1、Abbreviations:LANLocal Area Network 局域網(wǎng)MANMetropolitan Area Network 城域網(wǎng)PANpersonal area network個人局域網(wǎng)WANWide Area Network廣域網(wǎng)VLANVirtual Local Area Network虛擬局域網(wǎng)OSI/RM Open System Interconnection Reference Model 開放系統(tǒng)互連參考模型UTPUnshielded Twisted Pair非屏蔽雙絞線SNRSigal-to-Noise Ratio信噪比FDMFrequency

2、Division Multiplexing頻分多路復(fù)用TDMTime Division Multiplexing 時分多路復(fù)用WDM Wave length Division Multiplexing 波分復(fù)用CDMACode Division Multiple Access碼分多址CSMA/CDCarrier Sense Multiple Access with Collision Detection載波偵聽多路訪問/沖突檢測CSMA/CA Carrier Sense Multiple Access with Collision Avoidance 載波偵聽多路訪問/沖突避免RTSReque

3、st To Send請求發(fā)送CTSClear To Send 清除發(fā)送MACMedia Access Control介質(zhì)訪問控制PCMPulse Code Modulation 脈沖編碼調(diào)制QAM Quadrature Amplitude Modulation正交振幅調(diào)制QPSK Quadrature Phase Shift Keying正交相移鍵控ADSLAsymmetric Digital Subscriber Line非對稱數(shù)字用戶環(huán)路CRCCyclic Redundancy Check循環(huán)冗余校驗VCVirtual Circuit虛電路ATMAsynchronous Transfer

4、Mode異步傳輸模式PPPPoint to Point Protocol 點對點協(xié)議HLDCHigh-Level Data Link Control高級數(shù)據(jù)鏈路控制RIPRouting Information Protocol路由信息協(xié)議RTTRound-Trip Time往返時延CIDRClassless Inter-Domain Routing無類別域間路由IPInternet Protocol網(wǎng)絡(luò)之間互連的協(xié)議TTLTime To Live生存時間MTUMaximum Transmission Unit最大傳輸單元ICMPInternet Control Message Pro

5、tocol網(wǎng)絡(luò)控制報文協(xié)議UDPUser Datagram Protocol用戶數(shù)據(jù)報協(xié)議ARPAddress Resolution Protocol地址解析協(xié)議DHCPDynamic Host Configuration Protocol動態(tài)主機配置協(xié)議NATNetwork Address Translation網(wǎng)絡(luò)地址轉(zhuǎn)換RPCRemote Procedure Call遠程過程調(diào)用OSPFOpen Shortest Path First開放式最短路徑優(yōu)先BGP Border Gateway Protocol邊界網(wǎng)關(guān)協(xié)議TCPTransmission Control Protoco

6、l傳輸控制協(xié)議RTPReal-time Transport Protocol實時傳輸協(xié)議FTPFile Transfer Protocol文件傳輸協(xié)議SMTPSimple Mail Transfer Protocol簡單郵件傳輸協(xié)議POP3Post Office Protocol - Version 3郵局協(xié)議版本3IAMPinternet message access protocol因特網(wǎng)信息報文存取協(xié)議?不知道對不對IMAPInternet Mail Access ProtocolInternet郵件訪問協(xié)議DNSDomain Name System域名系統(tǒng)URLUniform Reso

7、ure Locator統(tǒng)一資源定位器HTTPHyper Text Transfer Protocol超文本傳輸協(xié)議WWW World Wide Web萬維網(wǎng)1、 Basic and important concepts or notations l network architecture, layers, protocols l OSI/RM, TCP/IP reference model, main tasks of data link/network/transport layers, protocols at each layer of TCP/IP modell PAN, LAN,

8、WAN, VLAN, WLANl bandwidth, link capacityl multiplexing, channel allocationl Frame, Ethernet frame format, MAC addressl framing method：character-count, byte-stuffing, bit-stuffing, flag bytel Connection/connectionless service, circuit switching/packet switching, l Error Control coding, Hamming dista

9、ncel Hamming code, even-parity /odd-parityl CRC, generator polynomial, remainder l stop-and-wait protocols, sliding window, go-back-n, selective-repeat, piggyback l CSMA/CDl hidden station problem, exposed station problem, CSMA/CAl hub/repeater /switch/bridge/router/gateway.l routing algorithm, Dist

10、ance vector routing, link state routingl IP address, classful and special addressing, CIDR, subnetting/aggregation, mask/prefixl routing table/forwarding, l packet, IPv4 headerl NAT, 3 reserved private IP address ranges l TCP, port/port number, TCP header, three-way handshakel congestion control, EC

11、N/RED, AIMD/slow start, congestion window, thresholdl DNS system, domain name resolution2、 Computations l Nyquist law, Shannon law, PCMl FDM, CDMAl bit-stuffing, byte-stuffing, character-count l Hamming code, CRCl CSMA/CD，minimal frame sizel Distance vector routing algorithml CIDR, aggregationl forw

12、arding in routers, longest matching第五版：P106 11,16,2011 What is the principal difference between connectionless communication and connection-oriented communication?Connection-oriented communication has three phases. In the establishment phase, a request is made to set up a connection. Only after this

13、 phase has been successfully completed can the data transfer phase be started and data transported. Then comes the release phase. Connectionless communication does not have these phases. It just sends the data.16 把比特流轉(zhuǎn)化為幀OSI 數(shù)據(jù)鏈路層 TCP/IP ：鏈路層決定哪條路徑通過子網(wǎng)OSI 網(wǎng)絡(luò)層 TCP/IP 互聯(lián)網(wǎng)層20 What is the main differenc

14、e between TCP and UDP.Answer：The first one, TCP (Transmission Control Protocol), is a reliable connection-oriented protocol that allows a byte stream originating on one machine to be delivered without error on any other machine in the Internet.The second protocol, UDP (User Datagram Protocol), is an

15、 unreliable, connectionless protocol for applications that do not want TCP's sequencing or flow control and wish to provide their own.TCP是transmission control protocol傳輸控制協(xié)議，UDP是user data gram protocol用戶數(shù)據(jù)電報協(xié)議。TCP是面向連接的協(xié)議，允許字節(jié)流無錯誤地從一臺機器上互聯(lián)到別的機器上。UDP是不可靠的，面向無連接的協(xié)議，適用于不需要TCP協(xié)議的那種流量控制和自我實現(xiàn)的應(yīng)用程序。P20

16、7 2,3,4,26,27,28,41,442 A noiseless 4-kHz channel is sampled every 1 msec. What is the maximum data rate?If an arbitrary signal has been run through a low-pass filter of bandwidth B, the filtered signal can be completely reconstructed by making only 2B (exact) samples per second.A noiseless channel

17、can carry an arbitrarily large amount of information, no matter how often it is sampled. Just send a lot of data per sample. For the 4 kHz channel, make 8000 samples/sec. If each sample is 16 bits, the channel can send 128 kbps. If each sample is 1024 bits, the channel can send 8.192 Mbps. The key w

18、ord here is noiseless. With a normal 4 kHz channel, the Shannon limit would not allow this.PCM (Pulse Code Modulation)脈碼調(diào)制它將4千赫茲的電話信道的模擬信號轉(zhuǎn)化為數(shù)字信號。一條普通的電話線路帶寬為4KHz，根據(jù)奈奎斯特定理：只要每秒4000*2=8000次的采樣頻率就能獲取一個4KHz的信道中的全部信息，再高的采樣頻率也無意義，所以PCM的采樣時間為1/8000=125us。3 If a binary signal is sent over a 3-kHz channel w

19、hose signal-to-noise ratio is 20 dB, what is the maximum achievable data rate?20dB=10lgS/N S/N=100 Shannon limit=Blog2(1+S/N)=3000*log2(1+100)= 19.975 KbpsNyquist limit=2Blog2V=2*3000*log22=6KbpsDate rate19.975Kbps and Date rate6Kbps （同時滿足）4 We have a channel with a 1 MHz bandwidth. The SNR for this

20、 channel is 63; what is the appropriate bit rate and signal level?Shannon limit=Blog2(1+S/N)=106*log2(1+63)= 6 MbpsNyquist limit=2Blog2V=2*106*log2V=6MbpsV=826 What is PCM? Why has the PCM sampling time been set at 125usec?PCM means pulse code modulation. It is the analog-to-digital signal 模擬數(shù)字信號 co

21、nverting technique for transmitting information from the 4-KHz telephone channel. 電話通道According to the Nyquist theorem, the sampling frequency needed to capture all the information in a 4-kHz channel is 8 KHz. That is to say, a sampling time is 1sec/8000 = 125sec.27 28 41 Suppose that A, B, and C ar

22、e simultaneously transmitting 0 bits, using a CDMA system with the chip sequences of Fig. 2-45. What is the resulting chip sequence?編碼方法：用要發(fā)送的bit串轉(zhuǎn)化成雙極性碼，”1”對應(yīng)”+1”, ”0”對應(yīng)”-1”, 不發(fā)送的用全“0”表示，然后按順序同各個站的碼片序列相乘，最后將各站的結(jié)果序列（混合）相加，就是在空中發(fā)射的信號。A: “0”à”-1”, -1*(-1 -1 -1 +1 +1 -1 +1 +1) = (+1 +1 +1 -1 -1 +1

23、 -1 -1) ; B: “0”à”-1”, -1 *(-1 -1 +1 -1 +1 +1 +1 -1)= (+1 +1 -1 +1 -1 -1 +1 +1); C: “0”à”-1”, -1 * (-1 +1 -1 +1 +1 +1 -1 -1)= (+1 -1 +1 -1 -1 -1 +1 +1) D：不發(fā)送，(0 0 0 0 0 0 0 0); 4個序列相加得：（+3 +1 +1 -1 -3 -1 -1 +1）.44 A CDMA receiver gets the following chips: (-1 +1 -3 +1 -1 -3 +1 +1). Assumin

24、g the chip sequences defined in Fig. 2-45, which stations transmitted, and which bits did each one send?解碼方法：用接收到的碼片同各個站的碼片序列進行“點積”(又稱為“內(nèi)積”)，也就是按位相乘，結(jié)果序列的所有位相加，除以碼片序列的長度，”1”意味著原先編碼時是”1”，”-1”表示編碼”0”,“0”表示沒有靜默，沒發(fā)送。A: (-1 +1 -3 +1 -1 -3 +1 +1) (-1 -1 -1 +1 +1 -1 +1 +1) =(1-1+3+1-1+3+1+1)/8=8/8=1, 即： A發(fā)

25、送了bit “1”;B: (-1 +1 -3 +1 -1 -3 +1 +1) (-1 -1 +1 -1 +1 +1 +1 -1) =(1-1-3-1-1-3+1-1)/8=-8/8=-1，即： B發(fā)送了bit “0”;C: (-1 +1 -3 +1 -1 -3 +1 +1) (-1 +1 -1 +1 +1 +1 -1 -1) =(1+1+3+1-1-3-1-1)/8=0/8=0，即： C沒有發(fā)送信息。P272 2,7,8,14,152 A是錯的 應(yīng)為 000001017 An 8-bit byte with binary value 10101111 is to be encoded usin

26、g an odd-parity Hamming code. What is the binary value after encoding? p1 p2 a3 p4 a5 a6 a7 p8 a9 a10 a11 a12=? ? 1 ? 0 1 0 ? 1 1 1 1P1+a3+a5+a7+a9+a11=奇 ？+1+0+0+1+1=奇P2+a3+a6+a7+a10+a11=奇 ？+1+1+0+1+1=奇P4+a5+a6+a7+a12=奇 ？+0+1+0+1=奇P8+a9+a10+a11+a12=奇 ？+1+1+1+1=奇得漢明碼為011001011111。若傳輸后，接收方收到的為01111101

27、1111，檢驗：P1：0+1+1+0+1+1=偶 p2：1+1+1+0+1+1=奇P4：1+1+1+0+1=偶 p8：1+1+1+1+1=奇8 ?14 15 A bit stream 10011101 is 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

28、transmission. Show that this error is detected at the receiver's end.生成式為1001。消息比特加上3個0后成為10011101000，除以1001后余式為100，因此實際傳輸?shù)谋忍卮疄?0011101100。左起第三比特顛倒后接收到的多項式為10111101100，除以生成式，余式為100，而不是0，這就使接收方明白發(fā)生了錯誤。P369 13,14,15,16,2213 14 15 A 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed

29、 of 200 m/µsec. Repeaters are not allowed in this system. Data frames are 256 bits long, including 32 bits of header, checksum, and other overhead. The first bit slot after a successful transmission is reserved for the receiver to capture the channel in order to send a 32-bit acknowledgement fr

30、ame. What is the effective data rate, excluding overhead, assuming that there are no collisions?(長為1km的CSMA/CD局域網(wǎng)（不是802.3網(wǎng)），數(shù)據(jù)率為10Mbps，傳播速度為200 m/µs。系統(tǒng)中不用中繼器。數(shù)據(jù)幀長256比特，其中包括32比特頭部、檢驗和和其它開銷。一次成功傳輸后的第一個時隙保留給接收方捕獲信道，以便讓它發(fā)送一個32比特的確認幀。假設(shè)不發(fā)生沖突，扣除開銷，有效的數(shù)據(jù)率為多大？)CSMS/CD局域網(wǎng)是分時隙的（time slotted）。每次發(fā)送之前必須偵聽信道

31、，至少 花費一個時隙(time slot) 才能確保不沖突 ！時隙長度取決于最大往返傳播延時。Single-trip takes 1km/ 200 m/µsec=5usec, round-trip propagation delay is = 5X2=10usec, so the length of time slot should be 10usec.信號從電纜一端傳導(dǎo)到另一端用時1000/ (200/10-6)=5us, 往返需5*2=10us，因此時隙長度為10us。transmitter seizes cable (10 sec): before transmission,

32、the transmitter checks the cable, and if the cable is idle, starts to transmit at the beginning of next time slot 發(fā)前“聽信道”要用一個時隙transmit data (25.6 sec) 傳輸延時Delay for last bit to get to the end (5.0 sec) 單程傳播延時receiver seizes cable (10 sec): similar to the transmitter同1.acknowledgement sent (3.2 sec)

33、傳輸延時Delay for last bit to get to the end (5.0 sec)單程傳播延時The sum of these is 58.8 sec. In this period, 224 data bits are sent, for a rate of about 3.8 Mbps.16 Consider building a CSMA/CD network running at 1Gbps over 1km cable with no repeater.The signal speed in the cable is 200,000 km/secWhat is mi

34、nimum frame size?22 In Fig. 4-27, four stations, A, B, C, and D, are shown. Which of the last two stations do you think is closest to A and why?Textbook：From the information provided in the RTS request, C can estimate how long the sequence will take, including the ACK, so it asserts a kind of virtua

35、l channel busy for itself, indicated by NAV (Network Allocation Vector). D does not hear the RTS, but it does hear Bs CTS, so it also asserts the NAV signal for itself. Note that the NAV signals are not transmitted; they are just internal reminders to keep quiet for a certain period of time P508 5,2

36、3,24,25,27,28,305 Consider the subnet of Fig. 5-12(a). Distance vector routing is used, and the following vectors have just come in to router C: from B: (5, 0, 8, 12, 6, 2); from D: (16, 12, 6, 0, 9, 10); and from E: (7, 6, 3, 9, 0, 4). The measured delays to B, D, and E, are 6, 3, and 5, respective

37、ly. What is C's new routing table? Give both the outgoing line to use and the cost.來自鄰居們的消息B: (A5, B0, C8, D12, E6, F2);D: (A16, B12, C 6, D0, E 9, F10);E: ( A7, B6 , C3, D9, E0, F4). 自己測量的delays from C to B, D, and E are 6, 3, and 5, Going via B gives (A11,B 6, C14, D18, E12, F8).Going via D gi

38、ves (A19, B15, C9, D3, E9, F10).Going via E gives (A 12, B11, C8, D14, E5, F9).Taking the minimum for each destination except C givesC: (11, 6, 0, 3, 5, 8).(B, B, , D, E, B) These are outgoing lines.路由器C更新后的路由表說明：到A，最近距離為11，從B走；到B，距離為6，從B走；到C嗎？就是自己這里；到F距離為8，從B走最近。23Suppose that instead of using 16 b

39、its for the network part of a class B address originally, 20 bits had been used. How many class B networks would there have been?( 假設(shè)B類地址用20bit作為網(wǎng)絡(luò)號，而不是16bit，總共有多少個B類網(wǎng)？)With a 2-bit prefix, there would have been 18 bits left over to indicate the network.去除2bit類別號（“10”），還有20-2=18bit代表網(wǎng)絡(luò)號Consequently,

40、 the number of networks would have been 218 or 262,144. However, all 0s and all 1s are special, so only 262,142 are available. 218=262144,全0和全1是特殊地址，不分配給網(wǎng)絡(luò)用，故共有262142個B類網(wǎng)2425 A network on the Internet has a subnet mask of . What is the maximum number of hosts it can handle?(因特網(wǎng)上一個子網(wǎng)的掩碼為

41、，它最多能容納多少臺主機？)=11111111.11111111.11110000.00000000 20個1對應(yīng)網(wǎng)絡(luò)號和子網(wǎng)號；12個0對應(yīng)主機編號。The mask is 20 bits long, so the network part is 20 bits. The remaining 12 bits are for the host, so 212-2=4094 host addresses exist另一題：A class B network on the Internet has a subnet mask of 255.255

42、.240.0. How many subnets in it? And What is the maximum number of hosts each subnet can handle?  (Internet 上的一個B類網(wǎng)絡(luò)的子網(wǎng)掩碼為，問該網(wǎng)絡(luò)劃分了多少個子網(wǎng)？每個子網(wǎng)最多可以有多少臺主機？) 無答案。已知一臺主機的IP地址為6，子網(wǎng)掩碼為92。求：該主機所在的網(wǎng)絡(luò)分了多少個子網(wǎng)？22=4 ? 到底是2還是4?它是位于哪個子網(wǎng)中？編號為01的子網(wǎng)主機號是多少？ 這是C類網(wǎng)C類網(wǎng)前24bit為網(wǎng)絡(luò)號，即

43、默認掩碼為 =1111111,11111111,1111111,00000000 與現(xiàn)在用的子網(wǎng)掩碼比較，多的2bit“1”是子網(wǎng)號。22=4，減去全“0”和全“1”的組合，共2個子網(wǎng)（01和10）。26=64,每個子網(wǎng)可容納62臺主機。當前IP地址對應(yīng)第2個子網(wǎng)的第2臺主機。27 A large number of consecutive IP address are available starting at . Suppose that four organizations, A, B, C, and D, request 4000, 200

44、0, 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.Answer：To start with, all the requests are rounded up to a power of two. 4000à4096, 2000à2048.The starti

45、ng address, ending address, and mask are as follows: A: 198.16.15. 255 written as /20B: 198. 16. 23. 255 written as /21If C start at 28 A router has just received the following new IP addresses: /21, /21, /21, and 5

46、/21. If all of them use the same outgoing line, can they be aggregated? If so, to what? If not, why not?30 A router has the following (CIDR) entries in its routing table:Address/mask Next hop/22 Interface 0/22 Interface 1/23 Router 1 Default Router 2For each

47、 of the following IP addresses, what does the router do if a packet with that address arrives?(a) 0(b) 4(c) (d) (e) 3、 Concepts Exercises a) Draw the TCP/IP reference model in a table, and list two or more protocols in each layer.OSIApplicationP

48、resentationSessionTransportNetworkData linkPhysicalTCP/IP 體系A(chǔ)pplicationTransportInternetLinkb) What layers are there in the OSI reference model and the TCP/IP model? Please sketch the correspondence between the layers of the two models like this:c) What is the main difference between virtual circuit

49、 network and datagram network?Answer: Virtual circuit network is connection-oriented while datagram network is connectionless. In a virtual circuit network, a path from the source router to the destination router must be established before any data packets can be sent. That path is called “virtual c

50、ircuit”. It is tore down after no data is to be sent. Datagram network routes each packet as a separate unit, independent of all others. 虛擬電路網(wǎng)絡(luò)和數(shù)據(jù)報網(wǎng)絡(luò)的區(qū)別：虛電路網(wǎng)絡(luò)是面向連接的而數(shù)據(jù)報網(wǎng)絡(luò)是面向無連接的。在虛電路網(wǎng)絡(luò)里，在傳輸信息包裹之前，一條路徑必須在起始路由器和目標路由器之間建立。那條路徑就叫做“虛電路”。數(shù)據(jù)傳輸結(jié)束之后，要拆除建立的連接路徑。而數(shù)據(jù)報網(wǎng)絡(luò)將信息分為多個包裹單元來傳輸，包裹之間相互獨立。d) Which of the OS

51、I layers divides the transmitted bit stream into frames? Do frames encapsulate packets or do packets encapsulate frames? Why?4、 幀封裝包。因為幀是工作在數(shù)據(jù)鏈路層，而數(shù)據(jù)包工作在網(wǎng)絡(luò)層，在數(shù)據(jù)傳輸時，上一層的的內(nèi)容有下一層的內(nèi)容傳輸，上層向下層進行封裝。a) What is the main tasks of the 3rd layer of OSI/RM. 網(wǎng)絡(luò)層的主要功能是控制子網(wǎng)的運行過程，將分組從源端送到目的端，向傳輸層提供兩類服務(wù)：面向連接的和面向無連接的。

52、b) What is the principal difference between connectionless communication and connection-oriented communication?面向連接通信分為三個階段，第一是建立連接，發(fā)出一個建議連接的請求。只有在連接成功建立之后，才能開始數(shù)據(jù)傳輸，這里第二階段。接著，數(shù)據(jù)傳輸完畢之后，必須釋放連接。而無連接通信沒有這么多階段，它直接進行數(shù)據(jù)傳輸。面向連接通信也具有數(shù)據(jù)的保序性，而無連接通信不能保證數(shù)據(jù)的順序與發(fā)送數(shù)據(jù)的順序一致。c) What is PCM? Why has the PCM sampling ti

53、me been set at 125usec?Answer: 脈沖編碼調(diào)制PCM means pulse code modulation. It is the analog-to-digital signal 模擬數(shù)字信號 converting technique for transmitting information from the 4-KHz telephone channel. 電話通道According to the Nyquist theorem, the sampling frequency needed to capture all the information in a

54、4-kHzchannel is 8 KHz. That is to say, a sampling time is 1sec/8000 = 125sec.脈碼調(diào)制它將4千赫茲的電話信道的模擬信號轉(zhuǎn)化為數(shù)字信號。一條普通的電話線路帶寬為4KHz，根據(jù)奈奎斯特定理：只要每秒4000*2=8000次的采樣頻率就能獲取一個4KHz的信道中的全部信息，再高的采樣頻率也無意義，所以PCM的采樣時間為1/8000=125us。d) Describe the main steps of CSMA/CD.工作原理：一旦檢測到有沖突發(fā)生，就不再繼續(xù)傳送幀，節(jié)省時間和帶寬。應(yīng)用：在LAN中的MAC子層中。工作步驟：

55、發(fā)送之前，進行載波偵聽。若無載波，采用“1”堅持，立即發(fā)送。邊發(fā)送邊檢測。若有沖突，立即停止，發(fā)送警告等待隨即時長，重新開始。若數(shù)據(jù)錯誤則采用二元指數(shù)退避法或者選擇重發(fā)送的方法。e) There are exposed terminal problem and hidden terminal problems in wireless LAN. In the case of Figure 4-1, which are exposed terminals? Which are hidden terminals? Figure 4-1Answer: When B sends to A while C

56、 sends to D, B and C are exposed stations.If A wants to send to B and C wants to send to B at the same time, C is the hidden station for A, and vice versa. Similarly, when B and D transmit to C at the same time, B and D are hidden stations 當B要發(fā)送給A而C要發(fā)送給D，B和C都是暴露站。如果A和C想同時發(fā)送給B，C對于A來說是隱藏站，反之亦然。同樣的，當B和

57、D想同時發(fā)送給C，B和D就是隱藏站。f) What is the hidden station problem?Answer： In wireless LAN, consider the figure below, where four wireless stations are illustrated. For our purposes, it does not matter which are base stations and which are notebooks. The radio range is such that A and B are within each other's range and can potentially interfe