c) Delay-sensitive;directoriesaretypicallyofmodestsize.
(d) Delay-sensitive; a file’s attributesare typicallymuch smaller than the file
itself.
16. (a) On a 100 Mbpsnetwork, each bit takes 1/108 = 10 ns to transmit. One
packetconsistsof12000bits, andso isdelayedduetobandwidth(serial-
ization) by 120µs along each link. The packet is also delayed 10µs on
eachofthetwolinksduetopropagationdelay,foratotalof260µs.
(b) Withthreeswitchesandfourlinks,thedelayis
4 120µs+4 10µs=520µs
× ×
(c) With cut-through,the switch delays the packetby 200 bits = 2µs. There
is still one 120µs delay waiting for the last bit, and 20µs of propagation
delay,sothetotalis142µs. Toputitanotherway,thelastbitstillarrives
120µs after the first bit; the first bit now faces two link delays and one
switchdelaybutneverhastowaitforthelastbitalongtheway.
17. Theanswerisinthebook.
18. (a) The effective bandwidth is 100Mbps; the sender can send data steadily
at this rate and the switches simply stream it along the pipeline. We are
assuming here that no ACKs are sent, and that the switches can keep up
andcanbufferatleastonepacket.
(b) Thedatapackettakes520µsasin16(b)abovetobedelivered;the400bit
ACKstake4µs/linktobesentback,pluspropagation,foratotalof4 4µs
×
+4 10µs=56µs; thusthetotalRTTis576µs. 12000bitsin576µsis
×
about20.8Mbps.
(c) 100 4.7 109bytes/12hours=4.7 1011bytes/(12 3600s) 10.9MBps
× × × × ≈
=87Mbps.
19. (a) 100 106bps 10 10−6sec=1000bits=125bytes.
× × ×
Chapter1 4
(b) The first-bit delay is 520µs through the store-and-forward switch, as in
16(a).100 106bps 520 10−6sec=52000bits=650bytes.
× × ×
(c) 1.5 106bps 50 10−3sec=75,000bits=9375bytes.
× × ×
(d) The path is through a satellite, i.e. between two ground stations, not to
asatellite;thisground-to-satellite-to-groundpathmakesthetotalone-way
travel distance 2 35,900,000 meters. With a propagation speed of c =
×
3 108meters/sec,theone-waypropagationdelayisthus2 35,900,000/c
× ×
=0.24sec. Bandwidth delayisthus1.5 106bps 0.24sec=360,000
× × ×
bits 45KBytes
≈
20. (a) Per-link transmit delay is 104bits / 108bps = 100 µs. Total transmission
timeincludinglinkandswitchpropagationdelays=2 100+2 20+35=
× ×
275 µs.
(b) Whensendingastwopackets,thetimetotransmitonepacketiscutinhalf.
Hereisatableoftimesforvariousevents:
T=0 start
T=50 Afinishessendingpacket1,startspacket2
T=70 packet1finishesarrivingatS
T=105 packet1departsforB
T=100 Afinishessendingpacket2
T=155 packet2departsforB
T=175 bit1ofpacket2arrivesatB
T=225 lastbitofpacket2arrivesatB
Thisissmallerthantheanswertopart(a)becausepacket1startstomake
itswaythroughtheswitchwhilepacket2isstillbeingtransmittedonthe
firstlink,effectivelygettinga50µsheadstart. Smallerisfaster,here.
21. (a) Withoutcompression the total time is 1MB/bandwidth. When we com-
pressthefile,thetotaltimeis
compression time+compressed size/bandwidth
Equatingtheseandrearranging,weget
bandwidth=compression size reduction/compression time
=0.5MB/1sec=0.5MB/secforthefirstcase,
=0.6MB/2sec=0.3MB/secforthesecondcase.
(b) Latencydoesn’taffecttheanswerbecauseitwouldaffectthecompressed
anduncompressedtransmissionequally.
22. Thenumberofpacketsneeded,N,is 106/D ,whereDisthepacketdatasize.
⌈ ⌉
Given that overhead= 50 N and loss = D (we have already counted the lost
×
packet’sheaderintheoverhead),wehaveoverhead+loss=50 106/D +D.
×⌈ ⌉
D overhead+loss
1000 51000
10000 15000
20000 22500
Chapter1 5
Theoptimalsizeis10,000byteswhichminimizestheabovefunction.
23. Comparisonofcircuitsandpacketsresultasfollows:
(a) Circuitspayanup-frontpenaltyof1024bytesbeingsentononeroundtrip
for a total data count of 2048+n, whereas packets pay an ongoing per
packetcostof24bytesforatotalcountof1024 n/1000.Sothequestion
×
really asks how many packet headers does it take to exceed 2048 bytes,
whichis86. Thusforfiles86,000bytesorlonger,usingpacketsresultsin
moretotaldatasentonthewire.
(b) The total transfer latency for packets is the sum of the transmit delays,
wheretheper-packettransmittimetisthepacketsizeoverthebandwidthb
(8192/b),introducedbyeachofsswitches(s t),totalpropagationdelay
×
forthelinks((s+2) 0.002),theperpacketprocessingdelaysintroduced
×
byeachswitch(s 0.001),andthetransmitdelayforallthepackets,where
×
the total packet count c is n/1000, at the source (c t). Resulting in a
×
totallatencyof(8192s/b)+0.003s+0.004+(8.192n/b)=(0.02924+
0.000002048n)seconds.Thetotallatencyforcircuitsisthetransmitdelay
forthewholefile(8n/b),thetotalpropagationdelayforthelinks,andthe
setup cost for the circuit which is just like sending one packet each way
on the path. Solving the resulting inequality 0.02924 + 8.192(n/b) >
0.076576+8(n/b)fornshowsthatcircuitsachievealowerdelayforfiles
largerthanorequalto987,000B.
(c) Only the payload to overhead ratio size effects the number of bits sent,
andtheretherelationshipissimple. Thefollowingtableshowthelatency
resultsofvaryingtheparametersbysolvingforthenwherecircuitsbecome
faster, as above. This table does not show how rapidly the performance
diverges;forvaryingpitcanbesignificant.
s b p pivotaln
5 4Mbps 1000 987000
6 4Mbps 1000 1133000
7 4Mbps 1000 1280000
8 4Mbps 1000 1427000
9 4Mbps 1000 1574000
10 4Mbps 1000 1721000
5 1Mbps 1000 471000
5 2Mbps 1000 643000
5 8Mbps 1000 1674000
5 16Mbps 1000 3049000
5 4Mbps 512 24000
5 4Mbps 768 72000
5 4Mbps 1014 2400000
(d) Manyresponsesare probablyreasonablehere. The modelonlyconsiders
thenetworkimplications,anddoesnottakeintoaccountusageofprocess-
ing or state storage capabilities on the switches. The model also ignores
thepresenceofothertrafficorofmorecomplicatedtopologies.
Chapter1 6
24. The time to send one 12000-bitpacket is 12000bits/100Mbps = 120 µs. The
lengthofcableneededtoexactlycontainsuchapacketis120 µs 2 108m/sec
× ×
=24,000meters.
12000bitsin24000metersis50bitsper100m.Withanextra10bitsofdelayin
each100m, we havea totalof60bits/100mor0.6bits/m. A 12000-bitpacket
nowfills12000/(.6bits/m)=20,000meters.
25. For music we would need considerably more bandwidth, but we could toler-
ate high (but bounded) delays. We could not necessarily tolerate higher jitter,
though;seeSection6.5.1.
We mightacceptan audibleerrorin voice traffic everyfew seconds; we might
reasonably want the error rate during music transmission to be a hundredfold
smaller. Audible errors would come either from outright packet loss, or from
jitter(apacket’snotarrivingontime).
Latency requirements for music, however, might be much lower; a several-
seconddelaywouldbeinconsequential.Voicetraffichasatleastatenfoldfaster
requirementhere.
26. (a) 640 480 3 30bytes/sec=26.4MB/sec
× × ×
(b) 160 120 1 5=96,000bytes/sec=94KB/sec
× × ×
(c) 650MB/75min=8.7MB/min=148KB/sec
(d) 8 10 72 72pixels=414,720bits=51,840bytes. At14,400bits/sec,
× × ×
thiswouldtake28.8seconds(ignoringoverheadforframingandacknowl-
edgments).
27. Theanswerisinthebook.
28. (a) A file server needs lots of peak bandwidth. Latency is relevant only if
itdominatesbandwidth;jitterandaveragebandwidthareinconsequential.
Nolostdataisacceptable,butwithoutreal-timerequirementswecansim-
plyretransmitlostdata.
(b) A print server needs less bandwidth than a file server (unless images are
extremelylarge).Wemaybewillingtoaccepthigherlatencythan(a),also.
(c) Afileserverisadigitallibraryofasort,butingeneraltheworldwideweb
getsalongreasonablywellwith muchless peakbandwidththanmostfile
serversprovide.
(d) Forinstrumentmonitoringwedon’tcareaboutlatencyorjitter.Ifdatawere
continually generated, rather than bursty, we might be concerned mostly
withaveragebandwidthratherthanpeak,andifthedatareallywereroutine
wemightjustacceptacertainfractionofloss.
(e) For voice we need guaranteedaverage bandwidth and bounds on latency
and jitter. Some lost data might be acceptable; e.g. resulting in minor
dropoutsmanysecondsapart.
Chapter1 7
(f) Forvideoweareprimarilyconcernedwithaveragebandwidth.Forthesim-
plemonitoringapplicationhere,relativelymodestvideoofExercise26(b)
might suffice; we could even go to monochrome (1bit/pixel), at which
point160 120 5frames/secrequires12KB/sec. Wecouldtoleratemulti-
× ×
secondlatencydelays; theprimaryrestrictionis thatif themonitoringre-
vealedaneedforinterventionthenwestillhavetimetoact. Considerable
loss,evenofentireframes,wouldbeacceptable.
(g) Full-scaletelevisionrequiresmassivebandwidth. Latency,however,could
behours.Jitterwouldbelimitedonlybyourcapacitytoabsorbthearrival-
time variations by buffering. Some loss would be acceptable, but large
losseswouldbevisuallyannoying.
29. In STDM the offeredtimeslices are always the same length, and are wasted if
they are unused by the assigned station. The round-robin access mechanism
would generally give each station only as much time as it needed to transmit,
or noneif the station hadnothingto send, andso networkutilizationwouldbe
expectedtobemuchhigher.
30. (a) Intheabsenceofanypacketlossesorduplications,whenweareexpecting
the Nth packet we get the Nth packet, and so we can keep track of N
locallyatthereceiver.
(b) The scheme outlined here is the stop-and-wait algorithm of Section 2.5;
asisindicatedthere,a headerwithatleastonebitofsequencenumberis
needed(todistinguishbetweenreceivinganewpacketandaduplicationof
thepreviouspacket).
(c) With out-of-orderdeliveryallowed, packetsup to 1 minuteapartmustbe
distinguishablevia sequence number. Otherwise a very old packetmight
arriveandbeacceptedascurrent. Sequencenumberswouldhavetocount
ashighas
bandwidth 1minute/packet size
×
31. Ineachcaseweassumethelocalclockstartsat1000.
(a) Latency:100.Bandwidth:highenoughtoreadtheclockevery1unit.
1000 1100
1001 1101
1002 1102
1003 1104 tinybitofjitter: latency=101
1004 1104
(b) Latency=100; bandwidth: only enough to read the clock every 10 units.
Arrivaltimesfluctuateduetojitter.
1000 1100
1020 1110 latency=90
1040 1145
1060 1180 latency=120
1080 1184
Chapter1 8
(c) Latency=5;zerojitterhere:
1000 1005
1001 1006
1003 1008 welost1002
1004 1009
1005 1010
32. Generally, with MAX PENDING =1, one or two connectionswill be accepted
andqueued;thatis,thedatawon’tbedeliveredtotheserver. Theotherswillbe
ignored;eventuallytheywilltimeout.
Whenthefirstclientexits,anyqueuedconnectionsareprocessed.
34. Note that UDP accepts a packet of data from any source at any time; TCP re-
quires an advance connection. Thus, two clients can now talk simultaneously;
theirmessageswillbeinterleavedontheserver.