《计算机网络英文答案.pdf》由会员分享，可在线阅读，更多相关《计算机网络英文答案.pdf（119页珍藏版）》请在taowenge.com淘文阁网|工程机械CAD图纸|机械工程制图|CAD装配图下载|SolidWorks_CaTia_CAD_UG_PROE_设计图分享下载上搜索。
1、Computer Networking:A Top-Down ApproachFeaturing the Internet,4th EditionSolutions to Review Questions and ProblemsVersion Date:October 29,2007This document contains the solutions to review questions and problems for the 4thedition of Computer Networking:A Top-Down Approach Featuring the Internet by
2、 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 this solution manual o
3、urselves to anyone who asks.All material copyright 1996-2007 by J.F.Kurose and K.W.Ross.All rights reservedChapter 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,Intern
4、et-connected PDAs,WebTVs,etc.2.Suppose Alice,an ambassador of country A wants to invite Bob,an ambassador ofcountry B,over for 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
5、 not available that particular date,but he is availableanother 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 als
6、o allow for either Alice or Bob to politely cancel the engagement 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 reque
7、sts and receives services from the serverprogram.4.1.Dial-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 bandwidt
8、h is shared among the users.On the downstream channel,all 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 Mbp
9、s,1 Gbps and 10 Gbps.For an X Mbps Ethernet(where X=10,100,1,000 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 t
10、wisted-pair copper wire and thin coaxial cable.It also 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;H
11、FC,downstream channel is 10-30 Mbpsand upstream channel 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)w
12、ithin a radius of few tens of meters.The base station istypically 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
13、,with the base station thus being managed by atelecommunications 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 bandwidthfor the duration of a call.Most packet-switche
14、d networks today(including theInternet)cannot make any end-to-end guarantees for 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.
15、At time to the sending host begins to transmit.At time t=L/R,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=。+U
16、Rz,the router completestransmission and the entire packet is received at the receiving host(again,nopropagation delay).Thus,the end-to-end delay is L/Ri+UR.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 o
17、r more tier-1.15.a)2 users can be supported because 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
18、 no queuing delay before thelink.Whereas,if three 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=0.2d)Probability t
19、hat all three users are transmitting simultaneously=一 p)3=(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
20、 processing delays,transmission delays,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,i
21、t adds headergenerating multiple packets from the file.The header in each packet includes theaddress of the destination:end system B.The packet switch uses the destinationaddress to determine the outgoing link.Asking which road to take is analogous to apacket asking which outgoing link it should be
22、forwarded on,given the packefsaddress.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 l
23、ayers in the Internet protocol stack 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 ont
24、othe transport layer;transport-layer 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-l
25、ayer header.25.Routers process layers 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 in
26、teraction to spread.Classic example:E-mailviruses.b)WormsNo user replication needed.Worm in infected host scans IP addresses and portnumbers,looking for vulnerable processes to infect.c)Trojan horseHidden,devious part of some otherwise useful software.27.Creation of a botnet requires an attacker to
27、find vulnerability in some application orsystem(e.g.exploiting the buffer overflow vulnerability that might exist in anapplication).After finding the vulnerability,the attacker needs to scan for hosts thatare vulnerable.The target is basically to compromise a series of systems byexploiting that part
28、icular vulnerability.Any system that 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 beco
29、mespossible fbr the attacker to issue a command to all the nodes,that target a singlenode(fbr 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(a
30、nd vice-versa)and partially or completelymodify 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$10,000”.Furthermore,Trudy can even drop the packets that are being sent by Bob to Alice(and vise-versa),even if the packets f
31、rom Bob to Alice are encrypted.Chapter 1 Problems:Problem 1.There is no single right answer to this question.Many protocols would do the trick.Here*s a simple answer below:Messages from ATM machine to ServerMsg name purposeHELO Let server know that there is a card in theATM machineATM card transmits
32、 user ID to ServerPASSWD User enters PIN,which is sent to serverBALANCE User requests balanceWITHDRAWL amount User asks to withdraw moneyBYE user all doneIn situation when theres not enough money:Messages from Server to ATM machine(display)Msg namepurposePASSWDAsk user for PIN(password)OKlast reques
33、ted operation(PASSWD,WITHDRAWL)OKERRlast requested operation(PASSWD,WITHDRAWL)in ERRORAMOUNT sent in response to BALANCE requestBYEuser done,display welcome screen at ATMCorrect operation:clientserverHELO(userid)-(check if valid userid)-PASSWDPASSWD -(check password)-AMOUNT WITHDRAWL -check if enoug
34、h$to coverwithdrawl-PASSWD -WITHDRAWL -withdrawl(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
35、 with predictable smooth bandwidthrequirements.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,
36、which are amortized over the lengthy duration 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.Howe
37、ver,sinceeach link offers sufficient bandwidth 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
38、 through the switch in the upper-right-hand comer 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 second
39、s.a)There are ten cars.It takes 120 seconds,or 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 repeat
40、s itself for traveling between thesecond and third 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)dprop=m ls seconds.b)dtrans=L IR
41、seconds.C)dend_lo_end=(m/s+LIR)seconds.d)The bit is just leaving 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)Want加，S=(2.5x IO*%893km.R 28 x ICP Problem 6Consider the first bit in a packet.Before this bit can be transmitted,all of the bits in
42、thepacket must be generated.This requires48-8，-7 sec=6msec.64xl03The time required to transmit the packet is48-8IxlO6sec=384 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 user
43、s can be supported because each user requires one tenth of the bandwidth.b)p=0.1.o f40V1(i-Pr.In j9,40、d)i-Z P(1-P严.We use the central limit theorem to approximate this probability.Let X.be independentrandom variables such that P(X y=1)=p.(4 0、P(11 or more users)=1-P X y,4 0、p ZXj wio=P740-0.1-0.9-6
44、V40 0.1-0.97P z=P(Z telnet utopia.poly.edu 80Since the index.html page in this web server was not modified since Fri,18 May2007 09:23:34 GMT,the following output was displayed when the abovecommands were issued on Sat,19 May 2007.Note that the first 4 lines are theGET message and header lines input
45、by the user and the next 4 lines(startingfrom HTTP/1.1 304 Not Modified)is the response from the web server.15.FTP uses two parallel TCP connections,one connection for sending controlinformation(such as a request to transfer a file)and another connection fbractually transferring the file.Because the
46、 control infbnnation is not sent over thesame connection that the file is sent over,FTP sends control information out ofband.16.Message is sent from Alices host to her mail server over HTTP.Alices mailserver then sends the message to Bobs mail server over SMTP.Bob thentransfers the message from his
47、mail server to his host over POP3.17.Received:Received:Received:Message-ID:Received:From:To:Bcc:Subject:Date:from 65.54.246.203(EHLO bay0-omc3-)(65.54.246.203)by with SMTP;Sat,19May 2007 16:53:51-0700from (65.55.135.106)by bay0-omc3-with Microsoft SMTPSVC(6.0.3790.2668);Sat,19 May 2007 16:52:42-0700
48、from mail pickup service by with Microsoft SMTPSVC;Sat,19 May 2007 16:52:41-0700from 65.55.135.123 by with HTTP;Sat,19 May 2007 23:52:36 GMTMprithula dhungeF*Test mailSat,19 May 2007 23:52:36+0000Mime-Version:1.0Content-Type:Text/html;fbrmat=flowedReturn-Path:Figure:A sample mail message headerRecei
49、ved:This header field indicates the sequence in which the SMTP serverssend and receive the mail message including the respective timestamps.In this example there are 4 Received:header lines.This means the mailmessage passed through 5 different SMTP servers before being delivered to thereceivers mail
50、 box.The last(forth)“Received:header indicates the mailmessage flow from the SMTP server of the sender to the second SMTP server inthe chain of servers.The senders SMTP server is at address 65.55.135.123 andthe second SMTP server in the chain is .The third“Received:header indicates the mail message