【命题要点】英语词语是最基础的部分，由于网络技术是不断更新的领域，不断有新的思想涌现，也伴随着新的词汇出现。由于新出现的词汇和网络技术的专业性，往往造成理解上的偏差，因此需要应试者在基本了解网络技术中英语专业词汇的基础上，将英语词汇和汉语词汇在功能和语义上相对应形成正确的理解。应试者需要准确掌握词语的意义，区分同义词在意义和使用上的差别；准确掌握名词单、复数形式，以及由单、复数形式带来不同语义的解释；准确掌握关系代词、关系副词、联系词在语句乃至整个语篇中的逻辑意义。在词汇复习中，尤其还需要注意专业词汇的缩写，这些缩写往往是某些技术、设备或协议的代称，在整个试题中是核心词汇。复习时应多读一些计算机网络方面的英语时文。解题时，一般先考虑语义，后考虑语法。一、考题格式上午科目的71?75题一般是完形填空的形式。主要考查应试者结合计算机专业技术知识对全文综合理解的程度，和联系上下文的能力；应试者语法知识和对句法结构的辨识能力；应试者的词汇量和词汇运用能力。具体而言，完形填空主要考查应试者对语篇中句法、词语和短语的把握能力，具有较强的测试性。每一个空处都要通过上下文进行综合考虑，仅仅依靠一个单句往往无法确立正确选项。语篇的内容往往是对网络技术中协议、通信过程、设备、最新技术等相关知识的描述，需要应试者对于这些内容有一定的了解。二、答题要领首先，通过首句或出现的核心词汇来推断全文的信息。短文的首句往往是主题句，或出现了核心词汇，能为理解文章的大意和主要内容提供必要线索。一般首句还提供背景资料，因此要特别注意首句，抓住整个段落的纲要。其次，把握文章发展的基本线索。文章总是按照一定思路发展起来的，而不同的逻辑关系主要依靠使用逻辑连接词来表达的，文章如果没有出现内在的逻辑关系，就会出现语义不清，逻辑混乱。所以通过表示逻辑关系的词汇把握文章发展的基本线索是至关重要的。借助语法知识和专业背景知识确定正确的词汇选项。计算机专业英语词汇的考查在试题中占一定比例，词汇选项的设计和文章难度的制定与语法都息息相关。应试者务必借助语法知识和专业背景知识来确定正确的词汇选项。同时注意填入的词汇和文中句子的结构要求相一致。三、答题步骤第一步，通读全文。完形填空是考查全面理解内容的基础上运用语言的能力，由于试题篇幅较短，完全有时间利用通读对全文内容有一个基本的了解。应试者要快速阅读段落，把握基本观点，通读时以浏览为主，可以忽略细节。第二步，复读答题。在通读的基础上，应试者最好能立即复读，并结合选项，从语法结构、语义、词义、固定搭配等方面结合专业知识来考虑选项。选定之后，还需要回读。在整个答题过程中，切记全文的整体意义，保持思路的连贯性，从而做出最合适的正确选择。第三步，重读检查。在确定了所有选项以后，一定还要重读全文，检查并核实每个选项在正篇文章中没有造成语义、结构、逻辑等方面的差错，确保短文是一个内容连贯、层次清晰、中心思想突出的整体。四、常用词汇下面列出常用的计算机网络技术基本术语，供考生复习时参考。accept 接受 access control 访问控制acknowledgement (ACK)确认adaptive routing 自适应路由address field 地址字段amplitude 振幅analog signal 模拟信号anonymous FTP 匿名 FTPapplication layer 应用层asynchronous 异步backbone 主干bandwidth 带宽baseband mode 基带模式baud rate波特率binary exponential backoff algorithm 二进制指数退避算法bit-oriented protocol 面向比特的协议 block check character 块校验字符Border Gateway Protocol 边界网关协议bridge网桥bridge protocol data unit 网桥协议数据单元broadcast address 广播地址 Brouter桥路器buffering 缓冲cable modem电缆调制解调器Caesar cipher凯撒密码 call confirmation 呼叫证实call request呼叫请求Carrier Sense Multiple Access (CSMA)载波侦听多路访问Carrier Sense Multiple Access with Collision Detection (CSMA/CD)带冲突检测的载波帧听多路访问carrier signal载波信号 Cell信元channel 信道 Checksum校验和Choke packet抑制分组 ciphertext 密文circuit switching 电路交换 client客户client/server model客户端/服务器模式 coaxial cable 同轴电缆code编码 common bus topology 公共总线拓扑结构Common Gateway公共网关接口 Common Management Information Protocol通用管理信息协议communications subnet 通信子网 compact disc 光盘compression 压缩 congestion 拥塞connection 连接contention 竞争contention protocol 竞争协议control bits 控制位control character 控制字符Data Circuit-Terminating Equipement (DCE)数据电路端接设备Data Encryption Standard (DES)数据加密标准data link layer数据链路层 datagram数据报decryption 解密 demodulation 解调destination address 目的地址 differential encoding 差分编码digital signal 数字信号 digital signature 数字签名directory service 目录服务 domain 域Domain Name System 域名系统 echo reply回送响应echo request回送请求 electronic mail (email)电子邮件encryption 加密 encryption key 密钥error control 差错控制 error correction 差错校正error detection 差错检测Ethernet以太网even parity 偶校验 exterior gateway protocol 外部网关协议fiber distributed data interface 光纤分布式数据接口file server文件服务器file transfer protocol 文件传输协议filter滤波器flow control流量控制fragment 段fragmentation 分段frame 帧 frequency 频率frequency-division multiplexing 频分多路复用full duplex 全双工 gateway 网关graded-index multimode fiber 级率多模光纤hacker黑客half duplex 半双工Hamming code 海明码handshaking 握手High-level Data Link Control (HDLC)高级数据链路控制协议host主机hypertext transfer protocol (HTTP)超文本传送infrared light 红外线 Internet因特网Internet Control Message Protocol (ICMP) 网际控制报文协议Internet Protocol 网际协议 Internet worm因特网蠕虫key exchange密钥交换 laser激光local area network (LAN)局域网local exchange本地交换局local loop本地回路Manchester code曼彻斯特编码medium access control (MAC)媒体访问控制message handling 报文处理message transfer protocol 报文传送协议modem 调制解调器modulation 调制multiplexer 多路复用器 network 网络Network Control Protocol 网络控制协议 network interface card (NIC)网络接口卡network layer protocol 网络层协议 network topology网络拓扑结构network virtual terminal 网络虚拟终端 Non-persistent CSMA 非坚持CSMANyquist theorem 奈奎斯特定理 octet字节odd parity奇校验 Open Systems Interconnect (OSI)开放系统互连packet分组packet header 分组头packet-switched network分组交换网络parity bit奇偶校验位path通路path control通路控制physical layer 物理层 pixels像素plaintext 明文 prefix前缀presentation layer 表示层 protocol 协议pulse amplitude modulation 脉冲幅度调制pulse code modulation 脉码调制radio 无线电收发机receiving window 接收窗口remote logins 远程登录 repeater 中继器reply 应答root bridge 根网桥root port 根端口route 路由router 路由器 routing algorithm 路由算法Routing Information Protocol路由信息协议routing table 路由表sampling frequency 采样率satellite人造卫星script脚本 security 安全 segment 段sequence number 序列号 serial transmission 串行传输server服务器 session 会话session control 会话控 session layer 会话层signal-to-noise ratio 信噪比 simplex 单工Simple Mail Transfer Protocol (SMTP)简单 邮件传输协议single-mode fiber 单模光纤sliding window 滑动窗口协议socket套接字source address 源地址source quench 源抑制spanning tree algorithm 生成树算法start bit起始位start of frame delimiter 帧起始定界符static routing 静态路由stop bit 停止位subnet子网 successor 后继switch交换机SYN character同步字符Synchronous 同步Synchronous Optical Network (SONET)同步光纤网telegraph 电报 terminal adapter终端适配器three-way handshake 三次握手 time exceeded 超时time to live 生存期 time-division 时分multiplexing 多路复用 timestamp reply 时戳应答timestamp request 时戳请求 token令牌 transceiver 收发器transmission rate 传输速率transparent bridge 透明网桥transport layer 运输层 twisted pair 双绞线tunneling 隧道Uniform Resource Locator (URL)统一资源定位器User Datagram Protocol (UDP)用户数据报协议 verification 验证virutalcircuit (route) 虚电路（路由） virus病毒wide area network (WAN)广域网 window 窗口World Wide Web 万维网 worm蠕虫【试题14-1】 2013年5月真题71~75Traditional IP packet forwarding analyzes the (71) IP address contained in the network layer header of each packet as the packet travels from its source to its final destination. A router analyzes the destination IP address independently at each hop in the network. Dynamic (72) protocols or static configuration builds the database needed to analyze the destination IP address (the routing table). The process of implementing traditional IP routing also is called hop-by-hop destination-based (73) routing. Although successful , and obviously widely deployed, certain restrictions, which have been realized for some time, exist for this method of packet forwarding that diminish its (74) .New techniques are therefore required to address and expand liie functionality of an IP-based network infrastructure. This first chapter concentrate on identifying these restrictions and presents a new architecture, known as multipleprotocol (75) switching, that provides solutions to some of these restrictions.(71) A.datagram B.destinationC.connectionD.service(72)A.routing B.forwarding C.transmissionD.management(73)A.anycast B.multicast C.broadcast D.unicast(74)A.reliabilityB.flexibilityC.stability D.capability(75)A.const B.cast C.mark D.label解析：当数据包从其来源到其最终目的地址时，传统的IP报文转发分析包含在每一个数据包的网络层 报头的目的IP地址。路由器独立地分析网络中的每一跳目标IP地址。动态路由协议或静态配置生成所需的数 据库来分析目标IP地址（路由表）。传统的IP路由实现的过程也被称为基于逐跳目的地的单播路由。虽然成功地广泛釆用，但是一些已经实现一段时间的限制仍存在，此方法存在报文的转发减少其灵活性。因此，需要新的技术，以处理和扩展一个基于IP的网络基础设施的功能。第一章着重识别这些限制，并提出了一种新的架构，被称为多协议标签交换，来提供解决这些限制的方法。【答案：（71) B、(72) A、(73) D、（74) B、(75) D】【试题14-2】 2013年11月真题71~75The de facto standard application program interface (API) for TCP/IP applications is the "sockets" interface. Although this API was developed for (71) in the early 1980s. it has also been implemented on a wide variety of non-Unix systems. TCP/EP (72) written using the sockets API have in the past enjoyed a high degree of portability and we would like the same (73) with IPv6 applications. But changes are required to the sockets API to support IPv6 and this memo describes these changes. These include a new socket address structure to carry IPv6 (74) ，new address conversion functions, and some new socket options. These extensions are designed to provide Access to the basic IPv6 features required by TCP and UDP applications, including multicasting,while introducing a minimum of change into the system and providing complete (75) for existing IPv4.(71) A.Windows B.Linux C.Unix D.DoS(72) A.applications B.networks C.protocols D.systems(73) A.portabilityB.availabilityC.capabilityD.reliability(74) A.connections B.protocols C.networks D.addresses(75) A.availability B.compatibility C.capability D.reliability解析：事实上的标准应用程序接口（API),用于TCP/ IP应用程序是“套接字”界面。虽然这个API 是在80年代初为Unix开发的。它也被在各种各样的非Unix系统的实施。使用套接字API编写的TCP/. IP应用程 序在过去享有的可移植性程度高，我们想同样的便携性与IPv6应用。但改变是必需的套接字API,支持IPv6和 本备忘录描述了这些变化。这包括一个新的套接字地址结构来进行IPv6地址，新地址转换功能，和一些新的套接字选项。这些扩展的目的是提供访问IPv6基本功能由TCP和UDP应用程序，包括多播要求，同时引入一个 最小变化到系统中，并提供给现有的IPv4应用程序完全兼容。【答案：（71) C、(72) A、(73) A、(74) D、(75) B】【试题14-3】 2012年5月真题71~75The TCP protocolis a (71) layer protocol. Each connection connects two TCPs that may be just one physical network apart or located on opposite sides of the globe. In other words, each connection creates a (72) witha length that may be totally different from another path created by another connection. This means that TCP cannot use the same retransmission time for all connections. Selecting a fixed retransmission time for all connections can result in serious consequences. If the retransmission time does not allow enough time for a (73) to reach the destination and an acknowledgment to reach the source, it can result in retransmission of segments that are still on the way. Conversely, if the retransmission time is longer than necessary for a short path, it may result in delay for the application programs .Even for one single connection, the retransmission time should not be fixed.A connection may be able to send segments and receive (74) faster during nontraffic period than during congested periods. TCP uses the dynamic retransmission time,a transmission time is different for each connection and which may be changed during the same connection. Retransmission time can be made (75) by basing it on the round-trip time (RTT). Several formulas are used for this purpose.(71) A.physical B.network C.transport D.application(72) A.path B.window C. response D.process(73) A.process B.segment C.program D.user(74) A.connectionsB.requestsC. acknowledgmentsD.datagrams(75) A.long B.short C.fixed D.dynamic解析：TCP是一种传输层协议，每个TCP连接都连接着两个TCP,这两个TCP可能是在一个物理网络里， 但是是分开的或者位于全局网络的对立面。换句话说，每个连接创建一个路径的长度可能完全不同于由另外一个连接创建的路径。这意味着TCP不能为所有的连接使用相同的转发时间。为所有的连接选择一个固定的 转发时间可能会导致严重的后果。如果这个转发时间并不为一个段提供足够的时间去到达目的地以确认获取 来源，将导致段转发的部分仍在路上。相反，如果转发时间超过了必要的短路径，它可能会导致延迟的应用程序，甚至一个单独的连接，转发时间也应该是不固定的，相比于拥挤的时期，一个连接在不拥挤的时期可 能能够更快地发送和接收确认。TCP使用动态转发时间，每个连接的传输时间是不同的，而且可能在相同的连接中变化。转发时间可能是动态的基于往返时间，有几个准则用于这一目的。【答案：（71) C； (72) A； (73) B； (74) C； (75) D】【试题14-4】 2012年11月真题71~75Let us now see how randomization is done when a collision occurs.After a (71) , time is divided into discrete slots whose length is equal to the worst-case round-trip propagation time on the ether (2t) .To accommodate the longest path allowed by Ethernet, the slot tome has been set tO 512 bit times,or 51.2(isec.After the first collision,each station waits either 0 or 1 (72) times before trying again.If two stations collide and each one picks the same random number,they will collide again.After the second collision,each one picks either 0，l，2,or3 at random and waits that number of slot times.If a third collision occurs (the probability of this happening is 0.25) ,then the next time the number of slots to wait is chosen at (73) from the interval 0 to 23-l.In general,after i collisions,a random number between 0 and 2*-l is chosen,and that number of slots is skipped.However, after ten collisions have been reached,the randomization (74) is frozen at a maximum of 1023 slots.After 16 collisions,the controller throws in the towel and reports failure back to the computer.Further recovery is up to (75) layers.(71) A.datagramB.collisionC.connectionD.service(72)A.slot B.switch C.process D.fire(73)A.rest B.randomC.once D.odds(74)A.unicastB.multicastC.broadcastD.interval(75)A.local B.next C.higher D.lower解析：现在让我们观察冲突发生时如何做随机处理。冲突发生后，时间被划分成离散的长度，等于最坏的往返传播时间的时槽。为了容纳以太网允许的最长路径，冲突时槽缩小为5.12μs。第一次冲突后，每个站在再次尝试前需要等待0或1时槽。如果每一站发生冲突，且每一个挑选相同的随机数，它们将再次发生冲突,第2次冲突之后，每一站随机选择0、1、2或3，即等待时槽的个数。如果第3次冲突发生（发生的概率为0.25)，则下次时槽的等待数量都是在0?23-1之间随机挑选的。总体来说，第i次冲突后，将在0~2i-1:之间挑选随机数，而且那个时槽数是被略过的。然而，当冲突次数达到10次，随机间隔被锁定在最高1023时槽。当发生16次冲突后，控制器丢弃报告而不再返回计算机。进一步恢复已经达到更高的层次。【答案：（71) B； (72) A； (73) B； (74) D； (75) C】【试题14-5】 2011年5月真题71-75Border Gateway Protocol(BGP) is inter-autonomous system (71) protocol. BGP is based on a routing method called path vector routing. Distance vector routing is not a good candidate for inter-autonomous system routing because there are occasions on which the route with the smallest (72) count is not the preferred route. For example, we may not want a packet through an autonomous system that is not secure even though it is shortest route. Also, distance vector routing is unstable due to the fact that the routers announce only the number of hop counts to the destination without defining the path that leads to that (73) . A router that receives a distance vector advertisement packet may be fooled if the shortest path is actually calculated through the receiving router itself. Link (74) routing is also not a good candidate for inner-autonomous system routing because an internet is usually too big for this routing method. To use link state routing for the whole internet would require each router to have a huge link state database. It would also take a long time for each router to calculate its routing (75) using the Dijkstra algorism.(71) A.routingB.switchingC.transmittingD.receiving(72) A.path B.hop C.route D.packet(73) A.connection B.window C.source D.destination(74) A.status B.search C.state D.research(75) A.table B.state C.metric D.cost参考译文：边界网关协议（BGP)是自治系统间的路由协议。BGP是基于路由的方法称为距离矢量路 由。距离矢量路由是自己系统路由的很好的候选者，因为跨自治系统路由场合上最小的跳数的路由不一定是最合适的路由。例如，我们可能不希望通过一个自治系统不安全的数据包，即使它是最短的路线。此外，距离矢量路由是不稳定的，路由器只宣布到目的地的跳数，但不指出到达目的地的路径。实际上，如果是通过接收路由器本身计算的最短路径，接收距离矢量通告报文的路由器可能被愚弄。链路状态路由在跨自治系统中也不是一个好的候选者，因为对于这种方法互联网是通常过大。要对整个互联网使用链接状态路由，就需要每个路由器有一个巨大的链路状态数据库。它也需要每个路由器用很长的时间来使用的Dijkstra算法计算自己的路由表。【答案:(71) A； (72) B； (73) D； (74) C； (75) A】【试题14-6】 2011年11月真题71~75A transport layer protocol usually has several responsibilties.One is to create a process-to-process communication UDP uses (71) numbers to accomplish this.Another responsibility is to provide control mechanisms at the transport level UDP does this task at a very minimal level.There is no flow control mechanism and there is no (72)for received packet.UDP,however,does provide error control to some extent.If UDP detects an error in the received packet,it will silently drop it.The transport layer also provides a connection mechanism for the processes.The (73) must be able to send streams of data to the transport layer.It is the responsibility of the transport layer at (74) station to make the connection with the receiver chop the stream into transportable units,number them，and send them one by one.it is the responsibility of the transport layer at the receiving end to wait until all the different units belonging to the same process have arrived,check and pass those that are (75) freehand deliver them to the receiving process as a stream.(71) A.hop B.port C.route D.packet(72) A.connectionB.windowC.acknowledgementD.destination(73) A.jobs B.processes C.programs D.users(74) A.sending B.routing C.switching D.receiving(75) A.call B.state C.cost D.error参考译文：传输层协议通常有多种责任。一个是创建进程间通信，UDP使用端口数来完成这项工作。另一个是在传输层提供控制机制。UDP以一个最低的级别完成这项工作，无流控制机制和接收包的应答机制。但是，UDP在一些范围上提供了差错控制。如果UDP在接收包中检测到了错误，它将静静地丢弃它。传输层同样为进程提供了一个连接机制。这个进程必须能发送数据流到传输层。而在发送站传输层的责任就是与接收站建立连接，将流放入传输单元，并编号，然后一个接一个地发送。接收站传输层的责任是等待一个相同进程中的不同的数据单元的到达，并且检查它们，将错误单元丢弃，并且以流的形式传递到接收进程中。【答案：（71) B； (72) C； (73) B； (74) A； (75) D】【试题14-7】 2010年5月真题71~75Although a given waveform may contain frequencies over a very broad range, as a practical matter any transmission system will be able to accommodate only a limited band of (71) . This, in turn, limits the data rate that can be carried on the transmission (72) . A square wave has an infinite number of frequency components andhence an infinite (73) . However, the peak amplitude of the kth frequency component, kf, is only 1/k, so most of the (74) in this waveform is in the first few frequency components. In general, any digital waveform will have (75) bandwidth. If we attempt to transmit this waveform as a signal over any medium, the transmission system will limit the bandwidth that can be transmitted.(71) A.frequencies B.connections C.diagrams D.resources(72) A.procedures B.function C.route D.medium(73) A.source B.bandwidth C.energy D.cost(74) A.frequency B.nergy C.amplitude D.phase(75) A.small B.limited C.infinite D.finite参考译文：虽然一个给定的波形包含了很宽的频率范围，但任何实际的传输系统只能通过有限的频率。这样，就限制了传输介质可以承载的数据速率。一个方波包含了无限多的频率成分，因而也具有无限的带宽。然而，第k个频率成分的峰值幅kf只是1/k，所以波形的大部分能量只是包含在前面的少数频率成分中。一般来说，任何数字波形都有无限带宽。如果我们试图在某种介质上传输这种波形信号，则传输系统实际上会限制可以发送的带宽。【答案：(71) A； (72) D； (73) B； (74) B； (75) C】【试题14-8】 2010年11月真题71~75The metric assigned to each network depends on the type of protocol.Some simple protocol, like RIP, treats each network as equals.The (71) of passing through each network is the same;it is one (72) count.So if a packet passes through 10 network to reach the destination, the total cost is 10 hop counts.Other protocols, such as OSPF, allow the administrator to assign a cost for passing through a network based on the type of service required.A (73)through a network can have different costs (metrics) .For example, if maximum (74) is the desired type of service, a satellite link has 狂 lower metric than a fiber-optic line.On the other hand, if minimum (75) is the desired type of service, a fiber-optic line has a lower metric than a satellite line.OSPF allow each router to have several routing table based on the required type of service.(71) A.numberB.connectionC.diagramD.cost(72) A.processB.hop C.route D.flow(73) A.flow B.window C.route D.cost(74) A.packet B.throughput C.error D.number(75) A.delay B.stream C.packet D.cost参考译文；每个网络釆用的度量标准依赖于协议类型。对于简单的协议，如RIP，将每个网络同等对待。 通过每个网络的代价都是一跳。因此，如果一个分组通过10个网络后到达目的地，则总的代价为10个其他协议，如OSPF，允许网络管理员基于服务为通过的网络设置代价。因此，通过一个网络的路由可能有不同的代价（度量标准）。例如，若期望服务类型是最大吞吐量，则卫星链路比光纤线路的度量小。如果期望的服务类型是最小时延，则光纤线路的度量小于卫星链路。OSPF允许每个路由器有几张基于服务类型的路由表。【答案：（71) D; (72) B； (73) C； (74) B； (75) A】