计算机网络第一次测试1
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Computer Network 2011-2012-1学期第一次测试
习题1 （快速完成，任选8个，5 * 8=40）
A．Internet protocol stack（互联网协议栈）包含哪五层？每层的主要功能是什么？
B．比较circuit switched networks 和 packet switched networks，各给出一个优点和缺点。
C．在packet switched networks中，影响end-to-end delay（端到端时延）的四个因素是什么？（5）D．DNS使用TCP还是UDP？Why？（5）
E．Discuss one advantage and one disadvantage of having a ‘stateful’ protocol for applications. F． Describe how Web caching can reduce the delay in receiving a requested object. Will Web caching reduce the delay for all objects requested by a user or for only some of the objects? Why? G．Push vs. Pull: Give examples of a push protocol and a pull protocol
H.POP3和IMAP之间的两个不同之处是什么？
习题2 （每个10分，共10）
A.For the following two scenarios（方案）, what contributes most to the total delay: queuing
delay（排队延时）, propagation delay （传播延时）or transmission delay（传输延时）? Scenario 1: A 1500 byte packet transmitted over a 10 Mbps lightly loaded Ethernet network. Scenario 2: A 1500 byte packet transmitted over a 10Gbps high-speed network from Sydney to Tokyo. 技能综合题3 (25+25，完成后发至zhuyachao@)
A.Ping程序用于检查到Internet上任一主机的往返时延RTT，请ping以下主机：(1) (Cambridge, MA),(2) (Austin, TX),(3) (Madison, WI),(4) (Los Angeles, CA), (5)msu.ru (Moscow,Russia),(6) (Paris, France),(7) www.u‐tokyo.ac.jp (Tokyo, Japan), (8)mu.ac.in(Mumbai,
India),(9)(Wuhan,China), (10)(Beijing, China),(11) .hk(Hongkong, China),(12).tw(Taipei, Taiwan, China)。
（1）请使用gnuplot或ploticus或excel绘图，其中横坐标为从柳州到各个城市之间的距离，纵坐标为RTT/light time。
RTT由ping程序得到，light time为光束在地球表面以最短距离到达目的地并返回的时间（光速以3*108m/计）。
地理距离的获得方式有两种：一是使用地理坐标（从网站
/latlong.html获得）；二是通过计算两个位置的距离的网站
（/projects-google-maps-distance-calculator.htm）获得。
（2）解释在（1）中绘出的图中为什么所以点的纵坐标值>=1。
（3）使用traceroute程序列出packet到达目的地时遇到的所有路由器（在Windows和Cygwin上使用“tracert”,Linux使用“traceroute”）；对题目中的所有主机都应用traceroute程序，并据获得的结果绘图，要求Y轴为路由器数目，X轴为地理距离。
从图中你观察到什么现象？解释之。
B.Suppose within your web browser, you click on a link. A 1kByte file (index.html) with
five1.5kByte embedded images is downloaded. Assume that all images are hosted on the same web server as the index page. The RTT (round-trip time) between your computer and the server is 50ms (constant). The delay incurred in resolving the IP address of the web server is negligible. Assume that the network path between your browser and web server is equivalent to a 1.5 Mbps link.
Note: Remember the difference between kByte (kilobyte) and kb (kilobit). 1KB = 8Kb = 8192 bits. Important Note: Marks were not deducted for minor differences in calculations. For example, if you assumed 1kB = 8000 bits, no marks were deducted.
1. What is the total download time with one non-persistent connection?
2. What is the total download time with two parallel non-persistent connections?
3. What is the total download time with a persistent connection?。
