计算机网络自顶向下第一章(答案)
《计算机网络自顶向下》课后习题答案答案(第一章中文版)
1复习题1.没有不同。
主机和端系统可以互换。
端系统包括PC，工作站，WEB服务器，邮件服务器，网络连接的PDA，网络电视等等。
2.假设爱丽丝是国家A的大使，想邀请国家B的大使鲍勃吃晚餐。
爱丽丝没有简单的打个电话说“现在我没一起吃晚餐吧”。
而是她先打电话给鲍勃建议吃饭的日期与时间。
鲍勃可能会回复说那天不行，另外一天可以。
爱丽丝与鲍勃不停的互发讯息直到他们确定一致的日期与时间。
鲍勃会在约定时间（提前或迟到不超过15分钟）出现再大使馆。
外交协议也允许爱丽丝或者鲍勃以合理的理由礼貌的退出约会。
3.联网（通过网络互联）的程序通常包括2个，每一个运行在不同的主机上，互相通信。
发起通信的程序是客户机程序。
一般是客户机请求和接收来自服务器程序的服务。
4.互联网向其应用提供面向连接服务（TCP）和无连接服务（UDP）2种服务。
每一个互联网应用采取其中的一种。
面相连接服务的原理特征是：①在都没有发送应用数据之前2个端系统先进行“握手”。
②提供可靠的数据传送。
也就是说，连接的一方将所有应用数据有序且无差错的传送到连接的另一方。
③提供流控制。
也就是，确保连接的任何一方都不会过快的发送过量的分组而淹没另一方。
④提供拥塞控制。
即管理应用发送进网络的数据总量，帮助防止互联网进入迟滞状态。
、无连接服务的原理特征：①没有握手②没有可靠数据传送的保证③没有流控制或者拥塞控制5.流控制和拥塞控制的两个面向不同的对象的不同的控制机理。
流控制保证连接的任何一方不会因为过快的发送过多分组而淹没另一方。
拥塞控制是管理应用发送进网络的数据总量，帮助防止互联网核心（即网络路由器的缓冲区里面）发生拥塞。
6.互联网面向连接服务通过使用确认，重传提供可靠的数据传送。
当连接的一方没有收到它发送的分组的确认（从连接的另一方）时，它会重发这个分组。
7.电路交换可以为呼叫的持续时间保证提供一定量的端到端的带宽。
今天的大多数分组交换网（包括互联网）不能保证任何端到端带宽。
《计算机网络自顶向下方法》原书第四版复习资料解析
物流信息技术复习秘籍TG与其他同学联合出品目录一、《物流信息技术》 (2)条码技术 (2)RFID (4)GPS (5)二、《计算机网络》 (7)第一章 (7)第二章 (7)第三章 (8)第四章 (9)第五章 (10)三、《计算机网络》补充题 (11)第一章 (11)第二章 (15)第三章 (19)第四章 (21)第五章 (25)人员分工： ...................................................... 错误！未定义书签。
一、《物流信息技术》条码技术书本第二章习题1选择题：（1）下列不属于一维条码制是 DA、交叉25码B、EANC、库德巴吗D、49码（3） A 码是美国统一代码委员会制定的一种商品用条码，主要是用于美国和加拿大地区。
A、UPCB、EANC、39D、93（4） A 二维条码形态上是由多行短截的一维条码堆叠而成，它在编码设计、校验原理、识读方式等方面继承了一维条码的一些特点，识读设备与条码印刷与一维条码技术兼容。
A、堆叠式/行排式B、矩阵式C、图像式D、数字式（6）从系统结构和功能上讲，条码识读系统由 D 等部分组成。
A、条码扫描和译码B、光学系统和探测器C、信号放大、滤波、波形整形D、扫描系统、信号整形、译码（7） B 可以识读常用的一维条码，还能识读行排式和矩阵式的二维条码。
A、光笔B、图形式条码识读器C、卡槽式条码识读器D、激光条码识读器（8）条码的编码方法中， A 是指条码符号中，条与空是由标准宽度的模块组合而成。
A、模块组合法B、宽度调节法C、矩阵法D、堆叠法2简答题（2）列举一维条码的主要码制。
一维条码的主要码制：UPC码 EAN码 25码交叉25码 39码库德巴码 128码以及93码（3）简述一维条码的结构。
一维条码的结构：一个完整的一维条码的组成次序依次为：静区（前）、起始符、中间分隔符、校验符、终止符静区（后）。
（4）简述条码识别系统的组成条码识别系统的组成：从系统结构和功能上讲，条码识读系统由扫描系统、信号整形、译码等部分组成。
《计算机网络·自顶向下方法》第七版第一章课后习题与问题答案
《计算机⽹络·⾃顶向下⽅法》第七版第⼀章课后习题与问题答案⾮官⽅答案，本⼈已尽最⼤努⼒，使结果正确，如有错误，请⼤佬指出正⽂：1.1节R1在计算机⽹络中，主机就是端系统举例：PC，⼿机，服务器，⽹络电视WEB服务器是⼀种端系统R2A protocol is a rule which describes how an activity should be performed, especially in the field of diplomacy. In diplomatic services and governmental fields of endeavor protocols are often unwritten guidelines. Protocols specify the proper and generally accepted behavior in matters of state and diplomacy, such as showing appropriate respect to a head of state, ranking diplomats in chronological order of their accreditation at court, and so on. One definition is:Protocol is commonly described as a set of international courtesy rules. These well-established and time-honored rules have made it easier for nations and people to live and work together. Part of protocol has always been the acknowledgment of the hierarchical standing of all present. Protocol rules are based on the principles of civility.⼤体意思就是说，外交协议是指在外交事务中，⼀种⾮书⾯形式的，被各国认为是适当的并普遍接受的⼀种国际礼节规则，有助于各国⼈民和谐共处R3如果两个端系统发送和接受信号的标准不同，双⽅可能并不能正常通信。
自顶向下计算机网络课后答案
第一章概述习题1-01 计算机网络的发展可划分为几个阶段?每个阶段各有何特点?答: 计算机网络的发展过程大致经历了四个阶段。
第一阶段：(20世纪60年代)以单个计算机为中心的面向终端的计算机网络系统。
这种网络系统是以批处理信息为主要目的。
它的缺点是：如果计算机的负荷较重，会导致系统响应时间过长；单机系统的可靠性一般较低，一旦计算机发生故障，将导致整个网络系统的瘫痪。
第二阶段：(20世纪70年代)以分组交换网为中心的多主机互连的计算机网络系统。
为了克服第一代计算机网络的缺点，提高网络的可靠性和可用性，人们开始研究如何将多台计算机相互连接的方法。
人们首先借鉴了电信部门的电路交换的思想。
所谓“交换”，从通信资源的分配角度来看，就是由交换设备动态地分配传输线路资源或信道带宽所采用的一种技术。
电话交换机采用的交换技术是电路交换(或线路交换)，它的主要特点是：①在通话的全部时间内用户独占分配的传输线路或信道带宽，即采用的是静态分配策略；②通信双方建立的通路中任何一点出现了故障，就会中断通话，必须重新拨号建立连接，方可继续，这对十分紧急而重要的通信是不利的。
显然，这种交换技术适应模拟信号的数据传输。
然而在计算机网络中还可以传输数字信号。
数字信号通信与模拟信号通信的本质区别在于数字信号的离散性和可存储性。
这些特性使得它在数据传输过程中不仅可以间断分时发送，而且可以进行再加工、再处理。
③计算机数据的产生往往是“突发式”的，比如当用户用键盘输入数据和编辑文件时，或计算机正在进行处理而未得出结果时，通信线路资源实际上是空闲的，从而造成通信线路资源的极大浪费。
据统计，在计算机间的数据通信中，用来传送数据的时间往往不到10%甚至1%。
另外，由于各异的计算机和终端的传输数据的速率各不相同，采用电路交换就很难相互通信。
为此，必须寻找出一种新的适应计算机通信的交换技术。
1964年，巴兰(Baran)在美国兰德(Rand)公司“论分布式通信”的研究报告中提出了存储转发(store and forward)的概念。
计算机网络-原理、技术与应用(第2版)部分习题参考答案(不外传!)
计算机网络-原理、技术与应用（第2版）部分习题参考答案第1章1.1答：计算机网络是通过传输介质、通信设施和网络通信协议，把分散在不同地点的计算机设备互联起来，实现资源共享和信息传输的系统。
涉及到的知识点：1、传输介质；2、通信协议；3、不同地点.；4、计算机设备；5、资源共享；6、数据传输；7、系统。
1.6答：1、数据通信。
计算机网络中的计算机设备，终端与计算机、计算机与计算机之间进行通信，数据传输，实现数据和信息的传输、收集和交换。
2、资源共享。
用户通过计算机网络可以共享系统内的硬件、软件、数据、文档信息，以及通过信息交流获取更多的知识。
3、给网络用户提供最好的性价比服务，减少重复投资。
4、提供大容量网络存储，不断增加新的多媒体应用。
5、提供分布式处理，使得协同操作为可能；平衡不同地点计算机系统的负荷，降低软件设计的复杂性，充分利用计算机网络系统内的资源，使得网格计算成为可能，提高计算机网络系统的效率。
6、对地理上分散的计算机系统进行集中控制，实现对网络资源集中管理和分配。
7、提供高可靠性的系统，借助在不同信息处理位置和数据存储地点的备份，通过传输线路和信息处理设备的冗余实现高可靠性。
1.13答：计算机网络中计算机进行通信、数据交换时需要制定双方都要遵守的通信规则和约定就是协议。
协议是按层次结构组织的，不同层次协议和网络层次的集合构成了协议体系结构。
网络协议层次结构包含两个基本内容：1、网络实现的功能分解到若干层次，每个功能用对等层协议实现，不同系统中的对等层要遵循对等层协议，通过对等层协议理解和完成该层的功能。
2、相邻层次之间通过接口交互必要的信息，构成下层为上次提供服务的关系，也成为接口关系。
网络服务靠服务原语进行描述，网络协议软件根据网络协议结构进行设计和开发。
1.20答：1）网络命令行程序2）网络协议分析工具3）网络仿真和模拟4）网络应用编程5）生活中的例子1.26答：与计算机网络和数据通信标准有关的国际组织有ISO、ITU-T（CCITT）、IEEE、ANSI、EIA、ACM等。
计算机网络自顶向下课后答案及解析
7第一章R11 L/R1 + L/R2R13a. 两个用户b. 每个用户需要1Mbps进行传输，若两个或更少用户同时进行传输，则带宽需求量最大为2Mbps，由于链路总带宽为2Mbps，所以无排队时延；若三个或更多用户同时进行传输，带宽需求超过3Mbps，多于链路总带宽，因此会出现排队时延。
c. 0.27d. 0.008；0.008R19a. 500kbpsb. 64sc. 100kbps；320sR23应用层：网络应用程序及应用层协议存留的地方；传输层：在应用程序端点之间传送应用层报文；网络层：将网络层分组（数据报）从一台主机移动到另一台主机；链路层：将分组从一个结点移动到路径上的下一个结点；物理层：将帧（链路层分组）中的一个一个比特从一个结点移动到下一个结点。
R25路由器：网络层，链路层，物理层链路层交换机：链路层，物理层主机：所有五层P3a. 电路交换网。
因为应用包含可预测的稳定带宽需求的长运行时间，由于传输率已知且非猝发，可在无明显浪费的情况下为每个应用周期预留带宽。
且建立与中断连接的总开销可被均摊在应用长时间的运行时间中。
b. 在最坏的情况下，所有应用同时经一条或多条链路传输。
然而由于每条链路都有足够带宽提供给所有应用，不会出现拥塞情况，因此不需要拥塞控制。
第二章R5目的主机的IP地址与目的进程套接字的端口号R12当用户首次访问网站时，服务器创建一唯一标识码，在其后端服务器中创建一入口，将该唯一标识码作为Cookie 码返回，该cookie 码储存在用户主机中，由浏览器管理。
在后来每次的访问与购买中，浏览器将cookie 码发送给网站，因此当该用户（准确地说，该浏览器）访问该网站时，网站会立即获知。
R15FTP 使用两平行TCP 连接，一条连接发送控制信息（例如文件传输请求），另一条连接用作实际传输文件。
由于控制信息不会通过与文件传输相同的连接发送，因此FTP 在“带外”发送控制信息。
R19是的，一个机构的邮件服务器和Web 服务器可以有完全相同的主机名别名。
计算机网络(自顶向下)
在数据传输前，不必建立端到端的连接 只要下一个节点空闲，即可传输
信道的使用率较高
数据的传输采用存储转发，延时不可估计 数据中必须包含地址域 接收到的分组不一定按序，可能还需重组 适合传输文本型数据
计算机网络和因特网概述
7
分组时延的4种来源
1. 节点处理: 检查比特差错 决定输出链路 2. 排队 等待输出链路传输的时间 取决于路由器拥塞的等级
考试时间：2014-04-27（周日）
下午1.30-3.10（100分钟）
1.考试题型
选择题（10*2）
名词性解释（10*2） 简答题（5*8） 综合题（2*10）
第1章 计算机网络和因特网概述
Computer Networks and the Internet
计算机网络：自顶向下方法 (原书第四版) 陈鸣译，机械工业出版社，2010年
邮件 服务器 用户 代理
输出 报文队列 用户邮箱
用户 代理
SMTP
用户代理 亦称为 “邮件阅读器” 写作、编辑、阅读邮件报文 例如Outlook, Apple Mail, Netscape Messenger
存储在服务器上的报文的达到
SMTP
SMTP
邮件 服务器
用户 代理 用户 代理
23
FTP: 文件传输协议
FTP FTP 用户接口 客户机 主机上 的用户 本地文件 系统 文件传输 FTP 服务器 远程文件系统
传输文件到远程主机或从远程主机获取文件
客户机/服务器模型
客户机: 发起传输一侧(到/来自远程之一) 服务器: 远程主机
ftp: RFC 959 ftp 服务器: 端口 21
计算机网络自顶向下(第七版)课后答案—英文
Computer Networking: A Top-Down Approach,7th EditionSolutions to Review Questions and Problems Version Date: December 2016This document contains the solutions to review questions and problems for the 7th edition of Computer Networking: A Top-Down Approach by Jim Kurose and Keith Ross. These solutions are being made available to instructors ONLY. Please do NOT copy or distribute this document to others (even other instructors). Please do not post any solutions on a publicly-available Web site. We’ll be happy to provide a copy (up-to-date) of this solution manual ourselves to anyone who asks.Acknowledgments:Over the years, several students and colleagues have helped us prepare this solutions manual. Special thanks goes to Honggang Zhang, Rakesh Kumar, Prithula Dhungel, and Vijay Annapureddy. Also thanks to all the readers who have made suggestions and corrected errors.All material © copyright 1996-2016 by J.F. Kurose and K.W. Ross. All rights reserved Chapter 1 Review Questions1.There is no difference. Throughout this text, the words “host” and “end system” areused interchangeably. End systems include PCs, workstations, Web servers, mail servers, PDAs, Internet-connected game consoles, etc.2.From Wikipedia: Diplomatic protocol is commonly described as a set of internationalcourtesy rules. These well-established and time-honored rules have made it easier for nations and people to live and work together. Part of protocol has always been the acknowledgment of the hierarchical standing of all present. Protocol rules are based on the principles of civility.3.Standards are important for protocols so that people can create networking systemsand products that interoperate.4. 1. Dial-up modem over telephone line: home; 2. DSL over telephone line: home orsmall office; 3. Cable to HFC: home; 4. 100 Mbps switched Ethernet: enterprise; 5.Wifi (802.11): home and enterprise: 6. 3G and 4G: wide-area wireless.5.HFC bandwidth is shared among the users. On the downstream channel, all packetsemanate from a single source, namely, the head end. Thus, there are no collisions in the downstream channel.6.In most American cities, the current possibilities include: dial-up; DSL; cable modem;fiber-to-the-home.7. Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps.8. Today, Ethernet most commonly runs over twisted-pair copper wire. It also can runover fibers optic links.9. Dial up modems: up to 56 Kbps, bandwidth is dedicated; ADSL: up to 24 Mbpsdownstream and 2.5 Mbps upstream, bandwidth is dedicated; HFC, rates up to 42.8 Mbps and upstream rates of up to 30.7 Mbps, bandwidth is shared. FTTH: 2-10Mbps upload; 10-20 Mbps download; bandwidth is not shared.10. There are two popular wireless Internet access technologies today:a)Wifi (802.11) In a wireless LAN, wireless users transmit/receive packets to/from anbase station (i.e., wireless access point) within a radius of few tens of meters. The base station is typically connected to the wired Internet and thus serves to connect wireless users to the wired network.b)3G and 4G wide-area wireless access networks. In these systems, packets aretransmitted over the same wireless infrastructure used for cellular telephony, with the base station thus being managed by a telecommunications provider. This provides wireless access to users within a radius of tens of kilometers of the base station.11. At time t0the sending host begins to transmit. At time t1 = L/R1, the sending hostcompletes transmission and the entire packet is received at the router (no propagationdelay). Because the router has the entire packet at time t 1, it can begin to transmit the packet to the receiving host at time t 1. At time t 2 = t 1 + L/R 2, the router completes transmission and the entire packet is received at the receiving host (again, no propagation delay). Thus, the end-to-end delay is L/R 1 + L/R 2.12. A circuit-switched network can guarantee a certain amount of end-to-end bandwidth for the duration of a call. Most packet-switched networks today (including the Internet) cannot make any end-to-end guarantees for bandwidth. FDM requires sophisticated analog hardware to shift signal into appropriate frequency bands .13. a) 2 users can be supported because each user requires half of the link bandwidth.b) Since each user requires 1Mbps when transmitting, if two or fewer users transmit simultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three users transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this case, there will be queuing delay before the link.c) Probability that a given user is transmitting = 0.2d) Probability that all three users are transmitting simultaneously = ()333133--⎪⎪⎭⎫ ⎝⎛p p = (0.2)3 = 0.008. Since the queue grows when all the users are transmitting, thefraction of time during which the queue grows (which is equal to the probability that all three users are transmitting simultaneously) is 0.008.14. If the two ISPs do not peer with each other, then when they send traffic to each other they have to send the traffic through a provider ISP (intermediary), to which they have to pay for carrying the traffic. By peering with each other directly, the two ISPs can reduce their payments to their provider ISPs. An Internet Exchange Points (IXP) (typically in a standalone building with its own switches) is a meeting point where multiple ISPs can connect and/or peer together. An ISP earns its money by charging each of the the ISPs that connect to the IXP a relatively small fee, which may depend on the amount of traffic sent to or received from the IXP.15. Google's private network connects together all its data centers, big and small. Traffic between the Google data centers passes over its private network rather than over the public Internet. Many of these data centers are located in, or close to, lower tier ISPs. Therefore, when Google delivers content to a user, it often can bypass higher tier ISPs. What motivates content providers to create these networks? First, the content provider has more control over the user experience, since it has to use few intermediary ISPs. Second, it can save money by sending less traffic into provider networks. Third, if ISPs decide to charge more money to highly profitable content providers (in countries where net neutrality doesn't apply), the content providers can avoid these extra payments.16. The delay components are processing delays, transmission delays, propagation delays,and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable.17. a) 1000 km, 1 Mbps, 100 bytesb) 100 km, 1 Mbps, 100 bytes18. 10msec; d/s; no; no19. a) 500 kbpsb) 64 secondsc) 100kbps; 320 seconds20. End system A breaks the large file into chunks. It adds header to each chunk, therebygenerating multiple packets from the file. The header in each packet includes the IP address of the destination (end system B). The packet switch uses the destination IP address in the packet to determine the outgoing link. Asking which road to take is analogous to a packet asking which outgoing link it should be forwarded on, given the packet’s destination address.21. The maximum emission rate is 500 packets/sec and the maximum transmission rate is350 packets/sec. The corresponding traffic intensity is 500/350 =1.43 > 1.Loss will eventually occur for each experiment;but the time when loss first occurs will be different from one experiment to the next due to the randomness in the emission process.22. Five generic tasks are error control, flow control, segmentation and reassembly,multiplexing, and connection setup. Yes, these tasks can be duplicated at different layers. For example, error control is often provided at more than one layer.23. The five layers in the Internet protocol stack are –from top to bottom –theapplication layer, the transport layer, the network layer, the link layer, and the physical layer. The principal responsibilities are outlined in Section 1.5.1.24. Application-layer message: data which an application wants to send and passed ontothe transport layer; transport-layer segment: generated by the transport layer and encapsulates application-layer message with transport layer header; network-layer datagram: encapsulates transport-layer segment with a network-layer header; link-layer frame: encapsulates network-layer datagram with a link-layer header.25. Routers process network, link and physical layers (layers 1 through 3). (This is a littlebit of a white lie, as modern routers sometimes act as firewalls or caching components, and process Transport layer as well.) Link layer switches process link and physical layers (layers 1 through2). Hosts process all five layers.26. a) VirusRequires some form of human interaction to spread. Classic example: E-mail viruses.b) WormsNo user replication needed. Worm in infected host scans IP addresses and portnumbers, looking for vulnerable processes to infect.27. Creation of a botnet requires an attacker to find vulnerability in some application orsystem (e.g. exploiting the buffer overflow vulnerability that might exist in an application). After finding the vulnerability, the attacker needs to scan for hosts that are vulnerable. The target is basically to compromise a series of systems by exploiting that particular vulnerability. Any system that is part of the botnet can automatically scan its environment and propagate by exploiting the vulnerability. An important property of such botnets is that the originator of the botnet can remotely control and issue commands to all the nodes in the botnet. Hence, it becomes possible for the attacker to issue a command to all the nodes, that target a single node (for example, all nodes in the botnet might be commanded by the attacker to send a TCP SYN message to the target, which might result in a TCP SYN flood attack at the target).28. Trudy can pretend to be Bob to Alice (and vice-versa) and partially or completelymodify the message(s) being sent from Bob to Alice. For example, she can easily change the phrase “Alice, I owe you $1000” to “Alice, I owe you $10,000”.Furthermore, Trudy can even drop the packets that are being sent by Bob to Alice (and vise-versa), even if the packets from Bob to Alice are encrypted.Chapter 1 ProblemsProblem 1There is no single right answer to this question. Many protocols would do the trick. Here's a simple answer below:Messages from ATM machine to ServerMsg name purpose-------- -------HELO <userid> Let server know that there is a card in theATM machineATM card transmits user ID to ServerPASSWD <passwd> User enters PIN, which is sent to server BALANCE User requests balanceWITHDRAWL <amount> User asks to withdraw moneyBYE user all doneMessages from Server to ATM machine (display)Msg name purpose-------- -------PASSWD Ask user for PIN (password)OK last requested operation (PASSWD, WITHDRAWL)OKERR last requested operation (PASSWD, WITHDRAWL)in ERRORAMOUNT <amt> sent in response to BALANCE requestBYE user done, display welcome screen at ATM Correct operation:client serverHELO (userid) --------------> (check if valid userid)<------------- PASSWDPASSWD <passwd> --------------> (check password)<------------- OK (password is OK)BALANCE --------------><------------- AMOUNT <amt>WITHDRAWL <amt> --------------> check if enough $ to coverwithdrawl<------------- OKATM dispenses $BYE --------------><------------- BYEIn situation when there's not enough money:HELO (userid) --------------> (check if valid userid)<------------- PASSWDPASSWD <passwd> --------------> (check password)<------------- OK (password is OK)BALANCE --------------><------------- AMOUNT <amt>WITHDRAWL <amt> --------------> check if enough $ to cover withdrawl<------------- ERR (not enough funds)error msg displayedno $ given outBYE --------------><------------- BYEProblem 2At time N*(L/R) the first packet has reached the destination, the second packet is stored in the last router, the third packet is stored in the next-to-last router, etc. At time N*(L/R) + L/R, the second packet has reached the destination, the third packet is stored in the last router, etc. Continuing with this logic, we see that at time N*(L/R) + (P-1)*(L/R) = (N+P-1)*(L/R) all packets have reached the destination.Problem 3a) A circuit-switched network would be well suited to the application, because theapplication involves long sessions with predictable smooth bandwidth requirements.Since the transmission rate is known and not bursty, bandwidth can be reserved for each application session without significant waste. In addition, the overhead costs of setting up and tearing down connections are amortized over the lengthy duration of a typical application session.b) In the worst case, all the applications simultaneously transmit over one or morenetwork links. However, since each link has sufficient bandwidth to handle the sum of all of the applications' data rates, no congestion (very little queuing) will occur.Given such generous link capacities, the network does not need congestion control mechanisms.Problem 4a)Between the switch in the upper left and the switch in the upper right we can have 4connections. Similarly we can have four connections between each of the 3 other pairs of adjacent switches. Thus, this network can support up to 16 connections.b)We can 4 connections passing through the switch in the upper-right-hand corner andanother 4connections passing through the switch in the lower-left-hand corner, giving a total of 8 connections.c) Yes. For the connections between A and C, we route two connections through B and two connections through D. For the connections between B and D, we route two connections through A and two connections through C. In this manner, there are at most 4 connections passing through any link.Problem 5Tollbooths are 75 km apart, and the cars propagate at 100km/hr. A tollbooth services a car at a rate of one car every 12 seconds.a) There are ten cars. It takes 120 seconds, or 2 minutes, for the first tollbooth to service the 10 cars. Each of these cars has a propagation delay of 45 minutes (travel 75 km) before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 47 minutes. The whole process repeats itself for traveling between the second and third tollbooths. It also takes 2 minutes for the third tollbooth to service the 10 cars. Thus the total delay is 96 minutes.b) Delay between tollbooths is 8*12 seconds plus 45 minutes, i.e., 46 minutes and 36 seconds. The total delay is twice this amount plus 8*12 seconds, i.e., 94 minutes and 48 seconds.Problem 6a) s m d prop /= seconds.b) R L d trans /= seconds.c) )//(R L s m d end to end +=-- seconds.d) The bit is just leaving Host A.e) The first bit is in the link and has not reached Host B.f) The first bit has reached Host B. g) Want()536105.2105612083=⨯⨯==s R L m km.Problem 7Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires31064856⨯⋅sec=7msec.The time required to transmit the packet is6102856⨯⋅sec=μ224sec.Propagation delay = 10 msec.The delay until decoding is7msec +μ224sec + 10msec = 17.224msecA similar analysis shows that all bits experience a delay of 17.224 msec.Problem 8a) 20 users can be supported.b) 1.0=p .c) ()n n p p n --⎪⎪⎭⎫ ⎝⎛1201120. d) ()∑=--⎪⎪⎭⎫ ⎝⎛-20012011201n n n p p n . We use the central limit theorem to approximate this probability. Let j X be independentrandom variables such that ()p X P j ==1.(P “21 or more users”)⎪⎪⎭⎫ ⎝⎛≤-=∑=2111201j j X P⎪⎪⎪⎭⎫ ⎝⎛⋅⋅≤⋅⋅-=⎪⎪⎭⎫ ⎝⎛≤∑∑==9.01.012099.01.0120122112011201j j j j X P X P ()74.2286.39≤=⎪⎭⎫ ⎝⎛≤≈Z P Z P 997.0=when Z is a standard normal r.v. Thus (P “21 or more users”)003.0≈.Problem 9a) 10,000b) ()∑+=--⎪⎪⎭⎫ ⎝⎛M N n n M n p p n M 11Problem 10The first end system requires L/R1 to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay of d proc; after receiving the entire packet, the packet switch connecting the first and the second link requires L/R2 to transmit the packet onto the second link; the packet propagates over the second link in d2/s2. Similarly, we can find the delay caused by the second switch and the third link: L/R3, d proc, and d3/s3.Adding these five delays givesd end-end = L/R1 + L/R2 + L/R3 + d1/s1 + d2/s2 + d3/s3+ d proc+ d procTo answer the second question, we simply plug the values into the equation to get 6 + 6 + 6 + 20+16 + 4 + 3 + 3 = 64 msec.Problem 11Because bits are immediately transmitted, the packet switch does not introduce any delay; in particular, it does not introduce a transmission delay. Thus,d end-end = L/R + d1/s1 + d2/s2+ d3/s3For the values in Problem 10, we get 6 + 20 + 16 + 4 = 46 msec.Problem 12The arriving packet must first wait for the link to transmit 4.5 *1,500 bytes = 6,750 bytes or 54,000 bits. Since these bits are transmitted at 2 Mbps, the queuing delay is 27 msec. Generally, the queuing delay is (nL + (L - x))/R.Problem 13a)The queuing delay is 0 for the first transmitted packet, L/R for the second transmittedpacket, and generally, (n-1)L/R for the n th transmitted packet. Thus, the average delay for the N packets is:(L/R + 2L/R + ....... + (N-1)L/R)/N= L/(RN) * (1 + 2 + ..... + (N-1))= L/(RN) * N(N-1)/2= LN(N-1)/(2RN)= (N-1)L/(2R)Note that here we used the well-known fact:1 +2 + ....... + N = N(N+1)/2b) It takes R LN / seconds to transmit the N packets. Thus, the buffer is empty when a each batch of N packets arrive. Thus, the average delay of a packet across all batches is the average delay within one batch, i.e., (N-1)L /2R .Problem 14a) The transmission delay is R L /. The total delay isI RL R L I R IL -=+-1/)1(b) Let R L x /=.Total delay = axx-1For x=0, the total delay =0; as we increase x, total delay increases, approaching infinity as x approaches 1/a.Problem 15Total delay aa R aL R L I R L -=-=-=-=μμμ1/1/1/1/1/.Problem 16The total number of packets in the system includes those in the buffer and the packet that is being transmitted. So, N=10+1.Because d a N ⋅=, so (10+1)=a*(queuing delay + transmission delay). That is, 11=a*(0.01+1/100)=a*(0.01+0.01). Thus, a=550 packets/sec.Problem 17a) There are Q nodes (the source host and the 1-Q routers). Let qproc d denote the processing delay at the q th node. Let q R be the transmission rate of the q th link and letq qtrans R L d /=. Let q prop d be the propagation delay across the q th link. Then[]∑=--++=Qq qprop q trans q proc end to end d d d d 1.b) Let qqueue d denote the average queuing delay at node q . Then[]∑=--+++=Qq qqueue q prop q trans q proc end to end d d d d d 1.Problem 18On linux you can use the commandtraceroute and in the Windows command prompt you can usetracert In either case, you will get three delay measurements. For those three measurements you can calculate the mean and standard deviation. Repeat the experiment at different times of the day and comment on any changes .Here is an example solution:Traceroutes between San Diego Super Computer Center and a)The average (mean) of the round-trip delays at each of the three hours is 71.18 ms,71.38 ms and 71.55 ms, respectively. The standard deviations are 0.075 ms, 0.21 ms,0.05 ms, respectively.b)In this example, the traceroutes have 12 routers in the path at each of the three hours.No, the paths didn’t change during any of the hours.c)Traceroute packets passed through four ISP networks from source to destination. Yes,in this experiment the largest delays occurred at peering interfaces between adjacent ISPs.Traceroutes from (France) to (USA).d)The average round-trip delays at each of the three hours are 87.09 ms, 86.35 ms and86.48 ms, respectively. The standard deviations are 0.53 ms, 0.18 ms, 0.23 ms,respectively. In this example, there are 11 routers in the path at each of the three hours. No, the paths didn’t change during any of the hours. Traceroute packets passed three ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs.Problem 19An example solution:Traceroutes from two different cities in France to New York City in United States a)In these traceroutes from two different cities in France to the same destination host inUnited States, seven links are in common including the transatlantic link.b)In this example of traceroutes from one city in France and from another city inGermany to the same host in United States, three links are in common including the transatlantic link.Traceroutes to two different cities in China from same host in United Statesc) Five links are common in the two traceroutes. The two traceroutes diverge before reaching ChinaProblem 20Throughput = min{R s , R c , R/M}Problem 21If only use one path, the max throughput is given by:}},,,m in{,},,,,min{},,,,m ax {m in{212222111211MN M M N N R R R R R R R R R .If use all paths, the max throughput is given by ∑=Mk kN k k R R R 121},,,min{ .Problem 22Probability of successfully receiving a packet is: p s = (1-p)N .The number of transmissions needed to be performed until the packet is successfully received by the client is a geometric random variable with success probability p s . Thus, the average number of transmissions needed is given by: 1/p s . Then, the average number of re-transmissions needed is given by: 1/p s -1.Problem 23Let’s call the first packet A and call the second packet B.a) If the bottleneck link is the first link, then packet B is queued at the first link waiting for the transmission of packet A. So the packet inter-arrival time at the destination is simply L/R s .b) If the second link is the bottleneck link and both packets are sent back to back, it must be true that the second packet arrives at the input queue of the second link before the second link finishes the transmission of the first packet. That is,L/R s + L/R s + d prop < L/R s + d prop + L/R cThe left hand side of the above inequality represents the time needed by the second packet to arrive at the input queue of the second link (the second link has not startedtransmitting the second packet yet). The right hand side represents the time needed by the first packet to finish its transmission onto the second link.If we send the second packet T seconds later, we will ensure that there is no queuing delay for the second packet at the second link if we have:L/R s + L/R s + d prop + T >= L/R s + d prop + L/R cThus, the minimum value of T is L/R c L/R s .Problem 2440 terabytes = 40 * 1012 * 8 bits. So, if using the dedicated link, it will take 40 * 1012 * 8 / (100 *106 ) =3200000 seconds = 37 days. But with FedEx overnight delivery, you can guarantee the data arrives in one day, and it should cost less than $100.Problem 25a)160,000 bitsb)160,000 bitsc)The bandwidth-delay product of a link is the maximum number of bits that can be inthe link.d)the width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meterslong, which is longer than a football fielde) s/RProblem 26s/R=20000km, then R=s/20000km= 2.5*108/(2*107)= 12.5 bpsProblem 27a)80,000,000 bitsb)800,000 bits, this is because that the maximum number of bits that will be in the linkat any given time = min(bandwidth delay product, packet size) = 800,000 bits.c).25 metersProblem 28a)t trans + t prop = 400 msec + 80 msec = 480 msec.b)20 * (t trans + 2 t prop) = 20*(20 msec + 80 msec) = 2 sec.c) Breaking up a file takes longer to transmit because each data packet and its corresponding acknowledgement packet add their own propagation delays.Problem 29Recall geostationary satellite is 36,000 kilometers away from earth surface. a) 150 msec b) 1,500,000 bits c) 600,000,000 bitsProblem 30Let’s suppose the passenger and his/her bags correspond to the data unit arriving to the top of the protocol stack. When the passenger checks in, his/her bags are checked, and a tag is attached to the bags and ticket. This is additional information added in the Baggage layer if Figure 1.20 that allows the Baggage layer to implement the service or separating the passengers and baggage on the sending side, and then reuniting them (hopefully!) on the destination side. When a passenger then passes through security and additional stamp is often added to his/her ticket, indicating that the passenger has passed through a security check. This information is used to ensure (e.g., by later checks for the security information) secure transfer of people.Problem 31a)Time to send message from source host to first packet switch = sec 4sec 10210866=⨯⨯With store-and-forward switching, the total time to move message from source host to destination host = sec 123sec 4=⨯hopsb)Time to send 1st packet from source host to first packet switch = . sec 5sec 10210164m =⨯⨯. Time at which 2nd packet is received at the first switch = time at which 1st packet is received at the second switch = sec 10sec 52m m =⨯c)Time at which 1st packet is received at the destination host = sec 153sec 5m hops m =⨯. After this, every 5msec one packet will be received; thus time at which last (800th ) packet is received = sec 01.4sec 5*799sec 15=+m m . It can be seen that delay in using message segmentation is significantly less (almost 1/3rd ). d)i. Without message segmentation, if bit errors are not tolerated, if there is asingle bit error, the whole message has to be retransmitted (rather than a single packet).ii. Without message segmentation, huge packets (containing HD videos, forexample) are sent into the network. Routers have to accommodate these hugepackets. Smaller packets have to queue behind enormous packets and suffer unfair delays.e)i. Packets have to be put in sequence at the destination.ii. Message segmentation results in many smaller packets. Since header size is usually the same for all packets regardless of their size, with message segmentation the total amount of header bytes is more.Problem 32Yes, the delays in the applet correspond to the delays in the Problem 31.The propagation delays affect the overall end-to-end delays both for packet switching and message switching equally.Problem 33There are F /S packets. Each packet is S=80 bits. Time at which the last packet is receivedat the first router is SFR S ⨯+80sec. At this time, the first F/S-2 packets are at thedestination, and the F/S-1 packet is at the second router. The last packet must then be transmitted by the first router and the second router, with each transmission taking R S 80+sec. Thus delay in sending the whole file is )2(80+⨯+=S FR S delay To calculate the value of S which leads to the minimum delay, F S delay dS d400=⇒=Problem 34The circuit-switched telephone networks and the Internet are connected together at "gateways". When a Skype user (connected to the Internet) calls an ordinary telephone, a circuit is established between a gateway and the telephone user over the circuit switched network. The skype user's voice is sent in packets over the Internet to the gateway. At the gateway, the voice signal is reconstructed and then sent over the circuit. In the other direction, the voice signal is sent over the circuit switched network to the gateway. The gateway packetizes the voice signal and sends the voice packets to the Skype user.。
计算机网络(自顶向下方法)第四版答案(中文版)
在目标攻击）.
28.特鲁迪能假装对爱丽丝 （和反之亦然）和部份地或完全地当鲍伯
修改从鲍伯被传送到爱丽丝的信息 （s）。举例来说，她能容易地
将片语 " 爱丽丝，我亏欠你＄1000" 换成 " 爱丽丝，我亏欠你＄10,000" 。
在情形中，当没有充足的钱：
HELO （在电子邮件地址中@符号之前的名称）－－－－－－－－－－－－－－＞ (检查是否有效的在电子邮件地址中@符号之前的名称）
＜－－－－－－－－－－－－－PASSWD
PASSWD ＜passwd＞－－－－－－－－－－－－－－＞ (检查密码）
＜－－－－－－－－－－－－－好（密码是好）
＜－－－－－－－－－－－－－再见
问题 2.
一）一个转变电路的网络会很好地被适合到描述的应用程序，因为
应用程序用可预期的平滑带宽包括长会议
需求。因为传输率被知道和不是 bursty，带宽能是
为每个应用程序会议电路，以没有重要的废料保留。此外，我们
需要不烦恼非常大约设定的在头上费用在而且上面扯裂下一
此外，特鲁迪能甚至放被鲍伯传送给爱丽丝的小包
（而且虎头钳－反之亦然）, 即使从鲍伯到爱丽丝的小包被编加密码。
第 1 章问题：
问题 1.
没有单一这一个疑问的正确答案。许多协定会达到预期效果。
一个简单的下面 回应在这里：
从非同步传输模式机器到服务器的信息
Msg 名字目的
－－－－－－－－－－－－－－－
电路被分期偿还过了冗长期间的连接一典型的
应用程序会议。
b）如有如此有雅量的联结容量，网络没有需要混杂控制
计算机网络——自顶向下方法于internet特色 第一章答案
第一章习题参考答案P63, No.4(a)共有10辆车，每辆车通过每个收费站需要12s ，则10辆车通过第一个和第二个收费站共需要240s ，车队以100kmph 完成200km 需要2小时，则总时延为2*60＋4=124分钟 (b)7辆车通过两个收费站共需要7*12*2=168s ，车队完成200km 需要2小时，则总时延为2*60分钟+168秒=122分钟48秒P63, No.5(a)VC 建立时间为t s ，每个包头部含h 比特，内容含F 比特，则每个包大小为F+h ，每段链路的速率为R bps ，则一个包通过每段链路需要(F+h)/R 秒，该包通过所有Q 段链路需要Q*(F+h)/R 秒，假设所传的文件为M 个包，则所有相应的时间可表示如下：t s 时刻： VC 连接建立t s +Q*(F+h)/R ： 第1个包到达终点（注意此时第2个包已经在倒数第2个节点） t s +(Q+1)*(F+h)/R ： 第2个包到达终点…t s +(Q+M-1)*(F+h)/R ： 第M 个包到达终点(b)网络改为包交换的数据报网络，每个包头部为2h 比特，内容含F 比特，则大小为F+2h 比特，一个包通过一段链路需要(F+2h)/R 秒，通过Q 段链路需要Q*(F+2h)/R 秒，因为使用无连接服务，所以不需要连接建立时间，假设所传的文件为M 个包，则所有相应的时间可表示如下：Q*(F+2h)/R ： 第1个包到达终点（注意此时第2个包已经在倒数第2个节点） (Q+1)*(F+2h)/R ： 第2个包到达终点…(Q+M-1)*(F+2h)/R ： 第M 个包到达终点(c)网络已改为电路交换网络，即该文件传输期内占用全部的链路资源，需要传输的内容为h+MF ，则该文件从起点到终点的传输时间为t s +(MF+h)/RP64，No.6(a)s m d prop /=(b)R L d trans /=(c)R L s m d end to end //+=−−(d) 此时最后1位刚刚离开主机A(e) 此时第1位处于链路中，尚未到达主机B(f) 此时第1位已经到达主机B (e) km S R L m 89310*5.210*2810083===P73, No.7对于包里的第1位而言，在被发送前，需要包的全部内容被转换，即需要610648483=⋅⋅msec ， 该包到达链路的另一端需要sec 3841018486μ=⋅⋅，加上传播时延2msec ，则总时延是 sec 384.8sec 2sec 384sec 6m m m =++μ。
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1复习题1.没有不同。
主机和端系统可以互换。
端系统包括PC，工作站，WEB服务器，邮件服务器，网络连接的PDA，网络电视等等。
2.假设爱丽丝是国家A的大使，想邀请国家B的大使鲍勃吃晚餐。
爱丽丝没有简单的打个电话说“现在我没一起吃晚餐吧”。
而是她先打电话给鲍勃建议吃饭的日期与时间。
鲍勃可能会回复说那天不行，另外一天可以。
爱丽丝与鲍勃不停的互发讯息直到他们确定一致的日期与时间。
鲍勃会在约定时间（提前或迟到不超过15分钟）出现再大使馆。
外交协议也允许爱丽丝或者鲍勃以合理的理由礼貌的退出约会。
3.联网（通过网络互联）的程序通常包括2个，每一个运行在不同的主机上，互相通信。
发起通信的程序是客户机程序。
一般是客户机请求和接收来自服务器程序的服务。
4.互联网向其应用提供面向连接服务（TCP）和无连接服务（UDP）2种服务。
每一个互联网应用采取其中的一种。
面相连接服务的原理特征是：①在都没有发送应用数据之前2个端系统先进行“握手”。
②提供可靠的数据传送。
也就是说，连接的一方将所有应用数据有序且无差错的传送到连接的另一方。
③提供流控制。
也就是，确保连接的任何一方都不会过快的发送过量的分组而淹没另一方。
④提供拥塞控制。
即管理应用发送进网络的数据总量，帮助防止互联网进入迟滞状态。
无连接服务的原理特征：①没有握手②没有可靠数据传送的保证③没有流控制或者拥塞控制5.流控制和拥塞控制的两个面向不同的对象的不同的控制机理。
流控制保证连接的任何一方不会因为过快的发送过多分组而淹没另一方。
拥塞控制是管理应用发送进网络的数据总量，帮助防止互联网核心（即网络路由器的缓冲区里面）发生拥塞。
6.互联网面向连接服务通过使用确认，重传提供可靠的数据传送。
当连接的一方没有收到它发送的分组的确认（从连接的另一方）时，它会重发这个分组。
7.电路交换可以为呼叫的持续时间保证提供一定量的端到端的带宽。
今天的大多数分组交换网（包括互联网）不能保证任何端到端带宽。
当发生拥塞等网络问题时，TDM中的数据丢失可能只会是一部分，而FDM中就可能是大部分或全部。
8.在一个分组交换网中，在链路上流动的来自不同来源的分组不会跟随任何固定的，预定义的模式。
在TDM电路交换中，每个主机从循环的TDM帧中获得相同的时隙。
9.t0时刻，发送主机开始传输。
在t1=L/R1时刻，发送主机完成发送并且整个分组被交换机接收（无传输时延）。
因为交换机在t1时刻接收到了整个分组，它可以在t1时刻开始向接收主机发送分组。
在t2=t1+L/R2时刻，交换机完成传输且接收主机收到了整个分组（同样，无传输时延）。
所以，端到端实验是L/R1+L/R2。
10.在一个虚电路网络中，每个网络核心中的分组交换机都对经过它传输的虚电路的连接状态信息进行维护。
有的连接状态信息是维护在一个虚电路数字传输表格中。
11.面向连接的VC电路的特点包括：a.建立和拆除VC电路是需要一个信令协议；b.需要在分组交换中维持连接状态。
有点方面，一些研究者和工程人员争论到：使用VC电路可以更容易提供QoS业务，如：保证最小传输率的业务，以及保证端到端的最大分组延时的业务。
12.a.电话线拨号上网：住宅接入；b.电话线DSL上网：住宅接入或小型办公；c.混合光纤同轴电缆：住宅接入；d.100M交换机以太网接入：公司；e.无线局域网：移动接入；f.蜂窝移动电话（如WAP）：移动。
13.一个第一层ISP与所有其它的第一层ISP相连；而一个第二层ISP只与部分第一层ISP相连。
而且，一个第二层ISP是一个或多个第一层ISP的客户。
14.POP是ISP网络中一个或多个路由器构成的一个组，其它ISP中的路由器也可以能连接到这个POP。
NAP是一个很多ISP（第一层，第二层，以及其它下层ISP）可以互联的局部网络。
15.HFC的带宽是用户间共享的。
在下行信道，所有的分组从头到尾由同一个源发出，因此在下行信道不会发生冲突。
16.以太网的传输速率有：10Mbps，100Mbps，1Gbps和10Gbps。
对于一个给定的传输速率，如果用户单独在线路上传输数据，则可以一直保持这个速率；但是如果有多个用户同时传输，则每个都不能达到所给定的速率（带宽共享）。
17.以太网通常以双绞线或者细的同轴电缆为物理媒体，也可以运行在光纤链路和粗同轴电缆上。
18.拨号调制解调器：最高56Kbps，带宽专用；ISDN：最高128Kbps，带宽专用；ADSL：下行信道5－8Mbps，上行信道最高1Mbps，带宽专用；HFC：下行信道10－30Mbps，上行信道一般只有几Mbps，带宽共享。
19.时延由处理时延、传输时延、传播时延和排队时延组成。
所有这些时延除了排队时延都是固定的。
20.5种任务为：错误控制，流量控制，分段与重组，复用，以及连接建立。
是的，这些任务可以由两层（或更多层）来执行，比如：经常在多于一个层次上执行错误控制。
21.英特网协议栈的5个层次从上倒下分别为：应用层，传输层，网络层，链路层，和物理层。
每一层的主要任务见1.7.1节。
应用层是网络应用程序及其应用层协议存留的地方；运输层提供了一个在应用程序的客户机和服务器之间传输应用层报文的服务；网络层负责将称为数据报的网络层分组从一台主机移动到另一台主机；链路层：通过一系列分组交换机（Internet中的路由器）在源和目的地之间发送分组；物理层：将该帧中的一个一个比特从一个节点移动到下一个节点。
22.应用层报文：应用程序要发出的在传输层上传递的数据；传输层报文段：将应用层报文加上传输层包头，由传输层管理和封装的信息；网络层数据报：将传输层报文段加上网络层包头之后封装；链路层帧：将网络层数据报加上链路层包头之后封装。
23.路由器处理第一层到第三层（这是一个善意的谎话？本人理解为“这样说不确切”因为现代路由器常常还要扮演防火墙或者高速缓存器的角色，一次也处理第四层）；链路交换机处理第一层和第二层；主机处理所有的5层。
习题1.本题不止一个答案，很多协议都能解决这个问题，下面是一个简单的例子：2. a.电路交换网更适合所描述的应用，因为这个应用要求在可预测的平滑带宽上进行长期的会话。
由于传输速率是已知，且波动不大，因此可以给各应用会话话路预留带宽而不会有太多的浪费。
另外，我们不需要太过担心由长时间典型会话应用积累起来的，建立和拆除电路时耗费的开销时间。
b.由于所给的带宽足够大，因此该网络中不需要拥塞控制机制。
最坏的情况下（几乎可能拥塞），所有的应用分别从一条或多条特定的网络链路传输。
而由于每条链路的带宽足够处理所有的应用数据，因此不会发生拥塞现象（只会有非常小的队列）。
3. a.因为这4对相邻交换机，每对之间可以建立n条连接，；因此最多可以建立4n条连接。
b.可以通过右上角的交换机建立n条连接，并且可以通过左下角交换机建立n条连接，因此最多可以建立2n条连接。
4.由于收费站间隔100km，车速100km/h，收费站以每12m通过一辆汽车的速度提供服务。
a)10辆车，第一个收费站要花费120s，即2分钟来处理。
每一辆车要达到第二个收费站都会有60分钟的传输延时，因此每辆车要花费62分钟才能达到第二个收费站，从第二个收费站到第三个收费站重复这一过程。
因此，（端到端）总延时为124分钟。
b)每两个收费站之间的延时为7×12秒＋60分＝61分24秒，（端到端）总延时＝3624×2＝7,248s，即112分48秒。
5.a)传输一个分组到一个链路层的时间是（L+h）/R。
Q段链路的总时间为：Q（L+h）/R。
所以发送文件所需要总的时间为：t s＋（L+h）/Q。
b)Q（L＋2h）/Rc)由于链路上没有存储转发延时，因此，总延时为：t s＋（L+h）/R。
6.a)传播时延d prop＝m/s秒b)传输时延d trans=L/R秒c)端到端时延d end－to－end＝（m/s＋L/R）秒d)该分组的最后一个bit刚刚离开主机A。
e)第一个比特在链路中，还没有到达B。
f)第一个比特已经到达B。
g)m=LS/R=100×2.5e8/28e3＝893km7.考虑分组中的第一个bit。
在这个bit被传输以前，先要收集这个分组中的其它bit，这个需要：48×8/64e3 ＝6e－3s＝6ms分组的传输延时：48×8/1e6=384e-6 s=0.384ms传播时延：2ms到该bit被解码的时延为：6＋0.384＋2＝8.384ms（英文答案中的那个“.”表示乘）8.a)由于每个用户需要十分之一的带宽，因此可以支持10个用户。
b)p=0.1c)d)我们用中心极限定理来求这个概率的近似解。
令X j表示J个用户同时传输的概率，如P（X j=1）=p,则：所以所求概率约为：0.0019.10.传输这N个分组需要LN/R秒。
当一批N个分组到达时，缓存器内是空的。
第一个分组没有排队时延，第二个分组的排队时延为L/R秒......第N个分组的排队时延为：（N-1）L/R秒，所以平均排队时延为：11.a)传输时延为L/R，总时延为b)令x=L/R，则总时延与x的函数为：总时延＝x/（1－ax）。
12.a)一共有Q个节点（源主机喝N-1个路由器）用表示第q个节点的处理时延。
用R q表示第q个节点处的传输速率，令。
用表示通过第q条链路的传播时延。
则：b)用表示节点q处的平均排队延时，则：13.实验题？不会考吧。
14.a)“带宽时延”积＝（1e7/2.5e8）*1e6=40,000bitb)40000bitc)一条链路的带宽时延积就是这条链路上具有的比特数的最大值。
d)1e7/4e4=250m,比一个足球场的长度还长。
e)s/R15.2.5e8/1e6=25bps16.a)（1e7/2.5e8）*1e9=40,000,000bitb)400,000bit（包长度）c)1e7/4e5＝25m17.a)传播时延＝1e7/2.5e8=40ms；传输时延＝4e5×250/2.5e8= 400ms因此总延时为：440msb)传播时延＝2×40＝80ms（发送及返回确认）；传输时延＝4e4×250/2.5e8=40ms，传送10个分组，总时延＝10×（80＋40）＝1200ms＝1.2s18.a)地球同步卫星距离地面3600km，因此该链路的传播时延＝3600e3/2.4e8=150msb)150e-3×10e6=1,500,000c)60×10e6＝6e8bit19.我们假设旅客和行李对应到达协议栈顶部的数据单元，当旅客检票的时候，他的行李也被检查了，行李和机票被加上标记。
这些信息是在包裹层被添加的（if Figure 1.20 that不知道怎么翻译......）允许在包裹层使服务生效或者在发送侧将旅客和行李分离，然后在目标测（如果可能的话）重新组合他们。