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1、Computer Networking:A Top-Down ApproachFeaturing the Internet,4th Edition计算机网络：自顶向下方法(第四版)Solutions to Review Questions and Problems课后习题解答Version Date:October 29,2007This document contains the solutions to review questions and problems fbr the 4thedition of Computer Networking:A Top-Down Approach Fe
2、aturing the Internet by JimKurose and Keith Ross.These solutions are being made available to instructors ONLY.Please do NOT copy or distribute this document to others(even other instructors).Pleasedo not post any solutions on a publicly-available Web site.Well be happy to provide acopy(up-to-date)of
3、 this solution manual ourselves to anyone who asks.Chapter 1 Review Questions1.There is no difference.Throughout this text,the words host and end system”areused interchangeably.End systems include PCs,workstations,Web servers,mailservers,Internet-connected PDAs,WebTVs,etc.2.Suppose Alice,an ambassad
4、or of country A wants to invite Bob,an ambassador ofcountry B,over fbr dinner.Alice doesnt simply just call Bob on the phone and say,“come to our dinner table now”.Instead,she calls Bob and suggests a date and time.Bob may respond by saying hes not available that particular date,but he is availablea
5、nother date.Alice and Bob continue to send messages“back and forth until theyagree on a date and time.Bob then shows up at the embassy on the agreed date,hopefully not more than 15 minutes before or after the agreed time.Diplomaticprotocols also allow for either Alice or Bob to politely cancel the e
6、ngagement if theyhave reasonable excuses.3.A networking program usually has two programs,each running on a different host,communicating with each other.The program that initiates the communication is theclient.Typically,the client program requests and receives services from the serverprogram.4.1.Dia
7、l-up modem over telephone line:residential;2.DSL over telephone line:residential or small office;3.Cable to HFC:residential;4.100 Mbps switchedEtherent:company;5.Wireless LAN:mobile;6.Cellular mobile access(for example,WAP):mobile5.HFC bandwidth is shared among the users.On the downstream channel,al
8、l packetsemanate from a single source,namely,the head end.Thus,there are no collisions inthe downstream channel.6.Current possibilities include:dial-up;DSL;cable modem;fiber-to-the-home.7.Ethernet LANs have transmission rates of 10 Mbps,100 Mbps,1 Gbps and 10 Gbps.For an X Mbps Ethernet(where X=10,1
9、00,1,000 or 10,000),a user cancontinuously transmit at the rate X Mbps if that user is the only person sending data.If there are more than one active user,then each user cannot continuously transmit atX Mbps.8.Ethernet most commonly runs over twisted-pair copper wire and“thirT coaxial cable.It also
10、can run over fibers optic links and thick coaxial cable.9.Dial up modems:up to 56 Kbps,bandwidth is dedicated;ISDN:up to 128 kbps,bandwidth is dedicated;ADSL:downstream channel is.5-8 Mbps,upstream channelis up to 1 Mbps,bandwidth is dedicated;HFC,downstream channel is 10-30 Mbpsand upstream channel
11、 is usually less than a few Mbps,bandwidth is shared.10.There are two most popular wireless Internet access technologies today:a)Wireless LANIn a wireless LAN,wireless users transmit/receive packets to/from a base station(wireless access point)within a radius of few tens of meters.The base station i
12、stypically connected to the wired Internet and thus serves to connect wireless usersto the wired network.b)Wide-area wireless access networkIn these systems,packets are transmitted over the same wireless infrastructureused for cellular telephony,with the base station thus being managed by atelecommu
13、nications provider.This provides wireless access to users within aradius of tens of kilometers of the base station.11.A circuit-switched network can guarantee a certain amount of end-to-end bandwidthfbr the duration of a call.Most packet-switched networks today(including theInternet)cannot make any
14、end-to-end guarantees fbr bandwidth.12.In a packet switched network,the packets from different sources flowing on a link donot follow any fixed,pre-defined pattern.In TDM circuit switching,each host getsthe same slot in a revolving TDM frame.13.At time to the sending host begins to transmit.At time
15、tj=L/Ri,the sending hostcompletes transmission and the entire packet is received at the router(no propagationdelay).Because the router has the entire packet at time tl,it can begin to transmit thepacket to the receiving host at time ti.At time 女 二力+L/Rz,the router completestransmission and the entir
16、e packet is received at the receiving host(again,nopropagation delay).Thus,the end-to-end delay is URi+L/Ri.14.A tier-1 ISP connects to all other tier-1 ISPs;a tier-2 ISP connects to only a few ofthe tier-1 ISPs.Also,a tier-2 ISP is a customer of one or more tier-1.15.a)2 users can be supported beca
17、use each user requires half of the link bandwidth.b)Since each user requires 1Mbps when transmitting,if two or fewer users transmitsimultaneously,a maximum of 2Mbps will be required.Since the availablebandwidth of the shared link is 2Mbps,there will be no queuing delay before thelink.Whereas,if thre
18、e users transmit simultaneously,the bandwidth requiredwill be 3Mbps which is more than the available bandwidth of the shared link.Inthis case,there will be queuing delay before the link.c)Probability that a given user is transmitting=0.2.(3、3/3-3d)Probability that all three users are transmitting si
19、multaneously=I p(1-pj=(0.2)3=0.008.Since the queue grows when all the users are transmitting,thefraction of time during which the queue grows(which is equal to the probabilitythat all three users are transmitting simultaneously)is 0.008.16.The delay components are processing delays,transmission dela
20、ys,propagation delays,and queuing delays.All of these delays are fixed,except for the queuing delays,which are variable.17.Java Applet18.10msec;d/s;no;no19.a)500 kbpsb)64 secondsc)100kbps;320 seconds20.End system A breaks the large file into chunks.To each chunk,it adds headergenerating multiple pac
21、kets from the file.The header in each packet includes theaddress of the destination:end system B.The packet switch uses the destinationaddress to detennine the outgoing link.Asking which road to take is analogous to apacket asking which outgoing link it should be forwarded on,given the packet9saddre
22、ss.21.Java Applet22.Five generic tasks are error control,flow control,segmentation and reassembly,multiplexing,and connection setup.Yes,these tasks can be duplicated at differentlayers.For example,error control is often provided at more than one layer.23.The five layers in the Internet protocol stac
23、k are-from top to bottom-theapplication layer,the transport layer,the network layer,the link layer,and thephysical layer.The principal responsibilities are outlined in Section 1.5.1.24.Application-layer message:data which an application wants to send and passed ontothe transport layer;transport-laye
24、r segment:generated by the transport layer andencapsulates application-layer message with transport layer header;network-layerdatagram:encapsulates transport-layer segment with a network-layer header;linklayer frame:encapsulates network-layer datagram with a link-layer header.25.Routers process laye
25、rs 1 through 3.(This is a little bit of a white lie,as modemrouters sometimes act as firewalls or caching components,and process layer four aswell.)Link layer switches process layers 1 through 2.Hosts process all five layers.26.a)VirusRequires some form of human interaction to spread.Classic example
26、:E-mailviruses.b)WormsNo user replication needed.Worm in infected host scans IP addresses and portnumbers,looking fbr vulnerable processes to infect.c)Trojan horseHidden,devious part of some otherwise useful software.27.Creation of a botnet requires an attacker to find vulnerability in some applicat
27、ion orsystem(e.g.exploiting the buffer overflow vulnerability that might exist in anapplication).After finding the vulnerability,the attacker needs to scan fbr hosts thatare vulnerable.The target is basically to compromise a series of systems byexploiting that particular vulnerability.Any system tha
28、t is part of the botnet canautomatically scan its environment and propagate by exploiting the vulnerability.Animportant property of such botnets is that the originator of the botnet can remotelycontrol and issue commands to all the nodes in the botnet.Hence,it becomespossible fbr the attacker to iss
29、ue a command to all the nodes,that target a singlenode(for example,all nodes in the botnet might be commanded by the attacker tosend a TCP SYN message to the target,which might result in a TCP SYN floodattack at the target).28.Trudy can pretend to be Bob to Alice(and vice-versa)and partially or comp
30、letelymodify the message(s)being sent from Bob to Alice.For example,she can easilychange the phrase Alice,I owe you$1000 to Alice,I owe you$10,000”.Furthennore,Trudy can even drop the packets that are being sent by Bob to Alice(and vise-versa),even if the packets from Bob to Alice are encrypted.Chap
31、ter 1 Problems:Problem 1.There is no single right answer to this question.Many protocols would do the trick.Heres a simple answer below:Messages from ATM machine to ServerMsg namepurposeHELO Let server know that there is a card in theATM machineATM card transmits user ID to ServerPASSWD User enters
32、PIN,which is sent to serverBALANCEUser requests balanceWITHDRAWL User asks to withdraw moneyBYEuser all doneMessages from Server to ATM machine(display)Msg name purposePASSWDAsk user for PIN(password)OKlast requested operation(PASSWD,WITHDRAWL)OKERRlast requested operation(PASSWD,WITHDRAWL)in ERRORA
33、MOUNT sent in response to BALANCE requestBYEuser done,display welcome screen at ATMCorrect operation:client server(check if valid userid)-PASSWDPASSWD-(check password)-AMOUNT WITHDRAWL-check if enough$to coverwithdrawlIn situation when theres not enough money:HELO(userid)-PASSWD -WITHDRAWL -withdraw
34、l-(check if valid userid)PASSWD(check password)OK(password is OK)AMOUNT check if enough$to coverERR(not enough funds)BYEProblem 2.a)A circuit-switched network would be well suited to the application described,becausethe application involves long sessions with predictable smooth bandwidthrequirements
35、.Since the transmission rate is known and not bursty,bandwidth can bereserved for each application session circuit with no significant waste.In addition,weneed not worry greatly about the overhead costs of setting up and tearing down acircuit connection,which are amortized over the lengthy duration
36、of a typicalapplication session.b)Given such generous link capacities,the network needs no congestion controlmechanism.In the worst(most potentially congested)case,all the applicationssimultaneously transmit over one or more particular network links.However,sinceeach link offers sufficient bandwidth
37、 to handle the sum of all of the applications*datarates,no congestion(very little queuing)will occur.Problem 3.a)We can n connections between each of the four pairs of adjacent switches.This givesa maximum of 4n connections.b)We can n connections passing through the switch in the upper-right-hand co
38、mer andanother n connections passing through the switch in the lower-left-hand comer,giving a total of 2n connections.Problem 4.Tollbooths are 100 km apart,and the cars propagate at 1 OOkm/hr.A tollbooth services acar at a rate of one car every 12 seconds.a)There are ten cars.It takes 120 seconds,or
39、 two minutes,for the first tollbooth toservice the 10 cars.Each of these cars has a propagation delay of 60 minutes beforearriving at the second tollbooth.Thus,all the cars are lined up before the secondtollbooth after 62 minutes.The whole process repeats itself for traveling between thesecond and t
40、hird tollbooths.Thus the total delay is 124 minutes.b)Delay between tollbooths is 7*12 seconds plus 60 minutes,i.e.,61 minutes and 24seconds.The total delay is twice this amount,i.e.,122 minutes and 48 seconds.Problem 5a)dp”=m/s seconds.b)dtrans=L/R seconds.C)da=(rn/s+L/R)seconds.d)The bit is just l
41、eaving Host A.e)The first bit is in the link and has not reached Host B.f)The first bit has reached Host B.g)Wantw=-5 =-(2.5x108)=893 km.R 28xIO?Problem 6Consider the first bit in a packet.Before this bit can be transmitted,all of the bits in thepacket must be generated.This requires*V c=6 m s e c.6
42、4xl03The time required to transmit the packet is48-8IxlO6sec=3844 sec.Propagation delay=2 msec.The delay until decoding is6msec+384 sec+2msec=8.384msecA similar analysis shows that all bits experience a delay of 8.384 msec.Problem 7a)10 users can be supported because each user requires one tenth of
43、the bandwidth.b)p =0.1.c)409H=0 n)We use the central limit theorem to approximate this probability.Let X j be independentrandom variables such thatP(XJ=l)=p.、(4 0P(4tl 1 or more users)=1-P X 107，40 P Z X/4 1 0 =P5)6_ _J40 0.L0。V40.0.1.0.97H P Z=P(ZProblem 9The first end system requires L/Ri to trans
44、mit the packet onto the first link;the packetpropagates over the first link in di/si；the packet switch adds a processing delay of dprocafter receiving the entire packet,the packet switch requires UR?to transmit the packetonto the second link;the packet propagates over the second link in d2k2.Adding
45、thesefive delays givesdend-end=L/Rj+L/R.2+d j/s j+d2k2+dprocTo answer the second question,we simply plug the values into the equation to get 8+8+16+4+1 =37 msec.Problem 10Because bits are immediately transmitted,the packet switch does not introduce any delay;in particular,it does not introduce a tra
46、nsmission delay.Thus,de nd-end=UR+d 1/S+d12k2For the values in Problem 9,we get 8+16+4=28 msec.Problem 11The arriving packet must first wait for the link to transmit 3,500 bytes or 28,000 bits.Since these bits are transmitted at 1 Mbps,the queuing delay is 28 msec.Generally,thequeuing delay is nL+(L
47、-x)/R.Problem 12The queuing delay is 0 for the first transmitted packet,L/R for the second transmittedpacket,and generally,(n-l)L/R for the nth transmitted packet.Thus,the average delay forthe N packets is(UR+2UR+.+(N-1)UR)/N=URN(1+2+.+(N-l)=LN(N-1)/(2RN)=(N-1)U(2R)Note that here we used the well-kn
48、own fact that1+2+.+N=N(N+l)/2Problem 13It takes LN IR seconds to transmit the N packets.Thus,the buffer is empty when abatch of N packets arrive.The first of the N packets has no queuing delay.The 2nd packet has a queuing delay ofL/R seconds.The ntb packet has a delay of(n-l)L/7?seconds.The average
49、delay isL(N-l)Problem 14a)The transmission delay is L/R.The total delay isIL L L/R-H-=-R 1-/b)Let x=L/R .xTotal delay=-I-a xProblem 15a)There are Q nodes(the source host and the N-1 routers).Let d晨 denote theprocessing delay at the qth node.Let Rq be the transmission rate of the qth link and letd黑$=
50、L/Rq.Let dprop be the propagation delay across the qth link.Thenend-to-endq+dg+d proc trans propb)Let 叫 denote the average queueing delay at node q.Thenend-to-endq+d 4 +dqproc trans prop queueProblem 16The command:traceroute-q 20 www.eurecom.frwill get 20 delay measurements from the issuing host to