下载文档原格式
《高等数学》习题参考资料第一篇 一元函数微积分第一章 极限与连续§1 函 数习 题1.确定下列初等函数的定义域：(1) 21)(2−−+=x x x x f ；(2)4)(2−=x x f ；(3) 21arcsin )(−=x x f ；(4)2)5lg()(x x x f −=；(5) 4lg )5lg()(2−−=x x x f ；(6)x x x f cos sin )(−=。
1． 【答案】(1) )},2()2,1()1,(|{:+∞∪−∪−−∞∈=x x D (2) )},2[]2,(|{:+∞∪−−∞∈=x x D (3) ]}3,1[|{:;−∈=x x D (4) )}5,0()0,(|{:∪−∞∈=x x D (5) ]}4,1[|{:∈=x x D (6)+ +∈=+∞−∞=U k k k x x D ππ452,412|:.2. 作出下列函数的图象：（1）|sin |sin )(x x x f −=；(2)|1|2)(−−=x x f ；（3）+−−=,1,1,21)(x x x x f .12,21,1||−<<−<<≤x x x 2 【答案】 (1)2(2)2 (3)3.判断下列函数的奇偶性：（1）x x x f ++−=11)(；（2）xxx f x x +−+−=11lg110110)(；（3）x x a a x f x x sin )(++=−；（4）)1lg()(2x x x f ++=。
3. 【答案】 (1) 偶函数; (2) 偶函数; (3) 偶函数; (4) 奇函数 .4.证明:两个奇函数的乘积是偶函数;一个奇函数与一个偶函数的乘积是奇函数。
4． 【答案】 设)(x f ,)(x h 是奇函数, )(x g 是偶函数,)()()(x h x f x f =,)()()(x g x f x G =, 于是)()()(x h x f x F −−=−))())(((x h x f −−=)()()(x F x h x f ==, 因此)(x F 是偶函数.)()()(x g x f x G −−=−)()(x g x f −=)(x G −=, 因此)(x G 是奇函数.5．设函数f 满足：D （f ）关于原点对称，且()xc x bf x af =+1)(，其中a ,b ,c 都是常数，||||b a ≠,试证明f 是奇函数。
传热学习题集第一章思考题1． 试用简练的语言说明导热、对流换热及辐射换热三种热传递方式之间的联系和区别。
答：导热和对流的区别在于：物体内部依靠微观粒子的热运动而产生的热量传递现象，称为导热；对流则是流体各部分之间发生宏观相对位移及冷热流体的相互掺混。
联系是：在发生对流换热的同时必然伴生有导热。
导热、对流这两种热量传递方式，只有在物质存在的条件下才能实现，而辐射可以在真空中传播，辐射换热时不仅有能 量的转移还伴有能量形式的转换。
2． 以热流密度表示的傅立叶定律、牛顿冷却公式及斯忒藩－玻耳兹曼定律是应当熟记的传热学公式。
试写出这三个公式并说明其中每一个符号及其意义。
答：① 傅立叶定律：，其中，－热流密度；－导热系数；－沿x方向的温度变化率，“－”表示热量传递的方向是沿着温度降低的方向。
② 牛顿冷却公式：，其中，－热流密度；－表面传热系数；－固体表面温度；－流体的温度。
③ 斯忒藩－玻耳兹曼定律：，其中，－热流密度；－斯忒藩－玻耳兹曼常数；－辐射物体的热力学温度。
3． 导热系数、表面传热系数及传热系数的单位各是什么？哪些是物性参数，哪些与过程有关？答：① 导热系数的单位是：W/(m.K)；② 表面传热系数的单位是：W/(m 2.K)；③ 传热系数的单位是：W/(m 2.K)。
这三个参数中，只有导热系数是物性参数，其它均与过程有关。
4． 当热量从壁面一侧的流体穿过壁面传给另一侧的流体时，冷、热流体之间的换热量可以通过其中任何一个环节来计算（过程是稳态的），但本章中又引入了传热方程式，并说它是“换热器热工计算的基本公式”。
试分析引入传热方程式的工程实用意义。
答：因为在许多工业换热设备中，进行热量交换的冷、热流体也常处于固体壁面的两侧，是工程技术中经常遇到的一种典型热量传递过程。
5． 用铝制的水壶烧开水时，尽管炉火很旺，但水壶仍然安然无恙。
而一旦壶内的水烧干后，水壶很快就烧坏。
试从传热学的观点分析这一现象。
第一章衍生金融工具概述复习思考题1.1.衍生金融工具的概念是什么? 它有哪些基本特征?1.2.如何理解衍生金融工具的高杠杆性?1.3.既然衍生金融工具是基于避险的目的产生的，为什么又说它具有高风险的特点呢? 1.4.衍生金融工具的主要种类有哪些?1.5.购买远期价格为100美元的远期合同与持有执行价格为100美元的看涨期权有什么不同?1.6.试述欧式期权与美式期权的差别。
1.7.期货合约与远期合约本质上相同，但也有差异，请解释。
1.8.投资者签订期货合约的成本和签订期权合约的成本相同吗?1.9.什么是互换协议? 举例说明如何用利率互换管理利率风险。
1.10.什么是场内市场? 试举出几种场内交易的衍生金融工具。
1.11.什么是场外市场? 试举出几种场外交易的衍生金融工具。
1.12.场内市场和场外市场有哪些区别?1.13.相对于传统金融工具，衍生金融工具的风险来自哪些方面?1.14.为什么早期的期货交易所主要从事农产品期货交易?1.15.试从风险管理的角度分析２０世纪７０年代以来衍生金融工具市场快速发展的原因。
1.16.1975年10月率先开办抵押债券期货，标志着利率期货的正式产生的交易所是( ) Ａ.芝加哥商业交易所Ｂ.芝加哥期货交易所Ｃ.纽约证券交易所Ｄ.美国堪萨斯农产品交易所1.17.1982年2月，推出第一个股指期货合约———价值线综合指数期货合约(Value Line)的交易所是( )Ａ.芝加哥商业交易所Ｂ.芝加哥期货交易所Ｃ.纽约证券交易所Ｄ.美国堪萨斯交易所1.18.２１世纪以来衍生金融工具市场呈现出什么样的发展趋势?1.19.我国的国债期货市场现在上市交易的有哪几种产品?1.20.目前在中国金融期货交易所交易的股指期货有哪几种合约?讨论题1.1.试从衍生金融工具与风险相互关系的角度分析２００８年金融危机后衍生金融工具的发展前景。
1.2.请问你如何看待衍生金融工具市场中的投机者? 为了金融市场的发展我们是否需要严格限制投机行为?1.3.我国衍生金融工具市场的发展现状如何? 结合国际衍生金融工具发展的动因，试分析我国衍生金融工具市场发展的潜力。
第一章政府与非营利组织会计概述一、单项选择题1.作为非营利组织会计的主体，我国的非营利组织是指（）A.政府机构B.事业单位C.民间非营利组织D.公立非营利组织2．根据社会组织类型与会计目标的不同，我国的会计体系划分为两大分支，一个分支是企业会计，另一分支是（）A．政府会计B．预算会计C．财务会计D．政府与非营利组织会计3．作为政府会计的主体，政府的概念是广义的，其范围主要包括（）A．立法机关、行政机关和司法机关B．行政机关、公益组织和司法机关C．立法机关、行政机关和国有企业D．行政机关、国有企业和司法机关4.下列有关政府与非营利组织会计的外部环境特征的说法不正确的是（）A.政府与非营利组织会计的外部环境相对复杂，集中体现在无所不在的利益相关者的广泛影响以及来自于法规制度的多重约束。
B.法律和行政的约束不是指导政府与非营利组织达到其目标的社会手段。
C.政府和非营利组织行为的影响无处不在，政府和非营利组织拥有“无所不在的利益相关者”。
D.企业只提供那些它们认为能够增加它们收益的商品和服务，在这方面，企业可以自行决策。
5.政府与非营利组织会计信息使用者对财务信息需求的目的包括（）A.为决算管理服务和为经济与财务管理服务B.为预算管理服务和为财务管理服务C.为预算管理服务和为经济管理服务D.为预算管理服务和为经济与财务管理服务6．我国的政府会计系统包括（）A.政府预算会计和政府财务会计B.政府预算会计和政府管理会计C.政府成本会计和政府财务会计D.政府财务会计和政府管理会计7．我国政府会计体系中，主要反映和监督政府会计主体的预算收支执行情况和结果的会计为（）。
A.政府财务会计B.政府管理会计C.政府预算会计D.政府成本会计8．我国政府会计体系中，主要反映和监督政府会计主体的财务状况、运行情况和现金流量等的会计为（）。
A.政府财务会计B.政府管理会计C.政府预算会计D.政府成本会计9．我国政府会计体系中，反映和监督各级各类国家机关、政党组织财务状况、预算执行情况及结果的专业会计为（）A.事业单位会计B.行政单位会计C.财政总预算会计D.非营利组织会计10．为了规范政府的会计核算，财政部颁布了《政府会计准则——基本准则》，其实施时间为（）。
第一章 绪论（一）基本知识类题型1-1． 什么是计量经济学？1-2． 简述当代计量经济学发展的动向。
1-3． 计量经济学方法与一般经济数学方法有什么区别？1-4．为什么说计量经济学是经济理论、数学和经济统计学的结合？试述三者之关系。
1-5．为什么说计量经济学是一门经济学科？它在经济学科体系中的作用和地位是什么？ 1-6．计量经济学的研究的对象和内容是什么？计量经济学模型研究的经济关系有哪两个基本特征？1-7．试结合一个具体经济问题说明建立与应用计量经济学模型的主要步骤。
1-8．建立计量经济学模型的基本思想是什么？1-9．计量经济学模型主要有哪些应用领域？各自的原理是什么？1-10．试分别举出五个时间序列数据和横截面数据，并说明时间序列数据和横截面数据有和异同？1-11．试解释单方程模型和联立方程模型的概念，并举例说明两者之间的联系与区别。
1-12．模型的检验包括几个方面？其具体含义是什么？1-13．常用的样本数据有哪些？1-14．计量经济模型中为何要包括随机误差项？简述随机误差项形成的原因。
1-15．估计量和估计值有何区别？哪些类型的关系式不存在估计问题？1-16．经济数据在计量经济分析中的作用是什么？1-17．下列假想模型是否属于揭示因果关系的计量经济学模型？为什么？⑴ S R t t =+1120012.. 其中S t 为第t 年农村居民储蓄增加额（亿元）、R t 为第t 年城镇居民可支配收入总额（亿元）。
⑵ S R t t -=+144320030.. 其中S t -1为第（1-t ）年底农村居民储蓄余额（亿元）、R t 为第t 年农村居民纯收入总额（亿元）。
1-18．指出下列假想模型中的错误，并说明理由：（1）RS RI IV t t t =-+83000024112...其中，RS t 为第t 年社会消费品零售总额（亿元），RI t 为第t 年居民收入总额（亿元）（城镇居民可支配收入总额与农村居民纯收入总额之和），IV t 为第t 年全社会固定资产投资总额（亿元）。
第一章课后习题答案1、5个女生，7个男生进行排列，(a) 若女生在一起有多少种不同的排列？(b) 女生两两不相邻有多少种不同的排列？(c) 两男生A和B之间正好有3个女生的排列是多少？解：(a) 若女生在一起，可将5个女生看作一个整体参与排列，有8!种方式，然后5个女生再进行排列，有5!种方式，根据乘法法则，共有8!5!种方式。
(b) 若女生两两不相邻，可将7个男生进行排列，有7!种方式，考虑到两个男生之间的6个位置和两头的2个位置，每个位置安排一个女生均符合题意，故从中选出5个位置，然后5个女生再进行排列，按顺序安排到这5个位置，有C(8, 5)5!种方式，根据乘法法则，共有7!C(8, 5)5!=7!P(8, 5)种方式。
(c) 若两男生A和B之间正好有3个女生，可以按照顺序操作如下：首先将女生分为两组，一组3人，一组2人，有C(5, 3)种方式；将男生A和B看作一个整体，加上其他5个男生，2人一组的女生进行排列，有8!种方式；将3人一组的女生安排到男生A和B之间进行排列，有3!种方式；男生A和B进行排列，有2!种方式。
根据乘法法则，所求的排列方式为8!C(5, 3)3!2!=8!P(5, 3)2!2、求3000到8000之间的奇整数的数目，而且没有相同的数字。
解：设介于3000到8000之间的奇整数表示为abcd，则a∈{3, 4, 5, 6, 7}, d∈{1, 3, 5, 7, 9}，对a进行分类如下：(1) 若a∈{3, 5, 7}，则d有4种选取方式，bc有P(8, 2)种方式，根据乘法法则，此类数字有3⨯4⨯P(8, 2)=672个(2) 若a∈{4, 6}，则d有5种选取方式，bc仍有P(8, 2)种方式，根据乘法法则，此类数字有2⨯5⨯P(8, 2)=560个根据加法法则，3000到8000之间数字不同的奇整数的数目为672+560=1232个3、证明nC(n-1, r)=(r+1)C(n, r+1)，并给出组合解释。
传热学习题集第一章思考题1． 试用简练的语言说明导热、对流换热及辐射换热三种热传递方式之间的联系和区别。
答：导热和对流的区别在于：物体内部依靠微观粒子的热运动而产生的热量传递现象，称为导热；对流则是流体各部分之间发生宏观相对位移及冷热流体的相互掺混。
联系是：在发生对流换热的同时必然伴生有导热。
导热、对流这两种热量传递方式，只有在物质存在的条件下才能实现，而辐射可以在真空中传播，辐射换热时不仅有能 量的转移还伴有能量形式的转换。
2． 以热流密度表示的傅立叶定律、牛顿冷却公式及斯忒藩－玻耳兹曼定律是应当熟记的传热学公式。
试写出这三个公式并说明其中每一个符号及其意义。
答：① 傅立叶定律：，其中，－热流密度；－导热系数；－沿x方向的温度变化率，“－”表示热量传递的方向是沿着温度降低的方向。
② 牛顿冷却公式：，其中，－热流密度；－表面传热系数；－固体表面温度；－流体的温度。
③ 斯忒藩－玻耳兹曼定律：，其中，－热流密度；－斯忒藩－玻耳兹曼常数；－辐射物体的热力学温度。
3． 导热系数、表面传热系数及传热系数的单位各是什么？哪些是物性参数，哪些与过程有关？答：① 导热系数的单位是：W/(m.K)；② 表面传热系数的单位是：W/(m 2.K)；③ 传热系数的单位是：W/(m 2.K)。
这三个参数中，只有导热系数是物性参数，其它均与过程有关。
4． 当热量从壁面一侧的流体穿过壁面传给另一侧的流体时，冷、热流体之间的换热量可以通过其中任何一个环节来计算（过程是稳态的），但本章中又引入了传热方程式，并说它是“换热器热工计算的基本公式”。
试分析引入传热方程式的工程实用意义。
答：因为在许多工业换热设备中，进行热量交换的冷、热流体也常处于固体壁面的两侧，是工程技术中经常遇到的一种典型热量传递过程。
5． 用铝制的水壶烧开水时，尽管炉火很旺，但水壶仍然安然无恙。
而一旦壶内的水烧干后，水壶很快就烧坏。
试从传热学的观点分析这一现象。
历史课后习题答案第一章1、帝国资本主义的入侵给中国带来了什么？1、积极作用：（1）政治：①鸦片战争打开了中国的大门，明清以来的“闭关锁国”政策被打破，中国一步步被卷入世界资本主义潮流之中。
②加速了中国封建社会的瓦解（2）经济：①中国的自然经济逐步解体②一些企业家试图以“实业救国”，民族资本主义得以发展起来。
③中国的经济逐渐融入世界（3）思想文化：帝国主义的入侵唤醒了一部分中国人，他们为救国图存开始奋斗，洋务运动，戊戌变法，五四运动，新文化运动等一系列思想文化运动蓬勃发展。
（4）科技：电报、电话、电影等传入中国。
2、消极作用：（1）政治：从鸦片战争开始，帝国主义列强对中国发动了一系列的侵略战争并强迫中国政府与之订立丧权辱国的不平等条约，在中国攫取了种种特权，从而使一个独立的中国逐渐沦为了一个半殖民地半封建社会的中国。
（2）经济：帝国主义势力操纵中国的经济命脉。
①帝国主义列强根据不平等条约，控制了中国一切重要的通商口岸，并把许多通商口岸划出一部分土地作为它们直接管理的租界。
它们控制了中国的海关和对外贸易，控制了中国的交通事业。
因此它们便能够大量地倾销它们的商品，把中国变成它们的工业品的市场，同时又使中国的农业生产服从于帝国主义的需要。
②在经济上中国逐步地形成了对帝国主义资本的依附而丧失了自己的独立性。
③资本主义国家一定程度上压制着民族资本主义的进一步发展，民族资本主义逐渐陷入困境。
（3）思想文化：①西方思想文化对中国传统思想文化的冲击。
②圆明园等宝贵的历史文化财富被帝国主义摧毁（4）对人民：大量中国人民惨遭杀害，是历史的巨大灾难。
2、反对外国侵略的斗争具有什么意义？第一，近代中国人民进行的反侵略战争，沉重打击了帝国主义侵华的野心，粉碎了他们瓜分中国和把中国变成完全殖民地的图谋。
帝国主义列强一次次对中国发动侵略战争，绝不仅仅是为了通商，而是为了掠夺和扩大殖民地，为了他们自身的殖民扩张利益。
每一次战争，都伴随着更大的贪梦目的和更多的利益要求。
第一章 X 射线物理习题一解答1-1 产生X 射线需要哪些条件？答：首先要有产生电子的阴极和被轰击的阳极靶，电子加速的环境条件即在阴极和阳极间建立电位差，为防止阴极和阳极氧化以及电子与中性分子碰撞的数量损失，要制造压强小于4-Pa 的真空环境，为此要有一个耐压、密封的管壳。
1-2 影响X 射线管有效焦点大小的因素有哪些？答：影响有效焦点大小的因素有：灯丝大小、管电压和管电流、靶倾角。
1-3 在X 射线管中，若电子到达阳极靶面的速度为1.5⨯810ms -1，求连续X 射线谱的最短波长和相应的最大光子能量。
答：此题的思路是由动能公式221v m 求出电子的最大动能，此能量也是最大的光子能量，从而求出最短波长。
但当速度可与光速c=3⨯810ms -1相比较时，必须考虑相对论效应，我们可以用下面公式求出运动中电子的质量此题的结果告诉我们，管电压为73.8KV 。
反过来，如果知道管电压，求电子到达阳极靶表面的电子速度时，同样需要考虑相对论效应。
1-4 下面有关连续X 射线的解释，哪些是正确的？A ．连续X 射线是高速电子与靶物质轨道电子相互作用的结果；B ．连续X 射线是高速电子与靶物质的原子核电场相互作用的结果；C ．连续X 射线的最大能量决定于管电压；D ．连续X 射线的最大能量决定于靶物质的原子序数；E ．连续X 射线的质与管电流无关。
正确答案：B 、C 、E1-5 下面有关标识X 射线的解释，哪些是正确的？A ．标识X 射线是高速电子与靶物质轨道电子相互作用的结果；B ．标识X 射线的质与高速电子的能量有关；C ．标识X 射线的波长由跃迁电子的能级差决定；D ．滤过使标识X 射线变硬；E ．靶物质原子序数越高，标识X 射线的能量就越大。
正确答案：A 、C 、E1-6 影响X 射线能谱的因素有哪些？答：电子轰击阳极靶产生的X 射线能谱的形状（归一化后）主要由管电压、靶倾角和固有滤过决定。
当然，通过附加滤过也可改变X 射线能谱的形状。
第一章习题一、选择题1.世界上第一台通用电子数字计算机诞生于( C )。
A、1950年B、1945年C、1946年D、1948年2.与二进制数(10111.101)2等值的十进制数是( A )。
A、23.625B、23.5C、39.5D、39.6253.与十进制数(101.1)10等值的二进制数是( D )。
A、5.5B、110010.00011C、11000101.0011D、1100101.000110011…4.与十六进制数(1AE.5D)16等值的八进制数是( C )。
A、(647.272)8B、(565.727)8C、(656.272)8D、(656.235)85.与二进制数(1111111111)2等值的十六进制数是( B )。
A、FF3HB、3FFHC、210-1D、1777O6. 在PC机中，1MB准确等于( C )。
A、1000×1000KBB、1024×1024KBC、1024×1024BD、1000×1000B7.已知真值X= +1101010，则其补码[X]补等于( B )。
A、00010110B、01101010C、10010110D、00101108.已知机器数[X]反=11111111，则其真值X为( D )。
A、00000000B、+0000000C、10000000D、-00000009.已知[X]原=10011110，则其对应的[X]补为( D )。
A、01100010B、11100001C、-0011110D、1110001010.已知A =01011101，B =11101010，则A○+B为( A )。
A、10110111B、01001000C、11111111D、1010001011.1MB等于( C )字节？A、10KB、100KC、1024KD、10000K12.把十进制数215转换成二进制数，结果为( D )。
1.-第一章课后习题及答案第一章1.(Q1) What is the difference between a host and an end system? List the types of end systems. Is a Web server an end system?Answer: 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, Internet-connected PDAs, WebTVs, etc.2.(Q2) The word protocol is often used to describe diplomatic relations. Give an example of a diplomatic protocol.Answer: Suppose Alice, an ambassador of country A wants to invite Bob, an ambassador of country B, over for dinner. Alice doesn’t simply just call Bob on the phone and say, come to our dinner table now”. Instead, she calls Bob and suggestsa date and time. Bob may respond by saying he’s not available that particular date, but he is available another date. Alice and Bob continue to send “messages” back and forth until they agree on a date and time. Bob then shows up at the embassy on the agreed date, hopefully not more than 15 minutes before or after the agreed time. Diplomatic protocols also allow for either Alice or Bob to politely cancel the engagement if they have reasonable excuses.3.(Q3) What is a client program? What is a server program? Does a server program request and receive services from a client program?Answer: A networking program usually has two programs, each running on a different host, communicating with each other. The program that initiates the communicationAnswer: Current possibilities include: dial-up (up to 56kbps); DSL (up to 1 Mbps upstream, up to 8 Mbps downstream); cable modem (up to 30Mbps downstream, 2 Mbps upstream.4.(Q7) What are some of the physical media that Ethernet can run over?Answer: Ethernet most commonly runs over twisted-pair copper wire and “thin” coaxial cable. It also can run over fibers optic links and thick coaxial cable.5.(Q8) Dial-up modems, HFC, and DSL are all used for residential access. For each of these access technologies, provide a range of transmission rates and comment on whether the transmission rate is shared or dedicated.Answer: Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to 1 Mbps, bandwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstream channel is usually less than a few Mbps, bandwidth is shared.6.(Q13) Why is it said that packet switching employs statistical multiplexing? Contrast statistical multiplexing with the multiplexing that takes place in TDM.Answer: In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.7.(Q14) Suppose users share a 2Mbps link. Also suppose each user requires 1Mbps when transmitting, but each user transmits only 20 percent of the time. (See the discussion of statistical multiplexing in Section 1.3.)a.When circuit switching is used, how many users can be supported?b.For the remainder of this problem, suppose packet switching is used. Why will there be essentially no queuing delay before the link if two or fewer users transmit at the same time? Why will there be a queuing delay if three users transmit at the same time?c.Find the probability that a given user is transmitting.d.Suppose now there are three users. Find the probability that at any given time, all three users are transmitting simultaneously. Find the fraction oftime during which the queue grows.Answer:a.2 users can be supported because each user requires half of the link bandwidth.b.Since each user requires 1Mbps when transmitting, if two or fewer users transmit simultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three users transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this case, there will be queuing delay before the link.c.Probability that a given user is transmitting = 0.2d.Probability that all three users are transmitting simultaneously=(3)p3(1−p)0=0.23=0.008. Since the 3queue 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 0.008.8.(Q16) Consider sending a packet froma source host to a destination host over a fixed route. List the delay components in the end-to-end delay. Which of these delays are constant and which are variable?Answer: 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.9.(Q19) Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has three links, of rates R1= 250 kbps, R2 = 500 kbps, and R3= 1 Mbps.a.Assuming no other traffic in the network, what is the throughput for the file transfer.b.Suppose the file is 2 million bytes. Roughly, how long will it take to transfer the file to Host B?c.Repeat (a) and (b), but now with R2 reduced to 200 kbps.Answer:a.250 kbpsb.64 secondsc.200 kbps; 80 seconds10.(P2) Consider the circuit-switched network in Figure 1.8. Recall that there are n circuits on each link.a.W hat is the maximum number ofsimultaneous connections that can be in progress at any one time in this network?b.S uppose that all connections arebetween the switch in the upper-left-hand corner and the switch in the lower-right-hand corner. What is the maximum number of simultaneous connections that can be in progress?Answer:a.We can n connections between eachof the four pairs of adjacent switches.This gives a maximum of 4n connections.b.We can n connections passingthrough the switch in the upper-right-hand corner and another n connections passing through the switch in the lower-left-hand corner, giving a total of 2n connections.11.(P4) Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 50 km/hour.a.S uppose the caravan travels 150 km,beginning in front of one tollbooth, passing through a second tollbooth, and finishing just before a third tollbooth.What is the end-to-end delay?b.R epeat (a), now assuming that thereare five cars in the caravan instead of ten.Answer: Tollbooths are 150 km apart, and the cars propagate at 50 km/hr, A tollbooth services a car at a rate of one car every 12 seconds.a.There are ten cars. It takes 120seconds, or two minutes, for the first tollbooth to service the 10 cars. Each of these cars has a propagation delay of 180 minutes before arriving at the second tollbooth. Thus, all the cars arelined up before the second tollbooth after 182 minutes. The whole process repeats itself for traveling between the second and third tollbooths. Thus the total delay is 364 minutes.b.Delay between tollbooths is 5*12seconds plus 180 minutes, i.e., 181minutes. The total delay is twice this amount, i.e., 362 minutes.12.(P5) This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B.a.E xpress the propagation delay, d prop ,in terms of m and s.b.D etermine the transmission time ofthe packet, d trans , in terms of L and R.c.I gnoring processing and queuingdelays, obtain an expression for the end-to-end delay.d.S uppose Host A begins to transmitthe packet at time t = 0. At time t =d trans , where is the last bit of the packet?e.S uppose d prop is greater than d trans . Attime t = d trans , where is the first bit of the packet?f.S uppose d prop is less than d trans . At timet = d trans , where is the first bit of the packet?g.S uppose s = 2.5*108, L = 100bits, and R= 28kbps. Find the distance m so thatd prop equals d trans .Answer:a.d prop = m/s seconds.b.d trans = L/R seconds.c.d end-to-end = (m/s + L/R) seconds.d.T he bit is just leaving Host A.e.T he first bit is in the link and has notreached Host B.f.T he first bit has reached Host B.g.W antm=LS=1003(2.5∗108) =893 km.13.(P6) In this problem we consider sending real-time voice from Host A to Host B over a packet-switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-Byte packets. There is one link between Host A and B; its transmission rate is 500 kbps and its propagation delay is 2 msec. As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receivesan entire packet, it converts the packet’s bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B)?Answer: Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires56∗8sec=7 msec3The time required to transmit the packet is56∗8sec=896 μsec3Propagation delay = 2 msec.The delay until decoding is7msec + 896μsec + 2msec = 9.896 msecA similar analysis shows that all bits experience a delay of 9.896 msec.14.(P9) Consider a packet of length Lwhich begins at end system A, travels overone link to a packet switch, and travelsfrom the packet switch over a second linkto a destination end system. Let d i, s i, and Rdenote the length, propagation speed, iand the transmission rate of link i, for i = 1, 2. The packet switch delays each packet by d proc. Assuming no queuing delays, in terms of d i, s i, R i, (i = 1, 2), and L, what is the total end-to-end delay for the packet? Suppose now the packet Length is 1,000 bytes, the propagation speed on both links is 2.5 * 108 m/s, the transmission rates of both links is 1 Mbps, the packet switch processing delay is 2 msec, the length of the first link is 6,000 km, and the length of the last link is 3,000 km. For these values, what is the end-to-end delay?Answer: The first end system requires L/R1to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay of d proc; after receiving the entire packet, the packet switch requires L/R2to transmit the packet onto the second link; the packet propagates over the second link in d2/s2. Adding these five delays givesdend-end = L/R1+ L/R2+ d1/s1+ d2/s2+ dprocTo answer the second question, we simply plug the values into the equation to get 8 + 8 + 24 + 12 + 2 = 54 msec.15.(P10) In the above problem, supposeR 1= R2= R and dproc= 0. Further suppose thepacket switch does not store-and-forward packets but instead immediately transmits each bit it receivers before waiting for the packet to arrive. What is the end-to-end delay?Answer: 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/s2For the values in Problem 9, we get 8 + 24 + 12 = 44 msec.16.(P11) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queued. Each packet is of length L and the link has transmission rate R. What is the average queuing delay for the N packets?Answer: 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)) = LN(N-1)/(2RN) = (N-1)L/(2R)Note that here we used the well-known fact that1 +2 + ....... + N = N(N+1)/217.(P14) Consider the queuing delay ina router buffer. Let I denote traffic intensity; that is, I = La/R. Suppose that the queuing delay takes the form IL/R (1-I) for I<1.a.Provide a formula for the total delay, that is, the queuing delay plus the transmission delay.b.Plot the total delay as a function of L/R.Answer:a.The transmission delay is L / R .The total delay isIL+ L=L/Rb.Let x = L / R.Total delay=x18.(P16) Perform a Traceroute between source and destination on the same continent at three different hours of the day.a.Find the average and standard deviation of the round-trip delays at each of the three hours.b.Find the number of routers in the path at each of the three hours. Did the paths change during any of the hours?c.Try to identify the number of ISP networks that the Traceroute packetspass through from source to destination. Routers with similar names and/or similar IP addresses should be considered as part of the same ISP. In your experiments, do the largest delays occur at the peering interfaces between adjacent ISPs?d.Repeat the above for a source and destination on different continents. Compare the intra-continent and inter-continent results.Answer: Experiments.19.(P18) Suppose two hosts, A and B, are separated by 10,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5•108 meters/sec.a.Calculate the bandwidth-delay product, R •d prop.b.Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximum number of bits that will be in the link at any given time?c.Provide an interpretation of the bandwidth-delay product.d.What is the width (in meters) of a bit in the link? Is it longer than a football field?e.Derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate R, and the length of the link m.Answer:a.d prop = 107 / 2.5•108 = 0.04 sec; so R •d prop = 80,000bitsb.80,000bitsc.The bandwidth-delay product of alink is the maximum number of bits thatcan be in the link.d.1 bit is 125 meters long, which islonger than a football field) = m / (R * m / s) =e.m / (R • dprops/R20.(P20) Consider problem P18 but now with a link of R = 1 Gbps.a.C alculate the bandwidth-delayproduct, R·d prop .b.C onsider sending a file of 400,000bits from Host A to Host B. Suppose the file is sent continuously as one big message. What is the maximum number of bits that will be in the link at any given time?c.W hat is the width (in meters) of abit in the link?Answer:a.40,000,000 bits.b.400,000 bits.c.0.25 meters.21.(P21) Refer again to problem P18.a.H ow long does it take to send the file,assuming it is sent continuously?b.S uppose now the file is broken up into10 packet is acknowledged by thereceiver and the transmission time of an acknowledgment packet is negligible.Finally, assume that the sender cannot send a packet until the preceding one is acknowledged. How long does it take to send the file?c.C ompare the results from (a) and (b).Answer:a.d trans + d prop = 200 msec + 40 msec = 240msecb.10 * (t trans + 2 t prop ) = 10 * (20 msec +80 msec) = 1.0 sec。
