计算机网络英文题库(附答案)chapter5

P63 #5 Consider sending a packet of F bits over a path of Q links. Each link transmits at R bps. The network is lightly loaded so that there are no queuing delays. Propagation delay is negligible.a.Suppose the network is a packet-switched virtual-circuit network. Denote the VC setup time by t s seconds. Suppose the sending layers add a total of h bits of header to the packet. How long does it take to send the file from source to destination?t s+[(F+h)/R]Qb.Suppose the network is a packet-switched datagram network and a connectionless service is used. Now suppose each packet has 2h bits of header. How long does it take to send the packet?[(F+2h)/R]Qc.Finally, suppose that the network is a circuit-switched network. Further suppose that the transmission rate of the circuit between source and destination is R bps. Assuming ts setup time and h bits of header appended to the packet, how long does it take to send the packet?t s+(F+h)/RP64 #6 This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters and suppose that the propagation speed along the link is s meters/sec. Host A sends a packet of size L bits to host B.[a] Express the propagation delay, d prop, in terms of m and s.[b] Determine the transmission time of the packet, d trans, in terms of L and R.[c] Ignoring processing and queueing delays, obtain an expression for the end-to-end delay.[d] Suppose Host A begins to transmit the packets at time t=0. At time t=d trans, where is the last bit of the packet?[e] Suppose d prop is greater than d trans. At time t=d trans, where is the first bit of the packet?[f] Suppose d prop is less than d trans. At time t=d trans, where is the first bit of the packet?[g] Suppose s=2.5 x 108, L=100 bits and R=28kbps. Find the distance m so that d prop = d trans.[a] d prop = m/s[b] d trans = L/R[c] end-to-end delay = d prop + d trans=m/s+L/R[d] The beginning position of the link.[e] On the channel between A and B.[f] On the host B.[g] m/s = L/R = > m = sL/R = > m = 892.86 kmP65 #10 Consider the queueing delay in a router buffer. Suppose that all packets are L bits, the transmission rate is R bps, and that N packets simultaneously arrive at the buffer every LN/R seconds. Find the average queueing delay of a packet (in terms of L, R and N). (Hint: The queueing delay for the first packet is zero; for the second packet L/R; for the third packet 2L/R. The Nth packet has already been transmitted when the second batch of packets arrives.)As the Nth packet has already been transmitted when the next batch of packets arrive, we only need to consider the delay for a single batch of packets.Average delay = Total delay / Number of packetsDelay for 1st packet = 0Delay for 2nd packet = L/RDelay for 3rd packet = 2L/R......Delay for Nth packet = (N-1)L/RTotal delay for N packets = (0 + 1 + 2 ... +(N-1) ) * (L/R)Using the formulas for sum of integer series, this can be written as: Total delay for N packets = (N-1) * (N/2) * (L/R)Therefore, average delay for N packets = ((N-1) * L) / 2RP170 #12 What is the difference between persistent HTTP with pipelining and persistent HTTP without pipelinning? Which of the two is used by HTTP/1.1?For the persistent connection without pipelining, the client issues a new request only when the previous has been received. In this case, the client experiences one RTT in order to request and receive each of the referenced objects.For the persistent connection with pipelining, the client issues a request as soon as it encounters a reference. It is possible for only RTT to be expended for all the referenced objects.P170 #14 Telnet into a Web server and send a multiline request message. Include in the request message theIf-modified-since: header line to force a response message with the 304 Not Modified status code.GET/somedir/exp.html HTTP/1.1Host: Connection: closeUser-agent: Mozilla/4.0If-Modified-Since: Thu, 30 May 2007 12:00:00 GMTAccept-language: frP172 #6 Suppose within your web browser you click on a link to obtain a web page. The IP address for the associated URL is not cached in your local host, so a DNS look-up is necessary to obtain the IP address. Suppose that n DNS servers are visited before your host receives the IP address from DNS; the successive visits incur an RTT (Round Trip Time) of RTT1, ... RTTn. Further suppose that the web page associated with the link contains exactly one object, consisting of a small amount of HTTP text. Let RTT0 denote the RTT between the local host and the remote server containing the object. Assuming zero transmission time of the object, how much time elapses from when the client clicks on the link until the client receives the object? (Hint: read pages 90 .. 93)Time to visit DNS servers and get IP address = RTT1 + RTT2 + ... + RTTnTime to establish TCP connection (SYN and SYNACK) = RTT0Time to send HTTP request and receive reply = RTT0Total time = 2 * RTT0 + (RTT1 + RTT2 + ... + RTTn)P171 #16 Suppose Alice with a Web-based e-mail account (such as Yahoo! Mail or Hotmail) sends a message to Bob, who accesses his mail from his mail server using POP3. Discuss how the message gets from Alice’s host to Bob’s host. Be sure to list the series of application-layer protocols that are used to move the message between the two hosts.The series of application-layer protocols: HTTP、SMTP、POP3Suppose that you send an e-mail message whose only data is a Microsoft Excel attachment. What might the header lines (including MIME lines) look like?From:***********To:***********Subject: helloMIME-Version: 1.0Content-Transfer-Encoding: base64Content-Type: Application/MS-ExcelP286 #5 Suppose host A sends two TCP segments back to back to host B over a TCP connection. The first segment has sequence number 90: the second has sequence number 110.a.How much data is in the first segment?a.20 bytesb.Suppose that the first segment is lost but the second segment arrives at B. In the acknowledgement that host B sends to host A, what will be the acknowledgement number?b.ACK90P291 #27 Consider the following plot of TCP window size as a function of time. (reproduced below for you) Assuming TCP Reno is the protocol experiencing the behavior shown above, answer the following questions. In all cases, you should provide a short discussion justifying your answer.a. Identify the intervals of time when TCP slow start is operating.b. Identify the intervals of time when TCP congestion avoidance is operating.c. After the 16th transmission round, is segment loss detected by a tripleduplicate ACK or by a timeout?d. After the 22nd transmission round, is segment loss detected by a triple duplicate ACK or by a timeout?e. What is the initial value of Threshold at the first transmission round?f. What is the value of Threshold at the 18th transmission round?g. What is the value of Threshold at the 24th transmission round?h. During what transmission round is the 70th segment sent?i. Assuming a packet loss is detected after the 26th round by the receipt of a triple duplicate ACK, what will be the values of the congestion-window size and of Threshold?Solution:a.1-6, 23-26b.6-16, 17-22c.a triple duplicate ACKd.timeoute.32f.21g.13h.7i.4, 4P293 #34 Consider sending an object of size O = 100 Kbytes from server to client. Let S = 536 bytes and RTT = 100 msec. suppose the transport protocol uses static windows with window size W. (See Section 3.7.2)a.For a transmission rate of 28 kbps, determine the minimum possible latency. Determine the minimum window size that achieves this latency.b.Repeat (a) for 100 kbps.tency=28.8s W=2tency=8.2s W=4P405 #8 Consider a datagram network using 8-bit host addresses. Suppose a router uses longest prefix matching and has t he following forwarding table:-----------------------------------------------------Prefix Match Interface-----------------------------------------------------00 001 110 211 3-----------------------------------------------------For each of the four interfaces, give the associated range of destination host addresses and the number of addresses in the range.6P407 #15 Consider sending a 3000-byte datagram into a link that has a MTU of 500 bytes. Suppose the original datagram is stamped with the identification number 422. How many fragments are generated? What are their characteristics?there are「2980/480」=7 fragments be generatedP408 #22 Consider the network shown in Problem 21 (reproduced below). Using Dijkstra’s algorithm, and showing your work using a table similar to Table 4.3, do the following:a. Compute the shortest path from s to all network nodesSteps D(t),P(t) D(u),P(u)D(v),P(v)D(w),P(w)D(x),P(x)D(y),P(y)D(z),P(z)0 1,s 4,s ∞∞∞∞∞1 3.t 10,t ∞∞5,t 3,t2 4,u 6,u ∞5,t 3,t3 4,u 6,u ∞5,t4 5,v 7,v 5,v5 6,w 5,v6 6,wPlease fill in the following tables using DV algorithm:For the node Z in the graph shown in the 22nd topic (P408), please fill in the following routing table in the router z about the initial distance-vector Destination node Next hop Current shortest distancevalue-DzS —∞T T 2U —∞V —∞W —∞X —∞Y Y 14Z Z 0following rout-ing table in the node z to update this routing tableDestination node Currentdistance-DyDestination node Current distance-DtS 5 S 1 T 4 T 0 U 2 U 2P493 #7 How big is the MAC address space？The IPv4 address space？The IPv6 address space?MAC address: 6 bytes, MAC address space 2^48IPV4 address: 4 bytes, IPV4 address space 2^32IPV6 address: 16 bytes, IPV6 address space 2^128P494 #4 Consider the 4-bit generator, G, shown in Figure 5.8, and suppose the D has the value 10101010. What is the value of R?G=1001, D=10101010, R=101。

.Chapter 2 Application Layer1. In the following four descriptions, which one is correct?A In C/S architecture, clients communicate with each other directly.B In C/S architecture, client has a fixed well-known address.C In P2P architecture, peers communicate with each other directly.D In P2P architecture, peer has a fixed well-known address.2.In the context of a communication session between a pair of process, theprocess that initiates the communication is labeled as the,the process that waits to be contacted to begin the session is labeled as the.A client process, server processB sending process, receivingprocessC input process,output processD communicating process, connecting process3.A socket is the interface between within a host.A the network layer and the link layerB the link layer and the physical layerC the application layer and the transport layerD the transport layer and the network layer4. In the following applications, which one is a loss-tolerant application?A E-mailB file transferC instant messagingD real-time audio5. In the following applications, which one is a bandwidth-sensitive application?.A E-mailB web applicationC real-time audioD file transfer6. The service of TCP can’tguarantee.A delivery of all data without errorB delivery of all data in theproper orderC the rate of deliveryD congestion control7. In the following applications, which one uses UDP?A E-mailB web applicationC file transferD DNS8.In the following descriptions about HTTP, which one is not correct?A HTTP uses non-persistent connections in its default mode.B HTTP uses TCP as its underlying transport protocol.C HTTP is a stateless protocol.D HTTP is client-server architecture.9.Suppose a web page consists of a base HTML file, 5 JEPG images and a java applet, and also suppose HTTP uses persistent connection without pipelining,the total response time is.A 2RTTB 8RTTC 12 RTTD 14RTT10. In HTTP response message, if the requested HTTP protocol version is not supported, the status code is.A 200B 301C 400D 5051~5 CACDC6~10 CDABD11. The port number of web application is.A25B20C80D5312. A network entity that satisfies HTTP requests on the behalf of an originweb server is.A server farmB server stackC proxy serverD edge server13. In the following descriptions about persistent connection, which one is not correct?A The server leaves the TCP connection open after sending a response.B Each TCP connection is closed after the server sending one object.C There are two versions of persistent connection: without pipelining and with pipelining.D The default mode of HTTP uses persistent connection with pipelining.14. FTP uses two parallel TCPconnections to transfer a file, there are.A control connection and data connectionB receiving connection and sending connectionC client connection and sever connectionD program connection and process connection15. In the following descriptions about FTP, which one is correct?A FTP is p2p architecture.B FTP sends its control informationout-of-band.C FTP uses persistent connection.D FTP is a stateless protocol.16. In the commandsof FTP, which one does not have parameter?A USERB PASSC LISTD STOR17. The Internet mail system has three components which they are.A user agent, SMTP, POP3B SMTP, POP3, IMAPC user agent, SMTP, IMAPD user agent, SMTP, mail server18. If the status code in HTTP response message is 404, it means.A Request succeeded.B The requested document doesn’texit on this server.C This is a generic error code indicating that the request could not be understood by the server.D Requested object has been permanently mover.19. Comparison HTTP with SMTP, the correct is.A HTTP is a push protocol, and SMTP is a pull protocol.B In the default mode, both of them use persistent connection.C HTTP places all of the message ’s objects into one message, and SMTP sendsevery object one by one.D HTTP requires each messageto be in 7-bit ASCII format,and SMTPdoesn’t impose this restriction.20. The headers in the MIME message must include except.A FromB Content-typeC Content-transfer-encodingD MIME version11~15CCBAB16~20 DDBBD21. In the following protocol, which one is stateless?A POP3B SMTPC FTPD IMAP22. DNS means.A Data Name SystemB Data National SystemC Domain Name SystemD Domain National System23. There are three classes of DNS server except.A Root DNS serverB Local DNS serverC TLD server D Authoritative DNS server24. DNS provides some services except.A Host aliasingB Mail server aliasingC Load distributionD A single point of failure25. There are three architectures for locating content in P2Pfile sharing,KaZaA uses.A Centralized Directory（Napster）B Query Flooding(Gnutella)C Exploiting Heterogeneity (p141)D Incentive Priorities26. There are three architectures for locating content in P2Pfile sharing, Napst er uses.A Centralized Directory (p137)B Query FloodingC Exploiting HeterogeneityD Incentive Priorities27. The following architectures in P2Pfile sharing, which is an overlay network?A Centralized DirectoryB Query Flooding(p139)C Exploiting HeterogeneityD Incentive Priorities28. The time it takes for a small packet to travel from client to server andthen back to the client is.A round-travel timeB next-hop timeC round-trip timeD prefix-matching time29.Suppose A ( with a Web-based e-mail account ) sends a message to B ( who accesses his mail server using POP3), which application-layer protocol is not used?A HTTPB SMTPC POP3D IMAP30.In the four following options, which protocol is included in Mail Access Protocol?A SMTPB DHCPC IMAPD FTP31. In FTP commands,is used to send user password to the server.A UserB PassC RetrD Stor32. The function of the additional header field in MIMEContent-Type is.A to convert the message body to its origin non-ASCII formB to determine what actions it should take on message bodyC to send an E-mail to the receiving user agentD to indicate what type the message is33.In the four following options, which application is organized as hybrid ofC/S and P2P architecture?A E-mailB OICQC File transferD Web application34. In the four following options, which is not a centralized DNSdesign ’s problem ?A a single point of failureB traffic volumeC distant centralized databaseD slow(maintenance)35. In the following options, from the application developer’s perspective, which is not correct?A the network architecture is fixedB the network architecture provides a specific set of services to applicationsC the application architecture is designed by the physical devicesD the application architecture dictates how the application is organized over the various end systems36. There are three predominant architectures used in modern network applications, which one is not included?A the client-server architectureB the P2P architectureC a hybrid of the client-server and P2P architectureD a hybrid of the client-server and browser-server architecture37.In the following options about C/S architecture, which is not correct?A In C/S architecture, there is an always-on host, called the server.B In C/S architecture, there is an always-on host, called the client.C The server has a fixed, well-known address, called IP address.D Clients do not directly communicate with each other.38.are often used to create a powerful virtual server in C/S architecture.A PeersB Server farmC server stackD local server39. A process sends messages into, and receives messages from, the network through its.A socketB programC clientD peer40.Which one is not defined by an application-layer protocol?A the types of messages exchangedB the syntax of various message typesC the semantics of the fieldsD rules for determining when and how to translate the socket41. HTTP can use two types of connections, which are.A persistent and non-persistent connectionB connection with pipelining and without pipeliningC TCP and UDPD parallel and serial connection42.takes for a small packet to travel from client to server and thenback to the client.A RDTB thresholdC RTTD overhead43. The default mode of HTTP uses.A non-persistent connection with pipeliningB non-persistent connection without pipeliningC persistent connection with pipeliningD persistent connection without pipelining44. In HTTP request messages, the request line has three fields,there are.A the method field, the URL field and the HTTP version fieldB the method, the connection and URL fieldC the user-agent, the method and HTTP version fieldD the user-agent, the URL and the HTTP version field45. In the header lines of HTTP request message, if the field of Connection isclose, it specifies.A the host on which object residesB what type of the user agentC that the browser wants the server to close the connection after sendingthe requested objectD which language can the browser receive46. In HTTP response message, if the status code is 404, it means.A request succeeded and the information is returned in the responseB requested object has been permanently movedC the requested HTTP protocol version is not supported by the serverD the requested document does not exist on this server47.is a network entity that satisfies HTTP requests on the behalfof an origin Web server.A proxy serverB local serverC DNS serverD Web server48. In the following four options about web cache, which one is not correct?A A web cache is both a server and a client at the same time.B A web cache is purchased and installed by an ISP.C A web cache can raise the response time for a client request.D A web cache can reduce traffic on an institution’s access link to the Internet.49. The request message in the conditional GET must include the headerline.A Last-ModifiedB Last-ReferencedC If-Modified-SinceD If–Referenced-Since50. FTP uses two parallel connections to transfer a file, they are.A TCP and UDP connectionB connection with pipelining and without pipeliningC control an data connectionD client-server and browser-server connection51.In FTP commands, which one is used to ask the server to send back alist of all files in the current remote directory?A USERB PASSC LISTD RETR52. In the Internet mail system,allow users to read, reply to, forward, save and compose message.A User agentsB mail serversC SMTPD TCP53. The two key MIME headers for supporting multimedia are.A Content-Type and MIME-VersionB Content-Type and Content-Transfer-EncodingC Content-Transfer-Encoding and MIME-VersionD MIME-Version and MIME-Type54. For Internet mail, the mail access protocol is used to.A transfer mail from the recipient ’s mail server to the recipient’s user agentB transfer mail from the sender’s mail server to the recipient’s mail serverC translate the mail from the sender’s mail serverD translate the mail into the recipient’s mail server55. POP3 progresses through three phases, which they are.A authorization, translation and transactionB authorization, translation and updateC authorization, transaction and updateD translation, transaction and update56. In the following four services, which one can not provide by DNS?A Host aliasingB Mail server aliasingC translate hostname to IP addressesD translate MAC addresses to IP addresses57. There are three classes of DNS servers, there are.A root DNS server, top-level domain DNS server and local DNS serverB root DNS server, top-level domain DNS server and authoritative DNS serverC root DNS server, local DNS server and authoritative DNS serverD root DNS server, local DNS server and top-level domain DNS server58.In the following four options about POP3, which one is not correct?A The user agent employed only three commands: List, Retr and QuitB The server does not carry state information across POP3 sessionsC The port number is 110D The POP3 protocol does not provide any means for a user to createremote folders and assign messages to folders.59. A resource record in DNSdistributed database is a four-tuple,which field ca n be ignored?(p132) A Name B Value C Type D TTL60. In the following four options about DNSresource record, which one is correct?A The meaning of Nameand Value depend on Type.B The meaning of Value and Type depend on Name.C If Type=A, then Nameis a domain and Value is the IP address for the hostname.D If Type=MX,then Nameis domain and Value is the IP address for the hostname.61.In the following four options about DNSmessages, which one is not correct?A There are only two kinds of DNSmessage.(p133)B Both query and reply messagehave the sameformat.C The header section in DNSmessagehas 12 bytes.D The authority section contains the resource records for the samethat wasorigi nally queried.62.In DNSmessage,contains information about the query that is being made.(p 133)A authority sectionB question sectionC answer sectionD additional sec tion63.There are three techniques are employed in most any P2Pfile-sharing systems, which one is not include?（p144）A Request queuingB incentive prioritiesC parallel downloadingD Responseq ueuing64. In the following four options about P2P file-sharing, which one isnot correct?A P2P file-sharing is highly scalable.B P2P file-sharing relies on P2P architecture.C The means for locating content in different P2P file-sharing are different.D P2P file-sharing systems not only share MP3s,but also videos, software,documents and images.65.In MIMEheader lines,specifies the nameof the SMTPserver that sent the m essage (from),the nameof the SMTPserver that received the message(by), and the time an which t he receiving server received the message.A ReceivedB FromC ToD MIME-Version66.If the header line Connection is close, it means that the client wants.A persistent connection with pipeliningB persistent connection without pipeliningC nonpersistent connectionD not connection67. In HTTP request message, the entity body is empty with the method,but is used with the method.A GET, POSTB POST,GETC GET, HEAD D POST, HEAD68. In HTTP response message, if the Date: header ;one indicates the time Fri.08 Aug. 2008 12:00:00 GMT, the Last-Modified: header line can not be.A Fri. 08 Aug. 2008 11:00:00 GMTB Fri. 08 Aug. 2008 11:30:00GMTC Fri. 08 Aug. 2008 12:00:00 GMTD Fri. 08 Aug. 2008 12:30:00 GMT69. In the following four options,which one is not the part of cookie technology?A Cookie header lines in the HTTP response message and request message.B One cookie header file kept on the user ’s end system and managedby the user ’s browser.C A network entity that satisfies HTTPrequests on the behalf of an origin Web server.D A back-end database at the Web site70. On-top of stateless HTTP,can be used to create a user session layer.A proxy serverB Web cacheC cookieD socket71. Processes communicate with each other by reading from and writing to.A programsB threadsC socketsD channels72.In the following four options about network architecture, which one isnot correct?A The network architecture is fixed.B The network architecture provides a specific set of services to application.C The network architecture is designed by application developer.D The network architecture dictates how the application is organized overspecial server.73. In Client-Server architecture,the clients visit the server ’s through.A client’s socketB client’s IP addressC server’s socketD server’s IP address74.can be thought of as a program that is running within end system.A processB threadC socketD context75. API means.A Application Program InterfaceB Application Process InterfaceC Appellation Program InterfaceD Appellation Process Interface76.One host can be running many network applications, so the system assignsthem differentto distinguish each other.A IP addressB port numberC hostnameD section77.In the following four applications, which one is both bandwith-sensitive and require tight timing constraints?A real-time audioB file transferC E-mailD Webdocuments78.The port number of the Web server is.A 25B 20C 80D 808079.The port number of the mail server is.A 25B 20C 80D 808080.Look the URL/rjxy/index.html,the object ’s path nameis.A B /rjxyC /rjxy/index.htmlD index.html81. Each URL has components, they are.A transport protocol and object’s path nameB host name and object’s path nameC transport protocol and host nameD client name and server name1. Consider an HTTP client will request a WEB page from a WEB server. Supposethe URLof the page is /somedepartment /somedir/exp.html. The client does not want to use persistent connections and want to receive Frenchversion of the object. The user agent is WindowsNT 5.1. Give the request message according to the given format.Request line:Header lines:.2.Telnet into a Web server and send a multiline request message. Includein the request message the If-modified-since: header line to force aresponse message with the 304 Not Modified status code.Solution:Request line:Header lines:3.Suppose within you Web browser you click on a link to obtain a Web page. The.look-up is necessary to obtain the IP address. Suppose that n DNS servers arevisited before your host receives the IP address from DNS; the successive visits incur an RTT of RTT, ⋯， RTT. Further suppose that the Webpage associated with1nthe link contains exactly one object,consisting of a small HTML text.Let RTT0 denote the RTT between the local host and the server containing the object.Assuming zero transmission time of the object, how much time elapses from whenthe client clicks on the link until the client receives the object?5. Suppose that you send an e-mail message whose only data is a microsoft excell attachment. What might the header lines (including MIME lines) look like?。

Chapter 5: Data CodingFill In the Blank5-1.It is much easier for a machine to process a code if all the characters in the code have _____.5-2.The number of possible combinations or characters in a coding system are called _____.5-3.The minimum number of bits that would be required to encode only the 26 characters of the alphabet is _____.5-4.The three different groups of character assignments that are normally made in a coding system are _____, _____, and _____.5-5.The coding system that has enough bit combinations to encode all of the characters in all of the worlds languages is called _____.5-6.Three major types of data compression are called _____, _____, and _____. 5-7.Arguably, the most important data coding scheme because it is the mostly widely used is _____.5-8.In an even parity system, a 1 bit is added to a character’s code when necessary to make the total number of 1 bits representing the character an _____ number.5-9.Character compression is also known as _____.5-10. Character stripping removes the leading and trailing _____ characters from a message and adds them back at the receiving end.5-11.R un length encoding replaces _____ characters or repetitive groups of characters with a different, much shorter group of characters.5-12. A measure of how few bits are required to accurately convey the meaning of a character is called _____ _____.5-13. Bits that are used to determine the code points in a code are called _____ _____.5-14. The _____ code is an 8-bit code that has 256 code points.True or False5-1.The term binary digit is abbreviated bit.5-2.Machines cannot send Morse code because the coded characters have different numbers of code elements.5-3.It is more difficult for a machine to process a code if all of the bits are of the same duration.5-4.When even parity is used, a 0 bit is added, if necessary, to make the total number of 0 bits in the character an even number.5-5.The escape character in a coding system is a different character than the one that is generated when you press the ESC key on the keyboard of a PC.5-6.ASCII is a 7-bit code.5-7.Unicode provides enough code points so that all characters in all languages of the world have a unique 16-bit code point.5-8.Code efficiency is a measure of how many extra bits are used beyond those required to convey the meaning of a character.5-9.Data compression is the process of reducing the number of bits used to represent a character, or reducing the number of characters before they are transmitted.5-10.C haracter stripping is the process of removing control characters from a message before it is transmitted.5-11.T he binary digits are 1 and 2.5-12.A code is a predetermined set of symbols that have variable meanings.5-13.M orse code is not considered a binary code, because its two elements are of different length.5-14.A binary code works well for machines communicating by electrical means because the 1 bits and 2 bits can be represented by a current flow that is either on or off.5-15.F or transmission efficiency, it is ideal to have a coding system that uses a minimum number of bits to represent each character.5-16.M ost coding systems include a technique called an escape character.5-17.T he necessity to support escape characters complicates the design of equipment that is to code and decode the data.5-18.T he ASCII code is the most widely used code in computers and communications networks today.5-19.U nicode is a 16-bit character code.Multiple Choice5-1. Characters that control the positioning of information on a workstation screen or paper are called _____.a. page control charactersb. format effector charactersc. device control charactersd. screen control characterse. a and b5-2. It is much easier for a machine to process a code if the code has the following attributes: _____.a. it is a true binary codeb. all of the characters have the same number of bitsc. all of the bits are perfectly formedd. all of the bits are the same duratione. a and cf. All of the above.5-3. The unique sequence of bits assigned to represent the various characters of a code are called _____.a. code pointsb. character assignmentsc. control charactersd. character algorithmse. None of the above.5-4.The purpose of escape characters is to _____.a. allow a program to abort a message prematurelyb. implement a priority system by allowing one message to be prematurelyended so another can be sent on the circuitc. act as an alternate escape control character when more than 8 bits areneededd. indicate that the characters that follow are to be interpreted as having analternate meaninge. indicate that a PDU has been received correctly5-5. The ASCII code is also known as the _____.a. International Telegraph Alphabet 5b. extended ASCII codec. 8-bit coded. PC codee. EBCDIC code5-6. The EBCDIC code _____.a. was developed by IBMb. is an 8-bit codec. has 256 code pointsd. is used on older IBM mainframe computerse. All of the above.f. None of the above.5-7. Unicode _____.a. is a 16-bit codeb. was developed by a group of companiesc. supports the characters of all languagesd. All of the above.e. None of the above.5-8. Code efficiency is calculated _____.a. by minimizing the number of bits transmitted for a given messageb. by dividing the information bits by the total number of bits in a characterc. by dividing the total number of bits by the information bits in a characterd. by measuring the error rate at the receiving end of a transmission5-9. Three major types of data compression are _____.a. character removal, run length encoding, and bit compressionb. bit stripping, double clocking, and character strippingc. character compression, Huffman coding, and character strippingd. character stripping, run length encoding, and character compressione. over clocking, transmission digitizing, and double ACKing5-10. A 10-bit code could represent how many characters?a. 100b. 210 -1c. 29d. 512e. 2105-11.The number of possible combinations or characters in a coding system is called the _____.a. character assignmentb. code pointsc. control characterd. code limite. binary code5-12.Many coding systems include an extra bit, called a _____ , for checking purposes.a. parity bitb. padding bitc. control bitd. binary digite. checking bit5-13.When an ESC character is sent as a part of the data, it means that _____.a. the transmission is being abortedb. this is the end of the transmissionc. the characters that follow are to be interpreted as having an alternate meaningd. the characters that follow are encryptede. the transmission should be ignored5-14. Graphic characters _____.a. can be displayed on a workstation screenb. are special characters only used by graphic artistsc. are only present in specialized code systemsd. can also be used as format effector characters5-15Code conversion _____.a. changes one character to another for security purposesb. changes the characters in one coding system to those of another coding systemc. can be readily performed by a computerd. All of the above.e. b and cChapter 6: Data Communication FundamentalsFILL IN THE BLANK6-1. The three attribute s of a sine wave are its _____ , _____ , and _____.6-2. The FM radio band in the United States extends from 88 MHz to 108 MHz. Its bandwidth is _____.6-3. A signal drop of 10 dB represents a drop in power of _____times.6-4. A radio signal with a frequency of 101 MHz falls within the _____radio band.6-5. Speech with frequencies above _____is attenuate d and not transmitted through the public telephone network.6-6. Changing a communication signal by alter ing its amplitude, frequency, or phase is called _____.6-7. The unit of signaling rate on a communication circuit is called _____.6-8. The unit of measure of a circuit's data carrying rate is called _____6-9. Sending a tribit on a circuit requires _____(how many) unique signal changes to be transmitted?6-10. A circuit that allows transmission in one direction only is called a _____ circuit.6-11. A circuit that allows transmission in either direction but not at the same time is called a _____circuit.6-12. A circuit that allows transmission in both directions simultaneously is called a _____circuit.6-13. A circuit that allows all of the bits in a character to be transmitted simultaneously is called a _____circuit.6-14. A circuit in which the bits of a character are transmitted one after the other on a single communication path is called a _____circuit.6-15. The mechanism by which character synchronization occurs in asynchronous transmission is called _____and _____bits.6-16. The means by which the receiver knows which bits to group together to form a character is called _____.6-17. When each block of data to be transmitted is surrounded with synchronizing and other control characters the transmission is called _____.6-18. When each character to be transmitted is surrounded with start and stop bits, the transmission is called _____.6-19. The _____condition is the idle condition on a line. Conventionally it is the same as the signal for a 1 bit.6-20. The purpose of the _____in synchronous transmission is to maintain synchronization between the transmitting node and the receiving node.6-21. _____modulation is better suited for high-speed analog transmission because the receiver more easily detects changes in _____than changes in amplitude or frequency.6-22. _____equalizers assume that the circuit meets a n average set of loss, noise, and distortion parameter s, and shape the transmitted signal accordingly.6-23. _____equalizers examine the actual parameters of the circuit being used, and shape the signal to match its specific characteristics.6-24. Modems that use the V.34 standard use a sophisticated modulation technique called _____, and they assume that most of the actual transmission will occur on _____circuits.6-25. High-speed modems achieve a high throughput rate by using sophisticated modulation and __________techniques.6-26. A _____ensures that the signal entering a digital circuit has properly shaped, square pulses, and is precisely timed.6-27. The time it takes the modems on an HDX circuit to reverse their function –the time for the transmitting modem to switch to being the receiver, and the receiving modem to switch to being the transmitter—is called _____.6-28. The advantages of _____transmission over _____transmission include better data integrity; higher capacity circuits; easier integration of voice, data and other signals; better security and privacy; and lower cost.6-29. The type of digital modulation that takes 8,000 samples per second but uses only 4 bits per sample to code the difference between the values of two samples is called _____.6-30. An _____converts analog signals to a digital format.6-31. The difference between the actual value (height) of an analog signal and the nearest integer value, which is assigned when the signal is digitized, is called_____.6-32. When the number of data bits in a character is divided by the total number of bits required to transmit the character, the __________is being calculated. TRUE OR FALSE6-1. Signaling is the propagation of a signal on a medium.6-2. Playing the A key above middle C on a piano results in a digital signal that has 880 pulses per second.6-3. Sound waves, electrical waves, and electromagnetic waves have essentially different characteristics.6-4. A digital signal level is measured in decibels.6-5. The loss of signal strength between two points on a communication circuit is called attenuation.6-6. The amplitude and phase of a voice signal remain constant while its frequency changes.6-7. ADPCM is accomplished by shifting each channel to a different part of the frequency spectrum.6-8. The speed of a circuit is measured in baud.6-9. A 2,400 baud signal can carry data at a maximum rate of 7,200 bps.6-10. When 2 bits of information are coded into one signal change, they are called bi-bits.6-11. The most common digital signals are unipolar.6-12. When transmitting digital signals on an analog circuit, a modem is used.6-13. An analog signal that has been digitized can be closely restored to its original analog shape by a D/A converter.6-14. PCM is the ITU-T’s recommended method for digitizing voice at 32 Kbps. 6-15. In a modem, equalizer circuitry compensates for the variability of the actual transmission line used.6-16. Line turnaround time in a modem can be a significant part of the total transmission time.6-17. Modems can never handle full-duplex transmission.6-18. Phase modulation is the technique of chang ing a digital signal’s phase in order to modulate it.6-19. V.34-bis modems assume that most data transmission occurs on digital lines.6-20. The transmission speed between a pair of V.90 modems is symmetric.6-21. Cable modems for each CATV system are unique.6-22. Modem eliminator s are a low-cost alternative to modems on LAN circuits.6-23. There are very few practical uses for simplex transmission.6-24. Parallel mode transmission is widely used on WANs to achieve high throughput.6-25. Asynchronous transmission is widely used today.6-26. Start bits precede every data block in synchronous transmission.6-27. To jam a spread spectrum signal, one simply broadcasts a jamming signal on the spread spectrum signal’s carrier frequency.MULTIPLE CHOICE6-1. The reasons that digital transmission is superior to analog transmission are _____.a. better data integrityb. higher capacityc. easier integrationd. better security and privacye. lower costf. All of the above.g. only a, b, c, and d6-2. A modem is a form of a _____.a. A/D converterb. D/A converterc. codecd. DTEe. None of the above.f. a and b6-3. The three attributes of an analog signal discussed in the chapter are _____.a. frequency, decibels, and modulationb. phase, frequency, and amplitudec. amplitude, carrier, and sined. bit rate, speed, and error ratee. unipolarity, quantization, and equilibrium6-4. Parallel transmission occurs _____.a. during spread spectrum transmissionb. when a PC sends data to a printer using a serial cablec. on most WAN circuitsd. All of the above.e. None of the above.6-5. Differential Manchester coding _____.a. is often used on WANsb. is almost never usedc. is subservient to Manchester codingd. requires an analog circuit to work properlye. None of the above.6-6. The most widely used interface standard between PCs and modems is _____.a. point-to-pointb. USBc. RS-232-Dd. RS-232-Ce. X.256-7. The X.21 interface standard _____.a. is a replacement for RS-232-Cb. defines the interface to a digital circuitc. uses a PL-259 connectord. All of the above.e. None of the above.6-8. A network interface card _____.a. provides the interface from a PC to a networkb. provides the interface from one network to anotherc. provides the interface from a modem to a networkd. converts signals from analog to digitale. None of the above.6-9. When two modems send signals to each other for the process of establishing transmission parameters, it is called _____.a. synchronizingb. call setupc. call establishmentd. MOST (acronym for modem startup)e. handshaking6-10. Hertz is the unit of measure for _____.a. baudb. bit ratec. trellis code modulationd. cycles per seconde. data ratef. None of the above.6-11. The difference between the exact height of an analog signal and the nearest integer value when a digitizing sample is taken is called _____.a. digitizing discrepancyb. digitizing errorc. quantizing noised. digitizing discernmente. a and cf. b and c6-12. QAM uses a combination of _____and _____to generate quadbits.a. phase changes and relative amplitudesb. phase changes and dibitsc. phase changes and frequency changesd. amplitude changes and frequency changese. amplitude changes and dibitsChapter 7: Data Link Control ProtocolsFILL IN THE BLANK7-1. A set of rules that define the exact format of messages exchanged between computers or between computers and people is called a _____.7-2. Data link protocols are necessary so that all of the devices on the network are operating by the same set of rules and are able to _____with each other.7-3. The part of a protocol consisting of the rules that specify the way the receiving terminal signals the sending terminal if it has received data correctly, how and under what circumstances the line will be turned around, and whether the receiving terminal can accept more data is called _____.7-4. The part of the protocol consisting of rules specifying what happens when an error is detected, what to do if communications suddenly and unexplainably cease, and the way communications are reestablished after they are broken is called _____.7-5. The part of a protocol consisting of rules for ending the communications under normal and abnormal circumstances is called _____.7-6. The ability of data terminal equipment (DTE) to be able to transmit and receive any bit pattern as data is called _____.7-7. Predetermined communication parameters are those that are set _____by switches or specified as parameters in software.7-8. When a central or control station asks each station on the circuit or network if it has traffic to send, it is using a line control technique called _____.7-9. When any station that has traffic looks to see if the circuit is free, and if it is, begins sending its traffic the line control technique is called _____.7-10. In a polling system, one way that priority can be given to a terminal is by listing its address in the polling list _____.7-11. The _____of a message contains and conveys information about the message, such as the destination node’s address, a sequence number, and perhaps a date and time.7-12. BISYNC is defined to only support the 6-bit transcode (SBT), EBCDIC, and _____.7-13. Character synchronization is accomplished in BISYNC by sending _____ characters at the beginning and periodically in the middle of each transmission.7-14. Compared to BISYNC, DDCMP and other byte-count protocols implement transparency in a much more _____manner.7-15. The HDLC ___ field is used to mark the beginning and ending of frames.7-16. The technique that ensures a flag’s uniqueness by not allowing any other sequence of six consecutive 1 bits in the data stream is called _____.7-17. A protocol that is primarily used by PC workstations to send IP overdial-up lines, typically when dial ing in to an ISP for connection to the Internet, is called _____.7-18. The _____protocol was developed by IBM and then generalize d to become HDLC.7-19. SDLC is a _____of HDLC.7-20. SDLC is a subset of _____.7-21. A _____changes one protocol to another, when, for example, a message is sent between two networks that use different protocols.7-22. A sophisticated check character at the end of each block of data is called a _____.7-23. A technique used to ensure that a fast transmitting node does not send data faster than a receiving node can receive and process it is called _____.7-24. The flow control technique that allows multiple frames to be in transit on the line at one time is called _____.TRUE OR FALSE7-1. Data link protocols are concerned with the transmission of data through an entire network.7-2. All protocols use the same bit patterns as control characters.7-3. Data transparency allows a protocol to be able to transmit and receive any bit patterns as data.7-4. A protocol is implemented by transmitting certain bit patterns or characters on the communication circuit.7-5. The code used for the data is independent of the protocol being used.7-6. Because of collision s, contention systems are rarely used for line access.7-7. Poll ing systems require several stations on the circuit to act as master stations.7-8. Hub polling is the most common implementation of a polling system.7-9. Polling systems are dependent on the data code being used.7-10. Token passing systems require one station on the circuit to take responsibility to ensure that there are always multiple tokens circulating on the circuit.7-11. Data messages normally consist of three parts: the header, text, and trailer. 7-12. The ACK character normally acknowledges the receipt of a block of data. 7-13. Stop-and-wait flow control is inherently efficient because of its simplicity. 7-14. Stop-and-wait flow control allows multiple blocks of data to be in transit on the line at one time.7-15. PPP has largely replaced the SLIP protocol because it has better error detection.7-16. KERMIT is error checked but is a very slow protocol.7-17. A bit-oriented protocol uses multiple control characters to mark the start and end of parts of a message.7-18. HDLC is a bit-oriented protocol.7-19. SLDC is a byte-count-oriented protocol.7-20. The best known bit-oriented protocol is BISYNC.7-21. SDLC was an outgrowth of the original work on HDLC.7-22. SDLC operates in one of three modes; however, one mode is rarely used. 7-23. The basic operational unit for HDLC is a PDU.7-24. HDLC uses a FRA character to mark the beginnings and ends of frames. 7-25. BISYNC supports only selected data codes.7-26. BISYNC supports transparency.7-27. DDCMP has largely been replaced by HDLC.7-28. Go-back-n is an implementation of sliding window flow control.7-29. The LAP protocols are variant s of HDLC.7-30. PPP is a widely used synchronous protocol for dialing into the Internet through an ISP.MULTIPLE CHOICE7-1. In HDLC, when a 0 bit is inserted after all strings of five consecutive 1 bits, the term applied is _____.a. zeroingb. bit stuffingc. synchronizingd. oneinge. string breaking7-2. When two stations on a circuit transmit at the same time, a(n)_____occurs.a. altercationb. divisionc. pollingd. bit stuffinge. collision7-3. In order to know what stations to poll and in what sequence, the master station uses a _____.a. message headerb. polling listc. start of header characterd. polling PDAe. cyclic redundancy check7-4. When a receiver must acknowledge every block of data before the next block is sent, the _____type of flow control is being used.a. stop-and-waitb. stop-and-checkc. stop-and-flopd. sliding framee. sliding window7-5. The name of the flow control protocol in which the sending station, on receipt of a NAK, resends the damaged or out of sequence frame and all frames after it is _____.a. selective rejectb. TCPc. selective repeatd. send all before ne. HDLC7-6. HDLC is an example of _____.a. sliding window flow controlb. a serial line interface protocolc. an asynchronous protocold. All of the above.e. None of the above.7-7. When an HDLC node receives a flag character, it knows that _____.a. an error has occurredb. it should signal the sender to stop sending trafficc. a frame is beginning or endingd. it should switch to transparent modee. None of the above.7-8. HDLC’s information frames _____.a. are used to send an NAK when a frame is received incorrectlyb. are used to establish the way the protocol will proceedc. are used to send information from one node to a router for flow controld. contain the data field that holds the information being transmittede. are not used in ABM mode7-9. The part of a message that contains the destination address is called the_____.a. headerb. address PDUc. textd. trailere. flag7-10. Software or hardware that changes one protocol to another is called a_____ .a. CODECb. protocol converterc. modemd. routere. encryptor7-11. Contention systems work best _____.a. on circuits that have only more than two DTEsb. when the speed of the circuit is relatively fastc. on circuits that have heavy message trafficd. All of the above.e. None of the above.7-12. Desirable attributes for a data link protocol to have are _____.a. transparencyb. efficiencyc. bit linearityd. a, b, and ce. a and b7-13. Techniques to ensure that a fast transmitting node does not send data faster than the receiving node can receive and process it are called _____ .a. token passingb. flow controlc. parity checkingd. cyclic redundancy checkinge. error control7-14. The LAPD protocol is used _____.a. between an X.25 DTE and a packet switching networkb. on an ISDN D channelc. on frame relay networksd. on value added networkse. as a substitute for TCP/IP7-15. Go-back-n and selective repeat are two common implementations of _____.a. bit oriented protocolsb. byte protocolsc. byte-count oriented protocolsd. sliding window flow controle. transmission control protocolsChapter 8: Transmission MediaFILL IN THE BLANK8-1. A type of media that provides some type of physical path, such as wire, cable, or optical fiber, along which the signal moves from end to end is called _______ or guided media.8-2. Air is an example of _______ media.8-3. Twisted pair wire is an example of _______ media.8-4. The best media to use in homes and offices where costs must be kept low and bandwidth requirements are moderate is ____.8-5. The most appropriate media to use when there is a need for very high bandwidth and high security is _____.8-6. The most appropriate media to use when there is a need for high bandwidth and to broadcast to a wide area is ____.8-7. Wire containing at least three twists per foot and that is about the same as normal telephone cable installed in most office buildings is called _____.8-8. _____ cable normally contains four pairs of wires and is the type that is most commonly being installed in new homes and businesses, particularly when it is known that a computer network will be used.8-9. The shielding in shielded twisted pair wire provides a degree of _____ _____ in electrically noisy environments.8-10. _____ is advantageous when tap s in the cable need to be made—it is easier to tap than optical fiber.8-11. The characteristic of _____ include: high bandwidth, difficult to tap and splice; not subject to electrical interference, cable is very small diameter and lightweight.8-12. A _____ is required to operate a microwave transmitter.8-13. _____ _____ is the delay that occurs because light and radio signals only travel at 186,000 miles per second.8-14. Satellites in _____ orbits circle the earth at 22,300 miles.8-15. Satellites in _____ orbits are approximately 6,000 miles high.8-16. _____ works by modulating a laser light beam instead of broadcasting a radio wave.8-17. _____ is a fiber that has a glass core approximately 9 microns in diameter. 8-18. Data are placed on an optical fiber with a light source, either an LED or a _____.8-19. When UTP is used in a data application it is normally terminate d with a _____ jack.8-20. A type of satellite commonly used to send television programming into homes is called a _____.TRUE OR FALSE8-1. Optical fiber is an example of an unguided medium.8-2. Guided media are also known as wireless media.8-3. Wireless media are the most common media used today.8-4. Optical fiber emits an electro-optical field when carrying communications signals.8-5. CAT 3 cable is most frequently used for telephone transmission in older buildings.8-6. CAT 5 cable is commonly being installed in new homes and businesses.8-7. An advantage of UTP wire is its relatively low cost.8-8. Coaxial cable can be tapped easily.8-9. Optical fiber can be tapped easily.。

计算机网络第五版答案完整版Computers have become an integral part of our lives, and computer networking plays a crucial role in connecting various devices and enabling communication between them. In the fifth edition of the book "Computer Networks," the authors provide comprehensive answers and solutions to various exercises and questions, giving readers a complete understanding of the concepts and principles of computer networking.1. Introduction to Computer NetworksComputer networks are essential for transmitting data and enabling communication between devices. In this chapter, the authors explain the basics of networks, including the OSI model, network topology, and types of networks. They also provide answers to exercises that help readers grasp the fundamental concepts of computer networking.2. Physical LayerThe physical layer is responsible for transmitting raw bits over a communication channel. The authors cover topics such as analog and digital signals, transmission media, and modulation techniques. Through detailed explanations and examples, readers gain a solid understanding of the physical layer's functions and mechanisms.3. Data Link LayerThe data link layer ensures reliable data transfer between two connected nodes. This chapter focuses on topics like error detection and correction, flow control, and media access control. The authors provide accurateanswers to questions related to these concepts, enabling readers to comprehend the data link layer's role in establishing error-free communication.4. Network LayerThe network layer facilitates the delivery of data packets across multiple networks. Addressing, routing algorithms, and internet protocols are among the key topics discussed in this chapter. By examining the provided answers, readers can enhance their knowledge of the network layer's functionalities and protocols.5. Transport LayerThe transport layer provides end-to-end communication between applications running on different hosts. This chapter explores topics like multiplexing, demultiplexing, reliable data transfer protocols, and congestion control. The authors offer complete solutions to exercises, allowing readers to grasp the complexities and mechanisms of the transport layer.6. Application LayerThe application layer enables network applications to communicate with each other. This chapter covers topics such as domain name system (DNS), email protocols, and World Wide Web (WWW) protocols. The authors present accurate and detailed answers, enabling readers to understand the application layer's role in facilitating various network services.7. Network SecurityNetwork security is essential to protect data from unauthorized access and malicious activities. This chapter discusses topics like symmetric and asymmetric encryption, public key infrastructure (PKI), and network security protocols. The authors provide comprehensive answers, helping readers to comprehend the importance of network security and the techniques used to safeguard data.8. Multimedia NetworkingIn the modern era, multimedia applications require robust networking capabilities. This chapter delves into topics such as streaming and real-time applications, multimedia protocols, and quality of service (QoS) mechanisms. By examining the provided answers, readers can gain a deeper understanding of the challenges and solutions in multimedia networking.9. Network ManagementEfficient network management is crucial for ensuring the smooth operation of computer networks. This chapter covers topics like Simple Network Management Protocol (SNMP), network monitoring, and network troubleshooting. The authors provide accurate and comprehensive answers, allowing readers to learn about the tools and techniques used in network management.By providing complete and accurate answers to exercises, the fifth edition of "Computer Networks" equips readers with the knowledge and understanding required to master the field of computer networking. The authors' attention to detail, clear explanations, and concise yet informative solutions make this book an invaluable resource for students, professionals, and anyone interested in computer networks.。

Solution of Selected Exercises from the End of Chapter ExercisesChapter 1 - Introduction And Overview1.4 To what aspects of networking does data communications refer? Answer: Data communications refers to the study of low-level mechanisms and technologies used to send information across a physical communication medium, such as a wire, radio wave, or light beam.b5E2RGbCAP1.5 What is packet-switching, and why is packet switching relevant tothe Internet?Answer: Packet switching divides data into small blocks, called packets, and includes an identification of the intended recipient in each packet. Packet switching changed networking in a fundamental way, and provided the basis for the modern Internet. Packet switching allows multiple senders to transmit data over a shared network. p1EanqFDPw1.8 What is a communication protocol? Conceptually, what two aspects ofcommunication does a protocol specify?Answer: A communication protocol refer to a specification for network communication.Major aspects of a protocol are syntax (format> and semantics (meaning> of the protocol. DXDiTa9E3d1.9 What is a protocol suite, and what is the advantage of a suite? Answer:protocols are designed in complete, cooperative sets called suites or families, instead of creating each protocol in isolation.Each protocol in a suite handles oneaspect of communication。

计算机网络双语试题及答案一、选择题（共20题，每题2分，共40分）1. 当使用HTTP协议进行通信时，下列哪个是无连接的协议？A. TCPB. IPC. UDPD. FTP2. 在计算机网络中，IP地址的作用是什么？A. 标识主机在网络中的唯一地址B. 实现数据的可靠传输C. 进行主机之间的通信D. 提供远程访问服务3. 下列哪个不属于网络拓扑结构的类型？A. 总线型拓扑B. 环形拓扑C. 星型拓扑D. 布线型拓扑4. 在TCP/IP协议中，下列哪个协议用于将IP地址转换为物理地址？B. DHCPC. FTPD. ICMP5. 在计算机网络中，HTTP和HTTPS协议之间的区别是？A. HTTP使用明文传输，HTTPS使用加密传输B. HTTP使用UDP传输，HTTPS使用TCP传输C. HTTP使用IP地址，HTTPS使用域名D. HTTP使用GET请求，HTTPS使用POST请求6. 某计算机的IP地址是192.168.0.1，子网掩码是255.255.255.0，那么该计算机所在的网络号是？A. 192.168.0B. 192.168.0.1C. 192.168.0.255D. 192.168.0.07. 在计算机网络中，下列哪个协议用于将域名解析为IP地址？A. DNSB. FTPC. DHCP8. 在TCP/IP协议中，下列哪个协议用于控制数据传输的可靠性？A. ICMPB. UDPC. FTPD. TCP9. 在计算机网络中，什么是反向代理服务器（Reverse Proxy Server）？A. 将外部网络请求转发给内部服务器的服务器B. 将内部网络请求转发给外部服务器的服务器C. 将HTTP请求转发给HTTPS服务器的服务器D. 将HTTPS请求转发给HTTP服务器的服务器10. 在计算机网络中，下列哪个协议用于电子邮件的发送和接收？A. SMTPB. POP3C. HTTPD. FTP11. 在计算机网络中，下列哪个协议用于文件传输？B. SMTPC. UDPD. TCP12. 以下哪个不是IPv6地址的特点？A. 128位长度B. 冒号分隔的十六进制C. 有固定的网络号和主机号D. 全球唯一的地址13. 在计算机网络中，下列哪个技术不属于无线局域网技术？A. Wi-FiB. BluetoothC. NFCD. 4G14. 在网络中，下列哪个设备用于将不同网段的数据转发到目的主机？A. 集线器B. 路由器D. 网关15. 下列哪个网络拓扑结构具有较高的可容错性和可拓展性？A. 星型拓扑B. 总线型拓扑C. 环形拓扑D. 树型拓扑16. 在计算机网络中，下列哪个协议用于互联网上的主机进行网络配置？A. DHCPB. DNSC. HTTPD. FTP17. 在TCP/IP协议中，下列哪个协议用于检测并纠正数据传输中的错误？A. TCPB. ARPC. ICMPD. UDP18. 下列哪个是网络安全的常见攻击方式之一？A. DDoS攻击B. 数据库攻击C. 剪贴板攻击D. 社会工程学攻击19. 下列哪个不是局域网（LAN）的特点？A. 覆盖范围较小B. 速度较快C. 价格较高D. 结构较简单20. 在计算机网络中，下列哪个协议用于向局域网中的所有主机广播消息？A. UDPB. TCPC. DHCPD. ICMP二、问答题（共5题，每题10分，共50分）1. 请简要介绍HTTP协议的工作原理。

Chapter 5 ProblemsProblem 1The rightmost column and bottom row are for parity bits.1 0 1 0 01 0 1 0 01 0 1 0 01 0 1 1 10 0 0 1 1Problem 2Suppose we begin with the initial two-dimensional parity matrix:0 0 0 01 1 1 10 1 0 11 0 1 0With a bit error in row 2, column 3, the parity of row 2 and column 3 is now wrong in the matrix below:0 0 0 01 1 0 10 1 0 11 0 1 0Now suppose there is a bit error in row 2, column 2 and column 3. The parity of row 2 is now correct! The parity of columns 2 and 3 is wrong, but we can't detect in which rows the error occurred!0 0 0 01 0 0 10 1 0 11 0 1 0The above example shows that a double bit error can be detected (if not corrected). Problem 3To compute the Internet checksum, we add up the values at 16-bit quantities: 00000000 0000000100000010 0000001100000100 0000010100000110 0000011100001000 00001001-------------------------00010100 00011001The one's complement of the sum is 11101011 11100110.Problem 4a) To compute the Internet checksum, we add up the values at 16-bit quantities: 00000001 0000001000000011 0000010000000101 0000011000000111 0000100000001001 00001010-------------------------00011001 00011110The one's complement of the sum is 11100110 11100001.b) To compute the Internet checksum, we add up the values at 16-bit quantities: 01000001 0100001001000011 0100010001000101 0100011001000111 0100100001001001 01001010-------------------------01011000 01011111The one's complement of the sum is 10100111 10100000c) To compute the Internet checksum, we add up the values at 16-bit quantities: 01100001 0110001001100011 0110010001100101 0110011001100111 0110011101101000 01101001-------------------------11111001 11111101The one's complement of the sum is 00000110 00000010.Problem 5If we divide 1001 into 10101010000 we get 10111101, with a remainder of R = 101.Problem 6a) If we divide 1001 into 10010001000 we get 10000001, with a remainder of R =001.b) If we divide 1001 into we get 10100011000 we get 10110101, with a remainderof R = 101.c) If we divide 1001 into 010********* we get 010111101, with a remainder of R =101.Problem 7 a)1)1()(--=N p Np p E21)1)(1()1()('------=N N p N Np p N p E ))1()1(()1(2----=-N p p p N NNp p E 1*0)('=⇒=b)NN NN N Np E NN N 11)11()11()11(1*)(11--=-=-=--1)11(lim =-∞→NN eNNN 1)11(lim =-∞→Thusep E N 1*)(lim =∞→Problem 8)1(2)1()(--=N p Np p E)3(2)2(2)1)(1(2)1()('------=N N p N Np p N p E))1(2)1(()1()3(2----=-N p p p N N121*0)('-=⇒=N p p E)1(2)1211(12*)(----=N N N N p Eee p E N 21121*)(lim =⋅=∞→Problem 9a) (1 – p(a))3 p(A) where,p(A) = probability that A succeeds in a slotp(A) = p(A transmits and B does not and C does not)= p(A transmits) p(B does not transmit) p(C does not transmit) = p(1 – p) (1 – p) = p(1 – p)2Hence, p(A succeeds for first time in slot 4) = (1 – p(a))3 p(A) = (1 – p(1 – p)2)3 p(1 – p)2b) p(A succeeds in slot 2) = p(1-p)2 p(B succeeds in slot 2) = p(1-p)2p(C succeeds in slot 2) = p(1-p)2p(either A or B or C succeeds in slot 2) = 3 p(1-p)2 (because these events are mutually exclusive)c) p(some node succeeds in a slot) = 3 p(1-p)2 p(no node succeeds in a slot) = 1 - 3 p(1-p)2Hence, p(first success occurs in slot 4) = p(no node succeeds in first 3 slots) p(some node succeeds in 4th slot) = (1 - 3 p(1-p)2)3 3 p(1-p)2d) efficiency = p(success in a slot) = 3 p(1-p)2Problem 10Problem 11The length of a polling round is)/(poll d R Q N +.The number of bits transmitted in a polling round is NQ . The maximum throughput therefore isQR d R d R Q N NQ poll poll +=+1)/(Problem 12a), b), c) See figure below.d)1. Forwarding table in A determines that the datagram should be routed to interface111.111.111.002.2. The adapter in A creates and Ethernet packet with Ethernet destination address22-22-22-22-22-22.3. The first router receives the packet and extracts the datagram. The forwardingtable in this router indicates that the datagram is to be routed to 122.222.222.003. 4. The first router then sends the Ethernet packet with the destination address of 55-55-55-55-55-55 and source address of 33-33-33-33-33-33 via its interface with IP address of 122.222.222.002.5. The process continues until the packet has reached Host F .e)ARP in A must now determine the LAN address of 111.111.111.002. Host A sends out an ARP query packet within a broadcast Ethernet frame. The first router receives the query packet and sends to Host A an ARP response packet. This ARP response packet is carried by an Ethernet frame with Ethernet destination address 00-00-00-00-00-00.Problem 13a) b) c) See figure below:d)1. Forwarding table in A determines that the datagram should be routed to interface111.111.111.002.2. The adapter in A creates and Ethernet packet with Ethernet destination address22-22-22-22-22-22.3. The router receives the packet and extracts the datagram. The forwarding table inthis router indicates that the datagram is to be routed via interface 122.222.222.002.4. The router then replaces the destination MAC address in the datagram it receivedfrom host A, with the MAC address of F (77-77-77-77-77-77) and sends the datagram over the interface 122.222.222.002.5. The switch (that is connected to the interface 122.222.222.002 of the router)forwards the datagram to its interface where F is connected. The datagram then reaches F.e) ARP in A must now determine the MAC address of 111.111.111.002. Host A sends out an ARP query packet within a broadcast Ethernet frame. The router receives the query packet and sends to Host A an ARP response packet. This ARP response packet is carried by an Ethernet frame with Ethernet destination address 00-00-00-00-00-00.Problem 14Wait for 51,200 bit times. For 10 Mbps, this wait is12.51010102.5163=⨯⨯bpsbits msecFor 100 Mbps, the wait is 512 μsec.Problem 15At 0=t A transmits. At 576=t , A would finish transmitting. In the worst case, B begins transmitting at time 224=t . At time 449225224=+=t B 's first bit arrives at A . Because 576449<, A aborts before completing the transmission of the packet, as it is supposed to do.Thus A cannot finish transmitting before it detects that B transmitted. This implies that if A does not detect the presence of a host, then no other host begins transmitting while A is transmitting.Problem 16Because A 's retransmission reaches B before B 's scheduled retransmission time, B refrains from transmitting while A retransmits. Thus A and B do not collide. Thus the factor 512 appearing in the exponential backoff algorithm is sufficiently large.Problem 17We want 5.)51/(1=+a or, equivalently, trans prop t t a /2.==. )108.1/(8⨯=d t prop m/sec and 576(=trans t bits 810/()bits/sec μ76.5)=sec. Solving for d we obtain 265=d meters. For the 100 Mbps Ethernet standard, the maximum distance between two hosts is 200 m.For transmitting station A to detect whether any other station transmitted during A 's interval, trans t must be greater than 26522⋅=prop t m 8108.1/⨯m/sec μ94.2=sec. Because76.594.2<, A will detect B 's signal before the end of its transmission.a)Let Y be a random variable denoting the number of slots until a success:1)1()(--==m m Y P ββ, where β is the probability of a success.This is a geometric distribution, which has mean β/1. The number of consecutive wasted slots is 1-=Y X thatββ-=-==11][][Y E X E x1)1(--=N p Np β11)1()1(1-----=N N p Np p Np xefficiency 11)1()1(1-----+=+=N N p Np p Np k kxk kb)Maximizing efficiency is equivalent to minimizing x , which is equivalent to maximizingβ. We know from the text that β is maximized at Np 1=.c)efficiency 11)11()11(1-----+=N N N N k k∞→N limefficiency 1/1/11-+=-+=e k k ee k kd) Clearly, 1-+e k k approaches 1 as ∞→k .a)bpsbits m m 681010204sec/102900⨯⋅+⋅sec5.12sec)108105.4(66μ=⨯+⨯=--b)∙ At time 0=t , both A and B transmit.∙ At time sec 5.12μ=t , A detects a collision.∙ At time sec25μ=t last bit of B 's aborted transmission arrives at A .∙At time sec 5.37μ=tfirst bit of A 's retransmission arrives at B .∙ At time sec 5.137********sec 5.376μμ=⨯+=bpsbits t A 's packet is completelydelivered at B .c) sec 5.512sec 1005sec 5.12μμμ=⋅+Problem 20i) from A to left router: Source MAC address: 00-00-00-00-00-00 Destination MAC address: 22-22-22-22-22-22 Source IP: 111.111.111.001 Destination IP: 133.333.333.003ii) from the left router to the right router: Source MAC address: 33-33-33-33-33-33 Destination MAC address: 55-55-55-55-55-55 Source IP: 111.111.111.001 Destination IP: 133.333.333.003iii) from the right router to F: Source MAC address: 88-88-88-88-88-88Destination MAC address: 99-99-99-99-99-99Source IP: 111.111.111.001Destination IP: 133.333.333.003Problem 21i) from A to switch: Source MAC address: 00-00-00-00-00-00Destination MAC address: 55-55-55-55-55-55Source IP: 111.111.111.001Destination IP: 133.333.333.003ii) from switch to right router: Source MAC address: 00-00-00-00-00-00Destination MAC address: 55-55-55-55-55-55Source IP: 111.111.111.001Destination IP: 133.333.333.003iii) from right router to F: Source MAC address: 88-88-88-88-88-88Destination MAC address: 99-99-99-99-99-99Source IP: 111.111.111.001Destination IP: 133.333.333.003Problem 22If all the 14 nodes send out data at the maximum possible rate of 100 Mbps, a total aggregate throughput of 14*100 = 1400 Mbps is possible.Problem 23Each departmental hub is a single collision domain that can have a maximum throughput of 100 Mbps. The links connecting the web server and the mail server has a maximum throughput of 100 Mbps. Hence, if the three collision domains and the web server and mail server send out data at their maximum possible rates of 100 Mbps each, a maximum total aggregate throughput of 500 Mbps can be achieved among the 14 end systems.Problem 24All of the 14 end systems will lie in the same collision domain. In this case, the maximum total aggregate throughput of 100 Mbps is possible among the 14 end sytems.Problem 25Problem 26The time required to fill 8⋅L bits is.sec 8sec 106483m L L =⨯⋅b) For ,500,1=L the packetization delay is.sec 5.187sec 81500m m =For ,48=L the packetization delay is.sec 6sec 848m m =c)Store-and-forward delay R L 408+⋅=For 500,1=L , the delay issec 77sec 15512sec 1015540815006μ≈≈⨯+⋅mFor ,48=L store-and-forward delay sec 1μ<.d) Store-and-forward delay is small for both cases for typical ATM link speeds. However, packetization delay for 1500=L is too large for real-time voice applications.Problem 27Problem 28Chapter 6 Review Questions1.In infrastructure mode of operation, each wireless host is connected to the largernetwork via a base station (access point). If not operating in infrastructure mode, a network operates in ad-hoc mode. In ad-hoc mode, wireless hosts have noinfrastructure with which to connect. In the absence of such infrastructure, thehosts themselves must provide for services such as routing, address assignment,DNS-like name translation, and more.2.a) Single hop, infrastructure-basedb) Single hop, infrastructure-lessc) Multi-hop, infrastructure-basedd) Multi-hop, infrastructure-less3.Path loss is due to the attenuation of the electromagnetic signal when it travelsthrough matter. Multipath propagation results in blurring of the received signal at the receiver and occurs when portions of the electromagnetic wave reflect offobjects and ground, taking paths of different lengths between a sender andreceiver. Interference from other sources occurs when the other source is alsotransmitting in the same frequency range as the wireless network.4.a) Increasing the transmission powerb) Reducing the transmission rate5.APs transmit beacon frames. An AP‟s beacon frames will be transmitted over oneof the 11 channels. The beacon frames permit nearby wireless stations to discover and identify the AP.6.False7.APs transmit beacon frames. An AP‟s beacon frames will be transmitted over oneof the 11 channels. The beacon frames permit nearby wireless stations to discover and identify the AP.8.False9.Each wireless station can set an RTS threshold such that the RTS/CTS sequenceis used only when the data frame to be transmitted is longer than the threshold.This ensures that RTS/CTS mechanism is used only for large frames.10.No, there would n‟t be any advantage. Suppose there are two stations that want totransmit at the same time, and they both use RTS/CTS. If the RTS frame is as long as a DATA frames, the channel would be wasted for as long as it would have been wasted for two colliding DATA frames. Thus, the RTS/CTS exchange is only useful when the RTS/CTS frames are significantly smaller than the DATA frames.11.Initially the switch has an entry in its forwarding table which associates thewireless station with the earlier AP. When the wireless station associates with the new AP, the new AP creates a frame with the wireless station‟s MAC address and broadcasts the frame. The frame is received by the switch. This forces the switch to update its forwarding table, so that frames destined to the wireless station are sent via the new AP.12.Any ordinary Bluetooth node can be a master node whereas access points in802.11 networks are special devices (normal wireless devices like laptops cannot be used as access points).13.False14.“Opportunistic Scheduling” refers to matching the physical layer protocol tochannel conditions between the sender and the receiver, and choosing the receivers to which packets will be sent based on channel condition. This allows the base station to make best use of the wireless medium.15.UMTS to GSM and CDMA-2000 to IS-95.16.No. A node can remain connected to the same access point throughout itsconnection to the Internet (hence, not be mobile). A mobile node is the one that changes its point of attachment into the network over time. Since the user is always accessing the Internet through the same access point, she is not mobile. 17.A permanent address for a mobile node is its IP address when it is at its homenetwork. A care-of-address is the one its gets when it is visiting a foreign network.The COA is assigned by the foreign agent (which can be the edge router in the foreign network or the mobile node itself).18.False19.The home network in GSM maintains a database called the home location register(HLR), which contains the permanent cell phone number and subscriber profile information about each of its subscribers. The HLR also contains information about the current locations of these subscribers. The visited network maintains a database known as the visitor location register (VLR) that contains an entry for each mobile user that is currently in the portion of the network served by the VLR.VLR entries thus come and go as mobile users enter and leave the network.The edge router in home network in mobile IP is similar to the HLR in GSM and the edge router in foreign network is similar to the VLR in GSM.20. Anchor MSC is the MSC visited by the mobile when a call first begins; anchorMSC thus remains unchanged during the call. Throughout the call‟s duration and regardless of the number of inter-MSC transfers performed by the mobile, the call is routed from the home MSC to the anchor MSC, and then from the anchor MSC to the visited MSC where the mobile is currently located.21. a) Local recoveryb) TCP sender awareness of wireless linksc) Split-connection approachesChapter 6 ProblemsProblem 1.Output corresponding to bit d 1 = [-1,1,-1,1,-1,1,-1,1]Output corresponding to bit d 0 = [1,-1,1,-1,1,-1,1,-1]Problem 2.Sender 2 output = [1,-1,1,1,1,-1,1,1]; [ 1,-1,1,1,1,-1,1,1]Problem 3.181111)1()1(111111)1()1(1112=⨯+⨯+-⨯-+⨯+⨯+⨯+-⨯-+⨯=d 181111)1()1(111111)1()1(1122=⨯+⨯+-⨯-+⨯+⨯+⨯+-⨯-+⨯=dProblem 4Sender 1: (1, 1, 1, -1, 1, -1, -1, -1)Sender 2: (1, -1, 1, 1, 1, 1, 1, 1)Problem 5a) The two APs will typically have different SSIDs and MAC addresses. A wirelessstation arriving to the café will associate with one of the SSIDs (that is, one of the APs). After association, there is a virtual link between the new station and the AP. Label the APs AP1 and AP2. Suppose the new station associates with AP1. When the new station sends a frame, it will be addressed to AP1. Although AP2 will also receive the frame, it will not process the frame because the frame is not addressed to it. Thus, the two ISPs can work in parallel over the same channel. However, the two ISPs will be sharing the same wireless bandwidth. If wireless stations in different ISPs transmit at the same time, there will be a collision. For 802.11b, the maximum aggregate transmission rate for the two ISPs is 11 Mbps.b)Now if two wireless stations in different ISPs (and hence different channels)transmit at the same time, there will not be a collision. Thus, the maximum aggregate transmission rate for the two ISPs is 22 Mbps for 802.11b.Problem 6Suppose that wireless station H1 has 1000 long frames to transmit. (H1 may be an AP that is forwarding an MP3 to some other wireless station.) Suppose initially H1 is the only station that wants to transmit, but that while half-way through transmitting its first frame, H2 wants to transmit a frame. For simplicity, also suppose every sta tion can hear every other station‟s signal (that is, no hidden terminals). Before transmitting, H2 will sense that the channel is busy, and therefore choose a random backoff value.Now suppose that after sending its first frame, H1 returns to step 1; that is, it waits a short period of times (DIFS) and then starts to transmit the second frame. H1‟s second frame will then be transmitted while H2 is stuck in backoff, waiting for an idle channel. Thus, H1 should get to transmit all of its 1000 frames before H2 has a chance to access the channel. On the other hand, if H1 goes to step 2 after transmitting a frame, then it too chooses a random backoff value, thereby giving a fair chance to H2. Thus, fairness was the rationale behind this design choice.Problem 7A frame without data is 32 bytes long. Assuming a transmission rate of 11 Mbps, thetime to transmit a control frame (such as an RTS frame, a CTS frame, or an ACK frame) is (256 bits)/(11 Mbps) = 23 usec. The time required to transmit the data frame is (8256 bits)/(11 Mbps) = 751DIFS + RTS + SIFS + CTS + SIFS + FRAME + SIFS + ACK= DIFS + 3SIFS + (3*23 + 751) usec = DIFS + 3SIFS + 820 usecProblem 8a) 1 message/ 2 slotsb) 2 messages/slotc) 1 message/slotd)i) 1 message/slotii) 2 messages/slotiii) 2 messages/slote)i) 1 message/4 slotsii) slot 1: Message A→ B, message D→ Cslot 2: Ack B→ Aslot 3: Ack C→ D= 2 messages/ 3 slotsiii)slot 1: Message C→ Dslot 2: Ack D→C, message A→ BRepeatslot 3: Ack B→ A= 2 messages/3 slotsProblem 10a)10 Mbps if it only transmits to node A. This solution is not fair since only A isgetting served. By “fair” it means that each of the four nodes should be allotted equal number of slots.b)For the fairness requirement such that each node receives an equal amount of dataduring each downstream sub-frame, let n1, n2, n3, and n4 respectively represent the number of slots that A, B, C and D get.Now,data transmitted to A in 1 slot = 10t Mbits(assuming the duration of each slot to be t)Hence,Total amount of data transmitted to A (in n1 slots) = 10t n1Similarly total amounts of data transmitted to B, C, and D equal to 5t n2, 2.5t n3, and t n4 respectively.Now, to fulfill the given fairness requirement, we have the following condition:10t n1 = 5t n2 = 2.5t n3 = t n4Hence,n2 = 2 n1n3 = 4 n1n4 = 10 n1Now, the total number of slots is N. Hence,n1+ n2+ n3+ n4 = Ni.e. n1+ 2 n1 + 4 n1 + 10 n1 = Ni.e. n1 = N/17Hence,n2 = 2N/17n3 = 4N/17n4 = 10N/17The average transmission rate is given by:(10t n1+5t n2+ 2.5t n3+t n4)/tN= (10N/17 + 5 * 2N/17 + 2.5 * 4N/17 + 1 * 10N/17)/N= 40/17 = 2.35 Mbpsc)Let node A receives twice as much data as nodes B, C, and D during the sub-frame.Hence,10tn1 = 2 * 5tn2 = 2 * 2.5tn3 = 2 * tn4i.e. n2 = n1n3 = 2n1n4 = 5n1Again,n1 + n2 + n3 + n4 = Ni.e. n 1+ n1 + 2n1 + 5n1 = Ni.e. n1 = N/9Now, average transmission rate is given by:(10t n1+5t n2+ 2.5t n3+t n4)/tN= 25/9 = 2.78 MbpsSimilarly, considering nodes B, C, or D receive twice as much data as any othernodes, different values for the average transmission rate can be calculated. Problem 11a)No. All the routers might not be able to route the datagram immediately. This isbecause the Distance Vector algorithm (as well as the inter-AS routing protocols like BGP) is decentralized and takes some time to terminate. So, during the time when the algorithm is still running as a result of advertisements from the new foreign network, some of the routers may not be able to route datagrams destined to the mobile node.b)Yes. This might happen when one of the nodes has just left a foreign network andjoined a new foreign network. In this situation, the routing entries from the oldforeign network might not have been completely withdrawn when the entries from the new network are being propagated.c)The time it takes for a router to learn a path to the mobile node depends on thenumber of hops between the router and the edge router of the foreign network for the node.Problem 12If the correspondent is mobile, then any datagrams destined to the correspondent would have to pass through the correspondent’s home agent. The foreign agent in the network being visited would also need to be involved, since it is this foreign agent that notifies the correspondent‟s home agent of the location of the correspond ent. Datagrams received by the correspondent‟s home agent would need to be encapsulated/tunneled between the correspondent‟s home agent and foreign agent, (as in the case of the encapsulated diagram at the top of Figure 6.23.Problem 13Because datagrams must be first forward to the home agent, and from there to the mobile, the delays will generally be longer than via direct routing. Note that it is possible, however, that the direct delay from the correspondent to the mobile (i.e., if the datagram is not routed through the home agent) could actually be smaller than the sum of the delayfrom the correspondent to the home agent and from there to the mobile. It would depend on the delays on these various path segments. Note that indirect routing also adds a home agent processing (e.g., encapsulation) delay.Problem 14First, we note that chaining was discussed at the end of section 6.5. In the case of chaining using indirect routing through a home agent, the following events would happen: ∙The mobile node arrives at A, A notifies the home agent that the mobile is now visiting A and that datagrams to the mobile should now be forwarded to thespecified care-of-address (COA) in A.∙The mobile node moves to B. The foreign agent at B must notify the foreign agent at A that the mobile is no longer resident in A but in fact is resident in Band has the specified COA in B. From then on, the foreign agent in A willforward datagrams it receives that are addressed to the mobile‟s COA in A to themobile‟s COA in B.∙The mobile node moves to C. The foreign agent at C must notify the foreign agent at B that the mobile is no longer resident in B but in fact is resident in C and has the specified COA in C. From then on, the foreign agent in B will forwarddatagrams it receives (from the foreign agent in A) that are addressed to themobile‟s COA in B to the mobile‟s COA in C.Note that when the mobile goes offline (i.e., has no address) or returns to its home network, the datagram-forwarding state maintained by the foreign agents in A, B and C must be removed. This teardown must also be done through signaling messages. Note that the home agent is not aware of the mobile‟s mobility beyond A, and that the correspondent is not at all aware of the mobile‟s mobility.In the case that chaining is not used, the following events would happen: ∙ The mobile node arrives at A, A notifies the home agent that the mobile is now visiting A and that datagrams to the mobile should now be forwarded to thespecified care-of-address (COA) in A.∙The mobile node moves to B. The foreign agent at B must notify the foreign agent at A and the home agent that the mobile is no longer resident in A but infact is resident in B and has the specified COA in B. The foreign agent in A canremove its state about the mobile, since it is no longer in A. From then on, thehome agent will forward datagrams it receives that are addressed to the mobile‟sCOA in B.∙The mobile node moves to C. The foreign agent at C must notify the foreign agent at B and the home agent that the mobile is no longer resident in B but in fact is resident in C and has the specified COA in C. The foreign agent in B canremove its state about the mobile, since it is no longer in B. From then on, thehome agent will forward datagrams it receives that are addressed to the mobile‟sCOA in C.When the mobile goes offline or returns to its home network, the datagram-forwarding state maintained by the foreign agent in C must be removed. This teardown must also bedone through signaling messages. Note that the home agent is always aware of the mobile‟s current foreign network. However, the correspondent is still blissfully unaware of the mobile‟s mobility.Problem 15Two mobiles could certainly have the same care-of-address in the same visited network. Indeed, if the care-of-address is the address of the foreign agent, then this address would be the same. Once the foreign agent decapsulates the tunneled datagram and determines the address of the mobile, then separate addresses would need to be used to send the datagrams separately to their different destinations (mobiles) within the visited network.Problem 16If the MSRN is provided to the HLR, then the value of the MSRN must be updated in the HLR whenever the MSRN changes (e.g., when there is a handoff that requires the MSRN to change). The advantage of having the MSRN in the HLR is that the value can be provided quickly, without querying the VLR. By providing the address of the VLR Rather than the MSRN), there is no need to be refreshing the MSRN in the HLR.Chapter 7 Review Questions1.Streaming stored audio/video: pause/resume, re-positioning, fast-forward; real-timeinteractive audio and video: people communicating and responding in real time.2.Camp 1: No fundamental changes in TCP/IP protocols; add bandwidth where needed;also use caching, content distribution networks, and multicast overlay networks.Camp 2: Provide a network service that allows applications to reserve bandwidth in the network. Camp 3, differentiated service: introduce simple classifying and policing schemes at the edge of the network, and give different datagrams different levels of service according to their class in the router queues.3.Uncompressed audio stored on CD has a bit rate of 1411.2 Kbps. mp3 files aretypically encoded in 128 Kbps or less, thereby giving them a compression ratio of almost 11. Image compression ratios are in the range of 10 to 100.4.Figure 6.1: simple, doesn‟t require meta file or streaming server; Figure 6.2: allowsmedia player t o interact directly with the web server, doesn‟t require a streaming server; Figure 6.3: media player interacts directly with a streaming server, which has been designed for the specific streaming application.5.End-to-end delay is the time it takes a packet to travel across the network from sourceto destination. Delay jitter is the fluctuation of end-to-end delay from packet to the next packet.6. A packet that arrives after its scheduled play out time can not be played out.Therefore, from the perspective of the application, the packet has been lost.7.First scheme: send a redundant encoded chunk after every n chunks; the redundantchunk is obtained by exclusive OR-ing the n original chunks. Second scheme: send a lower-resolution low-bit rate scheme along with the original stream. Interleaving does not increase the bandwidth requirements of a stream.8.The role of the DNS is to forward HTTP requests to DNS server managed by theCDN, which in turn redirects the request to an appropriate CDN server. The DNS does not have to be modified to support a CDN. A CDN should provide DNS with the host name and IP address of its authoritative name server (See Section 2.5.3).9.a)Models of traffic demand between network end pointsb)Well-defined performance requirementsc)Models to predict end-end performance for a given workload model, andtechniques to find a minimal high cost bandwidth allocation that will result in all user requirements being met.。

计算机⽹络英⽂版——提供给学⽣部分习题答案Solution of Selected Exercises from the End of Chapter ExercisesChapter 1 - Introduction And Overview1.4 To what aspects of networking does data communications refer?Answer:Data communications refers to the study of low-level mechanisms and technologies used to send information acrossa physical communication medium, such as a wire, radio wave, or light beam.1.5 What is packet-switching, and why is packet switching relevant to the Internet?Answer: Packet switching divides data into small blocks, called packets, and includes an identification of the intended recipient in each packet. Packet switching changed networking in a fundamental way, and provided the basis for the modern Internet. Packet switching allows multiple senders to transmit data over a shared network.1.8 What is a communication protocol? Conceptually, what two aspects of communication does a protocol specify? Answer: A communication protocol refer to a specification for network communication.Major aspects of a protocol are syntax (format) and semantics (meaning) of the protocol.1.9 What is a protocol suite, and what is the advantage of a suite?Answer:protocols are designed in complete, cooperative sets called suites or families, instead of creating each protocol in isolation. Each protocol in a suite handles one aspect of communication; together, the protocols in a suite cover all aspects of communication. The entire suite is designed to allow the protocols to work together efficiently. 1.11 List the layers in the TCP/IP model, and give a brief explanation of each.(See Textbook)1.14 Give a brief explain of the layers in the ISO Open System Interconnection model.(See Textbook)Chapter 3 - Internet Applications And Network Programming3.1 What are the two basic communication paradigms used in the Internet?Answer: There are various approaches, but according to textbook, we can specify them as Stream Paradigm and Message Paradigm.3.2 Give six characteristics of Internet stream communication.(See Textbook)3.3 Give six characteristics of Internet message communication.(See Textbook)3.4 If a sender uses the stream paradigm and always sends 1024 bytes at a time, what size blocks can the Internet deliverto a receiver?Answer: stream paradigm does not provide any guarantees for block sizes, so all depends on individual transfer.3.6 What are the three surprising aspects of the Internet’s message delivery semantics?Answer:The Internet’s message delivery has the followi ng undesirable characteristics:* Messages can be lost* Messages can be duplicated* Messages can be delivered out-of-order3.8 When two applications communicate over the Internet, which one is the server?Answer: T he application that waits for some other applications to contact is called server, and the application that contact other one is called client.3.14 What two identifiers are used to specify a particular server?Answer: A particular server is identified by the following identifiers:* An identifier for the computer on which a server runs (IP Address)* An identifier for a particular service on the computer (Port Number)Chapter 4 - Traditional Internet Applications4.1 What details does an application protocol specify?(See Textbook)4.3 What are the two key aspects of application protocols, and what does each include?(See Textbook)4.6 What are the four parts of a URL, and what punctuation is used to separate the parts?Answer: The URL into four components: a protocol, a computer name, a document name, and parameters. The computer name and protocol port are used to form a connection to the server on which the page resides. And the document name and parameters are used to request a specific page.4.7 What are the four HTTP request types, and when is each used?(See Textbook)4.12 When a user requests an FTP directory listing, how many TCP connections are formed? Explain.Answer: FTP uses two types of connections to perform its functionality, namely* A control connection is reserved for commands. Each time the server needs to download or upload a file, the server opens a new connection.* A data connection is used to transfer files.4.16 List the three types of protocols used with email, and describe each.(See Textbook)4.17 What are the characteristics of SMTP?(See Textbook)4.20 What are the two main email access protocols?Answer: Two major email access protocols are:* Post Office Protocol (POP)* Internet Mail Access Protocol (IMAP)Chapter 6- Information Sources and Signals6.4 State and describe the four fundamental characteristics of a sine wave.(See Textbook)6.9 What is the analog bandwidth of a signal?Answer: Analog bandwidth of signal can be defined as to be the difference between the highest and lowest frequencies of the constituent parts (i.e., the highest and lowest frequencies obtained by Fourier analysis)6.11 Suppose an engineer increases the number of possible signal levels from two to four. How many more bits can be sent in the same amount of time? Explain.Answer: The number of levels that can be represented by n bits is given by 2n . So if number of levels changes from 2→4, it means number of bits goes from 1→2612. What is the definition of baud?Answer: Baud is defined as the number of times that a signal can change per second.6.14 What is the bandwidth of a digital signal? Explain.Answer: According to the definition of analog bandwidth, a digital signal has infinite bandwidth because Fourier analysis of a digital signal produces an infinite set of sine waves with frequencies that grow to infinity.6.18 What is the chief advantage of a Differential Manchester Encoding?Answer: The most important property of differential encoding is that the encoding works correctly even if the two wires carrying the signal are accidentally reversed.6.20 If the maximum frequency audible to a human ear is 20,000 Hz, at what rate must the analog signal from a microphone be sampled when converting it to digital?Answer: The sampling rate = 2 × f max, so the signal should be sampled at 2x20,000 = 40,000 HzChapter 7 - Transmission Media7.2 What are the three energy types used when classifying physical media according to energy used?Answer: Three types of energy used when classifying physical media are electrical, electromechanical (radio), and light7.4 What three types of wiring are used to reduce interference form noise?(See Textbook)7.10 List the three forms of optical fiber, and give the general properties of each.(See Textbook)7.21 What is the relationship between bandwidth, signal levels, and data rate?Answer: If a transmission system uses K possible signal levels and has an anal og bandwidth B, Nyquist’s Theorem states that the maximum data rate in bits per second, D, is: D = 2 B log2K7.22 If two signal levels are used, what is the data rate that can be sent over a coaxial cable that has an analog bandwidthof 6.2 MHz?Answer: Using the D= 2 B log2 K relationship, D = 2*6.2*log22 = 2*6.2*1 = 12.4 Mbps7.24 If a system has an input power level of 9000, and an output power level of 3000, what is the difference when expressed in dB?Answer: Decibel is expressed as 10log10(P out/P in) → 10log10(3,000/9,000) = to be determined by reader7.23 If a system has an average power level of 100, an average noise level of 33.33, and a bandwidth of 100 MHz, whatis the effective limit on channel capacity?Answer: Shannon theorem specify the maximum data rate that could be achieved over a transmission system that experiences noise: C = Blog2 (1 + S/N) = 100,000,000 * log2 (1 + 100/33.33) = 100,000,000 * log24 = 200,000,000 = 200 Mbps7.25 If a telephone system can be created with a signal-to-noise ratio of 40 dB and an analog bandwidth of 3000 Hz, how many bits per second could be transmitted?Answer: First we should convert 40 dB to a real number, namely if 40 = 10 log10S/N→S/N = 10,000 , Using the Shannon’s capacity expression C = B log2(1 + S/N) → C = 3,000 log2 (1+ 10,000) = to be determined by readerCh 8 - Reliability And Channel Coding8.1 List and explain the three main sources of transmission errors.(See Textbook)8.3 In a burst error, how is burst length measured?Answer: For a burst error, the burst size, or length, is defined as the number of bits from the start of the corruption to the end of the corruption.8.4 What is a codeword?Answer: We can define the set of all possible messages to be a set of datawords, and define the set of all possible encoded versions to be a set of codewords. So each possible code sequence is considered to be a codeword.8.8 Compute the Hamming distance for the following pairs: (0000, 0001), (0101, 0001), (1111, 1001), and ( 0001, 1110). (See Textbook)8.11 Generate a RAC parity matrix for a (20, 12) coding of the dataword 100011011111.(See Textbook)8.15 Express the two values in the previous exercise as polynomials.Answer:X10+ X7 + X5 + X3 + XX4+ X2+ 1Ch 9 - Transmission Modes9.1 Describe the difference between serial and parallel transmission.Answer: Transmission modes can be divided into two fundamental categories:* Serial: one bit is sent at a time* Parallel: multiple bits are sent at the same time9.2 What are the advantages of parallel transmission? What is the chief disadvantage?Answer: A parallel mode of transmission has two chief advantages:* High speed: Because it can send N bits at the same time, a parallel interface can operate N times faster than an equivalent serial interface.* Match to underlying hardware: Internally, computer and communication hardware uses parallel circuitry.Thus, a parallel interface matches the internal hardware well.The main disadvantage of parallel transmission is number of cables required, for long distance communication, this is an important consideration.9.4 What is the chief characteristic of asynchronous transmission?Answer:Asynchronous transmission can occur at any time, with an arbitrary delay between the transmission of two data items, it allows the physical medium to be idle for an arbitrary time between two transmissions.Chapter 11 - Multiplexing And Demultiplexing11.2 What are the four basic types of multiplexing?(See Textbook)11.4 What is a guard band?Answer: For proper communication without interference, we should choose a set of carrier frequencies with a gap between them known as a guard band. The guard band reduces or eliminates the possible interference between neighboring carrier signals.11.8 Explain how a range of frequencies can be used to increase data rate.Answer:To increase the overall data rate, a sender divides the frequency range of the channel into K carriers, and sends 1 /K of the data over each carrier.11.12 Suppose N users compete using a statistical TDM system, and suppose the underlying physical transport can sendK bits per second. What is the minimum and maximum data rate that an individual user can experience?Answer: If we neglect the overhead generated by statistical TDM, a system will have two possibilities: * Minimum: If all channels have equal data then the rate will be K/N bps* Maximum: If only one channel active and the others are passive, then rate will be K bpsChapter 13 - Local Area Networks: Packets, Frames, And Topologies13.1 What is circuit switching, and what are its chief characteristics?Answer: The term circuit switching refers to a communication mechanism that establishes a path between a sender and receiver with guaranteed isolation from paths used by other pairs of senders and receivers. The circuit switching has the following main characteristics:* Point-to-point communication* Separate steps for circuit creation, use, and termination* Performance equivalent to an isolated physical path13.3 In a packet switching system, how does a sender transfer a large file?Answer: The packet switching system requires a sender to divide each message into blocks of data that are known as packets . The size of a packet varies; each packet switching technology defines a maximum packet size. So, a large file will be divided into smaller pieces and sent.13.5 What are the characteristics of LANs, MANs, and W ANs?Answer: There are lots of details that can be said and discussed for categorization of network types based on geography, few points are highlighted below:* Local Area Network (LAN): Least expensive; spans a single room or a single building* Metropolitan Area Network (MAN) Medium expense; spans a major city or a metroplex* Wide Area Network (WAN) Most expensive; spans sites in multiple cities13.6 Name the two sublayers of Layer 2 protocols defined by IEEE, and give the purpose of each.Answer: The Layer 2 protocols defined by IEEE defines two sub-layers as mentioned below:* Logical Link Control (LLC) Addressing and demultiplexing* Media Access Control (MAC) Access to shared media13.8 What are the four basic LAN topologies?Answer: The four basic LAN topologies are star, ring, mesh and bus.13.10 In a mesh network, how many connections are required among 20 computers?Answer: The expression to calculate number of connections in a mesh network is given by n (n-1)/2. So for 20 computers then number of connections required will be = 20 (20 – 1)/2 =19013.15 Give a definition of the term frame .Answer: In a packet-switched network, each frame corresponds to a packet processed at data link layer.Chapter 14 - The IEEE MAC Sub-Layer14.1 Explain the three basic approaches used to arbitrate access to a shared medium.(See Textbook)14.3 List the three main types of channelization and the characteristics of each.(See Textbook)14.6 What is a token, and how are tokens used to control network access?Answer: A special control message is called a token. In a token passing system, when no station has any packets to send, the token circulates among all stations continuously. When a station captures the token, it sends its data, and when transmission completed, it releases the token.14.8 Expand the acronym CSMA/CD, and explain each part.Answer: The acronym CSMA/CD stands for Carrier Sense Multi-Access with Collision Detection, which means the following: * Carrier Sense: Instead of allowing a station to transmit whenever a packet becomes ready, Ethernet requires each station to monitor the cable to detect whether another transmission is already in progress.* Multiple Access: The system allows multiple users/hosts to make use of a common/shared media* Collision Detection. A collision can occur if two stations wait for a transmission to stop, find the cable idle, and both start transmitting.14.10 Why does CSMA/CD use a random delay? (Hint: think of many identical computers on a network.)Answer: Randomization is used to avoid having multiple stations transmit simultaneously as soon as the cable is idle.That is, the standard specifies a maximum delay, d, and requires each station to choose a random delay less than d after a collision occurs. In most cases, when two stations each choose a random value, the station that chooses the smallest delay willChapter 15 - Wired LAN Technology (Ethernet And 802.3)15.1 How large is the maximum Ethernet frame, including the CRC?Answer: According to Fig. 15.1 a conventional Ethernet frame has the following fields:* Header: 14 bytes (fixed)* Payload: 46-1500 bytes (there is a minimum frame size because of collision detection)* CRC: 4 bytes (fixed)Accordingly an Ethernet frame will be maximum 1518 bytes and minimum 64 bytes15.3 In an 802.3 Ethernet frame, what is the maximum payload size?Answer: The 802.3 Ethernet makes use of 8-bytes of the original/conventional Ethernet for Logical Link Control / Sub-Network Attachment Point (LLC / SNAP) header instead of extending/increasing the header. This is for sake of backward compatibility. So the maximum pay load is reduced from 1500 bytes to 1492 bytes.15.6 How did a computer attach to a Thicknet Ethernet?Answer: Hardware used with Thicknet was divided into two major parts:* Transceiver: A network interface card (NIC) handled the digital aspects of communication, and a separate electronic device called a transceiver connected to the Ethernet cable and handled carrier detection, conversion of bits into appropriate voltages for transmission, and conversion of incoming signals to bits.* AUI: A physical cable known as an Attachment Unit Interface (AUI) connected a transceiver to a NIC in a computer. A transceiver was usually remote from a computer.15.7 How were computers attached to a Thinnet Ethernet?Answer: Thinnet Ethernet (formally named 10Base2) uses a thinner coaxial cable that was more flexible than Thicknet. The wiring scheme differed dramatically from Thicknet. Instead of using AUI connections between a computer and a transceiver, Thinnet integrates a transceiver directly on the NIC, and runs a coaxial cable from one computer to another.15.8 What is an Ethernet hub, and what wiring is used with a hub?Answer: An electronic device that serves as the central interconnection is known as a hub. Hubs were available in a variety of sizes, with the cost proportional to size. The hubs are becoming old-fashioned, and being replaced with switches.15.3 What category of twisted pair wiring is needed for a 10 Mbps network? 100 Mbps? 1000 Mbps?Answer: The three major categories of Ethernet and their wiring is listed below:* 10 Mbps: 10BaseT (Ethernet) Category 5* 100 Mbps: 100BaseT (Ethernet Fast) Category 5E* 1 Gbps: 1000BaseT (Gigabit Ethernet) Category 6Chapter 20 - Internetworking: Concepts, Architecture, and Protocols20.2 Will the Internet be replaced by a single networking technology? Why or why not?Answer: Incompatibilities make it impossible to form a large network merely by interconnecting the wires among networks. The beauty of the Internet is interconnection of wide range of technologies from various manufacturers.Diversity of the products and solutions is a richness instead of limitation as long as they all adopt the same set of protocols.20.3 What are the two reasons an organization does not use a single router to connect all its networks?Answer:An organization seldom uses a single router to connect all of its networks. There are two major reasons: * Because the router must forward each packet, the processor in a given router is insufficient to handle the traffic passing among an arbitrary number of networks.* Redundancy improves internet reliability. To avoid a single point of failure, protocol software continuously monitors internet connections and instructs routers to send traffic along alternative paths when a network or router fails.20.6 In the 5-layer reference model used with the TCP/IP Internet protocols, what is the purpose of each of the five layers?(See 1.11)Chapter 21- IP: Internet Addressing21.3 In the original classful address scheme, was it possible to determine the class of an address from the address itself? Explain.Answer:Yes, since in the classful addressing scheme initial bit(s) gives indication about the class being used.21.7 If an ISP assigned you a /28 address block, how many computers could you assign an address?Answer: When an organization is assigned /28 CIDR address, it means 28 bits out of 32 bits are fixed, so 32-28 = 4 bits available for user space. So number of users 24-2 = 4, since the all 0s and all 1s address are having special use and can’t be assigned to a user.21.8 If an ISP offers a / 17 address block for N dollars per month and a / 16 address block for 1.5 N dollars per month,which has the cheapest cost per computer?Answer: Number of addresses in /17 block 232-17 = 215Price per address: N /215 = N / 215Number of addresses in /16 block 232-16 = 216Price per address: 1.5N /216 = 0.75N/215 So /16 address block will be cheaper in comparison with the price given for /17 block.21.10 Suppose you are an ISP with a / 24 address block. Explain whether you accommodate a request from a customer who needs addresses for 255 computers. (Hint: consider the special addresses.)Answer: For a/24 address block, number of available addresses will be 232-24 = 28 = 256. However, a suffix with all 0s address is reserved for network ID and a suffix with all 1s address is reserved for broadcast address, so number of addresses that can be assigned to computers/hosts will be 256 -2 = 254.21.11 Suppose you are an ISP that owns a / 22 address block. Show the CIDR allocation you would use to allocateaddress blocks to four customers who need addresses for 60 computers each.Answer: The /22 address block can be assigned as follows:ddd.ddd.ddd.00/26ddd.ddd.ddd.01/26ddd.ddd.ddd.10/26ddd.ddd.ddd.11/26Chapter 22- Datagram Forwarding22.1 What are the two basic communication paradigms that designers consider when designing an internet?Answer:* Connection-oriented service * Connectionless service22.2 How does the Internet design accommodate heterogeneous networks that each have their own packet format?Answer: To overcome heterogeneity, the Internet Protocol defines a packet format that is independent of the underlying hardware. The result is a universal, virtual packet that can be transferred across the underlying hardware intact. The Internet packet format is not tied directly to any hardware. The underlying hardware does not understand or recognize an Internet packet.22.5 What is the maximum length of an IP datagram?In the current version of the Internet Protocol (IP version 4), a datagram can contain at most 64 K (65535) octets, including the header.22.7 If a datagram contains one 8-bit data value and no header options, what values will be found in header fields H.LEN and TOTAL LENGTH?Answer: H. LEN indicated header in 32-quantities, since no options, then this value will be 5. The TOTAL LENGTH indicated the number of bytes in a datagram including the header. This means 5x4 bytes + 1 (8-bits) = 21 bytesChapter 23 - Support Protocols And Technologies23.1 When a router uses a forwarding table to look up a next-hop address, the result is an IP address. What must happenbefore the datagram can be sent?Answer: Each router along the path uses the destination IP address in the datagram to select a next-hop address, encapsulates the datagram in a hardware frame, and transmits the frame across one network. A crucial step of the forwarding process requires a translation: forwarding uses IP addresses, and a frame transmitted across a physical network must contain the MAC address of the next hop.23.2 What term is used to describe the mapping between a protocol address and a hardware address?Answer: Translation from a computer’s IP address to an equivalent hardware address is known as address resolution, and an IP address is said to be resolved to the correct MAC address. The TCP/IP protocol being used for this is called Address Resolution Protocol (ARP). Address resolution is local to a network.23.5 How many octets does an ARP message occupy when used with IP and Ethernet addresses?Answer: According to Fig 23.3 an ARP message has 7-lines of each being 32-bit (4 bytes or octets), therefore,number of octets in an ARP can be determined as 7x4 = 28 octets23.10 What types of addresses are used in layers below ARP?Answer:ARP forms a conceptual boundary in the protocol stack; layers above ARP use IP addresses, and layers below ARP use MAC addresses.23.17 What is the chief difference between BOOTP and DHCP?Answer:The main difference is that the BOOTP protocol required manual administration. So before a computer could use BOOTP to obtain an address, a network administrator had to configure a BOOTP server to know the computer’s I P address. Chapter 24 - The Future IP (IPv6)24.3 List the major features of IPv6, and give a short description of each.(See Textbook)24.4 How large is the smallest IPv6 datagram header?Answer: IPv6 datagram header consists of a base header + zero or more extension header. Since, smallest header is being asked, we assume zero extension header and consider IPv6 will have only base header. If we look at IPv6 header format in Fig. 24.3, it shows that 10x4 bytes = 40 bytes.Chapter 26 - TCP: Reliable Transport Service26.2 List the features of TCP.(See Textbook)26.6 When using a sliding window of size N, how many packets can be sent without requiring a single ACK to be received?Answer: If the size of the window is N, then it means a sender can transmit up to N packets without waiting for an ACK, as long as other controls are in place.26.9 What is the chief cause of packet delay and loss in the Internet?Answer: The main cause of packet delay and loss in the Internet is congestion.Chapter 28 - Network Performance (QoS and DiffServ)28.1 List and describe the three primary measures of network performance.(See Textbook)28.2 Give five types of delay along with an explanation of each.(See Textbook)Chapter 30 - Network Security30.1 List the major security problems on the Internet, and give a short description of each.(See Textbook)30.2 Name the technique used in security attacks.(See Textbook)30.8 List and describe the eight basic security techniques.(See Textbook)。

HUST Examination Answer SheetCourse: Computer Networks (closed-book exam) Jan.2014 Department of Electronics and Information EngineeringSolution[Question 1] Network architecture(1) Draw the architecture diagrams of ISO/OSI reference model and TCP/IP network. (2) State the differences between the two architectures.(3) Mark the names of data unit and network equipment in the bottom 4 layers of ISO/OSI diagram.Solution:（1）ISO/OSI reference model TCP/IP modelDifferences:a) TCP/IP model is the actual architecture in the Internet, while OSI model is a theoreticalarchitecture.b) TCP/IP model does not imply strict layering, while OSI model implies strict layering. c) In TCP/IP model, IP serves as the focal point for the architecture.d) TCP/IP model allows for arbitrarily many different network technologies , ranging fromEthernet to wireless to single point-to-point links.e) TCP/IP model emphasize implementations of proposed protocols. (2)Physical : bit, repeater or hubData link : frame, Ethernet switch or bridge Internet : packet, routerTransport : message, gateway[Question 2] Principles of network designSelect ONE of the following principles, tell its main ideas and provide an example. (1) Keep it simple and stupid(2) Complex edge and simple core (3) Smart sender and dumb receiverSolution:(1) KISS: Keep It Simple and StupidIt means that you should make things simple in the designing. One example following it : Ethernet(2) Complex edge and simple coreIt means that the hosts are very complex and have many functions while the nodes are very simple and have few functions.One example following it : The design of router, or the functions of TCP and IP(3) Smart sender and dumb receiverIt means that the function of sender is more complex than that of the receiver, which is help to improve the robustness and performance of communication protocol. One example following it: The flow control of TCP protocol[Question 3] Error detection(1) Tell the main idea of error detection and error correction in communication.(2) Given the CRC polynomial x 4 + x 3 + 1, if the original message is 10110011010, what is the CRC message to send?(3) Suppose the first bit of the message in (2) is inverted due to the noise in transmission. How can the receiver detect it via CRC verification?Solution:(1)1087431()M x x x x x x x =+++++, 43()1C x x x =++. So k=4.a) Multiply M(x) by 2kto get 141211875()T x x x x x x x =+++++,b) Then divide T(x) by C(x) to get the remainder 0000.c) The message that should be transmitted is 101100110100000.(2)The message received is 001100110100000.Divide it by C(x), then the remainder is 1100.So it is not divisible by C(x).So the receiver knows that an error has occurred.(1) What are the essential components to realize reliable transmission?(2) Suppose two computers are communicated via Stop-and-wait protocol. The link bandwidth is 5kbps, and the one-way propagation delay is 20ms. To reach 80% or higher link utilization, what is the minimal frame size for this communication?(3) If it is upgraded to Sliding-Window protocol. To reach the same goal with (2), what is the minimal window size, how many bits are required to describe the frame sequence in window? (Suppose the frame size is 1 or 100Byte)Solution:(1)ACK, and timerActual_throughput = Data / Total_DelayTotal_Delay = RTT + Data / BWLink_Utilization = 100 * Actual_throughput / BWSo: Data/throughput = RTT + Data/BW(BW/throughput - 1) Data = BW*RTTData = BW*RTT/(1/Utilization -1)For the stop-and-wait protocol, for each RTT only one frame is sent,Thus Frame_size = Data = BW*RTT/(1/Utilization - 1) = 5kbps * 40ms / (1/0.8 - 1)= 200 bit / 0.25 = 800 bit = 100 ByteIf the students ignore the data transmission delay, their answer isframe_size = BW*RTT*Utilization = 5000 bit/sec * 40 / 1000 sec * 0.8 = 160 bit = 20 ByteIn this case, at least -2 score.(3)For the sliding-window protocol, in each RTT the data window can be transmitted at most.Thus Window_size = Data = BW*RTT/(1/Utilization - 1) = 5kbps * 40ms / (1/0.8 - 1)= 200 bit / 0.25 = 800 bit = 100 ByteIf the frame size is 100Byte, 1 frames are allowed.To indicate the frames in both sides of sender and receiver, the sequence number should describe 2 frames, thus the [log2(2) ]=1 bitIf the students ignore the data transmission delay, their answer iswindow_size = BW*RTT*Utilization = 5000 bit/sec * 40 / 1000 sec * 0.8 = 160 bit = 20 ByteIf the frame size is 100Byte, 1 frames are allowed.To indicate the frames in both sides of sender and receiver, the sequence number should describe 2 frames, thus the [log2(2) ]=1 bitIn this case, at least -2 score.(1) What is the main idea of CSMA/CD (Carrier Sense Multiple Access/Collision Detection) in traditional Ethernet? Can it be deployed for wireless local network, why?(2) Suppose one traditional Ethernet has 1km cable, the signal propagation speed is 2*105km/s, and the transmission rate is designed to be 100Mbps. What is the minimal frame size to support carrier sensing?Solution:(1) the main idea of CSMA/CD include two parts. One is carrier sensing, which means the node should detect the channel before sending any data. If the node finds the channel is idle, it begins to transmit. Otherwise, the node stop for next round. The second issues is collision detection, which means the node should detect the channel in the duration of its data transmission. If the node finds the channel become busy, or in other word, there is a collision, it should stop transmitting immediately.CSMA/CD cannot be deployed in wireless LAN, because the wireless radio transmitter and receiver can not work in dual mode. The wireless node cannot detect collision when it is transmitting data.(2)the minimal frame should be transmitted throughout the whole traditional Ethernet.Thus t = frame_size / transmit_rate = 2 * cable_length / prop_speed.Frame_size = 100 * 106 bits/sec * 2 * 103 m / (2 * 108)m/sec = 1000 bits = 125 Byte[Question 6] Switched network(1) What are the differences between circuit switching and packet switching?(2) For the following linear topology network, each link has 2ms propagation delay and 4 Mbps bandwidth.A B C DIf we use circuit switching, circuit setup requires a 1KB message to make one round-trip on the path, which incurs a 1ms delay at each switch after the message has been completely received. Then we can send the file as one contiguous bit stream. What is the delay for circuit switching to transmit n-byte from A to D?(3) If we use packet switching in the network of (2), we can break the file into 1KB packets, which has 24byte header and 1000byte payload. The switch takes 1ms process delay after receiving the packet, and then sent it continuously. What is the condition for packet switching to have less delay performance than circuit switching?Solution:(1)(2)T pkt = Packet_Size / Bandwidth = 1 KB / 4Mbps = 1024 *8 / (4 * 106) = 2.048 ms T p = 2 ms, T s = 1 ms,T t = n B / 4MbpsIn circuit switching, Total time duration:D = Singling_Delay + Transmission_Delay = 2 * Packet_Duration + Transmission_Delay = 2 * (T pkt * 2 + T p * 3 + T s * 2 + T pkt ) + (T p * 3 + T t ) = T pkt * 6 + T p * 9 + T s * 4 + T tThus, D = 2.048 * 6 + 2 * 9 + 1 * 4 + n * 8 bits / 4Mbps = 34.288 (ms) + 2n (us)(3)Main ideaThe reliability provided by endhost The reliability provided by thenetwork Information in packet Every packet has its dest-addrin header Every packet has its temp VCIlocally Forwarding action inswitch Every packet was treatedindependentlyThe packets are processed inthe manner of VCPackets received indestinationNot in sequenceIn sequenceT pkt = Packet_Size / Bandwidth = 1 KB / 4Mbps = 1024 *8 / (4 * 106) = 2.048 msT p = 2 ms,T s = 1 ms,T t = 1.024 * n B / 4MbpsIn packet switching, Total time duration:D = Delay_at_Switch * Switch_Num + Delay_at_last_hop= (T p + T pkt +T s)*2 + (T p + T t )= T pkt * 2 + T p * 3 + T s * 2 + T tThus, D = 2.048 * 2 + 2 * 3 + 1 * 2 + 1.024 * n B / 4Mbps= 12.096 (ms) + 2.048 n (us)In order to make34.288 (ms) + 2n (us) > 12.096 (ms) + 2.048 n (us)Thus 22.192(ms) > 0.048 n(us)n < 22.192 * 1000 / 0.048 = 462333 Byte = 451.5 KB[Question 7] Ethernet Switch(1) What are the differences between hub and switch in Ethernet?(2) Suppose one server and nine clients are connected via hub in 10Mbps Ethernet, what is the maximal bandwidth for the client-server connection?(3) If the hub is upgraded to switch, can the client-server connection obtain more bandwidth? if can, how much is it?Solution:(1)hub works in Layer2, it works as the shared media and relays the frames to all the nodes connecting to hub. Switch works in Layer2, it forwards the frames to the specific node according to the destination address embedded in the frame header.(2)the max bandwidth in client-server connection is 10Mbps/(1+9)= 1Mbps(3)the max bandwidth in client-server connection is 10Mbps/9 = 1.1111MbpsAdditional 0.1111 Mbps is obtained for each connection.[Question 8] Router(1) Somebody says that, ``the only difference between switch and router is that they do switch function based on the address in different layers.’’ Is it correct? Why?(2) If we obtain the following information from one router. What kind of routing protocol does itSolution:(1)it is not fully correct. The part talking about the forwarding function is correct, while it is not the only difference. Another but not the last difference is that, router has more functions on control plane, which do routing and find the paths for packets.(2)The routing protocol is RIP.The routing table is:Destination Next hop Interface192.168.1.0/24 * Fa 0/4192.168.2.0/24 * Fa 0/6192.168.10.0/24 192.168.1.1 Fa 0/4192.168.30.0/24 192.168.2.2 Fa 0/6[Question 9] Routing algorithm(1) State the main differences between distance vector routing and link state routing.(2) For the network given in the following figure, provide the steps of forward search in Dijkstra algorithm for node A finding the shortest path to node ESolution:(1) The idea of distance vector routing is to tell the neighbors about the learned topology to all the nodes in the network. The idea of link state routing is to tell all the nodes in network about the neighborhood topology.Step 1 2 3 4 5 6 7 8 9Confirmed (A,0,-) (A,0,-) (A,0,-)(D,2,D)(A,0,-)(D,2,D)(A,0,-)(D,2,D)(B,4,D)(A,0,-)(D,2,D)(B,4,D)(A,0,-)(D,2,D)(B,4,D)(E,6,D)(A,0,-)(D,2,D)(B,4,D)(E,6,D)(A,0,-)(D,2,D)(B,4,D)(E,6,D)(C,7,D)Tentative (B,5,B)(D,2,D)(B,5,B) (B,4,D)(E,7,D)(E,7,D)(E,6,D)(C,8,D)(C,8,D) (C,7,D)A→E A→D→EA→D→B→EA→D→B→E[Question 10] IP address allocationA company has one C class IP address of 200.1.1.*. It has four departments.(1) If each department has less than 25 computers. Provide a kind of IP address allocation. Give the network address, subnet mask, and the available IP address range for each department. (2) If the four departments have 72, 35, 34, 20 computers respectively. Provide the IP address allocation scheme again.Solution:(1)Since each department has less than 25 computers, even considering the additional two more IP address for gateway and broadcast, the 64-computer subnet is enough for them.One IP address allocation scheme is to even divide the 256 IP addresses into 4 subnets, each subnet allows 64 hosts.Another IP address allocation scheme is to even divide the 256 IP addresses into 4 subnets, each subnet allows 32 hosts.(2) Since one of the department has more requirements than 64, then the even distribution scheme[Question 11] TCP protocol(1) Somebody says that, ``because of the reliable transmission service in layer 2, there is no need to provide such service again within TCP protocol in layer 4’’. Is it correct? Why?(2) State the main rules of TCP connection setup according to the following figure. Explain every word and number in the figure.Solution:(1) It is wrong. TCP is based on the un-reliable IP layer, which only provides best effort service. If TCP wants to provide reliable transmission service, it has to realize this by itself.(2)TCP use three hand-shakes to setup the connection.According to the figure, there are two nodes. The sender is with IP address of 192.168.1.163 and the receiver 192.168.1.165 respectively.●At first, the sender send a request ``SYN’’to the receiver to setup the connection. Thismessage is with the sequence number of 424CF1DC;●Secondly, the receiver reply an acknowledgment message ``SYN/ACK’’ to the sender. Thismessage has two sequence numbers. The seq in the ACK is 424CF1DD, which is to confirm the last ``SYN’’ from sender. The seq in the SYN is 30318555, which is a new message from the receiver.●Thirdly, the sender reply an acknowledgment message ``ACK’’ to the receiver. The seq in theACK is 30318556, which is to confirm the last ``SYN’’ from receiver.In the end, both the sender and receiver knows that the other side is ready for this TCP connection.[Question 12] TCP congestion control(1) Both flow control and congestion control in TCP are realized by window based packet control. How can TCP get the window sizes in these two mechanisms?(2) Assume that TCP implements an extension that allows window sizes much larger than 64 KB. Suppose that you are using this extended TCP over a 1Gbps link with a latency of 150ms. TCP packet size is 1KB, and the max receive window is 1 MB. Suppose there is no real congestion and packet loss in transmission. How many RTTs does it take until slow start opens the send window to 1 MB? How long does it take to send the complete file? ( Suppose file size is 10MB )Solution:(1) In flow control, TCP sender knows the window size by the field of advertise-window replied from the receiver. In congestion control, TCP sender learns the window size adaptively by AIMD( Additive Increase and Multiplicative Decrease) mechanism responding to the packet lossevent.(2)When TCP realizes congestion control mechanism, its effective send window size will be min (CongestionWindow, AdvertizedWindow). In original design of TCP header, the field of AdvertizedWindow is 16 bit, which is 216=26*210=64 KB. So the maximum effective window of original TCP sender is 64KB. The assumption of the first sentence in this question relaxes such constraint for TCP.In slow start, the send window starts from w0=1 packet, which is 1 KB. For each RTT after a successful transmission, the window size will be doubled. After i RTT, it will be 2i * w0. Let 2i * 1KB = 1MB, soi = log2(1MB/1KB) = log2(210) = 10.It will take 10 RTTs to reach 1MB send window.Case 1: if the receiver window remains as 1MBIn the first 10 RTTs, total (1 + 2 + 4 + … + 210) * 1KB has been transmitted.Which is (211 - 1) * 1KB = 2 MB - 1 KB, the rest file is 10MB - (2MB - 1KB) = 8MB + 1KBIn the reset transmission, each RTT can only support 1MB transmission.Thus, additional 9 RTTs are required. Total 19 RTT = 19 * 150ms = 2.85 sCase 2: if the receiver window can be changed.Since there is no congestion and loss, the maximum send window will be the bandwidth * delay for this TCP connection.w max= 1G bps * 150ms = 109 * 150 * 10-3= 150 *106 bit = 18.75 * 106 byte = 17.88 MB.So, this 10 MB file can be transferred before reaching w max. In another word, it can be sent in its slow start phase. Assume x RTT is required to send this file, then:(1 + 2 + 4 + … + 2x) * 1KB ≥ 10 MB2 * 2x–1 ≥ 10 * 1024x ≥ log2(10241) – 1 = 12.3Thus x = 13, it will take 13 RTTs to transfer this file. 13RTT = = sTotal delay = 13RTT + Filesize/BW = 13 * 150ms+ 10MB / 1Gbps =1.95 + 0.08 * 1.024^2=2.03 s11。

Chapter 1 Computer Networks and the Internet1．The ( ) is a worldwide computernetwork, that is, a network that interconnectsmillionsofcomputing devices throughout theworld. ppt3 A public InternetB IntranetC switch netD television net2．Which kind of media is not a guidedmedia ( )A twisted-pair copper wireB a coaxial cableC fiber opticsD digital satellite channel3．Which kind of media is a guided media ( )A geostationary satelliteB low-altitude satelliteC fiber opticsD wireless LAN4．The units of data exchanged by a link-layer protocol are called ( ).A FramesB SegmentsC DatagramsD bit streams5．Which of the following optionbelongs to the circuit-switchednetworks ( )A FDMB TDMC VCnetworksD both A and B 6．( )makes sure that neither sideof a connection overwhelms theother side by sending too manypackets too fast.A Reliable data transferB Flow controlC Congestion controlD Handshaking procedure7．( ) means that the switch mustreceive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link.A Store-and-forward transmissionB FDMC End-to-end connectionD TDM8．Datagramnetworksandvirtual-circuit networks differ in that ( ).A datagram networks are circuit-switched networks, and virtual-circuit networks are packet-switched networks.B datagram networks are packet-switched networks, and virtual-circuit networks are circuit-switched networks.C datagram networks use destination addresses and virtual-circuit networks use VC. numbers to forward packets toward their destination.D datagram networks use VC. numbers and virtual-circuit networks use destination addresses to forward packetstoward their destination.9．In the following options, which one is not a guided media ( )A twisted-pairwireB fiber opticsC coaxial cableD satellite10．Processing delay does not include the time to ( ).A examine the packet ’s headerB wait to transmit the packet onto the linkC determine where to direct thepacketD check bit-error in the packet 11．In the following four descriptions, which one is correct ( )A The traffic intensity must begreater than 1.B The fraction of lost packetsincreases as the trafficintensity decreases.C If the traffic intensity isclose to zero, the averagequeuing delay will be close tozero.D If the traffic intensity isclose to one, the averagequeuing delay will be close toone.12．The Internet’s network layer is responsible for movingnetwork-layer packets known as( ) from one host to another.A frameB datagramC segmentD message13．The protocols of various layersare called ( ).A the protocol stackB TCP/IPC ISPD network protocol14．There are two classes ofpacket-switched networks: ( )networks and virtual-circuitnetworks.A datagramB circuit-switchedC televisionD telephone15．Access networks can be looselyclassified into threecategories: residential access,company access and ( ) access.A cabledB wirelessC campusD city areaQuestion 16～17Suppose, a is the average rate at which packets arrive at the queue,R is the transmission rate, and all packets consist of L bits, then thetraffic intensity is ( 16 ), and it should no greater than ( 17 ).16． A LR ／aB La ／RC Ra ／LD LR ／a 17．A 2B 1C 0D -118．In the Internet, the equivalentconcept to end systems is ( ).A hostsB serversC clientsD routers19．In the Internet, end systems are connected together by ( ).A copper wireB coaxial cableC communication linksD fiber optics 20．End systems access to the Internet through its ( ). A modemsB protocolsC ISPD sockets21．End systems, packet switches,and other pieces of the Internet, run ( ) that control thesending and receiving ofinformation within theInternet.A programsB processesC applicationsD protocols22．There are many private networks,such as many corporate andgovernment networks, whosehosts cannot exchange messages with hosts outside of the private network. These private networks are often referred to as ( ).A internetsB LANC intranetsD WAN23．The internet allows ( ) runningon its end systems to exchangedata with each other.A clients applicationsB server applicationsC P2P applicationsD distributed applications24．The Internet provides two services to its distributed applications: a connectionlessunreliable service and ()service.A flowcontrolB connection-oriented reliableC congestion controlD TCP25．It defines the format and theorder of messages exchangedbetween two or morecommunicating entities, as wellas the actions taken on thetransmission and/or receipt ofa message or other event. Thesentence describes ( ).A InternetB protocolC intranetD network26．In the following options, which does not define in protocol ( )A the format of messagesexchanged between two or morecommunicating entitiesB the order of messagesexchanged between two or morecommunicating entitiesC the actions taken on thetransmission of a message orother eventD the transmission signals aredigital signals or analogsignals27．In the following options, which is defined in protocol ( )A the actions taken on thetransmission and/or receiptof a message or other event B the objects exchanged between communicating entities C the content in the exchangedmessagesD the location of the hosts28．In the following options, whichdoes not belong to the network edge( )A end systemsB routersC clientsD servers29．In the following options, whichbelongs to the network core ( )A end systemsB routersC clientsD servers30．In the following options, whichis not the bundled with theInternet’sconnection-oriented service( )A reliable data transferB guarantee of thetransmission timeC flow controlD congestion-control31．An application can rely on theconnection to deliver all of itsdata without error and in theproper order. The sentencedescribes ( ). A flow controlB congestion-controlC reliable datatransferD connection-oriented service 32．It makes sure that neither sideof a connection overwhelms theother side by sending too manypackets too fast. The sentence describes ( ). A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer33．It helps prevent the Internetfrom entering a state of gridlock. When a packet switchbecomes congested, its bufferscan overflow and packet loss canoccur. The sentence describes( ).A flowcontrolB congestion-controlC connection-oriented serviceD reliable data transfer 34．TheInternet ’sconnection-oriented service has aname, it is ( ).A TCPB UDPC TCP/IPD IP35．In the following options, which service does not be provided to an application by TCP( )A reliable transportB flow controlC video conferencingD congestion control36．The Internet ’s connectionless service is called ( ).A TCPB UDPC TCP/IPD IP37．In the following options, which does not use TCP( )A SMTPB internet telephoneC FTPD HTTP38．In the following options, which does not use UDP( )A InternetphoneB video conferencingC streamingmultimediaD telnet39．There are two fundamentalapproaches to building a network core, ( ) and packet switching.A electrical current switchingB circuit switchingC data switchingD message switching40．In ( ) networks, the resourcesneeded along a path to provide for communication between the end system are reserved for theduration of the communicationsession.A packet-switchedB data-switchedC circuit-switchedD message-switched41．In ( ) networks, the resources are not reserved; a session’smessages use the resources ondemand, and as a consequence,may have to wait for access tocommunication link.A packet-switchedB data-switchedC circuit-switchedD message-switched42．In a circuit-switched network, if each link has n circuits, foreach link used by the end-to-endconnection, the connection gets( ) of the link’s bandwidthfor the duration of the connection.A a fraction 1/nB allC 1/2D n times43．For ( ), the transmission rateof a circuit is equal to theframe rate multiplied by thenumber of bits in a slot.A CDMAB packet-switched networkC TDMD FDM44．( ) means that the switch mustreceive the entire packetbefore it can begin to transmitthe first bit of the packet ontothe outbound link.A Queuing delayB Store-and-forwardtransmissionC Packet lossD Propagation45．The network that forwardspackets according to host destination addresses is called( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram46．The network that forwardspackets according tovirtual-circuit numbers iscalled ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram47．In the following entries, whichis not a kind of access network( )A residentialaccessB company accessC wirelessaccessD local access48．Suppose there is exactly onepacket switch between a sendinghost and a receiving host. The transmission rates between thesending host and the switch and between the switch and the receiving host are R 1 and R 2,respectively. Assuming that theswitch uses store-and-forwardpacket switching, what is thetotal end-to-end delay to senda packet of length L (Ignorequeuing delay, propagationdelay, and processing delay.) ( ) A L /R 1＋L /R 2B L /R 1C L /R 2D none of the above49．The time required to examine thepacket ’s header and determinewhere to direct the packet ispart of the ( ). A queuing delayB processing delayC propagation delayD transmission delay50．The time required to propagatefrom the beginning of the linkto the next router is ( ). A queuing delayB processing delayC propagation delayD transmission delay51．Consider sending a packet of3000bits over a path of 5 links. Eachlink transmits at 1000bps. Queuingdelays, propagation delay and processing delay are negligible. (6 points)(1).Suppose the network is apacket-switched virtual circuitnetwork. VC setup time is seconds.Suppose the sending layers add atotal of 500 bits of header to eachpacket. How long does it take to sendthe file from source to destination(2).Suppose the network is a packet-switched datagram networkand a connectionless service is used.Now suppose each packet has 200 bitsof header. How long does it take tosend the file(3).Suppose that the network is acircuit-switched network. Furthersuppose that the transmission rate of the circuit between source anddestination is 200bps. Assumingsetup time and 200 bits of header appended to the packet, how longdoes it take to send the packetS olution:(1).t=5*(3000+500)/1000+=( 2).t=5*(3000+200)/1000=16s( 3).t=(3000+200)/200+=。

Computing Essentials 2008课后练习题答案详解Chapter 1: Information Technology, The Internet, and Y ouCrossword Puzzle Answers: (p22)Across1 、The most essential part of an information system. People信息系统最基本的部分。

人7 、Coordinates computer resources. Operating System协调计算机资源。

操作系统10 、Modifies signals for processing. Modem修改信号以便处理。

调制解调器12 、Data that has been processed by the computer. Information计算机处理过的数据。

信息13 、Unprocessed facts. Data没处理过的事实。

数据14 、Notebook computer that accepts handwritten input.. Tablet PC可以用手写输入的笔记本电脑。

平板电脑DownNum. Clue Answer2 、Uses computers to become more productive. End User使用电脑变得更有效率。

终端用户3 、Rules or guidelines to follow when using software, hardware, and data. Procedures使用软件，硬件和数据时遵循的规则或指引。

指令说明书4 、Created by word processors. Document Files文字处理创建出来的。

文档文件5 、Specialized programs that allow input and output devices to communicate. Device Drivers 允许输入和输出设备通信的专业程序。

计算机网络英文题库附答案c h a p t e r精编W O R D版IBM system office room 【A0816H-A0912AAAHH-GX8Q8-GNTHHJ8】Chapter 1 Computer Networks and the Internet 1．The ( ) is a worldwide computer network, that is, a network that interconnects millions of computing devices throughout the world.ppt3A public InternetB IntranetC switch netD television net2．Which kind of media is not a guided media ( )A twisted-pair copper wireB a coaxial cableC fiber opticsD digital satellite channel3．Which kind of media is a guided media ( )A geostationary satelliteB low-altitude satelliteC fiber opticsD wireless LAN4．The units of data exchanged by a link-layer protocol are called ( ).A FramesB SegmentsC DatagramsD bit streams5．Which of the following option belongs to the circuit-switched networks ( )A FDMB TDMC VC networksD both A and B6．( )makes sure that neither side of aconnection overwhelms the other side bysending too many packets too fast.A Reliable data transferB Flow controlC Congestion controlD Handshaking procedure7．( ) means that the switch must receive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link.A Store-and-forward transmissionB FDMC End-to-end connectionD TDM8．Datagram networks and virtual-circuit networks differ in that ( ).A datagram networks are circuit-switchednetworks, and virtual-circuit networks arepacket-switched networks.B datagram networks are packet-switchednetworks, and virtual-circuit networks arecircuit-switched networks.C datagram networks use destinationaddresses and virtual-circuit networks useVC. numbers to forward packets towardtheir destination.D datagram networks use VC. numbersand virtual-circuit networks use destinationaddresses to forward packets toward theirdestination.9．In the following options, which one is not a guided media ( )A twisted-pair wireB fiber opticsC coaxial cableD satellite10．Processing delay does not include the time to ( ).A examine the packet’s headerB wait to transmit the packet onto the linkC determine where to direct the packetD check bit-error in the packet11．In the following four descriptions, which one is correct ( )A The traffic intensity must be greaterthan 1.B The fraction of lost packets increases asthe traffic intensity decreases.C If the traffic intensity is close to zero,the average queuing delay will be close tozero.D If the traffic intensity is close to one, theaverage queuing delay will be close to one.12．The Internet’s network layer isresponsible for moving network-layerpackets known as ( ) from one host toanother.A frameB datagramC segmentD message13．The protocols of various layers are called ( ).A the protocol stackB TCP/IPC ISPD network protocol14．There are two classes of packet-switched networks: ( ) networks and virtual-circuitnetworks.A datagramB circuit-switchedC televisionD telephone15．Access networks can be loosely classified into three categories: residential access,company access and ( ) access.A cabledB wirelessC campusD city areaQuestion 16～17Suppose, a is the average rate at which packets arrive at the queue, R is the transmission rate, and all packets consist of L bits, then the traffic intensity is ( 16 ), and it should no greater than ( 17 ).16．A LR／aB La／RC Ra／LD LR／a 17．A 2B 1C 0D -118．In the Internet, the equivalent concept to end systems is ( ).A hostsB serversC clientsD routers19．In the Internet, end systems are connected together by ( ).A copper wireB coaxial cableC communication linksD fiber optics20．End systems access to the Internet through its ( ).A modemsB protocolsC ISPD sockets21．End systems, packet switches, and other pieces of the Internet, run ( ) that controlthe sending and receiving of informationwithin the Internet.A programsB processesC applicationsD protocols22．There are many private networks, such as many corporate and government networks,whose hosts cannot exchange messageswith hosts outside of the private network.These private networks are often referredto as ( ).A internetsB LANC intranetsD WAN23．The internet allows ( ) running on its end systems to exchange data with each other.A clients applicationsB server applicationsC P2P applicationsD distributed applications24．The Internet provides two services to its distributed applications: a connectionlessunreliable service and () service.A flow controlB connection-oriented reliableC congestion controlD TCP25．It defines the format and the order of messages exchanged between two or morecommunicating entities, as well as theactions taken on the transmission and/orreceipt of a message or other event. Thesentence describes ( ).A InternetB protocolC intranetD network26．In the following options, which does not define in protocol ( )A the format of messages exchangedbetween two or more communicatingentitiesB the order of messages exchangedbetween two or more communicatingentitiesC the actions taken on the transmission ofa message or other eventD the transmission signals are digitalsignals or analog signals27．In the following options, which is defined in protocol ( )A the actions taken on the transmissionand/or receipt of a message or othereventB the objects exchanged between communicating entitiesC the content in the exchanged messagesD the location of the hosts28．In the following options, which does not belong to the network edge ( )A end systemsB routersC clientsD servers29．In the following options, which belongs to the network core ( )A end systemsB routersC clientsD servers30．In the following options, which is not the bundled with the Internet’s connection-oriented service( )A reliable data transferB guarantee of the transmission timeC flow controlD congestion-control31．An application can rely on the connection to deliver all of its data without error andin the proper order. The sentencedescribes ( ).A flow controlB congestion-controlC reliable data transferD connection-oriented service32．It makes sure that neither side of aconnection overwhelms the other side bysending too many packets too fast. Thesentence describes ( ).A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer33．It helps prevent the Internet from enteringa state of gridlock. When a packet switchbecomes congested, its buffers canoverflow and packet loss can occur. Thesentence describes ( ).A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer34．The Internet’s connection-oriented service has a name, it is ( ).A TCPB UDPC TCP/IPD IP 35．In the following options, which service does not be provided to an application by TCP( )A reliable transportB flow controlC video conferencingD congestion control36．The Internet’s connectionless service is called ( ).A TCPB UDPC TCP/IPD IP37．In the following options, which does not use TCP()A SMTPB internet telephoneC FTPD HTTP38．In the following options, which does not use UDP( )A Internet phoneB video conferencingC streaming multimediaD telnet39．There are two fundamental approaches to building a network core, ( ) and packetswitching.A electrical current switchingB circuit switchingC data switchingD message switching40．In ( ) networks, the resources needed along a path to provide forcommunication between the end systemare reserved for the duration of thecommunication session.A packet-switchedB data-switchedC circuit-switchedD message-switched41．In ( ) networks, the resources are not reserved; a session’s messages use theresources on demand, and as aconsequence, may have to wait for accessto communication link.A packet-switchedB data-switchedC circuit-switchedD message-switched42．In a circuit-switched network, if each link has n circuits, for each link used by theend-to-end connection, the connectiongets ( ) of the link’s bandwidth for theduration of the connection.A a fraction 1/nB allC 1/2D n times43．For ( ), the transmission rate of a circuit is equal to the frame rate multiplied by thenumber of bits in a slot.A CDMAB packet-switched networkC TDMD FDM 44．( ) means that the switch must receive the entire packet before it can begin totransmit the first bit of the packet onto theoutbound link.A Queuing delayB Store-and-forward transmissionC Packet lossD Propagation45．The network that forwards packetsaccording to host destination addresses iscalled ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram46．The network that forwards packetsaccording to virtual-circuit numbers iscalled ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram47．In the following entries, which is not a kind of access network()A residential accessB company accessC wireless accessD local access48．Suppose there is exactly one packet switch between a sending host and a receivinghost. The transmission rates between thesending host and the switch and betweenthe switch and the receiving host are R1and R2, respectively. Assuming that theswitch uses store-and-forward packetswitching, what is the total end-to-enddelay to send a packet of length L (Ignorequeuing delay, propagation delay, andprocessing delay.) ( )A L/R1＋L/R2B L/R1C L/R2D none of the above49．The time required to examine thepacket’s header and determine where todirect the packet is part of the ( ).A queuing delayB processing delayC propagation delayD transmission delay50．The time required to propagate from the beginning of the link to the next router is( ).A queuing delayB processing delayC propagation delayD transmission delay51．Consider sending a packet of 3000bits over a path of 5 links. Each link transmits at 1000bps. Queuing delays, propagation delay and processing delay are negligible. (6 points) (1).Suppose the network is a packet-switched virtual circuit network. VC setup time is 0.1 seconds. Suppose the sending layers add a total of 500 bits of header to each packet. How long does it take to send the file from source to destination?(2).Suppose the network is a packet-switched datagram network and a connectionless serviceis used. Now suppose each packet has 200 bitsof header. How long does it take to send the file?(3).Suppose that the network is a circuit-switched network. Further suppose that the transmission rate of the circuit between sourceand destination is 200bps. Assuming 0.02s setuptime and 200 bits of header appended to the packet, how long does it take to send the packet?So lution:(1).t=5*(3000+500)/1000+0.1=17.6s( 2).t=5*(3000+200)/1000=16s( 3).t=(3000+200)/200+0.02=16.02s。

第一章(1.2 1.3节)5.Calculate the total time required to transfer a 1,000-KB file in the following cases, assuming an RTT of 100 ms, a packet size of 1-KB data, and an initial 2 ×RTT of “handshaking” before data is sent.(a) The bandwidth is 1.5 Mbps, and data packets can be sent continuously.(b) The bandwidth is 1.5 Mbps, but after we finish sending each data packetwe must wait one RTT before sending the next.(c) The bandwidth is “infinite,”meaning that we take transmit time to bezero, and up to 20 packets can be sent per RTT.(d) The bandwidth is infinite, and during the first RTT we can send onepacket (21−1), during the second RTT we can send two packets (22−1),during the third we can send four (23−1), and so on. (A justification forsuch an exponential increase will be given in Chapter 6.)7. Consider a point-to-point link 2 km in length. At what bandwidth would propagation delay (at a speed of 2 × 108m/sec) equal transmit delay for 100-byte packets? What about 512-byte packets?13.How “wide” is a bit on a 1-Gbps link? How long is a bit in copper wire, where the speed of propagation is 2.3 × 108 m/s?15.Suppose a 100-Mbps point-to-point link is being set up between Earth and a new lunar colony. The distance from the moon to Earth is approximately 385,000 km, and data travels over the link at the speed of light—3 × 108 m/s.(a) Calculate the minimum RTT for the link.(b) Using the RTT as the delay, calculate the delay × bandwidth product forthe link.(c) What is the significance of he delay × bandwidth product computedin (b)?(d) A camera on the lunar base takes pictures of Earth and saves them in digitalformat to disk. Suppose Mission Control on Earth wishes to download themost current image, which is 25 MB. What is the minimum amount of time that will elapse between when the request for the data goes out andthe transfer is finished?18. Calculate the latency (from first bit sent to last bit received) for the following:(a) A 10-Mbps Ethernet with a single store-and-forward switch in the path,and a packet size of 5,000 bits. Assume that each link introduces a propaga-tion delay of 10 µs, and that the switch begins retransmitting immediatelyafter it has finished receiving the packet.(b) Same as (a) but with three switches.(c) Same as (a) but assume the switch implements “cut-through”switching: itis able to begin retransmitting the packet after the first 200 bits have beenreceived.第二章（除2.7 2.9 节）1.Show the NRZ, Manchester, and NRZI encodings for the bit pattern shown in Figure2.46. Assume that the NRZI signal starts out low.23.Consider an ARQ algorithm running over a 20-km point-to-point fiber link.(a) Compute the propagation delay for this link, assuming that the speed oflight is 2 × 108 m/s in the fiber.(b) Suggest a suitable timeout value for the ARQ algorithm to use.(c) Why might it still be possible for the ARQ algorithm to time out andretransmit a frame, given this timeout value?26.The text suggests that the sliding window protocol can be used to implement flow control. We can imagine doing this by having the receiver delay ACKs, that is, not send the ACK until there is free buffer space to hold the next frame. In doing so, each ACK would simultaneously acknowledge the receipt of the last frame and tell the source that there is now free buffer space available to hold the next frame. Explain why implementing flow control in this way is not a good idea.44.Let A and B be two stations attempting to transmit on an Ethernet. Each has steady queue of frames ready to send; A’s frames will be numbered A 1, A2 , and so on, and B’s similarly. Let T = 51.2 µs be the exponential backoff base unit. SupposeA andB simultaneously attempt to send frame 1, collide, and happen to choose backoff times of 0 × T and 1 × T, respectively, meaning A wins the race and transmits A 1 while B waits. At the end of this transmission, B will attempt to retransmit B1 while A will attempt to transmit A2 . These first attempts will collide, but now A backs off for either 0 × T or 1 × T, while B backs off for time equal to one of 0 × T, . . . , 3 × T.(a) Give the probability that A wins this second backoff race immediately after this firstcollision , that is, A’s first choice of backoff time k × 51.2 is less than B’s.(b) Suppose A wins this second backoff race. A transmits A 3 , and when it isfinished, A and B collide again as A tries to transmit A4 and B tries once more to transmit B1. Give the probability that A wins this third backoff race immediately after the first collision.(c) Give a reasonable lower bound for the probability that A wins all the re- maining backoff races.(d) What then happens to the frame B1?This scenario is known as the Ethernet capture effect.48. Repeat the previous exercise, now with the assumption that Ethernet is p -persistent with p = 0.33 (that is, a waiting station transmits immediately with probability p when the line goes idle, and otherwise defers one 51.2-µs slot time and repeats the process). Your timeline should meet criterion (1) of the previous problem, but in lieu of criterion (2), you should show at least one collision and at least one run of four deferrals on an idle line. Again, note that many solutions are possible.第三章（3.1 3.2节）ing the example network given in Figure 3.30, give the virtual circuit tables for all the switches after each of the following connections is established. Assume that the sequence of connections is cumulative, that is, the first connection is still up when the second connection is established, and so on. Also assume that the VCI assignment always picks the lowest unused VCI on each link, starting with 0.(a) Host A connects to host B.(b) Host C connects to host G.(c) Host E connects to host I.(d) Host D connects to host B.(e) Host F connects to host J.(f) Host H connects to host A.3.For the network given in Figure 3.31, give the datagram forwarding table for each node. The links are labeled with relative costs; your tables should forward each packet via the lowest-cost path to its destination.5. Consider the virtual circuit switches in Figure 3.33. Table 3.6 lists, for each switch, what port, VCI (or VCI, interface) pairs are connected to other. Connections are bidirectional. List all endpoint-to-endpoint connections.13. Given the extended LAN shown in Figure 3.34, indicate which ports are not selected by the spanning tree algorithm.15. Consider the arrangement of learning bridges shown in Figure 3.35. Assuming all are initially empty, give the forwarding tables for each of the bridges B1–B4 after the following transmissions:■ A sends to C.■ C sends to A.■ D sends to C.Identify ports with the unique neighbor reached directly from that port, thatis, the ports for B1 are to be labeled “A” and “B2.”17.Consider hosts X, Y, Z, W and learning bridges B1, B2, B3, with initially empty forwarding tables, as in Figure 3.36.(a) Suppose X sends to Z. Which bridges learn where X is? Does Y’s networkinterface see this packet?(b) Suppose Z now sends to X. Which bridges learn where Z is? Does Y’s network interface see this packet?(c) Suppose Y now sends to X. Which bridges learn where Y is? Does Z’s net-work interface see this packet?(d) Finally, suppose Z sends to Y. Which bridges learn where Z is? Does W’snetwork interface see this packet?第四章（4.1 4.2 4.3.1 4.3.5 4.5 节）4.Suppose a TCP message that contains 2,048 bytes of data and 20 bytes of TCP header is passed to IP for delivery across two networks of the Internet (i.e., from the source host to a router to the destination host). The first network uses 14-byte headers and has an MTU of 1,024 bytes; the second uses 8-byte headers with an MTU of 512 bytes. Each network’s MTU gives the size of the largest IP datagram that can be carried in a link layer frame. Give the sizes and offsets of the sequence of fragments delivered to the network layer at the destination host. Assume all IP headers are 20 bytes.21.Suppose a router has built up the routing table shown in Table 4.14. The router can deliver packets directly over interfaces 0 and 1, or it can forward packets to routers R2, R3, or R4. Describe what the router does with a packet addressed to each of the following destinations:(a) 128.96.39.10.(b) 128.96.40.12.(c) 128.96.40.151.(d) 192.4.153.17.(e) 192.4.153.90.45.Table 4.16 is a routing table using CIDR. Address bytes are in hexadecimal. The notation “/12” in C4.50.0.0/12 denotes a netmask with 12 leading 1 bits, that is, FF.F0.0.0. Note that the last three entries cover every address and thus serve in lieu of a default route. State to what next hop the following will be delivered.(a) C4.5E.13.87.(b) C4.5E.22.09.(c) C3.41.80.02.(d) 5E.43.91.12.(e) C4.6D.31.2E.(f) C4.6B.31.2E.第五章（5.1 5.2节）10. You are hired to design a reliable byte-stream protocol that uses a sliding window (like TCP). This protocol will run over a 1-Gbps network. The RTT of the network is 140 ms, and the maximum segment lifetime is 60 seconds. How many bits would you include in the AdvertisedWindow and SequenceNum fields of your protocol header?第六章（6.5节）第九章（9.1节）。

Unit 5-6单元测试题及答案第一题：问题：什么是计算机网络?答案：计算机网络是指由若干结点和连接这些结点的链路组成的，用于传输各种形式的信息的通信系统。

第二题：问题：计算机网络按照通信方式可分为哪几种?答案：计算机网络按照通信方式可分为点对点通信和广播通信两种方式。

第三题：问题：什么是TCP/IP协议?答案：TCP/IP协议是一种网络通信协议，它是由传输控制协议（TCP）和网际协议（IP）组成的，被广泛应用于互联网和局域网中。

第四题：问题：简述TCP和UDP协议的区别。

答案：TCP协议提供可靠的数据传输，保证数据的准确性和完整性，但是传输速度相对较慢。

UDP协议则着重于传输的效率，不对数据进行分包和排序，传输速度较快，但是不能保证数据的可靠性。

第五题：第六题：问题：什么是URL?答案：URL是统一资源定位符的缩写，用于标识互联网上资源的位置。

它由协议名、主机名和路径组成。

第七题：问题：什么是DNS?答案：DNS是域名系统的缩写，它是一种用于将域名解析为IP地址的系统。

通过DNS，用户可以通过域名访问互联网上的各种资源。

第八题：问题：简述TCP的三次握手过程。

答案：TCP的三次握手过程分为以下几个步骤：1. 客户端向服务器发送一个SYN包，请求建立连接。

2. 服务器收到SYN包后，向客户端发送一个SYN-ACK包，表示同意建立连接。

3. 客户端收到服务器的SYN-ACK包后，再向服务器发送一个ACK包，确认连接建立成功。

第九题：问题：简述TCP的四次挥手过程。

答案：TCP的四次挥手过程分为以下几个步骤：1. 客户端向服务器发送一个FIN包，表示不再发送数据。

2. 服务器收到FIN包后，向客户端发送一个ACK包，确认收到FIN包。

3. 服务器在准备关闭连接时，向客户端发送一个FIN包，请求关闭连接。

4. 客户端收到服务器的FIN包后，向服务器发送一个ACK包，确认关闭连接。

第十题：问题：什么是网络拓扑结构?答案：网络拓扑结构是指计算机网络中各个设备之间的连接关系。

第一章(1.2 1.3节)5.Calculate the total time required to transfer a 1,000-KB file in the following cases, assuming an RTT of 100 ms, a packet size of 1-KB data, and an initial 2 ×RTT of “handshaking”before data is sent.(a) The bandwidth is 1.5 Mbps, and data packets can be sent continuously.(b) The bandwidth is 1.5 Mbps, but after we finish sending each data packetwe must wait one RTT before sending the next.(c) The bandwidth is “infinite,”meaning that we take transmit time to bezero, and up to 20 packets can be sent per RTT.(d) The bandwidth is infinite, and during the first RTT we can send onepacket (21−1), during the second RTT we can send two packets (22−1),during the third we can send four (23−1), and so on. (A justification forsuch an exponential increase will be given in Chapter 6.)7. Consider a point-to-point link 2 km in length. At what bandwidth would propagation delay (at a speed of 2 ×108m/sec) equal transmit delay for 100-byte packets? What about 512-byte packets?13.How “wide”is a bit on a 1-Gbps link? How long is a bit in copper wire, where the speed of propagation is 2.3 ×108 m/s?15.Suppose a 100-Mbps point-to-point link is being set up between Earth and a new lunar colony. The distance from the moon to Earth is approximately 385,000 km, and data travels over the link at the speed of light—3 ×108 m/s.(a) Calculate the minimum RTT for the link.(b) Using the RTT as the delay, calculate the delay ×bandwidth product forthe link.(c) What is the significance of he delay ×bandwidth product computedin (b)?(d) A camera on the lunar base takes pictures of Earth and saves them in digitalformat to disk. Suppose Mission Control on Earth wishes to download themost current image, which is 25 MB. What is the minimum amount oftime that will elapse between when the request for the data goes out andthe transfer is finished?18. Calculate the latency (from first bit sent to last bit received) for the following:(a) A 10-Mbps Ethernet with a single store-and-forward switch in the path,and a packet size of 5,000 bits. Assume that each link introduces a propaga-tion delay of 10 µs, and that the switch begins retransmitting immediatelyafter it has finished receiving the packet.(b) Same as (a) but with three switches.(c) Same as (a) but assume the switch implements “cut-through”switching: itis able to begin retransmitting the packet after the first 200 bits have beenreceived.第二章（除2.7 2.9 节）1.Show the NRZ, Manchester, and NRZI encodings for the bit pattern shown in Figure2.46. Assume that the NRZI signal starts out low.23.Consider an ARQ algorithm running over a 20-km point-to-point fiber link.(a) Compute the propagation delay for this link, assuming that the speed oflight is 2 ×108 m/s in the fiber.(b) Suggest a suitable timeout value for the ARQ algorithm to use.(c) Why might it still be possible for the ARQ algorithm to time out andretransmit a frame, given this timeout value?26.The text suggests that the sliding window protocol can be used to implement flow control. We can imagine doing this by having the receiver delay ACKs, that is, not send the ACK until there is free buffer space to hold the next frame. In doing so, each ACK would simultaneously acknowledge the receipt of the last frame and tell the source that there is now free buffer space available to hold the next frame. Explain why implementing flow control in this way is not a good idea.44.Let A and B be two stations attempting to transmit on an Ethernet. Each has steady queue of frames ready to send; A’s frames will be numbered A 1, A2 , and so on, and B’s similarly. Let T = 51.2 µs be the exponential backoff base unit. Suppose A and B simultaneously attempt to send frame 1, collide, and happen to choose backoff times of 0 ×T and 1 ×T, respectively, meaning A wins the race and transmits A 1 while B waits. At the end of this transmission, B will attempt to retransmit B1 while A will attempt to transmit A2 . These first attempts will collide, but now A backs off for either 0 ×T or 1 ×T, while B backs off for time equal to one of 0 ×T, . . . , 3 ×T.(a) Give the probability that A wins this second backoff race immediately after this firstcollision , that is, A’s first choice of backoff time k ×51.2 is less than B’s.(b) Suppose A wins this second backoff race. A transmits A 3 , and when it isfinished, A and B collide again as A tries to transmit A4 and B tries oncemore to transmit B1. Give the probability that A wins this third backoffrace immediately after the first collision.(c) Give a reasonable lower bound for the probability that A wins all the re-maining backoff races.(d) What then happens to the frame B1?This scenario is known as the Ethernet capture effect.48. Repeat the previous exercise, now with the assumption that Ethernet is p -persistent with p = 0.33 (that is, a waiting station transmits immediately with probability p when the line goes idle, and otherwise defers one 51.2-µs slot time and repeats the process). Your timeline should meet criterion (1) of the previous problem, but in lieu of criterion (2), you should show at least one collision and at least one run of four deferrals on an idle line. Again, note that many solutions are possible.第三章（3.1 3.2节）ing the example network given in Figure 3.30, give the virtual circuit tables for all the switches after each of the following connections is established. Assume that the sequence of connections is cumulative, that is, the first connection is still up when the second connection is established, and so on. Also assume that the VCI assignment always picks the lowest unused VCI on each link, starting with 0.(a) Host A connects to host B.(b) Host C connects to host G.(c) Host E connects to host I.(d) Host D connects to host B.(e) Host F connects to host J.(f) Host H connects to host A.3.For the network given in Figure 3.31, give the datagram forwarding table for each node. The links are labeled with relative costs; your tables should forward each packet via the lowest-cost path to its destination.5. Consider the virtual circuit switches in Figure 3.33. Table 3.6 lists, for each switch, what port, VCI (or VCI, interface) pairs are connected to other. Connections are bidirectional. List all endpoint-to-endpoint connections.13. Given the extended LAN shown in Figure 3.34, indicate which ports are not selected bythe spanning tree algorithm.15. Consider the arrangement of learning bridges shown in Figure 3.35. Assuming all are initially empty, give the forwarding tables for each of the bridges B1–B4 after the following transmissions:■ A sends to C.■ C sends to A.■ D sends to C.Identify ports with the unique neighbor reached directly from that port, thatis, the ports for B1 are to be labeled “A”and “B2.”17.Consider hosts X, Y, Z, W and learning bridges B1, B2, B3, with initially empty forwarding tables, as in Figure 3.36.(a) Suppose X sends to Z. Which bridges learn where X is? Does Y’s networkinterface see this packet?(b) Suppose Z now sends to X. Which bridges learn where Zis? Does Y’s network interface see this packet?(c) Suppose Y now sends to X. Which bridges learn where Y is? Does Z’s net-work interface see this packet?(d) Finally, suppose Z sends to Y. Which bridges learn where Z is? Does W’snetwork interface see this packet?第四章（4.1 4.2 4.3.1 4.3.5 4.5 节）4.Suppose a TCP message that contains 2,048 bytes of data and 20 bytes of TCP header is passed to IP for delivery across two networks of the Internet (i.e., from the source host to a router to the destination host). The first network uses 14-byte headers and has an MTU of 1,024 bytes; the second uses 8-byte headers with an MTU of 512 bytes. Each network’s MTU gives the size of the largest IP datagram that can be carried in a link layer frame. Give the sizes and offsets of the sequence of fragments delivered to the network layer at the destination host. Assume all IP headers are 20 bytes.21.Suppose a router has built up the routing table shown in Table 4.14. The router can deliver packets directly over interfaces 0 and 1, or it can forward packets to routers R2, R3, or R4. Describe what the router does with a packet addressed to each of the following destinations:(a) 128.96.39.10.(b) 128.96.40.12.(c) 128.96.40.151.(d) 192.4.153.17.(e) 192.4.153.90.45.Table 4.16 is a routing table using CIDR. Address bytes are in hexadecimal. The notation “/12”in C4.50.0.0/12 denotes a netmask with 12 leading 1 bits, that is, FF.F0.0.0. Note that the last three entries cover every address and thus serve in lieu of a default route. State to what next hop the following will be delivered.(a) C4.5E.13.87.(b) C4.5E.22.09.(c) C3.41.80.02.(d) 5E.43.91.12.(e) C4.6D.31.2E.(f) C4.6B.31.2E.第五章（5.1 5.2节）10. You are hired to design a reliable byte-stream protocol that uses a sliding window (like TCP). This protocol will run over a 1-Gbps network. The RTT of the network is 140 ms, and the maximum segment lifetime is 60 seconds. How many bits would you include in the AdvertisedWindow and SequenceNum fields of your protocol header?第六章（6.5节）第九章（9.1节）。

（完整版）计算机⽹络英⽂题库（附答案）chapter5 Chapter 5 The Link Layer and Local Area Network1．A ( ) protocol is used to move a datagram over an individual link.A application-layerB transport-layerC network-layerD link-layer2．The units of data exchanged by a link-layer protocol are called ( ).A datagramsB framesC segmentsD messages3．Which of the following protocols is not a link-layer protocol? ( )A EthernetB PPPC HDLCD IP4．In the following four descriptions, which one is not correct? ( )A link-layer protocol has the node-to-node job of moving network-layer datagrams over a single link in the path.B The services provided by the link-layer protocols may be different.C A datagram must be handled by the same link-layer protocols on the different links in the path.D The actions taken by a link-layer protocol when sending and receiving frames include error detection, flow control and random access.5．Which of the following services can not offered by a link-layer protocol? ( )A congestion controlB Link AccessC Error controlD Framing6．( ) protocol serves to coordinate the frame transmissions of the many nodes when multiple nodes share a single broadcast link.A ARPB MACC ICMPD DNS7．In the following four descriptions about the adapter, which one is not correct? ( )A The adapter is also called as NIC.B The adapter is a semi-autonomous unit.C The main components of an adapter are bus interface and the link interface.D The adapter can provide all the link-layer services.8．Consider CRC error checking approach, the four bit generator G is 1011, and suppose that the data D is 10101010, then the value of R is( ).A 010B 100C 011D 1109．In the following four descriptions about random access protocol, which one is not correct? ( )A I n slotted ALOHA, nodes can transmit at random time.B I n pure ALOHA, if a frame experiences a collision, the node will immediately retransmit it with probability p.C T he maximum efficiency of a slotted ALOHA is higher than a pure ALOHA.D I n CSMA/CD, one node listens to the channel before transmitting.10．In the following descriptions about MAC address, which one is not correct? ( )A T he MAC address is the address of one node’s adapter.B N o two adapters have the same MAC address.C T he MAC address doesn’t change no matter where the adapter goes.D M AC address has a hierarchical structure.11．The ARP protocol can translate ( ) into ( ). ( )A h ost name, IP addressB h ost name, MAC addressC I P address, MAC addressD broadcast address, IP address12．The value of Preamble field in Ethernet frame structure is ( )A 10101010 10101010……10101010 11111111B 10101011 10101011……10101011 10101011C 10101010 10101010……10101010 10101011D 10101010 10101010……10101010 1010101013．There are four steps in DHCP, the DHCP server can complete ( ).A DHCP server discoveryB DHCP server offersC DHCP requestD DHCP response14．In CSMA/CD, the adapter waits some time and then returns to sensing the channel. In the following four times, which one is impossible? ( )A 0 bit timesB 512 bit timesC 1024 bit timesD 1028 bit times15．The most common Ethernet technologies are 10BaseT and 100BaseT. “10” and “100” indicate( ).A the maximum length between two adaptersB the minimum length between two adaptersC the transmission rate of the channelD the transmission rate of the node16．The principal components of PPP include but not( ).A framingB physical-control protocolC link-layer protocolD network-layer protocol17．In the following four options, which service can not be provided by switch? ( )A filteringB self-learningC forwardingD optimal routing18．In the following four services, which one was be required in PPP? ( )A packet framingB error detectionC error correctionD multiple types of link19．The ability to determine the interfaces to which a frame should be directed, and then directing the frame to those interfaces is( ).A filteringB forwardingC self-learningD optimal routing20．In ( ) transmission(s), the nodes at both ends of a link may transmit packets at the same time.A full-duplexB half-duplexC single-duplexD both full-duplex and half-duplex21．Consider the data D is 01110010001, if use even parity checking approach, the parity bit is( ① ), if use odd parity checking approach, the parity bit is( ② ). ( )A ①0 ②1B ①0 ②0C ①1 ②1D ①1 ②022．In the following four descriptions about parity checks, which one is correct? ( )A Single-bit parity can detect all errors.B Single-bit parity can correct one errors.C Two-dimensional parity not only can detect a single bit error, but also can correct that error.D Two-dimensional parity not only can detect any combination of two errors, but also can correct them.23．MAC address is ( ) bits long.A 32B 48C 128D 6424．Wireless LAN using protocol ( ).A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.1125．The following protocols are belonging to multiple access protocols except for ( ).A channel partitioning protocolsB routing protocolsC random access protocolsD taking-turns protocols26．Which of the following is not belonging to channel partitioning protocols? ( )A CSMAB FDMC CDMAD TDM27．In the following four descriptions about CSMA/CD, which one is not correct? ( )A A node listens to the channel before transmitting.B If someone else begins talking at the same time, stop talking.C A transmitting node listens to the channel while it is transmitting.D With CSMA/CD, the collisions can be avoided completely.28．( ) provides a mechanism for nodes to translate IP addresses to link-layer address.A IPB ARPC RARPD DNS29．A MAC address is a ( )address.A physical-layerB application-layerC link-layerD network-layer30．Which of the following is correct? ( )A No two adapters have the same MAC address.B MAC broadcast address is FF-FF-FF-FF-FF-FF.C A portable computer with an Ethernet card always has the same MAC address, no matter where the computer goes.D All of the above31．In the following four descriptions, which one is not correct? ( )A ARP resolves an IP address to a MAC address.B DNS resolves hostnames to IP addresses.C DNS resolves hostnames for hosts anywhere in the Internet.D ARP resolves IP addresses for nodes anywhere in the Internet.32．In the LAN, ( )protocol dynamically assign IP addresses to hosts.A DNSB ARPC DHCPD IP33．DHCP protocol is a four-step process: ①DHCP request. ②DHCP ACK. ③DHCP server discovery. ④DHCP server offer(s). The correct sequence is ( )A ①②③④B ③②①④C ③④①②D ①④③②34．In the Ethernet frame structure, the CRC field is ( )bytes.D 3235．In the Ethernet frame structure, the Data field carries the ( ).A IP datagramB segmentC frameD message36．In the following four descriptions, which one is not correct? ( )A Ethernet uses baseband transmission.B All of the Ethernet technologies provide connection-oriented reliable service to the network layer.C The Ethernet 10Base2 technology uses a thin coaxial cable for the bus.D The Ethernet 10BaseT technology uses a star topology.37．Ethernet’s multiple access protocol is ( ).A CDMAB CSMA/CDC slotted ALOHAD token-passing protocol38．In the following four descriptions about CSMA/CD, which one is not correct? ( )A An adapter may begin to transmit at any time.B An adapter never transmits a frame when it senses that some other adapter is transmitting.C A transmitting adapter aborts its transmission as soon as it detects that another adapter is also transmitting.D An adapter retransmits when it detects a collision.39．Which of the following descriptions about CSMA/CD is correct? ( )A No slots are used.B It uses carrier sensing.C It uses collision detection.D All of the above.40．The Ethernet 10BaseT technology uses( )as its physical media.A fiber opticsB twisted-pair copper wireC coaxial cableD satellite radio channel41．For 10BaseT, the maximum length of the connection between an adapter and the hub is ( )meters.D 1042．A ( )is a physical-layer device that acts on individual bits rather than on frames.A switchB hubC routerD gateway43．A hub is a ( )device that acts on individual bits rather than on frames.A physical-layerB link-layerC network-layerD ransport-layer44．A switch is a( )device that acts on frame.A physical-layerB link-layerC network-layerD transport-layer45．In the following four descriptions, which one is not correct? ( )A Switches can interconnect different LAN technologies.B Hubs can interconnect different LAN technologies.C There is no limit to how large a LAN can be when switches are used to interconnect LAN segments.D There is restriction on the maximum allowable number of nodes in a collision domain when hubs are used to interconnect LAN segments.46．The ability to determine whether a frame should be forwarded to some interface or should just be dropped is ( ).A f ilteringB f orwardingC s elf-learningD o ptimal routing47．Which of the following devices is not a plug and play device? ( )A hubB routerC switchD repeater48．Which of the following devices is not cut-through device? ( )D repeater49．In the following four descriptions, which one is not correct? ( )A Switches do not offer any protection against broadcast storms.B Routers provide firewall protection against layer-2 broadcast storms.C Both switches and routers are plug and play devices.D A router is a layer-3 packet switch, a switch is a layer-2 packet switch. 50．Which device has the same collision domain? ( )A HubB SwitchC RouterD Bridge51．IEEE802.2 protocol belong to ( )layerA networkB MACC LLCD physical52．IEEE802.11 protocol defines ( )rules.A Ethernet BusB wireless WANC wireless LAND Token Bus53．In data link-layer, which protocol is used to share bandwidth? ( )A SMTPB ICMPC ARPD CSMA/CD54．When two or more nodes on the LAN segments transmit at the same time, there will be a collision and all of the transmitting nodes well enter exponential back-off, that is all of the LAN segments belong to the same( ).A collision domainB switchC bridgeD hub55．( )allows different nodes to transmit simultaneously and yet have their respective receivers correctly receive a sender’s encoded data bits.C CSMA/CDD CSMA/CA56．Because there are both network-layer addresses (for example, Internet IP addresses) and link-layer addresses (that is, LAN addresses), there is a need totranslate between them. For the Internet, this is the job of ( ).A RIPB OSPFC ARPD IP57．PPP defines a special control escape byte, ( ). If the flag sequence, 01111110 appears anywhere in the frame, except in the flag field, PPP precedes that instance of the flag pattern with the control escape byte.A 01111110B 01111101C 10011001D 1011111058．The device ( ) can isolate collision domains for each of the LAN segment.A modemB switchC hubD NIC59．In the following four descriptions about PPP, which one is not correct? ( )A PPP is required to detect and correct errors.B PPP is not required to deliver frames to the link receiver in the same order in which they were sent by the link sender.C PPP need only operate over links that have a single sender and a single receiver.D PPP is not required to provide flow control.60．In the PPP data frame, the( ) field tells the PPP receivers the upper-layer protocol to which the received encapsulated data belongs.A flagB controlC protocolD checksum61．PPP’s link-control protocols (LCP) accomplish ( ).A initializing the PPP linkB maintaining the PPP linkC taking down the PPP linkD all of the above62．The PPP link always begins in the ( ) state and ends in the ( ) state. ( )A open, terminatingB open, deadC dead, deadD dead, terminating63．For( ) links that have a single sender at one end of the link and a single receiver at the other end of the link.A point-to-pointB broadcastC multicastD all of the above64．With ( )transmission, the nodes at both ends of a link may transmit packets at the same time.A half-duplexB full-duplexC simplex(单⼯)D synchronous65．With ( ) transmission, a node can not both transmit and receive at the same time.A half-duplexB full-duplexC simplex(单⼯)D synchronous66．Which of the following functions can’t be implemented in the NIC? ( )A encapsulation and decapsulationB error detectionC multiple access protocolD routing67．Which of the following four descriptions is wrong? ( )A The bus interface of an adapter is responsible for communication with the adapter’s parent node.B The link interface of an adapter is responsible for implementing the link-layer protocol.C The bus interface may provide error detection, random access functions.D The main components of an adapter are the bus interface and the link interface. 68．For odd parity schemes, which of the following is correct? ( )A 011010001B 111000110C 110101110D 00011011069．( )divides time into time frames and further divides each time frame into N time slots.A FDMB TMDC CDMAD CSMA70．With CDMA, each node is assigned a different ( )A codeB time slotC frequencyD link71．Which of the following four descriptions about random access protocol is not correct? ( )A A transmission node transmits at the full rate of the channelB When a collision happens, each node involved in the collision retransmits at once.C Both slotted ALOHA and CSMA/CD are random access protocols.D With random access protocol, there may be empty slots.72．PPP defines a special control escape byte 01111101. If the data is b1b201111110b3b4b5, the value is( )after byte stuffing.A b1b20111110101111110b3b4b5B b1b20111111001111101b3b4b5C b5b4b30111111001111101b2b1D b5b4b30111110101111110b2b173．MAC address is in ( ) of the computer.A RAMB NICC hard diskD cache74．Which of the following is wrong? ( )A ARP table is configured by a system administratorB ARP table is built automaticallyC ARP table is dynamicD ARP table maps IP addresses to MAC addresses75．NIC works in ( )layer.A physicalB linkC networkD transport76．In LAN, if UTP is used, the common connector is( ).A AUIB BNCC RJ-45D NNI77．The modem’s function(s) is(are) ( ).A translates digital signal into analog signalB translates analog signal into digital signalC both translates analog signal into digital signal and translates digital signal into analog signalD translates one kind of digital signal into another digital signal78．( )defines Token-Ring protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.279．( )defines Token-Bus protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.280．( ) defines CSMA/CD protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.281．The computer network that concentrated in a geographical area, such as in a building or on a university campus, is ( )A a LANB a MANC a WAND the Internet82．The MAC address is ( ) bits long.A 32B 48C 128D 25683．Which of the following four descriptions about MAC addresses is wrong? ( )A a MAC address is burned into the adapter’s ROMB No two adapters have the same addressC An adapter’s MAC address is dynamicD A MAC address is a link-layer address84．Which of the following four descriptions about DHCP is correct? ( )A DHCP is C/S architectureB DHCP uses TCP as its underlying transport protocolC The IP address offered by a DHCP server is valid foreverD The DHCP server will offer the same IP address to a host when the host requests an IP address85．The ( )field permits Ethernet to multiplex network-layer protocols.A preambleB typeC CRCD destination MAC address86．For 10BaseT, the maximum length of the connection between an adapter and the hub is ( ) meters.A 50B 100C 200D 50087．An entry in the switch table contains the following information excepts for ( )A the MAC address of a nodeB the switch interface that leads towards the nodeC the time at which the entry for the node was placed in the tableD the IP address of a node88．Consider the 4-bit generator , G is 1001, and suppose that D has the value 101110000. What is the value of R?89．Consider the following graph of the network. Suppose Host A will send a datagram to Host B, Host A run OICQ on port 4000, Host B run OICQ on port 8000. All of ARP tables are up to date. Enumerate all the steps when message “Hello” is sent from host A to host B.。