云南大学2015至2016学年上学期软件学院2014级《计算机网络原理》期中考试试卷（闭卷）答案满分：100分考试时间：100分钟任课教师：王世普第一题答题卡：第二题答题卡：1．（1）is the protocol suite for the current Internet..（1）A. NCP B. TCP/IP C.UNIX D.ACM2．A GIF image is sent as email ,What is the content-type （2） .（2）A．multipart/mixed B．multipart/imageC．image/JPEG D．image/gif3．A user want to send some forms（表单）to Web server using HTTP protocol, the request line method is （3）.（3）A．GET B．PA TCH C．MOVE D．POST4．If a TCP segment carries data along with an acknowledgment, this technology is called （4）acknowledgment.（4）A. backpacking B. piggybacking C. piggying D. mother’s help5．TCP is a （5）transport layer protocol that ensure data to be exchanged reliably by（6）. So it requires set up connection before data exchanged by ( 7 )-way handshaking.（5）A．connection B．connectionless C．join D．disconnection（6）A．datagrams B．acknowledgements C．data D．segment（7）A．one B．two C．three D．four6．A user requests a Web page that consists of a basic HTML file and 5 JPEG image files. d trans denoting the time to transfer a file. The total time is (8) to request the Web page inNonpersistent connections mode?（8）A. 6(2RTT+ d trans) B. 2RTT+6 d trans C. RTT+6 d trans D.6(RTT+ d trans)7．Host A sends a TCP segment (Seq = 1, ACK = 111) to host B and Host B replies with a TCP segment (Seq = 111, ACK = 81). The payload length of the TCP segment from host A to hostB is ( 9 ) .（9）A．80 bytes. B．81 bytes. C．82 bytes. D．unknown8．As a data packet moves from the lower to the upper layers, header are （10）.（10）A. modified B. added C. subtracted D. rearranged二、判断正误（每个问题1分，共5分。

正确的打“√”、错误的打“×”，请将判断结果填入第二题答题卡）1. The DNS defines a distributed, hierarchical database that provides only hostname to IP addressmappings.( × )2. In SR protocol, sender only resends pkts for which ACK not received. ( √ ) 3．If sequence number is k bit, then the send window maximum size of GBN protocol is equal to 2k-1. ( √ ) 4．The UDP header checksum is recomputed at every routers. ( × ) 5．Suppose that the last SampleRTT in a TCP connection is equal to 1 sec. Then Timeout for the connection will necessarily be set to a value >= 1 sec. ( × )三、回答下列问题（每个问题5分，共30分）1．For a communication session between two hosts, which host is the client and which is the server?要点：通信的发起者（请求者）是客户，通信的接受者（服务者）是服务器。

2．What is the difference between persistent HTTP with pipelining and persistent HTTP without pipelining?要点：前者在一个TCP连接期间可以传送多个WEB页面；后者在一个TCP 连接期间只能传送一个WEB页面。

3．What are the two types of services that the Internet provides to its applications? What are important characteristics of each of these services?要点：面向连接的服务（连接管理、正确可靠、按序交互、流控制、拥塞控制）；无连接服务（不需建立连接、不可靠、每个报文独立传送，无流控制和拥塞控制）4．Why is it said "out of band" (带外的) that FTP sends control information?要点：TCP 数据连接是主连接，不能提供FTP 命令的传输；TCP 控制连接仅仅是传输FTP 命令和执行状态，因而是数据通道之外的通道，故属于“带外”信道。

5．Please list the main transmission mediums used in the computer network.要点：铜线、光纤、（地面）无线、卫星6．What are the differences Congestion Control and flow control.要点：范围不同（全网；通信双方）、目的不同（防止接收端缓冲区溢出；防止网络拥塞）四、计算题（共25分）1. Suppose host A communicates with host B through TCP, sometime host A sends 120 bytes data to host B, then host B responds with 100bytes data. Analyze the necessary ACK numbers and the Sequence number in the segments sent between host A and host B . Please give the analysis procedure (8 分)要点：ACK 是期望接收的段的序号；当B 收到一个数据为120字节、序号为45的段，就意味着期望接收的下一个段的序号应该是45+120=165；当S 收到来自B 的一个数据为100字节、序号为108的段后，则意味着期望接收的段的序号是108+100=208.2． UDP and TCP use 1's complement sum for their checksums. Suppose you have the following three 16-bit words: 1100100110110101, 1010011101010101, 0110010111011101. What is the 1's complement sum of these words? （7分）结果：Sum 1101011011101000 Checksum0010100100010111计算过程：1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 1 + 1 0 1 0 0 1 1 1 0 1 0 1 0 1 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 1 0 1 0 1 0 1 1 1 0 0 0 1 0 0 0 0 1 0 1 1 + 0 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 0 1 1 0 1 0 1 1 0 1 1 1 0 1 0 0 0 + 0 sum 1 1 0 1 0 1 1 0 1 1 1 0 1 0 0 0 checksum 0 0 1 0 1 0 0 1 0 0 0 10 1 1 13．Consider two hosts, Host A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/second, and the size of packet is L bits. （10分）(1)Express the propagation delay, d prop in terms of m and s.d prop=m/s(2)Determine the transmission delay d trans in terms L and R.d trans =L/R(3)If R=1000 Mbps, m=270km, s=3 108m/s and L=1500 Bytes, please compute：d prop=(270×103m)/(3×108m/s)=9×104/108s=9×10-4s=0.9ms=900μsd trans=(1500×8)/(1000×106)s=12000/109s=12μs=0.012ms=1.2×10-5s五、综合题（共30分）1．Match the following tasks to one of the TCP/IP layers: （7分）(1)Transmission of Data Segment ( 传送层)(2)ASCII changed to EBCDIC ( 应用层)(3)Transmission of data frame ( 链路层)(4)Provides log-in and logout procedures ( 应用层)(5)Selects a destination network ( 网络层)(6)Provides services for the network applications ( 应用层)(7)Provides E-mail and file transfer services ( 应用层)3. Please answer the following questions.(1) Consider the following HTTP GET message. Answer the following questions, indicatingwhere in the message below you find the answer. ( 4分)GET /info/2012-10-18/0-1-411.html http/1.1Host: User-agent: Mozilla/4.0Accept-language: enKeep-Alive:300Connection:keep-alive<cr><lf>Questions:a.What is the URL of the document requested by the browser?b.What version of HTTP is the browser running?c.Does the browser request a non-persistent or a persistent connection?d.What is domain name of the server?要点：a. /info/2012-10-18/0-1-411.html (绝对URL)b. http 1.1c. persistent connectiond. or (2)The text below shows the reply sent from the server in response the to the HTTP GET messagein the question above(1). Answer the following questions, indicating where in the message below you find the answer. (4 points)HTTP/1.1 200 OKContent-Length: 19572Content-Type: text/htmlLast-Modified: Sun, 30 Sep 2012 07:23:15 GMTAccept-Ranges: bytesServer: Microsoft-IIS/6.0X-Powered-By: Date: Thu, 25 Oct 2012 21:26:48 GMTConnection: Keep-Alive<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="/1999/xhtml"><head><Much more document text following here(not shown)>……Questions:a.Was the server able to successfully find the document or not?b.What time was the document reply provided?c.When was the document last modified?d.How many bytes are there in the document being returned?要点：a. yesb. Thu, 25 Oct 2012 21:26:48 GMTc. Sun, 30 Sep 2012 07:23:15 GMTd. 19572 bytes3．The below figure is the Selective-Repeat protocol’s sender and receiver views of slide windows and the algorithm of sender of SR prot ocol. （15分）(1) Please analyse what does the SR protocol differ from the GBN protocol.(2) Please analyse what actions does the sender take when(a) timeout event occurring;(b) the sender received ACK packet from the receiver?(3) Please give the FSM of SR sender.Fig 1 views of slide windowsFig 2 the algorithm of sender of SR protocol要点：（1）窗口尺寸不同；定时方式不同（单独、最小）；超时重传不同（单独、回退重传）；接收缓冲不同；确认方式不同（单独、累计）（2）（a）重传超时的包；（b）标记对应的包为已确认的包；如果是窗口下沿的包的确认，则推进窗口到未被确认的序号最低的包。