计算机网络复习.doc

以下为计算机网络的复习内容，非考题答案This revision paper is intended as a guide to the type and format of questions that will appear in the forthcoming examination. Q1.The OSI 7-layer model is widely used as a way of describing the processes involved in transmitting data over a network.a) Briefly describe the purpose of each of the 7 layers.(简短的描述OSI七层中每一层的作用)b) How does TCP/IP differ from the 7-layer model?c) Briefly describe an example protocol from each layer. .(简短的描述每一层的协议)OSI Modeldescribeexample protocolApplication Layer（应用层）文件传输，电子邮件，文件服务，虚拟终端Telnet、FTP、HTTP、SNMPPresentation Layer（表示层）数据格式化、代码转换、数据加密N/ASession Layer（会话层）解除或建立与别的接点的关系N/ATransport Layer（传输层）提供端对端的接口TCP、UDPNetwork Layer（网络层）为数据包进行路由IP、ICMP、ARP、RARP、IGMPData Link Layer（数据链路层）传输有地址的帧及错误检测功能SLIP，CSLIP，PPP，MTUPhysical Layer（物理层）以二进制数据形式在物理媒体上传输数据ISO2110，IEEE802.2OSI ModelTCP/IP ModelApplication LayerApplication LayerPresentation LayerSession LayerTransport LayerTransport LayerNetwork LayerInternet LayerData Link LayerNetwork Access LayerPhysical LayerQ2.Voice over IP (VoIP) allows voice traffic to be sent over a network and replicates a traditional telephony system. a) Briefly describe the components required to implement such a system.(简要描述VoIP的系统必要组件要求)Media Gateway 媒体网关器：将语音信号转换成IP数据包Media Gateway Controller 媒体网关控制器：负责管理信号传输与转换工作Voice Server 语音服务器：提供电话不通、占线或忙线的语音相应服务Signaling Gateway 信号网关器：负责在交换过程中进行相关控制，以决定通话建立与否，提及提供相关应用的增值服务b) To provide a reliable and usable system it is necessary to implement QoS. What are the key QoS requirements for VoIP? 可用性（reliable）、吞吐量（throughput）、时延（delay）、时延变化（delay jitter）Q3.Analogue voice data is to be packetized and transmitted over the Internet. a) Assuming that the voice data has a maximum frequency of 4KHz , what would be the minimum sampling rate to meet Nyquists sampling theorem?b) Assuming the data is sampled at the Nyquist frequency and with an 8-bit quantisation, what would be the data rate expressed in bits per second?介个自己做去吧 亲们这是DSPQ4.Wired and Wireless Ethernet use the Carrier Sense Medium Access control mechanism(载波监听介质访问控制机制CSAM/CA) to gain access to the underlying medium. However, they differ in the approach taken to deal with potential data collisions. Briefly explain the two mechanisms deployed and how they differ.CSMA/CA协议的工作流程分为两个分别是:1.送出数据前，监听媒体状态，等没有人使用媒体，维持一段时间后，再等待一段随机的时间后依然没有人使用，才送出数据。由於每个设备采用的随机时间不同，所以可以减少冲突的机会。2.送出数据前，先送一段小小的请求传送报文(RTS : Request to Send)给目标端，等待目标端回应 CTS: Clear to Send 报文后，才开始传送。 利用RTS-CTS握手(handshake)程序，确保接下来传送资料时，不会被碰撞。 同时由於RTS-CTS封包都很小，让传送的无效开销变小。CSMA/CA通过这两种方式来提供无线的共享访问，这种显式的ACK机制在处理无线问题时非常有效。然而不管是对于802.11还是802.3来说，这种方式都增加了额外的负担，所以802.11网络和类似的Ethernet网比较总是在性能上稍逊一筹。Q5.Data can be transmitted across the network using either a connection-oriented or connectionless approach. a) Briefly describe the two mechanisms and explain the reason why there is a need for both approaches.（解释都需要两种协议的原因）TCP是一种面向连接的、可靠的、基于字节流的运输层（Transport layer）通信协议，包头大，占用资源多UDP 是User Datagram Protocol的简称， 中文名是用户数据包协议，是 OSI 参考模型中一种无连接的传输层协议，提供面向事务的简单不可靠信息传送服务。b) TCP and UDP are common transport layer protocols that are respectively connection-oriented and connectionless. Give an example when it would be appropriate to use TCP and UDP.在互联网中，网络不可靠，需要使用TCP协议在局域网中，网络相对稳定、可靠，可以使用UDP连接Q6.Domain Naming Service (DNS) is a protocol that allows users to map easily remembered names to IP addresses. a) Briefly describe the processes involved in resolving the IP address for a given name where the DNS server resides on the LAN.解释DNS原理，假设欲解释的域名为，先查询.cn下有没有.com，没有报错，有查.com下的test，有返回IP地址，没有报错b) How does DNS resolve the situation when a mapping does not exist on the local DNS?Q7.A PC has the following IP credentials:IP Address: 2Subnet Mask: Default Gateway: The PC is connected via an Ethernet LAN to a router which has a default route via its serial interface to its ISP. /Describe the step by step process from the packet being encapsulated into an Ethernet frame until it reaches the ISP.（描述如何封装到ISP）Q8.A video picture is to be transmitted over a network. The video picture has the following attributes:（没关注过多媒体传输，不会）Resolution: 800 x 600 pixelsPixel depth: 24 bitsFrame Rate: 50Hz800*600*24= 11520000bita) What would be the resultant data rate expressed in bits per second.b) What level of compression would be required to transmit this data across a fast Ethernet connection?Q9.Many organisations are considering the use of the Internet as