《网络原理 Computer Networks》期末(A)试卷.doc

上海应用技术学院20092010学年第1学期网络原理 Computer Networks期末（A）试卷课程代码: 学分: 考试时间: 分钟课程序号: 6668,6780, 6666 班级： 学号： 姓名: 我已阅读了有关的考试规定和纪律要求，愿意在考试中遵守考场规则，如有违反将愿接受相应的处理。题 号一二三四五六七八九十总 分得 分 试卷共 8 页，请先查看试卷有无缺页，然后答题。 I. Abbreviations: (10 points in total)The (everyday English) expansions of the following abbreviations are wanted: A1-10.(1 point each, 10 points in total)A1. WAN = Wide Area NetworkA2. TCP = Transmission Control ProtocolA3. DNS = Domain Name System/ServiceA4. OSPF = Open Shortest Path FirstA5. SMTP = Simple Mail Transfer ProtocolA6. PPP = Point-to-Point ProtocolA7. GPS = Global Positioning SystemA8. CSMA/CD = Carrier Sense Multiple Access with Collision DetectionA9. IP = Internet ProtocolA10. VLAN = Virtual LAN = Virtual Local Area NetworkII. Concepts: (30 points in total)Give the definition or description of the following concepts or names:B1-5 (6 points each, 30 points in total)B1. What is the main task of the network layer (in the OSI reference model)? Solution (c5-P00)The network layer is concerned with getting packets from the source machine all the way to the destination machine.B3. We have studied many channel allocation methods. Please name three of them.B4. What approaches do we take to do error control? (or what mechanisms do we use to do error control?)B5. Please draw a figure to show the OSI Reference Model. Solution: （c1）P37B6. In Fig II-1, one of the 802.11 MAC sublayer protocol is illustrated and four stations, A, B, C, and D, are shown. Which of the last two stations do you think is closest to A and why?III. Computation: (40 points in total)The process/procedure or explanation of computation and the correct results are all wanted. C1-5. (8 points each, 40 points in total).C1. 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 ESC FLAG B Awhen each of the following framing methods is used:(a) Character count.0000110 (b) Flag bytes with byte stuffing.(c) Starting and ending flag bytes, with bit stuffing.C2. A bit stream 10011101 is transmitted using the standard CRC method described in the text. The generator polynomial is x4+ x2+1. Show what code the receivers end receives if the receivers end will receive correct code.Solution:100111010000 / 10101 the remainder is 1000So the receiver will receive 100111011000C3. A router has the following (CIDR) entries in its routing table:Address/maskNext hop195.46.56.0/22Interface 0195.46.60.0/22Interface 1defaultRouter 1For each of the following IP addresses, what does the router do if a packet with that address arrives?a. 195.46.53.20b. 195.46.62.100c. 192.53.56.78SOLUTION: a. 53=00110101B Because mask is 252(=1111100B) solution is 00110100 =52So the router goes to Router 1 if a packet with address 195.46.53.20 arrives.b. 62=00111110B Because mask is 252(=11111100B) solution is 00111100 So the router goes to Interface 1 if a packet with address 145.46.62.100 arrives.c. router 1 b/c 192.53.C4. A large number of consecutive IP address are available starting at 198.16.0.0. 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. Solution(P478-40)To start with, all the requests are rounded up to a power of two. The startingaddress, ending address, and mask are as follows: A: 198.16.0.0 198.16.15.255 written as 198.16.0.0/20B: 198.16.16.0 198.16.23.255 written as 198.16.16.0/21C: 198.16.32.0 198.16.47.255 written as 198.16.32.0/20D: 198.16.64.0 198.16.95.255 written as 198.16.64.0/19C5. Consider building a CSMA/CD network running at 1 Gbps 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? Solution(P340-21) 对于1km 电缆，单程传播时间为1/200000 =510-6 s，即5，来回路程传播时间为2t =10。为了能够按照CSMA/CD 工作，最小帧的发射时间不能小于10。以1Gb/s 速率工作，10可以发送的比特数等于： 因此，最小帧是10 000 bit 或1250 字节长。 IV. Description: (20 points in total)a brief description with all principal points is wanted. D1-2. (10 points each, 20 points in total)D1. Two types of transmission technologies are in widespread use. They are broadcast links and point-to-point links. a) What do we mean by broadcast networks? Give the definition or a brief description.p15Broadcast 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.b) What do we mean by point-to-point networks? Give the definition or a brief description.p15point-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. Often multiple routes, of different lengths, are possible, so finding good ones is important in point-to-point networks.D2. The idea behind link state routing is simple and can be stated as five parts. What are the five parts or steps that each router must do?Soluti