04741计算机网络原理李全龙计算机网络原理,课后习题第一章

2022年4月高等教育自学考试全国统一命题考试计算机网络原理（课程代码04741）注意事项：1 .本试卷分为两部分，第一部分为选择题，第二部分为非选择题。

2 .应考者必须按试题顺序在答题卡（纸）指定位置上作答，答在试卷上无效。

3 .涂写部分、画图部分必须使用2B铅笔，书写部分必须使用黑色字迹签字笔。

第一部分选择题一、单项选择题：本大题共25小题，每小题1分，共25分.在每小题列出的备选项中只有一项是最符合题目要求的，请将其选出。

1 .通信链路的传输速率在计算机网络中也称为带宽，其单位是A. HzB. byteC. bpsD. baud带宽指最大传输速率，单位bps或bit/s2 .关于交换技术的说法错误的是A.报文交换事先需要建立连接B.分组交换中的分组长度有一个上限C.现代计算机网络没有采用报文交换技术的D.电路交换适用于语音和视频这类实时性强的业务报文交换与分组交换中的数据报方式类似，不需事先建立连接3 .在讨论网络总时间延迟时常常被忽略的是A.排队时延B.传输时延C.传播时延D.结点处理时延结点处理时延通常很小，并且对不同分组的结点处理时延变化也非常小，因此，在讨论网络总时间延迟时常常被忽略4 .主机A向主机B以存储转发的分组交换方式发送一个大文件，主机A到B的路径上有三段链路，速率分别为Rl=512kbit/s, R2=2Mbit/s, R3=lMbiVs,假设网络中没有其他流量，该文件传送的吞吐量是A. 512kbit/sB. 1 Mbit/sC. 2Mbit/sD. 3.5Mbit/s吞吐量指单位时间内源主机通过网络向目的主机实际送达的数据量，单位为bit/s或B/s本题主机A到B由三段链路组成，A到B的总链路速率取三段链路中速率最小的，所以吞吐量即为512kbit/s5. OSI参考模型中实现在相邻结点之间数据可靠而有效传输功能的是A.物理层B.数据链路层C.网络层D.传输层数据链路层的控制范围是相邻结点，传输层的控制范围是端到端6. WWW服务的应用层协议为A. SMTPB. FTPC.HTTPD. UDP7.从计算机网络应用体系结构的角度，WWW、FTP、电子邮件等应用属于A. C/S结构B.纯P2P结构C.层次结构D.混合结构WWW、FTP、电子邮件、DNS、DHCP等常见服务均属于C/S结构8.若主机要对因特网上的一个域名进行解析，而本地域名服务器无法解析时，首先求助于A.顶级域名服务器B.上层域名服务器C.代理域名服务器D.根域名服务器需掌握DNS解析过程，课本P569 .关于Cookie的说法中错误的是A. Cookie技术带来一定的安全问题B. Cookie是由服务器端生成的C. Cookie文件保存在服务器硬盘中D. Wed应用引入Cookie机制用于用户跟踪Cookie由服务器端生成，保存在客户端10 .有关电子邮件的说法中错误的是A. POP3是一个流行的电子邮件用户代理B. IMAP允许用户代理读取邮件的部分内容C. SMTP是实现邮件服务器间发送邮件的应用层协议D. SMTP不能直接传送可执行文件或其他二进制对象POP3是一个简单的邮件读取协议，功能有限。

2023 年4月高等教育自学考试全国统一命题考试计算机网络原理（课程代码04741）注意事项：本试卷分为两部分，第一部分为选择题，第二部分为非选择题。

应考者必须按试题顺序在答题卡（纸）指定位置上作答，答在试卷上无效。

涂写部分、画图部分必须使用2B 铅笔，书写部分必须使用黑色字迹签宇笔。

第一部分选择题一、单项选择题：本大题共25 小题，每小题1分，共25分。

在每小题列出的备选项中只有一项是最符合题目要求的，请将其选出。

1、定义实体彼此之间采用何种差错编码以及何种差错处理机制的网络协议要素是（）。

A语法 B.语义C.时序D.编码2、以“易于监控与管理，故障诊断与隔离容易”为主要优点的网络是（）。

A.星形拓扑结构网络B.总线型拓扑结构网络C.环形拓扑结构网络D.网状拓扑结构网络3、按照网络所采用的交换技术对计算机网络进行分类，其中不包括的类型是（）。

A.分组交换网络B.报文交换网络C.虚拟交换网络D.电路交换网络4、在讨论网络总时间延迟时常常被忽略的是（）。

A.传输时延B.排队时延C.传播时延D.结点处理时延5、最典型、最基本的网络应用结构是（）。

A.层次结构B.C/S结构C.纯P2P结构D.混合结构6、根据保存的域名信息及在域名解析过程中的作用对域名服务器分类，其中不包括（）。

A.顶级域名服务器B.权威域名服务器C.默认域名服务器D.中间域名服务器7、HTTP报文的四个组成部分中，由报文类型、功能等决定其有无的是（）。

A.起始行B.首部行C.空白行D.实体主体8、设码元速率为2400Baud，若采用曼彻斯特编码，无校验位，1位停止位的异步传输，30秒可传输数据（）。

A.3600BB.4000BC.7200BD.8000B9、Intemet标准的外部网关协议是（）。

A.OSPFB.BGPC.EGPD.RIP10、在OSI参考模型中，实现在相邻结点之间数据可靠而有效传输功能的是（）。

A.物理层B.数据链路层C.网络层D.传输层11、通常将TCP/IP参考模型传输层封装的数据包称为（）。

04741计算机网络原理试题及答案201404第一篇：04741计算机网络原理试题及答案201404绝密★考试结束前全国2014年4月高等教育自学考试计算机网络原理试题课程代码：04741 请考生按规定用笔将所有试题的答案涂、写在答题纸上。

选择题部分注意事项：1.答题前，考生务必将自己的考试课程名称、姓名、准考证号用黑色字迹的签字笔或钢笔填写在答题纸规定的位置上。

2.每小题选出答案后，用2B铅笔把答题纸上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其他答案标号。

不能答在试题卷上。

一、单项选择题(本大题共24小题，每小题1分，共24分)在每小题列出的四个备选项中只有一个是符合题目要求的，请将其选出并将“答题纸”的相应代码涂黑。

错涂、多涂或未涂均无分。

1.既可作为点—点线路通信子网又可作为广播信道通信子网的拓扑是 A.星形拓扑 C.树形拓扑B.网状形拓扑 D.总线形拓扑2.下列机构中，负责因特网RFC技术文件管理的是 A.ITU C.ANSIB.IETF D.ISO 3.按照网络的交换方式划分的网络类型中不包括．．．A.电路交换网C.分组交换网B.报文交换网 D.虚拟交换网4.计算机网络各层次结构模型及其协议的集合称为A.互联参考模型 C.网络体系结构B.网络概念框架 D.网络结构描述5.下列关于“面向连接服务”特点的叙述中错误的是．．A.传输数据前建立连接 C.收发数据的顺序不变B.每个分组要带目的节点地址 D.通信协议复杂通信效率不高6.下列网络协议中提供不可靠的用户数据报传送服务的是A.ARPC.DNSB.TCP D.UDP 7.TCP／IP参考模型中负责应用进程之间端—端通信的层次是 A.应用层C.互连层B.传输层 D.主机—网络层 8.规定DTE与DCE接口线的信号电平、发送器输出—阻抗的物理层特性属于 A.机械特性C.功能特性B.电气特性 D.规程特性9.蜂窝移动通信系统中的多址接入方法不包括．．．A.波分多址接入 C.时分多址接入B.频分多址接入 D.码分多址接入10.设码元速率为3600Baud，调制电平数为8，则数据传输速率为 A.1200bps C.10800bps11.TCP端口号的范围是 A.0~256 C.0~65535B.0~1023 D.1024~65535 B.7200bps D.14400bps 12.对模拟数据进行数字信号编码的最常用方法是 A.相位调制C.脉码调制13.ADSL服务采用的多路复用技术属于 A.频分多路复用 C.波分多路复用B.同步时分多路复用 D.异步时分多路复用 B.频率调制 D.振幅调制14.下列关于数据交换技术特点的说法中错误的是 A.报文交换不能满足实时通信要求 C.数据报方式中目的地要重组报文B.对于猝发式的通信电路交换最为合适D.虚电路方式数据传送前要设置虚电路15.在广域网中得到广泛应用的网络拓扑结构是 A.环形拓扑 C.网形拓扑B.星形拓扑 D.总线拓扑16.HDLC的帧类型中用于差错控制和流量控制的帧是 A.命令帧 C.无编号帧B.信息帧 D.监控帧17.既考虑拓扑结构又兼顾网络负载的静态路由算法是 A.最短路由选择算法 C.泛射路由选择算法B.基于流量路由算法 D.链路状态路由算法18.在数据通信过程中发现或纠正差错，把差错限制在尽可能小的范围内的技术和方法称为 A.差错控制 C.链路控制B.流量控制 D.拥塞控制19.下列关于IP协议基本任务的叙述中错误的是．．A.IP不保证服务的可靠性 C.IP不检查被丢弃的报文B.IP提供面向连接的数据报传输机制—无连接D.在主机资源不足时可能丢弃数据报20.在TCP／IP网络中，站点发送ARP请求所用的方式是 A.单播B.组播C.转播D.广播21.下列网络协议中，使用端口号为25的是 A.HTTP C.SMTPB.TELNET D.GOPHER 22.IEEE802.3MAC帧的前导码中每个字节的比特模式为 A.01111110C.10000001B.10101010 D.10101011 23.无线局域网中为实现传输介质共享而采用的协议是 A.CSMA／CAC.CSMA／CD 24.应对网络被动攻击通常采取的手段是 A.检测恢复 C.数据加密B.预防 D.认证 B.F'TP D.SDP非选择题部分注意事项：用黑色字迹的签字笔或钢笔将答案写在答题纸上，不能答在试题卷上。

04741计算机网络原理网络发展阶段：面向终端的计算机网络；计算机－计算机网络；开放式标准化网络；因特网广泛应用和高速网络技术发展。

三大网络：电信网络；广播电视网络；计算机网络。

网络发展趋势：宽带网络；全光网络；多媒体网络；移动网络；下一代网络。

电话系统组成：本地网络；干线；交换局。

ChinaNET：CHINAPAC；CHINADDN；PSTN。

文件共享、信息浏览、电子邮件、网络电话、视频点播、FTP、网上会议。

三网合一：把现在的传统电信网、广播电视网和计算机网互相融合，逐渐形成一个统一的网络系统，由一个全数字化的网络设施来支持包括数据、语音和图像在内的所有业务的通信。

高速网络技术表现：B-ISDN、ATM、高速局域网、交换局域网、虚拟网络。

宽带网络分为：宽带骨干网、宽带接入网。

骨干网：核心交换网，基于光纤通信系统的，能实现大范围的数据流传送。

接入类型：光纤、铜线、光纤同轴电缆混合接入、无线接入。

全光网络：以光节点取代现有网络的电节点，并用光纤将光节点互连成网，采用光波完成信号的传输、交换等功能，克服了现有网络在传输和交换时的瓶颈，减少信息传输的拥塞、延迟、提高网络呑吐量。

移动网络主要技术：蜂窝式数字分组数据通信平台；无线局域网；Ad hoc网络；无线应用协议WAP。

计算机网络：利用通信设备和线路将地理位置不同的、功能独立的多个计算机系统互连起来，以功能完善的网络软件实现网络中资源共享和信息传递的系统。

通信子网组成：网络节点、通信链路。

网络节点：也称转换节点、中间节点，作用是控制信息的传输和在端节点之间转发信息。

可以是：分组交换设备PSE、分组装配/拆卸设备PAD、集中器C、网络控制中心NCC、网间连接器G。

统称为接口信息处理机IMP。

存储－转发：信息在两端节点之间传输时，可能要经过多个中间节点的转发，这种方式称为存储转发。

计算机网络功能：硬件资源共享；软件资源共享；用户间信息交换。

计算机网络应用：办公自动化OA；远程教育；电子银行；证券及期货交易；校园网；企业网络；智能大厦和结构化综合布线系统。

第一章1.(Q1) What is the difference between a host and an end system? List the types ofendsystems. Is a Web server an end system?Answer: There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, Internet-connected PDAs, WebTVs, etc.2.(Q2) The word protocol is often used to describe diplomatic relations. Give an example of adiplomatic protocol.Answer: Suppose Alice, an ambassador of country A wants to invite Bob, an ambassador of country B, over for dinner. Alice doesn’t simply just call Bob on the phone and say, come to our dinner table now”. Instead, she calls Bob and sugges ts a date and time. Bob may respond by saying he’s not available that particular date, but he is available another date. Alice and Bob continue to send “messages” back and forth until they agree on a date and time. Bob then shows up at the embassy on the agreed date, hopefully not more than 15 minutes before or after the agreed time. Diplomatic protocols also allow for either Alice or Bob to politely cancel the engagement if they have reasonable excuses.3.(Q3) What is a client program? What is a server program? Does a server programrequestand receive services from a client program?Answer: A networking program usually has two programs, each running on a different host, communicating with each other. The program that initiates the communication is the client.Typically, the client program requests and receives services from the server program.4.(Q4) List six access technologies. Classify each one as residential access, company access, ormobile access.Answer:1. Dial-up modem over telephone line: residential; 2. DSL over telephone line: residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, 3G/4G): mobile5.(Q5) List the available residential access technologies in your city. For each type of access,provide the advertised downstream rate, upstream rate, and monthly price.Answer: Current possibilities include: dial-up (up to 56kbps); DSL (up to 1 Mbps upstream, up to 8 Mbps downstream); cable modem (up to 30Mbps downstream, 2 Mbps upstream.6.(Q7) What are some of the physical media that Ethernet can run over?Answer: Ethernet most commonly runs over twisted-pair copper wire and “thin” coaxial cable.It also can run over fibers optic links and thick coaxial cable.7.(Q8)Dial-up modems, HFC, and DSL are all used for residential access. For each of theseaccess technologies, provide a range of transmission rates and comment on whether the transmission rate is shared or dedicated.Answer:Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to1 Mbps, bandwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstreamchannel is usually less than a few Mbps, bandwidth is shared.8.(Q13)Why is it said that packet switching employs statistical multiplexing? Contraststatistical multiplexing with the multiplexing that takes place in TDM.Answer:In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.9.(Q14) Suppose users share a 2Mbps link. Also suppose each user requires 1Mbps whentransmitting, but each user transmits only 20 percent of the time. (See the discussion of statistical multiplexing in Section 1.3.)a.When circuit switching is used, how many users can be supported?b.For the remainder of this problem, suppose packet switching is used. Why will there beessentially no queuing delay before the link if two or fewer users transmit at the same time? Why will there be a queuing delay if three users transmit at the same time?c.Find the probability that a given user is transmitting.d.Suppose now there are three users. Find the probability that at any given time, allthree users are transmitting simultaneously. Find the fraction of time during which the queue grows.Answer:a. 2 users can be supported because each user requires half of the link bandwidth.b.Since each user requires 1Mbps when transmitting, if two or fewer users transmitsimultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three users transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this case, there will be queuing delay before the link.c.Probability that a given user is transmitting = 0.2d.Probability that all three users are transmitting simultaneously=33p3(1−p)0=0.23=0.008. Since the queue grows when all the users are transmitting, the fraction oftime during which the queue grows (which is equal to the probability that all three users are transmitting simultaneously) is 0.008.10.(Q16)Consider sending a packet from a source host to a destination host over a fixed route.List the delay components in the end-to-end delay. Which of these delays are constant and which are variable?Answer:The delay components are processing delays, transmission delays, propagation delays, and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable.11.(Q19) Suppose Host A wants to send a large file to Host B. The path from Host A to Host Bhas three links, of rates R1 = 250 kbps, R2 = 500 kbps, and R3 = 1 Mbps.a.Assuming no other traffic in the network, what is the throughput for the file transfer.b.Suppose the file is 2 million bytes. Roughly, how long will it take to transfer the file toHost B?c.Repeat (a) and (b), but now with R2 reduced to 200 kbps.Answer:a.250 kbpsb.64 secondsc.200 kbps; 80 seconds12.(P2)Consider the circuit-switched network in Figure 1.8. Recall that there are n circuits oneach link.a.What is the maximum number of simultaneous connections that can be in progress atany one time in this network?b.Suppose that all connections are between the switch in the upper-left-hand cornerand the switch in the lower-right-hand corner. What is the maximum number ofsimultaneous connections that can be in progress?Answer:a.We can n connections between each of the four pairs of adjacent switches. This gives amaximum of 4n connections.b.We can n connections passing through the switch in the upper-right-hand cornerandanother n connections passing through the switch in the lower-left-hand corner, giving a total of 2n connections.13.(P4) Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 50km/hour.a.Suppose the caravan travels 150 km, beginning in front of one tollbooth, passingthrough a second tollbooth, and finishing just before a third tollbooth. What is theend-to-end delay?b.Repeat (a), now assuming that there are five cars in the caravan instead of ten.Answer: Tollbooths are 150 km apart, and the cars propagate at 50 km/hr, A tollbooth services a car at a rate of one car every 12 seconds.a.There are ten cars. It takes 120 seconds, or two minutes, for the first tollbooth to servicethe 10 cars. Each of these cars has a propagation delay of 180 minutes before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 182 minutes. The whole process repeats itself for traveling between the second and third tollbooths. Thus the total delay is 364 minutes.b.Delay between tollbooths is 5*12 seconds plus 180 minutes, i.e., 181minutes. The totaldelay is twice this amount, i.e., 362 minutes.14.(P5) 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 the propagation speed along the link is s meters/sec. Host A is to send 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 queuing delays, obtain an expression for the end-to-enddelay.d.Suppose Host A begins to transmit the packet at time t = 0. At time t = d trans, where isthe 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 thepacket?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*108, L = 100bits, and R = 28kbps. Find the distance m so that dprop equals d trans .Answer:a. d prop = m/s seconds.b. d trans = L/R seconds.c. d end-to-end = (m/s + L/R) seconds.d.The bit is just leaving Host A.e.The first bit is in the link and has not reached Host B.f.The first bit has reached Host B.g.Wantm=LRS=10028∗1032.5∗108=893 km.15.(P6) In this problem we consider sending real-time voice from Host A to Host B over apacket-switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-Byte packets. There is one link between Host A and B; its transmission rate is 500 kbps and its propagation delay is 2 msec.As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packet’s bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded(as part of the analog signal at Host B)?Answer: Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in thepacket must be generated. This requires56∗8sec=7 msec64∗103The time required to transmit the packet is56∗8sec=896 μsec500∗103Propagation delay = 2 msec.The delay until decoding is7msec +896μsec + 2msec = 9.896msecA similar analysis shows that all bits experience a delay of 9.896 msec.16.(P9) Consider a packet of length L which begins at end system A, travels over one link to apacket switch, and travels from the packet switch over a second link to a destination end system. Let d i, s i, and R i denote the length, propagation speed, and the transmission rate of link i, for i= 1, 2. The packet switch delays each packet by d proc. Assuming no queuing delays, in terms of d i, s i, R i, (i= 1, 2), and L, what is the total end-to-end delay for the packet? Suppose now the packet Length is 1,000 bytes, the propagation speed on both links is 2.5 * 108m/s, the transmission rates of both links is 1 Mbps, the packet switch processing delay is 2 msec, the length of the first link is 6,000 km, and the length of the last link is 3,000 km. For these values, what is the end-to-end delay?Answer: The first end system requires L/R1to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay ofd proc; after receiving the entire packet, the packet switch requires L/R2to transmit the packetonto the second link; the packet propagates over the second link in d2/s2. Adding these five delays givesd end-end = L/R1 + L/R2 + d1/s1 + d2/s2 + d procTo answer the second question, we simply plug the values into the equation to get 8 + 8 +24 + 12 + 2= 54 msec.17.(P10) In the above problem, suppose R1 = R2 = R and d proc= 0. Further suppose the packetswitch does not store-and-forward packets but instead immediately transmits each bit it receivers before waiting for the packet to arrive. What is the end-to-end delay?Answer: Because bits are immediately transmitted, the packet switch does not introduce any delay;in particular, it does not introduce a transmission delay. Thus,d end-end = L/R + d1/s1 + d2/s2For the values in Problem 9, we get 8 + 24 + 12 = 44 msec.18.(P11) Suppose N packets arrive simultaneously to a link at which no packets are currentlybeing transmitted or queued. Each packet is of length L and the link has transmission rate R.What is the average queuing delay for the N packets?Answer:The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)L/R for the nth transmitted packet. Thus, the average delay for the N packets is(L/R + 2L/R + ....... + (N-1)L/R)/N = L/RN(1 + 2 + ..... + (N-1)) = LN(N-1)/(2RN) = (N-1)L/(2R) Note that here we used the well-known fact that1 +2 + ....... + N = N(N+1)/219.(P14) Consider the queuing delay in a router buffer. Let I denote traffic intensity; that is, I =La/R. Suppose that the queuing delay takes the form IL/R (1-I) for I<1.a.Provide a formula for the total delay, that is, the queuing delay plus the transmissiondelay.b.Plot the total delay as a function of L/R.Answer:a.The transmission delay is L / R . The total delay isILR(1−I)+LR=L/R1−Ib.Let x = L / R.Total delay=x 1−αx20.(P16) Perform a Traceroute between source and destination on the same continent at threedifferent hours of the day.a.Find the average and standard deviation of the round-trip delays at each of the threehours.b.Find the number of routers in the path at each of the three hours. Did the pathschange during any of the hours?c.Try to identify the number of ISP networks that the Traceroute packets pass throughfrom source to destination. Routers with similar names and/or similar IP addresses should be considered as part of the same ISP. In your experiments, do the largest delays occur at the peering interfaces between adjacent ISPs?d.Repeat the above for a source and destination on different continents. Compare theintra-continent and inter-continent results.Answer: Experiments.21.(P18) Suppose two hosts, A and B, are separated by 10,000 kilometers and are conn ectedby a direct link of R =2 Mbps. Suppose the propagation speed over the link is 2.5•108 meters/sec.a.Calculate the bandwidth-delay product, R•d prop.b.Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sentcontinuously as one large message. What is the maximum number of bits that will be in the link at any given time?c.Provide an interpretation of the bandwidth-delay product.d.What is the width (in meters) of a bit in the link? Is it longer than a football field?e.Derive a general expression for the width of a bit in terms of the propagation speed s,the transmission rate R, and the length of the link m.Answer:a.d prop= 107 / 2.5•108= 0.04 sec; so R •d prop= 80,000bitsb.80,000bitsc.The bandwidth-delay product of a link is the maximum number of bits that can be inthelink.d. 1 bit is 125 meters long, which is longer than a football fielde.m / (R •d prop ) = m / (R * m / s) = s/R22.(P20) Consider problem P18 but now with a link of R = 1 Gbps.a.Calculate the bandwidth-delay product,R·d prop .b.Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sentcontinuously as one big message. What is the maximum number of bits that will be inthe link at any given time?c.What is the width (in meters) of a bit in the link?Answer:a.40,000,000 bits.b.400,000 bits.c.0.25 meters.23.(P21) Refer again to problem P18.a.How long does it take to send the file, assuming it is sent continuously?b.Suppose now the file is broken up into 10 packet is acknowledged by the receiver andthe transmission time of an acknowledgment packet is negligible. Finally, assumethat the sender cannot send a packet until the preceding one is acknowledged. Howlong does it take to send the file?pare the results from (a) and (b).Answer:a. d trans + d prop = 200 msec + 40 msec = 240 msecb.10 * (t trans + 2 t prop ) = 10 * (20 msec + 80 msec) = 1.0sec。

04741计算机网络原理复习题及答案计算机网络原理复习资料（课程代码04741）一、单项选择题1．双绞线绞合的目的是【】A．增大传输距离B．减少相互间的辐射电磁干扰C．提高传送速度D．增大抗拉强度2．按照网络的地理分布范围来分类，下列哪种网络类型不属于这种分类范畴。

【】A．广域网(WAN) B．城域网(MAN)C．因特网 (Internet) D．局域网(LAN)3．下列不属于网络协议要素的是，【】A．语义B．定时C．语法D．状态4．在划分网络的层次体系结构时，一般要遵循的原则中不包含下列哪一条【】A．每层的功能应是明确的B．层间接口必须清晰，跨越接口的信息量应尽可能少C．尽可能满足以有的网络产品特点D．层数应适中5．下面陈述中，正确的是( )【】A．转发器和网桥都能够提高网络带宽，扩大网络地理范围B．转发器和网桥都不能提高网络带宽，但能扩大网络地理范围C．转发器不能提高网络带宽，网桥能够提高网络带宽D．转发器能够提高网络带宽，网桥不能提高网络带宽6．面向连接服务与无连接服务比较，下面哪一条不是面向连接的优点。

【】A．各分组不需要携带目的节点地址B．数据传输的可靠性高C．数据传输的收发数据顺序不变D．协议简单，通信效率高7．TCP/IP协议簇中和IP协议配套的三个协议是【】A．DNS，RARP，ICMP B．DNS，ARP，ICMPC．ICMP，RARP，ARP D．UDP，ARP，ICMP8．TCP/IP协议模型与OSI模型比较，TCP/IP协议模型中不包含下面哪一层【】A．网络层B．传输层C．应用层D．表示层9．网络的物理层接口协议设计不包含下面哪项特性【】A．机械的B．兼容的C．电气的D．规程性的10．在IPV4因特网协议中，合法的32位IP地址是【】A．202∶110∶112∶50 B．202、110、112、50C．202，110，112，50 D．202. 110. 112. 5011．要使模拟消息能够在数字信道上传输，须使用（）技术。

自考04741《计算机网络原理》课后习题答案第一章PSE：分组交换设备PAD：分组装配、拆卸装备NCC：网络控制中心FEP：前端处理机IMP：接口信息处理机PSTN：电话交换网ADSL：非对称用户环路DDN：数字数据网FR：帧中继ATM：异步转移模式ISDN：综合服务数字网VOD：电视点播WAN：广域网LAN：局域网MAN：城域网OSI：开放系统互连基本模型ITU：国际电信联盟IETF：英特网工程特别任务组第2章1.说明协议的基本含义，三要素的含义与关系。

答：为计算机网络中进行数据交换而建立的规则、标准或约定的集合就称为协议。

协议三要素：(1)语义：涉及用于协调与差错处理的控制信息。

(2)语法：涉及数据及控制信息的格式、编码及信号电平等。

(3)定时：涉及速度匹配和排序等。

3.计算机网络采用层次结构模型的理由是什么？有何好处？答：计算机网络系统是一个十分复杂的系统。

将一个复杂系统分解为若干个容易处理的子系统，然后“分而治之”逐个加以解决，这种结构化设计方法是工程设计中常用的手段。

分层就是系统分解的最好方法之一。

分层结构的好处在于使每一层实现一种相对独立的功能。

每一层的功能相对简单而且易于实现和维护。

具有很大的灵活性。

分层结构有利于交流、理解和标准化。

6.请比较面向连接服务和无连接服务的异同点。

答：面向连接服务和电话系统的工作模式相类似。

数据传输过程前必须经过建立连接、维护连接和释放连接的3个过程；在数据传输过程中，各分组不需要携带目的的节点的地址。

面向连接数据传输的手法数据顺序不变，传输可靠性好，需通信开始前的连接开销，协议复杂，通信效率不高。

无连接服务与邮政系统的信件投递过程相类似。

每个分组都是要携带完整的目的节点的地址，各分组在通信子网中是独立传送的。

数据传输过程不需要经过建立连接、维护连接和释放连接的3个过程；目的节点接收到的数据分组可能出现乱序、重复与丢失的现象。

可靠性不是很好，通信协议相对简单、效率较高。

04741计算机网络原理选择题及答案第一篇：04741计算机网络原理选择题及答案计算机网络原理选择题1.在Internet与Intranet之间起到检查网络服务请求合法性的设备是（A）A．防火墙 B．防病毒软件 C．网卡 D．网桥 2.网络协议主要要素为（C）A.数据格式.编码.信号电平B.数据格式.控制信息.速度匹配C.语法.语义.同步D.编码.控制信息.同步 3.TCP/IP体系结构中的TCP和IP所提供的服务分别为（D）。

A.链路层服务和网络层服务B.网络层服务和传输层服务C.传输层服务和应用层服务D.传输层服务和网络层服务4.在TCP/IP协议簇中，UDP协议(用户数据报协议)工作在（B）。

A.应用层B.传输层C.网络互联层D.网络接口层5.对等层间交换的数据单元称之为协议数据单元，其英文缩写为（C）。

A.SDUB.IDUC.PDUD.ICI 6.当一台计算机从FTP服务器下载文件时，在该FTP服务器上对数据进行封装的五个转换步骤是（B）A.比特，数据帧，数据包，数据段，数据B.数据，数据段，数据包，数据帧，比特C.数据包，数据段，数据，比特，数据帧D.数据段，数据包，数据帧，比特，数据7.在计算机网络中，一般局域网的数据传输速率要比广域网的数据传输速率（A）A.高B.低C.相同D.不确定8.IP地址是一个32位的二进制，它通常采用点分（C）。

A.二进制数表示B.八进制数表示C.十进制数表示D.十六进制数表示 9.传输层可以通过（B）标识不同的应用。

A.物理地址B.端口号C.IP地址D.逻辑地址 10.创建用户帐户是在（B）建立的A．活动目录控制台B．“用户属性”对话框C．“组属性”对话框中成员属于选项卡D．“添加组策略”对话框 11.DNS的作用是（B）。

A.用来将端口翻译成IP地址B.用来将域名翻译成IP地址C.用来将IP地址翻译成硬件地址D.用来将MAC翻译成IP地址12.浏览器与Web服务器之间使用的协议是（C）。

计算机网络原理课后答案（前四章）第一章1.P12.资源子网和通信子网.P83.P124.P165.P166.P17.7.P108.略。

第二章1.P192. 协议是控制两个对等实体进行通信的规则的集合。

在协议的控制下，两个对等实体间的通信使得本层能够向上一层提供服务，而要实现本层协议，还需要使用下面一层提供服务。

协议和服务的概念的区分：1、协议的实现保证了能够向上一层提供服务。

本层的服务用户只能看见服务而无法看见下面的协议。

下面的协议对上面的服务用户是透明的。

2、协议是“水平的”，即协议是控制两个对等实体进行通信的规则。

但服务是“垂直的”，即服务是由下层通过层间接口向上层提供的。

上层使用所提供的服务必须与下层交换一些命令，这些命令在OSI中称为服务原语。

3.P204.P205.P266.P237.P278.P279.P27..第三章1.P292.P343.P414.P515.6KHZ6.P557.P478.P639.( 50*10^3)/2*10^8=2.5*10^ 4秒800/x=5/2*10^410. 用直通交换，交换机延迟分组128位，即0.128μs。

在这种情况下仍然有1个5μs的发送延迟，2个10μs的传播延迟，再加上0.128μs的交换机转发延迟，因此总的延迟等于：5×1+10×2+0.128=25.128μs （1个发送延迟，2个传播延迟，1个转发延迟）如果像（b）那样有3个交换机，那么总的延迟将会等于：5×1+10×4+0.128×3=45.384μs。

（1个发送延迟，4个传播延迟，3个转发延迟）11. 1)55*10^9/3*10^8 (2)(5*10^3/128)+(55*10^9/3*10^8)12. (1)1920*1080*24*30 (2)8000*8Bit=64000 B/秒=64 KB/秒(3)260*50第四章1.帧同步功能，差错控制功能，流量控制功能，链路管理功能2.P713.由躁声引起，两大类：随机热噪声，冲击噪声。

⾃考04741《计算机⽹络原理》课后习题答案第１章节计算机⽹络概述1．计算机⽹络的发展可以分为哪⼏个阶段？每个阶段各有什么特点？A ⾯向终端的计算机⽹络：以单个计算机为中⼼的远程联机系统。

这类简单的“终端—通信线路—计算机”系统，成了计算机⽹络的雏形。

B 计算机—计算机⽹络：呈现出多处中⼼的特点。

C 开放式标准化⽹络：OSI/RM 的提出，开创了⼀个具有统⼀的⽹络体系结构，遵循国际标准化协议的计算机⽹络新时代。

D 因特⽹⼴泛应⽤和⾼速⽹络技术发展：覆盖范围⼴、具有⾜够的带宽、很好的服务质量与完善的安全机制，⽀持多媒体信息通信，以满⾜不同的应⽤需求，具备⾼度的可靠性与完善的管理功能。

2．计算机⽹络可分为哪两⼤⼦⽹？它们各实现什么功能？通信⼦⽹和资源⼦⽹。

资源⼦⽹负责信息处理，通信⼦⽹负责全⽹中的信息传递。

3．简述各种计算机⽹络拓扑类型的优缺点。

星形拓扑结构的优点是：控制简单；故障诊断和隔离容易；⽅便服务，中央节点可⽅便地对各个站点提供服务和⽹络重新配置。

缺点是：电缆长度和安装⼯作量客观；中央节点的负担较重形成“瓶颈”；各站点的分布处理能⼒较低。

总线拓扑结构的优点是：所需要的电缆数量少；简单⼜是⽆源⼯作，有较⾼的可靠性；易于扩充增加或减少⽤户⽐较⽅便。

缺点是：传输距离有限，通信范围受到限制；故障诊断和隔离较困难；分布式协议不能保证信息的及时传输，不具有实时功能。

树形拓扑结构的优点是：易于扩展、故障隔离较容易，缺点是：各个节点对根的依赖性太⼤。

环形拓扑结构的优点是：电缆长度短；可采⽤光纤，光纤的传输率⾼，⼗分适合于环形拓扑的单⽅向传输；所有计算机都能公平地访问⽹络的其它部分，⽹络性能稳定。

缺点是：节点的故障会引起全⽹故障；环节点的加⼊和撤出过程较复杂；环形拓扑结构的介质访问控制协议都采⽤令牌传递的⽅式，在负载很轻时，信道利⽤率相对来说就⽐较低。

混合形拓扑结构的优点是：故障诊断和隔离较为⽅便；易于扩展；安装⽅便。