计算机网络课后作业以及答案(中英文对照)
计算机网络课后题答案
计算机网络课后题答案1. 简答题1.什么是计算机网络？计算机网络是指将地理位置不同的计算机和其他设备通过通信链路互相连接起来，以共享资源、传输数据和信息，实现数据交换和通信的技术系统。
2.简述计算机网络的基本组成部分。
计算机网络的基本组成部分包括：终端设备、通信链路、交换设备和协议。
•终端设备：指连接到网络中的计算机、手机或其他网络设备，例如个人计算机、路由器、交换机等。
•通信链路：指物理媒介或无线信道，用于连接终端设备。
通信链路可以是有线的，如双绞线、光纤等；也可以是无线的，如无线局域网、蓝牙等。
•交换设备：指用于在网络中转发信息的设备，如交换机、路由器等。
交换设备主要负责将信息从一个端点传输到另一个端点。
•协议：指网络通信中规定了双方之间通信行为的标准，包括物理层、数据链路层、网络层、传输层和应用层等。
3.简述OSI参考模型。
OSI参考模型是一个由国际标准化组织（ISO）制定的网络通信协议参考模型，它将计算机网络通信划分为七个不同的层次，每层负责不同的功能：1.物理层：负责物理设备之间的原始比特流传输，如电缆和接口。
2.数据链路层：负责传输数据帧，错误检测和纠错等。
3.网络层：负责数据分组的路由和转发，实现不同网络之间的通信。
4.传输层：负责端到端的可靠数据传输，提供流控制和差错恢复。
5.会话层：负责建立、管理和终止会话。
6.表示层：负责数据的格式化、加密和压缩。
7.应用层：提供与用户交互的网络服务和应用程序。
OSI参考模型的分层设计使得不同的网络功能可以独立地开发和升级，提高了网络的可扩展性和互操作性。
2. 计算题1.假设一台主机A的IP地址为192.168.1.10，子网掩码为255.255.255.0，那么主机A所在的网络号是多少？主机A的IP地址为192.168.1.10，子网掩码为255.255.255.0。
根据子网掩码的规则，将IP地址和子网掩码进行AND运算，可得主机A所在的网络地址为192.168.1.0。
(完整版)计算机网络课后作业以及答案(中英文对照)
Chapter11-11.What are two reasons for using layered protocols?(请说出使用分层协议的两个理由)答：通过协议分层可以把设计问题划分成较小的易于处理的片段。
分层意味着某一层的协议的改变不会影响高层或低层的协议。
1-13. What is the principal difference between connectionless communication and connection-oriented communication?(在无连接通信和面向连接的通信两者之间，最主要的区别是什么？)答：主要的区别有两条。
其一：面向连接通信分为三个阶段，第一是建立连接，在此阶段，发出一个建立连接的请求。
只有在连接成功建立之后，才能开始数据传输，这是第二阶段。
接着，当数据传输完毕，必须释放连接。
而无连接通信没有这么多阶段，它直接进行数据传输。
其二：面向连接的通信具有数据的保序性，而无连接的通信不能保证接收数据的顺序与发送数据的顺序一致。
1-20. A system has an n-layer protocol hierarchy. Applications generate messages of length M bytes. At each of the layers, an h-byte header is added. What fraction of the network bandwidth is filled with headers?(一个系统有n层协议的层次结构。
应用程序产生的消息的长度为M字节。
在每一层上需要加上一个h字节的头。
请问，这些头需要占用多少比例的网络带宽)答：hn/(hn+m)*100%1-28. An image is 1024 x 768 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet?(一幅图像的分辨率为1024 x 768像素，每个像素用3字节来表示。
计网课后习题答案(英文)
Review Questions第一章1.Which characteristics of a multiterminal system make it different from a computer network?A processing power of multiterminal systems remained fully centralized, whilecomputer networks have a distributed processing power.一个多端系统处理能力仍然完全集中，而计算机网络具有分布式处理能力。
2.When were the first important results achieved in the field of joining computers using long-haul links?In the late 60s在60年代后期3.What is ARPANET?A.A network of supercomputers belonging to military organizations and research institutes in the United StatesB.An international scientific research networkC.The technology of creating WANsA is correct4.When did the first network operating systems appear?In the late 60s60年代后期5.In what order did the events listed here take place?A.The invention of WebB.The development of standard LAN technologiesC.The start of voice transmission in digital form through telephone networksThe invention of World Wide Web互联网的诞生6.Which of the events stimulated LAN development?Appearance of large-scale integrated circuits (LSI devices) resulted in invention of microcomputers which in its turn stimulated a research activity in the area of local computer networks (LANs). The adoption of personal computers was a powerful incentive for the development of LANs.大规模集成电路（LSI器件出现）导致这反过来刺激了研究活动在本地计算机网络（LAN）的微型计算机的发明。
计算机网络中英文互译
DWDM (Deｎｓe WDM) 密 集 波 分 复 用
AＲQ (Autｏmaｔic Rｅpeａt reQｕest) 自 动 请 求 重 发
ATM (Ａsynchronous Ｔransｆeｒ Moｄe) 异 步 传 递 方 式
ATU (Aｃceｓs Tｅrminatioｎ Unit) 接 入 端 接 单 元
ATU-C (Accesｓ Ｔermination Unit Ceｎtraｌ Ofｆicｅ ) 端 局 接 入 端 接 单 元
ＣCIＴT (Cｏnsultatiｖe Commｉｔtee, Intｅrnａtional Tｅlｅgraph and Tｅlｅpｈone）国际电报电话咨询委员会
CCＰ 通信控制处理机
CＤＭ (Cｏde Division Mulｔiplexinｇ) 码 分 复 用
CDMA (Ｃoｄe Ｄivisioｎ Ｍuｌtipｌex Aｃｃｅｓs) 码 分 多 址 ﻫCNNＩC (Neｔwｏｒｋ Information Centeｒ of China) 中 国 互 联 网 络 信 息 中 心
CＡP (Cａｒrierｌｅss Amplituｄe Phaｓe) 无 载 波 振 幅 相 位 调 制 ﻫCATV (Comｍunｉty Antenna TV, CAble ＴV) 有 线 电 视
CＢＲ ( Constant Bit Ｒatｅ ) 恒 定 比 特 率 ﻫCCIR （Ｃｏｎsultatiｖe Ｃommiｔｔｅe，Ｉnternational Rａdio） 国 际 无 线 电 咨 询 委 员 会
计算机网络课后习题答案完整版
的大型机工作特点是所有处理都发生在主机上。
（2）、浏览器/服务器(B／S)模式：主要特点是它与软硬件平台的无关性，把应用逻辑和业
务处理规则放在服务器一侧。
（3）、对等网或称为对等式的网络：对等网可以不要求具备文件服务器，特别是应用在一组
答：边缘部分由所有连接在因特网上的主机组成。这部分是用户直接使用的，用来进行通信（传送数据、音频或视频）和资源共享。核心部分由大量网络和连接这些网络的路由器组成。这部分是为边缘部分提供服务的（提供连通性和交换）。在网络边缘的端系统中运行的程序之间的通信方式通常可划分为两大类：客户服务器方式（C/S
方式）即Client/Server方式，对等方式（P2P方式）即Peer-to-Peer
switching)的关键构件，其任务是转发收到的分组，这是网络核心部分最重要的功能
1-13客户服务方式与对等通信方式的主要区别是什么？有没有相同的地方？
答：客户服务器方式是一点对多点的，对等通信方式是点对点的。被用户调用后运行，在打算通信时主动向远地服务器发起通信（请求服务）。因此，客户程序必须知道服务器程序的地址。系统启动后即自动调用并一直不断地运行着，被动地等待并接受来自各地的客户的通信请求。因此，服务器程序不需要知道客户程序的地址。对等连接方式从本质上看仍然是使用客户服务器方式，只是对等连接中的每一个主机既是客户又同时是服务器。对等连接也需要知道对方的服务器地址。
（3）分组交换分组交换实质上是在“存储——转发”基础上发展起来的。它兼有电路交换和报文交换的优点。分组交换在线路上采用动态复用技术传送按一定长度分割为许多小段的数据——分组。每个分组标识后，在一条物理线路上采用动态复用的技术，同时传送多个数据分组。把来自用户发端的数据暂存在交换机的存储器内，接着在网内转发。到达接收端，再去掉分组头将各数据字段按顺序重新装配成完整的报文。分组交换比电路交换的电路利用率高，比报文交换的传输时延小，交互性好。
计算机网络英文版——提供给学生部分习题答案
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)。
计算机网络(第四版)课后习题(英文)+习题答案(中英文)
ANDREW S. TANENBAUM由于请求和应答都必须通过卫星，因此传输总路径长度为和真空中的光速为300，000 公里/秒，因此最佳的传播延迟为160,000/300，000medium-speed line, a low-speed line, or no line. If it takes 100 ms of computer timeto generate and inspect each topology, how long will it take to inspect all ofthem?(E)将路由器称为A，B，C，D 和 E.则有10 条可能的线路；AB, AC, AD, AE, BC, BD, BE, CD, CE,和DE每条线路有 4 种可能性(3 速度或者不是线路)，拓扑的总数为410 = 1,048,576。
检查每个拓扑需要100 ms，全部检查总共需要104,857. 6 秒，或者稍微超过29 个小时。
9. A group of 2n - 1 routers are interconnected in a centralized binary tree, with a router at each tree node. Router i communicates with router j by sending a message to the root of the tree. The root then sends the message back down to j. Derive an approximate expression for the mean number of hops per message for large n, assuming that all router pairs are equally likely.(H)这意味着，从路由器到路由器的路径长度相当于路由器到根的两倍。
《计算机网络》中英词汇对照表
《计算机网络》中英文对照表Chapter 11.1Internet：因特网Computer network ：计算机网络Host: 主机End system: 终端系统Packet switching: 分组交换Route: 路径Internet service provider (ISP): 因特网服务提供商Protocol: 协议Transmission Control Protocol (TCP)：传输控制协议1.2Client: 客户端Server: 服务器Peer: 对等机Reliable data transfer: 可靠数据传输Flow control: 流量控制Congestion-control: 拥塞控制User Datagram Protocol (UDP): 用户数据报协议1.3Circuit switching: 电路交换/线路交换Packet switching: 分组交换Frequency-division multiplexing (FDM): 频分多路复用Time-division multiplexing (TDM): 时分多路复用Bandwidth: 带宽Time slot: 时隙Frame: 帧Message: 报文:Packet: 分组Store-and-forward: 存储转发Datagram network: 数据报网络Virtual-circuit network: 虚电路网络1.4Router: 路由器Modem: 调制解调器Local area network (LAN): 局域网Ethernet: 以太网Wireless LAN: 无线局域网Guided media: 导向型介质Unguided media: 非导向型介质Twisted-pair copper wire: 双绞线Unshielded twisted pair(UTP): 非屏蔽双绞线Coaxial cable: 同轴电缆Fiber optics: 光线/光缆1.6Nodal processing delay: 结点处理延迟Queuing delay: 排队延迟Transmission delay: 发送延迟Propagation delay: 传播延迟Traffic intensity: 流通强度End-to-end delay: 端到端延迟1.7Layer: 层次Protocol stack: 协议栈Application layer: 应用层Transport layer: 传输层Network layer: 网络层Link layer: 链路层Physical layer: 物理层Encapsulation: 封装Message: 报文Segment: 报文段Datagram: 数据报Frame: 帧Chapter 22.1Client-server architecture: 客户端－服务器体系结构；C/S结构P2P architecture: 对等结构Processes: 进程Socket: 套接字Application programming interface (API): 应用程序编程接口IP address: IP地址Prot number: 端口号Syntax: 语法Semantics: 语义Full-duplex: 全双工Handshaking: 握手Real-time application: 实时应用2.2The World Wide Web: 万维网HyperText Transfer Protocol (HTTP): 超文本传输协议Web page: 网页Object: 对象HyperText Markup Language (HTML): 超文本标记语言URL：统一资源定位符Browser: 浏览器Persistent connection: 持久连接Non-persistent connection: 非持久连接Round-trip time (RTT): 往返时间Without pipelining: 非流水线方式With pipelining: 流水线方式Web cache: web 缓存Proxy server: 代理服务器2.3File Transfer Protocol (FTP): 文件传输协议Control connection: 控制连接Data connection: 数据连接Out-of-band: 带外In-band: 带内2.4Electronic Mail: 电子邮件User agent: 用户代理Mail server: 邮件服务器Simple Mail Transfer Protocol (SMTP): 简单邮件传输协议Mailbox: 邮箱Multipurpose Internet Mail Extensions (MIME): 多用途因特网邮件扩展协议Post Office Protocol (POP): 邮局协议Internet Mail Access Protocol (IMAP): Internet 邮件访问协议2.5Domain Name System (DNS): 域名系统Hostname: 主机名Host aliasing: 主机别名Mail server aliasing: 邮件服务器别名Load distribution: 负载分配Root DNS server: 根DNS服务器Top-Level Domain (TLD) servers: 顶级域DNS服务器Authoritative DNS servers: 授权DNS服务器；权威DNS服务器Local DNS server: 本地DNS服务器Database: 数据库Chapter 33.1Logical communication: 逻辑通讯3.2Multiplexing: 多路复用Demultiplexing: 多路分解Well-known port number: 众所周知的端口号3.3UDP segment: UDP报文段Checksum: 校验和；检查和Wrapped around: 回卷3.4Channel: 通道；信道Positive acknowledgement : 肯定应答Negative acknowledgement: 否定应答ARQ （automatic repeat request）: 自动重传请求Feedback: 反馈Retransmission: 重传Stop-and-wait protocol: 停止－等待协议Duplicate packets: 冗余分组Sequence number: 顺序号Timer: 定时器Alternating-bit protocol: 比特交替协议Utilization: 利用率Go-back-N (GBN): 回退N步Window size: 窗口大小Sliding-window protocol: 滑动窗口协议Cumulative acknowledgement: 累积确认Timeout: 超时Selective Repeat (SR): 选择重传3.5Connection-oriented: 面向连接Point-to-point: 点到点Three-way handshake: 三次握手Maximum segment size (MSS): 最大报文段大小Maximum transmission unit (MTU): 最大传输单元Piggybacked: 捎带Sample RTT: 样本RTTFast retransmit: 快速重传Selective acknowledgement: 选择确认Flow-control: 流量控制Receive window: 接收窗口3.7Congestion control: 拥塞窗口Self-clocking: 自定时的Additive-increase, multiplicative-decrease: 加性增，乘性减Slow star: 慢启动Congestion avoidance: 拥塞避免Threshold: 阈值Fast recovery: 快速恢复Bottleneck: 瓶颈Latency: 延迟Chapter 44.1Forwarding: 转发Routing: 路由Routing algorithm: 路由算法Forwarding table: 转发表Router: 路由器Jitter: 抖动Best-effort service: 尽力而为的服务4.2Virtual-circuit (VC) network: 虚电路网络Datagram network: 数据报网络Prefix: 前缀Longest prefix matching rule: 最长前缀匹配规则4.3Input port: 输入端口Switching fabric: 交换结构Routing processor: 路由处理器Crossbar: 交叉结构4.4Time-to-live (TTL) :生存时间Fragmentation: 分片；片段Dotted-decimal notation: 点分十进制表示法Subnet: 子网Subnet mask: 子网掩码Classless Interdomain Routing (CIDR): 无类别域际路由选择Dynamic Host Configuration Protocol（DHCP）：动态主机配置协议Plug-and-play: 即插即用Network address translation (NA T): 网络地址转换Internet Control Message Protocol (ICMP): 因特网控制报文协议Dual-stack: 双栈Tunneling: 隧道4.5Default router: 默认路由器Graph: 图A global routing algorithm : 全局路由算法A decentralized routing algorithm : 分布式路由算法Static routing algorithm: 静态路由算法Dynamic routing algorithm : 动态路由算法Link-State (LS): 链路状态Distance-Vector(DV): 距离向量Routing table: 路由表Autonomous system (AS): 自治系统Intra-autonomous system routing protocol: 自治系统内路由协议Inter-AS routing protocol: 自治系统间路由协议4.6Interior gateway protocol: 内部网关协议Routing Information Protocol (RIP): 路由信息协议Open Shortest Path First (OSPF): 开放最短路径优先协议Advertisement: 公告Hop: 跳Border Gateway Protocol (BGP): 边界网关协议4.7Broadcast: 广播Multicast: 多播Chapter 55.1Node: 结点Link: 链路Frame: 帧Medium access control (MAC): 介质访问控制Full-duplex: 全双工Half-duplex: 半双工Adapter: 适配器Network interface card (NIC): 网卡Interface: 接口5.2Parity check: 奇偶校验Odd: 奇数Even: 偶数Cyclic redundancy check (CRC): 循环冗余校验Polynomial: 多项式5.3Collide: 冲突Multiple access protocol: 多路访问协议Channel partitioning protocol: 信道划分协议Random access protocol: 随机访问协议Taking-turns protocol: 轮转协议Code division multiple access (CDMA): 码分多址访问Carrier sensing: 载波侦听Collision detection: 冲突检测Polling protocol: 轮询协议Token-passing protocol: 令牌传递协议Token: 令牌Local Area Network (LAN): 局域网Token-ring: 令牌环Fiber distributed data interface (FDDI): 光纤分布式数据接口Metropolitan Area Network (MAN): 城域网5.4Address Resolution Protocol (ARP): 地址解析协议Dynamic Host Configuration Protocol (DHCP): 动态主机配置协议5.5Ethernet: 以太网Preamble: 前导码Manchester encoding: 曼彻斯特编码5.6Hub: 集线器Collision domain: 冲突域Switch: 交换机Filtering: 过滤Forwarding: 转发Switch table: 交换表Self-learning: 自学习Plug-and-play devices: 即插即用设备Cut-through switching: 直通式交换5.7Point-to-point: (PPP): 点到点。
计算机网络中英文名词对照
计算机⺴⽹网络关键字英⽂文中⽂文对照第⼀一章-协议（Protocol）-⺴⽹网络边缘（Network Edge）-有线⺴⽹网络，⽆无线⺴⽹网络（wired，wireless communication links ）-拨号调制解调器（dial-up Modem）-数字⽤用户线（digit subscriber line，DSL）-电缆调制解调器（cable modem）-混合光纤同轴电缆（hybrid fiber coax，HFC）-双绞线（Twisted Pair）-同轴电缆（Coaxial cable）-光缆（Fiber optic cable）-电路交换（circuit switching）-分组交换（packet switching）-端到端连接（end-to-end connection）-频分多路复⽤用（Frequency-Division Multiplexing，FDM）-时分多路复⽤用（Time-Division Multiplexing，TDM）-统计多路复⽤用（statistical multiplexing）-第⼀一层ISP（tier-1 ISP）-时延（delay）、节点处理时延（nodal processing delay）、排队时延（queuing delay）、传输时延（transmission delay）、传播时延（propagation delay）、节点总时延（total nodal delay）-丢包（package loss）-吞吐量（Troughput）-应⽤用层（application layer）、表⽰示层（presentation）、会话层（session）、传输层（transport layer）、⺴⽹网络层（network layer）、链路层（link layer）、物理层（physical layer）-报⽂文（message）、报⽂文段（segment）、数据报（datagram）、帧（frame）-HTTP——Hyper-Text Transfer Protocol-FTP——File Transfer Protocol-SMTP——Simple Mail Transfer Protocol-DNS——Domain Name Service-TCP——Transmission Control Protocol•⾯面向连接——Connection-Oriented（三次握⼿手）•数据可靠——Reliable data transport•拥塞控制——Congestion Control——抑制发送进程•流量控制——Flow Control•如email（SMTP）、远程终端访问（Telnet）、Web（HTTP）、⽂文件传输（FTP）、流媒体（HTTP、RTP）⺴⽹网络层使⽤用TCP协议-UDP——User Datagram Protocol•不可靠数据传输服务•如流媒体（HTTP、RTP）、因特⺴⽹网电话（SIP、RTP）通常⺴⽹网络层使⽤用UDP-IP——Internet Protocol-⽊木⻢马（Trojan horse）、病毒（Virus）、蠕⾍虫（Worm）、DDoS（Distributed Deny of Service）第⼆二章应⽤用层-客户机-服务器（Client-Server）、P2P-client process ：process that initiates communication-server process：Process that waits to be contacted-套接字（Socket）-传输层协议提供的服务•Reliable data transfer•Throughput•Timing•Security-HTTP•⽆无状态协议（stateless）•⾮非持久连接、短连接（non-persistent connection）•持久连接、⻓长连接（persistent connection）•pull protocol-往返时间（Round-Trip Time，RTT）-Cookie-server agent-SMTP•push protocol-多⽤用途因特⺴⽹网邮件扩展（Multipurpose Internet Mail Extension，MIME）-DNS•运⾏行在UDP上，使⽤用53号端⼝口•规范和别名•负载分配•主机名->IP地址转换-P2P•直接在对等⽅方传输•⾼高度的可拓展能⼒力•使⽤用客户机-服务器模型第三章运输层-逻辑通信（logical communication）-多路复⽤用（transport-layer multiplexing）与多路分解（demultiplexing）-源端⼝口号字段（source port number field）-⺫⽬目的端⼝口号字段（destination port number field）-周知端⼝口号（well-known port number）-Reliability data transport•完全可靠信道上的可靠数据传输 rdt1.0•具有⽐比特差错信道上的可靠数据传输 rdt2.0——stop-wait protocol•具有⽐比特差错的丢包信道上的可靠数据传输 rdt3.0——alternating-bit protocol•流⽔水线可靠数据协议-Go-Back_N——窗⼝口⼤大⼩小最⼤大2^k-1-选择性重传（selection repeat）窗⼝口⼤大⼩小最⼤大2^(k-1)-拥塞控制（congestion control）-流量控制（flow control）-第四章-转发（forwarding）-选路（routing）-虚电路（Virtual-Circuit，VC）⺴⽹网络：仅在⺴⽹网络层提供连接服务的计算机⺴⽹网络，如ATM、帧中继。
计算机网络课后习题答案完整版
信息资源和硬件资源，组网的计算机一般类型相同，甚至操作系统也相同，这种网络方式灵
活方便，但是较难实现集中管理与控制，安全性也低。
7、按企业和公司管理分类：
（1）、内部网：一般指企业内部网，自成一体形成一个独立的网络。
（2）、内联网：一般指经改造的或新建的企业内部网，采用通用的TCP／IP作为通信协议，
答：计算机网络向用户提供的最重要的功能有两个，连通性和共享。
1-2试简述分组交换的特点
答：分组交换实质上是在“存储——转发”基础上发展起来的。它兼有电路交换和报文交换的优点。分组交换在线路上采用动态复用技术传送按一定长度分割为许多小段的数据——分组。每个分组标识后，在一条物理线路上采用动态复用的技术，同时传送多个数据分组。把来自用户发端的数据暂存在交换机的存储器内，接着在网内转发。到达接收端，再去掉分组头将各数据字段按顺序重新装配成完整的报文。分组交换比电路交换的电路利用率高，比报文交换的传输时延小，交互性好。
对等网可以不要求具备文件服务器特别是应用在一组面向用户的pc机每台客户机都可以与其他每台客户机实现平等对话操作共享彼此的信息资源和硬件资源组网的计算机一般类型相同甚至操作系统也相同这种网络方式灵活方便但是较难实现集中管理与控制安全性也低
《计算机网络》课后习题答案<完整版>
第一章概述
1-1计算机网络向用户可以提供哪些服务？
具，其工作特点是文件从服务器被下载到工作站上，然后在工作站上进行处理，而基于主机
的大型机工作特点是所有处理都发生在主机上。
（2）、浏览器/服务器(B／S)模式：主要特点是它与软硬件平台的无关性，把应用逻辑和业
计算机网络英文版习题答案
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。
计算机网络英文版习题答案
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。
计算机网络系统方法英文版课后习题及解答
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）。
计算机网络专业词汇中英对照
《计算机网络》中英文对照Chapter 1End system P28 端系统Modem P29 调制解调器（俗称：猫）Base station P29 基站Communication link P30 通信链路Physical media P30 物理介质Coaxial cable P30 同轴电缆Fiber optics P30 光纤Radio spectrum P30 射频频谱Transmission rate P30 传输速率Packets P30 （数据）包，或分组Routers P30 路由器Link-layer switches P30 链路层交换机Path P30 路径ISP (Internet Service Provider) P30 网络服务提供商TCP (Transmission Control Protocol) P31 传输控制协议IP ( Internet Protocol) P31 网际协议Intranets P31 内网API (Application Programming Interface) P32 应用程序编程接口Network edge P35 网络边缘Access Networks P38 接入网Ethernet P42 以太网Network core P48 网络核心Circuit Switching P50 电路转换Packet Switching 分组交换FDM (frequency-division multiplexing) P50 频分多路复用TDM (time-division multiplexing) P50 时分多路复用Statistical Multiplexing 统计复用Store-and-forward 存储转发Queuing delays P53 排队延迟Transmission delay P60 传输延迟，或发送延迟Propagation delay P60 传播延迟Throughput P59 吞吐量Internet backbone P57 骨干网Delay P59 延迟，或时延Loss P59 丢包Packet-Switched Network P59 分组交换网络Nodal processing delay P60 节点处理延迟End-to-end delay P66 端到端延迟Instantaneous throughput P68 瞬时吞吐量Network interface card P74 网络接口卡（即网卡）Message P75 消息，或报文Segment P75 （报文）段Datagram P75 数据报Frames P75 帧Packet sniffer P82 数据包监听器Protocol Stack 协议栈Peer entities 对等实体Chapter 2 应用层Server farm P110 服务器集群Infrastructure P110 基础设施，或基础架构Self-scalability P111 自扩展性Timing P114 实时性Bandwidth-sensitive applications P115带宽敏感应用Connection-oriented service P117 面向连接的服务Directory service P121 目录服务Base HTML file P122 基本HTML文件Stateless protocol P124 无状态协议RTT (round-trip time ) P126 往返时间Web proxy caches P128 网页代理缓存Status line P130 状态行Out-of-band P141 （频）带外（的）In-band P141 带内（的）User agents P144 用户代理Mail servers P144 邮件服务器Pull protocol P148 拉式协议Push protocol p148 推式协议Host aliasing P158 主机别名Canonical hostname P158 规范主机名Mail server aliasing P158 邮件服务器别名Load distribution P158 负载分配Top-level domain (TLD) servers P161 顶级域名服务器Authoritative DNS servers P161 权威域名服务器Iterative queries P168 迭代查询Resource records (RRs) P165 资源记录Overlay network P179 覆盖网Nonpersistent HTTP 非持久HTTP，或非坚持HTTP Persistent HTTP 持久性HTTP，或坚持的HTTPPeer-to-Peer (P2P) Network 对等网络Socket programming 套接字编程Chapter 3 传输层Multiplexing and demultiplexing P226 复用与分用Unidirectional data transfer P241 单向数据传送Finite-state machine (FSM) P242 有限状态机Positive acknowledgments P243 肯定确认Negative acknowledgments P243 否定确认Countdown timer P250 （倒数）计时器Cumulative acknowledgment P258 累积确认Receive buffer P269 接收缓冲区，或接收缓存Resource-management cells 资源管理单元Source (port number) 源端口号Destination (port number) 目的端口号Checksum 校验和Pipelined protocols 流水线（型）协议Go-back-N 回退NSelective Repeat 选择重传Timeout （定时器）超时Fast Retransmit 快速重传Flow Control 流量控制Three way handshake 三次握手sequence number 序列号（简写为seq）acknowledgement number 确认号（简写为ack；注意与大小的ACK不同）Congestion Control 拥塞控制additive increase, multiplicative decrease 加性增乘性减Slow Start 慢启动congestion-avoidance 拥塞避免fast recovery 快速恢复duplicate (ACK) 冗余（ACK）Random Early Detection 随机早期检测Chapter 4 网络层Forwarding table P338 转发表Virtual-circuit networks P343 虚电路网络Datagram networks P343 数据报网络Signaling message P346 信令报文Content Addressable Memory P354 内容可寻址存储器Crossbar switch P356 纵横开关Active queue management 主动队列管理Head-of-the-line (HOL) 队头Classless interdomain routing (CIDR) P371 无类域间路由Plug-and-play P376 即插即用Anycast P386 任播Interior gateway protocols P414 内部网关协议Routing information Protocol P414 路由信息协议（RIP）Open shortest Path First OSPF P414 开放最短路径优先Area border routers P419 区域边界路由器Sequence-number-controlled flooding P430 序列号控制的洪泛，或带序列号的受控洪泛Reverse path forwarding (RPF) P431 逆向路径转发Rendezvous point P433 汇聚点Longest prefix matching 最长前缀匹配Scheduling 调度Fragmentation 分片，或分段Fragment Offset 报文段偏移量Network Address Translation (NAT) 网络地址转换NAT traversal NAT穿越Multicast 组播，或多播Unicast 单播Tunneling 隧道技术Link-State Routing Algorithm 链路状态路由算法Distance Vector Routing Algorithm 距离向量路由算法Count to Infinity Problem 无穷计数问题Hierarchical Routing 分层路由autonomous systems 自治系统BGP (Border Gateway Protocol) 边界网关协议in-network duplication 网内复制broadcast storm 广播风暴spanning tree 生成树redundant packets 冗余数据包Chapter 5 数据链路层，或链路层Broadcast channels P461 广播信道Trailer fields P464 尾部字段Link access P464 链路接入，或链路访问Network interface card P466 网络接口卡（即网卡）Parity checks P469 奇偶校验Forward error correction (FEC) P471 前向纠错Cyclic Redundancy Check 循环冗余校验Polynomial code P472 多项式码（即CRC码）Multiple access P475 多路接入Random access protocols P477 随机接入协议CSMA/CD P484 带冲突检测的载波侦听多路访问CSMA/CA 带冲突避免的载波侦听多路访问Token passing protocol P487 令牌传递协议ARP P491 地址解析协议Preamble P497 前导（字段）Exponential backoff P502 指数回退，或指数退避Repeater P504 中继器Virtual-channel identifier P520 虚拟信道标识Cell-loss priority P520 信元丢失优先权Label-switched router P524 标签交换路由器Framing （封装）成帧error detection 误差检测，或检错Channel Partitioning 信道分割式（MAC协议）Taking turns MAC protocol 轮流式MAC协议Collision 冲突，或碰撞Time Slot 时隙Slotted ALOHA 时隙ALOHAUnslotted ALOHA 无时隙ALOHANonpersistent CSMA 非坚持CSMA1-persistent CSMA 1坚持CSMAp-persistent CSMA p坚持CSMAToken Ring 令牌环(Wireless) LAN （无线）局域网Hub 集线器Collision domain 冲突域Bridge 网桥欢迎您的下载，资料仅供参考！致力为企业和个人提供合同协议，策划案计划书，学习资料等等打造全网一站式需求。
计算机网络课后作业答案
计算机网络课后作业答案1. OSI参考模型计算机网络通常使用开放系统互联参考模型（Open System Interconnection，简称OSI模型）作为标准。
它将通信过程分为七个层次，每个层次都负责特定的功能。
下面是对每个层次的解释：1) 物理层（Physical Layer）：主要负责传输比特流，将数字数据转化为电信号传输。
常见的物理层协议有以太网（Ethernet）和无线传输技术。
2) 数据链路层（Data Link Layer）：负责将原始的比特流转化为数据帧，并通过物理层传输。
数据链路层还负责错误检测和纠正。
常见的数据链路层协议有以太网协议、点对点协议（PPP）和局域网协议（LAN）。
3) 网络层（Network Layer）：负责将数据包从源主机传输到目标主机。
该层主要处理路由选择和分组转发。
常见的网络层协议有互联网协议（IP）和网际控制报文协议（ICMP）。
4) 传输层（Transport Layer）：负责在通信端点之间建立可靠的数据传输连接。
传输层协议一般分为面向连接的传输协议（如传输控制协议TCP）和无连接传输协议（如用户数据报协议UDP）。
5) 会话层（Session Layer）：负责建立、管理和终止会话连接。
这些会话连接可以是单向的或双向的。
6) 表示层（Presentation Layer）：负责数据格式的转换和数据加密解密。
该层的功能确保应用层可读格式的交换数据。
7) 应用层（Application Layer）：最顶层的应用层负责处理特定的网络应用，例如电子邮件、文件传输和远程登录。
常见的应用层协议有超文本传输协议（HTTP）、文件传输协议（FTP）和域名系统（DNS）。
2. TCP/IP协议族TCP/IP协议族是互联网通信的基础，它由许多协议组成，其中最重要的是传输控制协议（TCP）和互联网协议（IP）。
1) 互联网协议（IP）是TCP/IP协议族的核心协议，负责为数据包提供路由功能，将数据包从源主机传输到目标主机。
计算机网络技术课后习题及答案
计算机网络技术课后习题及答案一、选择题1. Which of the following is not a type of network topology?A. StarB. BusC. RingD. CameraAnswer: D2. What is the purpose of a router in a computer network?A. To connect multiple devices within a local area networkB. To connect multiple local area networks togetherC. To provide wireless connectivity to devicesD. To store and retrieve dataAnswer: B3. Which protocol is commonly used for email transmission?A. FTPB. HTTPC. SMTPD. SNMPAnswer: C4. What is the primary function of a firewall in a network?A. To prevent unauthorized access to the networkB. To increase network speed and performanceC. To store and manage network dataD. To provide wireless connectivity to devicesAnswer: A5. Which of the following is an example of a public IP address?A. 192.168.0.1B. 10.0.0.1C. 172.16.0.1D. 63.245.123.45Answer: D二、填空题1. The process of converting an IP address into a domain name is called ________.Answer: DNS resolution2. The maximum data rate of a channel is defined by its ________.Answer: bandwidth3. ________ is a protocol used for securely transmitting data over the internet.Answer: HTTPS4. The process of breaking a larger network into smaller subnetworks is called ________.Answer: subnetting5. A ________ address is assigned to a device by a router when it connects to a network.Answer: dynamic三、简答题1. What is the difference between TCP and UDP protocols?Answer: TCP (Transmission Control Protocol) is a connection-oriented protocol that ensures reliable data transmission between two devices. It guarantees that data is received in the correct order and checks for errors. UDP (User Datagram Protocol) is a connectionless protocol that does not guarantee reliable data transmission. It is faster but less reliable than TCP.2. Explain the concept of IP addressing.Answer: IP addressing is a method of assigning unique addresses to devices on a network. It allows for the identification and routing of data packets between devices. There are two versions of IP addresses: IPv4, which uses a 32-bit address format, and IPv6, which uses a 128-bit addressformat to accommodate the growing number of devices connected to the internet.3. What is a subnet mask and how is it used?Answer: A subnet mask is a 32-bit number used to divide an IP address into network and host portions. It determines which part of an IP address is used to identify the network and which part is used to identify the host. The subnet mask is applied bitwise to the IP address, allowing devices to determine if another device is on the same local network or a different network.四、编程题1. Write a program in Python to create a socket and establish a basic client-server connection.```python# Server codeimport socketserver_socket = socket.socket(socket.AF_INET,socket.SOCK_STREAM)server_address = ('localhost', 12345)server_socket.bind(server_address)server_socket.listen(1)print("Server is listening for incoming connections...")while True:client_socket, client_address = server_socket.accept()print("Connection established with:", client_address)message = "Welcome to the server!"client_socket.send(message.encode())client_socket.close()``````python# Client codeimport socketclient_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)server_address = ('localhost', 12345)client_socket.connect(server_address)message = client_socket.recv(1024)print("Server message:", message.decode())client_socket.close()```以上是关于计算机网络技术的课后习题及答案。
计算机网络英文答案
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.。
计算机网络英文版习题答案
XXXX有限公司信息系统账号和密码管理规定第一章总则第一条为保障XXXX有限公司信息系统的安全，确保合理访问和修改XXXX有限公司数据资源，根据《信息安全技术信息系统安全等级保护基本要求》（GB/T22239-2008），结合XXXX有限公司实际，制定本规定。
第二条本管理规定确立了包括建立、监控、修改、注销和删除公司信息系统账号的规则。
第三条本管理规定适用于XXXX有限公司信息系统的设备、平台等，适用对象是XXXX有限公司内部人员、第三方人员以及其他任何授权使用XXXX有限公司信息系统资源的人员。
第四条信息安全领导小组办公室负责制定信息系统用户账号名策略和密码策略。
第五条平台研发部下属系统运维部门负责XXXX有限公司信息系统账号及密码的分配和统一管理。
第二章账号管理第六条账号名（一）每个系统的账号名需能代表某个系统或应用的用户。
每个账号需要有一个所属人；（二）不允许共享账号身份或账号组身份；（三）账号名基本规则为员工工号。
第七条账号的申请（一）应根据业务的需求申请账号，并根据所属人的权限开通相关账号的功能；（二）账号的申请必须得到业务部门、人力资源部门、系统运维部门负责人关于访问该系统设备或服务的批准；（三）账号的权限应被严格控制。
账号尤其是特别权限的账号应被限制在职责范围内所需工作的最低权限。
超过普通用户的权限，必须基于业务需求，并得到系统运维部门负责人的批准；（四）账号和密码提供给员工之前，需要确认用户的身份。
推荐使用较严格的方式（如，递交给部门经理）来提供拥有特权的账号信息；（五）第三方人员申请信息系统账号时由接口的内部员工代为办理，并需明确其责任和义务。
第八条账号申请流程（一）首先由员工填写《系统账号管理申请单》提出书面申请，详细列出涉及的信息系统设备、所需权限、用途，由其部门负责人和分管领导审批；（二）由人力资源部门及系统运维部门负责人审核申请内容的合理性、安全性；（三）在以上各审批环节中，如任何一个审批人不同意该申请，则退回用户的申请。
计算机网络第四版（课后练习+答案）
计算机⽹络第四版（课后练习+答案）第 1 章概述1.假设你已经将你的狗Berníe 训练成可以携带⼀箱3 盒8mm 的磁带，⽽不是⼀⼩瓶内哇地. (当你的磁盘满了的时候，你可能会认为这是⼀次紧急事件。
)每盒磁带的窑最为7GB 字节;⽆论你在哪⾥，狗跑向你的速度是18km/h 。
请问，在什么距离范围内Berníe的数据传输速率会超过⼀条数据速率为150Mbps的传输线?答：狗能携带21千兆字节或者168千兆位的数据。
18 公⾥/⼩时的速度等于0.005 公⾥/秒，⾛过x公⾥的时间为x / 0.005 = 200x 秒，产⽣的数据传输速度为168/200x Gbps或者840 /x Mbps。
因此，与通信线路相⽐较，若x<5.6 公⾥，狗有更⾼的速度。
6. ⼀个客户·服务器系统使⽤了卫星⽹络，卫星的⾼度为40 000km. 在对⼀个请求进⾏响应的时候，最佳情形下的延迟是什么?答：由于请求和应答都必须通过卫星，因此传输总路径长度为160,000千⽶。
在空⽓和真空中的光速为300，000 公⾥/秒，因此最佳的传播延迟为160,000/300，000秒，约533 msec。
9.在⼀个集中式的⼆叉树上，有2n -1 个路出器相互连接起来:每个树节点上都布⼀个路由器。
路由器i 为了与路由器j 进⾏通信，它要给树的根发送⼀条消息。
然后树根将消息送下来给j 。
假设所有的路由器对都是等概率出现的，请推导出当n很⼤时，每条消息的平均跳数的⼀个近似表达式。
答：这意味着，从路由器到路由器的路径长度相当于路由器到根的两倍。
若在树中，根深度为1，深度为n，从根到第n层需要n-1跳，在该层的路由器为0.50。
从根到n-1 层的路径有router的0.25和n-2跳步。
因此，路径长度l为：18.OSI 的哪⼀层分别处理以下问题？答：把传输的⽐特流划分为帧——数据链路层决定使⽤哪条路径通过⼦⽹——⽹络层.28.⼀幅图像的分辨率为1024X 768 像素，每个像素⽤3 字节来表⽰。
计算机网络系统方法(英文版)课后习题及解答
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）。
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Chapter11-11.What are two reasons for using layered protocols?(请说出使用分层协议的两个理由)答：通过协议分层可以把设计问题划分成较小的易于处理的片段。
分层意味着某一层的协议的改变不会影响高层或低层的协议。
1-13. What is the principal difference between connectionless communication and connection-oriented communication?(在无连接通信和面向连接的通信两者之间，最主要的区别是什么？)答：主要的区别有两条。
其一：面向连接通信分为三个阶段，第一是建立连接，在此阶段，发出一个建立连接的请求。
只有在连接成功建立之后，才能开始数据传输，这是第二阶段。
接着，当数据传输完毕，必须释放连接。
而无连接通信没有这么多阶段，它直接进行数据传输。
其二：面向连接的通信具有数据的保序性，而无连接的通信不能保证接收数据的顺序与发送数据的顺序一致。
1-20. A system has an n-layer protocol hierarchy. Applications generate messages of length M bytes. At each of the layers, an h-byte header is added. What fraction of the network bandwidth is filled with headers?(一个系统有n层协议的层次结构。
应用程序产生的消息的长度为M字节。
在每一层上需要加上一个h字节的头。
请问，这些头需要占用多少比例的网络带宽)答：hn/(hn+m)*100%1-28. An image is 1024 x 768 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet?(一幅图像的分辨率为1024 x 768像素，每个像素用3字节来表示。
假设该图像没有被压缩。
请问，通过56kbps的调制解调器信道来传输这幅图像需要多长时间？通过1Mbps的电缆调制解调器呢？通过10Mbps的以太网呢？通过100Mbps的以太网呢?)答：The image is 1024*768*3 bytes or 2359296 bytes.This is 18874368 bit. At 56,000 bits/sec, it takes about 337.042 sec. At 1,000,000 bits/sec, it takes about 18.874 sec. At 10,000,000 bits/sec, it takes about 1.887 sec. At 100,000,000 bits/sec, it takes about 0.189 sec.Chapter22-2. A noiseless 4-kHz channel is sampled every 1 msec. What is the maximum data rate?(一条无噪声4kHz信道按照每1ms一次进行采样，请问最大数据传输率是多少？)答：无噪声信道最大数据传输率公式：最大数据传输率=2Hlog2V b/s。
因此最大数据传输率决定于每次采样所产生的比特数，如果每次采样产生16bits，那么数据传输率可达128kbps；如果每次采样产生1024bits，那么可达8.2Mbps。
注意这是对无噪声信道而言的，实际信道总是有噪声的，其最大数据传输率由香农定律给出。
2-4. If a binary signal is sent over a 3-kHz channel whose signal-to-noise ratio is 20 dB, what is the maximum achievable data rate?(如果在一条3kHz的信道上发送一个二进制信号，该信道的信噪比为20dB，则最大可达到的数据传输率为多少？)答：信噪比为20 dB 即 S/N =100由于 log101≈6.658，由香农定理，该信道2的信道容量为3log(1+100)=19.98kbps。
2又根据乃奎斯特定理，发送二进制信号的3kHz 信道的最大数据传输速率为2*3*log22=6kbps。
所以可以取得的最大数据传输速率为6kbps。
2-5. What signal-to-noise ratio is needed to put a T1 carrier on a 50-kHz line?(在50kHz的线路上使用1T线路需要多大的信噪比？)答：为发送T1 信号，我们需要所以，在50kHz 线路上使用T1 载波需要93dB 的信噪比。
2-34. A signal is transmitted digitally over a 4-kHz noiseless channel with one sample every 125 µsec. How many bits per second are actually sent for each of these encoding methods?(a) CCITT 2.048 Mbps standard.(b) DPCM with a 4-bit relative signal value.(c) Delta modulation.（一个信号在4kHz的无噪声信道上以数字方式进行传输，没125us采样一次。
请问，按照以下的编码方法，每秒钟实际发送多少位？（a）CCITT 2.048Mbps标准。
（b）有4位相对信号值的DPCM。
（c）增量调制。
）答：a．CCITT 2.048Mbps 标准用32 个8 位数据样本组成一个125的基本帧，30 个信道用于传信息，2 个信道用于传控制信号。
在每一个4kHz 信道上发送的数据率就是8*8000=64kbps。
b．差分脉码调制（DPCM）是一种压缩传输信息量的方法，它发送的不是每一次抽样的二进制编码值，而是两次抽样的差值的二进制编码。
现在相对差值是4 位，所以对应每个4kHz 信道实际发送的比特速率为4*8000=32bps。
c．增量调制的基本思想是：当抽样时间间隔s t 很短时，模拟数据在两次抽样之间的变化很小，可以选择一个合适的量化值作为阶距。
把两次抽样的差别近似为不是增加一个就是减少一个。
这样只需用1bit 二进制信息就可以表示一次抽样结果，而不会引入很大误差。
因此，此时对应每个4kHz 信道实际发送的数据速率为1*8000=8kHz。
2-43. Suppose that x bits of user data are to be transmitted over a k-hop path in a packet-switched network as a series of packets, each containing p data bits and h header bits,with x >> p+h.The bit rate of the lines is b bps and the propagation delay is negligible.What value of p minimizes the total delay?(假定x位用户数据将以一系列分组形式，在一个分组交换网络中沿着一条共有k跳的路径向前传播，每个分组包含p位数据和h位的头，这里x>>p+h。
线路的传输率为b bps，传播延迟忽略不计。
请问，什么样的p值使总延迟最小？) 答：所需要的分组总数是x /p，因此总的数据加上头信息交通量为(p+h)*x/p 位。
源端发送这些位需要时间为(p+h)*x/pb中间的路由器重传最后一个分组所花的总时间为(k-1)*(p+h)/ b因此我们得到的总的延迟为对该函数求p 的导数，得到令得到因为p＞0，所以故时能使总的延迟最小。
2-53. A CDMA receiver gets the following chips: (-1 +1 -3 +1 -1 -3 +1 +1). Assuming the chip sequences defined in Fig. 2-45(b), which stations transmitted, and which bits did each one send?(一个CDMA接收器得到了下面的时间片：(-1 +1 -3 +1 -1 -3 +1 +1)。
假设时间片序列如图2.45（b）中所定义，请问哪些移动站传输了数据？每个站发送了什么位?)答：Just compute the four normalized inner products:（1+1 3+1 1 3+1+1）d（1 1 1+1+1 1+1+1）/8=1（1+1 3+1 1 3+1+1）d（1 1+1 1+1+1+1 1）/8=1（1+1 3+1 1 3+1+1）d（1+1 1+1+1+1 1 1）/8=0（1+1 3+1 1 3+1+1）d（1+1 1 1 1 1+1 1）/8=1The result is that A and D sent 1 bits, B sent a 0 bit, and C was silent. Chapter33-3.The following data fragment occurs in the middle of a data stream for which the byte-stuffing algorithm described in the text is used: A B ESC C ESC FLAG FLAG D. What is the output after stuffing?(数据片断（A B ESC C ESC FLAG FLAG D）出现在一个数据流的中间，而成帧方式采用的是本章介绍的字节填充算法，请问经过填充之后的输出时什么？) 答：After stuffing, we get A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D.3-15. A bit stream 10011101 is transmitted using the standard CRC method described in the text. The generator polynomial is x3 + 1. Show the actual bit string transmitted. Suppose the third bit from the left is inverted during transmission. Show that this error is detected at the receiver's end.(利用本章中介绍的标准多项式CRC方法来传输位流1001101。