计算机网络期中考试题 B卷_Ans

计算机网络期中考试题B卷

基础概念题(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 receiver’s.

2.What are the two primary causes of packet loss?

Electrical noise that corrupts the signal and 2) buffer overflow in an intermediate node due to congestion.

3. A network's components can be divided into 3 parts? What are they?

Ans: 网络接入，网络边缘，网络核心

https://www.360docs.net/doc/e210952326.html, three email access protocols (that is, from user agent to SMTP server) for email

download. （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 user’s host, managed by user’s browser

4) back-end database at Web 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 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 is 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 following 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 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 web 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 flows. 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 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 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 with 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-oriented 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 UDP? 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

·应用层能更好地控制要发送的数据和发送时间（没有阻塞控制）

·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 & GBN（10 points）

What's the different between Go-back-N and Selective Repeat protocols.

a)GBN：

·累积ACK，如果有个gap，不会发ACK

·可以一次发最多N个包

·有一个最远未回复的包的计时器，维护一个发送窗口；如果超时，重发之前所有的包。

·报文信息简单，占用带宽多。

b)SR：

·每发一个包回复一个ACK

·每一个未ACK的包都有计时器，超时之后发还没回复ACK的包，会重新传输DNS—The Internet’s Directory Service (10 points)

1. please refer to various DNS servers below and explain the mechanism of recursive query.

Ans:

recursive query:

puts burden of name resolution on contacted name server

能正确写出具体流程即可。

2. Which server has the heaviest burden during recursive query?

Ans: root DNS server

流量控制与拥塞控制(20 points)

Assuming TCP Reno is the protocol experiencing the behavior shown above, answer the following questions. In all cases, you should provide a short discussion justifying your answer.

a.Give two reasons why slow start is used, and explain why it does a better job than

congestion avoidance for that function.

Ans:

1 TCP链接刚建立的时候需要发送大量的报文，慢启动可以防止建立连接初期单个

TCP链接发送太多报文占用太大带宽；

2慢启动有利于TCP从故障中恢复，并快速回到高效稳定的传输阶段慢启动比阻塞控制好的原因是慢启动在带宽空闲的时候可以快速增长TCP的窗口大小，TCP传输可以快速达到高效。

b.Identify the intervals of time when TCP congestion avoidance is operating. Why should

congestion avoidance be used instead of slow start during these intervals? Please clearly identify on specific reason.

Ans:

TCP congestion avoidance is operating in the intervals [6,16] and [17,22]

这些阶段TCP的窗口大小超过了设定的阈值，继续慢启动可能导致带宽占满，丢包，超时等不良结果。

c.After the 22nd transmission round, is segment loss detected by a triple duplicate ACK

or by a timeout?

Ans: After the 22nd transmission round, segment loss is detected due to timeout, and

hence the congestion window size is set to 1.

d.What is the value of Threshold at the 18th transmission round?

Ans: The threshold is set to half the value of the congestion window when packet loss is detected. When loss is detected during transmission round 16, the congestion windows size is 42. Hence the threshold is 21 during the 18th transmission round.

e.During what transmission round is the 50th segment sent?

Ans: During the 1st transmission round, packet 1 is sent; packet 2-3 are sent in the 2nd transmission round; packets 4-7 are sent in the 3rd transmission round; packets 8-15 are sent in the 4th transmission round; packets 16-31 are sent in the 5th transmission round; packets 32-63 are sent in the 6th transmission round; Thus packet 50 is sent in the 6th transmission round.

f.Suppose TCP Tahoe is used (instead of TCP Reno), and assume that triple duplicate

ACKs are received at the 16th round. What are the threshold and the congestion

window size at the 17th round?

Ans: Threshold is 21, and congestion window size is 1.

附加题：

(10pts)Appendix A contains the RDT sender and receiver FSMs as developed in the textbook. Suppose an application that uses RDT 3.0 somehow forces the receiver to keep sending back alternating ACK 0 and ACK 1 even if the corresponding data have not arrived at the receiver (so, this would be a channel with very low utilization given normal stop-and-wait operation). Would this change result in higher channel utilization? Why? Would there be any side-effects? Why? 2.5 each (10 pts total)

Ans:Yes, this actually causes the sender to send a number of pipelined data into the channel without waiting for a corresponding ACK. Yes, there are side effects. If data segments are lost in the channel, then the sender of rdt 3.0 won’t re-send those segments, unless there are some additional mechanism in the application to recover from loss.

Appendix A: