通信网基本概念与主体结构(第二版)答案Ch1
通信网——基本概念与主体结构ch2要点
面向连接的服务(connection-oriented- service)包含
3个步骤:
在两个n层SAP问建立连接。该建立过程包括协商连 接参数和初始化“状态信息”,如序号、流量控制变 量与缓存位置等。
利用n层协议实际传送n—SDU。 断开连接,释放用于该连接的各种资源。
HTTP驻留程序在TCP连接上发送文件,然后断开连接. 其间客户机接收文件并加以显示。为了取出图像,浏览 器必须启动另一TCP连接,用于GET交互作用。
HTTP与Web页浏览
在HTTP情况下,需双向连接,以 正确的次序无误地传送字节流。
TCP协议能够提供这种通信业务 , 每 个 HTTP 进 程 都 将 报 文 插 入 一缓存器内,由TCP以信息块(段 )形式将其传送到另一TCP进程, 每 段 包 含 端 口 号 , 加 到 HTTP 报 文 信 息 中 。 HTTP 利 用 底 层 TCP 提 供 服 务 , 这 样 , HTTP 客 户 机 与服务器间报文的传送实际上是
OSI参考模型
网络层(network layer)提供经通信网的分组(packet)
数据传送,其关键问题是为从源端设备到目的端设备的 分组选择路由,所谓路由选择,就是为通过网络的分组 选择路径的方法。网络中的各节点必须协同工作,以有 效地完成路由的选择。这一层还涉及由于分组信息流的 剧增而引起的拥塞问题。
图2.1对等通信
OSI参考模型 在OSI术语中,第n层的进程称作第n层实体(layer n entity)。第n层实体间通过交换协议数据单元 (protocol data unit,PDU)进行通信。每个PDU包 括一个头部,而头部中含有协议控制信息。通常用户 信息为服务数据单元(service data unit SDU)格式。n 层实体的行为由一组规则或约定进行管理,通常将这 些规则与约定称作第n层协议(layer n protoc0l)。
计算机网络(第二版)课后习题答案第一章
计算机网络(第二版)课后习题答案第一章1. 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:计算机网络的发展主要分为一下四个阶段:1)以单计算机为中心的联机系统2)计算机-计算机网络3)体系结构标准化网络4)Internet时代各个阶段特点参加课本。
2. 说明为什么在网络中使用通信控制处理机CCP?答:在网络中使用通信控制处理机CCP的主要目的是为了将通信功能从主机中分离出来,以减轻主机的负荷。
通信控制处理机负责网络上各主机间的通信控制和通信处理,它们组成的通信子网是网络的内层,为资源子网提供信息传输服务。
3. 计算机网络由哪些部分组成,什么是通信子网和资源子网?试述这种层次结构观的特点以及各层的作用是什么?答:通信控制处理机构成的通信子网是网络的内层,或骨架层,是网络的重要组成部分。
网上主机负责数据处理,是计算机网络资源的拥有者,它们组成了网络的资源子网,是网络的外层,通信子网为资源子网提供信息传输服务,资源子网上用户间的通信是建立在通信子网的基础上。
没有通信子网,网络不能工作,而没有资源子网,通信子网的传输也失去了意义,两者合起来组成了统一的资源共享的两层网络。
将通信子络的规模进一步扩大,使之变成社会公有的数据通信网,4. 为什么要研究网络体系结构与网络协议的标准化问题?答:为了保证计算机间及网络间方便、快捷的互连互通。
5. 一个完整的计算机网络的定义应包含哪些内容?答:1)物理结构:通过通信线路、通信设备将地理上分散的计算机连成一个整体2)逻辑结构:在网络协议控制下进行信息传输3)主要目的:资源共享6. 说明在公用网络以及私用网络中计算机网络主要有哪些功能?答:服务于企业的网络•资源共享•提高可靠性•节约经费•通信手段服务于公众的网络•访问远程信息•个人间通信•交互式娱乐7. 什么叫耦合度?按耦合度的不同如何区分计算机网络和其它计算机系统?答:耦合度是处理机之间连接的紧密程度。
它可用处理机之间的距离及相互连接的信号线数目来表示,下表说明了按耦合度的不同计算机网络和其它计算机系统的区别。
计算机网络基础(第二版)习题参考答案
计算机网络基础(第二版)习题参考答案计算机网络基础(第二版)习题参考答案1. 数据通信基础计算机网络是一种将分布在不同地理位置的计算机系统连接在一起,实现信息交换和资源共享的系统。
数据通信是计算机网络的基础,它指的是在计算机网络中传输和交换数据的过程。
1.1 数据通信的基本概念数据通信是指通过某种媒介传输数据,将计算机系统中的信息从一个地方发送到另一个地方。
数据通信的基本概念包括发送方、接收方、传输介质、传输模式等。
1.2 数据通信的基本组成数据通信的基本组成包括发送器、信道、接收器等。
发送器将信息转换为适合传输的形式,并通过信道将信息传输到接收器进行处理。
1.3 数据通信的基本模型数据通信的基本模型是指数据在传输过程中的几个基本要素,包括发送方、接收方、传输介质和传输协议。
发送方将信息分段转换为数据包,通过传输介质传输给接收方,接收方按照相应的传输协议进行解析和处理。
2. 计算机网络的概述计算机网络是指将多台计算机系统通过通信线路或其他传输介质连接起来,实现信息交换和资源共享的系统。
2.1 计算机网络的分类根据规模和范围的不同,计算机网络可以分为局域网(LAN)、城域网(MAN)和广域网(WAN)等不同类型。
2.2 计算机网络的体系结构计算机网络的体系结构包括物理层、数据链路层、网络层、传输层和应用层等不同层次。
每一层负责不同的功能,通过协议进行通信和交互。
3. TCP/IP协议世界TCP/IP是互联网的核心协议,由传输控制协议(TCP)和互联网协议(IP)组成。
3.1 TCP/IP协议的基本原理TCP/IP协议是一种面向连接的协议,通过将数据分割成数据段并进行封装,使用IP地址将数据段传输到目的地,再通过TCP协议实现数据的可靠传输。
3.2 TCP/IP协议的层次结构TCP/IP协议的层次结构包括网络接口层、网络层、传输层和应用层等不同层次。
每一层负责不同的功能,通过协议进行通信和交互。
3.3 TCP/IP协议的应用TCP/IP协议广泛应用于互联网和局域网等不同的网络环境中。
线务工程(第2版)第一章第一部分 通信工程基础知识第二部分 通信工程基础建设第三部分 通信工程施工
• 1.1 通信网 • 1.2 通信工程
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1.1 通 信 网
• 1.1.1 通信网的基本概念
• 通信网(telecommunicationnetwork)是构成多个用户相互通信的多 个电信系统互联的通信体系,是人类实现远距离通信的重要基础设施, 利用电缆、无线、光纤或者其他电磁系统,传送、发射和接收标识、 文字、图像、声音或其他信号。电信网由终端设备、传输链路和交换 设备三要素构成,运行时还应辅之以信令系统、通信协议以及相应的 运行支撑系统。
• 1.1.3 通信网的基本结构
• 任何通信网都具有信息传送、信息处理、信令机制、网络管理功能。 一个完整的通信网,尤其是面对普通用户运营的大型通信网络,一般可 以分为业务网、传输网和支撑网三个部分。
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1.1 通 信 网
• 1.1.4 通信网的拓扑结构
• 通信网的拓扑结构有网状网、星型网、环型网、总线型网、复合型网 等。
网、传真通信网、图像通信网、有线电视网等。 • 按服务区域范围可分为:本地电信网、农村电信网、长途电信网、移
动通信网、国际电信网等。 • 按传输媒介种类可分为:架空明线网、电缆通信网、光缆通信网、卫
星通信网、用户光纤网、低轨道卫星移动通信网等。 • 按交换方式可分为:电路交换网、报文交换网、分组交换网、宽带交
信工程项目,地市局工程项目。 • (4)四类工程 • 县局工程项目,其他小型项目。 • 2. 按项目建设范围划分 • (1)一般施工项目 • 一般施工项目是指按照单独的设计文件,单独进行施工的通信项目建
设工程。一般施工项目通常是雇主与施工队伍相互配合、协作,施工 团队根据雇主的设计文件进行施工。
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(完整版)通信网络课后题答案
第1章通信网络概论及数学基础1.1通信网络有哪些基本要素组成?试举例列出五种常用的通信网络。
1.2常用的通信链路有哪些?其主要特征是什么?1.3试简述分组交换网的要点。
1.4什么叫做虚电路?它与传统电话交换网中的物理链路有何差异?1.5 A TM信元与分组有何差别?ATM网络是如何支持不同种类业务的?1.6分层的基本概念是什么?什么是对等层?1.7试述OSI七层模型和TCP/IP协议体系的区别和联系。
1.8一个典型的通信网络可由哪些物理子网构成?路由器在该网络中的作用是什么?1.9通信网络要研究的基本理论问题有哪些?1.10 设随机过程定义为:,其中Y是离散随机变量,且。
试求该过程在时的均值,和时的自相关函数值。
1.11 设随机过程是一个随机相位信号,即,式中A和w c为常量, 是一个均匀分布的随机变量,其概率密度函数为。
试求的均值函数和自相关函数。
并讨论其平稳性和各态历经性。
1.12 试求Poisson过程的均值函数,方差函数和相关函数。
1.13 设到达某商店的顾客组成强度为的Poisson流,每个顾客购买商品的概率为p,各顾客是否购买商品与其它顾客无关,分别用和表示购买商品顾客和未购买商品顾客的顾客流过程,请证明他们分别是强度为和的Poisson流。
1.14 设某办公室来访的顾客数组成Poisson流,平均每小时到访的顾客数为3人,求:(1)一上午(8到12点)没有顾客来访的概率;(2)下午(2点到6点)第一个顾客到达的时间分布。
图1-25习题1-16图1.15 设有三个黑球和三个白球,把这六个球任意分给甲乙两人,并把甲拥有的白球数定义为该过程的状态,则有四种状态0,1,2,3。
现每次从甲乙双方各取一球,然后相互交换。
经过n次交换后过程的状态记为,试问该过程是否是马氏链?如是,试计算其一步转移概率矩阵,并画出其状态转移图。
1.16分别利用Prim-Dijkstra算法和Kruskal算法求解图1-25中的最小重量生成树。
通信网-基本概念与主体结构第二版教学设计
通信网-基本概念与主体结构第二版教学设计一、课程简介本课程是通信工程专业的一门基础课程,旨在使学生掌握通信网的基本概念、结构及相关技术。
通过本课程的学习,学生将了解通信网的建设、运行和维护,并具有一定的实际操作能力。
二、教学目标1. 知识与理解•掌握通信网的基本概念和基础知识,包括通信网的结构、网络拓扑、传输介质等;•理解通信网的安全性、稳定性、高可靠性的重要性;•了解现代通信技术的新进展。
2. 技能与能力•掌握通信网的设计、部署、调试、维护;•掌握通信网络规划、布线、接入方式、网络安全等基本技能;•能够进行网络故障诊断和维护工作。
3. 情感态度和价值观•培养工程师的职业道德和职业精神;•提高学生综合素质和团队协作能力;•培养学生的探究精神和创新意识。
1. 通信网基础概念•通信核心技术•通信网络基本概念和拓扑•通信介质特点和相互作用•通信网络的分类和特征2. 通信网的基本组成•通信网的结构•通信协议栈•通信子系统和接口•通信设备的分类和功能3. 通信网络与互联网的关系和发展趋势•通信网络的发展历程•互联网的产生和发展•通信网络与互联网的优化和融合•未来通信网络的发展方向和趋势四、教学方法本课程采用“理论教学+实践操作”相互结合的教学方法。
教师将通过多种途径传递通信网络的基本概念和原理,并引导学生进行全方位、实地的实验操作。
1. 学生的要求•具备一定的数学功底和物理知识;•具备基本计算机应用能力;•具备较强的组织协调和沟通能力。
2. 教师的要求•具备扎实的专业基础和丰富的实践经验;•熟练掌握教学内容,拥有良好的教学能力;•能够激发学生的学习兴趣,培养学生的实践操作能力。
3. 设备要求•课堂需要用到计算机、服务器、交换机、路由器等设备;•给学生提供实验室环境和相应的软硬件设施。
六、教学评价授课结束后,将通过以下方式对学生的学习情况进行评价:•课堂小测验:通过简单的测试,了解学生对课程知识的掌握程度;•实验考核:对学生在实验室中的表现进行考核;•期末考试:考验学生对整个课程的掌握情况;•个人项目:通过个人项目的完成情况评价学生的综合能力和实践操作能力。
通信网—基本概念与主体结构AlbertoLeon-GarciaIndraWidjaja)课程设计 (
通信网—基本概念与主体结构课程设计一、课程概述本课程旨在介绍通信网的基本概念和主体结构,深入探讨通信网络中所涉及的各种技术和协议,并且提供相应的实践措施。
通信网作为现代社会的核心基础设施,对于信息交流和经济发展具有重要意义。
因此,了解通信网的基本概念和主体结构是非常必要的。
二、课程目标1.了解通信网络的基本概念,包括计算机网络、数据通信、网络拓扑结构等。
2.掌握通信网络中所使用的各种协议和技术,包括TCP/IP协议、HTTP协议、路由选择算法等。
3.熟悉通信网络的主体结构,包括物理结构、逻辑结构和分层结构等。
4.能够应用所学知识,完成相应的通信网络实践,包括局域网的构建、应用程序的开发等。
三、课程大纲第一章:引言在本章中,将介绍通信网络的基本概念,包括计算机网络、数据通信、网络拓扑结构等。
第二章:协议和技术在本章中,将深入探讨通信网络中所使用的各种协议和技术,包括TCP/IP协议、HTTP协议、路由选择算法等。
第三章:通信网络主体结构在本章中,将详细介绍通信网络的主体结构,包括物理结构、逻辑结构和分层结构等。
第四章:通信网络的实践在本章中,学员将进行相应的通信网络实践,包括局域网的构建、应用程序的开发等。
四、考核要点1.学员需要在考试中准确地描述通信网络的基本概念和主体结构。
2.学员需要具备掌握各种协议和技术的能力,并且在实践中加以应用。
3.学员需要进行相应的通信网络实践,包括局域网的构建、应用程序的开发等。
五、参考资料1.周志华,《计算机网络》。
2.Andrew S. Tanenbaum,《计算机网络》。
3.W. Richard Stevens,《TCP/IP详解(卷一)》。
六、总结本课程介绍了通信网络的基本概念和主体结构,并且深入探讨了通信网络中所涉及的各种技术和协议。
学员在学完本课程后,应该具备了一定的通信网络应用程序开发能力,并且可以在实践中加以应用。
通信网基本概念与主体结构(第二版)答案Ch3
8. How many HDTV channels can be transmitted simultaneously over the optical fiber transmission systems in Table 3.3?
Solution:
Suppose that an optical fiber carries 1600 x 109 bps, and an HDTV channel is about 38 Mbps, then the fiber can carry about 1600000/38 = 40,000 HDTV channels.
4. Explain the difference between facsimile, GIF, and JPEG coding. Give an example of an image that is appropriate to each of these three methods.
Solution:
9. Comment on the properties of the sequence of frame images and the associated bit rates in the following examples.
计算机网络与通信(第2版)课后习题参考答案
解答:
以传递邮件为例。首先,E-mail数据通过本地计算机(信息源)发送给Modem(变换器),完成数字信号到模拟信号的转换,然后通过电话线(信道)传送到目的地的Modem
(反变换器),完成模拟信号到数字信号的转换,然后送给目的地计算机(受信者),从而完成整个数据的传输。
#include <stdio.h>
#include <graphics.h>
#include <stdlib.h>
#include <conio.h>
/*函数:初始化图形显示模式。*/
void InitPaint(void)
{
int gdriver = DETECT, gmode, errorcode;
(3)报文交换采用"存储-转发"方式进行传送,无需事先建立线路,事后更无需拆除。它的优点是:线路利用率高、故障的影响小、可以实现多目的报文;缺点是:延迟时间长且不定、对中间节点的要求高、通信不可靠、失序等,不适合计算机网络。
(4)分组交换中数据以短分组的形式传输,分组长度一般为1000字节。如果发送端有更长的报文需要发送,那么这个报文被分割成一个分组序列,每个分组由控制信息和用户数据两部分组成。分组交换适用于计算机网络,在实际应用中有两种类型:虚电路方式和数据报方式。分组交换的优点是:高效、灵活、迅速、可靠、经济,但存在如下的缺点:有一定的延迟时间、额外的开销会影响传输效率、实现技术复杂等。
目前光纤的最大传输速率可达10Gbps的数量级。
2.3 无线电通信频率范围是多少?无线电广播频谱和微波频谱各是多少?
解答:
无线电通信频率范围为3KHz~300GHz,无线电广播频谱为30 MHz~1 GHz,微波频谱为2~40 GHz。
通信网基本概念与主体结构(第二版)答案Ch6
通信网基本概念与主体结构(第二版)答案Ch2
Solutions to Chapter 2 (Note: solution to Problem 62 to be added)1. Explain how the notion of layering and internetworking make the rapid growth of applications such as the World Wide Web possible.Solution:Internetworking allows many component networks each with different underlying technology and operation to work together and form one large network. As new network technologies are introduced they can be readily incorporated into the Internet. This provides the ubiquitous connectivity for applications like WWW.The layering concept hides the specific underlying network technology from the upper layers and provides a common networking platform. Using the communication service provided by the layers below, new applications can be introduced independently and at a rapid rate.2. (a) What universal set of communication services is provided by TCP/IP?Solution:The TCP/IP protocol stack provides two basic types of communications services through its two transport layer protocols: TCP provides reliable connection-oriented transfer of a byte stream; UDP provides for best-effort connectionless transfer of individual messages. TCP/IP provides withglobally unique logical addressing that enables machines connected to the Internet to access these two services. The IP addressing scheme is very scalable because of its hierarchical structure.2. (b)How is independence from underlying network technologies achieved?Solution:The two basic communications services provided by TCP and UDP are built on the connectionless packet transfer service provided by the Internet Protocol (IP). Many network interfaces are defined tosupport IP. The salient part of the above figure is that all of the higher layer protocols access the network interfaces through IP. This is what provides the ability to operate over multiple networks.2. (c) What economies of scale result from (a) and (b)?Solution:Once a network interface for IP is defined for a given network technology, then hosts connected using the given network technology can connect to the Internet. This allows the reach of the Internet to grow rapidly, leveraging multiple coexisting networks technologies. Thus investment in new network technologies extends the reach of the Internet.3. What difference does it make to the network layer if the underlying data link layer provides a connection-oriented service versus a connectionless service?Solution:If the data link layer provides a connection-oriented service to the network layer, then the network layer must precede all transfer of information with a connection setup procedure. If the connection-oriented service includes assurances that frames of information are transferred correctly and insequence by the data link layer, the network layer can then assume that the packets it sends to its neighbor traverse an error-free pipe.On the other hand, if the data link layer is connectionless, then each frame is sent independently through the data link, probably in unconfirmed manner (without acknowledgments orretransmissions). In this case the network layer cannot make assumptions about the sequencing or correctness of the packets it exchanges with its neighbors.The Ethernet local area network provides an example of connectionless transfer of data link frames.The transfer of frames using "Type 2" service in Logical Link Control (discussed in Chapter 6)provides a connection-oriented data link control example.4. Suppose transmission channels become virtually error-free. Is the data link layer still needed?Solution:The data link layer is still needed for framing the data and for flow control over the transmissionchannel. In a multiple access medium such as a LAN, the data link layer is required to coordinate access to the shared medium among the multiple users.5. Why is the transport layer not present inside the network?Solution:Some of the functions provided by the transport layer can be provided inside the networks, but other functions cannot. For example, the transport layer provides functions at the end-system tocompensate for the limitations and impairments of the network layer, in order to meet requirements(e.g. QoS) of the upper layer. For example in TCP/IP, IP provides only best effort service. To providethe reliable service required by some applications - that is, to compensate for the shortcomings of best effort service - TCP establishes connections and implements error control on an end-to-end basis. One can imagine that a service provider could incorporate this error control function at the edge of its network. On the other hand, one of the main purposes of the transport layer is to allow multiple processes in the end systems to share a network service. This cannot be achieved inside the network.6. Which OSI layer is responsible for the following?(a)Determining the best path to route packets.The network layer is concerned with the selection of paths across the network.(b)Providing end-to-end communications with reliable service.The transport layer is concerned with providing reliable service on an end-to-end basis across the network.(c)Providing node-to-node communications with reliable service.The data link layer provides for the reliable transfer of information between adjacent nodes in anetwork.7. Should connection establishment be a confirmed service or an unconfirmed service? What about data transfer in a connection-oriented service? Connection release?Solution:In general, the establishment of a connection needs to be confirmed before information transfer can commence across a connection. Therefore connection establishment should be a confirmed service.A connection-oriented service is usually reliable so confirmation of data delivery is not necessary. Incertain situations, however, it is possible that the transfer across a connection is not reliable; in this case confirmation of correct data transfer may be required.In general it is desirable that the release of a connection be confirmed by the parties involved. We will see in Chapter 8, section 5, that sometimes it is not easy to confirm that a connection has beenclosed. Consequently, many protocols attempt to confirm the closing of a connection several times, and then give up and simply stop transmitting.8. Does it make sense for a network to provide a confirmed, connectionless packet transfer service?Solution:Yes. Connectionless packet transfer is often unreliable, that is, packets may be lost or discarded inside a network. Certain applications, for example, signaling in connection setup, requireconfirmation to acknowledge the receipt of packets.9. Explain how the notion of multiplexing can be applied at the data link, network, and transport layers. Draw afigure that shows the flow of PDUs in each multiplexing scheme.Solution:Transport Layer: Multiple application layers processes can share the service provided by UDP.When a UDP PDU arrives from the network layer, the destination port number in the PDU is used to deliver the SDU to the appropriate application layer process. Multiple application layer processes also share the service provided by TCP. In this case, when a TCP segment arrives, the TCPconnection ID, consisting of (source port #, source IP address, destination port #, destination IP address), is used to determine which application process to deliver the SDU to.Network Layer: The packet transfer service provided by IP can be used by all transport layersoperating in a machine. Each transfer layer passes SDUs to the IP layer which prepares IP packets with appropriate source and destination IP addresses for transfer across the Internet. Upon receiving an IP packet, a machine examines the protocol type field to determine which transport layer service to deliver the SDU to. We can also view all transport layer PDUs as sharing the IP packet transfer service between a source machine and a destination machine.Data Link Layer: Network layer packets from different protocols (IP, IPX, Appletalk, etc) can share a data link (such as PPP or Ethernet). We can also view packet flows that traverse a data linkbetween two routers as sharing the link.10. Give two features that the data link layer and transport layer have in common. Give two features in which they differ. Hint: Compare what can go wrong to the PDUs that are handled by these layers.Solution:Features they have in common:•Both layers can provide recovery from transmission errors.•Both layers can provide flow control.•Both layers can support multiplexing.Features in which they differ:•The transport layer is end to end and involves the interaction of peer processes across the network. The data link layer involves the interaction of peer-to-peer processes that areconnected directly. In general, the time that elapses in traversing a data link is much smaller than the time traversing a network, where packets can become trapped in temporary routing loops.Consequently, transport layer protocols must be able to deal with out-of-sequence PDUs and amuch larger backlog of PDUs than data link layers.•The data link layer is concerned with framing and the transport layer is not.•The data link layer may be concerned with medium access control, the transport layer does not have this concern.11(a). Can a connection-oriented, reliable message transfer service be provided across a connectionless packet network? Explain.Solution:Yes. To provide a connection-oriented service, the transport layer can establish a logical connection across the connectionless packet network by setting up state information (for example, packetsequence number) at the end systems. During the connection setup, the message is broken into separate packets, and each packet is assigned a sequence number.Using the sequence numbers, the end-system transport-layer entities can acknowledge received packets, determine and retransmit lost packets, delete duplicate packets, and rearrange out-of-order packets. The original message is reassembled as packets arrive at the receiving end.For example, TCP provides a connection-oriented reliable transfer service over IP, a connectionless packet transfer service.11b. Can a connectionless datagram transfer service be provided across a connection-oriented network?Solution:Yes. The connectionless datagram transfer service can be implemented by simply setting up aconnection across the network each time a datagram needs to be transferred. Alternatively, all nodes can have permanent connections to a “connectionless server” that has the function of relayingdatagrams in connectionless fashion.12. An internet path between two hosts involves a hop across network A, a packet-switching network, to a router and then another hop across packet-switching network B. Suppose that packet switching network A carries the packet between the first host and the router over a two-hop path involving one intermediate packet switch. Suppose also that the second network is an Ethernet LAN. Sketch the sequence of IP and non-IP packets and frames that are generated as an IP packet goes from host 1 to host 2.Solution:The IP layer in Host 1 generates an IP packet addressed to the destination host on the destination network and sends it to the router. The network interface in the host encapsulates the IP packet into the packet PDU used by network A. This packet is encapsulated in a frame that traverses data link 1 to the packet switch. The packet is recovered and then forwarded inside a frame along data link 2.The data link at the router recovers the Network A packet, and the IP network interface at the router recovers the IP packet and determines that the next hop is on Network B. The router encapsulates the IP packet into an Ethernet frame, puts the host 2 Ethernet physical address in the frame and sends it to the LAN. The Ethernet card on the host captures the frame and extracts the IP packet and passes it to the host.13. Does Ethernet provide connection-oriented or connectionless service?Solution:Ethernet provides connectionless transfer service of information frames.14. Ethernet is a LAN so it is placed in the data link layer of the OSI reference model.(a)How is the transfer of frames in Ethernet similar to the transfer of frames across a wire? How is itdifferent?The transfer of frames in Ethernet occurs directly over a transmission medium and in this sense is similar to direct transmission over a wire. The sequence of frames into Ethernet arrive in thesame order they are transmitted. However multiple stations can transmit in Ethernet which differs from direct transmission over a wire.(b)How is the transfer of frames in Ethernet similar to the transfer of frames in a packet-switching network?How is it different?Ethernet supports the transfer of frames among multiple end systems and in this sense is similar to a packet switching network. Ethernet does not involve routing which is a feature of packetswitching. Ethernet depends on broadcasting and/or bridging which differs from packet networks.15. Suppose that a group of workstations is connected to an Ethernet LAN. If the workstations communicate only with each other, does it make sense to use IP in the workstations? Should the workstations run TCP directly over Ethernet? How is addressing handled?Solution:Ethernet supports the exchange of frames between stations and can support the direct exchange of information. Using Ethernet without IP would result in an inflexible and difficult to managesystem. Ethernet addresses are fixed and tables need to be changed whenever a machine ismoved, while IP addresses are logical and can be changed whenever a machine is moved. ATCP connection uses the IP addresses in its connection ID so Ethernet addresses could not beused.16. Suppose two Ethernet LANs are interconnected by a box that operates as follows. The box has a table that tells it the physical addresses of the machines in each LAN. The box listens to frame transmissions on each LAN. If a frame is destined to a station at the other LAN, the box retransmits the frame onto the other LAN, otherwise the box does nothing.Solutions follow questions:a.Is the resulting network still a LAN? Does it belong in the data link layer or the network layer?The resulting network is a local area network that has been extended. The extended LAN transfers frames, and so it still belongs in the data link layer.b.Can the approach be extended to connect more than two LANs? If so, what problems arise as the number ofLANs becomes large?Yes, more than two LANs can be connected using the above approach to form an extended LAN. As the number of LANs becomes large, the number of physical addresses stored in the bridge grows and becomes unmanageable. Each time a machine is added the addresses in all the boxes need to be updated. Serious problems arise if boxes are connected so that loops can occur.17. Suppose all laptops in a large city are to communicate using radio transmissions from a high antenna tower. Is the data link layer or network layer more appropriate for this situation?Solution:The data link layer is concerned with the transfer of frames of information across a single hop. The network layer involves the transfer of information across a network using multiple hops per path in general. The connection from a radio antenna to the laptops is direct, and thus a data link layerprotocol is more suitable for this situation.Now suppose the city is covered by a large number of small antennas covering smaller areas. Which layer is more appropriate?A number of areas each covered by small antennas can be interconnected using the "bridging"approach of problem 16, which remains in the data link layer. However, the network layer may be more appropriate because it provides for the transfer of data in the form of packets across thecommunication network. A key aspect of this transfer is the routing of the packets from the source machine to the destination machine, typically traversing a number of transmission link and network nodes where routing is carried out.18. Suppose that a host is connected to a connection-oriented packet-switching network and that it transmits a packet to a server along a path that traverses two packet switches. Suppose that each hop in the path involves a point-to-point link, that is, a wire. Show the sequence of network layer and data link layer PDUs that are generated as the packet travels from the host to the server.Solution:Assume that a network connection has already been set up between the host machine and thenetwork machine. When the host generates an IP packet for transfer to the server, the IP packet will be transferred using the network connection as follows.•The IP packet is encapsulated into a network packet that has a connection ID in its header.The packet may then be encapsulated into a frame that traverses data link 1 and arrives atswitch 1.•The network packet is recovered from the data link 1 frame. The connection ID in the packet is used to determine the outgoing port from switch 1. The connection ID may need to bemapped into a corresponding connection ID over data link 2. The packet is encapsulated intoa frame that traverses data link 2.•The network packet is recovered from the data link 2 frame. The connection ID in the packet determines the outgoing port from switch 1 and the next connection ID. The packet isencapsulated into a frame that traverses data link 3.•The network packet is recovered from the data link 3 frame. The connection ID in the arriving packet indicates that this is the destination node. The IP packet is recovered.The connection-oriented network in this example could correspond to ATM or to frame relay.19. Suppose an application layer entity wants to send an L-byte message to its peer process, using an existing TCP connection. The TCP segment consists of the message plus 20 bytes of header. The segment is encapsulated into an IP packet that has an additional 20 bytes of header. The IP packet in turn goes inside an Ethernet frame that has 18 bytes of header and trailer. What percentage of the transmitted bits in the physical layer correspond to message information, if L = 100 bytes, 500 bytes, 1000 bytes?Solution:TCP/IP over Ethernet allows data frames with a payload size up to 1460 bytes. Therefore, L = 100, 500 and 1000 bytes are within this limit.The message overhead includes:•TCP: 20 bytes of header•IP: 20 bytes of header•Ethernet: total 18 bytes of header and trailer.ThereforeL = 100 bytes, 100/158 = 63% efficiency.L = 500 bytes, 500/558 = 90% efficiency.L = 1000 bytes, 1000/1058 = 95% efficiency.20. Suppose that the TCP entity receives a 1.5 megabyte file from the application layer and that the IP layer is willing to carry blocks of maximum size 1500 bytes. Calculate the amount of overhead incurred from segmenting the file into packet-sized units.Solution:1500 - 20 -20 = 1460 bytes1.5 Mbyte / 1460 byte = 1027.4, therefore 1028 blocks are needed to transfer the file.Overhead = ((1028 x 1500 - 1.5M)/1.5M) x 100 = 2.8%21. Suppose a TCP entity receives a digital voice stream from the application layer. The voice stream arrives at a rate of 8000 bytes/second. Suppose that TCP arranges bytes into block sizes that result in a total TCP and IP header overhead of 50 percent. How much delay is incurred by the first byte in each block?Solution:Assume the stream is segmented as shown below, where the white cells represent data and the shaded cells represent the TCP header overhead.Therefore, block size = 80 bytes and the payload size = 40 bytes.Assume zero processing delay due to data arrangement and segmenting.The delay incurred by the first byte of each block = 40/8000 = 0.5 ms.22. How does the network layer in a connection-oriented packet-switching network differ from the network layer ina connectionless packet-switching network?Solution:The network layer in connection-oriented networks maintains state information about everyconnection. It can allocate resources at the switches through admission control. The network layer in connectionless networks has no knowledge of "connections", and instead deals independently with each packet.The network layer in connection-oriented networks performs routing on a per connection basis. Each packet is routed based on a connection identifier of some sort and packets of the same connection have the same identifier value. In a connectionless network, routing is performed on per packet basis;each packet is routed independently based on information carried in the packet header, for example, the destination address.In connection-oriented networks, the network layer forwarding table is set up by a signaling procedure during the connection establishment. In connectionless networks, the routers may execute adistributed algorithm to share network state information and dynamically calculate the routing table continuously.In case of failure, the connection must be re-established in connection-oriented networks, whereas in connectionless networks, the packets are re-routed. The network layer in connectionless networks is more robust against failures.Summary of differences:Connection-oriented Connectionless Maintain state information about every connection No knowledge of the "connection"Allocate resources to connections at switches No resource allocationAdmission control No admission controlPer connection routing Per packet routingRoute packet based on identifier Route packet based on destination address.Forwarding table specifies the output port and outgoing identifier value as function of the incoming identifier value Routing table specifies the output port depending on the destination addressForwarding table set up by signaling during connection establishment.Router executes distributed algorithm to share network state information and dynamically calculate the routing tableConnection must be re-established in cases of failure Packets are rerouted around failures, robust against failures23. Identify session layer and presentation layer functions in the HTTP protocol.Solution:Presentation layer functions:The request message and the response message headers include information about the content type of the documents (e.g. text/html, image/gif).Session layer functions:The HTTP protocol defines the client/server interaction in three steps:1. Client sends the request for a file2. Server replies with the file or error message if file is not found.3. Server closes the TCP connection after some timeout period.24. Suppose we need a communication service to transmit real-time voice over the Internet. What features of TCP and what features of UDP are appropriate?Solution:TCP is desirable in that it provides a connection for the transfer of a stream of information, which characterizes a digital voice stream. However, to provide reliable service TCP uses acknowledgments and retransmissions that result in packet delay and jitter that can not be tolerated by real-time traffic.UDP provides connectionless service and delivers packets quickly. In case of packet loss, UDP does not provide retransmission, but some degree of packet loss can be tolerated by voice.25. Consider the end-to-end IP packet transfer examples in Figure 2.15. Sketch the sequences of IP packets and Ethernet and PPP frames that are generated by the three examples of packet transfers: from the workstation to the server; from the server to the PC, and from the PC to the server. Include all relevant header information in the sketch.Solution:Workstation to Server:IP packet headerEthernet FrameThe Ethernet frame is broadcast over the LAN. The server's NIC card recognizes that the frame is intended for its host, so it captures the frame and examines it. It finds that the protocol type is set to IP, so it passes the IP datagram up to the IP entity.Server to PC:IP packet headerEthernet FrameThe Ethernet frame is broadcast over the LAN. The router examines frame and passes IPdatagram to its IP entity which discover that the IP datagram is not for itself, but is to be routed on. The routing tables at the router show that the machine with address (2,2) is connecteddirectly on the other side of the point-to-point link. The router encapsulates the IP datagram in a PPP frame.IP packet headerPPP FrameThe PPP receiver at the PC receives the frame, checks the protocol type field and passes the IP datagram to its IP entity.PC to Server:The PC IP entity generates the IP packet shown below. The PPP transmitter at the PC encapsulates the IP packet into a PPP frame sends it to the point-to-point link. There's no need for a physical address specificationIP datagramIP packet headerPPP FrameThe router examines the PPP frame and passes the IP datagram to its IP entity which discoversthat the IP datagram is not for itself, but is to be routed on. The routing table at the router showsthat the machine with address (1,1) is connected in the other side of the Ethernet network. Therouter then encapsulates the IP datagram into an Ethernet frame that is broadcast in the LAN.The server's NIC card recognizes that the frame is intended for its host, so it captures the frameand examines it. It finds that the protocol type is set to IP, so it passes the IP datagram up to the IP entity.26. Suppose a user has two browser applications active at the same time, and suppose that the two applications are accessing the same server to retrieve HTTP documents at the same time. How does the server tell the difference between the two applications?Solution:A client application generates an ephemeral port number for every TCP connection it sets up. AnHTTP request connection is uniquely specified by the five parameters: (TCP, client IP address,ephemeral port #, server IP address, 80). The two applications in the above situations will havedifferent ephemeral port #s and will thus be distinguishable to the server.27. Consider the operation of non-persistent HTTP and persistent HTTP.(a)In non-persistent HTTP (version 1.0): Each client-server interaction involves setting up a TCP connection,carrying out the HTTP exchange, and closing the TCP connection. Let T be the time that elapses fromwhen a packet is sent from client to server to when the response is received. Find the rate at which HTTPexchanges can be made using non-persistent HTTP.(b)In persistent HTTP (version 1.1) the TCP connection is kept alive. Find the rate at which HTTP exchangescan be made if the client cannot send an additional request until it receives a response for each request.(c)Repeat part (b) if the client is allowed to pipeline requests, that is, it does not have to wait for a responsebefore sending a new request.Solution:(a) Each HTTP exchange involves: 1. a three-way handshake to set up the TCP connection; 2. anHTTP request-response interaction; and 3. a TCP close. The client can send its request after the first two handshakes in part 1 (which takes up T seconds). The request and response then take an additional T second. A new request can be initiated with an associated new TCP connection even while the previous TCP connection is being closes. Thus a maximum of one HTTP exchange per 2T seconds is possible.(b) Since each exchange is completed in T seconds, after the connection is setup, the exchange rateis 1/T.(c) The rate depends on how long it takes to send a request and how late it takes to compose aresponse. Considering the maximum of these to be t seconds, exchange rate can be up to 1/t.28. What is the difference between a physical address, a network address, and a domain name?Solution:The physical address is the unique hardware address that identifies an interface of a machine on a physical network such as a LAN. Physical addresses are used in the data link layer.A network address is a machine's logical address on a network. The network address is used in thenetwork layer. The network address used on the Internet is the IP address.Domain names are used as an aid to identify hosts and networks in the Internet, since names are easier to remember than numbers. The DNS system is used to translate between domain names and IP addresses. The domain name for the network address 128.100.132.30 is .29. Explain how a DNS query proceeds if the local name server does not have the IP address for a given host when the following approaches are used. Assume an example where four machines are involved in ultimately resolving a given query.(a)When a machine B cannot resolve an address in response to a query from A, machine B sends the query toanother machine in the chain. When B receives the response, it forwards the result to B.(b)When a machine B cannot resolve an address in response to a query from A, machine B sends a DNS replyto A with the IP address of the next machine in the chain, and machine A contacts that machine.Solution:(a) Host A sends a query to a name server B. B cannot resolve an address, therefore sends the queryto C. C cannot resolve an address either, and send the query to D. Similarly, D cannot resolve an address and sends the query to E, where finally an address is resolved and returned to D. D replies the address to C, C replies it to B, and finally B passes it to the host. In this scenario each server should remember the state of the query and its source.(b) Host A sends a query to name server B. B cannot resolve an address, replies to A with the IPaddress of C. Host A send a query to C this time. C cannot resolve an address, and replies to A with the IP address of D. A sends a query to D. D cannot resolve an address, and replies with the IP address of E. A sends a query to E, E finally resolves an address and returns it to A. In this scenario。
通信网基本概念与主体结构第二版教学设计
通信网基本概念与主体结构第二版教学设计一、教学目的本教学设计旨在全面介绍通信网的基本概念、主体结构以及相关技术,培养学生对通信网的理论和实践能力。
二、教学内容1. 通信网络基础•通信网络的基本概念及分类•计算机网络与通信网络的关系2. 通信网络体系结构•OSI七层模型及TCP/IP协议•传输介质及其特点3. 通信网络技术•无线通信技术及其应用•互联网技术及其应用4. 通信网络安全•通信网络攻击及其类型•通信网络安全技术及应用三、教学方法本课程采用多种教学方法,如教师讲解、案例分析、小组讨论、实验演示等,以深入浅出的方式使学生理解和掌握所学知识。
四、教学内容与时间分配教学内容时间分配通信网络基础2小时通信网络体系结构4小时通信网络技术6小时通信网络安全2小时五、教材和参考书目教材:《通信网基础与体系结构》参考书目:《计算机网络》《TCP/IP协议族》六、实验设计本课程实验旨在对所学知识进行实际操作和应用。
包括:•基本网络模拟实验•无线传输实验•网络安全实验七、考核方式•平时成绩:出勤、作业、实验等,占总成绩的30%•期末考试:占总成绩的70%八、教学成果与效果评估通过学生课堂表现、实验操作情况、作业成绩、期末考试成绩、论文等多个方面评估本课程教学效果,以此为依据进行教学成果评估。
九、教学实施方案本课程由一名主讲教师主持教学,辅助教师和实验室管理员进行实验指导和操作。
实验室按照教学进度提前进行准备,保证实验设备、实验材料等正常运转。
每周进行一次课堂教学,一次实验操作。
同时,组织小组讨论、课堂提问等方式使学生积极参与讨论,提高教学效果。
十、总结通过本次教学设计,旨在使学生掌握通信网络的基本概念、体系结构、技术应用等方面的知识,同时能够进行实际操作和应用,提高所学知识的熟练程度和理论实践能力。
同时,不断改进和更新教学内容,保持与时俱进的理念,不断提高教学质量和水平。
(完整版)通信网课后答案
第一章1、构成现代通信网的要素有哪些?它们各自完成什么功能?它们之间相互通信通过什么机制实现?答:(1)从硬件结构来看:由终端节点、变换节点、业务节点、传输系统构成。
功能:完成接入交换网的控制、管理、运营和维护。
(2)从软件结构来看:它们有信令、协议、控制、管理、计费等。
功能:完成通信协议以及网络管理来实现相互间的协调通信。
(3)通过保持帧同步和位同步、遵守相同的传输体制。
2、在通信网中交换节点主要完成哪些功能?无连接网络中交换节点实现交换的方式与面向连接的网络中交换节点的实现方式有什么不同?分组交换型网络与电路交换型网络节点实现交换的方式有什么不同?答:(1)完成任意入线的信息到指定出线的交换功能(2)无连接型网络不用呼叫处理和记录连接状态,但是面向连接的网络需要。
(3)电路交换的交换节点直接在预先建立的连接上进行处理、时延小,分组交换以“存储—转发”方式工作,时延大。
3、现代通信网为什么要采用分层结构?画出对等层之间的通信过程?答:(1)降低网络设计的复杂度、方便异构网络间的相互连通、增强网络的可升级性、促进了竞争和设备制造商的分工。
(2)图略第二章1.简述几种主要传输介质的特点及应用场合.双绞线:便宜易安装,抗干扰能力差,复用度不高,带宽窄。
应用场合:电话用户线,局域网中。
同轴电缆:抗干扰强于双绞线,适合高频宽带传输,成本高,不易安装埋设。
应用场合:CATV,光纤同轴混合接入网。
光纤:大容量,体积小,重量轻,低衰减,抗干扰能力强,安全保密性好。
应用场合:接入网,局域网,城域网,广域网。
无线介质:1.无线电:长距离传输,能穿越建筑物,其传输特性与频率有关。
应用场合:公众无线广播,电视发射,无线专用网。
2.微波:在空间沿直线传输。
应用场合:卫星通信,陆地蜂窝,无线接入网,专用网络等.3.红外线:不能穿越同体,短距离,小范围内通信。
应用场合:家电产品,通信接口等。
2.SDH的帧结构由那几部分组成,各起什么作用?由段开销SDH,管理单元指针AU-PTR,STM净负荷组成。
通信网基础课后答案
第一章通信网概述1.1简述通信系统模型中各个组成部分的含义,并举例说明。
答:通信系统的基本组成包括:信源,变换器,信道,噪声源,反变换器和信宿六部分。
信源:产生各种信息的信息源。
变换器:将信源发出的信息变换成适合在信道中传输的信号。
信道:按传输媒质分有线信道和无线信道,有线信道中,电磁信号或光电信号约束在某种传输线上传输;无线信道中,电磁信号沿空间传输。
反变换器:将信道上接收的信号变换成信息接收者可以接收的信息。
信宿:信息的接收者。
噪声源:系统内各种干扰。
1.2现代通信网是如何定义的?答:由一定数量的节点和连接这些节点的传输系统有机地组织在一起的,按约定信令或协议完成任意用户间信息交换的通信体系。
适应用户呼叫的需要,以用户满意的效果传输网内任意两个或多个用户的信息。
1.3试述通信网的构成要素及其功能。
答:通信网是由软件和硬件按特定方式构成的一个通信系统。
硬件由:终端设备,交换设备和传输系统构成,完成通信网的基本功能:接入、交换和传输;软件由:信令、协议、控制、管理、计费等,它们完成通信网的控制、管理、运营和维护,实现通信网的智能化。
1.4分析通信网络各种拓扑结构的特点。
(各种网络的拓扑结构图要掌握)答:基本组网结构:Ø网状网:优点:①各节点之间都有直达线路,可靠性高;②各节点间不需要汇接交换功能,交换费用低;缺点:①各节点间都有线路相连,致使线路多,建设和维护费用大;②通信业务量不大时,线路利用率低。
如网中有N个节点,则传输链路数H=1/2*N(N-1)。
Ø星形网:优点:①线路少,建设和维护费用低;②线路利用率高;缺点:①可靠性低,②中心节点负荷过重会影响传递速度。
如网中有N个节点,则传输链路数H=N-1。
Ø环形网:同样节点数情况下所需线路比网状网少,可靠性比星形网高。
如网中有N个节点,则传输链路数H=N。
Ø总线形网:优点:①节点接入方便②成本低,缺点:①传输时延不稳定②若传输总线损坏,整个网络会瘫痪。
复习-Ch1 通信网络概述
1.2 通信基本概念
1.模拟信号
模拟信号是指幅值连续、时间上也是连续的信号。其 特点是幅度连续 (连续的含义是在某一取值范围内可以取 无限多个数值)。
2. 数字信号
数字信号是指幅值和时间两个方面都离散的信号。其 特点是幅值离散(离散的含义是在某一取值范围内可以取 有限多个数值)。如常见的脉冲信号。
1.2 通信基本概念
在具体介绍PCM 30/32路系统前先了解几个基本概念: 1.时隙: 很小的时间片断,在PCM 30/32路系统中每个时隙 长度为3.9微秒,每个时隙中可以传输8个bit。 2. 帧: 由若干个时隙所组成的结构,在PCM 30/32路系统 中每个帧的时间长度是125微秒,每帧中有32个时隙。 3. 抽样频率: 在PCM中每秒抽样次数。在PCM 30/32路系统中抽 样频率为8000次/S,每次抽样间隔为125微秒。
1.2 通信基本概念
1.2 通信基本概念
七. 通信模型
具体通信系统中涉及大量具体设备,但所有通信系统可 以抽象为一个通信模型,其涵盖了所有通信系统的特征。
噪声源
信源
变换器
信道
反变换器
信宿
1.2 通信基本概念
八. 通信网络结构
1. 通信网的组成 通信网是由若干用户终端A,B,C,…,并通过传输 系统链接起来。用户终端之间通过一个或多个节点链接, 在节点处提供交换、处理网络管理等功能。
三. 多路复用技术
多路复用技术是通信系统常用的技术,通常在有线通 信中使用。 1. 概念 为了提高信道利用率,使多路信号沿同一信道传输而 互不干扰,称多路复用。 最常用的多路复用技术是 频分多路复用和时分多路复用 另外还有统计时分多路复用和波分多路复用技术。
1.2 通信基本概念
ch01通信网概述
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第二代通信网
• 从第一代到第二代通信网,通信技术由模拟通信技术转向数字通 信技术,发生了革命性的变化,实现了更新换代。由于第二代通 信网的业务(电话、电报)类型和业务的工作方式(面向连接,资源 独占)都没有发生变化,特别是用户环路与终端设备没有发生改变 (仍为模拟工作方式),因而过渡进展得很顺利,我国仅用数年就 完成了全部通信网的数字化工程,建成了第二代通信网。
• 传送网是指数字信号传送网,包括骨干传送网和接入网。 • 支撑网包括信令网、数字同步网和电信管理网。
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ch01通信网概述
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ch01通信网概述
三大业务网
• 电信部门:话音、数据 • 广电部门:同轴入户(700M),但单向 • 互联网:
• 发展活跃 • 文字、数据、图像、话音 • 问题:速度瓶颈,实时性差
• 通信业务划分:电话网、数据网、有线电视网 • 通信范围划分:公用网 、局部网、专用网 • 信号形式划分:模拟网、数字网、综合网 • 信道划分:有线网、无线网、移动网
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ch01通信网概述
通信网的划分
• 根据网络功能从水平方向划分:用户驻地网CPN、接入网AN和核 心网CN。
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ch01通信网概述
第一代通信网
• 以模拟通信技术为核心技术的通信网,通信网中的全部设备都是采用模拟通 信技术来设计、制造的。
• 这一代通信网的业务种类简单(电话、电报),业务的全程全网是模拟信息的 处理。电信业务是采用面向连接的工作方式,为每一对用户连接提供的资源 都是独占的(一个频段和一组交换触点),从而可保证电信业务的服务质量。
通信网基本概念与主体结构答案
通信网基本概念与主体结构答案每一个抄通信网的学生都是头疼的天使----Laofuzi Communication Networks 2nd Edition Chapter 10 Solutions Solutions to Chapter 10 10.1. Can all traffic in an IntServ router be of guaranteed service type Explain. Solution: All the traffic can be of guaranteed service only if all the flows are shaped and if they strictly comply with the traffic specifications. However in practice the total amount of traffic that can be admitted will have some limit. The reason is that to bound the delay the maximum queue length in the router must be below a certain threshold. If all of the flows are well behaved and without bursts the queue size will always be reasonably small. However if the flows are bursty the queue may grow to levels that can cause large unacceptable delays. To guarantee the delay bounds extra bandwidth should be assigned based to accommodate the burstiness of the flows. This overbooking can sometimes lead to the queue becoming empty and hence result in underutilization of the output link. Thus it is reasonable to fill the output bandwidth during such periods with best effort traffic. So in summary to efficiently utilize the bandwidth and achieve higher throughputs it is necessary to admit best effort traffic alongside guaranteed service traffic. 10.2. Develop a packet-scheduling approach to provide guaranteed service controlled load service and best effort service in a router. Can any mix of these three traffic types be supported Explain. Solution: Separate queues must be provided for the packets from each type of service in order for their packets to be serviced appropriately. The set of queues belonging to guaranteed service should always receive service with a rate R which should be larger than the total guaranteed service traffic. The aggregate rate R is guaranteed and controlled by giving priority to the guaranteed service queues and controlling the service they receive by a leaky bucket shaper with a sustained rate of R. Individual queues within the guaranteed service group are serviced in a round-robin fashion while for each queue its service is controlled by a dedicated leaky bucket shaper with the parameters set based on the requirements of the flow assigned to that queue. The set of controlled load service queues are serviced during the periods that guaranteed service queues are not serviced. The capacity in excess of R is available to controlled load traffic. Flows are assigned to different queues and weighted fair queueing can be used within the group to service the queues depending on their requirements. Best effort traffic is given the lowest priority. The queue is serviced only if the other two sets of queues are empty. 10.3. RSVP signaling is very different from ATM signaling. Discuss the differences and list the advantages and disadvantages of each protocol. Solution: RSVP uses Receiver-Initiated Reservation rather than ATM’s per-hop path request to reserve network resources. RSVP is intended for use with multicasting and is receiver oriented. ATM is designed for unicast traffic and is sender oriented. RSVP receiver-initiated reservation Advantage: interoperates with non-RSVP routers soft state provides robustness with respect to faults requests can be merged reducing the reservation load on the routers. Disadvantage: soft state – requires periodic updates receiver needs to know the path for connection. Leon-Garcia/Widjaja 1每一个抄通信网的学生都是头疼的天使----Laofuzi Communication Networks 2nd Edition Chapter 10 Solutions ATM per-hop path request Advantage: hard state –reserve only once. Disadvantage: Individual VCC is required for each host in multicast application. Source requests the connections so it needs to know the requirements of the receiver. 10.4. Consider the merging of reservations in RSVP. Suppose that there is a single sender and 2n receivers that are reached through a path that consists of a depth n binary tree that has a router located at each node. What is the total number of reservation request messages generated by the receivers Solution: The total number of reservation request messages generated by thereceivers and routers are equivalent to finding the branches of a binary tree with a depth of n. Let Nn be the number of branches in a binary tree with depth n. At level n there 2n new branches and so we have the recursion Nn Nn–1 2n. Draw a picture of a tree to verify this. It is easy to then verify that Nn total number of request messages is 2n1 – 1. 10.5. Suppose RSVP is used to set up a virtual private network to interconnect n users across an Internet. Discuss the appropriateness of the three RSVP reservation styles for this application. RSVP has three reservation styles Wildcard Filter Fixed Filter and Shared Explicit. Wild card filter WF creates a single reservation shared by all senders. It is designed for applications that are unlikely to have multiple senders transmitting simultaneously. It is not appropriate for VPN applications because WF cannot provide the receiver requests if more than one sender is transmitting at the same time. Fixed filter FF creates a distinct reservation for each sender. It is an appropriate solution for VPN requirements. Each sender VPN has its own resource reservation and this resource is shared by number of receivers hosts located in different location belong to the same VPN. Shared explicit SE creates a single reservation shared by a set of explicit senders. It may also be appropriate in VPN applications under certain conditions. The merging of the reservation between senders can result in a more efficient use of network resources. 10.6. Select and justify a choice an RSVP reservation style that is appropriate for the following situations: Solutions follow questions: a A multiuser application where a video stream from each participant must be delivered to every other participant. This situation is a multipoint to multipoint application so the Fixed Filter FF is the appropriate choice. FF creates a distinct reservation for each participant sender and it can merge downstream requests and forward the sum of flow spec upstream. It is an appropriate reservation style because each participant sender will only receive a single reservation request that has been merged by a multiple of other participants. This simplifies the processing overhead for each sender and allows the processing power to focus on accepting multiple video streams for other participants. b An audioconferencing application that is self-limiting in that more than two participants are unlikely to speak simultaneously. This audio conferencing application is best suited with the Wild card Filter WF style. The WF-style is suitable for applications that are unlikely to have multiple senders transmitting simultaneously. In this application the self-limiting feature prevents two participants from speaking simultaneously. Leon-Garcia/Widjaja 2每一个抄通信网的学生都是头疼的天使----Laofuzi Communication Networks 2nd Edition Chapter 10 Solutions c A multiuser application in which each participant must receive explicit permission before transmitting information. The Shared Explicit SE reservation style is best suited for this multiuser application. The SE style creates a single reservation shared by a set of explicit senders. In this application only the participant that receives an explicit permission is allowed to join a single reservation. Any participant that does not receive permission is not allowed to use this single reserved resource.10.7. What information needs to be carried by Resv messages in RSVP to set up guaranteed IP service with a specific bandwidth and a given delay-bound requirement How does a router process this information Solution: The information that must be carried by Resv message in RSVP to set up guaranteed IP service with a specific bandwidth and a given delay bound is: the common header SESSION RSVP_HOP TIME_V ALUES STYLE and Fixed Filter flow descriptor along with flow spec and filter spec. When a router receives this Resv message it reserves the necessary resources based on the flow specification which in turn governs the bandwidth and delay-bound requirements. Next the router exams the filter spec to determine the data packets that correspond to the defined flow spec. 10.8. Discuss the interplay between a QoS routing algorithm RSVP andintegrated services IP. Solution: Integrated services IP provides QoS packet delivery between the source and destination pair. The QoS parameters are dependent on the source requirements for example bandwidth requirement and delay-bound requirement. To provide integrated services it is necessary for the routing protocol network layer to provide QoS routing. QoS routing provides for the selection of paths with specific attributes that can support specified QoS requirements. RSVP provides the means for reserving resources along paths identified by QoS routing. 10.9. Explain how the soft-state feature of RSVP allows it to adapt to failures in the network. Solution: The RSVP soft-state approach requires each node to refresh its reservation periodically by using Path and Resv messages. When a state is not refreshed within a certain time-out period the state is deleted. Upon a link failure within the network a PathErr message is sent to the ingress node source node. The ingress node simply defines a new route by refreshing the Path message to a different next hop. The old failure path will eventually tear down after the timer is expired. This approach provides error recovery upon failure however it is necessary for the ingress node to keep updating the status of the link-state topology and determine the next possible route in case of the primary route fails. 10.10. Discuss whether RSVP provides a scalable means to set up connections to a session that consists of the broadcasting of a given TV program. Solution: Broadcasting a TV program is a point to multipoint multicast application. RSVP does provide a scalable means to set up connections to a session in multicast application. RSVP performs the resource reservation for multicast applications adapting dynamically to changing group membership and changing route. Also RSVP requires the receiver to initiate and maintain the resource reservation rather than obtaining the bandwidth requirement from each receiver. This process avoids an implosion problem and minimizes the processing overhead to the sender. Finally the RSVP routers provide different reservation styles and the requests can merge in several ways according to Leon-Garcia/Widjaja 3每一个抄通信网的学生都是头疼的天使----Laofuzi Communication Networks 2nd Edition Chapter 10 Solutions the reservation style. This prevents duplicate resource reservations if multiple downstream receivers belong to the same multicast group. 10.11. Explain what happens if RSVP is used to set up small-bandwidth flows on a large-bandwidth link How well does RSVP scale in this situation Solution: If RSVP is used to set up small-bandwidth flows on a large-bandwidth link the router will have to update the available resource bandwidth every time there are small-bandwidth flow request for resources. Moreover the router needs to keep track of the available resources and reject resource reservation requests when the available bandwidth cannot support a particular requirement. The router has to keep track of every single flow resource usage during the setup and tear down procedure. All of these activities consume a lot of processing power when the link has large bandwidth being shared by large numbers of small-bandwidth flows. 10.12. Obtain a packet capture file for RSVP either in the lab or downloaded from a web site. Examine the details of Path Resv and refresh messages. Solution: The following text shows the details of the first PATH message for MPLS path setup shown in Figure 2.27. The file is obtained by printing the capture data from Ethereal. We highlight in color the main fields and interesting information. In particular note the explicit route identified in the message and the LABEL REQUEST. The TIME V ALUES field is set to 30000 ms that is 30 second refresh period. Frame 1 306 bytes on wire 306 bytes captured Arrival Time: Feb 10 2000 08:49:03.806994000 Time delta from previous packet: 0.000000000 seconds Time relative to first packet: 0.000000000 seconds Frame Number: 1 Packet Length: 306 bytes Capture Length: 306 bytes Ethernet II Src: 00:90:92:9d:94:01 Dst: 00:d0:63:c3:b8:47 Internet Protocol Src Addr:17.3.3.3 17.3.3.3 Dst Addr: 16.2.2.2 16.2.2.2 Resource ReserVation Protocol RSVP: PATH Message. SESSION: IPv4-LSP Destination 16.2.2.2 Tunnel ID 1 Ext ID 11030303. SENDER TEMPLATE: IPv4-LSP Tunnel Source: 17.3.3.3 LSP ID: 1. RSVP Header. PA TH Message. RSVP Version: 1 Flags: 00 Message Type: PATH Message. 1 Message Checksum: 0xdb58 correct Sending TTL: 254 Message length: 264 SESSION: IPv4-LSP Destination 16.2.2.2 Tunnel ID 1 Ext ID 11030303. Length: 16 Class number: 1 - SESSION object C-type: 7 - IPv4 LSP Destination address: 16.2.2.2 16.2.2.2 Tunnel ID: 1 Extended Tunnel ID: 285410051 17.3.3.3 HOP: IPv4 210.0.0.1 Length: 12 Class number: 3 - HOP object C-type: 1 - IPv4 Neighbor address: 210.0.0.1 Leon-Garcia/Widjaja 4每一个抄通信网的学生都是头疼的天使----Laofuzi Communication Networks 2nd Edition Chapter 10 Solutions Logical interface: 0 TIME V ALUES: 30000 ms Length: 8 Class number: 5 - TIME V ALUES object .。
通信网基础课后答案
第一章通信网概述1.1简述通信系统模型中各个组成部分的含义,并举例说明。
答:通信系统的基本组成包括:信源,变换器,信道,噪声源,反变换器和信宿六部分。
信源:产生各种信息的信息源。
变换器:将信源发出的信息变换成适合在信道中传输的信号。
信道:按传输媒质分有线信道和无线信道,有线信道中,电磁信号或光电信号约束在某种传输线上传输;无线信道中,电磁信号沿空间传输。
反变换器:将信道上接收的信号变换成信息接收者可以接收的信息。
信宿:信息的接收者。
噪声源:系统内各种干扰。
1.2现代通信网是如何定义的?答:由一定数量的节点和连接这些节点的传输系统有机地组织在一起的,按约定信令或协议完成任意用户间信息交换的通信体系。
适应用户呼叫的需要,以用户满意的效果传输网内任意两个或多个用户的信息。
1.3试述通信网的构成要素及其功能。
答:通信网是由软件和硬件按特定方式构成的一个通信系统。
硬件由:终端设备,交换设备和传输系统构成,完成通信网的基本功能:接入、交换和传输;软件由:信令、协议、控制、管理、计费等,它们完成通信网的控制、管理、运营和维护,实现通信网的智能化。
1.4分析通信网络各种拓扑结构的特点。
(各种网络的拓扑结构图要掌握)答:基本组网结构:Ø网状网:优点:①各节点之间都有直达线路,可靠性高;②各节点间不需要汇接交换功能,交换费用低;缺点:①各节点间都有线路相连,致使线路多,建设和维护费用大;②通信业务量不大时,线路利用率低。
如网中有N个节点,则传输链路数H=1/2*N(N-1)。
Ø星形网:优点:①线路少,建设和维护费用低;②线路利用率高;缺点:①可靠性低,②中心节点负荷过重会影响传递速度。
如网中有N个节点,则传输链路数H=N-1。
Ø环形网:同样节点数情况下所需线路比网状网少,可靠性比星形网高。
如网中有N个节点,则传输链路数H=N。
Ø总线形网:优点:①节点接入方便②成本低,缺点:①传输时延不稳定②若传输总线损坏,整个网络会瘫痪。
ch01通信网概述
通信范围划分:公用网 、局部网、专用网
信号形式划分:模拟网、数字网、综合网
信道划分:有线网、无线网、移动网
2020/11/7
通信网基础
9
通信网的划分
根据网络功能从水平方向划分:用户驻地网 CPN、接入网AN和核心网CN。
2020/11/7
通信网基础
推动第一代向第二代发展的动力,来自对提高业务质 量的要求,特别是对提高长途电话质量的要求。在技 术上,“纳奎斯特”定理证明了可以从两倍于模拟频 率的抽象中完全无失真地恢复原模拟信号,奠定了数 字通信的基础。半导体技术的发展使数字通信技术的 应用成为可能,于是第二代通信网蓬勃发展起来。
2020/11/7
10
通信网的划分
根据网络功能从垂直方向划分:业务网、传送 网和支撑网(ITU-T 网络分层的概念)
2020/11/7
通信网基础
11
业务网,传送网和支撑网
业务网是指向公众提供电信业务的网络,包括 固定电话网、移动电话网、互联网、IP电话网、 数据通信网、智能网、窄带综合业务数字网 (N-ISDN)、宽带综合业务数字网(B-ISDN) 等。
2020/11/7
通信网基础
19
第三代通信网 (NGN)
基于IP的分组交换通信网,通信网中的全部设备都是 用IP分组的数字通信技术来设计、制造的。
第三代通信网中的所有设备均是用于产生、存储、交 换、传输数字化电信信息,实现电信信息的通信设备。 下一代通信网这个概念并不是现在才有的,而是在上 世纪80年代末、90年代初就已经提出来了。电信专家 为此作出了不屈不挠的努力,在摸索中不断前进。经 过10余年的研究和实践,认识到基于IP分组的数字通 信技术才是第三代通信网的核心技术。
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Solutions to Chapter 11a. Describe the step-by-step procedure that is involved from the time you deposit a letter in a mailbox to the time the letter is delivered to its destination. What role do names, addresses and mail codes (such as ZIP codes or postal codes) play? How might the letter be routed to its destination? To what extent can the process be automated?Solution:The steps involved in mailing a letter are:1. The letter is deposited in mailbox.2. The letter is picked up by postal employee and placed in sack.3. The letter is taken to a sorting station, where it is sorted according to destination, asdetermined by the mail code and grouped with other letters with the same destination mailcode. (If there is no mail code, then it is determined by the largest geographical unit, forexample, country (if specified), otherwise state (if specified), otherwise city (if specified).)4. The letter is shipped to the post office that handles the mail for the specific mail code (orcountry or city).5. The letter is then sorted by street address.6. The letter is picked up at the post office by the postal worker responsible for delivering to thespecified address.7. The letter is delivered according to the number and street.The name is not really used, unless the street address is missing or incorrect. The name is at the destination to determine who the letter belongs to. (Unless of course the letter is being sent to a small town, where most inhabitants are known to the postal worker.)The mail delivery process can be automated by using optical recognition on the mail code. The letter can then be sorted and routed to the destination postal station, and even to the destination neighborhood, depending on the amount of geographical detail built into the mail code.1b. Repeat part (a) for an e-mail message. At this point, you may have to conjecture different approaches about what goes on inside the computer network.Solution:The steps involved in e-mailing a message are:1. The message is sent electronically by clicking 'Send'. (In Chapter 2 we see that the SimpleMail Transfer Protocol (SMTP) is used to do this.)2. The mail provider of the sender sends a request to a name server for the network address ofthe mail provider of the recipient. The mail provider is determined by the information followingthe @ symbol.3. If the mail provider finds the network address of the recipient's mail provider, then it sends themessage to that address.4. Upon receipt of the message the mail provider either informs the recipient automatically thatthere is mail or waits until mail is requested by the recipient, depending on how the mailservice is set up. The recipient is determined by the information before the @ symbol.1c. Are the procedures in parts (a) and (b) connection-oriented or connectionless?Solution:The transfer service of letters in a postal system is connectionless. Users do not set upconnections prior to the transfer of each letter. (Note however that the postal system may have regularly scheduled shipments of bundles of mail from city to city, because the volume of mailbetween cities is predictable. These regular shipments can be viewed as pre-arranged"connections.")E-mail transfer is also a connectionless service in that the user does not set up end-to-endconnections to send a message. We will see in Chapter 2, however, that e-mail protocols useconnection-oriented procedures to exchange messages.NOTE: For parts (a) and (b) other step-by-step procedures are possible.2a. Describe what step-by-step procedure might be involved inside the network in making a telephone connection.Solution:1. The telephone number specifies an "address" where the receiver is located. In the NorthAmerican telephone numbering system the first three digits are the area code which specifiesthe main geographical region of the receiver; the next three digits specify a particulartelephone office in the North American network. The final four digits identify the particularlocation of the receiver.2. When a telephone number is dialed equipment at the other end of the telephone line uses thesequence of dialed numbers to determine a route across the telephone network from the call-originating phone to the destination phone. A circuit is established between the originatingand destination phone along the identified route.3. A ringing tone is then applied at the destination to indicate that there is an incoming call. If thedestination party (a person, an answering machine, or some other device) is ready to answer,the call is completed through the lifting of the phone set or some equivalent action. Thesetting up of telephone calls is discussed in Chapter 4.2b. Now consider a personal communication service that provides a user with a personal telephone number. When the number is dialed, the network establishes a connection to wherever the user is located at the given time. What functions must the network now perform in order to implement this service?Solution:1. The key difference here is that the personal telephone number is not tied to a specificlocation. Instead the personal number is associated with one or more pieces of equipment,for example, a cell phone, that can request service from various points in the network. Thisnecessitates the translation of the personal telephone number to a number that correspondsto a specific location in the network at a given time.2. When the telephone number is dialed, a message requesting a connection setup is sent tothe "home" location of the personal number.3. The home location must somehow be able to redirect the connection setup process to thecurrent location of the user. For example, the user may register one or more forwarding"addresses" to which calls are to be redirected. The signaling system is used to locate theuser. In cellular telephony, for example, requests for connections to a given mobiletelephone are broadcast over specific signaling channels.4. Once the location of the destination is identified a connection is established.3. Explain how the telephone network might modify the way calls are handled to provide the following services:Solutions follow questions:a.Call Display: the number and/or name of the calling party is listed on a screen before the call is answered.Along with the request for connection setup, the system sends identifying information regarding the originator's phone.b.Call Waiting: a special sound is heard when the called party is on the line and another user is trying toreach the called party.When a second caller tries to access a busy line, the system applies an audible sound, for example, a "beep" or a tone, to the voice signal that is being sent to the called party. The tone alerts the called party that there is an incoming call and provides the choice of answering the second call without disconnecting the first call.c.Call Answer: if the called party is busy or after the phone rings a prescribed number of times, the networkgives the caller the option of leaving a voice message.If the call is not answered after so many rings, the system establishes a connection to anothernumber (that is, forwards the call) that is associated with the answering system. The call is then automatically answered by an answering system.d.Three-Way-Calling: allows a user to talk with two other people at the same time.After the establishment of a call between two parties, the caller can then indicate to the telephone system that he or she wants to call a third party. Without disconnecting the first pair, the system sets up a connection to the second called party, and once established, transmits both conversations to all parties simultaneously. Special equipment is required to combine and distribute the voice signals of the three parties.4a. Suppose that the letter in problem 1 is sent by fax. Is this mode of communications connectionless or connection-oriented? Real-time or non-real time?Solution:In order to send the letter by fax, a telephone connection must first be established. Therefore the mode of communications is connection-oriented. The transfer of information across the network occurs in real-time.4b. Repeat part (a) if a voice mail message is left at a given telephone.Solution:Again, the process of leaving a voice mail message is connection-oriented, since there must be anend-to-end connection between the caller and the receiver with the answering service. The leaving of voice mail is also done in real-time. However, the mode of communication between the person that leaves the voice mail and the person that listens to the voice mail is connectionless and non-real-time. The people involved do not need to establish a simultaneous connection with each other in order to communicate; also, they record and listen to the messages at different times.5. Suppose that network addresses are scarce, so addresses are assigned so that they are not globally unique; in particular suppose that the same block of addresses may be assigned to different organizations. How can the organizations make use of these addresses? Can users from two such organizations communicate with each other?Solution:To make the example concrete suppose that two organizations are assigned the same set oftelephone numbers. Clearly, users within each organization can communicate with each other as long as they have a unique address within the organization. However, communications outside anorganization poses a problem since any given address is no longer unique across multipleorganizations.A possible approach to enabling communications between users in different organizations is to use atwo-step procedure as follows. Each organization has a special gateway to communicate outside the organization. Internal users contact the gateway to establish calls to other organizations. Gateways have procedures to establish connections with each other. This enables gateways to establishconnections between their internal users and users in other networks.6a. Describe the similarities and differences in the services provided by (1) a music program delivered over broadcast radio and (2) music delivered by a dedicated CD player.Solution:Both broadcast radio and a dedicated CD player provide users with similar types of information and with almost the same performance. However the broadcast radio offers its service in a real-time fashion with no interaction with users. A CD player, on the other hand, stores its information on a CD and delivers it on demand.6b. Describe how these services might be provided and enhanced by providing them through a communications network.Solution:Broadcast radio is typically transmitted in real time using radio waves "over the air." Broadcast audio programs can also be transmitted over any communication network. Furthermore, when stored in servers that are attached to a network, "broadcast" audio programs can be retrieved for listening at a later time. Indeed a server can also store CD audio material and retrieve it on demand. If the network and server are sufficiently responsive, it may also be possible to provide the interactivity of a personal CD player through a network-based service.7a. Use the World Wide Web to visit the sites of several major newspapers. How are these newspapers changing the manner in which they deliver news services?Solution:Traditionally, newspapers deliver news in printed "hard-copy" form in various editions during the day.The information is primarily in text form, supplemented by still pictures and graphics. The information in a newspaper is organized in several sections: main page, business, local, sports, classifieds, etc.Some of the revenue of newspapers is from subscriptions, but most of the newspaper revenue is usually from advertising.In the traditional model, getting the news from a particular newspaper required getting the physical printed form. The WWW and the Internet allow newspapers to deliver information in electronic form and at any time that a user requests it. Newspaper information on the Web still consists mostly of text, images, and graphics that can be retrieved and displayed readily by a Web browser. In most cases, subscription is not required and advertising is displayed along with news information. In order to be competitive, newspapers can no longer work in terms of editions, especially for late-breaking news. Instead, news items are updated continuously day and night. The use of a Web browser allowsa user to quickly find the news items of interest. In addition, newspapers can provide more in-depthcoverage in their web page on topics of special interest than is possible in print medium.7b. Now visit the websites of several major television networks. How are they changing the manner in which they deliver news over the Internet? What differences, if any, are there to the approach taken by television networks and newspapers?Solution:Traditionally, television networks deliver news using audio, video, and graphics, with very little text.News is traditionally delivered in programs that are broadcast at various times of the day. Again, the information is organized in several sections: international, business, sports, entertainment, etc. The revenue in television news is almost entirely from advertising. The WWW and the Internet allow users to retrieve news information in electronic form and at any time. However, the transmission of audio and video over the Internet poses a challenge, so this information must be transmitted in compressed form and in relatively low quality. Consequently, more of the information is provided in text form. The dependence on audio and video also required the development of applications for the decompression and display of such information. Again, in order to be competitive, news networks update their Web pages continuously day and night as news develops. The higher cost involved in providing video service has led to the introduction of subscription fees.8. Discuss the advantages and disadvantages of transmitting fax messages over the Internet instead of the telephone network.Solution:The comparison of fax transmission over the Internet and over the telephone network reflect the differences between e-mail and telephone calls. The transmission of a fax message as an attachment to e-mail means that the transfer of the message is not in real time and that delivery is not necessarily confirmed. The transmission of a fax message over the telephone network, on the other hand, is real time and is confirmed with fairly high certainty. On the other hand, the cost of transmission of a fax message over the Internet is very inexpensive and is not distance-sensitive. The transmission of fax messages over telephone networks may involve long distance charges.9a. Suppose that an interactive video game is accessed over a communication network. What requirements are imposed on the network if the network is connection-oriented? connectionless?Solution:We suppose that the game involves the interaction between a player and a server across a network.To support an interactive video game over a communications network, the network, whetherconnection-oriented or connectionless, must provide real-time delivery of the player's commands to the server, and of the server's responses to the player. With a connection-oriented network,connections between the player and the servers transfer the sequence of commands and responsesthroughout the game with very little delay. In a connectionless network, user commands may be delivered to the other end with variable delay, out-of-sequence, or not at all. The user’s network software is responsible for ensuring the ordered and correct delivery of game commands. In-time delivery of commands cannot be assured.9b. Repeat part (a) if the game involves several players located at different sites.Solution:The requirements on the network depend on how the game is implemented. In the centralizedapproach the players interact through a central server that processes the commands from all of the players, maintains a view of the state of the overall system, and issues appropriate responses to all the players. Alternatively, the game could be implemented in a decentralized fashion, where each player receives commands from some or all of the players, maintains a local view of the system state, and transmits responses to some or all of the players.In the centralized approach, the network requirements are essentially the same as those in part (a). In the decentralized approach, the real-time response requirement may apply only to a subset of players when they happen to be interacting with each other. If players are located at different sites, thenetwork could provide multicast capability so messages can be exchanged among the server and all the players.9c. Repeat part (b) if one or more of the players is in motion, for example, kids in the back of the van during a summer trip.Solution:If one or more players is in motion, then the network must also be able to locate each mobile player and then deliver and receive information to and from such player. Additional delay and loss issues come into play when a mobile user is “handed-off” from one radio coverage area to another.10. Discuss the similarities between the following national transportation networks and a communications network. Is the transportation system more similar to a telephone network or to a packet network?Solution:Transportation networks are designed to transfer people and goods; communications networks are designed to transfer information. Like communication networks, all transportation systems involve links, in the form of roads, rails, or air corridors, and switching points, in the form of stations, airports, and highway interchanges. Addressing and geographical names are used in transportation networks to identify destinations, and routing of various forms is required to direct goods to their destinations.Transportation resources, in the forms of cars or airplanes, are shared or "multiplexed" amongvarious goods that traverse common parts of the network.a.Railroad network.Railroad network: Goods or people arrive at train stations and are loaded for transport to specific places. Tickets are purchased for a given destination, usually guaranteeing that there will be a place in a car from the starting station all the way to the destination station. This is similar to establishing a connection across a telephone network. Alternatively, a passenger may choose to purchase a ticket at each station along the way. This corresponds more closely to a connectionless packet network model.b.Airline network.Airline network: In this case passengers purchase tickets that guarantee a seat all the way to the destination, even if transfers are made at intermediate airports. This is similar to the establishment of connections across a telephone network.c.Highway system.Highway systems: Trucks or cars enter the highway without making reservations ahead of time and without informing any central authority of their destination or route. This mode of operationcorresponds closely to the operation of a connectionless packet network.bination of (a), (b), and (c).Combination of transportation systems: The combination of air, rail, and highway transportationsystems can and are used jointly for the transfer of people and goods. Each transportation system is different in how it is organized and how transfers are accomplished. Nevertheless, it is possible to use these systems in combination to provide a higher degree of connectivity between sources anddestinations of people and goods. We will see that the Internet plays a role similar to the combined transportation systems in that the Internet enables the transfer of information across multipledissimilar networks that may differ in how they are organized and how they operate.11. In the 1950s, standard containers were developed for the transportation of goods. These standard containers could fit on a train car, on a truck, or in specially designed container ships. The standard size of the containers makes it possible to load and unload them much more quickly than using non-standard containers of different sizes. Draw an analogy to packet switching communications networks. In your answer identify what might constitute a container and speculate on the advantages that may come from standard-size information containers.Solution:In a packet-switching communications network, the component that is analogous to the container is a constant-size packet (preferably of short length) that can be used for the transport of information.Transmission systems and networks of various types can be designed to transfer information of the given standardized size, much like trucks, trains, and ships can be designed to carry standardcontainers. Packing and unpacking of fixed-size units is simpler than packing and unpacking ofvariable-length units. Consequently, it is simpler to schedule the transfer of packets across switches that use constant-size packets than across switches that make use of variable-length packets.12. The requirements of world commerce led to the building of the Suez and Panama canals. What analogous situations might arise in communication networks?Solution:The Suez and Panama canals were built to provide a shorter path between geographical areas with major trade relationships. The analogous situation arises in communication networks when the path between two areas in the network are too "long" either in terms of excessive delay or inadequate capacity to transfer the required volumes of traffic. The addition of network resources to provide shorter paths for lower delay or wider paths for higher capacity corresponds to the building of a canal.13. Two musicians located in different cities wish to have a jam session over a communications network. Find the maximum possible distance between the musicians if they are to interact in real-time, in the sense of experiencing the same delay in hearing each other as if they were 10 meters apart. The speed of sound is approximately 330 meters/second. Assume that the network transmits the sound at the speed of light in cable, 2.3 x 108 meters/second.Solution:The first step is to find the delay for the sound when the musicians are 10 meters apart: t10 = 10/330 = 30.30 millisecondsThe maximum distance is the time required for a real-time 'experience' times the cable speed:d = (2.3 x 108) x (30.30 x 10-3) = 6,969,000 meters = 6969 kilometers14. The propagation delay is the time that is required for the energy of a signal to propagate from one point to another.a. Find the propagation delay for a signal traversing the following networks at the speed of light in cable (2.3 x 108 meters/second):• a circuit board 10 cm• a room 10 m• a building 100 m• a metropolitan area 100 km• a continent 5000 km•up and down to a geostationary satellite 2 x 36000 kmSolution:To find the propagation delay, divide distance by the speed of light in cable. Thus we have:• a circuit board t prop = 4.347 x 10-10 seconds• a room t prop = 4.3478 x 10-8 seconds• a building t prop = 4.3478 x 10-7 seconds• a metropolitan area t prop = 4.3478 x 10-4 seconds• a continent t prop = 0.02174 seconds•up and down to a geostationary satellite t prop = 0.31304 seconds14b. How many bits are in transit during the propagation delay in the above cases, if bits are entering the above networks at the following transmission speeds: 10,000 bits/second; 1 megabit/second; 100 megabits/second; 10 gigabits/second.Solution:The number of bits in transit is obtained by multiplying the transmission rate R by the propagation delay:Distance (m)10 Kbps 1 Mbps100 Mbps10 Gbps0.1 4.347 x 10−6 4.347 x 10−40.04347 4.347810 4.3478 x 10−40.043478 4.3478434.780100 4.3478 x 10−30.4347843.4784347.800100 000 4.3478434.7843478 4.3478x1065 000 000217.4217402174000 2.174x10872 000 0003130.431304031304000 3.1304x10915. In problem 14, how long does it take to send an L-byte file and to receive a 1-byte acknowledgment back? Let L=1, 103, 106, and 109 bytes.Solution:The total time required to send a file and receive an acknowledgment of its receipt is given by:t total = L message/R + L ack/R +2* t prop = L message/R + L ack/R + 2*d/cwhere L message is the message length in bits, L ack is the acknowledgment length in bits, R is thetransmission bit rate, d is the distance traversed, and c is the speed of light.The above equation shows that there are two main factors that determine total delay:1. Message and ACK transmission time, which depends on the message length and thetransmission bit rate;2. Propagation delay, which depends solely on distance.When the propagation delay is small, message and ACK transmission times determine the total delay. On the other hand, when the bit rate becomes very large, the propagation delay provides a delay component that cannot be reduced no matter how fast the transmission rate becomes.The tables below show the two main components of the total delay in microseconds. The message transmission time is shown in red and the propagation delay is shown in blue. The entries in the total delay tables are colored according to which delay component is dominant.Message + ACK delay @10 kbps Message +ACK delay@1 Mbpsmessage+ACK delay@100 Mbpsmessage+ACK delay@10 Gbps1600160.160.0016Distance (meters) 2 * prop. delay(microseconds)Total delay@10 kbpstotal delay@1 Mbpstotal delay@100 Mbpstotal delay@10 Gbps0.1 0.00087 1600.0008716.00086960.160870.0024710 0.086957 1600.08695716.08695650.2469570.088557100 0.869565 1600.86956516.8695652 1.0295650.871165 100000 869.5652 2469.565217885.565217869.7252869.5668 5000000 43478.26 45078.2608743494.260943478.4243478.26 72000000 626087 627686.9565626102.957626087.1626087 Table 1: Message length = 1 byteMessage + ACK delay @10 kbps message +ACK delay@1 MbpsMessage+ACK delay@100 Mbpsmessage+ACK delay@10 Gbps800800800880.080.8008Distance (meters) 2 * prop. delay(microseconds)Total delay@10 kbpstotal delay@1 MbpsTotal delay@100 Mbpstotal delay@10 Gbps0.1 0.00087 800800.00098008.0008780.080870.8016710 0.086957 800800.0878008.0869680.166960.887757100 0.869565 800800.86968008.8695780.94957 1.670365 100000 869.5652 801669.56528877.56522949.6452870.366 5000000 43478.26 844278.260951486.260943558.3443479.06 72000000 626087 1426886.957634094.957626167626087.8 Table 2: Message length = 1000 bytesMessage + ACK delay @10 kbps message +ACK delay@1 Mbpsmessage+ACK delay@100 Mbpsmessage+ACK delay@10 Gbps8.00E+088.00E+0680000.08800.0008distance (meters) 2 * prop. delay(microseconds)Total delay@10 kbpstotal delay@1 Mbpstotal delay@100 Mbpstotal delay@10 Gbps0.1 0.00087 8.00E+088.00E+0680000.08800.001710 0.086957 8.00E+088.00E+0680000.17800.0878100 0.869565 8.00E+088.00E+0680000.95800.8704 100000 869.5652 8.00E+088.00E+0680869.651669.566 5000000 43478.26 8.00E+088.04E+06123478.344278.26 72000000 626087 8.01E+088.63E+06706087626887 Table 3: Message length = 106 bytes。