2010级爱恩学院网络管理名词解释和简答部分by_Gerry

1、Explain the following termsUnicode - The coding system that has enough bit combinations to encode all of the characters in all of the worlds languages Bandwidth - The difference between the upper and lower frequency is called the bandwidth6. flow control: Techniques to ensure that a fast transmittin

2、g node does not send data faster than the receiving node can receive and process it are called flow control7. Contention When any station that has traffic looks to see if the circuit is free, and if it is, begins sending its traffic the line control technique is called contention8. Switch link-layer

3、 device: smarter than hubs, take active rolestore, forward Ethernet framesexamine incoming frames MAC address, selectively forward frame to one-or-more outgoing links when frame is to be forwarded on segment, uses CSMA/CD to access segment9. Protocol a set of rules that define the exact format of me

4、ssages exchanged between computers or between computers and people 10 Node A node on a network is a point of connection into a network, or a point at which one or more transmission lines (circuits) interconnect, such as where a router or a switch connects into the network.11 Topology refers to the p

5、hysical layout of its computers, cables, and other resources, and also to how those components communicate with each other12. Network architecture is a set of principles used as the basic for the design and implementation of a communications network13 Subnet mask determines which part of address den

6、otes network portion and which denotes host14 Summary of Layer Functions15 Design Guidelines of internetThe four initial requirements confronting the Internets architects were :Multiplexing, note can share medium Survivability, get through network, no matter what happenedservice generality Highest p

7、ossible speed, while others require the highest possible reliability supporting diverse network technologies16 PROTOCOL FUNCTIONS Encapsulation Fragmentation and reassemblyConnection controlOrdered deliveryFlow controlError controlAddressingMultiplexingTransmission services17.Throughput- rate (bits/

8、time unit) at which bits transferred between sender/receiver18 What is the difference between a virus, a worm and a Trojan horse?a) VirusRequires some form of human interaction to spread. Classic example: E-mail viruses.b)WormsNo user replication needed. Worm in infected host scans IP addresses and

9、port numbers, looking for vulnerable processes to infect.Trojan horseHidden, devious part of some otherwise useful software.19 For the client-server application over TCP, why must the server program be executed before the client program? For the TCP application, as soon as the client is executed, it

10、 attempts to initiate a TCP connection with the server. If the TCP server is not running, then the client will fail to make a connection. 20 Describe why an application developer might choose to run an application over UDP rather than TCP?An application developer may not want its application to use

11、TCPs congestion control, which can throttle the applications sending rate at times of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCPs congestion control. Also, some applications do not need the

12、 reliable data transfer provided by TCP.21 In our rdt protocols, why did we need to introduce sequence numbers?Sequence numbers are required for a receiver to find out whether an arriving packet contains new data or is a retransmission.22 In our rdt protocols, why did we need to introduce timers?To

13、handle losses in the channel. If the ACK for a transmitted packet is not received within the duration of the timer for the packet, the packet is assumed to have been lost. Hence, the packet is retransmitted 23 What is the difference between routing and forwarding?Forwarding is about moving a packet

14、from a routers input link to the appropriate output link. Routing is about determining the end-to-routes between sources and destinations. PARITY CHECKING Bits 1010010 (no parity bit)With even parity 10100101With odd parity 10100100Longest prefix matchingDestination Address Range 11001000 00010111 0

15、0010* * 11001000 00010111 00011000 *11001000 00010111 00011* *otherwise DA: 11001000 00010111 00011000 10101010 Examples:DA: 11001000 00010111 00010110 10100001 Which interface?Which interface?when looking for forwarding table entry for given destination address, use longest address prefix that matc

16、hes destination address.Longest prefix matchingLink interface0123ExerciseWhich classes are the following IP address?6 hosts What is the start address of this block?What is the end address of this block?六、子网划分的实例An C class network 192.168.6. 0，100 hosts in this network, please divide the network into

17、 2 subnet:结论：不能3032683能6264224不能021步骤如下：1、确定用几位来标识子网号和每个子网中的 主机台数能否满足要求 每个子网中实际可用主机数 每个子网中主机数 实际可用子网个数 子网个数 子网位数 用第4字节中的前两位来表示子网号，每 个子网最多可有62台主机2、确定子网掩码C类网络 192. 168 . 6. 0子网掩码：255. 255. 255. 192192.168.6.01 000000192.168.6.10 000000网络地址3、确定每个子网的网络地址子网络号原有的网络号4、确定每个子网上所使用的主机地址范围子网一192.168.6.01 00000

18、1192.168.6.01 111110主机地址范围为 192.168.6.65192.168.6.126子网二192.168.6.10 000001192.168.6.10 111110主机地址范围为 192.168.6.129192.168.6.190划分子网后的效果图下列图给出了划分子网后的网络效果图，并且对每个子网各台主机的地址进行了配置。 Exercisehow to divided IP 200.10.10.0 into 2 subnets? Write down the subnet mask, how many bits need to borrow?3 subnets? Wr

19、ite down the subnet mask, how many bits need to borrow?4 subnets? Write down the subnet mask, how many bits need to borrow?10 subnets? Write down the subnet mask, how many bits need to borrow?1. the computer IP address is 195.169.20.25, Subnet mask is 255.255.255.240, please write this computers net

20、work address, subnet number and host number.SubnetsHow many?IP datagram formatverlength32 bitsdata (variable length,typically a TCP or UDP segment)16-bit identifierheader checksumtime tolive32 bit source IP addressIP protocol versionnumberheader length (bytes)max numberremaining hops(decremented at

21、each router)forfragmentation/reassemblytotal datagramlength (bytes)upper layer protocolto deliver payload tohead.lentype ofservice“type of data flgsfragment offsetupper layer32 bit destination IP addressOptions (if any)E.g. timestamp,record routetaken, specifylist of routers to visit.how much overhead with IP?20 bytes of IP1.What is the IP address of your computer? The IP address of my computer is 192.168.1.46 2.Within the IP packet header, what is the value in the up