编程第一课

计算机网络课程设计报告----组建校园局域网.pdf

试 题：《计算机网络课程设计》设计任务和考查内容包含以下四部分。 一、数据抓包和网络协议分析（要求写出抓包的主要截图、数据传输的分析过程及关键实验操作步骤） 1、 捕获ARP请求，ARP应答数据包，分析其组成特征。总结ARP协议运行的基本过程。 2、 捕获ICMP数据包，对其进行分析研究。(icmp请求数据包,icmp应答数据包) 3、 捕获TCP“三次握手”建立连接的数据包并分析；捕获TCP“2次两次握手”释放连接的数据包并分析。 a) tcp连接的建立 b) tcp连接的释放 4、 捕获HTTP，DNS，DHCP数据包，分析其构成。 a) http数据包的捕获与分析 b) DNS数据包的捕获与分析 5、 通过捕捉smtp协议包分析邮箱密码，并分析其工作过程。 提示:有些邮箱直接登录操作无法捕捉，如qq、163邮箱等。需要Web页面登录我的网易或QQ等邮箱，捕获数据并分析出密码。也可采用命令方式收发邮件状态下抓取账号密码数据。 二、网络编程设计（以下三题选择完成一个即可，要求有程序代码和运行结果） 1、 数据包的分析程序设计（不需图形界面，结果输出在命令行下）。 2、 简单的网络通信程序（包括服务器端程序，客户端程序，能实现客户端到服务器的主动连接，发送信息，服务器能否对信息做一次回应）。 3、 编写一个程序，能够根据输入的原始数据，及所选择的多项式，自动生成CRC校验码。 三、网络设备配置 从“CCNA 实验手册201806.doc”中自选两个项目，在机架或模拟器上完成，并写出主要的配置过程、实验数据、 相关截图和结论。其中所选实验要求包含交换机的Vlan配置和路由器的RIP、OSPF配置部分。 四、网络系统工程方案设计 以文华学院为背景，了解文华学院现有校园网络的构成。主要从网速提高,网络安全,无线网络构建等几个方面进行分析,并提出合理的改进意见，设计出更优的校园网络工程方案。 1、需求分析 以文华学院为背景，通过实地调查、现场访谈、书面调查等形式了解文华学院的组织结构、网络建设的背景，对文华学院的网络工程有一个全面的感性的认识。建设文华学院需要重新建立一个校园网，在前面了解的知识背景之上，明确网络需求和网络性能的评价标准。具体地，包括网络建设的目的与原则、投资规模、现有网络的问题与不足等；网络系统中所包含的信息点的数量、分布及信息流量、应用程序的类型及对QoS的要求、是否需要提供广域网接入和网络安全上的考虑因素等。需求分析完成后需要提交需求分析调研报告。 2、方案设计 根据需求分析，以层次化的网络设计方法，选择合适的网络技术，设计一个性能价格比相对优化的网络解决方案，网络要提供尽可能高的可靠性、有效性、可扩展性和安全性。设计内容包括、网络拓扑设计、IP寻址模式、路由规划、安全设计、网络管理、服务器相关配置等。 3、方案文档的撰写 撰写一个网络设计方案的书面文档，在该文档中要包括需求分析概述、网络建设的目标与原则、技术选择与技术设计、网络管理与安全、投资预算和设备清单等。 4、说明：完成此部分课程设计,请大家查阅相关资料了解下列网络术语,并尽可能将这些技术应用于你的设计之中；并按实验指导资料，完成校园网模拟配置实验。例如包括： 交换以太网、快速以太网、千兆以太网、VLAN、第三层交换技术、防火墙、无线局域网、DMZ技术等设备技术的选型配置 网络拓扑设计和地址规划等 接入层：交换机堆叠、交换机端口安全、ACL（访问控制列表）、802.1x 汇聚层：链路聚合、动态主机配置 核心层：骨干路由设备、服务器等 出口：NAT、防火墙等

计算机网络复习题 一、填空题 1．从计算机网络系统组成的角度看，计算机网络可以分为　通信　子网和　资源　子网 。 2．在Internet与Intranet之间，由　防火墙　　负责对网络服务请求的合法性进行检查 。 3．差错控制技术主要包括前向差错控制和　自动反馈重发/ARQ　； 4．根据IEEE802模型的标准将数据链路层划分为__LLC__子层和___MAC__子层。 5．同轴电缆按阻抗可分为50欧姆和75欧姆两种，50欧姆同轴电缆主要用于传输_数字信 号_信号，此类同轴电缆叫做_基带_同轴电缆。而75欧姆同轴电缆主要用于传输_模拟信 号_，此类同轴电缆又称为宽带同轴电缆。 6．决定局域网特征的主要技术有拓扑结构 、 传输介质、介质访问控制方法。 7．计算机网络按作用范围（距离）可分为__局域网/LAN__、___城域网/MAN___和__广域 网/WAN___； 8．调制解调器的作用是实现__模拟/数字_信号和__数字/模拟__信号之间的转变；数字 数据在数字信道上传输前需进行__编码______，以便在数据中加入时钟信号； 9．在一个IP网络中负责主机IP地址与主机名称之间的转换

网络安全课程设计，文档47页，附拓扑图、eNSP直接运行。 目 录 第1章 绪 论 1 1.1 课题研究背景和意义 1 1.2 国外现状分析 1 1.3 国内现状分析 2 1.4 课题研究的主要内容 3 第2章 设计分析 4 2.1 校园网的特点 4 2.2 校园网网络结构 4 2.3 校园网络安全体系问题分析 6 2.3.1 总体需求分析 6 2.3.2 网络安全平台需求分析 7 2.3.3 应用环境的安全需求分析 7 2.3.4 管理安全需求分析 8 第3章 团队协作 9 3.1 团队组成 9 3.2 团队分工 9 第4章 校园网络安全体系构建的设计与实现 11 4.1 校园网网络安全体系模型P-WPDRRC的建立 11 4.2 防火墙与入侵检测技术结合实现安全防范 14 4.3 主动防御技术在校园网安全体系中的应用 15 4.4 校园网络安全体系的实现 16 4.4.1 物理安全的实现 16 4.4.2 网络安全的实现 17 4.4.3 系统安全的实现 17 4.4.4 应用安全的实现 18 4.4.5 管理安全的实现 19 第5章 设计结果与分析 21 .......

大三计算机网络的课程设计项目 报告中相关命令代码写的很清楚了，拓扑图用思科模拟器直接打开就行，拓扑图都不用自己搭建

感谢你的观看 计算机网络课程设计报告 题目： 组建小型企业局域网 学生姓名： 学 号： 班 级： 指导教师： 2012年 05 月20 日 摘 要 随着计算机及局域网络应用的不断深入，特别是各种计算机软件系统被相继应用在实际工作中，在公司内部，需要使各部门相互间真正做到高效的信息交换、资源的共享，为员工提供准确、可靠、快捷的各种数据和信息，充分发挥公司现有计算机设备的功能，从而加强公司内各部门的业务和技术联系，提高工作效率，实现资源共享，降低运作及管理成本，因此，企业网络的建设是企业向信息化发展的必然选择，企业网网络系统是一个非常庞大而复杂的系统，它不仅为现代化企业综合信息管理和办公自动化等一系列应用提供基本操作平台，而且能提供多种应用服务，使信息能及时、准确地传送给各个系统。而企业网工程建设中主要应用了网络技术中的重要分支局域网技术来建设与管理的。本课程设计将从小型企业局域网的建设需求、建设原则、拓扑结构、子网划分以及路由协议等方面，展示一个企业网的组建过程，详细介绍关于交换机和路由器等的相关知识。 关键词：局域网，交换机，路由器，资源共享 目录 摘要………………………………………………………………1 1.课程设计目的及要求…………………………………………3 2.组建小型企业局域网…………………………………………3 3.课程设计的内容 1.需求分析…………………………………………………….5 2.设计原则…………………………………………………...6 4.拓扑结构图……………………………………………………7 5.详细步骤 1.设计总述……………………………………………………7 2.子网划分…………………………………………………..8 3.路由协议的选择…………………………………………8 6.配置的代码……………………………………………………9 7.最终的结果…………………………………………………..15 8.课程设计总结………………………………………………..17 一、课程设计目的及要求： 通过一周的课程设计，培养进一步理解和掌握网络组网的过程及方案设计，为今后从事实际工作打下基础；熟练掌握子网划分及路由协议的配置，熟练掌握路由器和交换机的基本配置。 要求能根据实际问题绘制拓扑结构图，拓扑结构图可以是树形、星形、网状形、环状形及混合形结构的之一，清晰的描述接口，进行路由器或交换机的代码配置实现，并且每个方案的需有以下几部分的内容： 需求特点描述； 设计原则； 解决方案设计，其中必须包含：（1）设备选型；（2）综合布线设计；（3）拓扑图；（4）IP地址规划；（5）子网划分； （6）路由协议的选择； (7) 路由器配置。 4.课程设计要求： 现有50台计算机，设计一个小型企业网络方案，要求： 1．资源共享，网络内的各个桌面用户可共享数据库、共享打印机，实现办公自动化系统中的各项功能； 2．通信服务，最终用户通过广域网连接可以收发电子邮件、实现Web应用、接入互联网、进行安全的广域网访问； 3．划分网段； 4．选择路由协议，配置路由，并为路由器设置密码。 二、组建小型企业局域网： "组网"是我们经常讨论的一个话题，如何让组网的费用降到最低，同时又保持网络的稳定和可靠。 首先，我们确定了这个网络规模拥有50台计算机，那么这个网络是拿来做什么用的呢？也就是我们要对这个网络进行需求分析，从50台计算机的规模来看，比较适合用在办工作环境中，在一般的办公环境下，组网主要是为了方便内部传输资料和访问互联网络，带宽要求不高，完全可以选择ADSL宽带接入，既省钱，效率又高。关于选择多少兆的ADSL宽带接入，那就要看情况了。不过从50台计算机的规模来看，选择8MB ADSL接入比较合适，价格不贵，在这里我就不说宽带收费问题了。 在确定宽带接入方式后，我们就要开始选择组网的模式。组网模式主要有两个大的类别，一种是有线，一种是无线。如果组建一个无线局域网，虽然省去了布线的麻烦，但考虑到办公环境中大部分还是用的台式机，如果每台计算机配一个无线网卡，少说也要好几千，成本太高，不符合小型办公网络的标准。所以，我们还是选择有线网络。但有线网络也有两种方案适合组建一个50台计算机的网络，一种是采用带路由功能的ADSL Modem加上一个超过50口的交换机，实现ADSL共享上网，但这种方式组建的网络性能较差。另外一种就是采用路由器加交换机组网，这种方式虽然投入的成本高一些，但性能却很好，所以，我们选择路由器加交换机的方式来组网。 在确定组网方式后，我们就要看看办公室内计算机分布的位置，因为这将直接决定我们使用的网络设备摆放位置，这一点对组网以及管理都非常重要。在通常情况下，一间办公室内能保证10人办公，也就是说一间办公室内拥有10台计算机，那么50台计算机大约分布

随着 Internet 的飞速发展，网络已是人类生活不可缺少的重要成分，信息时代给 人类的生活带来极大便利的同时，也对人类的生活造成了破坏，日益突出的信息安全问 题也越来越引起大家的注意，公共数据的安全问题日益成为人们关注的焦点，而防火墙 作为最早出现的网络安全产品也是目前最基本最有效的信息安全防护手段，正日益受到 用户和研发机构的青睐,得到公司和个人的广泛应用。 经过详细调查，考虑到现有硬件设备的限制，因包过滤防火墙的优势更明显，速度 快且效率高，功能强大，不耗内存，并可对数据进行细致的控制，自己设定过滤规则， 方便测试。在保证满足实验要求的环境下又能尽可能地简化了实验环境，因此本论文实 现的是基于主机设计的防火墙配置系统，故只需要一个联网的主机加虚拟机即可进行实 验测试。所以最终选择在 Linux 环境下基于 netfilter/iptables 防火墙设计技术实现 包过滤型软件防火墙的设计与应用。

华中科技大学-网络空间安全学院-网络安全课程设计，Linux 下状态检测防火墙的设计与实现+源代码+文档说明+实验报告 - 小白不懂运行，下载完可以私聊问，可远程教学 该资源内项目源码是个人的课程设计，代码都测试ok，都是运行成功后才上传资源，答辩评审平均分达到96分，放心下载使用！ ## 项目备注 1、该资源内项目代码都经过测试运行成功，功能ok的情况下才上传的，请放心下载使用！ 2、本项目适合计算机相关专业(如计科、人工智能、通信工程、自动化、电子信息等)的在校学生、老师或者企业员工下载学习，也适合小白学习进阶，当然也可作为毕设项目、课程设计、作业、项目初期立项演示等。 3、如果基础还行，也可在此代码基础上进行修改，以实现其他功能，也可用于毕设、课设、作业等。 下载后请首先打开README.md文件（如有），仅供学习参考, 切勿用于商业用途。 --------

图书名称： 计算机网络技术与应用 出 版 社： 研究出版社 出版日期： 2009年5月 作 者： 韩德志 吴彩虹 刘昊 查东辉 王畅 本书是根据普通高等教育“十一五”国家级规划教材的指导精神而编写的。 本书系统地介绍了计算机网络的基础知识，内容包括计算机网络概述、网络数据通信基础、计算机网络体系结构与协议、网络拓扑结构、网络设备与网络互联、计算机局域网技术、计算机广域网技术、网络操作系统、Internet技术及其应用、计算机网络安全、网络管理与维护以及实践与操作。 本书既注重基础知识的介绍，让读者对计算机网络理论有所认识，又注重实用性和先进性，以增强读者的实践能力。本书图文并茂、通俗易懂，便于初学者学习和掌握，可作为高等院校相关专业的教材和网络技术的培训教材，也可作为网络技术人员的参考资料。

计算机网络（第4版）（美）特南鲍姆（Tanenbaum,A.S.）著，潘爱民 译 本书是国内外使用最广泛的计算机网络经典教材。全书按照网络协议模型（物理层、数据链路层、介质访问控制层、网络层、传输层和应用层），自下而上系统地介绍了计算机网络的基本原理，并给出了大量实例。在讲述各网络层的同时，还融合了近年来迅速发展起来的各种网络技术，如Internet、SONET、ADSL、CDMA、WLAN和蓝牙等。另外，针对当前计算机网络的发展现状以及计算机安全的重要性，本书用了一整章的篇幅对计算机安全进行了深入讨论。\r\n 本书的适用对象很广泛。对于学习计算机网络课程的本科生以及研究生，本书都可以作为教材或教学参考书。每一章后面的大量练习题，可用于课程作业或复习要点。对于从事网络研究、网络工程以及使用和管理网络的科研和工程技术人员，本书也是一本很有价值的参考读物。

计算机网络相关知识、网络结构等等。 序言 前言 第1章 引言 1.1 计算机网络的产生和发展 1.2 计算机网络的功能 1.3 计算机网络分类 1.3.1 局域网 1.3.2 城域网 1.3.3 广域网 1.3.4 互联网 1.3.5 无线网 1.4 网络体系结构 1.4.1 协议分层 1.4.2 服务类型 1.4.3 服务原语 1.5 ISO/OSI参考模型 1.5.1 参考模型 1.5.2 模型评价 1.6 本书的结构 第一部分 数据通信 第2章 数据通信基础知识 2.1 基本概念 2.1.1 信号与通信 2.1.2 模拟通信 2.1.3 数字通信 2.2 数据通信基础理论 2.2.1 信号的频谱和带宽 2.2.2 信道的截止频率与带宽 2.2.3 信道的最大数据传输率 2.3 传输介质 2.3.1 双绞线 2.3.2 同轴电缆 2.3.3 光纤 2.3.4 无线介质 2.4 多路复用 2.4.1 频分多路复用 2.4.2 波分多路复用 2.4.3 时分多路复用 2.5 数据交换技术 2.5.1 电路交换 2.5.2 报文交换 2.5.3 分组交换 2.6 调制解调器 2.6.1 调制方式 2.6.2 Modem标准 2.6.3 Modem分类 2.6.4 工作模式 2.7 小结 习题 第3章 物理层接口 3.1 RS-232-C接口 3.1.1 机械特性 3.1.2 电气特性 3.1.3 功能特性 3.1.4 过程特性 3.1.5 空Modem电缆 3.2 其他接口 3.2.1 RS-449接口 3.2.2 RS-530接口 3.3 小结 习题 第二部分 底层物理网络 第4章 广域网 4.1 广域网结构 4.1.1 虚电路和数据报 4.1.2 两者比较 4.2 广域网实例 4.2.1 PSTN 4.2.2 X.25 4.2.3 DDN 4.2.4 帧中继 4.2.5 SMDS 4.2.6 B-ISDN/ATM 4.3 各种广域网的比较 4.4 小结 习题 第5章 局域网 5.1 介质访问控制协议 5.1.1 ALOHA协议 5.1.2 CSMA协议 5.1.3 CSMA/CD协议 5.2 以太网和IEEE 802.3 5.2.1 物理层标准 5.2.2 MAC协议 5.2.3 性能分析 5.3 令牌环网和IEEE 802.5 5.3.1 MAC协议 5.3.2 管理与维护 5.3.3 性能分析 5.4 网桥 5.4.1 透明网桥 5.4.2 源选径网桥 5.5 小结 习题 第6章 高速局域网 6.1 FDDI网络 6.1.1 与OSI的关系 6.1.2 帧格式 6.1.3 MAC协议 6.1.4 工作原理 6.1.5 拓扑结构 6.1.6 网络容错 6.1.6 技术指标 6.2 快速以太网 6.3 千兆位以太网 6.4 局域网交换机 6.5 小结 习题 第三部分 网络互联 第7章 网络互联与TCP/IP 7.1 网络互联层次 7.1.1 应用级互联 7.1.2 网络级互联 7.2 TCP/IP参考模型 7.3 TCP/IP参考模型的特点 7.3.1 TCP/IP的两大边界 7.3.2 IP层的地位 7.3.3 TCP/IP的可靠性思想 7.3.4 TCP/IP模型的特点 7.4 TCP/IP与ISO/OSI 7.5 小结 习题 第8章 IP 8.1 IP数据报 8.1.1 数据报格式 8.1.2 地址格式 8.2 IPv6 8.2.1 固定头部格式 8.2.2 IPv6地址 8.2.3 扩展头部 8.3 小结 习题 第9章 ARP、RARP和ICMP 9.1 ARP和RARP 9.1.1 ARP 9.1.2 RARP协议 9.1.3 报文格式 9.2 ICMP 9.2.1 ICMP报文类型 9.2.2 ICMP报文格式 9.2.3 ICMP差错报文 9.2.4 ICMP控制报文 9.2.5 ICMP请求/应答报文 9.3 小结 习题 第10章 IP路由协议 10.1 路由器与路由选择 10.1.1 路由器 10.1.2 路由选择 10.2 Internet结构 10.3 基本路由算法 10.3.1 D-V路由算法 10.3.2 L-S路由算法 10.4 IGP：内部网关协议 10.4.1 RIP 10.4.2 IGRP 10.4.3 OSPF协议 10.5 外部网关协议EGP 10.6 Internet组播 10.6.1 组播协议 10.6.2 组的维护 10.6.3 组播范围 10.7 移动IP路由 10.8 无类域间路由 10.9 小结 习题 第11章 传输层协议：TCP和UDP 11.1 传输层基本原理 11.1.1 服务质量 11.1.2 传输层端口 11.2 UDP 11.2.1 UDP报文格式 11.2.2 UDP伪头部 11.2.3 UDP多路复用 11.3 TCP 11.3.1 TCP报文格式 11.3.2 TCP连接端点 11.3.3 TCP可靠传输 11.3.4 TCP流量控制 11.3.5 TCP拥塞控制 11.3.6 TCP连接建立 11.3.7 TCP连接删除 11.3.8 TCP紧急数据传输 11.4 小结 习题 第四部分 网络应用 第12章 客户/服务器模型与套接字编程接口 12.1 网络间进程通信 12.2 客户/服务器模型 12.3 套接字编程接口 12.3.1 套接字基本概念 12.3.2 套接字系统调用 12.3.3 其它系统调用 12.4 基于客户/服务器模型的套接字编程举例 12.4.1 客户/服务器模型流程图 12.4.2 套接字实现机制 12.4.3 Unix环境下的套接字编程举例 12.5 小结 习题 第13章 DNS域名系统 13.1 域名结构 13.1.1 平面型命名机制 13.1.2 层次型命名机制 13.1.3 层次型名字管理 13.1.4 TCP/IP域名 13.2 域名解析 13.2.1 TCP/IP域名服务器 13.2.2 域名解析 13.2.3 逆向域名解析 13.2.4 域名解析的效率 13.3小结 习题 第14章 远程登录Telnet和Rlogin 14.1 为什么要引入远程登录？ 14.2 TELNET协议 14.2.1 Telnet工作原理 14.2.2 网络虚终端 14.2.3 Telnet选项 14.3 Rlogin 14.4 小结 习题 第15章 文件传输与访问 15.1 FTP：文件传输协议 15.1.1 FTP特点 15.1.2 FTP工作原理 15.1.3 FTP连接建立 15.1.4 FTP访问控制 15.2 TFTP：简单文件传输协议 15.3 NFS：网络文件系统 15.4 小结 习题 第16章 电子邮件 16.1 电子邮件系统体系结构 16.1.1 ISO/OSI电子邮件系统 16.1.2 TCP/IP电子邮件系统 16.2 TCP/IP电子邮件地址 16.3 TCP/IP电子邮件标准 16.3.1 TCP/IP电子邮件格式 16.3.2 MIME：多用途Internet邮件扩展 16.3.3 SMTP：简单邮件传输协议 16.4 邮箱访问 16.5 小结 习题 第17章 万维网 17.1 Web页面浏览 17.1.1 浏览器和服务器 17.1.2 HTTP 17.1.3 HTML语言 17.2 交互式动态页面 17.2.1 CGI 17.2.2 Java 17.2.3 ASP 17.3小结 习题 第五部分 网络管理与安全 第18章 网络管理 18.1 互连网管理 18.2 SNMP网管体系 18.2.1 管理员/代理模型 18.2.2 MIB：管理信息库 18.2.3 ASN.1 18.2.4 SNMP 18.3 小结 习题 第19章 网络安全 19.1 基本概念 19.2 网络安全攻击 19.3 安全策略 19.4 安全机制 19.4.1 加密 19.4.2 鉴别 19.4.3 数字签名 19.5 防火墙 19.5.1 包过滤 19.5.2 应用级网关 19.6 小结 习题 第20章 网络技术的未来发展 20.1 新型网络应用技术 20.2 宽带网络技术 20.3 无线接入技术 20.4 统一网络技术 20.5 网络安全技术 20.6 主动网络技术 20.7 小结 参考文献

个人在学习该课程时候自己做的复习笔记，从物理层一直到应用层都有，一共85页，文字图片搭配合理

计算机网络：自顶向下方法（原书第6版）第1版于12年前出版，首创采用自顶向下的方法讲解计算机网络的原理和协议，出版以来已被几百所大学和学院选用。是业界经典的计算机网络教材之一。 《计算机网络：自顶向下方法（原书第6版）》继续保持了以前版本的特色，为计算机网络教学提供了一种新颖和与时俱进的方法。 同时也进行了相当多的修订和更新：第1章更多地关注时下，更新了接入网的论述；第2章用Python替代了Java来介绍套接字编程；第3章补充了用于优化云服务性能的TCP分岔知识；第4章有关路由器体系结构的内容做了大量更新；第5章重新组织并新增了数据中心网络的内容；第6章更新了无线网络的内容以反映其全新进展；第7章进行了较大修订，深入讨论了流式视频，包括了适应性流和CDN的讨论；第8章进一步讨论了端点鉴别；等等。另外，书后习题也做了大量更新。

该PPT是谢希仁《计算机网络》教材对应的PPT，全面系统地介绍了计算机网络的发展和原理体系结构、物理层、数据链路层等内容。

计算机网络微课堂课件（湖科大教书匠） - 深入浅出，掌握网络世界的关键知识！ 专栏笔记地址：https://blog.csdn.net/qq_51646682/category_11661071.html?spm=1001.2014.3001.5482 课件亮点： 1. 全面覆盖：从基础理论到实践应用，涵盖计算机网络的各个方面，适合不同水平的学习者。 2. 生动案例：结合最新的技术动态和经典案例，使学习内容生动、实用。 3. 深度解析：对复杂的网络协议和技术进行深入浅出的讲解，帮助你轻松理解难点。 4. 互动练习：包含大量的习题和实验操作，加强理论与实践的结合。 适合人群： - 计算机专业学生 - IT行业从业者 - 网络技术爱好者 课件内容涵盖： - 网络基础：OSI模型、TCP/IP模型 - 核心协议：IP、TCP、UDP、HTTP等 - 网络安全：加密技术、防火墙等 - 最新技术：SDN、物联网、5G网络等

计算机网络笔记pdf - 掌握网络技术的钥匙！ 笔记特色： 1. 精炼要点：笔记精心整理，突出重点，便于快速回顾和理解计算机网络的核心概念。 2. 实战技巧：分享实用的网络配置技巧和故障排查方法，提升你的实战能力。 3. 最新动态：融入当前网络技术的最新发展，让你与时俱进。 4. 图解辅助：大量图解和示例，帮助你形象理解复杂的网络原理和协议。 适合人群： - 计算机专业学生及考研人士 - 网络工程师和IT技术人员 - 对计算机网络感兴趣的自学者 笔记内容覆盖： - 网络架构：深入浅出讲解OSI七层模型与TCP/IP模型 - 关键协议：详解IP、TCP、UDP、HTTP等协议的工作原理 - 前沿技术：物联网、云计算、5G等新兴技术的网络应用

这套课程参考了大家都在用的教材《计算机网络（第 8 版）》，课程里的内容按照教材编著者谢希仁教授的编排顺序，先从物理层开始介绍，包括物理层的传输媒体、信道复用技术等，再到数据链路层、网络层、运输层和应用层。每一层我都介绍了该层的主流协议，比如数据链路层的 ppp 协议、网络层的 ip 协议，运输层的 TCP、UDP 协议和应用层的 HTTP 协议。在介绍协议的时候，我不仅介绍了协议的用途，还讲解了协议的内部执行逻辑。从点到面，逐步深入，相信，有这样丰富的课程内容，再配上我生动的课程讲解，一定会让你轻松掌握计算机网络课程相关的知识。 课程目标： 通过本套课程的学习，你将具备： - 掌握计算机网络的行业发展历程 - 掌握 TCP/IP 五层网络体系结构 - 掌握计算机网络核心协议内容

计算机网络、现代通信组网相关的教程课件&案例&相关项目。 计算机网络和现代通信组网是信息技术领域的重要分支，涉及数据传输、网络协议、网络安全、无线通信等多个方面。以下是一些关于计算机网络和现代通信组网的教程课件、案例以及相关项目资源的描述： 一、基础理论与实践教学 1. 课程设置 网络基础理论：涵盖OSI七层模型、TCP/IP模型、IP地址分类与子网划分等基础知识点。 数据通信原理：介绍数据编码、调制解调技术、多路复用技术及差错控制方法。 2. 实验环节 网络配置实验：使用Cisco Packet Tracer或GNS3进行路由器和交换机的基础配置实验。 网络协议分析：利用Wireshark进行数据包捕获和分析，理解TCP/IP通讯过程。 网络安全实验：设置防火墙策略，进行网络攻击与防御的模拟实验。 二、高级网络设计与管理 1. 课程内容 网络架构设计：教授如何设计可扩展和高可用的网络架构。 网络管理技术：包括SNMP简单网络管理协议、网络监控和性能评估工具的使用。 2. 实战训练 虚拟化网络部署：使用VMware或Hyper-V部署虚拟网络，实践云网络技术。 企业级网络方案设计

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TCP协议分析实验指导 计算机网络实验节选

大学物联网通信技术期末复习ppt 全书共12章，分为三大部分：第一部分讲述物联网的基本知识，包括第1章物联网概述和第2章物联网体系架构；第二部分讲述感知层通信技术，包括第3章工业控制网络技术、第4章短距离无线通信技术和第5章无线传感器网络；第三部分讲述网络层通信技术，包括第6章接入网技术、第7章无线局域网技术、第8章电话通信网技术、第9章移动通信技术、第10章传送网技术、第11章虚拟专用网技术和第12章计算机网络技术。每章后都配有思考题。 本书注重选材，内容翔实，层次清楚，编写方法新颖，既可作为高等学校物联网专业以及信息、通信、电子、计算机、工程管理等专业本科生的教材，也可作为从事物联网研究的专业技术人员、管理人员，特别是从事物联网研究和设置的人员的参考书。

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计算机网络实验报告 一、实验名称： 简单以太网的组建 二、实验内容 1．观察教学机房，了解计算机网络结构，并画出计算机网络拓扑结构图。 2．了解计算机网络中的网络设备，并了解每台计算机上使用的网络标识、网络协议。 3．制作 2 根直通双绞线和 2 根交叉线，并测试。 4．分别用制作好的直通线、交叉线以及串口线、并口线，连接两台计算机。 三、实验步骤 1、教学机房网络拓扑结构 （1）记录联网计算机的数量、配置、使用的操作系统、网络拓扑结构、网络建成的时间 等数据。 "名称 "数量 "配置 "操作系统 " "教学机房（建成于"40 "CPU:Inet I5-6500 "Windows7 " "2016年10月） " "四核3.2GHz;内存：4G " " " " "DDR4，硬盘：500GB，显卡： " " " " "NVIDIA GeForceGTX750Ti 2GB，" " " " "显示器21.5寸 " " " " "LED，主板集成千兆网卡。 " " （2）认识并记录网络中使用的其他硬件设备的名称、用途和连接的方法。 "设备名称 "用途 "连接方法 " "交换机 "交换机是一种用 "交换机主要作为局域网设备集中连接使用" " "于电信号转发的 "，所以总的来说，在硬件连接方面相对要" " "网络设备。它可 "简单许多，通常只需把相应的传输介质接" " "以为接入交换机 "头插入到相应的交换机接口上即可。即将" " "的任意两个网络 "双绞线水晶头连接至交换机RJ-45口。 " " "节点提供独享的 " " " "电信号通路。 " " （3）根据以上数据及观察结果画出拓扑结构图。 （4）分析网络使用的结构及其所属类型。 教学机房所使用的是星型拓扑结构； 在星型拓扑结构中，网络中的各节点通过点到点的方式连接到一个中央节点（又称中央 转接站，一般是集线器或交换机）上，由该中央节点向目的节点传送信息。中央节点执 行集中式通信控制策略，因此中央节点相当复杂，负担比各节点重得多。在星型网中任 何两个节点要进行通信都必须经过中央节点控制。 优点： （1）控制简单。任何一站点只和中央节点相连接，因而介质访问控制方法简单，致使访 问协议也十分简单。易于网络监控和管理。 （2）故障诊断和隔离容易。中央节点对连接线路可以逐一隔离进行故障检测和定位，单 个连接点的故障只影响一个设备，不会影响全网。 （3）方便服务。中央节点可以方便地对各个站点提供服务和网络重新配置。 缺点： （1）需要耗费大量的电缆，安装、维护的工作量也骤增。 （2）中央节点负担重，形成"瓶颈"，一旦发生故障，则全网受影响。 （3）各站点的分布处理能力较低。 总的来说星型拓扑结构相对简单，便于管理，建网容易，局域网普遍采用的一种拓扑结 构。采用星型拓扑结构的局域网，一般使用双绞线或光纤作为传输介质，符合综合布线 标准，能够满足多种宽带需求。 （5）教学机房学生机主要网络配置参数见下图。 2、直通线的制作 （1）利用网线钳剪下所需要的双绞线长度，接着用双绞线网线钳把双绞线的一端剪齐， 再把剪齐的一端插入到网线钳用于剥线的缺口中。顶住网线钳后面的挡位以后，稍微握 紧网线钳慢慢旋转一圈，让刀口划开双绞线的保护胶皮并剥除外皮。 （2）剥除外包皮后会看到双绞线的4对芯线，可以看到每对芯线的颜色各不相同。将绞 在一起的芯线分开，按照橙白、橙、绿白、蓝、蓝白、绿、棕白、棕的颜色一字排列， 并用网线钳将线的顶端剪齐。 （3）将按照上述线序排列的每条芯线分别对应RJ- 45插头的1、2、3、4、5、6、7、8针脚。 （4）使RJ-45插头的弹簧卡朝下，然后将正确排列的双绞线插入RJ- 45插头中。在插的时候一定要将各条芯线都插到底部。由于RJ- 45插头是透明的，因此可以观察到每条芯线插入的位置。 （5）完成双绞线一端的制作工作后，重复2- 4步完成另一端的制作。注意双绞线两端的芯线排列顺序要完全一致。 （6）将双绞线的两端分别插入网线测试仪的RJ-45接口，并接通测试仪电源。如图 3- 5所示。如果测试仪上的8个绿色指示灯都顺利闪过，说明制作成功。如果其中某个指示 灯未闪烁，则说明插头中存在断路或者接触不良的现象。此时应再次对网线两端的RJ- 45插头用力压一次并重新测试，如果依然不能通过测试，则只能重新制作。 图 3-5 测线仪 3、交叉线的制作 交叉线的制作和直通线的基本步骤是一样的，所不同的就是芯线的排列规则了，并且两 个头的排列规则也不一样。一头与直通线一样，从左到右依次是：白橙、橙、白绿、蓝 、白蓝、绿、白棕、棕(这种排序称这568B排序），另一头的排列规则为：白绿、绿、白 橙、蓝、白蓝、橙、白棕、棕（这种排序称这568A排序）。 4、双绞线的测试 直通线：测线仪指示灯1、2、3、4、

　　单选题 　　1.使网络服务器中充斥着大量要求回复的信息，消耗带宽，导致网络或系统停止正常服务，这属于什么攻击类型? (A) 　　A、拒绝服务 　　B、文件共享 　　C、BIND漏洞 　　D、远程过程调用 　　2.为了防御网络监听，最常用的方法是 (B) 　　A、采用物理传输(非网络) 　　B、信息加密 　　C、无线网 　　D、使用专线传输 　　3.向有限的空间输入超长的字符串是哪一种攻击手段?(A) 　　A、缓冲区溢出; 　　B、网络监听 　　C、拒绝服务 　　D、IP欺骗 　　4.主要用于加密机制的协议是(D) 　　A、HTTP 　　B、FTP 　　C、TELNET 　　D、SSL 　　5.用户收到了一封可疑的电子邮件,要求用户提供银行账户及密码,这是属于何种攻击手段? 2021网络安全知识竞赛题库及答案全文共9页，当前为第1页。　　A、缓存溢出攻击; 2021网络安全知识竞赛题库及答案全文共9页，当前为第1页。 　　B、钓鱼攻击 　　C、暗门攻击; 　　D、DDOS攻击 　　6.Windows NT 和Windows 2000系统能设置为在几次无效登录后锁定帐号,这可以防止(B) 　　A、木马; 　　B、暴力攻击; 　　C、IP欺骗; 　　D、缓存溢出攻击 　　7.在以下认证方式中，最常用的认证方式是：(A) 　　A基于账户名/口令认证 　　B基于摘要算法认证 ; 　　C基于PKI认证 ; 　　D基于数据库认证 　　8.以下哪项不属于防止口令猜测的措施? (B) (B) 　　A、严格限定从一个给定的终端进行非法认证的次数; 　　B、确保口令不在终端上再现; 　　C、防止用户使用太短的口令; 　　D、使用机器产生的口令 　　9.下列不属于系统安全的技术是(B) 　　A、防火墙 　　B、加密狗 　　C、认证 　　D、防病毒 　　A、不用生日做密码 2021网络安全知识竞赛题库及答案全文共9页，当前为第2页。　　B、不要使用少于5位的密码 2021网络安全知识竞赛题库及答案全文共9页，当前为第2页。 　　C、不要使用纯数字 　　D、自己做服务器 　　11.不属于常见的危险密码是( D ) 　　A、跟用户名相同的密码 　　B、使用生日作为密码 　　C、只有4位数的密码 　　D、10位的综合型密码 　　12.不属于计算机病毒防治的策略的是( D ) 　　A. 确认您手头常备一张真正"干净"的引导盘 　　B. 及时、可靠升级反病毒产品 　　C. 新购置的计算机软件也要进行病毒检测 　　D. 整理磁盘 　　13.针对数据包过滤和应用网关技术存在的缺点而引入的防火墙技术，这是( 　　A、包过滤型 　　B、应用级网关型 　　C、复合型防火墙 　　D、代理服务型 　　14.在每天下午5点使用计算机结束时断开终端的连接属于( A ) 　　A、外部终端的物理安全 　　B、通信线的物理安全 　　C、偷听数据 　　D、网络地址欺骗 　　15.2003年上半年发生的较有影响的计算机及网络病毒是什么(B) 　　(A)SARS 　　(B)SQL杀手蠕虫 (D) )防火墙的特点。 2021网络安全知识竞赛题库及答案全文共9页，当前为第3页。　　(C)手机病毒 2021网络安全知识竞赛题库及答案全文共9页，当前为第3页。 　　(D)小球病毒 　　16.SQL 杀手蠕虫病毒发作的特征是什么(A) 　　(A)大量消耗网络带宽 　　(B)攻击个人PC终端 　　(C)破坏PC游戏程序 　　(D)攻击手机网络 　　17.当今IT 的发展与安全投入，安全意识和安全手段之间形成(B) 　　(A)安全风险屏障 　　(B)安全风险缺口 　　(C)管理方式的变革 　　(D)管理方式的缺口 　　18.我国的计算机年犯罪率的增长是(C) 　　(A)10% 　　(B)160% 　　(C)60% 　　(D)300% 　　19.信息安全风险缺口是指(A) 　　(A)IT 的发展与安全投入，安全意识和安全手段的不平衡 　　(B)信息化中，信息不足产生的漏洞 　　(C)计算机网络运行，维护的漏洞 　　(D)计算中心的火灾隐患 　　20.信息网络安全的第一个时代(B) 　　(A)九十年代中叶 　　(B)九十年代中叶前 　　(C)世纪之交 2021网络安全知识竞赛题库及答案全文共9页，当前为第4页。　　(D)专网时代 2021网络安全知识竞赛题库及答案全文共9页，当前为第4页。 　　21.信息网络安全的第三个时代(A) 　　(A)主机时代, 专网时代, 多网合一时代 　　(B)主机时代, PC机时代, 网络时代 　　(C)PC机时代,网络时代,信息时代 　　(D)2001年,2002年,2003年 　　22.信息网络安全的第二个时代(A) 　　(A)专网时代 　　(B)九十年代中叶前 　　(

网络安全题库 网络安全题库全文共14页，当前为第1页。网络安全题库全文共14页，当前为第1页。 网络安全题库全文共14页，当前为第1页。 网络安全题库全文共14页，当前为第1页。 网络安全复习题 第7题[单选题] 1. 关于多重防火墙技术错误的是 用专家系统实现对入侵检测， 需将有关入侵的知识转化为 （） A：组合主要取决于网管中心向用户提供什么样的服务，以 结构。 及网管中心能接受什么等级风险。 A：条件结构 B：设置的周边网络被攻破后，内部网络也就被攻破了。 B：then 结构 C：多重防火墙的组合方式主要有两种：叠加式和并行式。 C：if-then 结构 D：新旧设备必须是不同种类、不同厂家的产品。 D：while 循环结构 参考答案： B 参考答案： C 第 2 题[单选题] 第8题[单选题] 背包公钥密码系统利用的是 关于被屏蔽子网错误的是 A：背包问题的多解性 A：如果攻击者试图完全破坏防火墙，他可以重新配置连接 B：背包问题的 NP性 三个网的路由器， 既不切断连接又不要把自己锁在外面， 同 C：欧拉定理 时又不使自己 发现。 D：概率加密技术 B：在内部网络和外部网络之间建立一个被隔离的子网，用 参考答案： B 两台分组过滤路由器将这一子网分别与内部网络和外部网 第 3 题[单选题] 络分开。 下列哪项不是 RAS可靠性的内容？ C：在很多实现中，两个分组过滤路由器放在子网的两端， A：可靠性 在子网内构成一个"非军事区" DMZ。 B：可恢复性 D：这种配置的危险带仅包括堡垒主机、子网主机及所有连 C：可维护性 接内网、外网和屏蔽子网的路由器。 D：可用性 参考答案： A 参考答案： B 第9题[单选题] 第 4 题[单选题] 以下不属于软件的保护方式的是 下面哪一项是计算机网络里最大的安全弱点 A：破坏磁盘保护法 A：网络木马 B：网卡序列号及 CPU序列号：只认随机带的网卡或者 CPU B：计算机病毒 C：软件注册（注册码，序列号，注册文件） C：用户帐号 D：压缩 D：网络连接 参考答案： D 参考答案： C 第 10题[单选题 ] 第 5 题[单选题] 计算机犯罪起源于： 用户策略我们集中归纳为以下几点，哪个是错误的？ A：1940' A：只有用户拥有数据库通道。 B：1950' B：其他用户在没有管理员允许的情况下不能读取或更改文 C：1960' 件。 D：1970' C：管理员应该保证所有用户数据的完整性，机密性和有效 参考答案： A 性。 第 11题[单选题 ] D：用户应该知道他进入的命令，或者那些进入为了他的利 以下哪个不是安全操作系统的基本特征？ 益。 A：弱化特权原则 参考答案： C B：自主访问控制和强制访问控制 第 6 题[单选题] C：安全审计功能 能够提供"审慎的保护"的安全等级是： D：安全域隔离功能 A：A类 参考答案： A B：B类 第 12题[单选题 ] C：C类 下列哪些不属于计算机安全中的硬件类危险： D：D类 A：灾害 参考答案： C B：人为破坏 最新范本 ,供参考！ 网络安全题库全文共14页，当前为第2页。网络安全题库全文共14页，当前为第2页。 网络安全题库全文共14页，当前为第2页。 网络安全题库全文共14页，当前为第2页。 C：操作失误 C：计算机系统 硬件的保护机构不起作用。 或不能提供软件 D：数据泄漏 保护，操作系统没有安全保护机构，造成信息泄漏。 参考答案： D D：通讯网络 终端安装在不安全的环境， 产生电磁辐射， 以 第13题[单选题] 及通讯网线路上的泄漏。 KerberosSSP 能够提供三种安全性服务 , 不是的是 参考答案： A A：认证：进行身份验证 第 19题[单选题 ] B：代理：代替用户访问远程数据 以下加解密方法中，不能够泄露加密算法的是 C：数据完整性：保证数据在传送过程中不被篡改 A：E（ M）=C， D （C）=M D：数据保密性：保证数据在传送过程中不被获取 B：EK（M） =C， DK（C） =M 参考答案： B C：EK1（M）=C， DK2 （C）=M 第14题[单选题] D：EK（M） =C， DF（K）（C） =M Debug 的单步执行是跟踪调试软件技术的基石。 以下不能有 参考答案： A 效破坏 debug 的基本方法是 第 20题[单选题 ] A：抑制跟踪中断 黑客将完成某一动作的程序依附在某一合法用户的正常程 B：封锁键盘输入 序中，这种攻击方式是 C：设置显示器的显示性能 A：口令入侵术 D：屏蔽中断 B：特洛伊木马 参考答案： D C：Email 病毒 第15题[单选题] D：监听术 目录的访问模式的最小集合是？ 参考答案： B A： read 与 write-expand 第 21题[单选题

华中科技大学网络安全课程设计项目，基于Netfilter、Netlink的Linux传输层状态检测防火墙+源代码+文档说明 - 小白不懂运行，下载完可以私聊问，可远程教学 该资源内项目源码是个人的课程设计，代码都测试ok，都是运行成功后才上传资源，答辩评审平均分达到96分，放心下载使用！ ## 项目备注 1、该资源内项目代码都经过测试运行成功，功能ok的情况下才上传的，请放心下载使用！ 2、本项目适合计算机相关专业(如计科、人工智能、通信工程、自动化、电子信息等)的在校学生、老师或者企业员工下载学习，也适合小白学习进阶，当然也可作为毕设项目、课程设计、作业、项目初期立项演示等。 3、如果基础还行，也可在此代码基础上进行修改，以实现其他功能，也可用于毕设、课设、作业等。 下载后请首先打开README.md文件（如有），仅供学习参考, 切勿用于商业用途。 --------

什么是对称密码的本质成分？明文、加密算法、密钥、密文、解密算法。 分组密码和流密码的区别是什么？ 流密码是加密的数字数据流的一个位或一次一个字节。块密码是明文块被视为一个整体，用来产生一个相同长度的密文块...... 分组密码每次处理输入的一组分组，相应的输出一组元素。流密码则是连续地处理输入元素，每次输出一个元素。

Computer Networking: A Top-Down Approach, 6th Edition Solutions to Review Questions and Problems Version Date: May 2012 This document contains the solutions to review questions and problems for the 5th 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 1996-2012 by J.F. Kurose and K.W. Ross. All rights reserved Chapter 1 Review Questions There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, PDAs, Internet-connected game consoles, etc. From Wikipedia: Diplomatic 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. Standards are important for protocols so that people can create networking systems and products that interoperate. 1. Dial-up modem over telephone line: home; 2. DSL over telephone line: home or small 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. HFC bandwidth is shared among the users. On the downstream channel, all packets emanate from a single source, namely, the head end. Thus, there are no collisions in the downstream channel. 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 run over fibers optic links. 9. Dial up modems: up to 56 Kbps, bandwidth is dedicated; ADSL: up to 24 Mbps downstream 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: Wifi (802.11) In a wireless LAN, wireless users transmit/receive packets to/from an base 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. 3G and 4G wide-area wireless access networks. In these systems, packets are transmitted 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 t0 the sending host begins to transmit. At time t1 = L/R1, the sending host completes transmission and the entire packet is received at the router (no propagation delay). Because the router has the entire packet at time t1, it can begin to transmit the packet to the receiving host at time t1. At time t2 = t1 + L/R2, 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/R1 + L/R2. 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 = 0.2 d) Probability that all three users are transmitting simultaneously = = (0.2)3 = 0.008. Since the queue grows when all the users are transmitting, the fraction 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 bytes b) 100 km, 1 Mbps, 100 bytes 18. 10msec; d/s; no; no 19. a) 500 kbps b) 64 seconds c) 100kbps; 320 seconds 20. End system A breaks the large file into chunks. It adds header to each chunk, thereby generating 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 is 350 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 – the application 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 onto the 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 little bit 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) Virus Requires some form of human interaction to spread. Classic example: E-mail viruses. b) Worms No user replication needed. Worm in infected host scans IP addresses and port numbers, looking for vulnerable processes to infect. 27. Creation of a botnet requires an attacker to find vulnerability in some application or system (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 completely modify 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 Problems Problem 1 There 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 Server Msg name purpose -------- ------- HELO <userid> Let server know that there is a card in the ATM machine ATM card transmits user ID to Server PASSWD <passwd> User enters PIN, which is sent to server BALANCE User requests balance WITHDRAWL <amount> User asks to withdraw money BYE user all done Messages from Server to ATM machine (display) Msg name purpose -------- ------- PASSWD Ask user for PIN (password) OK last requested operation (PASSWD, WITHDRAWL) OK ERR last requested operation (PASSWD, WITHDRAWL) in ERROR AMOUNT <amt> sent in response to BALANCE request BYE user done, display welcome screen at ATM Correct operation: client server HELO (userid) --------------> (check if valid userid) <------------- PASSWD PASSWD <passwd> --------------> (check password) <------------- OK (password is OK) BALANCE --------------> <------------- AMOUNT <amt> WITHDRAWL <amt> --------------> check if enough $ to cover withdrawl <------------- OK ATM dispenses $ BYE --------------> <------------- BYE In situation when there's not enough money: HELO (userid) --------------> (check if valid userid) <------------- PASSWD PASSWD <passwd> --------------> (check password) <------------- OK (password is OK) BALANCE --------------> <------------- AMOUNT <amt> WITHDRAWL <amt> --------------> check if enough $ to cover withdrawl <------------- ERR (not enough funds) error msg displayed no $ given out BYE --------------> <------------- BYE Problem 2 At 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 3 a) A circuit-switched network would be well suited to the application, because the application 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 more network 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 4 Between the switch in the upper left and the switch in the upper right we can have 4 connections. 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. We can 4 connections passing through the switch in the upper-right-hand corner and another 4 connections passing through the switch in the lower-left-hand corner, giving a total of 8 connections. 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 5 Tollbooths 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 6 a) seconds. b) seconds. c) 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 km. Problem 7 Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires sec=7msec. The time required to transmit the packet is sec= sec. Propagation delay = 10 msec. The delay until decoding is 7msec + sec + 10msec = 17.224msec A similar analysis shows that all bits experience a delay of 17.224 msec. Problem 8 a) 20 users can be supported. b) . c) . d) . We use the central limit theorem to approximate this probability. Let be independent random variables such that . “21 or more users” when is a standard normal r.v. Thus “21 or more users” . Problem 9 10,000 Problem 10 The 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 dproc; 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, dproc, and d3/s3. Adding these five delays gives dend-end = L/R1 + L/R2 + L/R3 + d1/s1 + d2/s2 + d3/s3+ dproc+ dproc To 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 11 Because bits are immediately transmitted, the packet switch does not introduce any delay; in particular, it does not introduce a transmission delay. Thus, dend-end = L/R + d1/s1 + d2/s2+ d3/s3 For the values in Problem 10, we get 6 + 20 + 16 + 4 = 46 msec. Problem 12 The 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 13 The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)L/R for the nth 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)/2 It takes seconds to transmit the packets. Thus, the buffer is empty when a each batch of 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 14 The transmission delay is . The total delay is Let . Total delay = For x=0, the total delay =0; as we increase x, total delay increases, approaching infinity as x approaches 1/a. Problem 15 Total delay . Problem 16 The total number of packets in the system includes those in the buffer and the packet that is being transmitted. So, N=10+1. Because , 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 17 There are nodes (the source host and the routers). Let denote the processing delay at the th node. Let be the transmission rate of the th link and let . Let be the propagation delay across the th link. Then . Let denote the average queuing delay at node . Then . Problem 18 On linux you can use the command traceroute www.targethost.com and in the Windows command prompt you can use tracert www.targethost.com 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 www.poly.edu 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. 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. 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 www.stella-net.net (France) to www.poly.edu (USA). The average round-trip delays at each of the three hours are 87.09 ms, 86.35 ms and 86.48 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 19 An example solution: Traceroutes from two different cities in France to New York City in United States In these traceroutes from two different cities in France to the same destination host in United States, seven links are in common including the transatlantic link. In this example of traceroutes from one city in France and from another city in Germany 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 States Five links are common in the two traceroutes. The two traceroutes diverge before reaching China Problem 20 Throughput = min{Rs, Rc, R/M} Problem 21 If only use one path, the max throughput is given by: . If use all paths, the max throughput is given by . Problem 22 Probability of successfully receiving a packet is: ps= (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 ps. Thus, the average number of transmissions needed is given by: 1/ps . Then, the average number of re-transmissions needed is given by: 1/ps -1. Problem 23 Let’s call the first packet A and call the second packet B. 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/Rs. 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/Rs + L/Rs + dprop < L/Rs + dprop + L/Rc The 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 started transmitting 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/Rs + L/Rs + dprop + T >= L/Rs + dprop + L/Rc Thus, the minimum value of T is L/Rc  L/Rs . Problem 24 40 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 25 160,000 bits 160,000 bits The bandwidth-delay product of a link is the maximum number of bits that can be in the link. the width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters long, which is longer than a football field s/R Problem 26 s/R=20000km, then R=s/20000km= 2.5*108/(2*107)= 12.5 bps Problem 27 80,000,000 bits 800,000 bits, this is because that the maximum number of bits that will be in the link at any given time = min(bandwidth delay product, packet size) = 800,000 bits. .25 meters Problem 28 ttrans + tprop = 400 msec + 80 msec = 480 msec. 20 * (ttrans + 2 tprop) = 20*(20 msec + 80 msec) = 2 sec. Breaking up a file takes longer to transmit because each data packet and its corresponding acknowledgement packet add their own propagation delays. Problem 29 Recall geostationary satellite is 36,000 kilometers away from earth surface. 150 msec 1,500,000 bits 600,000,000 bits Problem 30 Let’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 31 Time to send message from source host to first packet switch = With store-and-forward switching, the total time to move message from source host to destination host = Time to send 1st packet from source host to first packet switch = . . Time at which 2nd packet is received at the first switch = time at which 1st packet is received at the second switch = Time at which 1st packet is received at the destination host = . After this, every 5msec one packet will be received; thus time at which last (800th) packet is received = . It can be seen that delay in using message segmentation is significantly less (almost 1/3rd). Without message segmentation, if bit errors are not tolerated, if there is a single bit error, the whole message has to be retransmitted (rather than a single packet). Without message segmentation, huge packets (containing HD videos, for example) are sent into the network. Routers have to accommodate these huge packets. Smaller packets have to queue behind enormous packets and suffer unfair delays. Packets have to be put in sequence at the destination. 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 32 Yes, 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 33 There are F/S packets. Each packet is S=80 bits. Time at which the last packet is received at the first router is sec. At this time, the first F/S-2 packets are at the destination, 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 sec. Thus delay in sending the whole file is To calculate the value of S which leads to the minimum delay, Problem 34 The 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.   Chapter 2 Review Questions The Web: HTTP; file transfer: FTP; remote login: Telnet; e-mail: SMTP; BitTorrent file sharing: BitTorrent protocol Network architecture refers to the organization of the communication process into layers (e.g., the five-layer Internet architecture). Application architecture, on the other hand, is designed by an application developer and dictates the broad structure of the application (e.g., client-server or P2P). The process which initiates the communication is the client; the process that waits to be contacted is the server. No. In a P2P file-sharing application, the peer that is receiving a file is typically the client and the peer that is sending the file is typically the server. The IP address of the destination host and the port number of the socket in the destination process. You would use UDP. With UDP, the transaction can be completed in one roundtrip time (RTT) - 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. One such example is remote word processing, for example, with Google docs. However, because Google docs runs over the Internet (using TCP), timing guarantees are not provided. a) Reliable data transfer TCP provides a reliable byte-stream between client and server but UDP does not. b) A guarantee that a certain value for throughput will be maintained Neither c) A guarantee that data will be delivered within a specified amount of time Neither d) Confidentiality (via encryption) Neither SSL operates at the application layer. The SSL socket takes unencrypted data from the application layer, encrypts it and then passes it to the TCP socket. If the application developer wants TCP to be enhanced with SSL, she has to include the SSL code in the application. A protocol uses handshaking if the two communicating entities first exchange control packets before sending data to each other. SMTP uses handshaking at the application layer whereas HTTP does not. 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. When the user first visits the site, the server creates a unique identification number, creates an entry in its back-end database, and returns this identification number as a cookie number. This cookie number is stored on the user’s host and is managed by the browser. During each subsequent visit (and purchase), the browser sends the cookie number back to the site. Thus the site knows when this user (more precisely, this browser) is visiting the site. Web caching can bring the desired content “closer” to the user, possibly to the same LAN to which the user’s host is connected. Web caching can reduce the delay for all objects, even objects that are not cached, since caching reduces the traffic on links. Telnet is not available in Windows 7 by default. to make it available, go to Control Panel, Programs and Features, Turn Windows Features On or Off, Check Telnet client. To start Telnet, in Windows command prompt, issue the following command > telnet webserverver 80 where "webserver" is some webserver. After issuing the command, you have established a TCP connection between your client telnet program and the web server. Then type in an HTTP GET message. An example is given below: Since the index.html page in this web server was not modified since Fri, 18 May 2007 09:23:34 GMT, and the above commands were issued on Sat, 19 May 2007, the server returned "304 Not Modified". Note that the first 4 lines are the GET message and header lines inputed by the user, and the next 4 lines (starting from HTTP/1.1 304 Not Modified) is the response from the web server. FTP uses two parallel TCP connections, one connection for sending control information (such as a request to transfer a file) 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. The message is first 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. 17. Received: from 65.54.246.203 (EHLO bay0-omc3-s3.bay0.hotmail.com) (65.54.246.203) by mta419.mail.mud.yahoo.com with SMTP; Sat, 19 May 2007 16:53:51 -0700 Received: from hotmail.com ([65.55.135.106]) by bay0-omc3-s3.bay0.hotmail.com with Microsoft SMTPSVC(6.0.3790.2668); Sat, 19 May 2007 16:52:42 -0700 Received: from mail pickup service by hotmail.com with Microsoft SMTPSVC; Sat, 19 May 2007 16:52:41 -0700 Message-ID: <BAY130-F26D9E35BF59E0D18A819AFB9310@phx.gbl> Received: from 65.55.135.123 by by130fd.bay130.hotmail.msn.com with HTTP; Sat, 19 May 2007 23:52:36 GMT From: "prithula dhungel" <prithuladhungel@hotmail.com> To: prithula@yahoo.com Bcc: Subject: Test mail Date: Sat, 19 May 2007 23:52:36 +0000 Mime-Version: 1.0 Content-Type: Text/html; format=flowed Return-Path: prithuladhungel@hotmail.com Figure: A sample mail message header Received: This header field indicates the sequence in which the SMTP servers send and receive the mail message including the respective timestamps. In this example there are 4 “Received:” header lines. This means the mail message passed through 5 different SMTP servers before being delivered to the receiver’s mail box. The last (forth) “Received:” header indicates the mail message flow from the SMTP server of the sender to the second SMTP server in the chain of servers. The sender’s SMTP server is at address 65.55.135.123 and the second SMTP server in the chain is by130fd.bay130.hotmail.msn.com. The third “Received:” header indicates the mail message flow from the second SMTP server in the chain to the third server, and so on. Finally, the first “Received:” header indicates the flow of the mail messages from the forth SMTP server to the last SMTP server (i.e. the receiver’s mail server) in the chain. Message-id: The message has been given this number BAY130-F26D9E35BF59E0D18A819AFB9310@phx.gbl (by bay0-omc3-s3.bay0.hotmail.com. Message-id is a unique string assigned by the mail system when the message is first created. From: This indicates the email address of the sender of the mail. In the given example, the sender is “prithuladhungel@hotmail.com” To: This field indicates the email address of the receiver of the mail. In the example, the receiver is “prithula@yahoo.com” Subject: This gives the subject of the mail (if any specified by the sender). In the example, the subject specified by the sender is “Test mail” Date: The date and time when the mail was sent by the sender. In the example, the sender sent the mail on 19th May 2007, at time 23:52:36 GMT. Mime-version: MIME version used for the mail. In the example, it is 1.0. Content-type: The type of content in the body of the mail message. In the example, it is “text/html”. Return-Path: This specifies the email address to which the mail will be sent if the receiver of this mail wants to reply to the sender. This is also used by the sender’s mail server for bouncing back undeliverable mail messages of mailer-daemon error messages. In the example, the return path is “prithuladhungel@hotmail.com”. 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 (office PC, home PC, etc.). 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 (including very old messages). 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. You should be able to see the sender's IP address for a user with an .edu email address. But you will not be able to see the sender's IP address if the user uses a gmail account. It is not necessary that Bob will also provide chunks to Alice. Alice has to be in the top 4 neighbors of Bob for Bob to send out chunks to her; this might not occur even if Alice provides chunks to Bob throughout a 30-second interval. Recall that in BitTorrent, a peer picks a random peer and optimistically unchokes the peer for a short period of time. Therefore, Alice will eventually be optimistically unchoked by one of her neighbors, during which time she will receive chunks from that neighbor. The overlay network in a P2P file sharing system consists of the nodes participating in the file sharing system and the logical links between the nodes. There is a logical link (an “edge” in graph theory terms) from node A to node B if there is a semi-permanent TCP connection between A and B. An overlay network does not include routers. Mesh DHT: The advantage is in order to a route a message to the peer (with ID) that is closest to the key, only one hop is required; the disadvantage is that each peer must track all other peers in the DHT. Circular DHT: the advantage is that each peer needs to track only a few other peers; the disadvantage is that O(N) hops are needed to route a message to the peer that is closest to the key. 25. File Distribution Instant Messaging Video Streaming Distributed Computing With the UDP server, there is no welcoming socket, and all data from different clients enters the server through this one socket. With the TCP server, there is a welcoming socket, and each time a client initiates a connection to the server, a new socket is created. Thus, to support n simultaneous connections, the server would need n+1 sockets. For the TCP application, as soon as the client is executed, it attempts to initiate a TCP connection with the server. If the TCP server is not running, then the client will fail to make a connection. For the UDP application, the client does not initiate connections (or attempt to communicate with the UDP server) immediately upon execution Chapter 2 Problems Problem 1 a) F b) T c) F d) F e) F Problem 2 Access control commands: USER, PASS, ACT, CWD, CDUP, SMNT, REIN, QUIT. Transfer parameter commands: PORT, PASV, TYPE STRU, MODE. Service commands: RETR, STOR, STOU, APPE, ALLO, REST, RNFR, RNTO, ABOR, DELE, RMD, MRD, PWD, LIST, NLST, SITE, SYST, STAT, HELP, NOOP. Problem 3 Application layer protocols: DNS and HTTP Transport layer protocols: UDP for DNS; TCP for HTTP Problem 4 The document request was http://gaia.cs.umass.edu/cs453/index.html. The Host : field indicates the server's name and /cs453/index.html indicates the file name. The browser is running HTTP version 1.1, as indicated just before the first <cr><lf> pair. The browser is requesting a persistent connection, as indicated by the Connection: keep-alive. This is a trick question. This information is not contained in an HTTP message anywhere. So there is no way to tell this from looking at the exchange of HTTP messages alone. One would need information from the IP datagrams (that carried the TCP segment that carried the HTTP GET request) to answer this question. Mozilla/5.0. The browser type information is needed by the server to send different versions of the same object to different types of browsers. Problem 5 The status code of 200 and the phrase OK indicate that the server was able to locate the document successfully. The reply was provided on Tuesday, 07 Mar 2008 12:39:45 Greenwich Mean Time. The document index.html was last modified on Saturday 10 Dec 2005 18:27:46 GMT. There are 3874 bytes in the document being returned. The first five bytes of the returned document are : <!doc. The server agreed to a persistent connection, as indicated by the Connection: Keep-Alive field Problem 6 Persistent connections are discussed in section 8 of RFC 2616 (the real goal of this question was to get you to retrieve and read an RFC). Sections 8.1.2 and 8.1.2.1 of the RFC indicate that either the client or the server can indicate to the other that it is going to close the persistent connection. It does so by including the connection-token "close" in the Connection-header field of the http request/reply. HTTP does not provide any encryption services. (From RFC 2616) “Clients that use persistent connections should limit the number of simultaneous connections that they maintain to a given server. A single-user client SHOULD NOT maintain more than 2 connections with any server or proxy.” Yes. (From RFC 2616) “A client might have started to send a new request at the same time that the server has decided to close the "idle" connection. From the server's point of view, the connection is being closed while it was idle, but from the client's point of view, a request is in progress.” Problem 7 The total amount of time to get the IP address is . Once the IP address is known, elapses to set up the TCP connection and another elapses to request and receive the small object. The total response time is Problem 8 . . Problem 9 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:  = (850,000 bits)/(15,000,000 bits/sec) = .0567 sec The traffic intensity on the link is given by =(16 requests/sec)(.0567 sec/request) = 0.907. Thus, the average access delay is (.0567 sec)/(1 - .907)  .6 seconds. The total average response time is therefore .6 sec + 3 sec = 3.6 sec. The traffic intensity on the access link is reduced by 60% since the 60% of the requests are satisfied within the institutional network. Thus the average access delay is (.0567 sec)/[1 – (.4)(.907)] = .089 seconds. The response time is approximately zero if the request is satisfied by the cache (which happens with probability .6); the average response time is .089 sec + 3 sec = 3.089 sec for cache misses (which happens 40% of the time). So the average response time is (.6)(0 sec) + (.4)(3.089 sec) = 1.24 seconds. Thus the average response time is reduced from 3.6 sec to 1.24 sec. Problem 10 Note that each downloaded object can be completely put into one data packet. Let Tp denote the one-way propagation delay between the client and the server. First consider parallel downloads using non-persistent connections. Parallel downloads would allow 10 connections to share the 150 bits/sec bandwidth, giving each just 15 bits/sec. Thus, the total time needed to receive all objects is given by: (200/150+Tp + 200/150 +Tp + 200/150+Tp + 100,000/150+ Tp ) + (200/(150/10)+Tp + 200/(150/10) +Tp + 200/(150/10)+Tp + 100,000/(150/10)+ Tp ) = 7377 + 8*Tp (seconds) Now consider a persistent HTTP connection. The total time needed is given by: (200/150+Tp + 200/150 +Tp + 200/150+Tp + 100,000/150+ Tp ) + 10*(200/150+Tp + 100,000/150+ Tp ) =7351 + 24*Tp (seconds) Assuming the speed of light is 300*106 m/sec, then Tp=10/(300*106)=0.03 microsec. Tp is therefore negligible compared with transmission delay. Thus, we see that persistent HTTP is not significantly faster (less than 1 percent) than the non-persistent case with parallel download. Problem 11 Yes, because Bob has more connections, he can get a larger share of the link bandwidth. Yes, Bob still needs to perform parallel downloads; otherwise he will get less bandwidth than the other four users. Problem 12 Server.py from socket import * serverPort=12000 serverSocket=socket(AF_INET,SOCK_STREAM) serverSocket.bind(('',serverPort)) serverSocket.listen(1) connectionSocket, addr = serverSocket.accept() while 1: sentence = connectionSocket.recv(1024) print 'From Server:', sentence, '\n' serverSocket.close() Problem 13 The MAIL FROM: in SMTP is a message from the SMTP client that identifies the sender of the mail message to the SMTP server. The From: on the mail message itself is NOT an SMTP message, but rather is just a line in the body of the mail message. Problem 14 SMTP uses a line containing only a period to mark the end of a message body. HTTP uses “Content-Length header field” to indicate the length of a message body. No, HTTP cannot use the method used by SMTP, because HTTP message could be binary data, whereas in SMTP, the message body must be in 7-bit ASCII format. Problem 15 MTA stands for Mail Transfer Agent. A host sends the message to an MTA. The message then follows a sequence of MTAs to reach the receiver’s mail reader. We see that this spam message follows a chain of MTAs. An honest MTA should report where it receives the message. Notice that in this message, “asusus-4b96 ([58.88.21.177])” does not report from where it received the email. Since we assume only the originator is dishonest, so “asusus-4b96 ([58.88.21.177])” must be the originator. Problem 16 UIDL abbreviates “unique-ID listing”. When a POP3 client issues the UIDL command, the server responds with the unique message ID for all of the messages present in the user's mailbox. This command is useful for “download and keep”. By maintaining a file that lists the messages retrieved during earlier sessions, the client can use the UIDL command to determine which messages on the server have already been seen. Problem 17 a) C: dele 1 C: retr 2 S: (blah blah … S: ………..blah) S: . C: dele 2 C: quit S: +OK POP3 server signing off b) C: retr 2 S: blah blah … S: ………..blah S: . C: quit S: +OK POP3 server signing off C: list S: 1 498 S: 2 912 S: . C: retr 1 S: blah ….. S: ….blah S: . C: retr 2 S: blah blah … S: ………..blah S: . C: quit S: +OK POP3 server signing off Problem 18 For a given input of domain name (such as ccn.com), IP address or network administrator name, the whois database can be used to locate the corresponding registrar, whois server, DNS server, and so on. NS4.YAHOO.COM from www.register.com; NS1.MSFT.NET from ww.register.com Local Domain: www.mindspring.com Web servers : www.mindspring.com 207.69.189.21, 207.69.189.22, 207.69.189.23, 207.69.189.24, 207.69.189.25, 207.69.189.26, 207.69.189.27, 207.69.189.28 Mail Servers : mx1.mindspring.com (207.69.189.217) mx2.mindspring.com (207.69.189.218) mx3.mindspring.com (207.69.189.219) mx4.mindspring.com (207.69.189.220) Name Servers: itchy.earthlink.net (207.69.188.196) scratchy.earthlink.net (207.69.188.197) www.yahoo.com Web Servers: www.yahoo.com (216.109.112.135, 66.94.234.13) Mail Servers: a.mx.mail.yahoo.com (209.191.118.103) b.mx.mail.yahoo.com (66.196.97.250) c.mx.mail.yahoo.com (68.142.237.182, 216.39.53.3) d.mx.mail.yahoo.com (216.39.53.2) e.mx.mail.yahoo.com (216.39.53.1) f.mx.mail.yahoo.com (209.191.88.247, 68.142.202.247) g.mx.mail.yahoo.com (209.191.88.239, 206.190.53.191) Name Servers: ns1.yahoo.com (66.218.71.63) ns2.yahoo.com (68.142.255.16) ns3.yahoo.com (217.12.4.104) ns4.yahoo.com (68.142.196.63) ns5.yahoo.com (216.109.116.17) ns8.yahoo.com (202.165.104.22) ns9.yahoo.com (202.160.176.146) www.hotmail.com Web Servers: www.hotmail.com (64.4.33.7, 64.4.32.7) Mail Servers: mx1.hotmail.com (65.54.245.8, 65.54.244.8, 65.54.244.136) mx2.hotmail.com (65.54.244.40, 65.54.244.168, 65.54.245.40) mx3.hotmail.com (65.54.244.72, 65.54.244.200, 65.54.245.72) mx4.hotmail.com (65.54.244.232, 65.54.245.104, 65.54.244.104) Name Servers: ns1.msft.net (207.68.160.190) ns2.msft.net (65.54.240.126) ns3.msft.net (213.199.161.77) ns4.msft.net (207.46.66.126) ns5.msft.net (65.55.238.126) d) The yahoo web server has multiple IP addresses www.yahoo.com (216.109.112.135, 66.94.234.13) e) The address range for Polytechnic University: 128.238.0.0 – 128.238.255.255 f) An attacker can use the whois database and nslookup tool to determine the IP address ranges, DNS server addresses, etc., for the target institution. By analyzing the source address of attack packets, the victim can use whois to obtain information about domain from which the attack is coming and possibly inform the administrators of the origin domain. Problem 19 The following delegation chain is used for gaia.cs.umass.edu a.root-servers.net E.GTLD-SERVERS.NET ns1.umass.edu(authoritative) First command: dig +norecurse @a.root-servers.net any gaia.cs.umass.edu ;; AUTHORITY SECTION: edu. 172800 IN NS E.GTLD-SERVERS.NET. edu. 172800 IN NS A.GTLD-SERVERS.NET. edu. 172800 IN NS G3.NSTLD.COM. edu. 172800 IN NS D.GTLD-SERVERS.NET. edu. 172800 IN NS H3.NSTLD.COM. edu. 172800 IN NS L3.NSTLD.COM. edu. 172800 IN NS M3.NSTLD.COM. edu. 172800 IN NS C.GTLD-SERVERS.NET. Among all returned edu DNS servers, we send a query to the first one. dig +norecurse @E.GTLD-SERVERS.NET any gaia.cs.umass.edu umass.edu. 172800 IN NS ns1.umass.edu. umass.edu. 172800 IN NS ns2.umass.edu. umass.edu. 172800 IN NS ns3.umass.edu. Among all three returned authoritative DNS servers, we send a query to the first one. dig +norecurse @ns1.umass.edu any gaia.cs.umass.edu gaia.cs.umass.edu. 21600 IN A 128.119.245.12 The answer for google.com could be: a.root-servers.net E.GTLD-SERVERS.NET ns1.google.com(authoritative) Problem 20 We can periodically take a snapshot of the DNS caches in the local DNS servers. The Web server that appears most frequently in the DNS caches is the most popular server. This is because if more users are interested in a Web server, then DNS requests for that server are more frequently sent by users. Thus, that Web server will appear in the DNS caches more frequently. For a complete measurement study, see: Craig E. Wills, Mikhail Mikhailov, Hao Shang “Inferring Relative Popularity of Internet Applications by Actively Querying DNS Caches”, in IMC'03, October 27­29, 2003, Miami Beach, Florida, USA Problem 21 Yes, we can use dig to query that Web site in the local DNS server. For example, “dig cnn.com” will return the query time for finding cnn.com. If cnn.com was just accessed a couple of seconds ago, an entry for cnn.com is cached in the local DNS cache, so the query time is 0 msec. Otherwise, the query time is large. Problem 22 For calculating the minimum distribution time for client-server distribution, we use the following formula: Dcs = max {NF/us, F/dmin} Similarly, for calculating the minimum distribution time for P2P distribution, we use the following formula: Where, F = 15 Gbits = 15 * 1024 Mbits us = 30 Mbps dmin = di = 2 Mbps Note, 300Kbps = 300/1024 Mbps. Client Server N 10 100 1000 u 300 Kbps 7680 51200 512000 700 Kbps 7680 51200 512000 2 Mbps 7680 51200 512000 Peer to Peer N 10 100 1000 u 300 Kbps 7680 25904 47559 700 Kbps 7680 15616 21525 2 Mbps 7680 7680 7680 Problem 23 Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of a rate of us/N. Note that this rate is less than each of the client’s download rate, since by assumption us/N ≤ dmin. Thus each client can also receive at rate us/N. Since each client receives at rate us/N, the time for each client to receive the entire file is F/( us/N) = NF/ us. Since all the clients receive the file in NF/ us, the overall distribution time is also NF/ us. Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of dmin. Note that the aggregate rate, N dmin, is less than the server’s link rate us, since by assumption us/N ≥ dmin. Since each client receives at rate dmin, the time for each client to receive the entire file is F/ dmin. Since all the clients receive the file in this time, the overall distribution time is also F/ dmin. From Section 2.6 we know that DCS ≥ max {NF/us, F/dmin} (Equation 1) Suppose that us/N ≤ dmin. Then from Equation 1 we have DCS ≥ NF/us . But from (a) we have DCS ≤ NF/us . Combining these two gives: DCS = NF/us when us/N ≤ dmin. (Equation 2) We can similarly show that: DCS =F/dmin when us/N ≥ dmin (Equation 3). Combining Equation 2 and Equation 3 gives the desired result. Problem 24 Define u = u1 + u2 + ….. + uN. By assumption us <= (us + u)/N Equation 1 Divide the file into N parts, with the ith part having size (ui/u)F. The server transmits the ith part to peer i at rate ri = (ui/u)us. Note that r1 + r2 + ….. + rN = us, so that the aggregate server rate does not exceed the link rate of the server. Also have each peer i forward the bits it receives to each of the N-1 peers at rate ri. The aggregate forwarding rate by peer i is (N-1)ri. We have (N-1)ri = (N-1)(usui)/u <= ui, where the last inequality follows from Equation 1. Thus the aggregate forwarding rate of peer i is less than its link rate ui. In this distribution scheme, peer i receives bits at an aggregate rate of Thus each peer receives the file in F/us. Again define u = u1 + u2 + ….. + uN. By assumption us >= (us + u)/N Equation 2 Let ri = ui/(N-1) and rN+1 = (us – u/(N-1))/N In this distribution scheme, the file is broken into N+1 parts. The server sends bits from the ith part to the ith peer (i = 1, …., N) at rate ri. Each peer i forwards the bits arriving at rate ri to each of the other N-1 peers. Additionally, the server sends bits from the (N+1) st part at rate rN+1 to each of the N peers. The peers do not forward the bits from the (N+1)st part. The aggregate send rate of the server is r1+ …. + rN + N rN+1 = u/(N-1) + us – u/(N-1) = us Thus, the server’s send rate does not exceed its link rate. The aggregate send rate of peer i is (N-1)ri = ui Thus, each peer’s send rate does not exceed its link rate. In this distribution scheme, peer i receives bits at an aggregate rate of Thus each peer receives the file in NF/(us+u). (For simplicity, we neglected to specify the size of the file part for i = 1, …., N+1. We now provide that here. Let Δ = (us+u)/N be the distribution time. For i = 1, …, N, the ith file part is Fi = ri Δ bits. The (N+1)st file part is FN+1 = rN+1 Δ bits. It is straightforward to show that F1+ ….. + FN+1 = F.) The solution to this part is similar to that of 17 (c). We know from section 2.6 that Combining this with a) and b) gives the desired result. Problem 25 There are N nodes in the overlay network. There are N(N-1)/2 edges. Problem 26 Yes. His first claim is possible, as long as there are enough peers staying in the swarm for a long enough time. Bob can always receive data through optimistic unchoking by other peers. His second claim is also true. He can run a client on each host, let each client “free-ride,” and combine the collected chunks from the different hosts into a single file. He can even write a small scheduling program to make the different hosts ask for different chunks of the file. This is actually a kind of Sybil attack in P2P networks. Problem 27 Peer 3 learns that peer 5 has just left the system, so Peer 3 asks its first successor (Peer 4) for the identifier of its immediate successor (peer 8). Peer 3 will then make peer 8 its second successor. Problem 28 Peer 6 would first send peer 15 a message, saying “what will be peer 6’s predecessor and successor?” This message gets forwarded through the DHT until it reaches peer 5, who realizes that it will be 6’s predecessor and that its current successor, peer 8, will become 6’s successor. Next, peer 5 sends this predecessor and successor information back to 6. Peer 6 can now join the DHT by making peer 8 its successor and by notifying peer 5 that it should change its immediate successor to 6. Problem 29 For each key, we first calculate the distances (using d(k,p)) between itself and all peers, and then store the key in the peer that is closest to the key (that is, with smallest distance value). Problem 30 Yes, randomly assigning keys to peers does not consider the underlying network at all, so it very likely causes mismatches. Such mismatches may degrade the search performance. For example, consider a logical path p1 (consisting of only two logical links): ABC, where A and B are neighboring peers, and B and C are neighboring peers. Suppose that there is another logical path p2 from A to C (consisting of 3 logical links): ADEC. It might be the case that A and B are very far away physically (and separated by many routers), and B and C are very far away physically (and separated by many routers). But it may be the case that A, D, E, and C are all very close physically (and all separated by few routers). In other words, a shorter logical path may correspond to a much longer physical path. Problem 31 If you run TCPClient first, then the client will attempt to make a TCP connection with a non-existent server process. A TCP connection will not be made. UDPClient doesn't establish a TCP connection with the server. Thus, everything should work fine if you first run UDPClient, then run UDPServer, and then type some input into the keyboard. If you use different port numbers, then the client will attempt to establish a TCP connection with the wrong process or a non-existent process. Errors will occur. Problem 32 In the original program, UDPClient does not specify a port number when it creates the socket. In this case, the code lets the underlying operating system choose a port number. With the additional line, when UDPClient is executed, a UDP socket is created with port number 5432 . UDPServer needs to know the client port number so that it can send packets back to the correct client socket. Glancing at UDPServer, we see that the client port number is not “hard-wired” into the server code; instead, UDPServer determines the client port number by unraveling the datagram it receives from the client. Thus UDP server will work with any client port number, including 5432. UDPServer therefore does not need to be modified. Before: Client socket = x (chosen by OS) Server socket = 9876 After: Client socket = 5432 Problem 33 Yes, you can configure many browsers to open multiple simultaneous connections to a Web site. The advantage is that you will you potentially download the file faster. The disadvantage is that you may be hogging the bandwidth, thereby significantly slowing down the downloads of other users who are sharing the same physical links. Problem 34 For an application such as remote login (telnet and ssh), a byte-stream oriented protocol is very natural since there is no notion of message boundaries in the application. When a user types a character, we simply drop the character into the TCP connection. In other applications, we may be sending a series of messages that have inherent boundaries between them. For example, when one SMTP mail server sends another SMTP mail server several email messages back to back. Since TCP does not have a mechanism to indicate the boundaries, the application must add the indications itself, so that receiving side of the application can distinguish one message from the next. If each message were instead put into a distinct UDP segment, the receiving end would be able to distinguish the various messages without any indications added by the sending side of the application. Problem 35 To create a web server, we need to run web server software on a host. Many vendors sell web server software. However, the most popular web server software today is Apache, which is open source and free. Over the years it has been highly optimized by the open-source community. Problem 36 The key is the infohash, the value is an IP address that currently has the file designated by the infohash.   Chapter 3 Review Questions Call this protocol Simple Transport Protocol (STP). At the sender side, STP accepts from the sending process a chunk of data not exceeding 1196 bytes, a destination host address, and a destination port number. STP adds a four-byte header to each chunk and puts the port number of the destination process in this header. STP then gives the destination host address and the resulting segment to the network layer. The network layer delivers the segment to STP at the destination host. STP then examines the port number in the segment, extracts the data from the segment, and passes the data to the process identified by the port number. The segment now has two header fields: a source port field and destination port field. At the sender side, STP accepts a chunk of data not exceeding 1192 bytes, a destination host address, a source port number, and a destination port number. STP creates a segment which contains the application data, source port number, and destination port number. It then gives the segment and the destination host address to the network layer. After receiving the segment, STP at the receiving host gives the application process the application data and the source port number. No, the transport layer does not have to do anything in the core; the transport layer “lives” in the end systems. For sending a letter, the family member is required to give the delegate the letter itself, the address of the destination house, and the name of the recipient. The delegate clearly writes the recipient’s name on the top of the letter. The delegate then puts the letter in an envelope and writes the address of the destination house on the envelope. The delegate then gives the letter to the planet’s mail service. At the receiving side, the delegate receives the letter from the mail service, takes the letter out of the envelope, and takes note of the recipient name written at the top of the letter. The delegate then gives the letter to the family member with this name. No, the mail service does not have to open the envelope; it only examines the address on the envelope. Source port number y and destination port number x. An application developer may not want its application to use TCP’s congestion control, which can throttle the application’s sending rate at times of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCP’s congestion control. Also, some applications do not need the reliable data transfer provided by TCP. Since most firewalls are configured to block UDP traffic, using TCP for video and voice traffic lets the traffic though the firewalls. Yes. The application developer can put reliable data transfer into the application layer protocol. This would require a significant amount of work and debugging, however. Yes, both segments will be directed to the same socket. For each received segment, at the socket interface, the operating system will provide the process with the IP addresses to determine the origins of the individual segments. For each persistent connection, the Web server creates a separate “connection socket”. Each connection socket is identified with a four-tuple: (source IP address, source port number, destination IP address, destination port number). When host C receives and IP datagram, it examines these four fields in the datagram/segment to determine to which socket it should pass the payload of the TCP segment. Thus, the requests from A and B pass through different sockets. The identifier for both of these sockets has 80 for the destination port; however, the identifiers for these sockets have different values for source IP addresses. Unlike UDP, when the transport layer passes a TCP segment’s payload to the application process, it does not specify the source IP address, as this is implicitly specified by the socket identifier. Sequence numbers are required for a receiver to find out whether an arriving packet contains new data or is a retransmission. To handle losses in the channel. If the ACK for a transmitted packet is not received within the duration of the timer for the packet, the packet (or its ACK or NACK) is assumed to have been lost. Hence, the packet is retransmitted. A timer would still be necessary in the protocol rdt 3.0. If the round trip time is known then the only advantage will be that, the sender knows for sure that either the packet or the ACK (or NACK) for the packet has been lost, as compared to the real scenario, where the ACK (or NACK) might still be on the way to the sender, after the timer expires. However, to detect the loss, for each packet, a timer of constant duration will still be necessary at the sender. The packet loss caused a time out after which all the five packets were retransmitted. Loss of an ACK didn’t trigger any retransmission as Go-Back-N uses cumulative acknowledgements. The sender was unable to send sixth packet as the send window size is fixed to 5. When the packet was lost, the received four packets were buffered the receiver. After the timeout, sender retransmitted the lost packet and receiver delivered the buffered packets to application in correct order. Duplicate ACK was sent by the receiver for the lost ACK. The sender was unable to send sixth packet as the send win

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1． 计算机网络的发展可以分为哪几个阶段？每个阶段各有什么特点？ A 面向终端的计算机网络：以单个计算机为中心的远程联机系统。这类简单的“终端—通 信线路—计算机”系统，成了计算机网络的雏形。 B 计算机—计算机网络：呈现出多处中心的特点。 C 开放式标准化网络：OSI/RM 的提出，开创了一个具有统一的网络体系结构，遵循国际 标准化协议的计算机网络新时代。 D 因特网广泛应用和高速网络技术发展：覆盖范围广、具有足够的带宽、很好的服务质量 与完善的安全机制，支持多媒体信息通信，以满足不同的应用需求，具备高度的可靠性与完 善的管理功能。 2． 计算机网络可分为哪两大子网？它们各实现什么功能？ 通信子网和资源子网。资源子网负责信息处理，通信子网负责全网中的信息传递。 3． 简述各种计算机网络拓扑类型的优缺点。 星形拓扑结构的优点是：控制简单；故障诊断和隔离容易；方便服务，中央节点可方便 地对各个站点提供服务和网络重新配置。缺点是：电缆长度和安装工作量客观；中央节点的 负担较重形成“瓶颈”；各站点的分布处理能力较低。 总线拓扑结构的优点是：所需要的电缆数量少；简单又是无源工作，有较高的可