计网第二章

计算机网络第四版参考答案第二章第二章物理层（P66）1、物理层要解决哪些问题？物理层的主要特点是什么？答：（1）物理层要解决的主要问题：①物理层要尽可能屏蔽掉物理设备、传输媒体和通信手段的不同，使上面的数据链路层感觉不到这些差异的存在，而专注于完成本层的协议与服务。

②给其服务用户（数据链路层）在一条物理的传输媒体上传送和接收比特流（一般为串行按顺序传输的比特流）的能力。

为此，物理层应解决物理连接的建立、维持和释放问题。

③在两个相邻系统之间唯一地标识数据电路。

（2）物理层的主要特点：①由于在OSI之前，许多物理规程或协议已经制定出来了，而且在数据通信领域中，这些物理规程已被许多商品化的设备所采用。

加之，物理层协议涉及的范围广泛，所以至今没有按OSI的抽象模型制定一套新的物理层协议，而是沿用已存在的物理规程，将物理层确定为描述与传输媒体接口的机械、电气、功能和规程特性。

②由于物理连接的方式很多，传输媒体的种类也很多，因此，具体的物理协议相当复杂。

4、、物理层的接口有哪些方面的特性？各包含什么内容？答：(1)机械特牲说明接口所用接线器的形状和尺寸、引线数目和排列、固定和锁定装置等等。

(2)电气特性说明在接口电缆的哪条线上出现的电压应为什么范围。

即什么样的电压表示1或0。

(3)功能特性说明某条线上出现的某一电平的电压表示何种意义。

(4)规程特性说明对于不同功能的各种可能事件的出现顺序。

5、奈氏准则与香农公式在数据通信中的意义是什么？比特和波特有何区别？答：奈氏准则与香农公式的意义在于揭示了信道对数据传输率的限制，只是两者作用的范围不同。

奈氏准则给出了每赫带宽的理想低通信道的最高码元的传输速率是每秒2个码元。

香农公式则推导出了带宽受限且有高斯白噪声干扰的信道的极限信息传输速率C=Wlog2（1+S/N），其中W为信道的带宽（以赫兹为单位），S为信道内所传信号的平均功率，N为信道内部的高斯噪声功率。

比特和波特是两个完全不同的概念，比特是信息量的单位，波特是码元传输的速率单位。

Chapter 2注：括弧之中红色标注为与课本习题对应的编号2.1，2.2节后作业：1.(R2)For a communication session between a pair of processes, which process is the client andwhich is the server?答：The process is always-on host called the server.The process which initiates the communication is the client.2.(R3)What is the difference between network architecture and application architecture?答：Network architecture refers to the organization of the communication process into layers.Application architecture, on the other hand, is designed by an application developer and dicates the broad structure of the application.3.(R4)What information is used by a process running on one host to identify a process runningon another host?答：The IP address of the destination host and the port number of the destination socket.4.(R7)Suppose you wanted to do a transaction from a remote client to a server as fast aspossible. Would you use UDP or TCP? Why?答：We would use UDP. With UDP, the transaction can be completed in one roundtrip time – the client sends the transaction request into a UDP socket, and the server sends the reply back to the client’s UDP socket. With TCP, a minimum of two RTTs are needed – one to set-up the TCP connection, and another for the client to send the request, and for the server to send back the reply.5.(R8)List the four broad classes of services that a transport protocol can provide. For each ofthe service classes, indicate if either UDP or TCP (or both) provides such a service.答：(1)Reliable Data Transfer: TCP provides a reliable byte-stream between client and server butUDP does not.(2)Throughput: A guarantee that a certain value for throughput will be maintained; Neither.(3)Timing: A guarantee that data will be delivered within a specified amount of time;Neither.(4)Security: Neither.6.(R9)Why do HTTP, FTP, SMTP, and POP3 run on top of TCP rather than on UDP?答：The applications associated with those protocols require that all application data be received in the correct order and without gaps. TCP provides this service whereas UDP does not.7.(R11)What is meant by a handshaking protocol?答：A protocol uses handshaking if the two communicating entities first exchange control packetsbefore sending data to each other. SMTP uses handshaking at the application layer whereas HTTP does not.2.3，2.4节后作业8.(R15)Suppose Alice, with a Web-based e-mail account (such as Hotmail or gmail), sends amessage to Bob, who accesses his mail from his mail server using POP3. Discuss how the message gets from Alice's host to Bob's host. Be sure to list the series of application-layer protocols that are used to move the message between the two hosts.答：Message is sent from Alice’s host to her mail server over HTTP. Alice’s mail server then sends the message to Bob’s mail server over SMTP. Bob then transfers the message from his mail server to his host over POP3.9.(R17)From a user's perspective, what is the difference between the download-and-deletemode and the download-and-keep mode in POP3?答：With download and delete, after a user retrieves its messages from a POP server, the messages are deleted. This poses a problem for the nomadic user, who may want to access the messages from many different machines. In the download and keep configuration, messages are not deleted after the user retrieves the messages. This can also be inconvenient, as each time the user retrieves the stored messages from a new machine, all of non-deleted messages will be transferred to the new machine.10.(R18)Is it possible for an organization's Web server and mail server to have exactly the samealias for a hostname (for example,foo. com)? What would be the type for the RR that contains the hostname of the mail server?答：Yes, an organization’s mail server and Web server can have the same alias for a host name.The MX record is used to map the mail server’s host name to its IP address.11.(R19)Why is it said that FTP sends control information "out-of-band"?答：FTP uses two parallel TCP connections, one connection for sending control information and another connection for actually transferring the file. Because the control information is not sent over the same connection that the file is sent over, FTP sends control information out of band.2.5，2.6节后作业12.(P1)True or false?a. A user requests a Web page that consists of some text and three images. For this page, the client will send one request message and receive four response messages.b. Two distinct Web pages (for example, /research.Html and /students. html) can be sent over the same persistent connection.c. With nonpersistent connections between browser and origin server, it is possible for a single TCP segment to carry two distinct HTTP request messages.d. The Date: header in the HTTP response message indicates when the object in the response was last modified.答：a. Fb. Tc. Fd. F13. (P7)Suppose within your Web browser you click on a link to obtain a Web page. The IPaddress for the associated URL is not cached in your local host, so a DNS lookup is necessary to obtain the IP address. Suppose that n DNS servers are visited before your host receives the IP address from DNS; the successive visits incur an RTT of n RTT RTT ,...,1. Further suppose that the Web page associated with the link contains exactly one object, consisting of a small amount of HTML text. Let 0RTT denote the RTT between the local host and the server containing the object. Assuming zero transmission time of the object, how much time elapses from when the client clicks on the link until the client receives the object?答：The time to get IP address is RTT 1+RTT 2+……+RTT n , once the IP address is known, the response time is 2RTT 0+RTT 1+……+RTT n .14. (P8)Referring to Problem P7, suppose the HTML file references three very small objects onthe same server. Neglecting transmission times, how much time elapses witha. Non-persistent HTTP with no parallel TCP connections?b. Non-persistent HTTP with parallel connections?c. Persistent HTTP (with pipelining)?答：a. RTT 1+RTT 2+……+RTT n +2RTT 0+3×2RTT 0=8RTT 0+RTT 1+……+RTT n .b. RTT 1+RTT 2+……+RTT n +2RTT 0+2RTT 0=4RTT 0+RTT 1+……+RTT n .c. RTT 1+RTT 2+……+RTT n +2RTT 0+RTT 0=3RTT 0+RTT 1+……+RTT n .2.7，2.8节后作业15.(P9)Consider Figure2.12, for which there is an institutional network connected to the Internet.Suppose that the average object size is 900,000 bits and that the average request rate from the institution’s browsers to the origin servers is 10 requests per second. Also suppose that the amount of time it takes from when the routers on the Internet side of the access link forwards an HTTP request until it receives the response is two seconds on average (see Section 2.2.5).Model the total average response time as the sum of the average access delay (that is, the delay from Internet router to institution router) and the average Internet delay. For the average access delay, use△/(1-△β), where △is the average time required to send an object over the access link andβis the arrival rate of objects to the access link.a.Find the total average response time.b.Now suppose a cache is installed in institution LAN. Suppose the hit rate is 0.6. Find thetotal response time.答：a.The time to transmit an object of size L over a link or rate R is L/R.The average time is the average size of the object divided by R : △=(900000 bits)/(15000000bits/sec)=0.06secThe traffic intensity on the link is ( 10 requests/sec)(0.06msec/request )=0.6The average access delay is ( 0.06sec )/( 1-0.6 )=0.15secThe total average response time is therefore 0.6sec+2sec=2.15sec.b.The average access delay is ( 0.06sec )/( 1-0.6×0.4 )=0.0456secWith cache and the hit rate is 0.6 , when miss the cache the average response time is0.0456sec+2sec=2.0456secThe average response time is (1-0.6)×2.0456sec=0.832sec16.(P20)In our coverage of an overlay network using query flooding in Section 2.6, we describein some detail how a new peer joins the overlay network. In this problem we want to explore what happens when a peer leaves the overlay network. Suppose every participating peer maintains TCP connections to at least four distinct peers at all times. Suppose Peer X, which has five TCP connections to other peers, wants to leave.a.First consider the case of a graceful departure, that is, Peer X explicitly close itsapplication, thereby graceful closing its five TCP connections. What actions would each of the five formerly connected peers take?b.Now suppose that Peer X abruptly disconnects from the Internet without notifying itsfive neighbors that is closing the TCP connections. What would happen?答：a.Each of the five Gnutella clients immediately learns that it has one less neighbor. Theone of these five clients called A. Suppose A has only three neighbors after X drops out .Then A needs to establish a TCP connection with another peer . A should have a fresh list of active peers; he sequentially contacts peers on this list until one accepts his TCP connection attempt.b.We still use A. A doesn’t immediately know that X has departed. A will only learn aboutX’s departure when it attempts to send a message (query or ping) to X. When A attempts to send a message, A’s TCP will make several unsuccessful attempts to send the message to B. A’s TCP will then inform the Gnutella client that X is down. A will try to establish a TCP connection with a new peer to rebuild a fifth connection.17.(P21)Consider query flooding, as discussed in Section 2.6. Suppose that each peer isconnected to at most N neighbors in the overlay network. Also suppose that the node-count field is initially set to K. Suppose Alice makes a query. Find an upper bound on the number of query messages that are sent into the overlay network.答：The maximum number of query messages is N+NM+NM2+……+NM(K-1=N[(N-1)K-1]/(N-2))附加选做题18.在TCP协议中,为了使通信不致发生混乱,引入了所谓套接字的概念,这里,套接字由( )和IP地址两部分组成。

计算机网络技术第二章知识点在计算机网络技术的学习中，第二章通常会涵盖一些关键的概念和技术，这些知识对于我们深入理解网络的运行原理和工作方式至关重要。

首先，我们来谈谈数据通信基础。

数据在网络中传输，就像是货物在运输线上流动。

数据通信的基本任务就是要将这些“货物”准确、快速地从源点传送到目的地。

这涉及到信号的类型，比如模拟信号和数字信号。

模拟信号就像连续的波浪，是平滑变化的；而数字信号则是离散的，像一个个台阶。

在数据传输中，还有一个重要的概念是带宽。

带宽可以理解为信息公路的宽度，它决定了在单位时间内能够传输的数据量。

带宽越大，就像公路越宽，能同时通过的车辆越多，数据传输速度也就越快。

数据传输方式也有多种。

比如串行传输和并行传输。

串行传输是一位一位地依次传输数据，就像单人依次通过狭窄的通道；并行传输则是多位数据同时传输，如同多人同时通过宽敞的大门。

但并行传输在距离较长时，由于信号同步等问题，成本较高，所以在长距离传输中，串行传输更为常见。

接着，我们说一说信道复用技术。

这就好比在一条道路上，通过巧妙的安排，让多个用户能够同时使用，提高道路的利用率。

常见的信道复用技术有时分复用、频分复用和波分复用。

时分复用是按照时间划分，不同用户在不同的时间段使用信道；频分复用则是根据频率来划分，每个用户使用不同的频率范围；波分复用主要用于光通信，是按照光的波长来划分信道。

差错控制也是第二章的一个重要知识点。

在数据传输过程中，由于各种干扰和噪声，可能会出现错误。

差错控制技术的目的就是检测和纠正这些错误，确保数据的准确性。

常见的差错控制方法有检错码和纠错码。

检错码只能检测出错误，但不能纠正；而纠错码不仅能检测错误，还能纠正错误，但实现起来通常更加复杂。

然后是数据交换技术。

这类似于在不同的地点之间传递物品，有不同的传递方式。

电路交换就像是建立一条专用的通道，在通信期间一直占用；报文交换则是把要发送的信息整体作为一个报文，通过存储转发的方式进行传输；分组交换则是把报文分成一个个较小的分组进行传输，提高了传输效率和灵活性。