派斯第一章(导论)练习题

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高等数学第一章课后习题答案

高等数学（本）第一章函数与极限1. 设 ⎪⎩⎪⎨⎧≥<=3||,03|||,sin |)(ππϕx x x x , 求).2(446ϕπϕπϕπϕ、、、⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛6sin )6(ππϕ=21=224sin )4(==ππϕ ()0222)4sin()4(==-=-ϕππϕ2. 设()x f 的定义域为[]1,0，问：⑴()2x f ； ⑵()x f sin ； ⑶()()0>+a a x f ； ⑷()()a x f a x f -++ ()0>a 的定义域是什么？（1）][；，－的定义域为所以知－11)(,111022x f x x ≤≤≤≤[]ππππ)12(,2)(sin ),()12(21sin 0)2(+∈+≤≤≤≤k k x f Z k k x k x 的定义域为所以知由][a a a x f ax a a x -+-≤≤≤+≤1,)(110)3(－的定义域为所以知－由][φ时，定义域为当时，定义域为当从而得－知由211,210111010)4(>-≤<⎩⎨⎧+≤≤-≤≤⎩⎨⎧≤-≤≤+≤a a a a a x a ax a a x a x班级姓名学号3. 设()⎪⎩⎪⎨⎧>-=<=111011x x x x f ，()x e x g =，求()[]x g f 和()[]x f g ，并做出这两个函数的图形。

⎪⎪⎩⎪⎪⎨⎧>=<==⎪⎩⎪⎨⎧>-=<=⎪⎩⎪⎨⎧>-=<=-1,1,11,)]([.)20,10,00,1)]([1)(,11)(,01)(,1)]([.)11)(x e x x e e x f g x x x x g f x g x g x g x g f x f 从而得4. 设数列{}n x 有界, 又,0lim =∞→n n y 证明: .0lim =∞→n n n y x{}结论成立。

第一章测试题(参考答案)

一、填空题1．各种物理量可分为矢量和标量两大类，只有大小特征的量称为标量，既有大小又有方向特征的量称为矢量，因此，温度是标量，速度是矢量。

2．求标量场2xy z φ=在点P （2,1,1）处的最大变化率值及_____沿ˆˆˆ2l xyz =++的方向导数_______9/______。

3．如果矢量场所在的空间中，0A =⨯∇，则这种场中不可能存在____漩涡源____，因而称为____无旋场____；如果矢量场所在的空间中，0=⋅∇A，则这种场中不可能存在____通量源____，因而称为___管形场_____； 4．ˆˆˆr xxyy zz =++，r '∇⋅= 0 ，r ∇= ˆr。

5．求解曲面坐标系矢量运算：()ˆˆˆˆz A A zA ρφρρφρ∂++=∂ /A ρρ∂∂ 。

二、判断题1．空间中心点的散度为0，则该空间矢量场的通量为0（ × ）；三、选择题1. 假设某介质表面的法向为ˆˆnz =，位于介质表面上的某点的电场强度为ˆˆˆ345xy z =++E ，则它的切向电场强度为： A. ˆˆ34xy =+E B. ˆ5z=E C.ˆˆ34y x =+E D.ˆˆ34y x =-E三、计算题如图1所示，随机粗糙面的高度起伏满足(,)z f x y =，其中，x,y为0-1之间的随机数，请计算该粗糙面上一个面元中心为A （0.1,0.6,0.5）的面元的法向。

图1[解]：令(,)f x y z φ=-，则曲面(,)z f x y =是标量场0φ=的等值面。

因而其法向为：ˆˆˆf fxy z x yφ∂∂∇=+-∂∂ 在A 点处， 0000,,ˆˆˆ|A x y x y f f xyzxyφ∂∂∇=+-∂∂||A φ∇=000,ˆˆˆ||x y x A A ffxyx ynφφ∂∂+∂∂=∇∇=。

第一章产业经济学导论答案

第一章产业经济学导论工商管理精英班张曦文学号PB12203158习题一、名词解释1.产业：（Industry）是一系列相关企业的集合。

2.马歇尔冲突：在追求规模经济和由此引起的垄断扼杀竞争活力构成了一对难分难解的矛盾，这就是“马歇尔冲突”。

3.可竞争市场：（Contestable market）进入完全自由以及退出没有成本的市场。

4.产业实验室研究方法：利用计算机在实验室内观察现实市场中无法观察到的某些变量，然后通过控制部分变量来考察研究者最为关心的变量之间的因果关系。

5.一般均衡分析：将所有相互联系的各个市场看成一个整体加以研究。

一般均衡分析是关于整个经济体系的价格和产量结构的一种研究方法，是一种比较周到和全面的分析方法。

二、单项选择1.现代产业组织理论的奠基人是（C）A.亚当·斯密B.马歇尔C.爱德华·H·张伯伦D.乔安·罗宾逊夫人2.有效竞争理论由（A）提出A.克拉克B.爱德华·H·张伯伦C.乔安·罗宾逊夫人D.霍夫曼3.下列哪本书的出版，标志着哈佛学派正式形成（D）A.张伯伦的《垄断竞争理论》B.施蒂格勒的《产业组织》C.罗宾逊夫人的《不完全竞争经济学》D.贝恩的《产业组织》4.下列哪一个分析范畴不属于实证分析( C )A.“是什么”B.“为什么”C.“该怎么办”D.“怎么样”5.产业组织中新奥地利学派主要采用的分析方法是( B )A.边际分析方法B.行为主义分析方法C.经济模型法D.统计计量研究三、多项选择题1.产业经济学产生的物质基础有（AB）A.三次社会大分工B.四次产业革命C.张伯伦的垄断竞争理论D.封建社会的瓦解2.根据霍夫曼的产业分类法，把产业分为（ACD）A.消费资料产业B.生产生产资料的产业C.资本资料产业D.其他产业3.以下属于三次产业分类法中的第一产业的有（ACDE）A.农业B.采矿业C.畜牧业D.林业E.渔业4.经济学中均衡方法有( ABD )A.局部均衡B.纳什均衡C.动态均衡D.一般均衡5.下面研究方法属于实证分析的有( ABCD )A.案例研究方法B.跨部门交叉分析法C.统计计量分析法D.普劳特的实验方法四、改错题1.张伯伦认为，现实的市场既存在竞争因素，也存在垄断因素，介于完全竞争和纯粹垄断之间，形成了“垄断竞争”格局。

第一章 1 习题解答

第一章 1 习题解答习题1－1 (第7页)A 组1.由两点式写直线的方程为35x ＋36y －41＝0.2.直线x 6＋y 4＝－2与x 轴、y 轴的交点坐标以及直线的斜率分别为(－12，0)、(0，－8)、－23. 3.解 △ABC 是以∠A 为直角的直角三角形，且AB 平行于x 轴，AC 平行于y 轴.∴∠A 的平分线的斜率为1，所在直线方程为x －y ＋1＝0.BC 所在直线的方程为4x ＋3y －29＝0，解⎩⎨⎧x －y ＋1＝0，4x ＋3y －29＝0，得⎩⎪⎨⎪⎧x ＝267，y ＝337.∠A 的平分线的长为1227. 4.解法一由两点式写出直线AB 的方程为3x ＋y －6＝0.将点C (4，－6)代入方程3×4＋(－6)－6＝0，点C 在直线AB 上，∴A 、B 、C 在同一条直线上.法二 ∵k AB ＝－3，k BC ＝－3∴A 、B 、C 三点在同一条直线上.5.解与x 轴交点令y ＝0，2x －10＝0，x ＝5，与y 轴交点令x ＝0，－5y －10＝0，y ＝－2，S △＝12×5×2＝5. 6.证明如图：矩形OABC .设OA ＝a ，OC ＝b ，以O 为原点建立如图所示的直角坐标系.则O 、A 、B 、C 的坐标分别为(0，0)，(a ，0)，(a ，b )，(0，b )|OB |＝a 2＋b 2，|AC |＝b 2＋（－a ）2＝a 2＋b 2，设圆的方程为(x －a )2＋(y －b )2＝r 2则⎩⎨⎧（－4－a ）2＋b 2＝r 2，（4－a ）2＋b 2＝r 2，a 2＋（5－b ）2＝r 2，得a ＝0，b ＝910，r 2＝412100，∴圆方程为x 2＋⎝ ⎛⎭⎪⎫y －9102＝1 681100.9.解 |A 1F 1|＋|A 2F 1|＝2＋14＝16＝2a ，a ＝8，F 1(－6，0)，F 2(6，0)，c ＝6，∴b 2＝28.∴椭圆标准方程为x 264＋y 228＝1.10.解 (1)由题意知a 2＝8，b 2＝5，椭圆方程为x 28＋y 25＝1.(2)由题意知a ＝3b当焦点在x 轴上时a ＝3，b ＝1，椭圆方程：x 29＋y 21＝1；当焦点在y 轴上时b ＝3，a ＝9，椭圆方程：x 29＋y 281＝1.(3)由题意知c ＝23，设椭圆方程为x 2a 2＋y 2b 2＝1，P (5，－6)在椭圆上.∴⎩⎪⎨⎪⎧5a 2＋6b 2＝1，a 2－b 2＝12，解得a 2＝20，b 2＝8，∴椭圆方程为x 220＋y 28＝1.11.略B 组1.证明 ∵圆直径的端点是A (x 1，y 1)，B (x 2，y 2)∴圆心坐标为⎝ ⎛⎭⎪⎫x 1＋x 22，y 1＋y 22，半径为（x 1－x 2）2＋（y 1－y 2）22∴圆的方程为⎝ ⎛⎭⎪⎫x －x 1＋x 222＋⎝⎛⎭⎪⎫y －y 1＋y 222 ＝（x 1－x 2）2＋（y 1－y 2）24， x 2－x (x 1＋x 2)＋（x 1＋x 2）24＋y 2－y (y 1＋y 2)＋（y 1＋y 2）42＝（x 1－x 2）2＋（y 1－y 2）24， x 2－x (x 1＋x 2)＋（x 1＋x 2）24－（x 1－x 2）24＋y 2－y (y 1＋y 2)＋（y 1＋y 2）24－（y 1－y 2）24＝0， x 2－x (x 1＋x 2)＋x 1x 2＋y 2－y (y 1＋y 2)＋y 1y 2＝0，(x －x 1)(x －x 2)＋(y －y 1)(y －y 2)＝0，∴圆的方程为(x －x 1)(x －x 2)＋(y －y 1)(y －y 2)＝0.2.解由⎩⎨⎧（x －3）2＋（y －5）2＝4，⎝ ⎛⎭⎪⎫x －322＋（y －5）2＝1得x －54＝0， ∴直线方程为x －54＝0. 3.解以地球球心与距地最近点所在直线为x 轴，以最近点与最远点的中点为原点建立平面直角坐标系.则2a ＝6 636＋8 196＝14 832，a ＝7 416，a 2＝54 997 056，c ＝8 196－7 416＝780，∴b 2＝54 388 656.∴椭圆方程为x 254 997 056＋y 254 388 656＝1.。

(完整版)第一章练习题及答案

第一章一、单项选择题(每小题1分) 1．一维势箱解的量子化由来( d )a. 人为假定b. 求解微分方程的结果c. 由势能函数决定的d. 由微分方程的边界条件决定的。

2．指出下列哪个是合格的波函数(粒子的运动空间为 0→+∞)( b )a. sinxb. e -xc. 1/(x-1)d. f(x) = e x( 0≤ x ≤ 1); f(x) = 1 ( x > 1)3.首先提出微观粒子的运动满足测不准原理的科学家是( c. ) a.薛定谔 b. 狄拉克 c. 海森堡 c.波恩4．立方势箱中22810ma h E <时有多少种状态( c )a. 11b. 3c. 7d. 25.立方势箱在22812ma h E ≤的能量范围内，能级数和状态数为( c)a.5，20b. 6，6c. 5，11d. 6，176．立方势箱中2287ma h E <时有多少种状态( c )a. 11b. 3c. 4d. 27．立方势箱中2289ma h E <时有多少种状态( c )a. 11b. 3c. 4d. 28.已知xe 2是算符x Pˆ的本征函数，相应的本征值为( d ) a.i h 2b.ih 4 c. 4ih d.πi h9．已知2e 2x 是算符xi ∂∂-的本征函数，相应的本征值为( d ) a. -2 b. -4i c. -4ih d. -ih/π 10.下列条件不是品优函数必备条件的是( c ) a. 连续 b. 单值 c. 归一 d. 有限或平方可积11．一维谐振子的势能表达式为221kx V =，则该体系的定态Schrodinger 方程中的哈密顿算符为( d ) a.221kx b.222212kx m +∇ c.222212kx m -∇-d.222212kx m +∇- 二、多项选择题(每小题2分)1. 下列哪些条件并非品优波函数的必备条件( a c )a. 归一化b. 连续c.正交性d. 单值e. 平方可积三、填空题(每小题1分)1.德布罗意关系式为___________。

第一章l.p及单纯形法练习题答案

第一章 L.P 及单纯形法练习题答案一、判断下列说法是否正确1. 线性规划模型中增加一个约束条件，可行域的范围一般将缩小，减少一个约束条件，可行域的范围一般将扩大。

(✓)2. 线性规划问题的每一个基解对应可行域的一个顶点。

(✗)3. 如线性规划问题存在某个最优解，则该最优解一定对应可行域边界上的一个点。

(✓)4. 单纯形法计算中，如不按最小比值原则选取换出变量，则在下一个基可行解中至少有一个基变量的值为负。

(✓)5. 一旦一个人工变量在迭代中变为非基变量后，该变量及相应列的数字可以从单纯形表中删除，而不影响计算结果。

(✓)6. 若1X 、2X 分别是某一线性规划问题的最优解，则1212X X X λλ=+也是该线性规划问题的最优解，其中1λ、2λ为正的实数。

(✗)7. 线性规划用两阶段法求解时，第一阶段的目标函数通常写为ai iMinZ x =∑(x ai 为人工变量)，但也可写为i ai iMinZ k x =∑，只要所有k i 均为大于零的常数。

(✓)8. 对一个有n 个变量、m 个约束的标准型的线性规划问题，其可行域的顶点恰好为m n C 个。

(✗)9. 线性规划问题的可行解如为最优解，则该可行解一定是基可行解。

(✗)10. 若线性规划问题具有可行解，且其可行域有界，则该线性规划问题最多具有有限个数的最优解。

(✗)二、求得L.P 问题=+++=⎧⎪++=⎪⎨++=⎪⎪≥=⎩1122111122312112224231132253j MaxZ c x c x a x a x x b a x a x x b a x a x x b x 0;j 1,2,,5 的解如下： X ⑴=(0,3,2,16,0)T ;X ⑵=(4,3,-2,0,0)T ;X ⑶=(3.5,2,0.5,2,4)T ;X ⑷=(8,0,0,-16,12)T ; =(4.5,2,-0.5,-2,4)T ; X ⑹=(3,2,1,4,4)T ; X ⑺=(4,2,0,0,4)T 。

运筹学习题答案(第一章)

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School of Management
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运筹学教程
第一章习题解答
当c/d在3/10到5/2之间时最优解为图中的A点；当 c/d大于5/2且c大于等于0时最优解为图中的B点；当c/d 小于3/10且d大于0时最优解为图中的C点；当c/d大于 5/2且c小于等于0时或当c/d小于3/10且d小于0时最优解为图中的原点。
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page 9 15 June 2013
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School of Management
运筹学教程
第一章习题解答
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page 19 15 June 2013
School of Management
运筹学教程
第一章习题解答
max Z 4 x 1 x 2 3 x1 x 2 3 4 x1 3 x 2 x 3 6 st x 2 x2 x4 4 1 x j 0（ j 1, , 4） ,

高数第一章练习题

B. f ( x ) 在 x0 点的极限值不一定存在 D. f ( x ) 在点 x0 必连续）
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1.-第一章课后习题及标准答案

第一章1.(Q1) What is the difference between a host and an end system? List the types of endsystems. 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 adiplomatic 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 suggests a 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 requestand 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 communication is the client.Typically, the client program requests and receives services from the server program.4.(Q4) List six access technologies. Classify each one as residential access, company access, ormobile access.Answer:1. Dial-up modem over telephone line: residential; 2. DSL over telephone line: residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, 3G/4G): mobile5.(Q5) List the available residential access technologies in your city. For each type of access,provide the advertised downstream rate, upstream rate, and monthly price.Answer: 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.6.(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.7.(Q8) Dial-up modems, HFC, and DSL are all used for residential access. For each of theseaccess 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 to1 Mbps, bandwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstreamchannel is usually less than a few Mbps, bandwidth is shared.8.(Q13) Why is it said that packet switching employs statistical multiplexing? Contraststatistical 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.9.(Q14) Suppose users share a 2Mbps link. Also suppose each user requires 1Mbps whentransmitting, 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 beessentially 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, allthree users are transmitting simultaneously. Find the fraction of time 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 transmitsimultaneously, 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=(33)p3(1−p)0=0.23=0.008. Since the queue grows when all the users are transmitting, the fraction oftime during which the queue grows (which is equal to the probability that all three users are transmitting simultaneously) is 0.008.10.(Q16) Consider sending a packet from a 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.11.(Q19) Suppose Host A wants to send a large file to Host B. The path from Host A to Host Bhas 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 toHost B?c.Repeat (a) and (b), but now with R2 reduced to 200 kbps.Answer:a.250 kbpsb.64 secondsc.200 kbps; 80 seconds12.(P2) Consider the circuit-switched network in Figure 1.8. Recall that there are n circuits oneach link.a.What is the maximum number of simultaneous connections that can be in progress atany one time in this network?b.Suppose that all connections are between the switch in the upper-left-hand cornerand the switch in the lower-right-hand corner. What is the maximum number ofsimultaneous connections that can be in progress?Answer:a.We can n connections between each of the four pairs of adjacent switches. This gives amaximum of 4n connections.b.We can n connections passing through the switch in the upper-right-hand corner andanother n connections passing through the switch in the lower-left-hand corner, giving a total of 2n connections.13.(P4) Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 50km/hour.a.Suppose the caravan travels 150 km, beginning in front of one tollbooth, passingthrough a second tollbooth, and finishing just before a third tollbooth. What is theend-to-end delay?b.Repeat (a), now assuming that there are 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 120 seconds, or two minutes, for the first tollbooth to servicethe 10 cars. Each of these cars has a propagation delay of 180 minutes before arriving at the second tollbooth. Thus, all the cars are lined 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*12 seconds plus 180 minutes, i.e., 181minutes. The totaldelay is twice this amount, i.e., 362 minutes.14.(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.Express the propagation delay, d prop , in terms of m and s.b.Determine the transmission time of the packet, d trans , in terms of L and R.c.Ignoring processing and queuing delays, obtain an expression for the end-to-enddelay.d.Suppose Host A begins to transmit the packet at time t = 0. At time t = d trans , whereis the last bit of the packet?e.Suppose d prop is greater than d trans . At time t = d trans , where is the first bit of thepacket?f.Suppose d prop is less than d trans . At time t = d trans , where is the first bit of thepacket?g.Suppose s = 2.5*108, L = 100bits, and R = 28kbps. Find the distance m so that d propequals 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.The bit is just leaving Host A.e.The first bit is in the link and has not reached Host B.f.The first bit has reached Host B.g.Wantm=LRS=10028∗103(2.5∗108)=893 km.15.(P6) In this problem we consider sending real-time voice from Host A to Host B over apacket-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 linkbetween 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 receives an 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 msec64∗103The time required to transmit the packet is56∗8sec=896 μsec500∗103Propagation 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.16.(P9) Consider a packet of length L which begins at end system A, travels over one link to apacket switch, and travels from the packet switch over a second link to a destination end system. Let d i, s i, and R i denote the length, propagation speed, and 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 * 108m/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 ofd proc; after receiving the entire packet, the packet switch requires L/R2to transmit the packetonto the second link; the packet propagates over the second link in d2/s2. Adding these five delays givesd end-end = L/R1 + L/R2 + d1/s1 + d2/s2 + d procTo answer the second question, we simply plug the values into the equation to get 8 + 8 +24 + 12 + 2 = 54 msec.17.(P10) In the above problem, suppose R1 = R2 = R and d proc= 0. Further suppose the packetswitch 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,d end-end = L/R + d1/s1 + d2/s2For the values in Problem 9, we get 8 + 24 + 12 = 44 msec.18.(P11) Suppose N packets arrive simultaneously to a link at which no packets are currentlybeing 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)/219.(P14) Consider the queuing delay in a 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 transmissiondelay.b.Plot the total delay as a function of L/R.Answer:a.The transmission delay is L / R . The total delay isILR(1−I)+LR=L/R1−Ib.Let x = L / R.Total delay=x 1−αx20.(P16) Perform a Traceroute between source and destination on the same continent at threedifferent hours of the day.a.Find the average and standard deviation of the round-trip delays at each of the threehours.b.Find the number of routers in the path at each of the three hours. Did the pathschange during any of the hours?c.Try to identify the number of ISP networks that the Traceroute packets pass throughfrom 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 theintra-continent and inter-continent results.Answer: Experiments.21.(P18) Suppose two hosts, A and B, are separated by 10,000 kilometers and are connectedby 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 sentcontinuously 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 a link is the maximum number of bits that can be in thelink.d. 1 bit is 125 meters long, which is longer than a football fielde.m / (R •d prop ) = m / (R * m / s) = s/R22.(P20) Consider problem P18 but now with a link of R = 1 Gbps.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 sentcontinuously as one big message. What is the maximum number of bits that will be inthe link at any given time?c.What is the width (in meters) of a bit in the link?Answer:a.40,000,000 bits.b.400,000 bits.c.0.25 meters.23.(P21) Refer again to problem P18.a.How long does it take to send the file, assuming it is sent continuously?b.Suppose now the file is broken up into 10 packet is acknowledged by the receiver andthe transmission time of an acknowledgment packet is negligible. Finally, assumethat the sender cannot send a packet until the preceding one is acknowledged. Howlong does it take to send the file?pare the results from (a) and (b).Answer:a. d trans + d prop = 200 msec + 40 msec = 240 msecb.10 * (t trans + 2 t prop ) = 10 * (20 msec + 80 msec) = 1.0 sec。

第一章参考答案

(2) 由 13 题可知该分块矩阵可逆. 根据 13 题证明中的下述等式 ) ) ( )( ( A B A B I 0 , = 0 D − CA−1 B C D −CA−1 I 再由本题 (1) 中的结论可知 (以下记 G = D − CA−1 B ) ( ( = A B C D A−1 0 ) ) )−1 ( )−1 ( ( I 0 A B A B I 0 −1 ) = −CA−1 I 0 G 0 G −CA−1 I ) ( −1 ) )( A + A−1 BG−1 CA−1 −A−1 BG−1 −A−1 BG−1 I 0 = . −G−1 CA−1 G−1 G−1 −CA−1 I )−1 ( =(
故由上题 (2) 可知，X −1 = A−1 + A−1 BG−1 CA−1 ，其中 G = I − CA−1 B (由本题证明过程可知 G 是可逆矩阵). ) ( 0 In 为 2n 阶分块矩阵. 一个 2n 阶复矩阵 M 称为是辛辛矩阵如 16. 设 Ω = −In 0 果 M T ΩM = Ω. 证明: (1) 2n 阶辛矩阵的全体构成一个群, 即辛矩阵的逆矩阵仍是辛矩阵, 两个辛矩阵的乘积仍是辛矩阵; (2) 任何辛矩阵的行列式均为 1. (提示: 利用分块矩阵.) 证明：(1) 显然. (2) 由 M T ΩM = Ω 可知 |M |2 = 1, 从而 |M | = ±1. 但证明 |M | = 1 较难，？？？？ 17. 求下列各矩阵的满秩分解: 1 2 3 0 (1) A = 0 2 1 −1 ; 1 0 2 1 答案： ) ( 1 0 1 2 3 0 ; (1) A = 0 −1 0 −2 −1 1 1 1 1 −1 ( ) −1 1 1 0 0 0 (2) A = −1 −1 0 1 −1 −1 ; . 1 1 1 −1 1 1 −1 1 −1 −1 . (2) A = −1 −1 1 1 1 1 −1 −1

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派斯第一章（导论）练习题
一、判断题
1.社会经济统计的研究对象是社会经济现象总体的各个方面。（× ）

2.在全国工业普查中，全国企业数是统计总体，每个工业企业是总体单位。（×）
3.总体单位是标志的承担者，标志是依附于单位的。（√ ）
4.答案是文字的变量是品质变量。（ √ ）
5.参数估计、回归分析都属于描述统计学的内容。（×）
6. 有统计学之名、无统计学之实的学派是国势学派。（ √ ）
7.总体和总体单位的确定，要视研究目的和研究范围而定。（ √ ）
8.三个同学的成绩不同，因此存在三个变量。（ × ）
9.学校数和建筑面积都属于离散变量。（× ）
二、单项选择题
1.以下属于推断统计学的研究范围的是（ D ）。
A.数据调查与收集 B.数据的计算 C.数据汇总 D.相关与回归分析
2.调查某企业职工的健康状况，总体单位是（ C ）。
A.这个企业 B.所有的职工 C.每个职工 D.每个职工的健康状况
3.威廉·配第是（ B ）的代表人物。
A.记述学派 B.政治算术学派 C.社会学派 D.数理统计学派。
4. 数量指标表现为（ C ）。
A.相对数 B.平均数 C.绝对数 D.变异数
5.2个大学生的身高分别为165厘米、172厘米，则165、172是（ D ）。
A.2个变量 B.2个标志 C.2个指标 D.2个数据
6.总体与总体单位的确定（ A ）。
A.与研究目的有关 B.与研究目的无关
C.与总体范围大小有关 D.与研究方法有关
7.通过有限数量的种子发芽试验结果来估计整批种子的发芽率，这种统计方法属
于（ A ）。
A.推断统计学 B.描述统计学 C.数学 D.逻辑学
8.统计学是一门关于研究客观事物数量方面和数量关系的（ C）
A.社会科学 B.自然科学 C.方法论科学 D.实质性科学
2

9.通过测量或计算取得其值的是（ B ）。
A.品质变量 B.连续变量 C.离散变量 D.以上都不对
10.统计学的创始人是（ A ）。
A.威廉.配第 B.阿亨瓦尔 C.凯特勒 D.恩格尔
11.指标是说明总体特征的，标志是说明总体单位特征的，所以（B ）
A.标志和指标之间的关系是固定不变的
B.标志和指标之间的关系是可以变化的
C.标志和指标都是用数值表示的
D.只有指标才可以用数值表示
12.一个统计总体（ D ）。
A.只能有一个标志 B.只能有一个指标
C.可以有多个标志 D.可以有多个指标
13.在重庆地区经济普查中，总体单位是（D ）
A.重庆地区所有经济单位 B.重庆地区每个区县
C.重庆地区每个经济单位负责人 D.重庆地区每个经济单位
三、多项选择题
1.下面属于推断统计学研究内容的是（BCD ）。

A.数据收集 B.抽样调查 C.相关分析
D.假设检验 E.指数
2.下列指标不属于绝对数的有（ABD）。
A.合格率 B.价格 C.产量
D.出勤率 E.星球个数
3. “统计”一词有三层含义（ BCD ）。
A.统计设计 B.统计工作 C.统计资料
D.统计科学 E.统计图表
4.以下关于描述统计学和推断统计学的叙述中，正确的有（ ACD ）。
A.描述统计学是现代统计学的基础 B.推断统计学是现代统计学的基础
C.推断统计学是现代统计学的核心 D.两者相辅相成缺一不可
E.两者各自独立，毫不相干
3

5.下列标志属于数量标志的有（ACD ）。
A.GDP B.经济类型 C.考试分数 D.在校学生数 E.颜色
6.调查某学院的教师情况，则（ BD）。
A.教师总数是总体 B.某教师是湖北人，则“湖北”是定类数据
C.男教师比重是数量指标 D.每一个教师是总体单位
E.教师平均年龄是数量标志，也是数值变量
7. 下列指标属于质量指标的有（ ABD ）。
A.合格率 B.价格 C.产量
D.出勤率 E.星球个数
8.以下属于连续变量的有(BCE )。
A.职称 B.价格 C.温度 D.人数 E.土地面积
9.以下属于离散变量的有( AE)。
A.毕业生人数 B.性别 C.利润率 D.身高 E.手机部数
10.要了解某地区全总成年人口的就业情况，那么（ABE ）。
A.全部成年人是总体 B.成年人口总数是统计指标
C.成年人口就业率是标志 D．“职业”是数量标志
E.某人职业是“教师”，这里的“教师”是数据