《數(shù)據(jù)與計算機通信》第七版課后習(xí)題答案

《數(shù)據(jù)與計算機通信》

（第七版）

課后習(xí)題參考答案

第2章的參考答案

2.1答案：

2.1Theguesteffectivelyplacestheorderwiththecook.Thehostcommunicatesthis

ordertotheclerk,whoplacestheorderwiththecook.Thephonesystemprovides

thephysicalmeansfortheordertobetransportedfromhosttoclerk.Thecook

givesthepizzatotheclerkwiththeorderform(actingasaheader'tothepizza).

Theclerkboxesthepizzawiththedeliveryaddress,andthedeliveryvanencloses

alloftheorderstobedelivered.Theroadprovidesthephysicalpathfordelivery.

2.2答案

a.

TblephoxeLine

ThePMsspeakasiftheyarespeakingdirectlytoeachother.Forexample,when

theFrenchPMspeaks,headdresseshisremarksdirectlytotheChinesePM.

However,themessageisactuallypassedthroughtwotranslatorsviathephone

system.TheFrenchPM'stranslatortranslateshisremarksintoEnglishand

telephonesthesetotheChinesePMstranslator,whotranslatestheseremarksinto

Chinese.

b.

b.

Anintermediatenodeservestotranslatethemessagebeforepassingiton.

2.4答案：

2.4No.Thereisnowaytobeassuredthatthelastmessagegetsthrough,exceptby

acknowledgingit.Thus,eithertheacknowledgmentprocesscontinuesforever,or

onearmyhastosendthelastmessageandthenactwithuncertainty.

2.7答案：

2.7a.No.Thiswouldviolatetheprincipleofseparationoflayers.Tolayer(N-1),

theN-levelPDUissimplydata.The(N-1)entitydoesnotknowaboutthe

internalformatoftheN-levelPDU.ItbreaksthatPDUintofragmentsand

reassemblesthemintheproperorder.

b.EachN-levelPDUmustretainitsownheader,forthesamereasongivenin(a).

第3章的參考答案

3.13答案

3.13a.(30pictures/s)(480x500pixels/picture)=7,2x106pixels/s

Eachpixelcantakeononeof32valuesandcanthereforeberepresentedby5

bits:

R=7.2x106pixels/sx5bits/pixel=36Mbps

b.Weusetheformula:C=Blog2(1+SNR)

B=4.5x106MHz=bandwidth,and

SNRdB=35=10log10(SNR),hence

SNR=1()35/10=io3-5,andtherefore

C=4.5x106log2(1+IO35)=4.5x106xIog2(3163)

C=(4.5x106x11.63)-52.335x106bps

c.Alloweachpixeltohaveoneoftenintensitylevelsandleteachpixelbeoneof

threecolors(red,blue,green)foratotalof10x3=30levelsforeachpixel

element.

3.15答案

3.15UsingShannon'sequation:C=Blog2(1+SNR)

WehaveW=300Hz(SNR)dB=3

Therefore,SNR=IO03

03

C=300log2(1+IO)=300log2(2.995)=474bps

316答案

3.16UsingNyquistsequation:C=2Blog2M

WehaveC=9600bps

a.log7M=4,becauseasignalelementencodesa4-bitword

Therefore,C=9600=2Bx4,andB=1200Hz

b.9600=2Bx8,andB=600Hz

3.19答案

已知C=20Mbps,B=3Mbps.

根據(jù)香農(nóng)定理直=6log2(1+SNR),

C=20x1。6<=3x1O'xlog?(1+SNR)

則

log2"SNR)>=6.67

SNR>=101

另解：

3.19C=Blog2(1+SNR)

66

20x10=3x10xlog2(l+SNR)

log2(l+SNR)=6.67

1+SNR=102

SNR=101

▲補充作業(yè)：

設(shè)采用異步傳輸，1位起始位，2位終止位，1位奇偶位，每一個信號碼源2

位，對下述速率，分別求出相應(yīng)的有效數(shù)據(jù)速率(b/s)：

(l)300Baud(2)600Baud(3)I200Baud(4)4800baud

答：

異步傳輸?shù)臄?shù)據(jù)效率為7/11,而每一個信號碼源2位，

714

R=81og2M，所以/?=打乂28=H8

714

(1)R=—x2B=——B=38l.8b/s

1111

714

(2)R=—x28=—8=763.66/s

1111

714

(3)R=—x2B=—B=\52n.3bls

1111

714

(4)R=—x2B=—B=6109.1b/s

1111

第5章的參考答案

5.6答

NRZ-L

NRZI

Bipolar。、”

?~~???~~?????

Pseudoternary|J??|||?;

Manchester[EHjHHE1

?=；UTLTLhJmrLil

B8ZS3

HDB3

5.7答:

5.7WithManchester,thereisalwaysatransitioninthemiddleofabitperiod.

n_n_nj_i_n_ru_i_i_LT

|I|I|I|O|O|I|I|O|I|O|

5.8答:

11111011111

心________________________________________________n______

，-bJ-Ln"LTLTL

c.__________________|~|_____________________

5.9答:

5.9Theerrorisatbitposition7,wherethereisanegativepulse.ForAMI,positiveand

negativepulsesareusedalternatelyforbinary1.Thepulseinposition1represents

thethirdbinary1inthedatastreamandshouldhaveapositivevalue.

第6章的參考答案

6.1答:

6.1a.Eachcharacterhas25%overhead.For10,000characters,thereare20,000extra

bits.Thiswouldtakeanextra20,000/2400=8.33seconds.

b.Thefiletakes10framesor480additionalbits.Thetransmissiontimeforthe

additionalbitsis480/2400=0.2seconds.

c.Tentimesasmanyextrabitsandtentimesaslongforboth.

d.Thenumberofoverheadbitswouldbethesame,andthetimewouldbe

decreasedbyafactorof4=9600/2400.

另答：

(a)

1+1八八~

------=20%

8+1+1

.?.額外開銷率為20%

傳輸速率:2400b/s=240w/s

傳輸時間為3S=41.67s

240

(b)

48

0.59%

8000+48

額夕卜開銷：48x10=4808〃

傳輸一-幀:建￡=3.35s

2400

共擔(dān)也=1。

1000

總耗時：10x3.35=33.5s

(c)

異步、同步額外開銷不變。

耗時：異步:41.67x10=416.7s

同步:100x3.35=335s

(d)

10000

耗時：異步:------=104.2s

9600/10

同步：傳輸幀耗時：色生=0.8383s

9600

???共耗時：100x0.8383=83.83s

6.5答案

6.5LetthebitdurationbeT.Thenaframeis12Tlong.Letaclockperiodbe「.Thelast

bit(bit12)issampledat11.5T'.Forafastrunningclock,theconditiontosatisfyis

T11s

H-5F>117^-<-^=1.045=fclock<1.045加

Foraslowrunningclock,theconditiontosatisfyis

T115

11.57'<127=無>宣=0.958nfclock>^^fbit

Therefore,theoverallcondition:0.958<fclock<1.045fbit

另解：

不發(fā)生幀差錯，則8+1比特總誤差不超過50%,即小于50%/9=5.6%,精

確率在95%以上。

另解：設(shè)能夠容忍的時鐘精確率的百分比為X%,

.??(8+1+2)(100—幻=5%

?"=95?5

.?.能夠容忍的時鐘精確百分比為95.5%

6.10答:

(a)

1Q00100010011000

10001000000100001/10000000000000000000000000000000

10001000000100001:：：：；；；；；；；：：：：

looooooiooooioooo；：；：：；；：!!：

10001000000100001!；；；?；；；；；；

10010001100010000?：：：：：：

..........................................................

11001100110001666!!!!

loooioooooaiogo”；!!!

10001001101010010；；；

10001000000100001!!;

remainder=0001101110011000

(b)

||5||4||J||2

IXs

laput

Shift|ShiftRegister|Input

0000000000000000

100010000001000011

200100000010000100

301000000100001000

410000001000010000

500010010001100010

600100100011000100

/0100I000110001000

s10010001100010000

900110011001100010

1001100110011000100

1111001100110001000

1210001001101010010

1300000011011100110

1400000110111001100

1500001101110011000

16001101110110010

CRC

6.12答：

6.12

10110110

110011/1110001100000

110011：：：：：：：

-101111：：：：：

::::;

111000：::;

110011-III

"IOIIOO：：

noon：：

111110;

110011!

CRC="11010

中文答案：

R=11010

T=1110001111010

6.13答:

6.13a.

c3<————c2<—a<—c0

Input

b.Data=10011011100

M(X)=1+X3+X4+X6+X7+X8

X4M(X)-X12+x11+X10+X8+X7+X4

P(X)P(X)

R(X)-X2

T(X)=X4M(X)+R(X)=X12+Xn+X10+X8+X7+X4+X2

Code-001010011011100

c.Code-001010001011100

T(X)

；

—R—x)y>ieldsanonzeroremainder

第7章的參考答案

7.2答案

v=------->50%

1+2。

Jpmp_20ms

ci——

tfweU4k

,：a<0.5

A/>160

，幀長度應(yīng)大于160加

7.2LetLbethenumberofbitsinaframe.Then,usingEquation7.5ofAppendix7A:

PropagationDelay20x10-380

TransmissionTime一103VL

UsingEquation7.4ofAppendix7A:

U=>0.5

l+2a-l+(160/￡)

￡>160

Therefore,anefficiencyofatleast50%requiresaframesizeofatleast160bits.

7.3答案

270ms

a==270

1000/IMpbs

(a)u0.0018

l+2a

w7

⑸u==0.0126

1+2。一-1+540

w127

(c)u0.2286

1+2。-'1+540

w255

(d)u0.459

1+2/-1+540

PropagationDelay270x10-J)

ZM=1O3/1O6=270

a.U=1/(1+2a)=1/541=0.002

b.UsingEquation7.6:U=W/(l+2a)=7/541=0.013

c.U=127/541=0.23

d.U=255/541=0.47

7.5答案

7.5Roundtrippropagationdelayofthelink=2xLxt

Timetotransmitaframe=B/R

Toreach100%utilization,thetransmittershouldbeabletotransmitframes

continuouslyduringaroundtrippropagationtime.Thus,thetotalnumberof

framestransmittedwithoutanACKis:

N=-a1R~+1，whereisthesmallestintegergreaterthanorequaltoX

ThisnumbercanbeaccommodatedbyanM-bitsequencenumberwith:

M=「log2(N)]

中文答案：

當(dāng)窗口大小w?2a+l時，信道得利用率為100%

，應(yīng)取w=2a+l

'"<>pL,t1.p

a=———=-------=—LRt

“eB/RB

2

：.w=-LR-t+l=2"-I

B

o

???幀號字段的長度應(yīng)為n=[log(-A-7?-r+l)]

2B

7.7答案：

7.7Assumea2-bitsequencenumber:

1.StationAsendsframes0z1,2tostationB.

2.StationBreceivesallthreeframesandcumulativelyacknowledgeswithRR3.

3.Becauseofanoiseburst,theRR3islost.

4.Atimesoutandretransmitsframe0.

5.Bhasalreadyadvanceditsreceivewindowtoacceptframes3,0,1,2.Thusit

assumesthatframe3hasbeenlostandthatthisisanewframe0,whichit

accepts.

7.8答案:

7.8Usethefollowingformulas:

a0.11.10100

S&W(1-P)/1.2(1-P)/3(1-P)/21(1-P)/201

CBN⑺(l-P)/(l-?-0.2P)(1-P)/(1+2P)7(1-7(1-P)/201(l+6P)

P)/21(1+6P)

GBN(127)(l-P)/(l+0.2P)(1-P)/(1+2P)(1-P)/(l+20P)127(l-P)/201(l+126P)

SREJ⑺1-P1-P7(1-P)/217(1-P)/201

SREJ(127)1-P1-P1-P127(1-P)/201

Foragivenvalueofa,theutilizationvalueschangeverylittleasafunctionofP

overareasonablerange(say10'3to10-12).Wehavethefollowingapproximate

valuesforP=10"6:

a0.11.010100

Stop-and-wait0.830330.050.005

GBN⑺1.01.00.330.035

GBN(127)1.01.01.00.63

SREJ(7)1.01.00330.035

SREJ(127)1.01.01.00.63

7.9答案:

7.9a.

節(jié)點A節(jié)點B

7.10答案：

7.10AlostSREJframecancauseproblems.Thesenderneverknowsthattheframewas

notreceived,unlessthereceivertimesoutandretransmitstheSREJ.

▲問題在于接收方無法通知發(fā)送方是否收到了其補發(fā)的幀。

7.11答案

7.11Fromthestandard:'ASREJframeshallnotbetransmittedifanearlierREJ

exceptionconditionhasnotbeencleared(Todosowouldrequestretransmission

ofadataframethatwouldberetransmittedbytheREJoperation.)"Inother

words,sincetheREJrequiresthestationreceivingtheREJtoretransmitthe

rejectedframeandallsubsequentframes,itisredundanttoperformaSREJona

framethatisalreadyscheduledforretransmission.

Alsofromthestandard:"Likewise,aREJframeshallnotbetransmittedifoneor

moreearlierSREJexceptionconditionshavenotbeencleared.TheREJframe

indicatestheacceptanceofallframespriortotheframerejectedbytheREJframe.

ThiswouldcontradicttheintentoftheSREJframeorframes.

REJ：發(fā)送方重發(fā)第N(R)幀及其后的各幀，接收方丟棄N(R)及其以后的各幀;

SREJ：發(fā)送方重發(fā)第N(R)幀，接受方繼續(xù)接收并保存已收到的幀。

7.12答案

7.12Leth=timetotransmitasingleframe

1024bits

"-心bps=1.024msec

Thetransmittingstationcansend7frameswithoutanacknowledgment.Fromthe

beginningofthetransmissionofthefirstframe,thetimetoreceivethe

acknowledgmentofthatframeis:

t2=270++270=541.024msec

Duringthetimet2,7framesaresent.

Dataperframe=1024-48=976

.7x976bits,..

ihrouehput=-------------------------=126zkbps

541.024x10-3sec

▲另解：

假設(shè)控制字段長8bit,FCS長16bit,則在一幀中數(shù)據(jù)比例為

1024-4x8.6=95.3%

1024

假設(shè)該鏈路可用GO-BACK-N差錯控制，則窗口尺寸可達(dá)7

“=—^—=1.3%

1+2。

數(shù)據(jù)的比特吞吐量為：

lMx95.3%xl.3%=12389M/s

7.13答案

7.13No,becausethefieldisofknownfixedlength.

7.14答案

7.14Thefollowingenhancementsarepossibilities:

?Alwaystransmitanintegralnumberofoctets

?Includealengthfield

?Donotusethesameflagtocloseoneframeandopenanother

?Ignoreanyframecontainingfewerthan32bits

?Ignoreanyframeendinginsevenormoreones

Thelengthfieldisusedtocomparethenumberofoctetsreceivedwiththenumber

transmitted.Anydiscrepanciesresultindiscardingtheframe.Thelastthree

enhancementsallowtherejectionofframeswhentheclosingflaghasbeen

destroyed.

7.16答案

7.16N(R)=2.Thisisthenumberofthenextframethatthesecondarystationexpects

toreceive.

根據(jù)題意知窗口序號為3比特，以8為模。因可以連續(xù)發(fā)送6幀，可斷定采

用回退N幀ARQ而不是選擇拒絕ARQo

因是無差錯操作，當(dāng)發(fā)送的第6幀信息幀的輪詢位置1,從站將給予RR或

RNR應(yīng)答，由于發(fā)送6幀信息前主站的N(S)為3,之后發(fā)送的信息幀的N(S)從

4開始，因此，從站返回的N(R)計數(shù)值為2,表示已接收到了4、5、6、7、0和

1幀，可以接收的下一幀的序號是2。

N(R)=010

7.17答案

7.17Oneexampleofsuchaschemeisthemultilinkprocedure(MLP)definedaspartof

layer2ofX.25.ThesameframeformatasforLAPBisused,withoneadditional

field:

FlagAddressControl|MLC一Packet|FCSFlag

Themultilinkcontrol(MLC)fieldisa16-bitfieldthatcontainsa12-bitsequence

numberthatisuniqueacrossalllinks.TheMLCandpacketfieldsformamultilink

protocol(MLP)frame.OnceanMLPframeisconstructed,itisassignedtoa

particularlinkandfurtherencapsulatedinaLAPBframe,asshownabove.The

LAPBcontrolfieldincludes,asusual,asequencenumberuniquetothatlink.

TheMLCfieldperformstwofunctions.First,LAPBframessentoutover

differentlinksmayarriveinadifferentorderfromthatinwhichtheywerefirst

constructedbythesendingMLP.ThedestinationMLPwillbufferincoming

framesandreorderthemaccordingtoMLPsequencenumber.Second,ifrepeated

attemptstotransmitaframeoveronelinkfails,theDTEorDCEwillsendthe

frameoveroneormoreotherlinks.TheMLPsequencenumberisneededfor

duplicatedetectioninthiscase.

7.18答案

7.18Theselective-rejectapproachwouldburdentheserverwiththetaskofmanaging

andmaintaininglargeamountsofinformationaboutwhathasandhasnotbeen

successfullytransmittedtotheclients;thego-back-Napproachwouldbelessofa

burdenontheserver.

該題實際上是問該B/S模式應(yīng)用究竟是回退N幀ARQ還是選擇拒絕ARQ

效率更高？

回退N幀ARQ會增加網(wǎng)上流量和服務(wù)器重傳的信息量，尤其當(dāng)線路質(zhì)量不

好時，服務(wù)器發(fā)送的數(shù)據(jù)量會劇增。

選擇拒絕ARQ會使接收和發(fā)送邏輯更復(fù)雜一些，尤其會加重服務(wù)器接收緩

沖的負(fù)擔(dān)。

WEB服務(wù)器因接收信息量小，發(fā)送信息量大，選擇拒絕ARQ的缺點對其影

響相對較小，倒是重傳信息量是主要問題。因此選擇拒絕ARQ對減輕WEB服

務(wù)器負(fù)擔(dān)可能更好一些。

第七章補充作業(yè):

1.若數(shù)據(jù)鏈路的發(fā)送窗口限度（尺寸）為4,在發(fā)送3號幀，并接受2號幀的確

認(rèn)幀后，發(fā)送方還可連續(xù)發(fā)幾幀？請給出可發(fā)幀的序號？

解答：

可以連續(xù)發(fā)送4幀，序號為3,4,5,6。

2.兩個相鄰的節(jié)點（A和B）通過后退N幀ARQ協(xié)議通信，幀順序為3位，窗口

大小為4。假定A正在發(fā)送，B正在接收，對下面兩種情況說明窗口的

位置：

①A開始發(fā)送之前

②A發(fā)送了0,1,2三個幀，而B應(yīng)答了0,1兩個幀

③A發(fā)送了3,4,5三個幀，而B應(yīng)答了第4幀

解答：

①窗口的位置在0

②窗口的位置在2

③窗口的位置在5

3.知數(shù)據(jù)幀長1024比特，其中幀頭24比特，數(shù)據(jù)速率500kb/s,線路的傳播延遲為

5ms,試計算：

①采用停等協(xié)議，求最大的信道利用率。

②設(shè)滑動窗口W>=2a+1,求窗口至少有多大？

解答：

①

5x10-3

a=----------=2.44

1024/5xlO5

?=—1―=0.17

1+2。

當(dāng)窗口大小卬22a+l時，信道得利用率為100%

應(yīng)取卬=2。+1

trop5x10-3

a二———=2.44

5

frame1024/5xlO

：.w=T-\=2a+\

???幀號字段的長度應(yīng)為n=[log2（2a+2）]=3

第8章的參考答案

8.1.請描述T1載波的原理。

答：

Bell系統(tǒng)的T1載波利用脈碼調(diào)制PCM和時分TDM技術(shù)，使24路采樣聲

音信號復(fù)用一個通道。每一個幀包含193位，每一幀用125us時間傳送。T1系

統(tǒng)的數(shù)據(jù)傳輸速率為1.544Mbps。

數(shù)據(jù)位控制信號幀編碼

82在T1載波中，由非用戶數(shù)據(jù)引入的開銷占百分比是多少？

答：

T1載波是把24個話音信道多路復(fù)用在一條高速信道上，每個信道包含7位的數(shù)

據(jù)和1位的控制信令位，此外加入一位幀同步位組成基本幀。其中，用戶的開

銷為24X1（控制位）+1（基本幀）=25b總開銷為：（7+1）X24+1=193b因此,

用戶的開銷所占的百分比為：25/193X100%^13%

第9章的參考答案

本章無作業(yè)

第10章的參考答案

10.4答案

10.4a.CircuitSwitching

T=Cj+C2where

C[=CallSetupTime

C2=MessageDeliveryTime

J=S=0.2

C2=PropagationDelay+TransmissionTime

=NxD+L/B

=4x0.001+3200/9600=0.337

T=0.2+0.337=0.537sec

DatagramPacketSwitching

T=D1+D2+D3+D4where

D]=TimetoTransmitandDeliverallpacketsthroughfirsthop

D->=TimetoDeliverlastpacketacrosssecondhop

D3=TimetoDeliverlastpacketacrossthirdhop

D4=TimetoDeliverlastpacketacrossforthhop

ThereareP-H=1024-16=1008databitsperpacket.Amessageof3200bits

requirefourpackets(3200bits/1008bits/packet=3.17packetswhichweroundup

to4packets).

D[=4xt+pwhere

t=transmissiontimeforonepacket

p=propagationdelavforonehop

D]=4x(P/B)+D

=4x(1024/9600)+0.001

=0.428

D)=D3=D4=t+p

=(P/B)+D

=(1024/9600)+0.001=0.108

T=0.428+0.108+0.108+0.108

=0.752sec

VirtualCircuitPacketSwitching

T=Vj+V2where

V]=CallSetupTime

V2=DatagramPacketSwitchingTime

T=S+0.752=0.2+0.752=0.952sec

b.CircuitSwitchingvs.DiagramPacketSwitching

Tc=End-to-EndDelay,CircuitSwitching

Tc=S+NxD+L/B

Td=End-to-EndDelay,DatagramPacketSwitching

rL-

Np=Numberofpackets=———

Td-D1+(N-1)D2

D]=TimetoTransmitandDeliverallpacketsthroughfirsthop

D2=TimetoDeliverlastpacketthroughahop

D]=Np(P/B)+D

D2=P/B+D

T=(Np+N-l)(P/B)+NxD

T=Td

S+L/B=(%+N-1)(P/B)

CircuitSwitchingvs.VirtualCircuitPacketSwitching

Tv=End-to-EndDelay,VirtualCircuitPacketSwitching

Tv=S+Td

TC=TV

L/B=(Np+N-1)(P/B)

Datagramvs.VirtualCircuitPacketSwitching

Td=TV-S

10.7答案

10.7Thelayer2flowcontrolmechanismregulatesthetotalflowofdatabetweenDTE

andDCE.Eithercanpreventtheotherfromoverwhelmingit.Thelayer3flow

controlmechanismregulatestheflowoverasinglevirtualcircuit.Thus,resources

ineithertheDTEorDCEthatarededicatedtoaparticularvirtualcircuitcanbe

protectedfromoverflow.

另解：

兩者都是必要的。

因為在第三層分組中采用的流控和差錯控制雖然在格式與處理上與HDLC

相似，但因其分組中具有的D字段可以實現(xiàn)對于本地的或者是端對端的流控。

而第二層的鏈路層則采用LAPB（HDLC的子集）來實現(xiàn)大多數(shù)的鏈路控制與數(shù)

據(jù)傳輸，但不提供分組層中D字段具有的功能。

10.8答案

10.8Transmission,processing,andqueuingdelays.

X.25的分組格式中確實沒有FCS字段，但它作為PDU被傳遞到鏈路層是

由鏈路層協(xié)議將其封裝為LAPB幀，從而加上了FSC字段，這樣可以確保傳輸

LAPB幀中的數(shù)據(jù)域，從而保證X.25分組被正確地傳遞了。

10.9答案

10.9Framerelayismuchsimplified,comparedtoX.25.Themajordifferencesarethat

framerelayusesout-of-channelsignalingwhileX.25usesallin-channelcontrol.In

framerelaythereisno"hop-by-hop"flowcontrolorerrorcontrolasthereisinX.25.

Ifaframeerrorisdetecteditisjustdroppedratherthanbeingretransmitted.

Similarly,onanend-to-endbasis,thereisnoerrorcontrolorflowcontrolexcept

whatisprovidedbyhigherlevelprotocolsoutsideofframerelay.Finally,frame

relayisatwolevel(physicalandlink)layerprotocolandthemultiplexingof

logicalchannelstakesplaceatLevel2ratherthanintheLevel3PacketLayerasin

X.25.

另解：

因為X.25允許一個DTE與另一個DTE之間通過一條物理DTE-DCE鏈路,