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(word完整版)101G09XT1[1].宁波大学计算机网络A.答案
一、 单选题,每题仅一个正确答案,每题1.5分,共30分:
[ ]1。 Which is not a network communication links?
D。Subway
[ ]2。Which is not the types of protocol data unit in TCP/IP?
C.datetime
[ ]3.In which state the delays of router become small?
A. La/R ~ 0
[ ]4.When congest occur in network, which delay become large?
B。queuing delay
[ ]5。Application protocol define the types of messages exchanged, which are:
A。request & response messages
[ ]6。Which version of hypertext transfer protocol are used in default mode now?
C。HTTP 1。1
[ ]7。The Http client works on which port:
C.TCP port above 1024
[ ]8.To indicates end of message in Http, we can use:
A。Carriage return and line feed
[ ]9。Which is not an email protocol:
C.Emule
[ ]10。 TCP retransmission can not be caused by:
D。 ACK
[ ]11. Which of the following measures the time from one segment transmission to ACK receipt:
A. sample RTT
[ ]12。 Which event will not be in TCP Receiver:
C。Arrival of out-of-order segment higher-than-expect seq. # .
No gap detected.
[ ]13.TCP Flow Control is a speed-matching service for matching:
A。the sender rate to the receiving application’s drain rate
[ ]14.Which one is not the mechanism of TCP Congestion Control:
C.FIFO
[ ]15。In RDT protocol, which can increase the use of physical resources:
C.Pipeline
[ ]16. Which one is not the Key Network-Layer Functions in today’s internet:
C。 Connection setup
[ ]17. Which of the following IP address belongs to category C address?
A. 210。98。3.4
[ ]18。Network address translation implement in NAT router, it must use:
C.NAT translation table
[ ]19。 ICMP do not error reporting about:
D. unknown username
[ ]20. In Ethernet’s CSMA/CD Jam Signal make sure:
A。other transmitters are aware of collision
二、多选题,每题有2—3个正确答案,请在正确项前打钩(全钩不给分),每题2分,共20分:
1。 The communication services be provided to applications, It can be:
[ ]A.connectionless
[ ]B.connection—oriented
2.Which are application architectures model in the internet?
[ ]A.client/server model
[ ]B。peer-peer model
[ ]C。browser/server model
3. User Agent is an interface with user “above” and network “below", such as:
[ ]A。Web: browser
[ ]B.E-mail: mail reader
[ ]D.Streaming audio/video: media player
4。Ftp protocol can work in which model that listed:
[ ]A。PASV model
[ ]C.Port model
5. The problems with centralized directory of P2P maybe:
[ ]A.Single point of failure
[ ]C.Performance bottleneck
[ ]D。Copyright infringement
6.Which services are not available in TCP:
[ ]A。delay guarantees
[ ]C。bandwidth guarantees
7. In TCP congestion control, there are those State in TCP Sender:
[ ]A. Congestion Avoidance
[ ]C. Slow Start
[ ]D. Timeout
8.In TCP three way handshake, it will be:
[ ]A.client host sends TCP SYN segment to server
[ ]C.server host receives SYN, replies with SYN/ACK segment
[ ]D.client receives SYN/ACK, replies with ACK segment
9。Those are the routing in the Internet:
[ ]A.RIP
[ ]B.OSPF
[ ]D。BGP
10。Those are the Random Access Protocols:
[ ]B.CSMA
[ ]C.CSMA/CD
三、名称缩写选择,请在匹配项上填写缩写编号,如果在下列18个选项中没有,
可以在19-20项中自主添加选项。共10分:
[ 1 ]往返时间 [ 20 ]可靠传输协议
[ 13 ]网络接口卡 [ 17 ]自治系统
[ 12 ]网络地址翻译协议 [ 5 ]用户数据报协议
[ 10 ]协议数据单元 [ 9 ]动态地址分配协议
[ 19 ]Internet协议 [ 3 ]生存时间
——-—-—-———--—--—--———-———--—----—-—-————-------—-—-—--—-——--——-—-————-—-—----——-——-———--—---—-—--——--—————---—--——
19。( IP ) 20。( RDT )
四、问答题(共30分)
1。(共10分)下图是运行RIP协议的一个网络:
1。1 请描述该路由表的含义:(3分)
答:该路由表说明了由路由器D到底各相邻网络和路由器的开销(距离),如
w/A/2 表示由A路由器到w网络为2跳
y/B/2 表示由B路由器到y网络为2跳
z/B/7 表示由B路由器到z网络为7跳
x/-/1 表示由D路由器到x网络为1跳(直接连接)
v/—/— 表示由D路由器到v网络为1跳(直接连接)
1。2 当网络上出现由A发给D的如下路由宣告后路由表的变化情况:(2分)
答:z/B/7 => z/A/5
1。3 接着又出现由B发给D的如下路由宣告后路由表的变化情况:(2分)
答:z/A/5 =〉 z/B/3
1.4 请问上述路由表中hops的含义是什么?(2分)
该数值包含了网络的带宽和延时等信息吗?(1分)
答:hops代表跳数(路由器个数)
不包含带宽和延时。
2、(共10分)如图所示为TCP协议头部结构,请对照该图回答如下问题:
2.1 一个基本的TCP头部占多少个字节?(1分)
比UDP的头部大还是小?(1分)
答:20字节
大(UDP为8字节)
2.2 TCP连接建立时用到那几个字段?(2分)
答:sp# dp# seq# A/S/ACK#等
2.3 TCP连接拆除时用到哪几个字段?(2分)
答:sp# dp# seq# F/S/ACK#等
2。4 TCP流量控制主要使用了哪个字段?(1分)
答:Receive window
2.5 图中出现了number字样,其数值的单位是什么?(1分)
两个number分别代表什么意义?(1分)
答:byte
SEQ#表示发送数据的第一个字节的编号
ACK#表示希望下次接收到的数据的第一个字节的编号
2.6 根据该结构TCP在发送数据时可以顺便确认对方刚才发送的数据,是如何实现的?(1分)
答:在发送数据的时候用SEQ#记录当前数据的第一个字节的序号,在ACK#中携带对方刚才发过来数据
的序号的
下一个字节,顺便完成确认。
3、(共10分)cache是信息技术中常用的一种方法,请回答:
3.1 cache的一般作用是什么?(1分)
答:缓存作用,提高性能和效率
3.2 在http协议中client cache的基本原理是什么?(4分)
答:利用条件get实现客户端缓存,当服务器端文件没有更新过就不需要重新下载,提高HTTP传输效率。
3。3 在http协议中利用server cache有何作用?(3分)
答:可以实现http代理(PROXY),提高局域网共享上网的效率、降低成本.
3.2 在http协议中的CDNs应用可以起到什么作用?(2分)
答:可以实现分布式内容发布服务。
六、文献分析(10分)
根据文献中提供的信息,回答以下问题:
1. What is the value of MTU in Ethernet? (2分)
答:1500
2. So, How many bytes can be carried by IP datagram?(2分)
答:1480
3. Where is the IP datagram fragment operating? (2分)
答:Router network layer
4. Where is the IP datagram reassembly operating? (2分)
答:Destination host (end system) network layer
5. Which field of IP can be examined to determine the datagrams
are actually fragments of the same bigger datagram?(2分)
答: ID# (identification number)
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