ch5 Computer network

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通信网概述.

通信网概述.
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时钟原理
• 原子钟和晶体钟
• 全球定位系统------GPS
• 通信大楼综合定时供给系统------BITS
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我国数字同步网的等级结构
第一级 长 途 网 第二 级 A 类 B 类 本 地 网 第三级 第四级 基准时钟 一级和二级长途交换中心、国际局的局内 综合定时供给设备时钟和交换设备时钟 三级和四级长途交换中心的局内综合定时 供给设备时钟和交换设备时钟 在大城市内有多个长 途交换中心时,应按 照它们在网中的等级 相应地设置时钟
电信网的基本组成
电信网是由用户终端设备、交换设备、传输设备组成
端局 用户 汇接局
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电信网的基本组成
• 电信网:交换设备+传输设备+终端设备
– 交换设备:完成信息的交换
– 传输设备:负责信息的传送
– 终端设备:负责信息的接入
• 电信网的硬件通常被划为接入网、交换网和传输网。
2
电信网的划分
按照实现网络业务功能划分: – 电信基本业务网
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广东省TMN管理现状
目前已开发的主要系统有: • 电话网路管理和集中操作维护系统 • 传输监控系统 • 112集中受理测试系统 • 七号信令网管系统 • 机房环境和动力监控系统等
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广东省交换网管系统结构
省网管中心系统 数据通信网 DCN
本地网管系统
本地网管系统
本地数据通信网
DCN
网管接口机 MD
COMPUTER
FAX
TELEPHONE
… ...
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ISDN的综合接入
ISDN—综合的接入
X.25 TERMINAL NETWORK ACCESS POINT

网路管理- Ch05 WAN Concepts

网路管理- Ch05 WAN Concepts

WAN Standardization (cont.)
Frame Relay §¨}-oiX ¨ X.25¨ ó w A - q°Tw é~è ¤¤e Ae o éù~ ±pCH X.25¨ ó w ] § t ¤ F í h 鱨 M ù~-×\à C °T°e
WAN Standardization (cont.)
WAN Standardization (cont.)
TCarrier P SONET A ó ì h h Frame Relay(°T°e )B ISDN (X A°ì )P ATM( D P ¨ B é )C
WAN Standardization (cont.)
WANபைடு நூலகம்Standardization (cont.)
Frame Relay (cont.)
¨ Frame Relay A s u ¨ êsu (Virtual Connection) G s u ¨ ê uOs¤W Frame Relay Y-`I A ¨ s u ¤ è ¤wwqn A¨±o ¨ § OH±Mu±s C
AT&T ¤ q ±q 1957 ~ } l o i TCarrier (Trunk CarrierA D F éCé ) é § N Aìoi O§±zL ìé§N Ab ¤@ ±éu¤W h-§Yy-q °T A H K z L ¤h¤u (Time Division Multiplexing, TDM)§N Pi hqDy- qC
WAN Standardization (cont.)
WAN Data Link Protocols (cont.) X.25/LAPB (cont.)
X.25 provides communication between two points over virtual circuits. A virtual circuits is a logical circuit and not a dedicated physical link.

计算机网络大题

计算机网络大题

第一章1. Consider sending a packet of F bits over a path of Q links. Each link transmits at R bps. The network is lightly loaded so that there are no queuing delays. Propagation delay is negligible.a. Suppose the network is a packet-switched virtual-circuit network. Denote the VC setup time by t s seconds. Suppose the sending layers add a total of h bits of header to the packet. How long does it take to send the file from source to destination?b. Suppose the network is a packet-switched datagram network and a connectionless service is used. Now suppose each packet has 2h bits of header. How long does it take to send the packet?c. Finally, suppose that the network is a circuit-switched network. Further suppose that the transmission rate of the circuit between source and destination is R bps. Assuming t s setup time and h bits of header appended to the packet, how long does it take to send the packet? 1、a. t s +(F+h)/R*Qb. (F+2h)/R*Qc. C.t s +(F+h)/R2. This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meter, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. a. Express the propagation delay, d prop , in terms of m and s .b. Determine the transmission time of the packet, d trans , in terms of L and R .c. Ignoring processing and queuing delays, obtain an expression for the end-to-end dealy.d. Suppose Host A begins to transmit the packet at time t=0. At time t=d trans , where is the last bit of the packet?e. Suppose d prop is greater than d trans . At time t= d trans , where is the first bit of the packet?f. Suppose d prop is less than d trans . At time t=d trans , where is the first bit of the packet?g. Suppose s=2.5·108, L =100bits, and R=28kbps. Find the distance m so that d prop equals d trans . 2、a) m/s b) L/Rc) m/s+ L/Rd) Interface of host A e) s*d trans =s*L/R f) Host Bg) m/s= L/R=> m=8.928*105第二章1. In the following figure, Alice sends an e-mail to Bob, and Bob gives a reply to Alice. If Alice and Bob both use mail user agent, please write what protocol is used in procedur e ①~⑥? If both Alice and Bob use Web browser, what protocol is used in procedure ①~⑥?(1)①SMTP; ②POP3 or IMAP; ③SMTP; ④SMTP; ⑤POP3 or IMAP; ⑥SMTP (2)①HTTP; ②HTTP; ③SMTP; ④SMTP; ⑤HTTP; ⑥HTTPAlice BobAlice ’s mail Bob ’s mail ① ② ③ ④⑤ ⑥2. A client requests a Web page /jwzx/index.do from a server through IE browser. The version of HTTP is 1.1 and default connection is persistent connection. The page is supported in Chinese. Please write the request message.If the server gives a reply to the client, the response message is created at 8:00:00 on 27-09-2011, and the object is modified at 00:00:00 on 18-09-2011. Please write the response message.The known-conditions are: server is apache/2.0.40; the object is a text file and consists 1796 bytes. If there is a proxy server between the client and the Web server, how does the proxy server check the page it caches is newer or older?(1)GET /jwzx/index.do HTTP/1.1Host: Connection: Keep-AliveUser-agent: Mozilla/4.0Accept-language: zh-cn(2)HTTP/1.1 200 OKDate: Tue, 27 Sep 2011 08:00:00 GMTServer: Apache/2.0.40Content-Length: 1796Content-Type: text/htmlLast-Modified: Sun, 18 Sep 2011 00:00:00 GMT(3)the proxy server uses conditional GETGET /jwzx/index.do HTTP/1.1Host: If-modified-since: Sun, 18 Sep 2011 00:00:00 GMT第三章1. Consider a TCP connection between Host A and Host B. Suppose that the TCP segmentstraveling from Host A to Host B have source port number x and destination port number y.What are the source and destination port numbers for the segments traveling from Host B to Host A?Source port number is y, destination port number is x.2. True of False?(1) Host A is sending Host B a large file over a TCP connection. Assume Host B has no data tosend Host A. Host B will not send acknowledgments to Host A because Host B can not piggyback the acknowledgments on data.×(2) The size of the TCP RcvWindow never changes throughout the duration of the connection.×(3) Suppose Host A is sending Host B a large file over a TCP connection. The number ofunacknowledged bytes that A sends cannot exceed the size of the receive buffer. ×(4) Suppose Host A is sending a large file to Host B over a TCP connection. If the sequencenumber for a segment of this connection is m, then the sequence number for the subsequent segment will necessarily be m+1. ×(5) The TCP segment has a field in its header for RcvWindow.√(6) Suppose that the last SampleRTT in a TCP connection is equal to 1 sec. Then the currentvalue of TimeoutInterval for the connection will necessarily be≥1 sec. ×(7) Suppose Host A sends over a TCP connection to Host B one segment with sequence number 38and 4 bytes of data. In this same segment the acknowledgment number is 42. ×(8) Consider congestion control in TCP. When the timer expires at the sender, the threshold is set to one half of its previous value. ×3. Suppose Host A sends two TCP segments back to back to Host B over a TCP connection. The first segment has sequence number 90; the second has sequence number 110. (1) How much data is in the first segment? 20 bytes(2) Suppose that the first segment is lost but the second segment arrives at B. In the acknowledgment that Host B sends to Host A, what will be acknowledgment number? 904. Suppose you have the following six 16-bit bytes: 01010000 11001101, 00110011 01100110, 0001001001001000, please compute the 16-bit checksum. 0110100110000100第四章1. There is a host A in the Internet. Its IP address is 210.100.89.204/20. Please answer the following questions: (1) Network address. (2)Broadcast address.(3) What is the subnet mask? (4) The fist host IP address. (5) The last host IP address.(6) How many hosts are there in the network? Solution:(1) 210.100.80.0 (2)210.100.95.255 (3) 255.255.240.0 (4) 210.100.80.1 (5) 210.100.95.254 (6) 232-20=212=40942. For the given topolo gy of the network, use Dijstra’s shortest path algorithm to compute the shortest path from A to all network nodes. Give a shortest path tree and node A’s routing table.N ‘ D(B),p(B) D(C),p(C) D(D),p(D) D(E),p(E) D(F),p(F) A5,A2,A1,A2,A ∞1 E A BD C F1 25 2 21532 11 E A BD C F 1 2 2 1AD 2,D 2,A 2,A 4,DADC 2,D 2,A 4,D/4,CADCE 2,D 3,EADCEB 3,EADCEBF第五章1. Please write the typical features of the following popular interconnection devices at trafficisolation, plug and play, optimal routing and cut-through.⑴hubs;⑵routers;⑶link-layer switchesRouters Switches Hubs Traffic isolateion Yes Yes NoPlug-and-play No Yes YesOptimal routing Yes No NoCut-through No Yes Yes2. There are two nodes going to communicate with each other. The sender and the receiver agree on a generator G as 1001. Consider the piece of data, D=10111000, that the sending node wants to send to the receiving node. Suppose the two communicating nodes use CRC method detecting errors.⑴Please write the bit-stream that the sending node transmits to the link.⑵Assume the receiving node receives the value 10111000110, what does the receiving node will do? Is the bit-stream correct or not?Answers:⑴ 1 0 1 0 1 1 0 11 0 0 1)1 0 1 1 1 0 0 0 0 0 01 0 0 11 0 1 01 0 0 11 1 0 01 0 0 11 0 1 01 0 0 11 1 0 01 0 0 11 0 1R=101,so the bit-stream that the sending node transmits to the link is 10111000101⑵ 1 0 1 0 1 1 0 11 0 0 1)1 0 1 1 1 0 0 0 1 1 01 0 0 11 0 1 01 0 0 11 1 0 01 0 0 11 0 1 11 0 0 11 0 1 01 0 0 11 1R=011, R≠0, so the bit-stream has bit-error.3. Draw lines.CSMA/CDrandom access protocols FDM point-to-point link channel partitioning protocols slotted ALOHAPPP protocol CDMA broadcast link taking-turns protocols polling protocolTDMToken-passing protocol Answer:TCP connection and tear downIn the following figure, there is a client initiates a TCP connectionto server via packet①. The connection is established after the clientreceives packet③. Suppose the client find it has no data need tosend. The client sends packet④to close the TCP connection.Please write down the Sequence number and Acknowledgmentnumber of each packet. Notice the special bit of each packet.In the following figure, there is a client and a servercommunicating with each other. The connection is established via packet①~③. Reliable data transfer works with Packet④~⑦. Packet⑧is sentfrom server to close the TCP connection.⑴Suppose, packet④carries 10 bytes data, and packet⑤has payload 20 bytes. Packet⑥carries 10 bytes data, and packet⑦has no payload.⑵Suppose, packet④carries 10 bytes data, and packet⑥carries 10 bytes data. Packet⑤and packet⑦ has no payload.The time-sequence diagram is shown on right, and the Sequence number of packet①is 179, the Sequence number of packet②is 531. Please write down the Sequence number and Acknowledgment number of each packet.计算机网络复习大纲CH1 Computer Networks and the Internet1.什么是协议,网络应用有哪几种工作模式,通信的双方如何定义角色2.电路交换网的几种类型,分组交换网的几种类型,什么是带宽3.分组交换的设备都有哪些4.网络接入的方法大致有几种类型,可用的物理媒体有哪些5.分组在网络中所经历的各种时延分别在什么位置发生,每种时延是由什么引起的,分组丢失在哪里发生?6.网络上的流通强度应该满足什么条件?CH2 Application Layer1.应用层进程之间通信,如何对进程进行寻址2.HTTP的工作原理,持久连接和非持久连接是如何工作的;两种工作方式在请求页面时,需要多少个RTT?3.HTTP报文格式:请求报文和响应报文,条件get报文的作用及其响应报文;Web cache 的作用4.FTP的工作原理5.邮件系统的组成部分,SMTP工作原理,邮件格式,MIME,邮件访问协议有哪些?6.DNS的功能,工作原理,DNS层次结构CH3 Transport Layer1.传输层提供的服务2.多路复用、多路分解分别表示什么意思3.TCP、UDP报文段格式中各字段的含义,checksum计算4.TCP的可靠数据传输是如何实现的5.TCP timer的超时间隔如何设置及设置依据6.TCP连接的3次握手过程与TCP连接断开的过程7.TCP流量控制中接收窗口是如何取值的,又是如何通知发送方的?8.TCP拥塞控制算法(Reno),TCP延迟计算(静态)CH4 The Network Layer1.网络层提供的服务2.网络层有哪些主要功能?2.工作在网络层的网络设备是什么,由哪些部分组成,有哪些功能,处理的数据单位及依据3.IPv4数据报格式及编址涉及的计算,获取地址的方法4.ICMP协议的功能5.IPv6数据报格式的变化,编址的变化及地址位数的变化6.LS选路算法7.RIP、OSPF、BGP协议CH5 The Link Layer and Local Area Networks1.链路层提供的服务2.网络链路的几种类型3.适配器的定义、功能4.奇偶校验、CRC检测5.多址访问协议有哪几种类型?6.什么是局域网7.MAC编址、ARP工作原理;DHCP工作原理8.Ethernet链路层帧的结构,CSMA/CD原理9.Ethernet技术,网段的划分;在10BaseT网络中使用hub及双绞线连接两台主机,这两台主机之间的最大距离;在其他网络中,使用hub、switch连接两台主机,这两台主机之间的最大距离。

CH0-me 课程简介

CH0-me 课程简介

②传输层:主要介绍传输层的基本功能、分布式进程通信、 客户/服务器模型、用户数据报协议(UDP)和传输控制协议(TCP) 等内容。
③应用层:应用层协议的基本概念、域名系统、Internet接 入技术、万维网、邮件系统、FTP服务以及Internet最新应用服务、 Intranet结构与组成等内容。


为培养学生在计算机网络系统的规划与构建,网络应 用系统的建立与开发等方面的能力打下坚实的基础。
4 教学内容

“以TCP/IP协议为主线的网络课程体系” 组织教学,将整个教学内容分为4大学习 模块9个知识大类 。

(1)第1个学习模块(通过本单元的学习为“以 TCP/IP协议为主线”打下基础)
①计算机网络概论:主要介绍计算机网络的基本 概念、计算机网络的发展与应用、网络计算研究与应 用的现状与趋势等内容。 ②网络体系结构:主要介绍网络体系结构与网络 协议的基本概念、ISO/OSI参考模型、TCP/IP协议参考 模型以及两个参考模型的比较等内容。

(2)第2个学习模块
①物理层:主要介绍物理层与物理层协议的基本概念、数据 通信的基本概念、数据编码技术、数据传输技术、多路复用技术 以及同步数字体系SDH等内容。
②数据链路层:主要介绍数据链路层的基本概念、差错控制 技术、数据链路层协议(BSC、HDLC)、Internet中的数据链路层 协议等内容。
(4)第4个学习模块
网络安全与网络管理技术:主要介 绍网络安全技术的基本知识、加密与认 证技术、入侵检测技术、防火墙技术、 网络防病毒技术、网络文件的备份与恢 复技术、网络管理的基本概念与网络管 理协议等内容。
5 考试形式

闭卷考试:70% 作业+上机实验+课堂出勤:30%

《Java程序设计》教材第五章练习题答案

《Java程序设计》教材第五章练习题答案

习题一、选择题1. 面向对象程序设计的基本特征是(BCD)。

(多选)A.抽象B.封装C.继承D.多态2.下面关于类的说法正确的是(ACD)。

(多选)A.类是Java 语言中的一种复合数据类型。

B.类中包含数据变量和方法。

C.类是对所有具有一定共性的对象的抽象。

D.Java 语言的类只支持单继承。

上机指导1.设计银行项目中的注册银行用户基本信息的类,包括账户卡号、姓名、身份证号、联系电话、家庭住址。

要求是一个标准Java类(数据私有,提供seter/getter),然后提供一个toString方法打印该银行用户的信息。

答:源代码请参见“CH05_LAB\src\com\inspur\ch05\BankUser.java”2.设计银行项目中的帐户信息,包括帐户卡号、密码、存款,要求如“练习题1”。

答:源代码请参见“CH05_LAB\src\com\inspur\ch05\Account.java”3.设计银行项目中的管理员类,包括用户名和密码。

要求如“练习题1”。

答:源代码请参见“CH05_LAB\src\com\inspur\ch05\Manager.java”4.创建一个Rectangle类。

添加两个属性width、height,分别表示宽度和高度,添加计算矩形的周长和面积的方法。

测试输出一个矩形的周长和面积。

答:源代码请参见“CH05_LAB\src\com\inspur\ch05\Rectangle.java”5.猜数字游戏:一个类A有一个成员变量v,有一个初值100。

定义一个类,对A类的成员变量v进行猜。

如果大了则提示大了,小了则提示小了。

等于则提示猜测成功。

答:源代码请参见“CH05_LAB\src\com\inspur\ch05\Guess.java”6.编写一个Java程序,模拟一个简单的计算器。

定义名为Computer的类,其中两个整型数据成员num1和num1,编写构造方法,赋予num1和num2初始值,再为该类定义加、减、乘、除等公有方法,分别对两个成员变量执行加减乘除的运算。

计算机网络英文题库(附答案)chapter5

计算机网络英文题库(附答案)chapter5

Chapter 5 The Link Layer and Local Area Network1.A ( ) protocol is used to move a datagram over an individual link.A application-layerB transport-layerC network-layerD link-layer2.The units of data exchanged by a link-layer protocol are called ( ).A datagramsB framesC segmentsD messages3.Which of the following protocols is not a link-layer protocol? ( )A EthernetB PPPC HDLCD IP4.In the following four descriptions, which one is not correct? ( )A link-layer protocol has the node-to-node job of moving network-layer datagrams over a single link in the path.B The services provided by the link-layer protocols may be different.C A datagram must be handled by the same link-layer protocols on the different links in the path.D The actions taken by a link-layer protocol when sending and receiving frames include error detection, flow control and random access.5.Which of the following services can not offered by a link-layer protocol? ( )A congestion controlB Link AccessC Error controlD Framing6.( ) protocol serves to coordinate the frame transmissions of the many nodes when multiple nodes share a single broadcast link.A ARPB MACC ICMPD DNS7.In the following four descriptions about the adapter, which one is not correct? ( )A The adapter is also called as NIC.B The adapter is a semi-autonomous unit.C The main components of an adapter are bus interface and the link interface.D The adapter can provide all the link-layer services.8.Consider CRC error checking approach, the four bit generator G is 1011, and suppose that the data D is 10101010, then the value of R is( ).A 010B 100C 011D 1109.In the following four descriptions about random access protocol, which one is not correct? ( )A I n slotted ALOHA, nodes can transmit at random time.B I n pure ALOHA, if a frame experiences a collision, the node will immediately retransmit it with probability p.C T he maximum efficiency of a slotted ALOHA is higher than a pure ALOHA.D I n CSMA/CD, one node listens to the channel before transmitting.10.In the following descriptions about MAC address, which one is not correct? ( )A T he MAC address is the address of one node’s adapter.B N o two adapters have the same MAC address.C T he MAC address doesn’t change no matter where the adapter goes.D M AC address has a hierarchical structure.11.The ARP protocol can translate ( ) into ( ). ( )A h ost name, IP addressB h ost name, MAC addressC I P address, MAC addressD broadcast address, IP address12.The value of Preamble field in Ethernet frame structure is ( )A 10101010 10101010……10101010 11111111B 10101011 10101011……10101011 10101011C 10101010 10101010……10101010 10101011D 10101010 10101010……10101010 1010101013.There are four steps in DHCP, the DHCP server can complete ( ).A DHCP server discoveryB DHCP server offersC DHCP requestD DHCP response14.In CSMA/CD, the adapter waits some time and then returns to sensing the channel. In the following four times, which one is impossible? ( )A 0 bit timesB 512 bit timesC 1024 bit timesD 1028 bit times15.The most common Ethernet technologies are 10BaseT and 100BaseT. “10” and “100” indicate( ).A the maximum length between two adaptersB the minimum length between two adaptersC the transmission rate of the channelD the transmission rate of the node16.The principal components of PPP include but not( ).A framingB physical-control protocolC link-layer protocolD network-layer protocol17.In the following four options, which service can not be provided by switch? ( )A filteringB self-learningC forwardingD optimal routing18.In the following four services, which one was be required in PPP? ( )A packet framingB error detectionC error correctionD multiple types of link19.The ability to determine the interfaces to which a frame should be directed, and then directing the frame to those interfaces is( ).A filteringB forwardingC self-learningD optimal routing20.In ( ) transmission(s), the nodes at both ends of a link may transmit packets at the same time.A full-duplexB half-duplexC single-duplexD both full-duplex and half-duplex21.Consider the data D is 01110010001, if use even parity checking approach, the parity bit is( ① ), if use odd parity checking approach, the parity bit is( ② ). ( )A ①0 ②1B ①0 ②0C ①1 ②1D ①1 ②022.In the following four descriptions about parity checks, which one is correct? ( )A Single-bit parity can detect all errors.B Single-bit parity can correct one errors.C Two-dimensional parity not only can detect a single bit error, but also can correct that error.D Two-dimensional parity not only can detect any combination of two errors, but also can correct them.23.MAC address is ( ) bits long.A 32B 48C 128D 6424.Wireless LAN using protocol ( ).A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.1125.The following protocols are belonging to multiple access protocols except for ( ).A channel partitioning protocolsB routing protocolsC random access protocolsD taking-turns protocols26.Which of the following is not belonging to channel partitioning protocols? ( )A CSMAB FDMC CDMAD TDM27.In the following four descriptions about CSMA/CD, which one is not correct? ( )A A node listens to the channel before transmitting.B If someone else begins talking at the same time, stop talking.C A transmitting node listens to the channel while it is transmitting.D With CSMA/CD, the collisions can be avoided completely.28.( ) provides a mechanism for nodes to translate IP addresses to link-layer address.A IPB ARPC RARPD DNS29.A MAC address is a ( )address.A physical-layerB application-layerC link-layerD network-layer30.Which of the following is correct? ( )A No two adapters have the same MAC address.B MAC broadcast address is FF-FF-FF-FF-FF-FF.C A portable computer with an Ethernet card always has the same MAC address, no matter where the computer goes.D All of the above31.In the following four descriptions, which one is not correct? ( )A ARP resolves an IP address to a MAC address.B DNS resolves hostnames to IP addresses.C DNS resolves hostnames for hosts anywhere in the Internet.D ARP resolves IP addresses for nodes anywhere in the Internet.32.In the LAN, ( )protocol dynamically assign IP addresses to hosts.A DNSB ARPC DHCPD IP33.DHCP protocol is a four-step process: ①DHCP request. ②DHCP ACK. ③DHCP server discovery. ④DHCP server offer(s). The correct sequence is ( )A ①②③④B ③②①④C ③④①②D ①④③②34.In the Ethernet frame structure, the CRC field is ( )bytes.A 2B 4C 8D 3235.In the Ethernet frame structure, the Data field carries the ( ).A IP datagramB segmentC frameD message36.In the following four descriptions, which one is not correct? ( )A Ethernet uses baseband transmission.B All of the Ethernet technologies provide connection-oriented reliable service to the network layer.C The Ethernet 10Base2 technology uses a thin coaxial cable for the bus.D The Ethernet 10BaseT technology uses a star topology.37.Ethernet’s multiple access protocol is ( ).A CDMAB CSMA/CDC slotted ALOHAD token-passing protocol38.In the following four descriptions about CSMA/CD, which one is not correct? ( )A An adapter may begin to transmit at any time.B An adapter never transmits a frame when it senses that some other adapter is transmitting.C A transmitting adapter aborts its transmission as soon as it detects that another adapter is also transmitting.D An adapter retransmits when it detects a collision.39.Which of the following descriptions about CSMA/CD is correct? ( )A No slots are used.B It uses carrier sensing.C It uses collision detection.D All of the above.40.The Ethernet 10BaseT technology uses( )as its physical media.A fiber opticsB twisted-pair copper wireC coaxial cableD satellite radio channel41.For 10BaseT, the maximum length of the connection between an adapter and the hub is ( )meters.A 100B 200C 500D 1042.A ( )is a physical-layer device that acts on individual bits rather than on frames.A switchB hubC routerD gateway43.A hub is a ( )device that acts on individual bits rather than on frames.A physical-layerB link-layerC network-layerD ransport-layer44.A switch is a( )device that acts on frame.A physical-layerB link-layerC network-layerD transport-layer45.In the following four descriptions, which one is not correct? ( )A Switches can interconnect different LAN technologies.B Hubs can interconnect different LAN technologies.C There is no limit to how large a LAN can be when switches are used to interconnect LAN segments.D There is restriction on the maximum allowable number of nodes in a collision domain when hubs are used to interconnect LAN segments.46.The ability to determine whether a frame should be forwarded to some interface or should just be dropped is ( ).A f ilteringB f orwardingC s elf-learningD o ptimal routing47.Which of the following devices is not a plug and play device? ( )A hubB routerC switchD repeater48.Which of the following devices is not cut-through device? ( )A hubB routerC switchD repeater49.In the following four descriptions, which one is not correct? ( )A Switches do not offer any protection against broadcast storms.B Routers provide firewall protection against layer-2 broadcast storms.C Both switches and routers are plug and play devices.D A router is a layer-3 packet switch, a switch is a layer-2 packet switch. 50.Which device has the same collision domain? ( )A HubB SwitchC RouterD Bridge51.IEEE802.2 protocol belong to ( )layerA networkB MACC LLCD physical52.IEEE802.11 protocol defines ( )rules.A Ethernet BusB wireless WANC wireless LAND Token Bus53.In data link-layer, which protocol is used to share bandwidth? ( )A SMTPB ICMPC ARPD CSMA/CD54.When two or more nodes on the LAN segments transmit at the same time, there will be a collision and all of the transmitting nodes well enter exponential back-off, that is all of the LAN segments belong to the same( ).A collision domainB switchC bridgeD hub55.( )allows different nodes to transmit simultaneously and yet have their respective receivers correctly receive a sender’s encoded data bits.A CDMAB CSMAC CSMA/CDD CSMA/CA56.Because there are both network-layer addresses (for example, Internet IP addresses) and link-layer addresses (that is, LAN addresses), there is a need totranslate between them. For the Internet, this is the job of ( ).A RIPB OSPFC ARPD IP57.PPP defines a special control escape byte, ( ). If the flag sequence, 01111110 appears anywhere in the frame, except in the flag field, PPP precedes that instance of the flag pattern with the control escape byte.A 01111110B 01111101C 10011001D 1011111058.The device ( ) can isolate collision domains for each of the LAN segment.A modemB switchC hubD NIC59.In the following four descriptions about PPP, which one is not correct? ( )A PPP is required to detect and correct errors.B PPP is not required to deliver frames to the link receiver in the same order in which they were sent by the link sender.C PPP need only operate over links that have a single sender and a single receiver.D PPP is not required to provide flow control.60.In the PPP data frame, the( ) field tells the PPP receivers the upper-layer protocol to which the received encapsulated data belongs.A flagB controlC protocolD checksum61.PPP’s link-control protocols (LCP) accomplish ( ).A initializing the PPP linkB maintaining the PPP linkC taking down the PPP linkD all of the above62.The PPP link always begins in the ( ) state and ends in the ( ) state. ( )A open, terminatingB open, deadC dead, deadD dead, terminating63.For( ) links that have a single sender at one end of the link and a single receiver at the other end of the link.A point-to-pointB broadcastC multicastD all of the above64.With ( )transmission, the nodes at both ends of a link may transmit packets at the same time.A half-duplexB full-duplexC simplex(单工)D synchronous65.With ( ) transmission, a node can not both transmit and receive at the same time.A half-duplexB full-duplexC simplex(单工)D synchronous66.Which of the following functions can’t be implemented in the NIC? ( )A encapsulation and decapsulationB error detectionC multiple access protocolD routing67.Which of the following four descriptions is wrong? ( )A The bus interface of an adapter is responsible for communication with the adapter’s parent node.B The link interface of an adapter is responsible for implementing the link-layer protocol.C The bus interface may provide error detection, random access functions.D The main components of an adapter are the bus interface and the link interface. 68.For odd parity schemes, which of the following is correct? ( )A 011010001B 111000110C 110101110D 00011011069.( )divides time into time frames and further divides each time frame into N time slots.A FDMB TMDC CDMAD CSMA70.With CDMA, each node is assigned a different ( )A codeB time slotC frequencyD link71.Which of the following four descriptions about random access protocol is not correct? ( )A A transmission node transmits at the full rate of the channelB When a collision happens, each node involved in the collision retransmits at once.C Both slotted ALOHA and CSMA/CD are random access protocols.D With random access protocol, there may be empty slots.72.PPP defines a special control escape byte 01111101. If the data is b1b201111110b3b4b5, the value is( )after byte stuffing.A b1b20111110101111110b3b4b5B b1b20111111001111101b3b4b5C b5b4b30111111001111101b2b1D b5b4b30111110101111110b2b173.MAC address is in ( ) of the computer.A RAMB NICC hard diskD cache74.Which of the following is wrong? ( )A ARP table is configured by a system administratorB ARP table is built automaticallyC ARP table is dynamicD ARP table maps IP addresses to MAC addresses75.NIC works in ( )layer.A physicalB linkC networkD transport76.In LAN, if UTP is used, the common connector is( ).A AUIB BNCC RJ-45D NNI77.The modem’s function(s) is(are) ( ).A translates digital signal into analog signalB translates analog signal into digital signalC both translates analog signal into digital signal and translates digital signal into analog signalD translates one kind of digital signal into another digital signal78.( )defines Token-Ring protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.279.( )defines Token-Bus protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.280.( ) defines CSMA/CD protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.281.The computer network that concentrated in a geographical area, such as in a building or on a university campus, is ( )A a LANB a MANC a WAND the Internet82.The MAC address is ( ) bits long.A 32B 48C 128D 25683.Which of the following four descriptions about MAC addresses is wrong? ( )A a MAC address is burned into the adapter’s ROMB No two adapters have the same addressC An adapter’s MAC address is dynamicD A MAC address is a link-layer address84.Which of the following four descriptions about DHCP is correct? ( )A DHCP is C/S architectureB DHCP uses TCP as its underlying transport protocolC The IP address offered by a DHCP server is valid foreverD The DHCP server will offer the same IP address to a host when the host requests an IP address85.The ( )field permits Ethernet to multiplex network-layer protocols.A preambleB typeC CRCD destination MAC address86.For 10BaseT, the maximum length of the connection between an adapter and the hub is ( ) meters.A 50B 100C 200D 50087.An entry in the switch table contains the following information excepts for ( )A the MAC address of a nodeB the switch interface that leads towards the nodeC the time at which the entry for the node was placed in the tableD the IP address of a node88.Consider the 4-bit generator , G is 1001, and suppose that D has the value 101110000. What is the value of R?89.Consider the following graph of the network. Suppose Host A will send a datagram to Host B, Host A run OICQ on port 4000, Host B run OICQ on port 8000. All of ARP tables are up to date. Enumerate all the steps when message “Hello” is sent from host A to host B.。

名词解释

名词解释

1.在大多数情况下,检索的目的是为了找到相关文献,而不是"答案"。

2.二八定律在期刊文献检索中的体现是:20%的期刊登载了80%的重要文献,体现这种特性的期刊是核心期刊。

3.当计算机访问范围受到限制时,可以通过代理服务器访问外部网络?4.PDF、VIP文件对应的打开程序分别为ACROBAT READER ,VIPBROWSER 或者维普浏览器。

5.下载比较大的文件和程序,经常使用的是FTP 协议。

6.检索式COMPUT*,其中*代表截词符,这种截断方式称为右截断,又称为前方一致截断。

7.如果希望在搜索引擎中使用短语检索方式检索COMPUTER NETWORK,检索式为:“computer network”。

8.使用网络蜘蛛(spider)对广域网的信息进行采集,提取关键词建立索引,主要通过关键词进行检索。

以上描述的搜索引擎属于索引型搜索引擎。

9.文献的特性有知识信息性、客观物质性、人工记录性、动态发展性。

10.文献的内容特征用于找出相关文献,外部特征用于获得特定文献。

11.文献获取与利用的过程中存在三重语言障碍,分别为:自然语言障碍、专业语言障碍、检索语言障碍。

12.检索工具的基本类型有:目录、题录、文摘、索引。

13.以单篇文献为著录对象,并附有摘要的检索工具为:文摘。

14.从内部特征进行检索的最重要的两种检索途径是:主题途径、分类途径。

15.检索工具中在文献来源项的著录中,常常将期刊名称按一定的缩写规则进行缩写,JOURNAL一般缩写为J. 。

16.文献检索语言是标引与检索共同使用的约定语言。

17.中国图书馆图书分类法简称为:中图法。

18.主题词语言具有三个基本特征:着眼于从内容特性方面去提示文献主题、采用文字符号、按字顺排列。

19.当查询关键词具有多个含义的时候,容易造成误检,使得查准率较低。

20.主题词的体现形式是叙词表。

21.任何一种截词检索,都隐含着布尔逻辑检索的逻辑或运算。

朗文快车五年级第二学期英语练习 CH5

朗文快车五年级第二学期英语练习 CH5

Primary Longman Express 5B Practice CH5Class________ Name__________ Signature _________Score _________I.Put the words into right order and add punctuations into the sentences then rewrite them.a hill its said important fire theres not to put if the out fire by grandpa yourself____________________________________________________________________ II.Fill in the blanks according to the sentences.1. Do you know how to make masks? Let me tell you. First __________ out a rabbit mask template from the computer. Next ____________ the mask. Then __________ it out around the ___________. After that cut a long piece of _____________. Finally ____________ the ends of the string ___________ the mask.2. I want to make a pen holder and put it on my desk. I need a piece of _____________,a piece of coloured paper, ______________________, some ______________ and some _____________ and a pair of _________________.III.Fill in the blanks.Use a shoe box to make a treasure box!What you○1____________:a shoe box, some crayons, some pieces of coloured paper, a○2____________________, a○3_______________________, some○4_____________ and some○5______________. What you○6____________:First○7___________________________the box and the lid. The○8_________________ __________________ the coloured paper. ○9___________use○10_________________ to join the lid to the box. After that ○11________________ short pieces of string. ○12 ___________ stick each piece of string onto the lid. Then you can ○13_____________ the strings together to close your treasure box.IV.Put the sentences in the correct order. Write the numbers in the blanks.1. Making ice cream cakes( ) a. Amy wants to eat ice cream and cake at the same time. She has no idea! ( ) b. After that she spreads a layer of ice cream on top.( ) c. Then she puts one piece on a plate.( ) d. First she cuts a cake into two pieces.( ) e. Next she puts the other piece of cake on top.( ) f. Finally she puts the cake in the fridge for two hours.2. Making banana strawberry milk shake( ) a. First she cuts a banana up into three pieces and puts them in a blender. ( ) b. After that she turns on the blender.( ) c. Finally she pours the milk shake into a glass.( ) d. Next she adds the strawberries, a cup of milk and some sugar.( ) e. Then she rinse six strawberries and cuts them in half.( ) f. Mrs Wong likes making banana strawberry milk shake.V.Read carefully. Answer the questions in complete sentences.1. Where do piñatas come from?___________________________________________________________2. Who plays with piñatas?___________________________________________________________3. What can you find inside pinatas?___________________________________________________________。

翻译网站大全,爱翻译的娃一定要收藏哦

翻译网站大全,爱翻译的娃一定要收藏哦

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Computerandnetworksecurity

Computerandnetworksecurity

Computer and Network Securityc Copyright2000R.E.NewmanComputer&Information Sciences&EngineeringUniversity Of FloridaGainesville,Florida32611-6120***********fl.eduHashes(Pfleeger Ch.3,KPS Ch.4)Art1Definitions1.1One-way FunctionA one-way function F is a function that is“hard”to invert.That is,ifF:A→B,then given some b∈B,it is hard tofind an a∈A for whichF(a)=b.By hard,we mean that no method much faster than trying elements of A using brute force is known to invert F effectively.This is collision-resistance.1.2Hash FunctionA hash H is a function that•takes inputs from a large set,A,and•maps them tofixed length elements in afinite set,B.1.3One-way or Secure Hash FunctionIf a hash function is also a one-way function,then it is a one-way hash or secure hash function.2Uses of HashesIf H is a one-way hash,then it may be used in many ways:1.for authentication2.as a MIC(message integrity check)3.as a MAC(message authentication check)4.as a PRNG(for key stream generation)5.for password security.3Hash Attacks3.1The Birthday ProblemIf there are23or more people in a room,the odds are better than.5that two of them will have the same birthday.•Assume that365days of the year are equally likely as birthdays•birthdays are random among people•With N people in the room,there areP=N(N−1)/2distinct pairs of people.•For each pair,there is a probability ofp=1/365that the two have the same birthday.•Expected number of matches is number of pairs times probability,E(matches)=P p.•For the expected number of matches to exceed.5,P>.5/p=365/2=183.•ThusP=N(N−1)/2>365/2henceN(N−1)>365√N>3.2So What?Well,the Birthday Problem tells us1.that tofind two messages with the same n-bit hash value,only2n/2candidates will have to be considered,on the average;2.if another message m′with the same n-bit hash value H(m)as some given message m is required,then onthe average,2n−1candidates will have to be tested.Since hashes must protect against intentional misuse(a user producing two messages with the same hash that mean very different things,then either•signing one and claiming that the other was actually sent later,or•getting someone else to sign the one and then sending the other with the signature of thefirst),they must be twice as long as we would otherwise need for security.4Why use a hash?One-way hashes are1.small and fast to compute;2.collision-resistant;3.may be used with public key systems for signatures much faster than signing the entire message or document;4.export a little better than pure cryptosystems do,yet can be used for encryption.5AuthenticationRather than using a challenge-response authentication method based on encrypting the challenge nonces,hashing may be used instead.A→B:r AB→A:H(K AB|r A)1.Alice sends Bob a nonce challenge,2.Bob replies with the hash of a common secret K AB concatenated with the nonce(R A)Alice can then verify the hash received,and no party that does not have access to K AB is likely to be able to reproduce that value.Bob can authenticate Alice similarly.NOTA BENE:r A must be used ONLY once by Alice,and there are a host of other subtleties that must be considered in authentication.6MICs and MACsUsing a one-way hash for a Message Integrity Code(MIC):•H(m)by itself does not suffice,since H is known to all•H(K|m)may be used,so that only those who know K can either generate or check the integrity of m.Hashing for MICsH(K|M) Figure1:Secure Hash for MICAttack on MICsH(K|M)H(K|M|P|A) Figure2:Attack on MIC•Unfortunately,several popular MD algorithms use as an intermediate value when computing the hash the same quantity that is used as the hash output at the end.This means that,given a message m and its hash,H(m),some addition A may be made to m and a valid hash computed for it by initializing the MD algorithm to have H(m)as its carried information.This is not a problem with MD2,and is easily solved in any of several ways.Two ways are as follows.e H(m|K)instead.e only half of the computed H(m)as the MICIf only one other party knows the secret K,then the MIC serves as a MAC as well,authenticating that the message was sent by the only other party to have the secret.7Encryption with HashesHash function output is unpredictable,so may be used as PRNG in a Vernam Cipher.7.1OFB•K0=IV,the initialization vector,which is the key•K i=H(K i−1),i=1,2,...•C i=P i+K i,i=1,2,...•P i=C i+K i,i=1,2,...7.2CFB•C0=IV•C i=P i+H(K,C i−1),i=1,2,...•P i=C i+H(K,C i−1),i=1,2,...7.3Hash in Feistel StructureThey may be used as round function in Feistel structure7.3.1Karn’s approach•Plaintext P=L|R(two halves,L and R,concatenated)•Key K=l|r(two halves,l and r,concatenated)•Ciphertext C=D|E where–E=R+H(L|l)–D=L+H(E|r)•Decryption given K=l|r and C=D|E,–L=D+H(E|r)–R=E+H(L|l)7.3.2Weakness in Karn’s ApproachLuby and Rackoff:Karn’s approach susceptible to attack if two plaintext messages P and P′had the same left half,P=L|R and P′=L′|R′where L=L′.If the attacker knows all the ciphertext,all of P and that L′=L,then it is easy tofind R′.7.3.3A Fix for Karn’s Approach•K=l|r as before•P=L|R as before•X=R+H(l|L)•Y=L+H(r|X)•Z=X+H(l|Y)•C=Z|Y=the ciphertext7.3.4Other Ways to Fix Karn’s ApproachEnsure that L′=L ever,by prepending•Timestamp(must protect time,resolution)•Random number(must never reuse)•Sequence number(must never wrap around,must remember)8Password Security with Hashes8.1Hashing PasswordsHow to store information to verify passwords?•in plaintext(bad idea)•encrypted(could be decrypted if key were compromised)•cryptographic hashes of passwordsIn Unix and some other systems,password is1.converted into the key for an encryption algorithm(modified version of DES in the case of Unix),2.key used to encrypt a constant(0in the case of Unix)3.encryption may also be modified by a salt(12bits in Unix)4.so same password with different salts has different hash values(modifies the DES data expansion algorithmin Unix)E "encrypted password"E "encrypted password"S|PP K CSaltHashing Passwords Figure3:Hashing Passwords for Storage8.2Lamport’s One-time PasswordsGoal:to provide a means for a central computer to verify the identity of a user in such a way that even if the verification information kept on the computer is read,the attacker cannot fool the system.Approach:Uses one-way hashes as follows.er U picks password P and lifetime N2.U passes P through hash function N times to produce P N:P N=H(H(H...H(P)...))N times=H N(P).3.P N is securely given to host C along with N;C storesU,N,P NWhen U wants to authenticate herself to C,1.U sends request to C giving her name,U2.C replies with the number N−13.U computes P N−1=H N−1(P)and sends this to C as message M4.C then computes H(M)and verifies thatH(M)=P N5.C then decrements N and replaces P N with P N−11.H is one-way,so only U could have produced the preimage of P N,even if P N and H are known.Lamport Hash Authentication- Method for authentication even if someone reads server database- A chooses password_A, n, computes H(password_A), H(H(password_A))...- Server holds(A, n, H n(password_A)=Y)A Sget password_A n-1decrement n, set Y=XFigure4:Lamport One-Time Hash Authentication2.Susceptible to man-in-the-middle attack3.Requires reinitialization when N=14.Must be modified if more than one host is used with same P9Using encryption as a hash function1.message broken into key-sized blocks2.blocks used as keys in the encryption fuction to encrypt repeatedly some constant(once for each key)3.If output is too small for a usable hash,then(a)use more than one constant,or(b)reverse the key order,etc.10Selected Hash Functions10.1GenerallyThere are a few steps that most hash functions must always take.1.Arbitrary(bit or byte)length input is padded to a multiple of the block length-padding must be consistentbut does not have to be removeable2.Some checksum may be added to the the padded input(MD2)3.Padded input is processed in blocksHashing Using EncryptionM:K1K2K3KnC E E E EH(M)Figure5:Hashing Using Encryption4.Blocks may have multiple passes made over them5.Output of preceding block is passed as initialization information for processing the next blockGeneric Hash Function OperationH(M)Figure6:Generic Hash Function10.2MD21.byte-oriented computations2.128-bit digest3.Arbitrary byte length input is padded to a multiple of16bytes.(last p bytes,1≤p≤16,all have value pin binary)4.A16-byte“checksum”is computed over the whole message and appended to its end5.The padded,checksummed message is then processed in16-byte chunks6.carrying forward128-bit intermediate results(p i)7.48bytes are processed per chunk(m i,p i,m i⊕p i)8.with18passes at each stage9.to produce16-byte digest10.3MD41.32-bit word computations2.128-bit digest3.(almost)arbitrary bit length input4.padded to64bits less than than a multiple of64bytes(512bits)5.64-bit lengthfield added to end6.processed in16-byte chunks,7.carrying forward128-bit intermediate results8.with3passes per stage9.with same constants for each message word in passes1and310.to produce16-byte digest10.4MD51.32-bit word computations2.128-bit digest3.(almost)arbitrary bit length input4.padded to64bits less than than a multiple of64bytes(512bits)5.64-bit lengthfield added to end6.processed in16-byte chunks,7.carrying forward128-bit intermediate results8.with4passes per stage9.with different constants for each message word in all passes10.to produce16-byte digestbyte n-1byte n pi-substitutionpass#(0-17)original messagep bytes, all = p p <= 16 bytesmultiple of 16 bytes (128 bits)3. Compute intermediate values for each 16-byte chunk1. Pad2. Append checksum16-byte chunkintermediatemessage chunkphantom byte(-1)(init to 0)byte 47byte 0+new intermediatediscarded4. Last intermediate value is MD2 hash value (digest)MD2Figure 7:MD2structure10.5SHA1.32-bit word computations2.160-bit digest3.(almost)arbitrary bit length input4.MD4/MD5padding5.processed in 16-byte chunks,6.carrying forward 160-bit intermediate results7.with 5passes per stageing complicated mangling functionk%264M L-1M 0M 1M 2A = 0x01234567B = 0x89abcdefC = 0xfedcba98MD51. Padkmessage m10000 (000). . .h = ABCD 02. Initializeii-1h h Figure 8:Overall structure of MD59.to produce 20-byte digest10.6SummaryHashes will be especially useful for 1.MICs2.MACs/Digital Signatureswhen public key cryptosystems are involved.ii-1h h MD5 (con’t.)H(X,Y,Z) = (X + Y + Z)I(X,Y,Z) = Y + (X | !Z)F(X,Y,Z) = (X & Y) | (!X & Z)G(X,Y,Z) = (X & Z) | (Y & !Z)X-select Z-select parityFigure 9:MD5internals。

地铁英语缩写翻译

地铁英语缩写翻译

城市轨道交通专业词汇缩写AC :信标/计轴 Axle CounterACS :计轴系统 Axle Counter SystemADC :自动关门 Auto Door CloseADO :自动开门 Auto Door OpenADM :系统管理器ADU :特征显示单元AF:音频AFC :自动售检票系统 Auto Fare CollectionAM :列车自动运行驾驶模式 Automatic ModelAMU :ATO 匹配单元AP :接入点、轨旁无线单元 /应用模块 Application P ⋯⋯APAM:ATO功率放大板块API :应用程序接口APR :绝对位置参考应答器、信标AR:自动折返驾驶 /列车自动折返模式ARS :列车进路设定AS :管理服务器 /接入交换机 Access Switch ASK :数字调幅、幅移键控ATB :自动折返按钮 Automatic Turnback Button ATC :列车自动控制系统ATI:列车到达时刻显示器ATO :列车自动运行ATP :列车自动防护ATR :列车自动调整‘精选文库ATS :列车自动监控 Automatic Train Supervision CBTC :基于通信的列车控制 Communication Based Train ControlAXC :计轴器CC :车载控制器 Carborne ControllerB&A :操作和显示CCTE :车载安全计算机(包括 ATP/ATO子系统)BAS :环境与设备监控系统CCTV :闭路电视 /电视监视器Bd :波特CD :载频检测模块bond :棒CDM :电码检测模块BS :骨干交换机 Backbone Switch CDTA :中央数据传输系统BUMA :总线控制板CE :控制设备CA: 控制中心自动控制模、中央自动模式CENELEC :欧洲电工标准委员会CAN :现场总线CESB :中央紧急停车按钮CAZ :冲突防护区域’CER :控制室CBI: 计算机联锁 Computer Based Interlocking CG :编码发生器CBN :通信系统CH :校核信号CI:计算机联锁 Computer Based InterlockingCLC :线路控制器CLOW :中央联锁工作站 Center Locking Workstation CM:编码人工驾驶模式COAST :惰行COM :通信服务器COTS :可购买的商用产品CPL :耦合器模块 CouplerCPISA :通信处理器CPS :条件电源块CPU :中央处理单元CRC :循环冗余校验CRT :阴极射线显示器精选文库CS :中央服务器 Center ServerCSEX :电码系统模拟器扩展CTC :调度集中CTS :光数据传输系统DAB :报警按钮(为了及时处理意外或临时事故而设置在车厢里的乘客报警按钮)DB :轨道数据库 Data BaseDCC ;元件接口模块/车辆段、停车场控制中心Depot Contral Center DCS: 数据通信系统 Data Communication SubsystemDCU :数据储存单元DCR :车站综合控制室DDS :数字频率合成技术、DDU :诊断和数据上载单元、诊断和数据更新单元DEBLIMO :闪光元件接口模块DSU :数据服务单元 Data Service UnitDEM :调节器DT : VCC 数据传输DESIMO :信号机元件接口模块DTC :数字轨道电路DEWEMO :道岔元件接口模块DTI :发车计时器、发车时间表示显示器 Departure Time Indicator DI:列车发车时刻显示器DTM :现场 LDTS 分机DID :目的地号 Destination Identification DTRO :无人驾驶列车折返运行DIOM :离散输入、输出板块DTS :光纤网、数据传输系统、光纤通信系统读点DOC :驱动输出模块EBR :紧急制动继电器DOT :倒换方向EB:紧急制动DPU :车辆段程序单元ECC :元件接口模块DS :模拟 MMI 、演示系统、数据服务器EFAST :列车控制元件接口模块DSP :数字信号处理技术EFID :入口馈电设备DSIT :接口控制模块EPROM :只读储存器ERC :人工取消进路 E ⋯⋯Route Cancel FSK :数字调频、频移键控ESB 、 ESP :紧急关闭按钮 Emergency Stop Button FTGS :西门子公司的遥供无绝缘音频轨道电路 /音频无绝缘轨道电路ESS :紧急车站停车系统GEBR :可保证的紧急制动率 Guaranteed Emergency Brake Rate ESIT :电子元件接口模块GO : ATP 速度命令选择和核准电路EU :电子单元HMI :人机接口 /人机界面Human-Machine InterfaceFAS :火灾自动报警系统IBP :综合后备盘 Integrated Backup PanelFEC :非向前纠错I/O :输入 /输出 Input/OutputFEP :前端处理器ICM :输入控制模板、输入模块FFT :快速傅立叶变换ICU :区域控制中心、控制单元、计算模块FID :馈电设备ID :识别FOTL :光纤传输线IEC :国际电工委员会FRONTAM :数据存储单元IFS :接口服务器 Interface ServerFSB :全常用制动 Full Service Braking ILC :联锁控制器 InterLocking ControllerIRU :接口继电器单元LEU :轨旁电子单元、信号接口JTC ;无绝缘轨道电路LFU :环路馈送单元KOMDA :开关量输出板LISTE :信号机元件接口板块LAN :局域网 Local area Net LIU :环线调谐单元LC :车站控制LMM :环路调制解调器板块LCC :本地控制台LOM :逻辑输出板块LCD :液晶显示器 Liquid Crystal Display LPU ;车站程序单元LCP :局域控制板(设于站控室内墙LCP控制盘上,需要扣车或取消时,LZB :连续式列车自动控制系统按压按钮扣车或取消扣车,当站台的紧急停车按钮被按动时,在 LCP 上报MAL :移动授权 Movement Authority Limit警应按取消报警按钮)MAZ :移动授权区域 Movement Authority Zone LCW :本地控制工作站Local Control Workstation MD :调频检测板块LDTS :现场数据传输系统MDC :手动关门 Manual Door CloseLED :发光二极管 Light Emitting Diode MDO :手动开门 Manual Door OpenME :存储互换模块 Memory Exchange MTIB :移动列车初始化信标MELDE: 开关量输入板MTO :无人驾驶MI:联锁单元MUX :接口电机MicroLok :微机联锁 /联锁设备NDO :非安全数字输出板MMI :人机界面NFS: 网络文件系统MMS ;维护管理系统NIC :网络接口卡MODEM :调制解调器NISAL: 数字集成安全保障逻辑MPM :主处理器板块NMS :网管系统 /网络管理工作站MR :车载无线设备NRM :非限制人工驾驶模式MSK :最小移频键控NRZI :不归零倒置MSS :最大安全速度NSS :网络支撑系统MWS :维护工作站 Maintenance Work Station NVI :非安全型输入MT :轨道联锁、城市轨道交通、NVLE :非安全逻辑模拟器工作站NVO :非安全型输出PB :停车制动OBE :车载设备PC :道岔控制OCC :运营控制中心Operational Contral Center PCB :控制器、印路电路板OCM :输出控制模块PCU :协议传输单元ODI :操作 /显示接口PD :多项式除法器OLM ;通信模块、光连接模块PEB :站台紧急按钮、OLP :光连接插头PF:工频OPG :速度脉冲发生器PI :站台显示器OTN :开放的传输网PID :乘客导向系统OVW :全线表示盘子系统PIIS :乘客信息显示器PAC :环路调制解调器PIS:乘客导向系统PAL :逻辑处理模块PL:运行等级/站到站的运行时间PAS :车站广播系统PM:道岔转辙机PROFI BUS :过程现场总线RCC ;远程通信控制器PROM :课编程计数器RCM :远程通信控制模块PSA :远方报警盘RI :继电器接口 Relay Interface/接口设备PSC :远台屏蔽门中央控制盘RM :限制人工驾驶PSD :站台屏蔽门RMO :限速模式PSL :就地控制盘RTOS :实时操作系统PSU :电源单元RTU :车站远程终端单元 Remote Terminal UnitPTI :列车识别系统RX :接收器PVID: 永久性车辆标识 Permernent Vehicle Identification SAN :存储区域网络PWD :梯形波调幅SB :脚踏阀、常用闸,行驶制动器 Servicebrake/常用制动 Service Braking RAMS :安全性SBD :安全制动距离 Safe Braking DistanceRB :重定位信标SBO :安全型单断输出RC:进路控制SC;运行图编辑子系统SCADA :电力监控系统SIR :安全联锁继电器SCC :车站控制计算机 /车站引导控制计算机SISIG :烙断器板SCEG :车站控制器紧急通路SLC :同步环线盒SCI :计算机联锁SLM :速度和位置模块SCR :车站控制室SM :列车自动防护驾驶、系统维护台、系统维护模块S&D :诊断服务、检修和诊断SMC :系统管理中心SD :安全装置SMSS :维护监测子系统SDM :联锁系统维护工作站SNMP :简单网络管理协议SDT :站停时间 Station Dwell Time SNOOPER :列车和事件监控器SER :信号设备室SO :维护操作台SICAS :西门子计算机辅助信号/微机联锁设备S— PC :模拟 PCSIL :安全完整度等级SPDI :瞬间接触开关SIOM :串行输入、输出模块SQL :结构化查询语言SRS :运行图TDT; 列车发车计时器STA ;天线TID :列车输入数据模块 /列车追踪号 Tracking Identification STC :车站控制器TM:室内控制柜STEKOP :现场接口计算机TMT :列车监督和追踪STIB :静态列车初始化信标TOD :司机显示盘、列车输出数据模块 /司机操作显示单元STS :厂家测试成套设备TR :分线柜 /接口设备SYN :同步天线TRC :列车进路计算机TAC; 测速电机出来模块TS :目标速度 Target Speed/终端服务器Terminal Server TC :轨道区段、轨道电路TSR :临时限速 Temporary Speed RestrictionTCM :轨道编码模块TTE :时刻表编辑器TCP/IP ;远程控制协议 /国际协议TTF :时刻表TD :列车位置检测TTT :列车跟踪 Train Tracking T⋯⋯TDB :线路数据库TU:调谐单元、轨道电路控制单元TVP :轨道空闲处理VICOS :车辆和基础集中控制操作系统TWC :车 -地通信 Traffic Wayside Communication VO :表决器模块 VoterTX :发送器VOBC :车载计算机、车载控制设备UPS :不间断电源VPI :安全型计算机联锁URM :非限制人工驾驶模式VR :列车调整 Vehicle RegulationVAS :车辆报告系统VRD :安全继电器驱动器VCC ;车辆控制中心VSC :安全型串行控制器VCS :车辆通信系统WEEZ Bond :小型调谐阻抗连接变压器VDI :安全数字输入板WCC :轨旁通信控制器VDO :安全数字输出板WE;轨旁设备VENUS :处理器板中断板WESTE :道岔接口模块VESUV :同步比较板ZC :区域控制器 Zone ControlerVHM :车况监视器名称全称中文意义FAS 1.1 Fire Alarm System火灾报警系统LAN Local Area Network局域网BAS Building Automation System建筑设备自动化系统WAN Wide Area Network广域网AFC Auto Fare Collection自动售检票系统OTN Open Transport Network开放传输网络ATP Automatic Train Protection列车自动防护Tc (A)Trailer Car拖车ATS Automatic Train Supervision列车自动监控Mp (B)Motor Car With Pantograph带受电弓的动车ATC Automatic Train Control列车自动控制M (C)Motor Car动车ATO Automatic Train Operation列车自动运行AW0空载SCADA Scan Control Alarm Database供电系统管理自动化AW1每位乘客都有座位OCC Operated Control Center控制中心AW2每平方米 6 人MMI Man Machine Interface人机接口AW3每平方米9 人UPS Uninterrupted Power Supply不间断电源供给CSC Contactless Smart Card非接触智能卡MOC Ministry Of construction建设部CST Contactless Smart Token非接触智能筹码IDC Intermodality Data Center清结算数据中心EOD Equipment Operating Data设备运行参数专业 :车辆专业MPU Main Processor Unit主控单元名称全称中文意义APU Audio Power Unit放大器单元LRU Line Replaceable Unit线路可替换单元VPI Visual Passenger Information可视乘客信息TBD To be Defined待定义,待规定VVVF Variable voltage Variable Frequency变压变频TBEx Trailer Bogie -External拖车外转向架专业 :信号系统TBIn Train Bogie -Intermediate拖车中间转向架名称全称中文意义TBU Tread Brake Unit踏面制动单元PTI Positive Train Identification列车自动识别WSP Wheel Speed Sensor轮速传感器SICAS Siemens Computer Aided Signaling西门子计算机辅助信号PB Powered Bogie动车转向架DTI Departure Time Indicator发车计时器FDU Frontal Display Unit前部显示单元PIIS Passenger Information and Indication System旅客向导系统IDU Internal Display Unit内部显示单元ADM Administrator Workstation系统工作管理站TIMS Train Integrated Management System列车综合管理系统RM Restricted Manual Mode ATP 限制允许速度的人工驾驶DVA Digital and Audio Announcements数字语音广播器AR Automatic Reversal自动折返ATT Automatic Train Tracking列车自动跟踪DxTiP Digital Exchange for TETRA TETRA数字交换机SIC Station Interface Case车站接口箱ISDN Integrated Services Digital Network综合业务数字网LCP Local Control Panel局部控制台OMS OTN Management System OTN 管理系统ARS Automatic Route Setting列车自动进路排列NCC Network Control Center网络控制中心ATR Automatic Train Regulation列车自动调整名称全称中文意义专业 :通信系统PCM Pulse Code Modulation脉冲编码调制名称全称中文意义TETRA Terrestial trunked Radio欧洲数字集群标准MDF Multiplex Distribution Frame综合配线架TDM Time Division Multiplexing时分复用TBS TETRA Base Station TETRA 基站PSTN Public Switched Telephone Network公用电话交换网PABX Private Automatic Branch Exchange专用自动小交换机CDD Configuration and Data Distribution Server配置及数字分配服DDF Digital Distribution Frame数字配线架务器ODF Optical Distribution Frame光配线架专业 :自动售检票系统VDF Audio Distribution Frame音频配线架名称全称中文意义2 2.1 File Transfer Protocol文件传输协议专业 :火灾报警TCP/IP Transmission Control Protocol/ Internet Protocol名称全称中文意义传输控制/网络协议GCC Graphic Control Computer图形监视计算机CPS Central Processing System中央计算机系统MTBF Mean Time Between Failures平均无故障运行时间SPS Station Processing System车站计算机系统EMC Electro Magnetic Compliance电磁兼容性PIN Personal Identification Number个人身份号码FACMCBF Mean Cycles Between Failure运行设备两次损坏之间的次数消防专项合格证书MTTR Mean Time To Repair维修耗时平均值I/O Input/Output输入 /输出TVM Ticket Vending Machine自动售票机专业 :环境监控SEMI-TVMManually Operated Ticket Vending Machine半自动售票名称全称中文意义机EMCS Electrical and Mechanical Control System车站设备监控系统PVU Portable Verifying Unit便携式验票机ECS Environment Control System环境控制系统GATE闸机DDC Dircct Digital Controller数字直接控制器PLC Programmable Logic Controller可编程逻辑控制器API Application Programming interfac应用程序接口。

CNI5e Ch .ppt

CNI5e Ch .ppt
• The advent and utility of networking has created dramatic economic shifts
– Network has made telecommuting available to individuals – It has changed business communication – An entire industry emerged that develops networking technologies,
• often by using technologies in new unconventional ways
– Computer networks seem complex
• because technologies can be combined and interconnected in many ways
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Topics Covered
• 1.1 Growth of Computer Networking • 1.2 Why Networking Seems Complex • 1.3 The Five Key Aspects of Networking • 1.4 Public and Private Parts of the Internet • 1.5 Networks, Interoperability, and Standards • 1.6 Protocol Suites and Layering Models • 1.7 How Data Passes Through Layers • 1.8 Headers and Layers • 1.9 ISO and the OSI Seven-Layer Reference Model • 1.10 The Inside Scoop • 1.11 Remainder of the Text • 1.12 Summary

基于FPGA_的5G_物理层LowPHY_设计

基于FPGA_的5G_物理层LowPHY_设计

doi:10.20149/ki.issn1008-1739.2024.02.010引用格式:范晓星.基于FPGA 的5G 物理层LowPHY 设计[J].计算机与网络,2024,50(2):150-155.[FAN Xiaoxing.Design of 5G Physical Layer LowPHY Based on FPGA[J].Computer and Network,2024,50(2):150-155.]基于FPGA 的5G 物理层LowPHY 设计范晓星1,2(1.河北远东通信系统工程有限公司,河北石家庄050200;2.专网通信设备与技术河北省工程研究中心,河北石家庄050200)摘㊀要:在对5G O-RAN 架构及物理层功能研究的基础上,设计实现了基于现场可编程门阵列(Field Programmable Gate Array,FPGA)的5G 物理层LowPHY 功能模块㊂通过对5G 物理层LowPHY 的功能进行分析,制定了总体设计方案,将功能模块划分为上行LowPHY 模块㊁下行LowPHY 模块及物理随机接入信道(Physical Random Access Channel,PRACH)LowPHY 模块㊂针对LowPHY 数据处理量大的特点,选用在数据处理方面具备高性能㊁低时延㊁低功耗优势的FPGA 器件来实现㊂将设计的LowPHY 功能模块与其他模块集成为O-RU 单元,在5G NR 系统中进行测试验证,测试结果表明本设计满足了项目需求,实现了设计目标㊂关键词:5G;物理层;现场可编程门阵列;LowPHY;物理随机接入信道;FFT;IFFT;循环前缀;正交频分复用中图分类号:TN929.5文献标志码:A 文章编号:1008-1739(2024)02-0150-06Design of 5G Physical Layer LowPHY Based on FPGAFAN Xiaoxing 1,2(1.Hebei Far-East Communication System Engineering Co.,Ltd.,Shijiazhuang 050200,China ;2.Hebei Engineering Research Center for Private Network Communication Equipment and Technology ,Shijiazhuang 050200,China )Abstract :Based on the research of 5G O-RAN architecture and physical layer function,the 5G physical layer LowPHY function module based on Field Programmable Gate Array (FPGA)is designed and implemented.Based on the analysis of the LowPHY function of 5G physical layer,the overall design scheme is developed,and the functional modules are divided into upstream LowPHY module,downstream LowPHY module and Physical Random Access Channel (PRACH)LowPHY module.Aiming at the characteristics of large data processing capacity of LowPHY,FPGA devices with high performance,low delay and low power consumption in data processing are selected.The designed LowPHY function module is integrated with other modules into O-RU unit,which is tested and verified in 5G NR system.The test results show that the design can meet the project requirements and realize the design objectives.Keywords :5G;physical Layer;FPGA;LowPHY;PRACH;FFT;IFFT;cyclic prefix;OFDM收稿日期:2023-12-210㊀引言随着移动通信技术的发展,第四代移动通信系统(4G)在现代社会获得了广泛应用,人们的生活方式发生了巨大变化㊂在智能终端普及与移动互联网发展的双重因素作用下,移动互联网的多媒体业务获得了长足发展,4G 已经无法满足未来用户更低时延㊁更快速率㊁更大容量的通信需求,因此具有更高性能指标要求的第五代移动通信系统(5G)的研究是通信发展的必然趋势[1-2]㊂5G 在高清视频传输[3]㊁物联网和智能制造[4]等方面已经获得了比较好的效果㊂物理层是5G 新空口(5G New Radio,5G NR)的核心,提供物理介质传输bit 信号流所需要的功能,包括编解码㊁调制解调㊁时频资源映射,并以传输信道的方式为上层提供服务㊂5G 物理层在4G 物理层基础上进行了一些全新的设计[5],采用正交频分复用(Orthogonal Frequency Division Multiplexing,OFDM)技术作为物理层设计基础,引入可扩展的OFDM 子载波间隔参数配置,采用循环前缀OFDM;采用新型信道编码方案低密度奇偶校验码(LDPC),运算量低㊁时延低且硬件实现方便;采用大规模MIMO 技术[6],在多天线处理中实现更高的空间解析度和更高的频谱效率㊂本文结合项目实际需求,设计了基于FPGA 器件实现的5G 物理层LowPHY 功能模块,应用在5G NR 系统中㊂1㊀O-RAN 架构分析O-RAN 是一种开放式无线接入网,允许不同设备商开发的网络设备间互操作㊂O-RAN 的理念是基于开放式体系结构㊁开放式接口和开放式空间,通过开放协议和接口来支持不同厂商的设备㊁软件和服务的互操作性㊂通过借助开放式体系结构和接口,推进无线通信产业的开放化㊁互联互通和创新,打破当前封闭式网络结构,降低部署㊁运维成本,推动数字经济的发展㊂O-RAN架构如图1所示㊂图1㊀O-RAN架构O-RAN架构是开放式无线接入网架构,将基带单元和射频单元分割为3个不同的功能模块和协议层[7],每层设备均可以由不同的厂商提供㊂O-CU (中央单元)和核心网之间的接口称为回传,O-DU (分布单元)和O-CU之间的接口称为中传,O-DU和O-RU(射频单元)之间的接口称为前传㊂各分层单元的实现功能如图2所示㊂O-RU负责处理射频和物理层下部(LowPHY);O-DU负责执行物理层上部(HighPHY)㊁介质访问控制(MAC)和无线电链路控制(RLC)的任务;O-CU负责管理分组数据聚合协议(PDCP)㊁服务数据适配协议(SDAP)和无线电资源控制(RRC)协议实体㊂图2㊀O-RAN分层功能2㊀总体方案设计2.1㊀LowPHY功能分析O-RAN联盟选择Option7-2x拆分选项将5G PHY分为HighPHY与LowPHY,5G PHY下行㊁上行传输链路需要完成的工作分别如图3与图4所示㊂图3㊀5G PHY下行功能框图图4㊀5G PHY上行功能框图如图3所示,HighPHY下行方向需要完成的功能是:同步信号及辅助同步信号(PSS/SSS)处理㊁物理广播信道(PBCH)处理㊁物理下行控制信道(PD-CCH)处理㊁物理共享数据信道(PDSCH)处理㊁参考信号(RS)处理及RE映射(Resource Element Map-ping)㊂RE映射后的数据进入LowPHY部分进行进一步下行处理,在LowPHY部分中,LowPHY主要完成的工作有:OFDM相位补偿㊁IFFT和插入循环前缀(Cyclic Prefix,CP)功能,LowPHY处理完成的下行数据进入射频处理部分发射出去㊂如图4所示,在上行方向,射频部分接收的数据进入5G PHY处理后,首先是LowPHY的数据信息处理,LowPHY对帧数据进行去CP操作,然后分为2路,一路进行OFDM相位补偿㊁FFT;另一路是物理随机接入信道(Physical Random Access Channel,PRACH)的数据处理,频率移位㊁DDC(数据下采样)和FFT㊂LowPHY 处理完成的数据进入HighPHY 部分进一步处理,HighPHY 上行方向主要完成的功能是:RE 解映射㊁上行控制信道(PUCCH)处理㊁上行共享信道(PUSCH)处理㊁随机接入信道(PRACH)处理和RS 处理㊂2.2㊀方案设计本设计以Xilinx 公司UltraScale +及以上系列产品作为设计芯片平台㊂UltraScale +系列采用最先进的16ns /20ns 纳米制程工艺,可以实现更快速㊁更高效和更低功耗的系统设计㊂同时,此系列产品具备丰富的DSP48E2和36kB RAM 资源,十分适合通信算法的实现,此外Xilinx 开发工具具有丰富的IP 资源,可以大大提高软件设计效率㊂将LowPHY 处理模块分为3个子模块进行设计[8],分别是DL LowPHY Process 模块㊁UL LowPHY Process 模块和PRACH LowPHY Process 模块㊂方案具体功能模块示意如图5所示㊂从工程实现角度出发,在协议规定的功能模块基础上,增加IQ_map 模块和IQ_Demap 模块,此2种模块是为了解决无线帧数据与FFT /IFFT 之间数据匹配重构而设计㊂图5㊀LowPHY FPGA 模块示意图5所示的LowPHY 数据处理模块处理的是一路天线数据,对于2T2R 或4T4R 设备来说,各天线的数据处理过程是独立的,在FPGA 工程中复制相应数量的模块即可完成多天线数据处理㊂3㊀FPGA 软件设计3.1㊀下行LowPHY 实现根据图5中下行LowPHY 子功能模块划分,按数据流方向依次实现其FPGA 设计[9-10]㊂相位补偿模块的功能是完成HighPHY 下发的相位补偿因子缓存㊁根据不同symbol 号查找不同的相位因子㊁将天线数据以复数乘法的方式进行相位补偿处理以及将复乘后的IQ 数据进行截位处理㊂对于补偿因子缓存及因子查找输出功能,以RAM 为核心实现,将相位因子缓存进RAM,后续不同的symbol 数据流到此处理过程时,相位因子选取逻辑通过符号数据其time 信息的symbol 号读取相应的相位因子供下级复数乘法处理使用㊂复数乘法器将IQ (I㊁Q 均16bit)数据与其相对应相位因子进行相乘,完成相位补偿处理,此时乘法器输出I /Q 数据为32bit,通过截位操作,丢弃低16bit,保留高16bit 数据来实现功率控制,完成相位补偿处理㊂在5G FR1中,有273个物理资源块(Physical RB,PRB ),每个PRB 有12个子载波,共3276(273ˑ12)个采样点,而5G IFFT 处理的采样点为4096,因此IQ 数据在进行IFFT 处理前需要进行补0及数据重构,此步通过2个RAM 以乒乓的方式控制读写来实现IFFT 输入数据的排序,命名为IQ_map 模块㊂具体数据重构逻辑如图6所示,将1个symbol 数据的3276个点缓存后,分为两部分,首先读出高部分1638个32bit 数据,在1639~2459位置填充0值,再读出低部分1638个32bit 数据㊂图6㊀下行数据重构格式示意IFFT IP 核定制部分截图如图7所示㊂图7㊀IFFT IP 定制部分配置IFFT 功能通过使用Xilinx 公司官方IP 核来实现,IP 定制的配置是:数据点4096,实现架构Pipe-lined,Streaming I /O,数据格式Fixed Point,Un-scaled,数据输入㊁系数输入均为16bit㊂经过IFFT 处理的数据,截位处理恢复到I /Q 均为16bit 的格式,根据其symbol 号来动态配置添加CP,至此下行帧数据完成了LowPHY 处理,频域数据转为时域数据,传递到下一阶段进行数字上变频(DUC)㊁波峰消减(CFR)和数字预失真(DPD)数字域处理,直至DAC 后在射频域发射至无线空间㊂3.2㊀上行LowPHY 实现上行是下行的逆过程,数据处理过程极为相似,部分逻辑代码可以移植复用㊂其中,OFDM 相位补偿子模块,上下行均在频域完成,功能一致,上行LowPHY 的FPGA 模块实现直接通过例化上述设计文件的方式实现㊂去CP 子模块,根据symbol 数据的time 信息,判读其CP 长度,以计数器为基准,将CP 数据剔除,最终输出4096点时域symbol 数据完成去CP 操作的symbol 数据进入FFT 模块,进行时域向频域的转换,FFT 功能与IFFT 功能一样,使用IP 核实现,配置保持一致,输出数据进行截位处理,恢复I /Q 均为16bit 的数据格式㊂上行数据重构格式示意图如图8所示㊂图8㊀上行数据重构格式示意IQ_Demap 是IQ_map 的逆过程,按照图8所示的处理方式将FFT 输出的4096点symbol 数据处理成3276点数据㊂实现逻辑依然是通过控制2个乒乓RAM 读写来实现数据重构㊂完成OFDM 相位补偿处理的上行帧数据就完成了LowPHY 处理,传递到HighPHY 进行进一步处理,就完成了5G 上行PHY 处理㊂3.3㊀PRACH LowPHY 实现由于功能相同,PRACH LowPHY 的去CP 模块,直接例化UL 方向的去CP HDL 文件㊂由于prach 的种类很多,不同种类的cp 长度㊁数据长度和重复次数不同,但处理流程相似,增加prach_ctrl 控制逻辑,用于完成对不同prach 场景的控制[11]㊂prach_ctrl 功能逻辑由10ms 帧位置定位㊁slot 号位置定位及symbol 位置定位提取模块构成㊂图9是控制逻辑状态机状态转移图㊂10ms 帧位置定位模块用来检测10ms 帧是否有prach,对应用prach 的10ms 帧的位置输出指示信号到slot 号位置定位模块,然后进一步精确到slot 位置,输出信息到symbol 位置定位提取模块,最终确定prach 的详细位置㊂图9㊀prach_ctrl 状态机状态转移DDC 模块目的是将IQ 数据流搬移至0频,对数据进行降采样操作,实现方式是使用Xilinx 官方FIR 滤波器IP 核,通过多级半带滤波器的滤波㊁抽取,达到设计目标;FFT 模块同样使用FFT IP 核来实现,与Ul LowPHY 中IP 定制参数一致;IQ_Demap模块根据prach 长度命令参数值的大小,通过选择与计数逻辑,输出数据,存至乒乓FIFO 组成的IQ_buf 模块,最终发送至上层HighPHY,完成PRACH LowPHY 的处理过程㊂4㊀功能测试验证LowPHY 功能各单元小模块在完成单元测试后需进行整体功能测试验证㊂作为O-RU 单元的组成部分,将LowPHY 功能与其他功能集成为整机,并在5G 基站系统中进行测试,更加能够测试出设计的性能,因此本设计使用系统测试环境进行功能验证测试㊂对于下行测试,第一种方式使用MAC 层下发测试向量的方式进行测试㊂MAC 层通过配置发送测试数据到HighPHY,HighPHY 将测试数据传输给LowPHY 处理,LowPHY 处理完成的数据经过后级数据处理模块进一步处理最终发射到空间㊂将O-RU 天线口通过衰减器连接至频谱分析仪,可以观测下行信号的EVM㊁星座图等射频指标,通过这些指标可以测试出设计的性能㊂图10是发送ETM3-1a 向量在频谱仪的显示结果,向量ETM3-1a 发送的是100MHz 带宽256QAM 调制数据,可以看到EVM 为2.80,小于标准4.50,满足3GPP 标准,星座图打点清晰,证明在幅度失衡㊁正交误差㊁相位噪声㊁相位误差和调制误差等指标方面,本设计的数据处理功能有比较优良的性能,对于QPSK㊁16QAM 和64QAM 调制数据的测试,结果同样满足标准规定㊂图10㊀频谱分析仪截图㊀㊀对于上行测试和PRACH LowPHY 功能测试,可以合并进行㊂将手机在5G 核心网完成开户,测试其能否入网,并在入网后在手机侧使用Iperf 工具向MAC 层灌包,观测MAC 层接收数据包速率,验证其功能及性能㊂同时MAC 层向手机侧灌包,进一步验证下行性能,为下行功能验证的第2种方式㊂测试系统设置为帧格式2.5ms 双周期,双流灌包㊂图11是2.5ms 双周期的帧结构,每5ms 里面包含5个全下行时隙㊁3个全上行时隙和2个特殊时隙㊂slot3和slot7等为特殊时隙,配比为10ʒ2ʒ2,理论峰值灌包速率,下行660Mb /s㊁上行330Mb /s㊂图11㊀ 2.5ms 双周期帧结构㊀㊀在完成入网的手机上设置上行Iperf 灌包,在MAC 层设置下行灌包,图12是上下行灌包速率统计过程的截图㊂从图中可以看到,左侧显示下行灌包速率约为660Mb /s㊁右侧显示上行灌包速率约为332Mb /s,证明测试手机不但在测试系统完成了入网,验证了PRACH LowPHY 功能,同时,上下行数据通道功能㊁性能良好,也验证测试了UL /DL LowPHY 功能㊂图12㊀上下行灌包速率结果截图㊀㊀以上测试结果表明,本文基于FPGA设计的5G LowPHY功能模块,实现了3GPP标准规定的功能,同时满足指标要求㊂5㊀结束语本文根据实际项目需求,设计实现了基于FPGA 的5G物理层LowPHY功能,并详细阐述了LowPHY 的数据处理过程划分㊁总体方案设计以及在FPGA平台的实现过程,最终通过将功能模块嵌入测试系统的方式对其功能㊁性能进行了测试㊂本文设计的功能模块具备较高的移植适应性,不但可以在多种FPGA型号上运行,还可以根据项目总体设计需要,在不同的O-RAN切分方式下,将功能部署在不同的实体硬件单元设备上,大大扩展了其应用场景㊂参考文献[1]㊀ANDREWS J G,BUZZI S,CHOI W,et al.What Will5GBe?[J].IEEE Journal on Selected Areas in Communica-tions,2014,32(6):1065-1082.[2]㊀尤肖虎,潘志文,高西奇,等.5G移动通信发展趋势与若干关键技术[J].中国科学:信息科学,2014,44(5):551-563.[3]㊀邝野.基于5G网络的无人机超清实时图传系统设计与实现[D].广州:广东工业大学,2022.[4]㊀陈斌,陈武军,樊忠文,等.5G+MEC专网智能制造工厂[J].通信技术,2021,54(1):215-223. [5]㊀王霁超.5G NR物理层软件无线电设计与实现[D].南京:东南大学,2021.[6]㊀章辰.面向5G的大规模MIMO物理层关键技术的研究和仿真[D].北京:北京邮电大学,2020.[7]㊀李剑.O-RAN网络中基于QEE优化的休眠策略研究[D].桂林:桂林电子科技大学,2021.[8]㊀龙凯.5G物理层比特级处理并行架构设计与实现[D].成都:电子科技大学,2020.[9]㊀黄旭.5G通信系统中载波同步帧同步系统的研究[D].南京:南京邮电大学,2019.[10]张越良.5G新空口下行同步和广播信道的仿真与FPGA实现[D].北京:北京邮电大学,2019.[11]朱恒恒.5G移动通信系统物理随机接入信道的实现[D].成都:电子科技大学,2022.作者简介范晓星㊀男,(1987 ),硕士,工程师㊂。

美国PAC声发射仪器

美国PAC声发射仪器

Video
video
Par 8 Waveform Optional Module 4Channel SPM Signal Processing Module
Mouse Keybord
KEYBOARD
PCI COMPUTER BOARD PIII or P4 CPU
USB Printer COM1
H Disk CDROM floppy
28v
AST 4 Channel AE Module
PCI-DSP Board Block Diagram
PCI-DSP4 Master (with 8 Parametrics)
PARAMETRICS
SENSOR INPUTS
Par 1
8 Channel Parametrics & Group Controller (In 1st Board Only) DSP Based Channel Controller
PCI-2特点
18位A/D,40MHZ采样; Streaming function 连续波形处理; 广义AE应用; 嵌入式结构—并行硬件; 开 放 式 系 统 —SDK , Lab View driver 最好的研发工具;
柱状容器定位
球面定位
AEWIN
NOESIS Pattern Recognition Software
Expansion Interface Connector For>56CH
DiSP 系统结构图
Personal computer or industrial PC chassis
ch. 1
ch. 2 ch. 3 ch. 4 8-ch. para. inputs

《计算机网络》英文缩写

《计算机网络》英文缩写

AACK (ACKnowledgement) 确认ADSL(Asymmetric Digital Subscriber Line)非对称数字用户线AES (Advanced Encryption Standard)先进的加密标准AF PHB (Assured Forwarding PerHop Behavior)确保转发每跳行为AH (Authntication Header)鉴别首部AIMD (Additive Increase Multiplicative Decrease)加法增大乘法减小AN (Access Network)接入网ANSI (American National Standards Institute)美国国家标准协会AP(Access Point)接入点AP(Application)应用程序API (Applicatin Programming Interface)应用编程接口APNIC (Asia Pacific Network Information Center)亚太网络信息中心ARIN (American Registry for Internet Numbers) 美国因特网好码注册机构ARP(Address Resolution Protocol) 地址解析协议ARPA(Advanced Research Project Agency)美国国防部远景研究规划局(高级研究计划署)ARQ (Automatic Repeat reQuest)自动重传请求AS (Authentication System)自治系统AS (Authentication Server)鉴别服务器ASCII (American Standard Code for Information Interchange)美国信息交换标准码ASN (Autonomous System Number)自治系统号ASN.I (Abstract Syntax Notation One)抽象语法记法ATM (Asynchronous Transfer Node)异步传递发式ATU (Access Termination Unit) 接入端接单元ATUC (Access Termination Unit Central Office)端局接入端接单元ATUR (Access Termination Unit Remote)远端接入端接单元AVTWG(Audio/Video Transport Working Group)音频/视频运输工作组AWT(Abstract WindowToolkit)抽象窗口工具箱BBER (Bit Error Rate)误码率BER (Basic Encoding Rule)基本编码规则BGP(Border Gateway Protocol) 边界网关协议BOOTP(BOOTstrap Protocol) 引导程序协议BSA(Basic ServiceArea )基本服务区BSS (Basic Service Set) 基本服务集BSSID (Basic Service Set ID)基本服务集标识符BT(BitTorrent) 一种P2P程序CCA(CertificationAuthority)认证中心CA(CollisionAvoidance)碰撞避免CATV(Community AntennaTV,Cable TV)有线电视CBT(Core Based Tree)基于核心的转发树CCIR(Consultative Committee ,International Radio)国际无线电咨询委员会CCITT(Consutative Committee, International Telegraphand Telephone)国际电报电话咨询委员会CDM(Code Division Multiplexing)码分复用CDMA(Code Division Multiplex Access)码分多址CE(Consumer Electronics)消费电子设备CFI(Canonical Format Indicator)规范格式指示符CGI(Common Gateway Interface)通用网关接口CHAP(Challenge Handshake Authentication Protocol)口令握手鉴别协议CIDR(Classless InterDomain Routing)无分类域间路由选择CNAME(Canonical NAME)规范名CNNIC(Network Information Center of China)中国互联网络信息中心CRC(Cyclic Redundancy Check)循环冗余检查CSACELP(Conjugate Structure Algebraic Code Excited Linear Prediction)共轭结构代数码激励线性预测(声码器)CSMA/CD(Carrier Sense Multiple Access/Collision Detection),载波监听多点接入/冲突检测CSMA/CA(Carrier Sense Multiple Access/Collision Avoidance),载波监听多点接入/冲突避免CSRC(Contributing SouRCe identifier)参与源标知符CTS(ClearTo Send)允许发送DDACS(Digital Access and Crossconnect System)数字交接系统DARPA(Defense Advanced Research Project Agency)美国国防部远景规划局(高级研究署)DCF(Distributed Coordination Function)分布协调功能DDoS(Distributed Denial of Service)分布式拒绝服务DES(Date Encryption Standard)数据加密标准DF(Don’t Fragment)不能分片DHCP(Dynamic Host Configuration Protocol)动态主机配置协议DiffServ(Distributed Coordination Identifier)数据链路连接标知符DIFS(Distributed Coordination Function IFS)分布协调功能祯间间隔DLCI(Data Link Connection Identifier)数据链路连接标知符DMT(Discrete Multi Tone)离散多音(调制)DNS(Domain Name System)域名系统DOCSIS(Data Over Cable Service Interface Specifications)电缆数据服务接口规约DoS(Denial of Service)拒绝服务DS(Distribution System)分配系统DS(Differentiated Services)区分服务(也写作DiffServ)DSCP(DifferentiatedServices Code Point)区分服务码点DSL(Digital Subscriber Line)数字用户线DSLAM(DSL Access Multiplexer)数字用户线接入复用器DSSS(Direct Sequence Spread Spectrum)直接序列扩展DVMRP(Distance Vector Multicast Routing Protocol)距离向量多播路由选择协议DWDM(Dense WDM)密集波分复用EEBCDIC(Extended Binary Coded Decimal Interchangfe Code)扩充的二、十进制交换码EDFA(Erbirm Dooped Fiber Amplifier)掺铒光纤放大器EFM(Ethernet in the First Mile)第一英里的以太网EFPHB(Expedited Forwarding PerHop Behavior)迅速转发每跳行为EGP(External Gateway Protocol)外部网关协议EIA(Electronic Industries Association)美国工业协会EOT(End Of Transmission)传输结束ESP(Encapsulating Security Payload)封装完全有效载荷ESS(Extended Service Set)扩展的服务集ETSI(European Telecommunications Standrards Institute)欧洲电信标准协会EUI(Extended Unique Identifier)扩展的唯一标识符FFC(Fibre Channel)光纤通道FCS(Frame Check Sequence)帧检验列FDDI(Fiber Distributed Data Interface)光纤分布式数据FDM(Frequency Division Multiplexing)频分复用FEC(Forwarding Equivalence Class) 转发等价类FFD(Full Function Device)全功能设备FHSS(Frequency Hopping Spread Spectrum)跳频扩频FIFO(First In First Out)先进先出FQ(Fair Queuing)公平排队FTP(File Transfer Protocol) 文件传送协议FTTB(Fiber To The Building)光纤到大楼FTTC(Fiber To The Curb)光纤到路边FTTD(Fiber To The Door)光纤到门户FTTF(Fiber To The Floor)光纤到楼层FTTH(Fiber To The Home)光纤到家FTTN(Fiber To The Neighbor)光纤到邻居FTTO(Fiber To The Office)光纤到办公室FTTZ(Fiber To The Zone)光纤到小区GGIF(Graphics Interchange Format)图形交换格式GSM(Graphics System for Mobile)全球移动通信系统,GSM体制HHDLC(High-level Data Link Control) 高级数据链路控制HDSL(High speed DSL)高速数字用户线HFC(Hybrid Fiber Coax)光纤同轴混合(网)HIPPI(High Performance Parallel Interface)高性能并行接口HRDSSS(High Rate Direct Sequence Spread Spectrum)高速直接序列扩频HSSG(High Speed Study Group)高速研究组HTML(Hyper Text Markup Language)超文本标记语言HTTP(Hyper Text Transfer Protocol)超文本传送协议IIAB(Internet Assigned Numbers Authority)因特网体系结构委员会IANA(Internet Corporation for Assigned Names and Numbers)因特网赋号管理局ICANN(Internet Corporation for Assigned Names and Numbers ) 因特网名字与号码指派公司ICMP(Internet Control Message Protocol)国际控制报文协议IDEA(Internet Data Encryption Algorithm)国际数据加密算法IEEE(Institute of Electrical and Electronic Engineering)(美国)电气和电子工程师学会IESG(Internet Engineering Steering Group)因特网工程指导小组IETF(Internet Engineering Task Force)因特网工程部IFS(Internet Frame Space)桢间间隔IGMP(Internet Group Management Protocol)网际组管理协议IGP(Interior Gateway Protocol)内部网关协议IM(Instant Messaging)即时传信IMAP(Internet Message Access Protocol)因特网报文存取协议IntServ(Integrated Services)综合服务IP(Internet Protocol)网际协议IPCP(IP Control Protocol)IP 控制协议IPng(IP Next Generation)下一代的IPIPRA(Intenet Policy Registration Authorrity)因特网政策登记管理机构IPsec(IP security)IP安全协议IPX(Internet Packet Exchange)Novel 公司的一种联网协议IR(InfraRed)红外技术IRSG(Internet Reseach Seering Group)因特网研究指导小组IRTF(Internet Reaserch Task Force)因特网研究部ISDN(Integrated Services Digital Network)综合业务数字网ISO(International Organization for Standardization)国际标准化组织ISOC(Internet Society)因特网协会ISM(Industrial,Scientific,and Medical)工业、科学与医药(频段)ISP(Internet Service Provider)因特网服务提供者ITU(Internation Telecommunication Union)国际电信联盟ITU-T(ITU Telecommunication Standardization Sector)国际电信联盟电信标准化部门JJPEG(Joint Photographic Expert Group)联合图像专家JVM(javaVirtual Machine)java虚拟机KKDC(Key Distributio Center)密钥分配中心LACNIC(Latin American & Caribbean NetworkInternetCenter)拉美与加勒比海网络信息中心LAN(Local Area Network)局域网LCP(Link Control Protocol)链路控制协议LDP(Label Distribution Protocol)标记分配协议LED(Light Emitting Diode)发光二极管LMDS(Local Multipoint Distribution System)本地多点分配系统LLC(Logical Link Control)逻辑链路控制LOS(Line of Sight)视距LPC(linear Prediction Coding)线性预测编码LSP(Label Switched Path)标记交换路径LSR(Label Switching Router)标记交换路由器MMAC(Medium Access Control)媒体接入控制MACA(Multiple Access with Collision Avoidance)具有碰撞避免的多点接入MAGIC(Mobile multimedia,Anytime/anywhere, Global mobility support,Integrated wireless and Customized personal service)移动多媒体、任何时间/地点、支持全球移动性、综合无线和定制的个人服务MAN(Metropolitan Area Network)城域网MANET(Mobile Adhoc NETworks)移动自组网络的工作组MBONE(Multicast Backbone On the Internet)多播主干网MCU(Multipoint Control Unit)多点控制单元MD(Message Digest)报文摘要MF(More Fragment)还有分片MFTP(Multisource File Transfer Protocol)多源文件传输协议MIB(Management Information Base)管理信息库MIME(Multipurpose Internet Mail Extensions)通用因特网邮件扩充MIPS(Million Instructions Per Second)百万指令每秒MMUSIC(Multiparty MUltimedia SessIon Control)多参与者多媒体会话控制MOSPF(Multicast extensions to OSPF)开放最短通路优先的多播扩展MP3(Mpeg1 Audio layer3)一种音频压缩标准MPEG(Motion Picture Experts Group)活动图像专家组MPLS(Multi Protocol Label Switching)多协议标记交换MPPS(Million Packets Per Second)百万分组每秒MRU(Maximum Receive Unit)最大接收单元MSL(Maximum Segment Lifetime)最长报文段寿命MSS(Maximum Segment Size)最长报文段MTU(Maximum Transfer Unit)最大传送单元NNAP(Network Access Point)网络接入点NAT(Network Address Translation)网络地址转换NAV(Network Allocation Vector)网络分配向量NCP(Network Control Protocol)网络控制协议NFS(Network File System)网络文件系统NGI(Next Generation Internet)下一代因特网NGN(Next Generation Network)下一代电信网NIC(Network Interface Card)网络接口卡、网卡NLA(Next Level Aggregation)下一级聚合NLRI (Network Layer Reachability Information)网络层可达性信息NOC (Network Operations Center)网络运行中心NSAP (Network Service Access Point) 网络层服务访问点NVT (Network Virtual Terminal) 网络虚拟终端OOC (Optical Carrier)光载波ODN (Optical Distribution Node)光分配结点OFDM (Orthogonal Frequency Division Multiplexing)正交频分复用OSI/RM (Open Systems Interconnection Reference Model) 开放系统互连基本参考模型OSPF (Open Shortest Path First) 开放最短通路优先OUI (Organizationally Unique Identifier)机构唯一标识符PP2P (PeertoPeer)对等方式PAN (Personal Area Network)个人区域网PAP (PasswordAuthentication Protocol) 口令鉴别协议PARC (PoloAlto Research Center)(美国施乐公司(XEROX)的)PARC研究中心PAWS (Protect Against Wrapped Sequence numbers) 防止序号绕回PCA (Policy Certification Authority)政策认证中心PCF (Point Coordination Function)点协调功能PCM (Pulse Code Modulation)脉码调制PCMCIA (Personal Computer Memory Card Interface Adapter)个人计算机存储器卡接口适配器PDA (Personal Digital Assistant) 个人数字助理PDU (Protocol Data Unit)协议数据单元PEM (Privacy Enhanced Mail)因特网的正式邮件加密技术PGP (Pretty Good Privacy)一种电子邮件加密技术PHB (PerHop Behavior)每跳行为PIFS (Point Coordination Function IFS)点协调功能桢间间隔PIMDM (Protocol Independent Multicast Dense Mode)协议无关多播密集方式PIMSM (Protocol Independent Multicast Sparse Mode)协议无关多播稀疏方式PING (Packet Internet Groper)分组网间探测,乒程序,ICMP的一种应用PK (public key)公钥,公开密钥PKI (Public Key Infrastructure)公钥基础结构PoP (Post Office Protocol) 汇接点POP (Post Office Protocol) 邮局协议POTS (Plain Old Telephone Service)传统电话PPP (Pointto Point Protocol over Ethernet) 点对点协议PPPoE(Point-to-Point Protocol over Ethernet)以太网上的点对点协议PS (POTS Splitter)电话分离器PTE (Path Terminating Element) 路径端接设备QQAM (Quadrature Amplitude Modulation)正交幅度调制QoS (Quality of Service)服务质量QPSK(Quarternary Phase Shift Keying)正交相移键控RRA(Registration Authority)注册管理机构RARP(Reverse Address ResolutionProtocol)逆地址解析协议RAS(Registration/Adminssion/Status)登记/接纳/状态RED(Random Early Detection)随机早期检测RED(Random Early Discard,Randomm Early Drop)随机早期丢弃RFC(Request For Comments)请求评论RG(Research Group)研究组RIP(Routing Information Protocol)路由信息协议RIPE(法文表示的EuropeanIP Network)欧洲的IP网络RPB(ReversePath Broadcasting )反向路由广播RSA(Rivest,Shamir andAdleman)用三个人名表示的一种公开密钥算法的名称RSVP(Resourcereservation Protocol)实时传送控制协议RTCP(Realtime Transfer Protocol) 实时传送控制协议RTO(Retransmission Time Out)超时重传时间RTP(Realtime Transfer Protocol)实时传送协议RTS(Request To Send)请求发送RTSP(Realtime Streaming Protocol)实时流式协议RTT(Round Trip Time)往返时间SSA(Security Association)安全关联SACK(Selective ACK)选择确认SCTP(Stream Control Transmission Protocol)流控制传输协议SDH(Synchronous Digital Hierarchy)同步数字系列SDSL(Singleline DSL)1 对线的数字用户线SDU(Service Data Unit)服务数据单元SET(Secure Electronic Transaction)安全电子交易SHA(Secure Hash Algorithm)安全散列算法SIFS(Short IFS)短帧间间隔SIP(Session Initiation Protocol)会话发起协议SK(Secret Key)密钥SLA(Service Level Agreement)服务等级协议SMI(Structure of Management Information)管理信息结构SMTP(Simple Mail Transfer Protocol)简单邮件传送协议SNA(System Network Archiecture)系统网络体系结构SNMP(Simple Network Management Protocol)简单网络管理协议SOH(Start Of Header)首部开始SONET(Synchronous Optical Network)同步光纤网SPI(Security Parameter Index)安全指数索引SRA(Seamless Rate Adaptation)无缝速率自适应技术SSID(Service Set IDentifier)服务集标识符SSL(Secure Socket Layer)安全插口层,或安全套接层SSRC(Synchronous SouRce identifier)同步源标实符STDM(StatisticTDM )统计时分复用STM(Synchronous Transfer Module)同步传输模块STP(ShielderTeisted Pair)屏蔽双绞线STS(Synchronous Transport Signal) 同步传送信号TTAG(TAG Swithcing)标记交换TCB(Transmission Control Protocol) 传输控制程序块TCP(Transmission Control Protocol) 传输控制协议TDM(Time Division Multiplexing)时分复用TELNET(TELetype NET work)电传机网络,一种因特网的应用程序TFTP(Trivial File Transfer Protocol) 简单文件传送协议TIA(Telecommunications Industries Association)电信行业协会TLA(Top Level Aggergation)顶级聚合TLD(Top Level Domain)顶级域名TLI(Transport Layer Interface)运输层接口TLS(Transport Layer Security)运输层安全协议TLV(Type Length Value)类型长度值TPDU(Transport Protocol Data Unit) 运输协议数据单元TSS(TelecommunicationStandardization Sector)电信标准化部门TTL(Time to Live)生存时间,或寿命UUA(User Agent) 用户代理UAC(User Agent Client) 用户代理客户UAS(User Agent Server)用户代理服务器UDP(User Datagram Protocol) 用户数据报协议UIB(User Interface Box)用户接口盒URL(Uniform Resource Locator)统一资源定位符UTP(Unshielder Twisted Pair)无屏蔽双绞线UWB( UltraWide Band)超宽带VVC(Virtual Circuit)虚电路VCI(Virtual Channel Identifier)虚拟路标识VDSL(Very high speed DSL)甚高速数字用户线VID(VLAN ID)标识符VLAN(Virtual LAN)虚拟局域网VLSM(Variable Length Subner Mask)变长子网掩码VoIP(Voice over IP)在IP上的话音VON(Voice On the Net) 在因特网上的话音VIP(Virtual Path Identifier)虚拟道标识符VPN (Virtual Private Network)虚拟专用网VSAT (Very Small Aperture Terminal) 甚小孔径地球站WWAN (WideArea Network)广域网WDM (Wavelength Division Multiplexing)波分复用WEP (Wired Equivalent Privacy)有线等效保密字段WFQ (Weighted Fair Queuing)加权公平排队WG (Working Group)工作组WiFi(Wireless Fidelity)无线保真度(无线局域网的同义词)WIMAX(Worldwide interoperability for Microwave Access) 全球微波接入的互操作性,即WMAN。

浅谈节约ip地址的方法

浅谈节约ip地址的方法

浅谈节约ip地址的方法浅谈节约ip地址的方法大学计算机网络基础课程研究报告Introduction of Computer Network 浅谈节约IP地址的方法陈家佳(人文学院广告0901 joycech@) 摘要:本文主要简单介绍了IP地址的概念和作用,介绍两种节约IP地址的方法:NAT和子网划分,以及两种方法的实现原理,并且指出解决IP地址匮乏的最终方案。

关键词:IP地址,NAT,子网划分,实现原理IP address of the Method of saving Chen Jia Jia (school of Humanities,Advertising0901,joycech@) Abstract: This paper briefly introduces the concept and role of IP addresses, describes two methods of saving IP addresses: NAT and subnetting, and the implementation principle of the two methods, and solve the IP address that lack of final plan. Key words: IP addresses, NAT, subnetting, racetrack 一、IP地址概述因特网是全世界范围内的计算机联为一体而构成的通信网络的总称,是一个庞大的网间网,其中包含了数以千万计的计算机,因此,要想实现这些计算机之间的通信,首先要解决的是给每台计算机一个唯一的标识,这就是IP地址。

根据I P协议的规定,每个I P 地址的长度为32位,分4 段,每段8位,常用十进制数字表示,每段数字范围为1~255,段与段之间用小数点分隔。

每段也可以用十六进制或二进制表示。

每个IP地址包括两部分,即网络号和主机号。

Exam4(Ch5)

Exam4(Ch5)

(50 minutes)Part I: Multiple Choice Questions: (1 pts each)1.Suppose array a is int[] a = {1, 2, 3}, what is a[0] - a[2]?a. 1b. 2c. 3d. None of the above2. An array can be used in which of the following ways?a. As a local variableb. As a parameter of a methodc. As a return value of a methodd. All of the above3. Which of the following are valid array declarations?a. char[] charArray = new char[26];b. int[] words = new words[10];c. char[] charArray = "Computer Science";d. double[3] nums = {3.5, 35.1, 32.0};e. None of the above4. Consider the following code fragment:int[] list = new int[10];for (int i = 0; i <= list.length; i++) {list[i] = (int)(Math.random() * 10);}Which of the following statements is true?a. list.length must be replaced by 10b. The loop body will execute 10 times, filling up the array with random numbers.c. The loop body will execute 10 times, filling up the array with zeros.d. The code has a runtime error indicating that the array is out of bound.5. Assume the signature of the method xMethod is as follows.public static void xMethod(double[] a)Which of the following could be used to invoke xMethod?a. xMethod(5);b. xMethod({3, 4});c. xMethod(new int[2]);e. None of the above.6. Given the following statementint[ ] list = new int[10];list.length has the valuea. 10b. 9c. The value depends on how many integers are stored in list.d. None of the above.7. Given the following statementint[ ] list = new int[10];a. The array variable list contains a memory address that refers to an array of 10 int values.b. The array variable list contains a memory address that refers to an array of 9 int values.c. The array variable list contains ten values of type int.d. The array variable list contains nine values of type int.e. None of the above.8. Given the following declaration:int[ ][ ] m = new int[5][6];Which of the following statements is true?a. The name m represents a two-dimensional array of 30 int values.b. m[2][4] represents the element stored in the 2nd row and the 4th column of m.c. m.length has the value 6.d. m[0].length has the value 5.9. In the following code, what is the printout for list2?class Test {public static void main(String[] args) {int[] list1 = {3, 2, 1};int[] list2 = {1, 2, 3};list2 = list1;list1[0] = 0; list1[1] = 1; list2[2] = 2;for (int i = list2.length - 1; i >= 0; i--)System.out.print(list2[i] + " ");}}a. 1 2 3b. 3 2 1c. 0 1 2d. 2 1 0e. 0 1 3.10. Analyze the following code:public class Test {public static void main(String[] args) {int[] x = {0, 1, 2, 3, 4, 5};}public static void xMethod(int[] x, int length) {for (int i = 0; i < length; i++)System.out.print(" " + x[i]);}}a. The program displays 0 1 2 3 4.b. The program displays 0 1 2 3 4 and then raises a runtime exception.c. The program displays 0 1 2 3 4 5.d. The program displays 0 1 2 3 4 5 and then raises a runtime exception.11. Analyze the following code:public class Test {public static void main(String[] args) {int[] x = new int[5];for (int i = 0; i < x.length; i++)x[i] = i;System.out.println(x[i]);}}a. The program displays 0 1 2 3 4.b. The program displays 4.c. The program has a runtime error because the last statement in the main method causes ArrayIndexOutOfBounds exception.d. The program has syntax error because i is not defined in the last statement in the main method.12. Assume double[] scores = {1, 2, 3, 4, 5}, what value does java.util.Arrays.binarySearch(scores, 3.5) return?a. 0b. -4c. 3d. -3e. 4Part II. (4 pts) Find and fix all errors in the following program:(A)public class Test {public static void main(String[] args) {double[4] list = {3, 4, 5, 6.3};xMethod(list);}public static double xMethod(int[] list) {double sum = 0;for (int i=0; i <= list.length; i++) {sum += list[i];}return sum;}(B) There is exactly one logic error in the following binary search method. Fix it./** Use binary search to find the key in the list */public static int binarySearch(int[] list, int key) {int low = 0;int high = list.length - 1;while (high > low) { // Fix error on this lineint mid = (low + high) / 2;if (key < list[mid])high = mid - 1;else if (key == list[mid])return mid;elselow = mid + 1;}return -1 - low;}Part III: Show the printout of the following code:a. (2 pts)public class Test {public static void main(String[] args) {int[] a = {1, 2};swap(a[0], a[1]);System.out.println("a[0] = " + a[0] + " a[1] = " + a[1]);}public static void swap(int n1, int n2) {int temp = n1;n1 = n2;n2 = temp;}}b.(2 pts)public class Test {public static void main(String[] args) {int[] a = {1, 2};swap(a);System.out.println("a[0] = " + a[0] + " a[1] = " + a[1]);}public static void swap(int[] a) {int temp = a[0];a[0] = a[1];a[1] = temp;}}c.(4 pts) Given the following program, show the values of the array in the following figure:public class Test {public static void main(String[] args) {int[] values = new int[5];for (int i = 1; i < 5; i++) {values[i] = i;}values[0] = values[1] + values[4];}}d.(1 pts) Show the printout of the following program:public class Test {public static void main(String[] args) {double[] numbers = {1, 4.3, 5.55, 3.4};double[] x = numbers;numbers[0] = 300;System.out.println(x[0]);}}Part IV:a. (3 pts): Write a method that computes the average of the values in an array of doubles. The header of the method is as follows:public static double average(double[] x)b.(3 pts) Write a method that returns the index of the smallest element in an array of integers. If there aremore than one such elements, return the smallest index.public static int minIndex(int[] list)c. (Note: This a very challenging program.) (7 pts): Write a method that returns a new array with no duplicates. The header of the method is as follows:public static int[] removeDuplicate(int[] x)For example, if x is {1, 3, 1, 2, 3, 5}, removeDuplicate(x) returns a new array {1, 3, 2, 5}(Hint: To be given in the class)。

ch5-图灵机1教程

ch5-图灵机1教程

图灵机接受的语言
L(M) = {ω │ω ∈T*且q0ω ├* α
1
p α
2
,p∈F, α 1α 2∈∑*}
图灵机接受的语言是输入字母表中这样一些字符串的集合, 初始时,这些字符串放在M的带上,M处于状态q0,且M的带头处 在最左单元上,这些字符串将使M进入某个终止状态。 假定:
当输入被接受时,图灵机将停止,没有下一个动作。
(1)当在搜寻b时,M找到了空白符B,则M停止,不接受该串。
(此时,a的个数大于b的个数) (2) 当将b改为y后,左边再也找不到a,此时,若右边再无b, 接受;若仍有b,则b的个数大于a的个数,不接受。
School of Computer Science & Technology, BUPT 10
School of Computer Science & Technology, BUPT
L 0/1 1/1 q2 B/B q4 B/0 L 0/0,1/B
16
1/1 R
例4:L= 0

m
m=2n, n 0
设计思路:对输入串w 1. 从左到右扫描带,隔一个消一个0; 2.若带上只剩唯一一个0,接受; 3.若带上不止一个0,且个数为奇数,拒绝; 4.让读写头返回带的最左端; 5. 转第一步。
School of Computer Science & Technology, BUPT 9
图灵机举例
例1:设语言 L={an bn│n>=1},设计图灵机接受L 。
思路:最初带上为
a a … a
n个a
b b… b B B B ……
n个b
首先用x替换M最左边的a,再右移至最左边的b用y替换之,左移 寻找最右的x,然后右移一单元到最左的a,重复循环。 如果

计算机一级英文缩写

计算机一级英文缩写

计算机一级英文缩写
摘要:
1.计算机一级英文缩写的概念
2.计算机一级英文缩写的具体内容
3.计算机一级英文缩写的应用领域
4.计算机一级英文缩写的重要性
正文:
计算机一级英文缩写是指在计算机领域中,一些常用术语和概念的英文缩写。

这些缩写通常是由单词的首字母组成,用以简化表达和提高沟通效率。

计算机一级英文缩写在计算机科学和信息技术领域具有广泛的应用,如编程语言、操作系统、网络通信等。

计算机一级英文缩写的具体内容包括很多方面。

例如,计算机领域的一些重要概念,如计算机硬件(Computer Hardware,CH)、计算机软件(Computer Software,CS)、计算机网络(Computer Network,CN)等。

此外,还有一些常用的计算机术语,如中央处理器(Central Processing Unit,CPU)、随机存取存储器(Random Access Memory,RAM)等。

这些缩写在计算机领域的学术论文、专业书籍和日常交流中被广泛使用。

计算机一级英文缩写在多个应用领域发挥着重要作用。

首先,在计算机教育和培训中,掌握这些缩写对于学生和从业者来说十分重要,可以提高学习效率和专业素养。

其次,在计算机领域的科研和开发工作中,使用这些缩写可以简化表达,降低沟通成本,提高工作效率。

最后,在计算机产品的市场推广和
普及过程中,了解这些缩写有助于提高用户的专业认知和产品接受度。

总之,计算机一级英文缩写在计算机科学和信息技术领域具有重要意义。

学习和掌握这些缩写,不仅可以提高计算机领域的学术和专业素养,还可以为科研、开发和市场推广等工作提供便利。

计算机网络英文题库(附答案)chapter5

计算机网络英文题库(附答案)chapter5

Chapter 5 The Link Layer and Local Area Network1.A ( ) protocol is used to move a datagram over an individual link.A application-layerB transport-layerC network-layerD link-layer2.The units of data exchanged by a link-layer protocol are called ( ).A datagramsB framesC segmentsD messages3.Which of the following protocols is not a link-layer protocol? ( )A EthernetB PPPC HDLCD IP4.In the following four descriptions, which one is not correct? ( )A link-layer protocol has the node-to-node job of moving network-layer datagrams over a single link in the path.B The services provided by the link-layer protocols may be different.C A datagram must be handled by the same link-layer protocols on the different links in the path.D The actions taken by a link-layer protocol when sending and receiving frames include error detection, flow control and random access.5.Which of the following services can not offered by a link-layer protocol? ( )A congestion controlB Link AccessC Error controlD Framing6.( ) protocol serves to coordinate the frame transmissions of the many nodes when multiple nodes share a single broadcast link.A ARPB MACC ICMPD DNS7.In the following four descriptions about the adapter, which one is not correct? ( )A The adapter is also called as NIC.B The adapter is a semi-autonomous unit.C The main components of an adapter are bus interface and the link interface.D The adapter can provide all the link-layer services.8.Consider CRC error checking approach, the four bit generator G is 1011, and suppose that the data D is 10101010, then the value of R is( ).A 010B 100C 011D 1109.In the following four descriptions about random access protocol, which one is not correct? ( )A I n slotted ALOHA, nodes can transmit at random time.B I n pure ALOHA, if a frame experiences a collision, the node will immediately retransmit it with probability p.C T he maximum efficiency of a slotted ALOHA is higher than a pure ALOHA.D I n CSMA/CD, one node listens to the channel before transmitting.10.In the following descriptions about MAC address, which one is not correct? ( )A T he MAC address is the address of one node’s adapter.B N o two adapters have the same MAC address.C T he MAC address doesn’t change no matter where the adapter goes.D M AC address has a hierarchical structure.11.The ARP protocol can translate ( ) into ( ). ( )A h ost name, IP addressB h ost name, MAC addressC I P address, MAC addressD broadcast address, IP address12.The value of Preamble field in Ethernet frame structure is ( )A 10101010 10101010……10101010 11111111B 10101011 10101011……10101011 10101011C 10101010 10101010……10101010 10101011D 10101010 10101010……10101010 1010101013.There are four steps in DHCP, the DHCP server can complete ( ).A DHCP server discoveryB DHCP server offersC DHCP requestD DHCP response14.In CSMA/CD, the adapter waits some time and then returns to sensing the channel. In the following four times, which one is impossible? ( )A 0 bit timesB 512 bit timesC 1024 bit timesD 1028 bit times15.The most common Ethernet technologies are 10BaseT and 100BaseT. “10” and “100” indicate( ).A the maximum length between two adaptersB the minimum length between two adaptersC the transmission rate of the channelD the transmission rate of the node16.The principal components of PPP include but not( ).A framingB physical-control protocolC link-layer protocolD network-layer protocol17.In the following four options, which service can not be provided by switch? ( )A filteringB self-learningC forwardingD optimal routing18.In the following four services, which one was be required in PPP? ( )A packet framingB error detectionC error correctionD multiple types of link19.The ability to determine the interfaces to which a frame should be directed, and then directing the frame to those interfaces is( ).A filteringB forwardingC self-learningD optimal routing20.In ( ) transmission(s), the nodes at both ends of a link may transmit packets at the same time.A full-duplexB half-duplexC single-duplexD both full-duplex and half-duplex21.Consider the data D is 01110010001, if use even parity checking approach, the parity bit is( ① ), if use odd parity checking approach, the parity bit is( ② ). ( )A ①0 ②1B ①0 ②0C ①1 ②1D ①1 ②022.In the following four descriptions about parity checks, which one is correct? ( )A Single-bit parity can detect all errors.B Single-bit parity can correct one errors.C Two-dimensional parity not only can detect a single bit error, but also can correct that error.D Two-dimensional parity not only can detect any combination of two errors, but also can correct them.23.MAC address is ( ) bits long.A 32B 48C 128D 6424.Wireless LAN using protocol ( ).A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.1125.The following protocols are belonging to multiple access protocols except for ( ).A channel partitioning protocolsB routing protocolsC random access protocolsD taking-turns protocols26.Which of the following is not belonging to channel partitioning protocols? ( )A CSMAB FDMC CDMAD TDM27.In the following four descriptions about CSMA/CD, which one is not correct? ( )A A node listens to the channel before transmitting.B If someone else begins talking at the same time, stop talking.C A transmitting node listens to the channel while it is transmitting.D With CSMA/CD, the collisions can be avoided completely.28.( ) provides a mechanism for nodes to translate IP addresses to link-layer address.A IPB ARPC RARPD DNS29.A MAC address is a ( )address.A physical-layerB application-layerC link-layerD network-layer30.Which of the following is correct? ( )A No two adapters have the same MAC address.B MAC broadcast address is FF-FF-FF-FF-FF-FF.C A portable computer with an Ethernet card always has the same MAC address, no matter where the computer goes.D All of the above31.In the following four descriptions, which one is not correct? ( )A ARP resolves an IP address to a MAC address.B DNS resolves hostnames to IP addresses.C DNS resolves hostnames for hosts anywhere in the Internet.D ARP resolves IP addresses for nodes anywhere in the Internet.32.In the LAN, ( )protocol dynamically assign IP addresses to hosts.A DNSB ARPC DHCPD IP33.DHCP protocol is a four-step process: ①DHCP request. ②DHCP ACK. ③DHCP server discovery. ④DHCP server offer(s). The correct sequence is ( )A ①②③④B ③②①④C ③④①②D ①④③②34.In the Ethernet frame structure, the CRC field is ( )bytes.A 2B 4C 8D 3235.In the Ethernet frame structure, the Data field carries the ( ).A IP datagramB segmentC frameD message36.In the following four descriptions, which one is not correct? ( )A Ethernet uses baseband transmission.B All of the Ethernet technologies provide connection-oriented reliable service to the network layer.C The Ethernet 10Base2 technology uses a thin coaxial cable for the bus.D The Ethernet 10BaseT technology uses a star topology.37.Ethernet’s multiple access protocol is ( ).A CDMAB CSMA/CDC slotted ALOHAD token-passing protocol38.In the following four descriptions about CSMA/CD, which one is not correct? ( )A An adapter may begin to transmit at any time.B An adapter never transmits a frame when it senses that some other adapter is transmitting.C A transmitting adapter aborts its transmission as soon as it detects that another adapter is also transmitting.D An adapter retransmits when it detects a collision.39.Which of the following descriptions about CSMA/CD is correct? ( )A No slots are used.B It uses carrier sensing.C It uses collision detection.D All of the above.40.The Ethernet 10BaseT technology uses( )as its physical media.A fiber opticsB twisted-pair copper wireC coaxial cableD satellite radio channel41.For 10BaseT, the maximum length of the connection between an adapter and the hub is ( )meters.A 100B 200C 500D 1042.A ( )is a physical-layer device that acts on individual bits rather than on frames.A switchB hubC routerD gateway43.A hub is a ( )device that acts on individual bits rather than on frames.A physical-layerB link-layerC network-layerD ransport-layer44.A switch is a( )device that acts on frame.A physical-layerB link-layerC network-layerD transport-layer45.In the following four descriptions, which one is not correct? ( )A Switches can interconnect different LAN technologies.B Hubs can interconnect different LAN technologies.C There is no limit to how large a LAN can be when switches are used to interconnect LAN segments.D There is restriction on the maximum allowable number of nodes in a collision domain when hubs are used to interconnect LAN segments.46.The ability to determine whether a frame should be forwarded to some interface or should just be dropped is ( ).A f ilteringB f orwardingC s elf-learningD o ptimal routing47.Which of the following devices is not a plug and play device? ( )A hubB routerC switchD repeater48.Which of the following devices is not cut-through device? ( )A hubB routerC switchD repeater49.In the following four descriptions, which one is not correct? ( )A Switches do not offer any protection against broadcast storms.B Routers provide firewall protection against layer-2 broadcast storms.C Both switches and routers are plug and play devices.D A router is a layer-3 packet switch, a switch is a layer-2 packet switch. 50.Which device has the same collision domain? ( )A HubB SwitchC RouterD Bridge51.IEEE802.2 protocol belong to ( )layerA networkB MACC LLCD physical52.IEEE802.11 protocol defines ( )rules.A Ethernet BusB wireless WANC wireless LAND Token Bus53.In data link-layer, which protocol is used to share bandwidth? ( )A SMTPB ICMPC ARPD CSMA/CD54.When two or more nodes on the LAN segments transmit at the same time, there will be a collision and all of the transmitting nodes well enter exponential back-off, that is all of the LAN segments belong to the same( ).A collision domainB switchC bridgeD hub55.( )allows different nodes to transmit simultaneously and yet have their respective receivers correctly receive a sender’s encoded data bits.A CDMAB CSMAC CSMA/CDD CSMA/CA56.Because there are both network-layer addresses (for example, Internet IPaddresses) and link-layer addresses (that is, LAN addresses), there is a need to translate between them. For the Internet, this is the job of ( ).A RIPB OSPFC ARPD IP57.PPP defines a special control escape byte, ( ). If the flag sequence, 01111110 appears anywhere in the frame, except in the flag field, PPP precedes that instance of the flag pattern with the control escape byte.A 01111110B 01111101C 10011001D 1011111058.The device ( ) can isolate collision domains for each of the LAN segment.A modemB switchC hubD NIC59.In the following four descriptions about PPP, which one is not correct? ( )A PPP is required to detect and correct errors.B PPP is not required to deliver frames to the link receiver in the same order in which they were sent by the link sender.C PPP need only operate over links that have a single sender and a single receiver.D PPP is not required to provide flow control.60.In the PPP data frame, the( ) field tells the PPP receivers the upper-layer protocol to which the received encapsulated data belongs.A flagB controlC protocolD checksum61.PPP’s link-control protocols (LCP) accomplish ( ).A initializing the PPP linkB maintaining the PPP linkC taking down the PPP linkD all of the above62.The PPP link always begins in the ( ) state and ends in the ( ) state. ( )A open, terminatingB open, deadC dead, deadD dead, terminating63.For( ) links that have a single sender at one end of the link and a single receiver at the other end of the link.A point-to-pointB broadcastC multicastD all of the above64.With ( )transmission, the nodes at both ends of a link may transmit packets at the same time.A half-duplexB full-duplexC simplex(单工)D synchronous65.With ( ) transmission, a node can not both transmit and receive at the same time.A half-duplexB full-duplexC simplex(单工)D synchronous66.Which of the following functions can’t be implemented in the NIC? ( )A encapsulation and decapsulationB error detectionC multiple access protocolD routing67.Which of the following four descriptions is wrong? ( )A The bus interface of an adapter is responsible for communication with the adapter’s parent node.B The link interface of an adapter is responsible for implementing the link-layer protocol.C The bus interface may provide error detection, random access functions.D The main components of an adapter are the bus interface and the link interface. 68.For odd parity schemes, which of the following is correct? ( )A 011010001B 111000110C 110101110D 00011011069.( )divides time into time frames and further divides each time frame into N time slots.A FDMB TMDC CDMAD CSMA70.With CDMA, each node is assigned a different ( )A codeB time slotC frequencyD link71.Which of the following four descriptions about random access protocol is not correct? ( )A A transmission node transmits at the full rate of the channelB When a collision happens, each node involved in the collision retransmits at once.C Both slotted ALOHA and CSMA/CD are random access protocols.D With random access protocol, there may be empty slots.72.PPP defines a special control escape byte 01111101. If the data is b1b201111110b3b4b5, the value is( )after byte stuffing.A b1b20111110101111110b3b4b5B b1b20111111001111101b3b4b5C b5b4b30111111001111101b2b1D b5b4b30111110101111110b2b173.MAC address is in ( ) of the computer.A RAMB NICC hard diskD cache74.Which of the following is wrong? ( )A ARP table is configured by a system administratorB ARP table is built automaticallyC ARP table is dynamicD ARP table maps IP addresses to MAC addresses75.NIC works in ( )layer.A physicalB linkC networkD transport76.In LAN, if UTP is used, the common connector is( ).A AUIB BNCC RJ-45D NNI77.The modem’s function(s) is(are) ( ).A translates digital signal into analog signalB translates analog signal into digital signalC both translates analog signal into digital signal and translates digital signal into analog signalD translates one kind of digital signal into another digital signal78.( )defines Token-Ring protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.279.( )defines Token-Bus protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.280.( ) defines CSMA/CD protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.281.The computer network that concentrated in a geographical area, such as in a building or on a university campus, is ( )A a LANB a MANC a WAND the Internet82.The MAC address is ( ) bits long.A 32B 48C 128D 25683.Which of the following four descriptions about MAC addresses is wrong? ( )A a MAC address is burned into the adapter’s ROMB No two adapters have the same addressC An adapter’s MAC address is dynamicD A MAC address is a link-layer address84.Which of the following four descriptions about DHCP is correct? ( )A DHCP is C/S architectureB DHCP uses TCP as its underlying transport protocolC The IP address offered by a DHCP server is valid foreverD The DHCP server will offer the same IP address to a host when the host requests an IP address85.The ( )field permits Ethernet to multiplex network-layer protocols.A preambleB typeC CRCD destination MAC address86.For 10BaseT, the maximum length of the connection between an adapter and the hub is ( ) meters.A 50B 100C 200D 50087.An entry in the switch table contains the following information excepts for ( )A the MAC address of a nodeB the switch interface that leads towards the nodeC the time at which the entry for the node was placed in the tableD the IP address of a node88.Consider the 4-bit generator , G is 1001, and suppose that D has the value 101110000. What is the value of R?89.Consider the following graph of the network. Suppose Host A will send a datagram to Host B, Host A run OICQ on port 4000, Host B run OICQ on port 8000. All of ARP tables are up to date. Enumerate all the steps when message “Hello” is sent from host A to host B.。

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Services Provided to the Transport Layer – Connectionless
• • • The network is inherently unreliable The hosts should do error control and flow control themselves Provide SEND PACKET and RECEIVE PACKET primitives Each packet carries full destination address The receiving hosts do packet ordering
• • • • • • • • • • The Optimality Principle Shortest Path Routing Flooding Distance Vector Routing Link State Routing Hierarchical Routing Broadcast Routing Multicast Routing Routing for Mobile Hosts Routing in Ad Hoc Networks
Implementation of Connectionless Service
• The transport-layer code adds a transport header to a message from a process on a host and hands it to the network layer It may be broken into packets – also called datagrams, each individually routed without advance set-up Every router has an internal table telling it where to send packets for each possible destination The routing table may change to take into account the prevailing network conditions – equipment failure, network congestion, etc.

Shortest Path Routing
The first 5 steps used in computing the shortest path from A to D. The arrows indicate the working node.
Dijkstra's algorithm
5-8 top
Flooding
• • An incoming packet is sent out on all other outgoing lines A packet may carry a hop counter, initialized to a reasonably large value On decrementing the hop count to zero, a packet is simply discarded Alternatively, a packet may be given a sequence number by the source router; duplicates are ignored Selective flooding – forward a packet on those lines going approximately in the right direction

• •
Flooding
• • • Robust Concurrent updates to distributed databases Wireless networks where a station can receive all messages within radio range Find the shortest path for benchmarking



The Optimality Principle
(a) A subnet. (b) A sink tree for router B.
Shortest Path Routing
• • Routers and communication lines are represented as nodes and arcs of a graph The length of an arc could be computed as a function of the distance, bandwidth, average traffic, communication cost, mean queue length, measured delay and other factors For computing the shortest path between two nodes, Dijkstra’s algorithm may be used

• பைடு நூலகம் •
Routing Algorithms
Conflict between fairness and optimality.
The Optimality Principle
• If router J is on the optimal path from router I to router K, then the optimal path from J to K also falls along the same route Otherwise, a better path from J to K can be concatenated to the optimal path from I to J, yielding a better path from I to K A sink tree is the set of optimal routes from all sources to a given destination forming a tree rooted at the destination All routing algorithms try to discover and use the sink trees for all routers
Chapter 5
The Network Layer
Network Layer Design Issues
• • • • • Store-and-Forward Packet Switching Services Provided to the Transport Layer Implementation of Connectionless Service Implementation of Connection-Oriented Service Comparison of Virtual-Circuit and Datagram Subnets
Routing Algorithms
• • • Correctness Simplicity Robustness – able to cope with changes in the topology and traffic Stability – a stable algorithm reaches equilibrium and stays there Fairness and optimality Non-adaptive algorithms – static routing Adaptive algorithms – routing decisions adapt to changes in the topology and traffic
• •
Services Provided to the Transport Layer – Connection-oriented
• • Quality of service is important It is difficult to achieve quality of service without connections in the network



Implementation of Connectionless Service
Routing within a diagram subnet.
Implementation of Connection-oriented Service
• At connection set-up, a route from source machine to destination machine is chosen and stored in tables inside the routers Each packet carries an identifier for the virtual circuit it belongs to

Implementation of Connection-Oriented Service
Routing within a virtual-circuit subnet.
Comparison of Virtual-Circuit and Datagram Subnets
5-4
Routing Algorithms
• • •
Store-and-Forward Packet Switching
fig 5-1
The environment of the network layer protocols.
Services Provided to the Transport Layer
• • • The services should be independent of router technology The transport layer should be shielded from the number, type and topology of the routers present The network addresses made available to the transport layer should use a uniform numbering plan Should the layer provide connection-oriented service or connectionless service?
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