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Assignment #1

Assignment #1
Assignment #1

Solutions to Homework Assignment #4

Due: October 30, 2007

1.What are the two most important network-layer functions in a datagram network?

What are the three most important functions in a virtual-circuit network?

2.What is the difference between routing and forwarding?

3.Suppose you purchase a wireless router and connect it to your cable modem. Also

suppose that your ISP dynamically assigns your connected device (that is, your

wireless router) one IP address. Also suppose that you have five PCs at home that use 802.11 to wirelessly connect to your wireless router. How are IP addresses

assigned to the five PCs? Does the wireless router use NAT? Why or why not?

4.Consider a datagram network using 8-bit host addresses. Suppose a router uses

longest prefix matching and has the following forwarding table:

For each of the four interfaces, give the associated range of destination host addresses and the number of addresses in the range.

5.Consider a router that interconnects three subnets: Subnet 1, Subnet 2, and Subnet

3. Suppose all of the interfaces in each of these three subnets are required to have

the prefix 223.1.17/24. Also suppose that Subnet 1 is required to support up to 125 interfaces, and Subnet 2 and 3 are each required to support up to 60 interfaces.

Provide three network addresses (of the form a.b.c.d/x) that satisfy these

constraints.

6.Consider a subnet with prefix 101.101.101.64/26. Give an example of one IP

address (of form xxx.xxx.xxx.xxx) that can be assigned to this network. Suppose an ISP owns the block of addresses of the form 101.101.128/17. Suppose it wants to create four subnets from this block, with each block having the same number of IP addresses. What are the prefixes (of form a.b.c.d/x) for the four subnets?

7.Consider the following network. With the indicated link costs, use Dijkstra’s

shortest-path algorithm to compute the shortest path from x to all network nodes.

Show how the algorithm works by computing a table similar to the one discussed in class (or Table 4.3 in the text).

8.Consider the network shown below, and assume that each node initially knows

the costs to each of its neighbors. Consider the distance-vector algorithm an show the distance table entries at node z.

9.Consider the network shown below. Suppose AS3 and AS2 are running OSPF for

their intra-AS routing protocol. Suppose AS1 and AS4 are running RIP for their intra-AS routing protocol. Suppose eBGP and iBGP are used for the inter-AS

routing protocol. Initially suppose there is no physical link between AS2 and AS4.

(a)Router 3c learns about prefix X from which routing protocol?

(b)Router 3a learns about X from which routing protocol?

(c)Router 1c learns about X from which routing protocol?

(d)Router 1d learns about X from which routing protocol?

chapter3-assignment

第三章 线性分组码 习题 1.证明[n ,k ]线性分组码的最大距离为n -k +1。 2.设一个[7,4]码的生成矩阵为 1000111010010100100110 00111 0G ????? ?=?????? (1)求出该码的全部码矢; (2)求出该码的一致校验矩阵; (3)作出该码的标准阵译码表。 3.证明定理3.1.3。 4.一个[8,4]系统码,它的一致校验方程为: c 0=m 1+m 2+m 3c 1=m 0+m 1+m 2c 2=m 0+m 1+m 2c 3=m 0+m 2+m 3 式中,m 0,m 1,m 2,m 3是信息位,c 0,c 1,c 2,c 3是校验位。找出该码的G 和H ,并证明该码的最小距离为4。 5.构造第4题中码的对偶码。 6.设H 1是[n ,k ]线性分组码C 1的校验矩阵,且有奇数最小距离为d 。作一个新的码C 2,它的校 验矩阵为 12000111 1H H ?? ????? ?=???????? M L (1)证明C 2是一个(n +1,k )分组码; (2)证明C 2中每一码字的重量为偶数; (3)证明C 2码的最小重量为d +1。 7.设C 1是一个有最小距离为d 1的[n 1,k ]线性系统码,生成矩阵为G 1=[P 1I k ]。C 2是一个有最小距离为d 2的[n ,k ]线性系统码,它的生成矩阵G 2=[P 2I k ]。对满足下述一致校验矩阵 212T k n n k T p H I P I ???? ? ?=??????

的[n 1+n 2,k]线性码,证明它有最小距离至少为d 1+d 2。 8.设一个二进制[n ,k ]码C 的G 矩阵不含全零列,将C 的所有码字排成2k ×n 的阵。 证明: (a )阵中不含有全零列; (b )阵中的每一列由2k -1个零和2k -1个1组成; (c )在一特定分量上为0的所有码字构成C 的一个子空间,问该子空间的维数是多少? 9.令是所有二进制[n ,k ]线性系统码的集合。证明非零二进制n 重V 或者恰巧含于的 ΓΓ(1)()2k n k ??个码中,或者不在的任一码中。 Γ10.证明二进制[23,12,7]Golay 码和三进制的[11,6,5]Golay 码是完备码。 11.若d 是码C 的最小重量,且为偶数,(1)/2t d =????? 。证明有两个重量均为t +1的矢量必在C 码的同一陪集中。 12.求出d =3,至多只有3个校验元的二进制码的码长n ;和d =5,至多只有8个校验元的二进 制码的码长n 。 13.计算二进制[24,12,8]扩张Golay 码的覆盖半径,及[8,4]RM 码的覆盖半径。 14.证明定理3.9.3。 15.构造三个二进制的[10,3,5]LUEP 码,其分离矢量分别为(8,2,2),(7,4,4), (6,4,4)。写出它们标准形式的G 和系统码形式的G 。 16.证明定理3.10.3。 17.构造一个具有最高码率的k =10,t =2的2-EC/AUED 码。 18.证明定理3.10.6。

财务分析

哈佛分析框架:财务战略分析新思维 2015-09-15中国管理会计网 管理会计微信号:china-cma 本文运用哈佛分析框架的基本原理,结合亚泰集团企业案例,通过分析其揭示其财务及经营状况并预测其发展前景。 财务分析一直以来在评价企业业绩过程中起到重要且不可替代作用,它通过对报表数据的各种变换计算,运用模型建立起一系列评价体系,从各个方面、各个角度分析企业经营成果并预测未来。利益相关者主要通过企业发布的财务报表来评价企业。因此,财务报表成为传统财务分析的根本。然而作为经济实体的主体——企业,并不是孤立存在的,它不仅受自身的经营及财务政策的影响,还受其所处的相关行业及宏观经济环境的影响。 有效地财务评价体系不仅应注重对企业财务数据的分析,而且应重视非财务信息。应将财务分析对象由财务报表扩展到与财务报表及企业经营相关的行业环境、政策环境、宏观经济环境等,站在战略的高度对企业进行评价。哈佛分析框架在一定程度上克服了传统财务分析的缺陷,鉴于此,本文利用哈佛分析框架对亚泰集团的行业状况和发展战略、会计质量、财务报表及发展前景进行了定性和定量的综合分析,这种分析方法有效地克服了传统财务报表分析的局限性,能够从整体上把握企业集团的整体经营状况,从而预测企业未来发展前景。 哈佛分析框架 哈佛分析框架由哈佛大学佩普(K.G.Palepu)、希利(P.M.Healy)和伯纳德(V.L.Bernard)三位学者提出,他们认为财务分析不应只分析报表数据,应该站在战略的高度,结合企业内外部环境并在科学预测的基础上为企业未来发展指明方向,哈佛分析框架主要包括企业战略分析、会计分析、财务分析及前景分析。

solutions for assignment3

Chapter 3 Assignments P13.Consider a reliable data transfer protocol that uses only negative acknowledgments. Suppose the sender sends data infrequently. Would a NAK-only protocol be preferable to a protocol that uses ACKs? Why? Now suppose the sender has a lot of data to send and the end-to-end connection experiences few losses. In this second case, would a NAK-only protocol be preferable to a protocol that uses ACKs? Why? Answer: In a NAK only protocol, the loss of packet x is only detected by the receiver when packet x+1 is received. That is, the receivers receives x-1 and then x+1, only when x+1 is received does the receiver realize that x was missed. If there is a long delay between the transmission of x and the transmission of x+1, then it will be a long time until x can be recovered, under a NAK only protocol. On the other hand, if data is being sent often, then recovery under a NAK-only scheme could happen quickly. Moreover, if errors are infrequent, then NAKs are only occasionally sent (when needed), and ACK are never sent – a significant reduction in feedback in the NAK-only case over the ACK-only case. P23. We have said that an application may choose UDP for a transport protocol because UDP offers finer application control (than TCP) of what data is sent in a segment and when. a. Why does an application have more control of what data is sent in a segment? b. Why does an application have more control on when the segment is sent? Answer: a) Consider sending an application message over a transport protocol. With TCP, the application writes data to the connection send buffer and TCP will grab bytes without necessarily putting a single message in the TCP segment; TCP may put more or less than a singe message in a segment. UDP, on the other hand, encapsulates in a segment whatever the application gives it; so that, if the application gives UDP an application message, this message will be the payload of the UDP segment. Thus, with UDP, an application has more control of what data is sent in a segment. b) With TCP, due to flow control and congestion control, there may be significant delay from the time when an application writes data to its send buffer until when the data is given to the network layer. UDP does not have delays due to flow control and congestion control.

商务交际英语任务(assignment)

商务交际英语(二) Assignment ● You work in the Sales Department of an international company. Manuela Garcia, an important client, is visiting your company for a day. There are some changes to the itinerary you sent her last week. ● Write a letter of 120 - 140 words to Ms Garcia, using the original itinerary and your handwritten notes, informing her of the changes. Proposed itinerary for one-day visit of Manuela Garcia Wednesday 20 October 11.00 John Sallis to meet Ms Garcia at airport 11.30 Arrival at company John Sallis to give Ms Garcia an introductory tour of company 12.30 14.00 ( Carol Snape Tom McAllister Sue Smith Manuela Garcia ) 15.30 Coffee break 16.00 18.00 John Sallis to take Ms Garcia to the airport 19.30 英语老师:周俐 2012年10月17日(第七周) (作业于10月26日前上交给我)

Assignment 1

Corporate Finance Assignment One 1. A stock has the following year-end prices and dividends: What are the arithmetic and geometric returns for the stock? 2.You find a certain stock that had returns of 4 percent, -5 percent, -15 percent, and 16 percent for four of the last five years. The average return of the stock for the 5-year period was 13 percent. What is the standard deviation of the stock's returns for the five-year period? 3.Consider the following information: a.Your portfolio is invested 30% each in A and C, and 40% in B. What is the expected return of the portfolio? b.What is the variance of the portfolio? The standard deviation? 4.You have $100000 to invest in a portfolio containing stock X and stock Y. Your goal is to create a portfolio that has an expected return of 18.5%. If stock X has expected return of 17.2% and a beta of 1.4, and stock Y has an expected return of 13.6% and a beta of 0.95, how much money will you invest in stock Y? How do you interpret your answer? What is the beta of your portfolio? 5.Suppose you observe the following situation: a.Calculate the expected return on each stock. b.Assuming the capital asset pricing model holds and stock A’s beta is greater than stock

assignment写作浅谈

在百度问答中留学论文宝老师发现,很多学生都会进行网上交流,比如学生会问到,留学论文宝在哪里?留学论文宝好不好?留学论文宝是不是跟网上说的一样好?我是澳洲的学生,某某学科的论文作业能不能辅导?很多这样的问题等到第二天再看的时候发现百度问题已经被接受过留学论文宝老师辅导的学生给刷屏了。感谢各位留学生对留学论文宝的信赖、支持和宣传。 既然还是有很多学生抱着怀疑得态度,那么留学论文宝老师就来介绍一下自己吧。留学论文宝是2009年成立在英国本地的一家正规论文辅导机构,在十多年的竞争中我们成功地走到了今天并且快速地发展着,直至今日,我们已经成功解决了超过30000留学生的论文课业负担。 留学论文宝的核心力量是辅导老师,作为专业的英文写作辅导老师都是从国外的知名大学毕业的研究生和博士生,因此我们的绝对优势在于: 1、辅导老师资质过硬,不仅师资力量雄厚,而且专业无敌,老师遍布各大高校,因此任何一个留学生都能找到对口的院校专业辅导老师,在英文表达上是绝对没有问题的,在写作技巧上也是非常娴熟的。 2、在辅导水平。在学历上可以分别是本科预科、本科、硕士预科、硕士,有关博士的我们可以提供专业的Proposal指导;在辅导学科上可以是文理科、工科、商科、医学、艺术等;在辅导范围上,可以是平常的期中期末作业、毕业生的Dissertation,留学文书(包括PS个人陈述、CV个人简历、Recommendation 推荐信、成绩单毕业证学位证等各种材料的翻译),还可以是学生的演讲稿、PPT 和调查问卷的制作等。 3、服务态度。辅导老师会根据学生提供的学校学历专业、论文要求、导师上课的课件和风格,学生掌握的知识程度,先列出一份写作大纲,学生满意后方动笔,最后在规定的时间内上交全稿。如果在中间过程学生有不满意的要及时提出来以便修正。最后的全稿会通过客服老师进行专业评估和使用专业的Turn it in检测系统检测抄袭率,保证在5%以下,有不少都是100%原创。所有学生的文章一律过关,还有不少取得了不错的分数呢! 在留学文书方面,留学论文宝老师因为都是出过国的,所以在对各大学的地理位置、办学特色,甚至对某一个具体知名的导师都是很熟悉的,再加上自己也制作过留学文书,因此事绝对没有问题的。凡是接受过留学论文宝辅导老师的文书制作服务的,都写出了非常满意的PS、CV和推荐信,最后也都拿到了自己梦想大学的offer,留学成功。 如果您想更多地了解留学论文宝,想知道Essay和Dissertation的写作方

Assignment3经原

经济学原理(双学位,2010年秋季学期) 作业3(第12-17章) 上交日期:2010年11月27日课上 第一部分:教材习题(注:括号内为第四版对应的题目号) 第12章,问题与应用,3,6,9,11 第13章,问题与应用,1,4(5),11(10),12(11) 第14章,问题与应用,1(2),4(5),6(7),12(11) (提示:第1题:考虑石油价格上升导致造船的边际成本上升相同的数量。) 第15章,问题与应用,1,6,11(13),13(14) 第16章,问题与应用,6(对应第四版第17 章第5题),7(第17章第7题) 第17章,问题与应用,1(16章3),3(16章4),6(16章6),8(16章9),10(16章11) 第二部分:补充英文题目 Supplemental Questions (Multiple-choice questions if marked with a-d, otherwise True/False questions.) 1. (Ch12) The deadweight loss associated with a tax on a commodity is generated by a. those consumers who still choose to consume the commodity, but pay a higher price that reflects the tax. b. those consumers who choose to not consume the commodity that is taxed. c. all citizens who are able to use services provided by government. d. those consumers who are unable to avoid paying the tax. 2.(Ch13) Adam Smith's example of the pin factory demonstrates that economies of scale result from specialization. (For Chapter 14) 1. The Wheeler Wheat Farm sells wheat to a grain broker in Seattle, Washington. Since the market for wheat is generally considered to be competitive, the Wheeler Farm a. does not choose the quantity of wheat to produce. b. does not have any fixed costs of production. c. is not able to earn an accounting profit. d. does not choose the price at which it sells its wheat. (For Chapter 15) 2.The De Beers Diamond company advertises heavily to promote the sale of all diamonds, not just its own. This is evidence that they have a monopoly position to some degree. (For Chapter 16) 3. A monopolistically competitive firm chooses its production level the same way as a(n) a. monopolist. b. oligopolist. c. perfectly competitive firm.

关于财务尽职调查的深度解析

关于尽职调查的10000字深度长文解析(财务篇) 概述 尽职调查概念 1、概念 尽职调查又称谨慎性调查, 是指投资人在与目标企业达成初步合作意向后,经协商一致,投资人对目标企业一切与本次投资有关的事项进行现场调查、资料分析的一系列活动。 财务尽职调查即由财务专业人员针对目标企业与投资有关财务状况的审阅、分析、核查等专业调查。 2、种类 尽职调查的种类包括四类: ■法律尽职调查■财务尽职调查 ■业务尽职调查■·其他尽职调查 尽职调查的目的 尽职调查就是要搞清楚: 1、他是谁?即交易对手实际控制人的底细和管理团队 2、他在做什么?即产品或服务的类别和市场竞争力 3、他做得如何?即经营数据和财务数据收集,尤其是财务报表反映的财务状况、经营成果、现金流量及纵向、横向(同业)比较 4、别人如何看?包括银行同业和竞争对手的态度 5、我们如何做?在了解客户的基础上进行客户价值分析,用经验和获得的信息设计授信方案和控制措施,把交流变成可行的交易。 简言之,即做好股东背景和管控结构、行业和产品、经营和财务数据、同业态度的调查,提供我们的做法。

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