计算机网络第七版 第九章

计算机网络第七版答案第一章概述1-01 计算机网络向用户可以提供那些服务？答：连通性和共享1-02 简述分组交换的要点。

答：（1）报文分组，加首部（2）经路由器储存转发（3）在目的地合并1-03 试从多个方面比较电路交换、报文交换和分组交换的主要优缺点。

答：（1）电路交换：端对端通信质量因约定了通信资源获得可靠保障，对连续传送大量数据效率高。

（2）报文交换：无须预约传输带宽，动态逐段利用传输带宽对突发式数据通信效率高，通信迅速。

（3）分组交换：具有报文交换之高效、迅速的要点，且各分组小，路由灵活，网络生存性能好。

1-04 为什么说因特网是自印刷术以来人类通信方面最大的变革？答：融合其他通信网络，在信息化过程中起核心作用，提供最好的连通性和信息共享，第一次提供了各种媒体形式的实时交互能力。

1-05 因特网的发展大致分为哪几个阶段？请指出这几个阶段的主要特点。

答：从单个网络APPANET向互联网发展；TCP/IP协议的初步成型建成三级结构的Internet；分为主干网、地区网和校园网；形成多层次ISP结构的Internet；ISP首次出现。

1-06 简述因特网标准制定的几个阶段？答：（1）因特网草案(Internet Draft) ——在这个阶段还不是RFC 文档。

（2）建议标准(Proposed Standard) ——从这个阶段开始就成为RFC 文档。

（3）草案标准(Draft Standard)（4）因特网标准(Internet Standard)1-07小写和大写开头的英文名internet 和Internet在意思上有何重要区别？答：（1）internet（互联网或互连网）：通用名词，它泛指由多个计算机网络互连而成的网络。

；协议无特指（2）Internet（因特网）：专用名词，特指采用TCP/IP 协议的互联网络。

区别：后者实际上是前者的双向应用1-08 计算机网络都有哪些类别？各种类别的网络都有哪些特点？答：按范围：（1）广域网WAN：远程、高速、是Internet的核心网。

第九章无线网络9-01．无线局域网都由哪几部分组成？无线局域网中的固定基础设施对网络的性能有何影响？接入点AP 是否就是无线局域网中的固定基础设施？答：无线局域网由无线网卡、无线接入点(AP)、计算机和有关设备组成，采用单元结构，将整个系统分成许多单元，每个单元称为一个基本服务组。

所谓“固定基础设施”是指预先建立起来的、能够覆盖一定地理范围的一批固定基站。

直接影响无线局域网的性能。

接入点AP 是星形拓扑的中心点，它不是固定基础设施。

9-02．Wi-Fi 与无线局域网WLAN 是否为同义词？请简单说明一下。

答：Wi-Fi 在许多文献中与无线局域网WLAN 是同义词。

802.11 是个相当复杂的标准。

但简单的来说，802.11 是无线以太网的标准，它是使用星形拓扑，其中心叫做接入点AP(Access Point)，在MAC 层使用CSMA/CA 协议。

凡使用802.11系列协议的局域网又称为Wi-Fi(Wireless-Fidelity,意思是“无线保真度”)。

因此，在许多文献中，Wi-Fi 几乎成为了无线局域网WLAN 的同义词。

9-03 服务集标示符SSID 与基本服务集标示符BSSID 有什么区别？答：SSID（Service Set Identifier）AP 唯一的ID 码，用来区分不同的网络，最多可以有32 个字符，无线终端和AP 的SSID 必须相同方可通信。

无线网卡设置了不同的SSID 就可以进入不同网络，SSID 通常由AP 广播出来，通过XP 自带的扫描功能可以相看当前区域内的SSID。

出于安全考虑可以不广播SSID，此时用户就要手工设置SSID 才能进入相应的网络。

简单说，SSID 就是一个局域网的名称，只有设置为名称相同SSID 的值的电脑才能互相通信。

BSS 是一种特殊的Ad-hoc LAN 的应用，一个无线网络至少由一个连接到有线网络的AP 和若干无线工作站组成，这种配置称为一个基本服务装置BSS (Basic Service Set)。

计算机网络第七版谜底欧阳家百（2021.03.07）第一章概述101 计算机网络向用户可以提供那些办事？答：连通性和共享102 简述分组交换的要点。

答：（1）报文分组，加首部（2）经路由器贮存转发（3）在目的地合并103 试从多个方面比较电路交换、报文交换和分组交换的主要优缺点。

答：（1）电路交换：端对端通信质量因约定了通信资源获得可靠包管，对连续传送年夜量数据效率高。

（2）报文交换：无须预约传输带宽，静态逐段利用传输带宽对突发式数据通信效率高，通信迅速。

（3）分组交换：具有报文交换之高效、迅速的要点，且各分组小，路由灵活，网络生存性能好。

104 为什么说因特网是自印刷术以来人类通信方面最年夜的变动？答：融合其他通信网络，在信息化过程中起核心作用，提供最好的连通性和信息共享，第一次提供了各种媒体形式的实时交互能力。

105 因特网的成长年夜致分为哪几个阶段？请指出这几个阶段的主要特点。

答：从单个网络APPANET向互联网成长；TCP/IP协议的初步成型建成三级结构的Internet；分为主干网、地区网和校园网；形成多条理ISP结构的Internet；ISP首次呈现。

106 简述因特网标准制定的几个阶段？答：（1）因特网草案(Internet Draft) ——在这个阶段还不是RFC 文档。

（2）建议标准(Proposed Standard) ——从这个阶段开始就成为RFC 文档。

（3）草案标准(Draft Standard)（4）因特网标准(Internet Standard)107小写和年夜写开头的英文名internet 和Internet在意思上有何重要区别？答：（1）internet（互联网或互连网）：通用名词，它泛指由多个计算机网络互连而成的网络。

；协议无特指（2）Internet（因特网）：专用名词，特指采取TCP/IP 协议的互联网络。

区别：后者实际上是前者的双向应用108 计算机网络都有哪些类别？各种类另外网络都有哪些特点？答：按规模：（1）广域网WAN：远程、高速、是Internet的核心网。

计算机网络谢希仁第七版课后答案完整版1. 概述计算机网络是当今社会发展不可或缺的一部分，它负责连接世界各地的计算机和设备，提供信息交流和资源共享的便利。

而谢希仁的《计算机网络》第七版是一本经典的教材，旨在帮助读者深入了解计算机网络的原理、技术和应用。

本文将提供《计算机网络谢希仁第七版》全部课后答案的完整版本，以便帮助读者更好地掌握该教材的知识点。

2. 第一章：绪论本章主要介绍了计算机网络的基本概念和发展历程。

通过学习本章，读者将了解到计算机网络的定义、功能和分类，以及互联网的起源和发展。

3. 第二章：物理层物理层是计算机网络的基础，它负责传输原始比特流。

本章对物理层的相关内容进行了全面的介绍，包括数据通信基础、传输媒介、信道复用技术等。

4. 第三章：数据链路层数据链路层负责将原始比特流划分为以太网帧等数据包进行传输。

本章详细介绍了数据链路层的各种协议和技术，如以太网、局域网、无线局域网等。

5. 第四章：网络层网络层是计算机网络中最关键的一层，它负责将数据包从源主机传输到目标主机。

本章对网络层的相关内容进行了深入研究，包括互联网协议、路由算法、IP地址等。

6. 第五章：传输层传输层负责提供端到端的可靠数据传输服务。

本章对传输层的相关知识进行了细致的讲解，包括传输层协议的设计原则、TCP协议、UDP协议等。

7. 第六章：应用层应用层是计算机网络中最高层的一层，它负责向用户提供各种网络应用服务。

本章详细介绍了应用层的相关内容，包括HTTP协议、DNS协议、电子邮件等。

8. 第七章：网络安全与管理网络安全和管理是计算机网络中不可忽视的重要方面。

本章对网络安全和管理的相关内容进行了全面的阐述，包括网络安全威胁、防火墙、入侵检测系统等。

9. 第八章：多媒体网络多媒体网络是指能够传输音频、视频等多种媒体数据的计算机网络。

本章介绍了多媒体网络的相关技术和应用，包括流媒体、语音通信、视频会议等。

10. 第九章：计算机网络的高级话题本章涵盖了计算机网络中的一些高级话题，如网络性能评价、网络协议的形式化描述方法、无线和移动网络等。

计算机网络第七版答案第一章概述1-01 计算机网络向用户可以提供那些服务？答：连通性和共享1-02 简述分组交换的要点。

答：（1）报文分组，加首部（2）经路由器储存转发（3）在目的地合并1-03 试从多个方面比较电路交换、报文交换和分组交换的主要优缺点。

答：（1）电路交换：端对端通信质量因约定了通信资源获得可靠保障，对连续传送大量数据效率高。

（2）报文交换：无须预约传输带宽，动态逐段利用传输带宽对突发式数据通信效率高，通信迅速。

（3）分组交换：具有报文交换之高效、迅速的要点，且各分组小，路由灵活，网络生存性能好。

1-04 为什么说因特网是自印刷术以来人类通信方面最大的变革？答：融合其他通信网络，在信息化过程中起核心作用，提供最好的连通性和信息共享，第一次提供了各种媒体形式的实时交互能力。

1-05 因特网的发展大致分为哪几个阶段？请指出这几个阶段的主要特点。

答：从单个网络APPANET向互联网发展；TCP/IP协议的初步成型建成三级结构的Internet；分为主干网、地区网和校园网；形成多层次ISP结构的Internet；ISP首次出现。

1-06 简述因特网标准制定的几个阶段？答：（1）因特网草案(Internet Draft) ——在这个阶段还不是 RFC 文档。

（2）建议标准(Proposed Standard) ——从这个阶段开始就成为 RFC 文档。

（3）草案标准(Draft Standard)（4）因特网标准(Internet Standard)1-07小写和大写开头的英文名internet 和Internet在意思上有何重要区别？答：（1） internet（互联网或互连网）：通用名词，它泛指由多个计算机网络互连而成的网络。

；协议无特指（2）Internet（因特网）：专用名词，特指采用 TCP/IP 协议的互联网络。

区别：后者实际上是前者的双向应用1-08 计算机网络都有哪些类别？各种类别的网络都有哪些特点？答：按范围：（1）广域网WAN：远程、高速、是Internet的核心网。

计算机⽹络⾃顶向下第七版第九章答案Computer Networking: A Top-Down Approach,7th Edition计算机⽹络⾃顶向下第七版Solutions to Review Questions and ProblemsChapter 9 Review Questions1.2.Spatial Redundancy: It is the redundancy within a given image. Intuitively, an imageconsists of mostly white space has a high degree of redundancy and can be efficiently compressed without significantly sacrificing image quality.Temporal Redundancy reflects repetition from image to subsequent image. If, for example, an image and the subsequent image are exactly the same, there is no reason to re-encode the subsequent image; it is instead more efficient simply to indicate during encoding that the subsequent image is exactly the same. If the two images are very similar, it may be not efficient to indicate how the second image differs from the first, rather than re-encode the second image.3.Quantizing a sample into 1024 levels means 10 bits per sample. The resulting rate ofthe PCM digital audio signal is 160 Kbps.4.Streaming stored audio/video: In this class of applications, the underlying medium isprerecorded video, such as a movie, a television show, or a prerecorded sportingevent. These prerecorded videos are played on servers, and users send requests to the servers to view the videos on demand. Many internet companies today providestreaming video, including YouTube, Netflix, and Hulu.Conversational Voice- and Video-over-IP: Real-time conversational voice over the Internet is often referred to as Internet telephony, since, from the user’s perspective, it is similar to the traditional circuit-switched telephone service. It is also commonly called Voice-over-IP (VOIP).Conversational video is similar except that it includes the video of the participants as well as their voices. Conversational voice and video are widely used inthe Internet today, with the Internet companies like Skype and Google Talk boasting hundreds of millions of daily users. Streaming Live Audio and Video: These applications allow users to receive a live radio or television transmission over the Internet. Today, thousands of radio and television stations around the world are broadcasting content over the internet.5.UDP Streaming: With UDP streaming, the server transmits video at a rate thatmatches the client’s video consumption rate by clocking out the video chunks over UDP at a steady rate.HTTP Streaming: In HTTP streaming, the video simply stored in an HTTP server as ordinary file with a specific URL. When a user wants to see the video, the client establishes a TCP connection with the server and issues an HTTP GET request for that URL. The server then sends the video file, within an HTTP response message, as quickly as possible, that is, as quickly as TCP congestion control and flow control will allow.Adaptive HTTP Streaming (DASH): In Dynamic Adaptive Streaming over HTTP, the video is encoded several different versions, with each version having a different bit rate and, correspondingly, a different quality level. The client dynamically requests the chunks of video segments of a few seconds in length from the different versions. When the amount of available bandwidth is high, the client naturally selects chunks from a high-rate version; and when the available bandwidth is low, itnaturally selects from a low-rate version.6.The three significant drawbacks of UDP Streaming are:1.Due to unpredictable and varying amount of available bandwidth between serverand client, constant-rate UDP streaming can fail to provide continuous play out.2.It requires a media control server, such as an RTSP server, to process client-to-server interactivity requests and to track client state for each ongoing clientsession.3.Many firewalls are configured to block UDP traffic, preventing users behindthese firewalls from receiving UDP video.7.No. On the client side, the client application reads bytes from the TCP receive bufferand places the bytes in the client application buffer.8.The initial buffering delay is t p = Q/x = 4 seconds.9.End-to-end delay is the time it takes a packet to travel across the network from sourceto destination. Delay jitter is the fluctuation of end-to-end delay from packet to the next packet.10.A packet that arrives after its scheduled play out time cannot be played out.Therefore, from the perspective of the application, the packet has been lost.11.First scheme: send a redundant encoded chunk after every n chunks; the redundantchunk is obtained by exclusive OR-ing the n original chunks. Second scheme: send a lower-resolution low-bit rate scheme along with the original stream. Interleaving does not increase the bandwidth requirements of a stream.12.RTP streams in different sessions: different multicast addresses; RTP streams in thesame session: SSRC field; RTP packets are distinguished from RTCP packets by using distinct port numbers.13.The role of a SIP registrar is to keep track of the users and their corresponding IPaddresses which they are currently using. Each SIP registrar keeps track of the users that belong to its domain. It also forwards INVITE messages (for users in its domain) to the IP address which the user is currently using. In this regard, its role is similar to that of an authoritative name server in DNS.Chapter 9 ProblemsProblem 1a)Client begins playout as soon as first block arrives at t1 and video blocks are to beplayed out over the fixed amount of time, d. So it follows that second video block should be arrived before time t1 + d to be played at right time, third block att1 + 2d and so on. We can see from figure that only blocks numbered 1,4,5,6 arrive at receiver before their playout times. b)Client begins playout at time t1 + d and video blocks are to be played out over thefixed amount of time, d. So it follows that second video block should be arrived before time t1 +2d to be played at right time, third block at t1 + 3d and so on. We can see from figure that video blocks numbered from 1 to 6 except 7 arrive atreceiver before their playout times.c)Maximum two video blocks are ever stored in the client buffer. Video blocksnumbered 3 and 4 arrive before t1 + 3d and after t1 + 2d, hence these two blocks are stored in the client buffer. Video block numbered 5 arrives before time t1 + 4d and after t1 + 3d, which is stored in the client buffer along with already stored videoblock numbered 4.d)The smallest playout at the client should be t1 + 3d to ensure that every block hasarrived in time.Problem 2a)During a playout period, the buffer starts with Q bits and decreases at rate r - x.Thus, after Q/(r - x) seconds after starting playback the buffer becomes empty.Thus, the continuous playout period is Q/(r - x) seconds. Once the buffer becomes empty, it fills at rate x for Q/x seconds, at which time it has Q bits and playback begins. Therefore, the freezing period is Q/x seconds.b)Time until buffer has Q bits is Q/x seconds. Time to add additional B - Q bits is(B - Q)/(x - r) seconds. Thus the time until the application buffer becomes full isQ x + B?Qx?rseconds.Problem 3a)The server’s average send rate is H/2 .b)This part (b) is an odd question and will be removed from the next edition. Afterplaying out the first frame, because x(t) < r, the next frame will arrive after the scheduled playout time of the next frame. Thus playback will freeze afterdisplaying the first frame.c)Let q(t) denote the number of bits in the buffer at time t. Playout begins when q(t)= Q. Let’s assume throughout this problem that HT/2 ≥ Q, so that q(t) = Q by the end of the first cycle for x(t). We haveq(t)=∫HT sdst=Ht2/2T.Therefore, q (t) = Q when t = √2QT/H= tp.d)At time t = T, q (t) = HT/2 = Q, so that playout begins. If subsequently there is nofreezing, we need q(t + T) > 0 for all t ≥ T, we haveq(t+T)=HT2?rt+∫ x(s)dst> H2(T?t)+∫ x(s)dstWith t = nT + ?, with 0 < ? < T, we have from above2(T?nT??)+nHT2+H?22T= H2(T??+?2T)Which is easily seen to be possible for all 0 < ? < T.e)First consider the [0, T]. We haveq(t)=H2Tt2?r(t?tp)for t p ≤ t≤ Tq (t) is minimized at t = rT/H. It can then be shown that q (rT/H) ≥ 0 if and only if t p≥ rT/2H. Furthermore, if t p = rT/2H, proof can be extended to show q (t) > 0 for all t ≥ T thus, tp < rT/2H and Q = r2T/8H.f)This is a very challenging problem. Assuming that B is reached before time T,then tf is solution to H2Tt2?r(t? √2QT/H)= B.Problem 4a) Buffer grows at rate x – r. At time E, (x - r)*E bits are in buffer and are wasted.b) Let S be the time when the server has transmitted the entire video. If S > E, buffer grows at rate x – r until time E, so the waste is again (x - r)*E. If S < E, then attime E there are T – E seconds of video still to be played in buffer. So the waste is r* (T - E).Problem 5a) N*N = N 2.b) N+N = 2NProblem 6a) h +160 bytes are sent every 20 msec. Thus the transmission rate isKbpsh Kbps h )4.64(208)160(+=?+ b)IP header: 20 bytes UDP header: 8 bytes RTP header: 12 bytesc) h=40 bytes (a 25% increase in the transmission rate!)a) Denote )(n d for the estimate after the n th sample.44)1(t r d -=))(1()(4433)2(t r u t r u d --+-=[]))(1()()1()(443322)3(t r u t r u u t r u d --+--+-=)()1()()1()(4423322t r u t r u u t r u --+--+-=)3(11)4()1()(d u t r u d -+-=)()1()()1()()1()(4433322211t r u t r u u t r u u t r u --+--+--+-=b))()1()()1(11)(n n n j j n j j n t r u t r u u d--+--=∑-=c))()1(11)(j j j j t r u u u d---=∑∞=∞)(9.911j j j jt r -=∑∞=The weight given to past samples decays exponentially. Problem 8a) Denote )(n v for the estimate after the n th sample. Let j j j t r -=?. )1(4)1(d v -?= (=0))1(4)2(3)2()1(d u d u v -?-+-?=)2()3(2)3()1(v u d u v -+-?=)1(42)2(3)3(2)1()1(d u d u u d u -?-+-?-+-?=)3()4(1)4()1(v u d u v -+-?=)2(32)3(2)4(1)1()1(d u u d u u d u -?-+-?-+-?=)1(43)1(d u -?-+[])2(32)3(2)4(1)1()1(d u d u d u -?-+-?-+-?=)1(43)1(d u -?-+ b))1()1(111)()1()1(d u d u u v n n j n j n j j n -?-+-?-=+--=-∑Problem 9a) r 1 – t 1 + r 2 - t 2 + …+r n-1-t n-1 = (n-1)d n-1Substituting this into the expression for d n givesn t r d n n d n n n n -+-=-11b) The delay estimate in part (a) is an average of the delays. It gives equal weight to recent delays and to “old” delays. The delay estimate in Section 6.3 gives more weight to recent delays; delays in the distant past have relatively little impact on the estimate.Problem 10The two procedures are very similar. They both use the same formula, thereby resulting in exponentially decreasing weights for past samples.One difference is that for estimating average RTT, the time when the data is sent and when the acknowledgement is received is recorded on the same machine. For the delay estimate, the two values are recorded on different machines. Thus the sample delay can actually be negative.Problem 11a)The delay of packet 2 is 7 slots. The delay of packet 3 is 9 slots. The delay ofpacket 4 is 8 slots. The delay of packet 5 is 7 slots. The delay of packet 6 is 9slots. The delay of packet 7 is 8 slots. The delay of packet 8 is > 8 slots.b)Packets 3, 4, 6, 7, and 8 will not be received in time for their playout if playoutbegins at t-8.c)Packets 3 and 6 will not be received in time for their playout if playout begins att=9.d)No packets will arrive after their playout time if playout time begins at t=10. Problem 12The answers to parts a and b are in the table below:Problem 13a)Both schemes require 25% more bandwidth. The first scheme has a playbackdelay of 5 packets. The second scheme has a delay of 2 packets.b)The first scheme will be able to reconstruct the original high-quality audioencoding. The second scheme will use the low quality audio encoding for the lost packets and will therefore have lower overall quality.c)For the first scheme, many of the original packets will be lost and audio qualitywill be very poor. For the second scheme, every audio chunk will be available at the receiver, although only the low quality version will be available for everyother chunk. Audio quality will be acceptable.Problem 14a)Each of the other N - 1 participants sends a single audio stream of rate r bps to theinitiator. The initiator combines this stream with its own outgoing stream to createa stream of rate r. It then sends a copy of the combined stream to each of the N -1other participants. The call initiator therefore sends at a total rate of (N-1)r bps, and the total rate aggregated over all participants is 2(N-1)r bps.b)As before, each of the other N - 1 participants sends a single video stream of rate rbps to the initiator. But because the streams are now video, the initiator can no longer combine them into a single stream. The initiator instead must send each stream it receives to N - 2 participants. The call initiator therefore sends at a total rate of (N-1)*(N-1)r bps, and the total rate aggregated over all participants is (N-1)r + (N-1)*(N-1)r = N (N-1) r bps.c)N *(N-1)r bpsProblem 15a)As discussed in Chapter 2, UDP sockets are identified by the two-tuple consistingof destination IP address and destination port number. So the two packets will indeed pass through the same socket.b)Ye s, Alice only needs one socket. Bob and Claire will choose different SSRC’s,so Alice will be able distinguish between the two streams. Another question we could have asked is: How does Alice’s software know which stream (i.e. SSRC) belongs to Bob and whic h stream belongs to Alice? Indeed, Alice’s software may want to display the sender’s name when the sender is talking. Alice’s software gets the SSRC to name mapping from the RTCP source description reports. Problem 16a)Trueb)Truec)No, RTP streams can be sent to/from any port number. See the SIP example inSection 6.4.3d)No, typically they are assigned different SSRC values.e)Truef)False, she is indicating that she wishes to receive GSM audiog)False, she is indicating that she wishes to receive audio on port 48753h)True, 5060 for both source and destination port numbersi)Truej)False, this is a requirement of H.323 and not SIP.Problem 17Problem 18Problem 19No. The answer still remains the same as in Problem 21. Problem 20 See figure below. For the second leaky bucket, .1r,=p=bFigure: Solution to problem 26Problem 21No.Problem 22Let τ be a time at which flow 1 traffic starts to accumulate in the queue. We refer to τ as the beginning of a flow-1 busy period. Let τ>t be another time in the same flow-1 busy period. Let ),(1t T τ be the amount of flow-1 traffic transmitted in the interval ],[t τ. Clearly,)(),(11ττ-≥∑t R W W t T jLet )(1t Q be the amount of flow-1 traffic in the queue at time t . Clearly,),()()(1111t T t r b t Q ττ--+=)()(111ττ-+-+≤∑t R W W t r b j--+=∑R W W r t b j 111)(τSince R W W r j∑<11, 11)(b t Q ≤. Thus the maximum amount of flow-1 traffic in the queue is 1b . The minimal rate at which this traffic is served is ∑jW RW 1.Thus, the maximum delay for a flow-1 bit is ./max 11d W R W b j=∑。

计算机网络谢希仁第七版课后答案完整版在学习计算机网络这门课程时，谢希仁教授所著的第七版教材是众多学子的重要参考资料。

而课后答案则对于我们理解和掌握知识点起着关键的作用。

以下将为您呈现一份完整的计算机网络谢希仁第七版课后答案。

第一章主要介绍了计算机网络的基本概念、组成和分类等。

课后习题中，对于网络的定义和功能的理解，答案强调网络是将地理位置不同的具有独立功能的多台计算机及其外部设备，通过通信线路连接起来，在网络操作系统、网络管理软件及网络通信协议的管理和协调下，实现资源共享和信息传递的计算机系统。

其功能包括数据通信、资源共享、分布式处理、提高可靠性和负载均衡等。

第二章探讨了物理层的相关知识。

对于信号的传输方式，答案解释了基带传输、宽带传输以及频带传输的特点和适用场景。

在涉及到信道复用技术的问题时，答案详细阐述了频分复用、时分复用、波分复用和码分复用的工作原理和优缺点。

第三章聚焦于数据链路层。

关于数据链路层的三个基本问题，即封装成帧、透明传输和差错检测，答案给出了清晰的解释和示例。

在滑动窗口协议方面，详细分析了停止等待协议、后退 N 帧协议和选择重传协议的工作流程和性能特点。

第四章讲述了网络层。

对于网络层提供的两种服务，即虚电路服务和数据报服务，答案对比了它们的差异和适用情况。

在路由算法的问题上，分别介绍了距离向量路由算法和链路状态路由算法的原理和计算过程。

关于 IP 地址的分类和子网掩码的使用，答案通过实例进行了详细的说明。

第五章涉及运输层。

在运输层的端口号相关问题上，答案解释了端口号的作用和分类，并说明了如何通过端口号来识别不同的应用进程。

对于 TCP 和 UDP 协议的特点和应用场景，答案进行了深入的比较和分析。

在 TCP 的拥塞控制算法方面，详细阐述了慢开始、拥塞避免、快重传和快恢复等算法的工作机制。

第六章阐述了应用层。

对于 DNS 系统的工作原理，答案说明了域名解析的过程以及各级域名服务器的作用。

计算机网络第七版答案第一章概述1-01 计算机网络向用户可以提供那些服务？答：连通性和共享1-02 简述分组交换的要点。

答：（1）报文分组，加首部（2）经路由器储存转发（3）在目的地合并1-03 试从多个方面比较电路交换、报文交换和分组交换的主要优缺点。

答：（1）电路交换：端对端通信质量因约定了通信资源获得可靠保障，对连续传送大量数据效率高。

（2）报文交换：无须预约传输带宽，动态逐段利用传输带宽对突发式数据通信效率高，通信迅速。

（3）分组交换：具有报文交换之高效、迅速的要点，且各分组小，路由灵活，网络生存性能好。

1-04 为什么说因特网是自印刷术以来人类通信方面最大的变革？答：融合其他通信网络，在信息化过程中起核心作用，提供最好的连通性和信息共享，第一次提供了各种媒体形式的实时交互能力。

1-05 因特网的发展大致分为哪几个阶段？请指出这几个阶段的主要特点。

答：从单个网络APPANET向互联网发展；TCP/IP协议的初步成型建成三级结构的Internet；分为主干网、地区网和校园网；形成多层次ISP结构的Internet；ISP首次出现。

1-06 简述因特网标准制定的几个阶段？答：（1）因特网草案(Internet Draft) ——在这个阶段还不是 RFC 文档。

（2）建议标准(Proposed Standard) ——从这个阶段开始就成为 RFC 文档。

（3）草案标准(Draft Standard)（4）因特网标准(Internet Standard)1-07小写和大写开头的英文名internet 和Internet在意思上有何重要区别？答：（1） internet（互联网或互连网）：通用名词，它泛指由多个计算机网络互连而成的网络。

；协议无特指（2）Internet（因特网）：专用名词，特指采用 TCP/IP 协议的互联网络。

区别：后者实际上是前者的双向应用1-08 计算机网络都有哪些类别？各种类别的网络都有哪些特点？答：按范围：（1）广域网WAN：远程、高速、是Internet的核心网。