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1、本文格式为Word版，下载可任意编辑计算机网络期中考试题 B卷 计算机网络期中考试题B卷 根基概念题(10 points) 1. What is the difference between congestion control and flow control? Congestion control prevents overrunning buffers in the network, while flow control prevents overflowing the receivers. 2. What are the two primary causes of packet los
2、s? Electrical noise that corrupts the signal and 2) buffer overflow in an intermediate node due to congestion. 3. A networks components can be divided into 3 parts? What are they? Ans: 网络接入，网络边缘，网络核心 4. Name three email access protocols (that is, from user agent to SMTP server) for email download. （
3、2pts total ,1 pt each item） POP, IMAP, and HTTP 5. Many Web sites use cookies. List the four components. four components: 1) cookie header line of HTTP response message 2) cookie header line in HTTP request message 3) cookie file kept on users host, managed by users browser 4) back-end database at W
4、eb site Delay: (10points)Name the four factors of delay for a packet. Then, identify which factor will most likely predominate (i.e., be the largest factor) for 1000 Byte packets in a flow on a 1) 10 Mb/s LAN segment between two PCs in the same building, 2) 1 Mb/s geosynchronous-orbit satellite link
5、 between the US and Nepal, 3) on the Internet between the US and Nepal. 1 pt each identified largest factor (3 pts total) (8pts)The four factors of packet delay are processing, transmission, propagation, and queueing. For (1) the predominant delay will be transmission in the 1 ms range (processing i
6、s microseconds, propagation is nanoseconds, and queueing in non-existent), For (2) the propagation will be the largest delay (in the 100s of milliseconds), and for (3) the largest delay will likely be queueing due to the many hops between the US and Nepal. Application layer (15 points) Answer the fo
7、llowing questions regarding the Application Layer. a) In general, an application layer protocol defines four things. What are these four things? 1 pt each thing (4pts) Ans:An application layer protocol defines 1) types of messages, 2) syntax of messages, 3) semantics of messages, and 4) timing (when
8、 and how) b) What is a web cache used for (i.e., what benefits are derived from using a web cache)? Where can a web cache be located or placed? 2 pt each subquestion (4pts) Ans：A web cache may be used to reduce response time as experienced by a user, reduce load on a link, and/or reduce load on a we
9、b server. A web cache may be placed in the user client (e.g., within the browser application), at “my” network edge (we call this a proxy server), and/or at the edge of the network containing the server (we call this a transparent cache) c) Describe how email works. Describe the key components and f
10、lows. Identify key standards that apply. Use figures as needed.(7pts) Ans: Email is an asynchronous communications medium based on sent and received text messages (may include non-text attachments). The three major components of email are user agents, mail servers, and the SMTP (Simple Mail Transfer
11、 Protocol). SMTP is used to transfer messages between mail servers using TCP/IP (client/server). SMTP uses ASCII commands and headers. Commands are not authenticated. Between a mail server and a user agent, POP (Post Office Protocol), IMAP (Internet Mail Access Protocol), or HTTP (HyperText Transfer
12、 Protocol) is used to access received mail stored in inboxes on the mail server. Mail servers have a fixed IP address and are always powered-on. The user agents need not always be powered-on and also need not be fixed in location or IP address. The figure shows a user agent accessing a mail server w
13、ith POP, two mail servers exchanging messages using SMTP, and another user agent accessing its mail server using IMAP. user agent server server user agent +-+ POP +-+ SMTP +-+ IMAP +-+ | |-| |-| |-| | | | | | +-+ +-+ +-+ +-+ |-| The Internet Transportation layer(15 points) TCP&UDP 1. In connection-o
14、riented demux, a TCP socket is identified by: Select one or more:( ) a. source IP address b.source port number c. ACK number. d. Sequence number e. dest port number f. dest IP address The correct answer is: source IP address, source port number, dest IP address, dest port number 2. Why is there a UD
15、P? List the 4 advantages. no connection establishment (which can add delay) simple: no connection state at sender, receiver small segment header no congestion control: UDP can blast away as fast as desired 应用层能更好地操纵要发送的数据和发送时间（没有阻塞操纵） 无需连接建立 无连接状态 分组首部开销小 3. Checksum calculation: given the 4 16-bit numbers in Hexadecimal(十六进制): 0x0001, 0xf203, 0xf4f5, 0xf6f7.Please give answer in Hexadecimal. Ans:把四个数累加，溢出的就回卷即可。 the sum is 0xddf2, and checksum is 0x220d. SR &
