网络原理期末复习整理.doc

第一章： UDP：The second protocol in this layer, UDP(User Datagram Protocol),is an unreliable, connectionless protocol for applications that do not want TCPs sequencing or flow control and wish to provide their own. UDP: 第二个协议是UDP(User Datagram Protocol，用户数据报协议)，它是一个不可靠的、无连接的协议，主要用于那些“不想要TCP的序列化或者流控制功能，而希望自己提供这些功能”的应用程序。 OSI TCP/IP7.Application应用层Application应用层6.Presentation表示层Not present在模型In the model中不存在5.Session会话层4.Transport传输层Transport3.Network网络层Internet互连网层2.Data link数据链路层Host-to-network主机至网络1.Physical物理层 The Physical Layer:物理层 The physical layer is concerned with transmitting raw bits over a communication channel. 物理层涉及到在通信信道上传输的原始数据位。The Dada Link Layer数据链路层 The main task of the data link layer is to transform a raw transmission facility into a line that appears free of undetected transmission errors to the network layer. 数据链路层的主要任务是将一个原始的传输设施转变成一条逻辑的传输线路，在这条传输线路上，所有未检测出来的传输错误也会反映到网络层上。The Network Layer网络层 The network layer controls the operation of the sunbet.A key design issue is determining how packets are routed from source to destination. 网络层控制子网的运行过程。一个关键的设计问题是确定如何将分组从源端路由到目标端。Network Hardware网络硬件Broadcast networks:广播网络 Broadcast networks have a single communication channel that is shared by all the machines on the network. Short messages ,called packets in certain contexts, sent by any machine are received by all the others. 广播网络只有一个通信信道，网络上所有的机器都共享该信道。在机器之间传递的是短消息（在有些上下文环境中称为分组或包，packet），任何一台机器发送的短消息都可以被其他所有的机器接收到。Point-to-point links点到点连接 In contrast, point-to-point networks consist of many connections between individual pairs of machines. To go from the source to the destination, a packet on this type of network may have to first visit one or more intermediate machines. 与此相反，点到点网络则是由许多连接构成的，每一对连接对应着一对机器。在这种网络中，为了将一个分组从源端传送到目的地，该组可能首先要经过一台或者多台中间机器。习题： 1-18Which of the OSI layers handles each of the following: (a)dividing the transmitted bit stream into frames. (b)determining which route through the subnet to use. OSI模型中的哪一层处理以下问题： (a)把传输的位流分成帧 (b)在通过子网的时候决定使用哪条路由路径。 Answer: (a):The data link layer数据链路层 (b):The network layer网络层 1-22.What is the main difference between TCP and UDP? TCP与UDP之间的主要区别是什么？Answer: TCP is a reliable connection-oriented protocol TCP是一个可靠的面向连接的协议。 UDP is an unreliable, connectionless protocol. UDP是一个不可靠的无连接协议。第二章最大数据传输率maximum data rate=2Hlog2Vbits/sec(H-带宽，V-离散级数)最大数据传输率maximum number of bits/sec=Hlog2(1+S/N)(包含有信噪比的)(S/N为信噪比，用公式10log10S/N=？dB计算出)习题： 2-3.Television channels are 6 MHz wide. How many bits/sec can be sent if four-level digital signals are used? Assume a noiseless channel. 电视频道的带宽是6MHz。如果使用4级数字信号，则每秒钟可以发送多少位？假设电视频道为无噪声信道。Answer: Maximum data rate = 2 H log2V =2*6 *log 2 4 =24Mbits/s 2-4.If a binary signal is sent over a 3-KHz channel whose signal-to-noise ratio is 20 dB, what is the maximum achievable data rate? 如果在一条3KHz的信道上发送一个二进制信号，该信道的信噪比为20dB，则最大可达到的数据传输率为多少？Answer: Maximum data rate = 2 H log2V= 2*3 log22= 6Kbps signal-to noise ratio is 20 Db . S/N=100 Maximum number of bits/sec= H log 2(1+S/N) =3* log 2(1+100)=3*6.658=19.98Kbps 2-31.Compare the maximum data rate of a noiseless 4-KHz channel using (a)Analog encoding (e.g., QPSK) with 2 bits per sample (b)The T1 PCM system 比较使用以下方案的4KHz无噪声信道的最大数据传输率：(a) 每次采样2位的模拟编码（比如QSPK）(b) T1 PCM系统Answer: QPSK:V=4 QAM-16:V=16 QAM-64:V=64 T1 PCM:V=4 (a)Maximum data rate=2Hlog2V=2*4*log227=56bits/s (noiseless: S/N=0) (b)Maximum data rate=2Hlog2V=2*4*log24=16bits/s第三章Error Control错误控制 The usual way to ensure reliable delivery is to provide the sender with some feedback about what is happening at the other end of the line. 确保可靠递交的常用方法是向发送方提供一些有关线路另一端状况的反馈信息。 This possibility is dealt with by introducing timers into the data link layer. 这种可能出现的问题可以通过在数据链路层中引入定时器来解决。 To prevent this from happening, it is generally necessary to assign sequence numbers to outgoing frames, so that the receiver can distinguish retransmissions from originals. 为了避免发生这样的情形，一般有必要给送出的帧分配序列号，这样接收方能够区别原始帧和重传帧。习题：3-2.The following character encoding is used in a data link protocol: A:01000111;B:11100011;FLAG:01111110;ESC:11100000 Show the bit sequence transmitted (in binary) for the four-character frame: A B ESC FLAG when each of the following framing methods are used:(a) Character count.(b) Flag bytes with byte stuffing.(c) Starting and ending flag bytes, with bit stuffing. 数据链路协议中使用了下面的字符编码： A:01000111;B:11100011;FLAG:01111110;ESC:11100000 为了传输一个包含4个字符的帧：A B ESC FLAG，请给出当使用下面的成帧方法时所对应的位序号（用二进制表示）(a) 字符计数(b) 包含字节填充的标志字节(c) 包含位填充的起始和结束标志Answer: A: 01000111 B: 11100011 FLAG: 01111110 ESC :11100000A B ESC FLAG(a) 4 A B ESC FLAG 00000100 01000111 11100011 11100000 01111110(b) FLAG A B ESC ESC ESC FLAG FLAG 01111110 01000111 11100011 11100000 1110000011100000 01111110 01111110 (C) FLAG A B ESC FLAG FLAG01111110 01000111110100011111000000011111010 011111103-9.Sixteen-bit messages are transmitted using a Hamming code. How many check bits are needed to ensure that the receiver can detect and correct single bit errors? Show the bit pattern transmitted for the message 1101001100110101. Assume that even parity is used in the Hamming code. 假设使用海明码来传输16位的报文。请问，需要多少个检查位才能确保接收方可以检测并纠正单个位错误？对于报文1101001100110101，请给出所传输的位模式。假设在海明码中使用了偶数位。Answer: m:信息位 r:检验位 m+r+1=2r16+r+1r的最小值为53-15.A bit stream 10011101is transmitted using the standard CRC method described in the text. The generator polynomial is x3+1. Show the actual bit string transmitted. Suppose the third bit from the left is inverted during transmission. Show that this error is detected at the receivers end. 利用本章中介绍的标准CRC方法来传输位流10011101。生成器多项式为x3+1。请给出实际被传输的位串。假设在传输过程中左边第三位变反了。请证明，这个错误可以在接收端被检测出来。Answer: x3+1=1001 算法见P198 10011101后面所加0的个数与X的最高次幂相同，此题加3个0，即1001/10011101000,(相同取0，相异取1)，得出余数remainder=100. 10011101100所得出的余数不是0，因此可以被检测。This error is detected.第四章Hidden station problem隐藏站问题 The problem of a station not being able to detect a potential competitor for the medium because the competitor is too far away is called the hidden station problem. 由于竞争者离得太远而导致一个站无法检测到潜在的介质竞争对手，这个问题称为隐藏站问题。Exposed station problem暴露站问题 If C senses the medium, it will hear an ongoing transmission and falsely conclude that it may not send to D, when in face such a transmission would cause bad reception only in the zone between B and C, where neither of the intended receivers is located. This is called the exposed station problem. 如果C正在检测介质，则它将会听到有一个传输正在进行，从而也会错误地得出结论：它不能向D发送数据，而实际上，它所听到的传输过程只会影响到B和C之间的区域中的接收过程，但是，不会影响到目标接收方（D）所在的区域。这个问题称为暴露站问题。 In the literature, broadcast channels are sometimes referred to as multi-access channels or random access channels. 在有些文献中，广播信道有时候也称为多路访问信道或者随机访问信道。习题： 4-17.Sketch the Manchester encoding for the bit stream: 0001110101. 画出位流0001110101的曼彻斯特编码。Answer: Bit stream: 0:低电平 1:高电平 Binary encoding: 0:先低后高 1:先高后低 4-18. Sketch the differential Manchester encoding for the bit stream of the previous problem. Assume the line is initially in the low state. 画出上一个问题中的位流的差分曼彻斯特编码。假设线路的初始电压为低电压。Answer: 0:与前面相同 1:与前面相反 4-21.Consider building a CSMA/CD network running at 1Gbps over a 1-km cable with no repeaters. The signal speed in the cable is 200,000 km/sec. What is the minimum frame size? 考虑在一条1km长的电缆（无中继器）上建立一个1Gbps速率的CSMA/CD网络。信号在电缆中的速度为200,000km/s。请问最小的帧长度为多少？Answer: t=1/200000=5*10-6s 2t=10*10-6s 10*10-6s*1*109bits/s=10000bits The minimum frame size is 10000bits. 4-23.Ethernet frames must be at least 64 bytes long to ensure that the transmitter is still going in the event of a collision at the far end of the cable. Fast Ethernet has the same 64-byte minimum frame size but can get the bits out ten times faster. How is it possible to maintain the same minimum frame size? 以太网帧必须至少64字节长，这样做的理由是，当电缆的另一端发生冲突的时候，传送方仍然还在发送过程中。快速以太网也有同样的64字节最小帧长度限制，但是，它可以以快10倍的速度发送数据。请问它如何有可能维持同样的最小帧长度限制？Answer: The maximum wire length in fast Ethernet is 1/10 as long as in Ethernet.第五章The Network Layer网络层 The network layer is concerned with getting packets from the source all the way to the destination. 网络层关注的是如何将分组从源端沿着网络路径送达目标端。Link State Routing链路状态路由 5个要点：1. Discover its neighbors and learn their network addresses.发现它的邻居节点，并知道其网络开销。 2. Measure the delay or cost to each of its neighbors. 测量各邻居节点的延迟和开销。3. Construct a packet telling all it has just learned. 构造分组，包含刚知道的消息。4. Send this packet to all other routers. 分组发给其他路由器。5. Compute the shortest path to every other router. 计算出到每一个其他路由器的最短路径。习题： 5-39. A network on the Internet has a subnet mask of . What is the maximum number of hosts it can handle? Internet上的一个网络的子网掩码为。请问它最多能够处理多少台主机？Answer: Subnet mask: 两个255代表network网络号;240代表subnet子网;0代表主机 Max hosts=(255-0+1)*(255-240+1)=4096 But all 0, and all 1 is special So max hosts=4096-2=4094. 5-40. A large number of consecutive IP address are available starting at . Suppose that four organizations, A, B, C, and D, request 4000, 2000, 4000, and 8000 addresses, respectively, and in that order. For each of these, give the first IP address assigned, the last IP address assigned, and the mask in the w. x. y. z/s notation. 假定从开始有大量连续的IP地址可以使用。现在4个组织A、B、C和D按照顺序依次申请4000, 2000, 4000, 和 8000个地址，对于每一个申请，请利用w. x. y. z/s的形式写出所分配的第一个IP地址、最后一个IP地址，以及掩码。Answer: IP: starting at (order:顺序) A:4000 B:2000 C:4000 D:8000 A:55 198.16一般不动，16后面的那个0按照2r的规律，最后一个0的最大值为255。 比4000大的2r=4096=256*16，然后把子网0加上16减去1，即 0+16-1=15，这个15作为last IP的子网，最后主机取最大255。 最后那个s则将first与last化成二进制相互比较，将相同位的个数作为s的值。So, mask=/20第六章The Transport Layer传输层 Its task is to provide reliable, cost-effective data transport from the source machine to the destination machine, independently of the physical network or networks currently in use. 它的任务是在源机器和目标机器之间提供可靠的、性价比合理的数据传输功能，并且与当前所使用的物理网络完全独立。The TCP Service Model TCP服务模型 TCP service is obtained by both the sender and receiver creating end points called sockets, as discussed in Sec.6.1.3. Each socket has a socket number (address) consisting of the IP address of the host and a 16-bit number local to that host, called a port. 正如我们在6.1.3节中所讨论的，要想获得TCP服务，发送方和接收方必须创建一种被称为套接字的端点。每个套接字有一个套接字号（地址），它是由主机的IP地址以及本地主机局部的一个16位数值组成的，此16位数值被称为端口(port)。第七章DNS域名系统 The essence of DNS is the invention of a hierarchical, domain-based naming scheme and a distributed database system for implementing this naming scheme. It is primarily used for mapping host names and e-mail destinations to IP address but can also be used for other purposes. DNS is defined in RFCs 1034 and 1035. DNS的本质是，它发明了一种分层次的、基于域的命名方案，并且用一个分布式数据库系统来实现此命名方案。DNS的主要用途是，将主机名和电子邮件目标地址映射成IP地址，但它还有其他的一些用途。RFC1034和1035定义了DNS。 Very briefly, the way DNS is used is as follows. To map a name onto an IP address, an application problem calls a library procedure called the resolver, passing it the name as a parameter. 简要地来说，DNS的使用方法如下所述。为了将一个名字映射成IP地址，应用程序调用一个名为解析器（resolve）的库过程，并将该名字作为参数传递给此过程。Architecture and Services结构与服务 Composition refers to the process of creating messages and answers. 撰写是指创建消息和回信的过程。 Transfer refers to moving messages from the originator to the recipient. 传输指的是把消息从发信人处传递到收信人处。 Reporti