计算机网络英文题库(附答案)chapter1

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计算机网络(双语)Andrew S.Tanenbaum第五版(第一章)CN-chapter1

2.2 Network hardware-scale
2.2.1 Network hardware-scale
Personal Area Networks
2.2.2 Network hardware-scale
Local Area Networks
2.2.3 Network hardware-scale
applicatictions
Home applications
• • • •
Twitter Youtube Facebook Ubiquitous computing
• • • •
Smart phone GPS M-commerce Wearable computer
Mobile users
Social issues
Since computer scientists like to have their cake and eat it, too, we will use the hybrid model of Fig. 1-23 as the framework for this book.
4.1 Reference models-OSI
4.1 Reference models-OSI
4.2 Reference models-TCP/IP
4.3 Reference models- Used in This Book
目前在计算机网络领域有影响的标准化组织： 1、国际电信联盟（ITU) 2、国际标准化组织（ISO） 3、电子工业协会（EIA） 4、电气与电子工程师协会（IEEE） 5、ATM论坛
Network hardware Network software
Reference models

(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。

计算机网络英文——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 across a physical communication medium, such as a wire, radio wave, or light beam.b5E2RGbCAP1.5 What is packet-switching, and why is packet switching relevant tothe Internet?Answer: Packet switching divides data into small blocks, called packets, and includes an identification of the intended recipient in each packet. Packet switching changed networking in a fundamental way, and provided the basis for the modern Internet. Packet switching allows multiple senders to transmit data over a shared network. p1EanqFDPw1.8 What is a communication protocol? Conceptually, what two aspects ofcommunication does a protocol specify?Answer: A communication protocol refer to a specification for network communication.Major aspects of a protocol are syntax (format> and semantics (meaning> of the protocol. DXDiTa9E3d1.9 What is a protocol suite, and what is the advantage of a suite? Answer:protocols are designed in complete, cooperative sets called suites or families, instead of creating each protocol in isolation.Each protocol in a suite handles oneaspect of communication。

计算机网络双语试题及答案

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

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)。

计算机网络试卷及答案-英文版

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

计算机网络英文题库附答案chater

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

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。

计算机网络英文题库(附答案)chapter1(推荐文档)

Chapter 1 Computer Networks and the Internet1．The ( ) is a worldwide computer network, that is, a network that interconnects millions of computing devices throughout the world.ppt3A public InternetB IntranetC switch netD television net2．Which kind of media is not a guided media? ( )A twisted-pair copper wireB a coaxial cableC fiber opticsD digital satellite channel3．Which kind of media is a guided media? ( )A geostationary satelliteB low-altitude satelliteC fiber opticsD wireless LAN4．The units of data exchanged by a link-layer protocol are called ( ).A FramesB SegmentsC DatagramsD bit streams5．Which of the following option belongs to the circuit-switched networks? ( )A FDMB TDMC VC networksD both A and B6．( )makes sure that neither side of a connection overwhelms the other side by sending too many packets too fast.A Reliable data transferB Flow controlC Congestion controlD Handshaking procedure7．( ) means that the switch must receive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link.A Store-and-forward transmissionB FDMC End-to-end connectionD TDM8．Datagram networks and virtual-circuit networks differ in that ( ).A datagram networks are circuit-switched networks, andvirtual-circuit networks are packet-switched networks.B datagram networks are packet-switched networks, andvirtual-circuit networks are circuit-switched networks.C datagram networks use destination addresses and virtual-circuitnetworks use VC. numbers to forward packets toward theirdestination.D datagram networks use VC. numbers and virtual-circuit networksuse destination addresses to forward packets toward their destination. 9．In the following options, which one is not a guided media? ( )A twisted-pair wireB fiber opticsC coaxial cableD satellite10．Processing delay does not include the time to ( ).A examine the packet’s headerB wait to transmit the packet onto the linkC determine where to direct the packetD check bit-error in the packet11．In the following four descriptions, which one is correct? ( )A The traffic intensity must be greater than 1.B The fraction of lost packets increases as the traffic intensitydecreases.C If the traffic intensity is close to zero, the average queuing delaywill be close to zero.D If the traffic intensity is close to one, the average queuing delaywill be close to one.12．The Internet’s network layer is responsible for moving network-layer packets known as ( ) from one host to another.A frameB datagramC segmentD message13．The protocols of various layers are called ( ).A the protocol stackB TCP/IPC ISPD network protocol14．There are two classes of packet-switched networks: ( ) networks and virtual-circuit networks.A datagramB circuit-switchedC televisionD telephone15．Access networks can be loosely classified into three categories: residential access, company access and ( ) access.A cabledB wirelessC campusD city areaQuestion 16～17Suppose, a is the average rate at which packets arrive at the queue, R is the transmission rate, and all packets consist of L bits, then the traffic intensity is ( 16 ), and it should no greater than ( 17 ).16． A LR／aB La／RC Ra／LD LR／a17．A 2B 1C 0D -118．In the Internet, the equivalent concept to end systems is ( ).A hostsB serversC clientsD routers19．In the Internet, end systems are connected together by ( ).A copper wireB coaxial cableC communication linksD fiber optics20．End systems access to the Internet through its ( ).A modemsB protocolsC ISPD sockets21．End systems, packet switches, and other pieces of the Internet, run ( ) that control the sending and receiving of information within theInternet.A programsB processesC applicationsD protocols22．There are many private networks, such as many corporate and government networks, whose hosts cannot exchange messages withhosts outside of the private network. These private networks are often referred to as ( ).A internetsB LANC intranetsD WAN23．The internet allows ( ) running on its end systems to exchange data with each other.A clients applicationsB server applicationsC P2P applicationsD distributed applications24．The Internet provides two services to its distributed applications: a connectionless unreliable service and () service.A flow controlB connection-oriented reliableC congestion controlD TCP25．It defines the format and the order of messages exchanged between two or more communicating entities, as well as the actions taken on thetransmission and/or receipt of a message or other event. The sentence describes ( ).A InternetB protocolC intranetD network26．In the following options, which does not define in protocol? ( )A the format of messages exchanged between two or morecommunicating entitiesB the order of messages exchanged between two or morecommunicating entitiesC the actions taken on the transmission of a message or other eventD the transmission signals are digital signals or analog signals 27．In the following options, which is defined in protocol? ( )A the actions taken on the transmission and/or receipt of a message orother eventB the objects exchanged between communicating entitiesC the content in the exchanged messagesD the location of the hosts28．In the following options, which does not belong to the network edge? ( )A end systemsB routersC clientsD servers29．In the following options, which belongs to the network core? ( )A end systemsB routersC clientsD servers30．In the following options, which is not the bundled with the Internet’s connection-oriented service? ( )A reliable data transferB guarantee of the transmission timeC flow controlD congestion-control31．An application can rely on the connection to deliver all of its data without error and in the proper order. The sentence describes ( ).A flow controlB congestion-controlC reliable data transferD connection-oriented service32．It makes sure that neither side of a connection overwhelms the other side by sending too many packets too fast. The sentence describes ( ).A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer33．It helps prevent the Internet from entering a state of gridlock. When a packet switch becomes congested, its buffers can overflow and packet loss can occur. The sentence describes ( ).A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer34．The Internet’s connection-oriented service has a name, it is ( ).A TCPB UDPC TCP/IPD IP35．In the following options, which service does not be provided to an application by TCP?( )A reliable transportB flow controlC video conferencingD congestion control36．The Internet’s connectionless service is called ( ).A TCPB UDPC TCP/IPD IP37．In the following options, which does not use TCP?( )A SMTPB internet telephoneC FTPD HTTP38．In the following options, which does not use UDP?( )A Internet phoneB video conferencingC streaming multimediaD telnet39．There are two fundamental approaches to building a network core, ( ) and packet switching.A electrical current switchingB circuit switchingC data switchingD message switching 40．In ( ) networks, the resources needed along a path to provide for communication between the end system are reserved for the duration of the communication session.A packet-switchedB data-switchedC circuit-switchedD message-switched41．In ( ) networks, the resources are not reserved; a session’s messages use the resources on demand, and as a consequence, may have to wait for access to communication link.A packet-switchedB data-switchedC circuit-switchedD message-switched42．In a circuit-switched network, if each link has n circuits, for each link used by the end-to-end connection, the connection gets ( ) of thelink’s bandwidth for the duration of the connection.A a fraction 1/nB allC 1/2D n times43．For ( ), the transmission rate of a circuit is equal to the frame rate multiplied by the number of bits in a slot.A CDMAB packet-switched networkC TDMD FDM44．( ) means that the switch must receive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link.A Queuing delayB Store-and-forward transmissionC Packet lossD Propagation45．The network that forwards packets according to host destination addresses is called ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram46．The network that forwards packets according to virtual-circuit numbers is called ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram47．In the following entries, which is not a kind of access network?( )A residential accessB company accessC wireless accessD local access48．Suppose there is exactly one packet switch between a sending host and a receiving host. The transmission rates between the sending host and the switch and between the switch and the receiving host are R1 and R2,respectively. Assuming that the switch uses store-and-forward packetswitching, what is the total end-to-end delay to send a packet of length L? (Ignore queuing delay, propagation delay, and processing delay.)( )A L/R1＋L/R2B L/R1C L/R2D none of the above49．The time required to examine the packet’s header and determine where to direct the packet is part of the ( ).A queuing delayB processing delayC propagation delayD transmission delay50．The time required to propagate from the beginning of the link to the next router is ( ).A queuing delayB processing delayC propagation delayD transmission delay51．Consider sending a packet of 3000bits over a path of 5 links. Each link transmits at 1000bps. Queuing delays, propagation delay and processing delay are negligible. (6 points)(1).Suppose the network is a packet-switched virtual circuit network. VC setup time is 0.1 seconds. Suppose the sending layers add a total of 500 bits of header to each packet. How long does it take to send the file from source to destination?(2).Suppose the network is a packet-switched datagram network and a connectionless service is used. Now suppose each packet has 200 bits of header. How long does it take to send the file?(3).Suppose that the network is a circuit-switched network. Further suppose that the transmission rate of the circuit between source and destination is 200bps. Assuming 0.02s setup time and 200 bits of header appended to the packet, how long does it take to send the packet?Solution: (1). t=5*(3000+500)/1 000+0.1=17.6s(2). t=5*(3000+200)/1 000=16s(3). t=(3000+200)/20 0+0.02=16.02s。

计算机网络英文题库附答案chapter定稿版

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

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)计算机网络第四版(课后练习+答案)Introduction:计算机网络是现代信息技术的基础，它涉及到计算机与计算机之间如何进行数据交换和通信。

《计算机网络第四版》是一本权威指南，提供了大量的课后练习和答案，帮助读者加强对计算机网络的理解。

本文将对《计算机网络第四版》课后练习和答案进行综述，以便读者更好地掌握网络通信的关键概念和原理。

Chapter 1: Introduction to Networking在第一章中，课后练习的内容涵盖了计算机网络的基本概念和发展历程。

学习者可以通过这些练习加深对网络通信的了解，例如描述计算机网络的基本组成部分、定义OSI模型的七层结构以及解释分组交换和电路交换的区别。

Chapter 2: Network Models第二章课后练习着重介绍了计算机网络的各种模型，包括OSI模型和TCP/IP模型。

练习题目涵盖了每个模型的层次结构和功能，同时还要求学习者能够比较这两个模型之间的异同点。

Chapter 3: Physical Layer and Media物理层和传输介质是计算机网络的基础，第三章课后练习起到了巩固和扩展这些概念的作用。

学习者将通过回答问题和解决实际情况的案例，深入理解诸如数据信号的调制和解调、传输介质的特性以及常见的物理层设备等内容。

Chapter 4: Data Link Layer数据链路层构建在物理层之上，并处理节点到节点之间的数据传输。

第四章的课后练习要求学习者熟练掌握数据链路层的基本概念，包括帧的封装和解封装、错误检测和纠正技术以及介绍局域网和广域网等。

Chapter 5: Network Layer网络层负责数据包的转发和路由选择，在第五章的练习题中，学习者需要回答关于IP地址的分配和路由表的设计的问题，深入理解网络层的功能和特性。

还会涉及到IP协议的各种细节，例如子网划分、地址转换和网络控制协议等。

Chapter 6: Transport Layer传输层提供端到端的可靠数据传输服务，第六章的课后练习通过设计案例和讨论问题的方式，帮助学习者掌握TCP和UDP协议的细节和应用场景。

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）。

计算机英语考试试题及答案

计算机英语考试试题及答案Introduction：计算机英语考试是测试学生在计算机领域具备的英语水平和技能的考试，旨在评估学生综合运用英语和计算机知识的能力。

本文将提供一些计算机英语考试的典型试题及其答案，以帮助读者更好地理解和准备此类考试。

Section 1: 填空题1. HTML是一种用于创建（1）的（2）标准。

答案：1. 网页； 2. 标记语言2. CSS是用于定义和（1）网页的外观和布局的样式（2）。

答案：1. 美化； 2. 表达语言Section 2: 选择题3. 在计算机网络中，TCP/IP代表的是：a) 网络协议 b) 网页设计语言c) 数据库管理系统 d) 电子邮件协议答案：a) 网络协议4. 下列哪项属于非编程语言？a) Java b) Pythonc) SQL d) HTML答案：d) HTMLSection 3: 阅读理解阅读下面的一段话，然后回答问题。

The cloud computing model allows users to access applications, information, data storage, and computing power via the internet. Instead of relying on local servers or personal devices, users can utilize cloud services provided by third-party providers. This model offers benefits such as scalability, flexibility, and cost-effectiveness. However, it also raises concerns regarding data security and privacy.5. What is the main benefit of cloud computing?答案：The main benefit of cloud computing is scalability, flexibility, and cost-effectiveness.6. The cloud computing model relies on ________.答案：The cloud computing model relies on accessing applications, information, data storage, and computing power via the internet.Section 4: 编程题7. 请写一个简单的Python程序来计算斐波那契数列的前n项，并将结果打印出来。

计算机网络英文题库（附答案）chapter1

Chapter 1 Computer Networks and the Internet 1．The The ( ( ) ) is is is a a a worldwide worldwide worldwide computer computer computer network, network, network, that that that is, is, is, a a a network network network that that interconnects millions of computing devices throughout the world. ppt3 A public Internet B Intranet C switch net D television net 2．Which kind of media is not a guided media? ( ) A twisted-pair copper wire B a coaxial cable C fiber optics D digital satellite channel 3．Which kind of media is a guided media? ( ) A geostationary satellite B low-altitude satellite C fiber optics D wireless LAN 4．The units of data exchanged by a link-layer protocol are called ( ). A Frames B Segments C Datagrams D bit streams 5．Which of the following option belongs to the circuit-switched networks? ( ) A FDM B TDM C VC networks D both A and B 6．( )makes sure that that neither neither neither side side side of of of a a a connection overwhelms connection overwhelms the the other other side by sending too many packets too fast. A Reliable data transfer B Flow control C Congestion control D Handshaking procedure 7．( ) means that the switch must receive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link. A Store-and-forward transmission B FDM C End-to-end connection D TDM 8．Datagram networks and virtual-circuit networks differ in that ( ). A datagram networks are circuit-switched networks, and virtual-circuit networks are packet-switched networks. B datagram networks are packet-switched networks, and virtual-circuit networks are circuit-switched networks. C datagram networks use destination addresses and virtual-circuit networks use VC. numbers to forward packets toward their destination. D datagram networks use VC. numbers and virtual-circuit networks use destination addresses to forward packets toward their destination. 9．In the following options, which one is not a guided media? ( ) A twisted-pair wire B fiber optics C coaxial cable D satellite 10．Processing delay does not include the time to ( ). A examine the packet ’s header B wait to transmit the packet onto the link C determine where to direct the packet D check bit-error in the packet 11．In the following four descriptions, which one is correct? ( ) A The traffic intensity must be greater than 1. B The fraction of lost packets increases as the traffic intensity decreases. C If the traffic intensity is close to zero, the average queuing delaywill be close to zero. D If the traffic intensity is close to one, the average queuing delay will be close to one. 12．The The Internet Internet Internet’’s s network network network layer layer layer is is is responsible responsible responsible for for for moving moving moving network-layer network-layer packets known as ( ) from one host to another. A frame B datagram C segment D message 13．The protocols of various layers are called ( ). A the protocol stack B TCP/IP C ISP D network protocol 14．There are two classes of packet-switched networks: ( ) networks and virtual-circuit networks. A datagram B circuit-switched C television D telephone 15．Access networks can be loosely classified into three categories: residential access, company access and ( ) access. A cabled B wireless C campus D city area Question 16～17 Suppose, a is the average rate at which packets arrive at the queue, R is the the transmission transmission transmission rate, rate, rate, and and and all all all packets packets packets consist consist consist of of L bits, bits, then then then the the the traffic traffic intensity is ( 16 ), and it should no greater than ( 17 ). 16． A LR ／aB La ／RC Ra ／L D LR ／a 17．A 2 B 1 C 0 D -1 18．In the Internet, the equivalent concept to end systems is ( ). A hosts B servers C clients D routers 19．In the Internet, end systems are connected together by ( ). A copper wire B coaxial cable C communication links D fiber optics 20．End systems access to the Internet through its ( ). A modems B protocols C ISP D sockets 21．End systems, packet switches, and other pieces of the Internet, run ( ) that control the sending and receiving of information within the Internet. A programs B processes C applications D protocols 22．There are many private networks, such as many corporate and government networks, whose hosts cannot exchange messages with hosts outside of the private network. These private networks are often referred to as ( ). A internets B LANC intranets D W AN 23．The internet allows ( ) running on its end systems to exchange data with each other. A clients applications B server applications C P2P applications D distributed applications 24．The Internet provides two services to its distributed applications: a connectionless unreliable service and () service. A flow control B connection-oriented reliable C congestion control D TCP 25．It defines the format and the order of messages exchanged between twoor more communicating entities, as well as the actions taken on the transmission and/or receipt of a message or other event. The sentence describes ( ). A Internet B protocol C intranet D network 26．In the following options, which does not define in protocol? ( ) A the format of messages exchanged between two or more communicating entities B the order of messages exchanged between two or more communicating entities C the actions taken on the transmission of a message or other evenD the transmission signals are digital signals or analog signals 27．In the following options, which is defined in protocol? ( ) A the actions taken on the transmission and/or receipt of a message oother event B the objects exchanged between communicating entities C the content in the exchanged messages D the location of the hosts 28．In the following options, which does not belong to the network edge( ) A end systems B routers C clients D servers 29．In the following options, which belongs to the network core? ( ) A end systems B routers C clients D servers 30．In the following options, which is not the bundled with the Internet ’s connection-oriented service? ( ) A reliable data transfer B guarantee of the transmission time C flow control D congestion-control 31．An application can rely on the connection to deliver all of its data without error and in the proper order. The sentence describes ( ). A flow control B congestion-control C reliable data transfer D connection-oriented service 32．It makes sure that neither side of a connection overwhelms the other side by sending too many packets too fast. The sentence describes ( ). A flow control B congestion-control C connection-oriented service D reliable data transfer 33．It helps prevent the Internet from entering a state of gridlock. When a packet switch becomes congested, its buffers can overflow and packet loss can occur. The sentence describes ( ). A flow control B congestion-control C connection-oriented service D reliable data transfer 34．The Internet ’s connection-oriented service has a name, it is ( ). A TCP B UDP C TCP/IP D IP 35．In In the the the following following following options, options, options, which which which service service service does does does not not not be be be provided provided provided to to to an an application by TCP?( ) A reliable transport B flow control C video conferencing D congestion control 36．The Internet ’s connectionless service is called ( ). A TCP B UDP C TCP/IP D IP 37．In the following options, which does not use TCP?( ) A SMTP B internet telephone C FTP D HTTP 38．In the following options, which does not use UDP?( ) A Internet phone B video conferencing C streaming multimedia D telnet 39．There are two fundamental approaches to building a network core, ( ) and packet switching. A electrical current switching B circuit switching C data switching D message switching 40．In ( ) networks, the resources needed along a path to provide for communication between the end system are reserved for the duration ofthe communication session. A packet-switched B data-switched C circuit-switched D message-switched 41．In ( ) networks, the resources are not reserved; a session’s messages use the resources on demand, and as a consequence, may have to wait for access to communication link. A packet-switched B data-switched C circuit-switched D message-switched 42．In a circuit-switched network, if each link has n circuits, for each link used by the end-to-end connection, the connection gets ( ) of the link link’’s bandwidth for the duration of the connection. A a fraction 1/n B all C 1/2 D n times 43．For ( ), the transmission rate of a circuit is equal to the frame rate multiplied by the number of bits in a slot. A CDMA B packet-switched network C TDM D FDM 44．( ) means that the switch must receive the entire packet before it canbegin to transmit the first bit of the packet onto the outbound link. A Queuing delay B Store-and-forward transmission C Packet loss D Propagation 45．The network that forwards packets according to host destination addresses is called ( ) network. A circuit-switched B packet-switched C virtual-circuit D datagram 46．The network that forwards packets according to virtual-circuit numbers is called ( ) network. A circuit-switched B packet-switched C virtual-circuit D datagram 47．In the following entries, which is not a kind of access network?( ) A residential access B company access C wireless access D local access 48．Suppose there is exactly one packet switch between a sending host and a receiving host. The transmission rates between the sending host and the switch and between the switch and the receiving host are R 1 and R 2, respectively. Assuming that the switch uses store-and-forward packet switching, what is the total end-to-end delay to send a packet of length L ? (Ignore queuing delay, propagation delay, and processing delay.) ( )A L /R 1＋L /R 2 B L /R 1C L /R 2D none of the above 49．The time required to examine the packet ’s header and determine where to direct the packet is part of the ( ). A queuing delay B processing delay C propagation delay D transmission delay 50．The time required to propagate from the beginning of the link to the next router is ( ). A queuing delay B processing delay C propagation delay D transmission delay 51．Consider sending a packet of 3000bits over a path of 5 links. Each link transmits transmits at at at 1000bps. 1000bps. 1000bps. Queuing Queuing Queuing delays, delays, delays, propagation propagation propagation delay delay delay and and and processing processing delay are negligible. (6 points) (1).Suppose (1).Suppose the the the network network network is is is a a a packet-switched packet-switched packet-switched virtual virtual virtual circuit circuit circuit network. network. network. VC VC setup time is 0.1 seconds. Suppose the sending layers add a total of 500 bits of header to each packet. How long does it take to send the file from source to destination? (2).Suppose the the network network network is is is a a a packet-switched packet-switched datagram datagram network network network and and and a a connectionless connectionless service service service is is is used. used. used. Now Now Now suppose suppose suppose each each each packet packet packet has has has 200 200 200 bits bits bits of of header. How long does it take to send the file? (3).Suppose that the network is a circuit-switched network. Further suppose that that the the the transmission transmission transmission rate rate rate of of of the the the circuit circuit circuit between between between source source source and and and destination destination destination is is 200bps. Assuming 0.02s setup time and 200 bits of header appended to the packet, how long does it take to send the packet? Solution: (1). t=5*(3000+500)/1000+0.1=17.6s(2). t=5*(3000+200)/1000=16s(3). t=(3000+200)/200+0.02=16.02s。

计算机网络自顶向下方法英文版答案

Computer Networking: A Top-Down Approach Featuring the Internet, 3rd EditionSolutions to Review Questions and Problems Note: These solutions are incomplete. Complete solutions will be available by 1 September 2005Version Date: July 1, 2004This document contains the solutions to review questions and problems for the 3rd edition of Computer Networking: A Top-Down Approach Featuring the Internet by Jim Kurose and Keith Ross. These solutions are being made available to instructors ONLY. Please do NOT copy or distribute this document to others (even other instructors). Please do not post any solutions on a publicly-available Web site. We’ll be happy to provide a copy (up-to-date) of this solution manual ourselves to anyone who asks.All material © copyright 1996-2004 by J.F. Kurose and K.W. Ross. All rights reservedChapter 1 Review Questions1.There is no difference. Throughout this text, the words “host” and “end system” areused interchangeably. End systems include PCs, workstations, Web servers, mail servers, Internet-connected PDAs, WebTVs, etc.2.Suppose Alice, an ambassador of country A wants to invite Bob, an ambassador ofcountry B, over for dinner. Alice doesn’t simply just call Bob on the phone and say, “come to our dinner table now”. Instead, she calls Bob and suggests a date and time.Bob may respond by saying he’s not available that particular date, but he is available another date. Alice and Bob continue to send “messages” back and forth until they agree on a date and time. Bob then shows up at the embassy on the agreed date, hopefully not more than 15 minutes before or after the agreed time. Diplomatic protocols also allow for either Alice or Bob to politely cancel the engagement if they have reasonable excuses.3. A networking program usually has two programs, each running on a different host,communicating with each other. The program that initiates the communication is the client. Typically, the client program requests and receives services from the server program.4.The Internet provides its applications a connection-oriented service (TCP) and aconnectionless service (UDP). Each Internet application makes use of one these two services. The two services will be discussed in detail in Chapter 3. Some of the principle characteristics of the connection-oriented service are:•Two end-systems first “handshake” before either starts to send application data to the other.•Provides reliable data transfer, i.e., all application data sent by one side of the connection arrives at the other side of the connection in order and without any gaps.•Provides flow control, i.e., it makes sure that neither end of a connection overwhelms the buffers in the other end of the connection by sending to many packets to fast.•Provides congestion control, i.e., regulates the amount of data that an application can send into the network, helping to prevent the Internet from entering a state of grid lock.The principle characteristics of connectionless service are:•No handshaking•No guarantees of reliable data transfer•No flow control or congestion control5.Flow control and congestion control are two distinct control mechanisms with distinctobjectives. Flow control makes sure that neither end of a connection overwhelms the buffers in the other end of the connection by sending to many packets to fast.Congestion control regulates the amount of data that an application can send into the network, helping to prevent congestion in the network core (i.e., in the buffers in the network routers).6.The Internet’s connection-oriented service provides reliable data transfer by usingacknowledgements and retransmissions. When one side of the connection doesn’t receive an acknowledgement (from the other side of the connection) for a packet it transmitted, it retransmits the packet.7. A circuit-switched network can guarantee a certain amount of end-to-end bandwidthfor the duration of a call. Most packet-switched networks today (including the Internet) cannot make any end-to-end guarantees for bandwidth.8.In a packet switched network, the packets from different sources flowing on a link donot follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.9.At time t0the sending host begins to transmit. At time t1 = L/R1, the sending hostcompletes transmission and the entire packet is received at the router (no propagation delay). Because the router has the entire packet at time t1, it can begin to transmit the packet to the receiving host at time t1. At time t2= t1 + L/R2, the router completes transmission and the entire packet is received at the receiving host (again, no propagation delay). Thus, the end-to-end delay is L/R1 + L/R2.10.In a VC network, each packet switch in the network core maintains connection stateinformation for each VC passing through it. Some of this connection state information is maintained to a VC-number translation table. (See page 25)11.The cons of VCs include (i) the need to have a signaling protocol to set-up and tear-down the VCs; (ii) the need to maintain connection state in the packet switches. For the pros, some researchers and engineers argue that it is easier to provide QoS services - such as services that guarantee a minimum transmission rate or services that guarantee maximum end-to-end packet delay – when VCs are used.12.1. Dial-up modem over telephone line: residential; 2. DSL over telephone line:residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, WAP): mobile13.A tier-1 ISP connects to all other tier-1 ISPs; a tier-2 ISP connects to only a few ofthe tier-1 ISPs. Also, a tier-2 ISP is a customer of one or more tier-114.A POP is a group of one or more routers in an ISPs network at which routers in otherISPs can connect. NAPs are localized networks at which many ISPs (tier-1, tier-2 and lower-tier ISPs) can interconnect.15.HFC bandwidth is shared among the users. On the downstream channel, all packetsemanate from a single source, namely, the head end. Thus, there are no collisions in the downstream channel.16.Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps.For an X Mbps Ethernet (where X = 10, 100, 1,000 or 10,000), a user can continuously transmit at the rate X Mbps if that user is the only person sending data.If there are more than one active user, then each user cannot continuously transmit at X Mbps.17.Ethernet most commonly runs over twisted-pair copper wire and “thin” coaxial cable.It also can run over fibers optic links and thick coaxial cable.18.Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps,bandwidth is dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to 1 Mbps, bandwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstream channel is usually less than a few Mbps, bandwidth is shared.19.The delay components are processing delays, transmission delays, propagation delays,and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable.20.Five generic tasks are error control, flow control, segmentation and reassembly,multiplexing, and connection setup. Yes, these tasks can be duplicated at different layers. For example, error control is often provided at more than one layer.21.The five layers in the Internet protocol stack are –from top to bottom –theapplication layer, the transport layer, the network layer, the link layer, and the physical layer. The principal responsibilities are outlined in Section 1.7.1.22.application-layer message: data which an application wants to send and passed ontothe transport layer; transport-layer segment: generated by the transport layer and encapsulates application-layer message with transport layer header; network-layer datagram: encapsulates transport-layer segment with a network-layer header; link-layer frame: encapsulates network-layer datagram with a link-layer header.23.Routers process layers 1 through 3. (This is a little bit of a white lie, as modernrouters sometimes act as firewalls or caching components, and process layer four as well.)Link layer switches process layers 1 through 2. Hosts process all five layers.Chapter 1 ProblemsProblem 1.There is no single right answer to this question. Many protocols would do the trick. Here's a simple answer below:Messages from ATM machine to ServerMsg name purpose-------- -------HELO <userid> Let server know that there is a card in theATM machineATM card transmits user ID to ServerPASSWD <passwd> User enters PIN, which is sent to server BALANCE User requests balanceWITHDRAWL <amount> User asks to withdraw moneyBYE user all doneMessages from Server to ATM machine (display)Msg name purpose-------- -------PASSWD Ask user for PIN (password)OK last requested operation (PASSWD, WITHDRAWL)OKERR last requested operation (PASSWD, WITHDRAWL)in ERRORAMOUNT <amt> sent in response to BALANCE requestBYE user done, display welcome screen at ATM Correct operation:client serverHELO (userid) --------------> (check if valid userid)<------------- PASSWDPASSWD <passwd> --------------> (check password)<------------- OK (password is OK)BALANCE --------------><------------- AMOUNT <amt>WITHDRAWL <amt> --------------> check if enough $ to coverwithdrawl<------------- OKATM dispenses $BYE --------------><------------- BYEIn situation when there's not enough money:HELO (userid) --------------> (check if valid userid)<------------- PASSWDPASSWD <passwd> --------------> (check password)<------------- OK (password is OK)BALANCE --------------><------------- AMOUNT <amt>WITHDRAWL <amt> --------------> check if enough $ to cover withdrawl<------------- ERR (not enough funds)error msg displayedno $ given outBYE --------------><------------- BYEProblem 2.a) A circuit-switched network would be well suited to the application described, because the application involves long sessions with predictable smooth bandwidth requirements. Since the transmission rate is known and not bursty, bandwidth can be reserved for each application session circuit with no significant waste. In addition, we need not worry greatly about the overhead costs of setting up and tearing down a circuit connection, which are amortized over the lengthy duration of a typical application session.b)Given such generous link capacities, the network needs no congestion control mechanism. In the worst (most potentially congested) case, all the applications simultaneously transmit over one or more particular network links. However, since each link offers sufficient bandwidth to handle the sum of all of the applications' data rates, no congestion (very little queueing) will occur.Problem 3.a)We can n connections between each of the four pairs of adjacent switches. This givesa maximum of 4n connections.b)We can n connections passing through the switch in the upper-right-hand corner andanother n connections passing through the switch in the lower-left-hand corner, giving a total of 2n connections.Problem 4.Tollbooths are 100 km apart, and the cars propagate at 100km/hr. A tollbooth services a car at a rate of one car every 12 seconds.(a) There are ten cars. It takes 120 seconds, or two minutes, for the first tollbooth to service the 10 cars. Each of these cars have a propagation delay of 60 minutes before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 62 minutes. The whole process repeats itself for traveling between the second and third tollbooths. Thus the total delay is 124 minutes.(b) Delay between tollbooths is 7*12 seconds plus 60 minutes, i.e., 61 minutes and 24 seconds. The total delay is twice this amount, i.e., 122 minutes and 48 seconds.Problem 5.a) The time to transmit one packet onto a link is R h L /)(+. The time to deliver the packet over Q links is R h L Q /)(+. Thus the total latency is R h L Q t s /)(++.b) R h L Q /)2(+c) Because there is no store-and-forward delays at the links, the total delay isR L h t s /)(++.Problem 6.a) s m d prop /= seconds.b) R L d trans /= seconds.c) )//(R L s m d end to end +=-- seconds.d) The bit is just leaving Host A.e) The first bit is in the link and has not reached Host B.f) The first bit has reached Host B.g) Want()893105.2102810083=⨯⨯==S R L m km.Problem 7.Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires31064848⨯⋅sec=6msec.The time required to transmit the packet is6101848⨯⋅sec=μ384sec.Propagation delay = 2 msec.The delay until decoding is6msec +μ384sec + 2msec = 8.384msecA similar analysis shows that all bits experience a delay of 8.384 msec.Problem 8.a) 10 users can be supported because each user requires one tenth of the bandwidth.b) 1.0=p .c) ()n n p p n --⎪⎪⎭⎫ ⎝⎛40140. d) ()∑=--⎪⎪⎭⎫ ⎝⎛-90401401n n n p p n . We use the central limit theorem to approximate this probability. Let j X be independent random variables such that ()p X P j ==1.(P “11 or more users”)⎪⎪⎭⎫ ⎝⎛≤-=∑=101401j j X P⎪⎪⎪⎭⎫ ⎝⎛⋅⋅≤⋅⋅-=⎪⎪⎭⎫ ⎝⎛≤∑∑==9.01.04069.01.040410401401j j j j X P X P ()16.36.36≤=⎪⎭⎫ ⎝⎛≤≈Z P Z P 999.0=when Z is a standard normal r.v. Thus (P “10 or more users”)001.0≈.Problem 9.a) 10,000b) ()∑+=--⎪⎪⎭⎫ ⎝⎛M N n n M n p p n M 11Problem 10.It takes R LN / seconds to transmit the N packets. Thus, the buffer is empty when a batch of N packets arrive.The first of the N packets has no queueing delay. The 2nd packet has a queueing delay of R L / seconds. The n th packet has a delay of R L n /)1(- seconds.The average delay is2)1(2)1(11/)1(1101-=-==-∑∑-==N R L N N N R L n N R L R L n N N n N n .Problem 11.a) The transmission delay is R L /. The total delay isIR L R L I R IL -=+-1/)1( b) Let R L x /=. Total delay = axx -1Problem 12.a) There are Q nodes (the source host and the 1-N routers). Let q proc d denote theprocessing delay at the q th node. Let q R be the transmission rate of the q th link and let q q trans R L d /=. Let q prop d be the propagation delay across the q th link. Then[]∑=--++=Q q q prop q trans q proc end to end d d d d 1.b) Let q queue d denote the average queueing delay at node q . Then[]∑=--+++=Q q q queue q prop q trans q proc end to end d d d d d 1.Problem 13.The command:traceroute -q 20 www.eurecom.frwill get 20 delay measurements from the issuing host to the host, www.eurecom.fr. The average and standard deviation of these 20 measurements can then be collected. Do you see any differences in your answers as a function of time of day?Problem 14.a) 40,000 bitsb) 40,000 bitsc) the bandwidth-delay product of a link is the maximum number of bits that can be inthe linkd) 1 bit is 250 meters long, which is longer than a football fielde) s/RProblem 15.25 bpsProblem 16.a) 40,000,000 bitsb) 400,000 bitsc) .25 metersProblem 17.a) t trans + t prop = 400 msec + 40 msec = 440 msecb) 10 * (t trans + 2 t prop ) = 10*(40 msec + 80 msec) = 1.2 secProblem 18.a) 150 msecb) 1,500,000 bitsc) 600,000,000 bitsProblem 19.Let’s suppose the passenger and his/her bags correspond to the data unit arriving to the top of the protocol stack. When the passenger checks in, his/her bags are checked, and a tag is attached to the bags and ticket. This is additional information added in theBaggage layer if Figure 1.20 that allows the Baggage layer to implement the service or separating the passengers and baggage on the sending side, and then reuniting them(hopefully!) on the destination side. When a passenger then passes through security, and additional stamp is often added to his/her ticket, indicating the at the passenger haspassed through a security check. This information is used to ensure (e.g., by later checks for the security information) secure transfer of people.Problem 20.a) time taken to send message from source host to first packet switch =sec 5sec 105.1105.766=⨯⨯. With store-and-forward switching, the total time to move message from source host to destination host = sec 153sec 5=⨯hopsb) time taken to send 1st packet from source host to first packet switch = .sec 1sec 105.1105.163m =⨯⨯. Time at which 2nd packet is received at the first switch = time at which 1st packet is received at the second switch = sec 2sec 12m m =⨯ c) time at which 1st packet is received at the destination host = .sec 33sec 1m hops m =⨯. After this, every 1msec one packet will be received, thus time at which last (5000th ) packet is received =sec 002.5sec 1*4999sec 3=+m m . It can be seen that delay in using message segmentation is significantly less (almost 1/3rd ).d) drawbacks:i. packets have to be put in sequence at the destination.ii. Message segmentation results in many smaller packets. Sinceheader size is usually the same for packets regardless of its size,with message segmentation total amount of header bytes sentincreases.Problem 21.Java AppletProblem 22.Time at which the 1st packet is received at the destination =240⨯+R S sec. After this, one packet is received by the destination every RS 40+sec. Thus delay in sending the whole file = )1(40)40()1(240+⨯+=+⨯-+⨯+=SF R S R S S F R S delay To calculate the value of S which leads to the minimum delay,F S R SS S R F S delay 4001)401(02=⇒=++-⇒=∂∂Chapter 2 Review Questions1.The Web: HTTP; file transfer: FTP; remote login: Telnet; Network News: NNTP; e-mail: SMTP.work architecture refers to organization of communication into layers (e.g., thefive-layer Internet architecture). Application architecture, on the other hand, is designed by an application developer and dictates how the application is (e.g., client-server or P2P)3.Instant Messaging involves the initiator to contact a centralized server to locate theaddress (IP address.) of the receiver: client server model. After this, the instant messaging can be peer to peer – message between the two communicating parties are sent directly between them.4.The process which initiates a service request is the client; the process that waits to becontacted is the server.5.No. As stated in the text, all communication sessions have a client side and a serverside. In a P2P file-sharing application, the peer that is receiving a file is typically the client and the peer that is sending the file is typically the server.6.The IP address of the destination hosts and the port numbers of the destination socket.7.You probably use a browser and a mail reader on a daily basis. You may also use anFTP user agent, a Telnet user agent, an audio/video player user agent (such as a Real Networks player), an instant messaging agent, a P2P file-sharing agent, etc.8.There are no good examples of an application that requires no data loss and timing. Ifyou know of one, send an e-mail to the authors.9. A protocol uses handshaking if the two communicating entities first exchange controlpackets before sending data to each other. SMTP uses handshaking at the application layer whereas HTTP does not.10.The applications that use those protocols require that all application data is receivedin the correct order and without gaps. TCP provides this service whereas UDP does not.11.In both cases, the site must keep a database record for the user. With cookies, the userdoes not explicitly provide a username and password each time it visits the site.However, browser identifies the user by sending the user’s cookie number each time the user accesses the site.12.In persistent HTTP without pipelining, the browser first waits to receive a HTTPresponse from the server before issuing a new HTTP request. In persistent HTTP withpipelining, the browser issues requests as soon as it has a need to do so, without waiting for response messages from the server.13.Web caching can bring the des ired content “closer” to the user, perhaps to the sameLAN to which the user’s host is connected. Web caching can reduce the delay for all objects, even objects that are not cached, since caching reduces the traffic on links. 15.FTP uses two parallel TCP connections, one connection for sending controlinformation (such as a request to transfer a file) and another connection for actually transferring the file. Because the control information is not sent over the same connection that the file is sent over, FTP sends control information out of band.16.Message is sent from Alice’s host to her mail server over HTTP. Alice’s mailserver then sends the message to Bob’s mail server over SMTP. Bob then transfers the message from his mail server to his host over POP3.18.With download and delete, after a user retrieves its messages from a POP server, themessages are deleted. This poses a problem for the nomadic user, who may want to access the messages from many different machines (office PC, home PC, etc.). In the download and keep configuration, messages are not deleted after the user retrieves the messages. This can also be inconvenient, as each time the user retrieves the stored messages from a new machine, all of non-deleted messages will be transferred to the new machine (including very old messages).19.Yes an organization’s mail server and Web server can have the same alias for a hostname. The MX record is used to map the mail server’s host name to its IP address 20.The overlay network in a P2P file sharing system consists of the nodes participatingin the file sharing system and the logical links between the nodes. There is a logical directed link (a “directed edge” in graph theory terms) from node A to node B if nodeA sends datagrams that are explicitly addressed to node B’s IP address. An overlaynetwork does not include routers. With Gnutella, when a node wants to join the Gnutella network, it first discovers (“out of band”) the IP address of one or more nodes already in the network. It then sends join messages to these nodes. When the node receives confirmations, it becomes a member of the of Gnutella network. Nodes maintain their logical links with periodic refresh messages.21.Three companies as of this writing (August 2002) are KaZaA, eDonkey, Bit Torrent.22.With the UDP server, there is no welcoming socket, and all data from different clientsenters the server through this one socket. With the TCP server, there is a welcoming socket, and each time a client initiates a connection to the server, a new socket is created. Thus, to support n simultaneous connections, the server would need n+1 sockets.23.For the TCP application, as soon as the client is executed, it attempts to initiate a TCPconnection with the server. If the TCP server is not running, then the client will fail to make a connection. For the UDP application, the client does not initiate connections (or attempt to communicate with the UDP server) immediately upon execution Chapter 2 ProblemsProblem 1.a) Fb) Tc) Fd) FProblem 2.Access control commands:USER, PASS, ACT, CWD, CDUP, SMNT, REIN, QUIT.Transfer parameter commands:PORT, PASV, TYPE STRU, MODE.Service commands:RETR, STOR, STOU, APPE, ALLO, REST, RNFR, RNTO, ABOR, DELE, RMD, MRD, PWD, LIST, NLST, SITE, SYST, STAT, HELP, NOOP. Problem 3.SFTP: 115, NNTP: 119.Problem 4.Application layer protocols: DNS and HTTPTransport layer protocols: UDP for DNS; TCP for HTTPProblem 5.Persistent connections are discussed in section 8 of RFC 2616 (the real goal of this question was to get you to retrieve and read an RFC). Sections 8.1.2 and 8.1.2.1 of the RFC indicate that either the client or the server can indicate to the other that it is going toclose the persistent connection. It does so by including the including the connection-token "close" in the Connection-header field of the http request/reply.Problem 6.The total amount of time to get the IP address isn RTT RTT RTT +++ 21.Once the IP address is known, O RTT elapses to set up the TCP connection and another O RTT elapses to request and receive the small object. The total response time isn o RTT RTT RTT RTT ++++ 212Problem 7.a)o o n RTT RTT RTT RTT 2321⋅++++n o RTT RTT RTT +++= 18.b)o o n RTT RTT RTT RTT 221++++n o RTT RTT RTT +++= 14.c)o o n RTT RTT RTT RTT ++++21n o RTT RTT RTT +++= 13.Problem 8.HTTP/1.0: GET, POST, HEAD.HTTP/1.1: GET, POST, HEAD, OPTIONS, PUT, DELETE, TRACE, CONNECT.See RFCs for explanations.Problem 9.a) The time to transmit an object of size L over a link or rate R is L/R . The average time is the average size of the object divided by R :∆= (900,000 bits)/(1,500,000 bits/sec) = .6 secThe traffic intensity on the link is (1.5 requests/sec)(.6 msec/request) = .9. Thus, the average access delay is (.6 sec)/(1 - .9) = 6 seconds. The total average response time is therefore 6 sec + 2 sec = 8 sec.b) The traffic intensity on the access link is reduced by 40% since the 40% of therequests are satisfied within the institutional network. Thus the average access delayis (.6 sec)/[1 – (.6)(.9)] = 1.2 seconds. The response time is approximately zero if the request is satisfied by the cache (which happens with probability .4); the averageresponse time is 1.2 sec + 2 sec = 3.2 sec for cache misses (which happens 60% of the time). So the average response time is (.4)(0 sec) + (.6)(3.2 sec) = 1.92 seconds. Thus the average response time is reduced from 8 sec to 1.92 sec.Problem 11.UIDL abbreviates “unique-ID listing”. When a POP3 client issues the UIDL command, the server responds with the unique message ID for all of the messages present in the users mailbox. This command is useful for “download and keep”. By keeping a file that lists the messages retrieved in earlier sessions, the client can use the UIDL command to determine which messages on the server have already been seen.Problem 13.a)For a given input of domain name (such as ), IP address or networkadministrator name, whois database can be used to locate the corresponding registrar, whois server, dns server, etc.f)An attacker can use the whois database and nslookup tool to determine the IP addressranges, DNS server addresses, etc. for the target institution.g)If under an attack a victim can analyze the source address of packets, the victim canthen use whois to obtain information about domain from which attack is coming and possibly inform the administrators of the origin domain.Problem 14.No. Because the link is full duplex, you have 128 kbps in each direction, and the uploading does not interfere with the downloading.Problem 15.There are N nodes in the overlay network. There are N(N-1)/2 edges.Problem 16.a) In this case, each of the five Gnutella clients immediately learns that it has one less neighbor. Consider one of these five clients, called, Bob. Suppose Bob has only three neighbors after X drops out. Then Bob needs to establish a TCP connection with another peer. Bob should have a fresh list of active peers; he sequentially contacts peers on this list until one accepts his TCP connection attempt.b) In this case, Bob does not immediately know that X has departed. Bob will only learn about X’s departure when it attempts to send a message (query or ping) to X. When Bob attempts to send a message, Bob’s TCP w ill make several unsuccessful attempts to send the message to B. Bob’s TCP will then inform the Gnutella client that X is down. Bob。

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Chapter 1 Computer Networks and the Internet1．The ( ) is a worldwide computer network, that is, a network that interconnects millions of computing devices throughout the world. ppt3A public InternetB IntranetC switch netD television net2．Which kind of media is not a guided media? ( )A twisted-pair copper wireB a coaxial cableC fiber opticsD digital satellite channel3．Which kind of media is a guided media? ( )A geostationary satelliteB low-altitude satelliteC fiber opticsD wireless LAN4．The units of data exchanged by a link-layer protocol are called ( ).A FramesB SegmentsC DatagramsD bit streams5．Which of the following option belongs to the circuit-switched networks? ( )A FDMB TDMC VC networksD both A and B6．( )makes sure that neither side of a connection overwhelms the other side by sending too many packets too fast.A Reliable data transferB Flow controlC Congestion controlD Handshaking procedure7．( ) means that the switch must receive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link.A Store-and-forward transmissionB FDMC End-to-end connectionD TDM8．Datagram networks and virtual-circuit networks differ in that ( ).A datagram networks are circuit-switched networks,and virtual-circuit networks are packet-switchednetworks.B datagram networks are packet-switched networks,and virtual-circuit networks are circuit-switchednetworks.C datagram networks use destination addresses andvirtual-circuit networks use VC. numbers to forwardpackets toward their destination.D datagram networks use VC. numbers andvirtual-circuit networks use destination addresses toforward packets toward their destination.9．In the following options, which one is not a guided media? ( )A twisted-pair wireB fiber opticsC coaxial cableD satellite10．Processing delay does not include the time to ( ).A examine the packet’s headerB wait to transmit the packet onto the linkC determine where to direct the packetD check bit-error in the packet11．In the following four descriptions, which one is correct? ( )A The traffic intensity must be greater than 1.B The fraction of lost packets increases as the trafficintensity decreases.C If the traffic intensity is close to zero, the averagequeuing delay will be close to zero.D If the traffic intensity is close to one, the averagequeuing delay will be close to one.12．The Internet’s network layer is responsible for moving network-layer packets known as ( ) from one host to another.A frameB datagramC segmentD message13．The protocols of various layers are called ( ).A the protocol stackB TCP/IPC ISPD network protocol14．There are two classes of packet-switched networks: ( ) networks and virtual-circuit networks.A datagramB circuit-switchedC televisionD telephone15．Access networks can be loosely classified into three categories: residential access, company access and ( )access.A cabledB wirelessC campusD city areaQuestion 16～17Suppose, a is the average rate at which packets arrive at the queue, R is the transmission rate, and all packets consist of L bits, then the traffic intensity is ( 16 ), and it should no greater than ( 17 ).16． A LR／aB La／RC Ra／LD LR／a17．A 2B 1C 0D -118．In the Internet, the equivalent concept to end systems is ( ).A hostsB serversC clientsD routers19．In the Internet, end systems are connected together by ( ).A copper wireB coaxial cableC communication linksD fiber optics20．End systems access to the Internet through its ( ).A modemsB protocolsC ISPD sockets21．End systems, packet switches, and other pieces of the Internet, run ( ) that control the sending and receiving of information within the Internet.A programsB processesC applicationsD protocols22．There are many private networks, such as many corporate and government networks, whose hosts cannot exchange messages with hosts outside of the private network. These private networks are often referred to as ( ).A internetsB LANC intranetsD WAN23．The internet allows ( ) running on its end systems to exchange data with each other.A clients applicationsB server applicationsC P2P applicationsD distributed applications24．The Internet provides two services to its distributed applications: a connectionless unreliable service and ()service.A flow controlB connection-oriented reliableC congestion controlD TCP25．It defines the format and the order of messages exchanged between two or more communicating entities, as well as the actions taken on the transmission and/or receipt of amessage or other event. The sentence describes ( ).A InternetB protocolC intranetD network26．In the following options, which does not define in protocol? ( )A the format of messages exchanged between two ormore communicating entitiesB the order of messages exchanged between two or morecommunicating entitiesC the actions taken on the transmission of a message orother eventD the transmission signals are digital signals or analogsignals27．In the following options, which is defined in protocol? ( )A the actions taken on the transmission and/or receipt ofa message or other eventB the objects exchanged between communicating entitiesC the content in the exchanged messagesD the location of the hosts28．In the following options, which does not belong to the network edge? ( )A end systemsB routersC clientsD servers29．In the following options, which belongs to the network core? ( )A end systemsB routersC clientsD servers30．In the following options, which is not the bundled with the Internet’s connection-oriented service? ( )A reliable data transferB guarantee of the transmission timeC flow controlD congestion-control31．An application can rely on the connection to deliver all of its data without error and in the proper order. The sentencedescribes ( ).A flow controlB congestion-controlC reliable data transferD connection-oriented service32．It makes sure that neither side of a connection overwhelms the other side by sending too many packets too fast. Thesentence describes ( ).A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer33．It helps prevent the Internet from entering a state of gridlock.When a packet switch becomes congested, its buffers can overflow and packet loss can occur. The sentence describes ( ).A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer 34．The Internet’s connection-oriented service has a name, it is ( ).A TCPB UDPC TCP/IPD IP35．In the following options, which service does not be provided to an application by TCP?( )A reliable transportB flow controlC video conferencingD congestion control36．The Internet’s connectionless service is called ( ).A TCPB UDPC TCP/IPD IP37．In the following options, which does not use TCP?( )A SMTPB internet telephoneC FTPD HTTP38．In the following options, which does not use UDP?( )A Internet phoneB video conferencingC streaming multimediaD telnet39．There are two fundamental approaches to building a network core, ( ) and packet switching.A electrical current switchingB circuit switchingC data switchingD message switching40．In ( ) networks, the resources needed along a path to provide for communication between the end system arereserved for the duration of the communication session.A packet-switchedB data-switchedC circuit-switchedD message-switched41．In ( ) networks, the resources are not reserved; a session’s messages use the resources on demand, and asa consequence, may have to wait for access tocommunication link.A packet-switchedB data-switchedC circuit-switchedD message-switched42．In a circuit-switched network, if each link has n circuits, for each link used by the end-to-end connection, theconnection gets ( ) of the link’s bandwidth for theduration of the connection.A a fraction 1/nB allC 1/2D n times43．For ( ), the transmission rate of a circuit is equal to the frame rate multiplied by the number of bits in a slot.A CDMAB packet-switched networkC TDMD FDM44．( ) means that the switch must receive the entire packet before it can begin to transmit the first bit of the packetonto the outbound link.A Queuing delayB Store-and-forward transmissionC Packet lossD Propagation45．The network that forwards packets according to host destination addresses is called ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram46．The network that forwards packets according to virtual-circuit numbers is called ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram47．In the following entries, which is not a kind of access network?( )A residential accessB company accessC wireless accessD local access48．Suppose there is exactly one packet switch between a sending host and a receiving host. The transmission rates between the sending host and the switch and between theswitch and the receiving host are R1 and R2, respectively.Assuming that the switch uses store-and-forward packet switching, what is the total end-to-end delay to send apacket of length L? (Ignore queuing delay, propagationdelay, and processing delay.) ( )A L/R1＋L/R2B L/R1C L/R2D none of the above49．The time required to examine the packet’s header and determine where to direct the packet is part of the ( ).A queuing delayB processing delayC propagation delayD transmission delay50．The time required to propagate from the beginning of the link to the next router is ( ).A queuing delayB processing delayC propagation delayD transmission delay51．Consider sending a packet of 3000bits over a path of 5 links. Each link transmits at 1000bps. Queuing delays, propagation delay and processing delay are negligible. (6 points)(1).Suppose the network is a packet-switched virtual circuit network. VC setup time is 0.1 seconds. Suppose the sending layers add a total of 500 bits of header to each packet. How long does it take to send the file from source to destination? (2).Suppose the network is a packet-switched datagram network and a connectionless service is used. Now suppose each packet has 200 bits of header. How long does it take to send the file?(3).Suppose that the network is a circuit-switched network. Further suppose that the transmission rate of the circuit between source and destination is 200bps. Assuming 0.02s setup time and 200 bits of header appended to the packet, how long does it take to send the packet?Solution: (1). t=5*(3000+50 0)/1000+0.1=17.6s(2). t=5*(3000+20 0)/1000=16s(3). t=(3000+200) /200+0.02=16.02s。