计算机网络复习重点重点终结版.docx

1、计算机网络复习重点重点终结版物理拓扑结构（physical topology）网状拓扑结构（mesh topology）各台设备之间都有一条专用的点到点的链路，每台设备必须有n-1个I/O端口星型拓扑结构（star topology）每台设备拥有一条仅与中央控制器连接的点到点专用链路总线拓扑结构（bus topology）环状拓扑结构（ring topology）协议和标准: 协议是用来管理数据通信的一组规则。协议规定了通信的内容、通信的方式和通信的时间。 标准是由标准化组织协商达成一致的规则。因特网模型（Internet model）物理层 ; 数据链路层 ; 网络层 ; 传输层 ; 应用层

2、An exchange using the Internet modelPhysical layerNote: The physical layer is responsible for transmitting individual bits from one node to the next.Data link layer Note: The data link layer is responsible for transmitting frames from one node to the next.Network layerNote: The network layer is resp

3、onsible for the delivery of packets from the original source to the final destination.Transport layerNote:The transport layer is responsible for delivery of a message from one process to another.Application layerNote: The application layer is responsible for providing services to the user.Summary of

4、 duties=Position of the physical layerNote: Frequency and period are inverses of each other.Note: A digital signal is a composite signal with an infinite bandwidth. The bit rate and the bandwidth are proportional to each other. The analog bandwidth of a medium is expressed in hertz; the digital band

5、width, in bits per second. Digital transmission needs a low-pass channel. Analog transmission can use a band-pass channel.noiseless channelnoisy channelAttenuationDistortionNoise=Unipolar:Unipolar encoding uses only one voltage level.Polar:Polar encoding uses two voltage levels (positive and negativ

6、e).NRZ-L: In NRZ-L the level of the signal is dependent upon the state of the bit.NRZ-I: In NRZ-I the signal is inverted if a 1 is encountered.RZ encodingA good encoded digital signal must contain a provision for synchronization.Manchester encodingIn Manchester encoding, the transition at the middle

7、 of the bit is used for both synchronization and bit representation.Differential Manchester encodingIn differential Manchester encoding, the transition at the middle of the bit is used only for synchronization. The bit representation is defined by the inversion or noninversion at the beginning of th

8、e bit.BipolarIn bipolar encoding, we use three levels: positive, zero, and negative.Bipolar AMI encodingFrom analog signal to PCM digital codeAccording to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequencyHow many bits should be sent for each sample?1 bit for the s

9、ign and X bits for the value.=Bit rate is the number of bits per second. Baud rate is the number of signal units per second. Baud rate is less than or equal to the bit rate.ASKRelationship between baud rate and bandwidth in ASKIn ASK the baud rate and bit rate are the same.An ASK signal requires a m

10、inimum bandwidth equal to its baud rateFSKIn FSK the baud rate and bit rate are the same.BW=fc1-fc0+NbaudPSKThe 4-PSK methodThe 8-PSK characteristicsRelationship between baud rate and bandwidth in PSKA telephone line has a bandwidth of almost 2400 Hz for data transmission.The total bandwidth require

11、d for AM can be determined from the bandwidth of the audio signal: BWt = 2 x BWm.The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BWt = 10 x BWm.The bandwidth of a stereo audio signal is usually 15 KHz. Therefore, an FM station needs at least a bandwidth

12、of 150 KHz. The FCC requires the minimum bandwidth to be at least 200 KHz (0.2 MHz).=multiplexingFDM is an analog multiplexing technique that combines signals.FDM process&FDM demultiplexing exampleWDMWDM is an analog multiplexing technique to combine optical signals (光信号) .TDMTDM is a digital multip

13、lexing technique to combine data.In a TDM, the data rate of the link is n times faster, and the unit duration is n times shorter. Radio waves are used for multicast communications, such as radio and television, and paging systems.Microwaves are used for unicast communication such as cellular telepho

14、nes, satellite networks, and wireless LANs.Infrared signals can be used for short-range communication in a closed area using line-of-sight propagation.=ADSL is an asymmetric communication technology designed for residential users; it is not suitable for businesses.The existing local loops can handle

15、 bandwidths up to 1.1 MHz.语音通信带宽为4kData Link Layer奇偶性校验 CRC校验(循环冗余码校验) 校验和校验Simple parity check can detect all single-bit errors. It can detect burst errors only if the total number of errors in each data unit is odd.Two-dimensional parityBinary division in a CRC generator ChecksumPositions of redun

16、dancy bits in Hamming codeStop-and-Wait ARQ Piggybacking Go-Back-N ARQGo-Back-N ARQ: sender window sizeIn Go-Back-N ARQ, the size of the sender window must be less than 2m; the size of the receiver window is always 1.Selective Repeat ARQIn Selective Repeat ARQ, the size of the sender and receiver wi

17、ndow must be at most one-half of 2m. HDLC高级数据链路控制HDLC frameBit stuffing(比特填充)is the process of adding one extra 0 whenever there are five consecutive 1s in the data so that the receiver does not mistake the data for a flag.Point-to-Point12.1 PPP frameTransition states Multiple-access protocolsFigure

18、 13.5 CSMA persistent strateg13.3 ChannelizationFDMA: In FDMA, the bandwidth is divided into channels. TDMA: In TDMA, the bandwidth is just one channel that is timeshared.CDMA: In CDMA, one channel carries all transmissions simultaneously.Figure 13.16 CDMA multiplexer=*=A repeater connects segments

19、of a LAN.A repeater forwards every frame; it has no filtering capability. A repeater is a regenerator, not an amplifier.HubsA bridge has a table used in filtering decisions.A bridge does not change the physical (MAC) addresses in a frame.AMPS (高级移动电话系统, 北美) is an analog cellular phone system using F

20、DMA.D-AMPS, or IS-136, is a digital cellular phone system using TDMA and FDMA.GSM is a digital cellular phone system using TDMA and FDMA.IS-95 is a digital cellular phone system using CDMA/DSSS and FDMA.GSMThe Iridium system has 66 satellites in six LEO orbits, each at an altitude of 750 km.Iridium

21、is designed to provide direct worldwide voice and data communication using handheld terminals, a service similar to cellular telephony but on a global scale.Teledesic has 288 satellites in 12 LEO orbits, each at an altitude of 1350 km.=Frame Relay operates only at the physical and data link layers.F

22、rame Relay does not provide flow or error control; they must be provided by the upper-layer protocols.Virtual connection identifiers in UNIs and NNIs18.21 An ATM cellNetwork LayerSwitching at the network layer in the Internet is done using the datagram approach to packet switching.Communication at t

23、he network layer in the Internet is connectionless. 分类Figure 19.17 Network addressIn classful addressing, the network address is the one that is assigned to the organization. A network address is different from a netid. A network address has both netid and hostid, with 0s for the hostid.Network Laye

24、r Protocols: ARP, IPv4, ICMPv4, IPv6, and ICMPv6An ARP request is broadcast; an ARP reply is unicast.Figure 20.7 IP datagramThere is no flow control or congestion control mechanism in IP.IPv6 addressIPv4-IPv6 Three transition strategies双协议栈 隧道技术 数据包头转换 RIP Updating AlgorithmReceive: a response RIP m

25、essage1. Add one hop to the hop count for each advertised destination.2. Repeat the following steps for each advertised destination: 1. If (destination not in the routing table) 1. Add the advertised information to the table. 2. Else 1. If (next-hop field is the same) 1. Replace entry in the table w

26、ith the advertised one. 2. Else 1. If (advertised hop count smaller than one in the table) 1. Replace entry in the routing table.3. Return.Figure 21.20 Shortest-path calculationDijkstra Algorithm1. Start with the local node (router): the root of the tree. 2. Assign a cost of 0 to this node and make

27、it the first permanent node.3. Examine each neighbor node of the node that was the last permanent node. 4. Assign a cumulative cost to each node and make it tentative.5. Among the list of tentative nodes 1. Find the node with the smallest cumulative cost and make it permanent. 2. If a node can be re

28、ached from more than one direction 1. Select the direction with the shortest cumulative cost.6. Repeat steps 3 to 5 until every node becomes permanent.Transport LayerUDP is a connectionless, unreliable protocol that has no flow and error control. It uses port numbers to multiplex data from the appli

29、cation layer.Application LayerDNS can use the services of UDP or TCP, using the well-known port 53.FTP uses the services of TCP. It needs two TCP connections. The well-known port 21 is used for the control connection, and the well-known port 20 is used for the data connection.HTTP uses the services of TCP on well-known port 80.RTP uses a temporary even-numbered UDP port. RTCP（实时传输控制协议） message typesRTCP uses an odd-numbered UDP port number that follows the port number selected for