An Efficient Data Structure for Advance Bandwidth Reservationson the Internet
An Ef?cient Data Structure for Advance Bandwidth Reservations
on the Internet
Andrej Brodnik Andreas Nilsson
May19,2002
Abstract
In this contribution we present a problem of resource reservation during some time.We show that the problem has a lower bound of per operation on average and also give a matching upper bound algorithm.
1Introduction
In Computer Communications we need to make a reservation of bandwidth over the Internet to provide Quality of Service(QoS)for the end users.The IETF(Internet Engineering Task Force)de?ned a standard for Integrated Services in routers([10,21])and the end-to-end reservation setup protocol RSVP([6]).The protocol had a?aw that it did not scale well([14]).Therefore IETF came up with a new approach,known as differentiated services (diffserv,[4]).Olov Schel′e n et.al.([17,18])used the diffserv to design a new QoS architecture.In this architecture they provide virtual leased lines using the differentiated services to perform admission control through the system of agents.The agents work on per-hop basis and they need to maintain a database of the reservations made on their hop.In the backbone of the Internet it will most likely be many reservations to administrate and hence the use of an ef?cient data structure will be required.Moreover,in the design of the agents the authors propose that a single agent administrates several hops to make it more attractive for the ISP’s(Internet Service Provider).Such a scenario puts even bigger need for use of an ef?cient data structure.
The bandwidth reservation problem is a special case of a more generic problem,where we need to administrate a limited resource over the time;https://www.wendangku.net/doc/a414654818.html,e of human resources,computational power of super-computer,pool of cars etc.Although the solution in this paper covers all these problems,we use the term bandwidth when we talk about the reserved resource.
De?nition1In the Bandwidth Reservation Problem we have a?xed amount of bandwidth to administer.Customers want to make reservations for a part of the bandwidth during a time interval(, the interval starts at time and ends at time,and it includes both end points).The operations to support,besides initialization and disposition,are:
,that reserves units of bandwidth for the time period,where.
,that frees the bandwidth during the interval.Note that freeing the bandwidth is the same as making a reservation with a negative bandwidth.
,that returns maximum reserved bandwidth during the interval.
For the sake of clarity,we sometimes use the subscripts for queries and for reservations.For example,a reservation interval.In the paper we also use the notation denoting a function returning the bigger of and.
In the literature we could not?nd any references to the bandwidth reservation problem where the points in time, when the reservation can be made,are arbitrary(not prede?ned).However,the problem is similar to problems we ?nd in other?elds of computer science that handle intervals on a real line(https://www.wendangku.net/doc/a414654818.html,putational geometry,dynamic
computation and geometric search[1,2,7,8,15]).These problems are generally solved using segment trees(
[16,19]),which were introduced by Bentley([3])as a solution to the Klee’s rectangle problem([12]).The limitation in all these problems is that the end-points of the intervals belong to a?xed set of points while in our
problem the set of points changes during the time.
Kuchem et.al.in([13])presented in a way similar data structure to ours,although it still deals with a?xed set
of points.They use the structure in a VLSI design.Bose et.al.independently developed in([5])a similar data
structure to solve a number of geometric problems.
Another pair of related problems are the well studied partial sum problem([9],brief in[11]),and the pre?x
sum problem([9]).In the pre?x sum problem we have an array on which we want to perform
these two operations:(1):;and(2):for arbitrary values of,and.In[9]Fredman shows a lower bound of for the problem under the comparison based model.
In the same paper Fredman also presents an algorithm with a matching upper bound.
In the rest of the paper we?rst show that the logarithmic lower bound carries over to the bandwidth reservation
problem.We continue with a presentation of a data structure we call BinSeT(binary segment tree)that gives us a
matching upper bound.We conclude the paper with?nal remarks.
2Lower bound
Theorem1The problem of bandwidth reservation requires at least comparisons per operation on the
average,where is a number of intervals we are dealing with.
Proof:Assume that we have a solution to the bandwidth reservation problem that requires time.We
will show how to use such a solution to solve the pre?x sum problem in time which contradicts the lower bound by Fredman([9]).
We translate the array of elements in the pre?x sum problem into end-points of intervals.More precisely,the element of the array is represented by the interval that starts at point and ends at the right most point:. Therefore,the reserved bandwidth at point is the sum of all reserved bandwidths for intervals starting at points, where.This gives us the following translation of pre?x-sum problem operations:
the operation into;and
the operation into a query.
This translation gives us an solution to the pre?x sum problem and hence contradicts the lower bound
by Fredman.
Note that,the pre?x sum as presented by Fredman([9])is also a static problem–i.e.the array of elements
neither expands nor shrinks.On the other hand,the solution we present in the following section does support insertion of new points(intervals)and deletion of points(intervals).Hence,by using the translation in the proof we also get a logarithmic solution to the dynamic version of the pre?x-sum problem.
3Upper bound
To prove an upper bound we present a data structure called BinSeT that supports the required operations in loga-rithmic time.Before going into detail presentation we describe how we represent the reservations.
3.1Representation of Intervals
We do not represent an reservation interval as a single entity,but we split it into two,what we call,reservation events.
A reservation event is point in time when an increase or decrease in the amount of a reserved bandwidth occurs. For example,we store a reservation as reservation events and.In other terms,we convert an interval into two semi-in?nite intervals and.
In the rest of the paper we will describe a data structure that handles the reservation events,or semi-in?nite intervals,ef?ciently.The operations mentioned in De?nition1are converted into:
into insertion of reservation events and;and
2
into insertion of reservation events and.
If we want to store extra information with each reservation we introduce an additional dictionary data structure to store this information and bind the reservation events to records in the dictionary.
3.2Data Structure
The binary segment tree BinSeT is a data structure that combines properties of a binary and a segment tree.The former gives us possibility to dynamically insert and delete reservation events and the later let us answer the queries about the maximum reserved bandwidth.In detail,the leaves represent and store information about the reservation events,while each internal node covers a segment(interval)and stores an information about the values(band-width)on that interval.We have the following invariance:
Invariance1The information stored with the interval is the maximum value on the interval and the change of the value on the interval.
Besides we split the interval into two sub-intervals in such a way that the number of reservation events in each of them differs for at most one.Such a split gives us the necessary and suf?cient condition to get a balanced data structure.Obviously,the internal node also stores the pointers to the nodes representing sub-intervals and the splitting time.The splitting time is in fact the start time of the right sub-interval.
The detail data structure is represented in Algorithm1.The structure is slightly different from the one described sBinSeT
tResource;
tResource;
tTime;
struct
sBinSeT*right;
tBinSeT;
in the previous paragraph since it does not store times and,but only the.The structure,as it will be seen in the description of operations,still ful?lls all the requirements.At this point we note two things:?rst,a node has either two sub-trees(an internal node)or none(a leaf);and second,a leaf stores in and in amount of the reserved bandwidth at the reservation event it represents and in the time of the event.As a consequence of the?rst observation we conclude,that the number of internal nodes is one less than the number of leaves.Since the number of leaves is at most,where is a number of reservation intervals,this proves the following lemma,under the RAM model:
Lemma1The size of the BinSeT storing reservation intervals is words.
3.3Operations
Finally we describe how to implement operations from De?nition1.All our solutions will be recursive and will start traversing the data structure from the root.We assume that we store with BinSeT also the time of the?rst() and the last()reservation event.These are also times and,respectively,for the root of the complete BinSeT. If we descend in the left subtree,then the and for this subtree become values and,respectively,of the root. We treat similarly the right subtree.This is also the reason why we need not store values and with a node.
We start with a query.Assuming Invariance1we prove:
Lemma2Let BinSeT be balanced(cf.[20])then there exists a worst case implementation of query .
Proof:The correctness of the proof uses induction.Due to the limited presentation space we give only a justi?cation of the induction step.Let the query be for the interval and let the node cover interval .If and,the answer is of the node.If()then the answer is the
3
same as answer to the query in the left subtree,(right subtree to which we add, ).In the third case when and(and),the answer is(.Finally,the fourth and the most general case is when.Here the answer is
(1)
To see that the running time of the query is logarithmic,i.e.proportional to the height of the BinSeT,observe that the fourth case occurs only once.
Next we describe insertion,but due to lack of space we leave out the detail description of the deletion.Note that the deletion is always implicit when the amount of reserved bandwidth at a leaf becomes0.
Lemma3There exists worst case implementation of updates and in BinSeT.
Proof:As explained in sect.3.1each of operations and is converted into a pair of insertions of reservation events.We will show that the reservation event insertion takes logarithmic time which will prove the lemma.
The insertion starts at the root and recursively descends to the leaves.We descend the BinSeT using and time of the reservation event to?nd the proper leaf.When we reach the leaf its can be the same as of the inserted event.If so,we add the values.At this point the value can become0and we delete the leaf and replace its ancestor with leaf’s sibling.If the leaf’s is different from,we insert a new internal node with, either or as appropriate,and make an old leaf as one of its leaves.Besides we create a new leaf with
and.
In the last phase we climb back up to the root and at nodes appropriately update values and.Values are always increased for.on the other hand the value of is set to,similarly as in eq.(1).
The tree obtained this way might be unbalanced,but we can re-balance it in in a logarithmic time(cf.A VL trees,[20]).Due to lack of space we leave proof to the reader.
Putting all results together brings us to the?nal theorem:
Theorem2There exists a solution to the Bandwidth Reservation Problem under the comparison based machine model that requires per operation and words of space.This is tight.
Obviously it is straight forward to adapt the solution to handle also queries of the minimum reserved bandwidth. Moreover,using the translation in Theorem1we also get a logarithmic time solution to the dynamic versions of partial sum problem and of pre?x sum problem.
4Conclusions
We showed that the data structure BinSeT(binary segment tree)solves the dynamic version of the Bandwidth Reservation Problem optimally(space-and time-wise)under the comparison based model.The solution requires time for the queries and updates and space.This is a substantial improvement over the previous solution which was static only–new intervals could not be inserted.
There are a number of open problems left.For example,what are lower and upper bounds under the cell probe model and bounded universe?
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5
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1998-Data mining-- statistics and more
Data Mining:Statistics and More? David J.H AND Data mining is a new discipline lying at the interface of statistics,database technology,pattern recognition,machine learning,and other areas.It is concerned with the secondary analysis of large databases in order to?nd previously un-suspected relationships which are of interest or value to the database owners.New problems arise,partly as a con-sequence of the sheer size of the data sets involved,and partly because of issues of pattern matching.However,since statistics provides the intellectual glue underlying the e?ort, it is important for statisticians to become involved.There are very real opportunities for statisticians to make signi?-cant contributions. KEY WORDS:Databases;Exploratory data analysis; Knowledge discovery. 1.DEFINITION AND OBJECTIVES The term data mining is not new to statisticians.It is a term synonymous with data dredging or?shing and has been used to describe the process of trawling through data in the hope of identifying patterns.It has a derogatory con-notation because a su?ciently exhaustive search will cer-tainly throw up patterns of some kind—by de?nition data that are not simply uniform have di?erences which can be interpreted as patterns.The trouble is that many of these “patterns”will simply be a product of random?uctuations, and will not represent any underlying structure.The object of data analysis is not to model the?eeting random pat-terns of the moment,but to model the underlying structures which give rise to consistent and replicable patterns.To statisticians,then,the term data mining conveys the sense of naive hope vainly struggling against the cold realities of chance. To other researchers,however,the term is seen in a much more positive light.Stimulated by progress in computer technology and electronic data acquisition,recent decades have seen the growth of huge databases,in?elds ranging from supermarket sales and banking,through astronomy, particle physics,chemistry,and medicine,to o?cial and governmental statistics.These databases are viewed as a re-source.It is certain that there is much valuable information in them,information that has not been tapped,and data min-ing is regarded as providing a set of tools by which that in-formation may be extracted.Looked at in this positive light, it is hardly surprising that the commercial,industrial,and David J.Hand is Professor of Statistics,Department of Statistics,The Open University,Milton Keynes,MK76AA,United Kingdom(E-mail: d.j.hand@https://www.wendangku.net/doc/a414654818.html,).economic possibilities inherent in the notion of extracting information from these large masses of data have attracted considerable interest.The interest in the?eld is demon-strated by the fact that the Third International Conference on Knowledge Discovery and Data Mining,held in1997, attracted around700participants. Super?cially,of course,what we are describing here is nothing but exploratory data analysis,an activity which has been carried out since data were?rst analyzed and which achieved greater respectability through the work of John Tukey.But there is a di?erence,and it is this di?erence that explains why statisticians have been slow to latch on to the opportunities.This di?erence is the sheer size of the data sets now available.Statisticians have typically not con-cerned themselves with data sets containing many millions or even billions of records.Moreover,special storage and manipulation techniques are required to handle data collec-tions of this size—and the database technology which has grown up to handle them has been developed by entirely di?erent intellectual communities from statisticians. 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Nexus仓库管理
Nexus仓库管理
管理仓库 为了管理Nexus仓库，以administrative用户登陆进Neuxs，然后点击Nexus 菜单左手边的Views/Repositories菜单中的Repositories。 Nexus提供了三种不同种类的仓库： ●Proxy Repositories ●Hostedrepositories ●Virtual repositories 1.代理仓库(Proxy Repository) 代理仓库是一个远程仓库的代理。默认的，Nexus附带下面已经配置的代理仓库： Apache Snapshots 这个仓库包含来ApacheSoftware Foundation发布的快照。 Codehaus Snapshots 这个仓库包含来Codehaus发布的快照。 Central 这是中央仓库包含发布组件。前身为Maven Central，它是Apache Maven默认的内置仓库，并且直接支持其它的构造工具，如Gradle, SBT或Ant/Ivy。Nexus通过HTTPS使用https://https://www.wendangku.net/doc/a414654818.html,/maven2/链接连接中央仓库。 2.宿主仓库(Hosted Repository) 主要用于部署无法从公共仓库获取的构件（如 oracle 的 JDBC 驱动）以及自己或第三方的项目构件。Nexus附带下面已经配置的宿主仓库： 3rd Party 这里存放的是某些在公共的Maven仓库中不存在的库，但依赖于第三方的仓库。这些依赖库可以是商业的，私有的库，例如Oracle JDBC驱动。 Releases
是你的组织将要发布的内部版本。这里存放我们自己项目中发布的构建, 通常是Release版本的, 比如我们自己做了一个FTP Server的项目, 生成的构件为ftpserver.war, 我们就可以把这个构建发布到Nexus的Releases本地仓库. 关于符合发布后面会有介绍. Snapshots 是你的组织将要发布的内部块状。这个仓库非常的有用, 它的目的是让我们可以发布那些非release版本, 非稳定版本, 比如我们在trunk下开发一个项目,在正式release之前你可能需要临时发布一个版本给你的同伴使用, 因为你的同伴正在依赖你的模块开发, 那么这个时候我们就可以发布Snapshot版本到 这个仓库, 你的同伴就可以通过简单的命令来获取和使用这个临时版本. 2.虚拟仓库(Virtual Repository) Virutual repository作为不同类型仓库的适配器。当前，Nexus支持转换Maven 1仓库和Maven 2仓库。另外，你可以公开任何仓库格式为NuGet或OBR 仓库。例如，Maven 2仓库可以包含OSGi Bundles。a Maven 2 repository can contain OSGi Bundles, which can be exposed as a OSGi Bundle repository with the virtual repository Provider set to OBR. By default it ships with a Central M1 shadow repository that exposes the Central repository in Maven 1 format. 配置仓库 Repository窗口允许你建立，更新和删除不同的仓库使用Add，Delete和Trash按钮。Refresh按钮更新显示仓库和仓库组列表。正当在Nexus用户界面之行任何的delete操作，Trash按钮允许清空那些已删除组件的备份垃圾文件夹。 默认的，Repository窗口显示仓库配置和管理。在Trash按钮的右边允许切换仓库列表和浏览被Nexus管理的仓库
Maven配置
maven安装+maven本地仓库配置+maven镜像 1.安装maven前我们电脑上要确保安装了JDK1.6以上的版本 2.下载Maven Maven官网下载地址是https://www.wendangku.net/doc/a414654818.html,/download.html 下载apache-maven-3.2.5后，解压到你想要解压的地方 我本地解压到D:\Program Series\Tools\Apache-maven这个路径下 3.然后在配置maven的环境变量： MAVEN_HOME D:\Program Series\Tools\Apache-maven\3.2.5 PATH %MAVEN_HOME%bin
配置好以后环境变量以后，打开cmd窗口，输入mvn -v，如果出现以下内容就表示安装成功。 4.然后我们给maven添加本地的仓库地址 解压后的maven目录下找到conf目录，其中就会有一个setting.xml文件 我本地的地址是D:\Program Series\Tools\Apache-maven\3.2.5\conf 然后通过编辑器找到第53行，把注释去掉，修改成自己想要设置的本地仓库地址就可以了。我本地设置在maven的安装目录同级目录下 mvn文件夹的名字可以随便取，但是一定要确保设置的仓库地址是有效的地址，因为maven 是不会自动创建这个mvn文件夹的。 5.实验一下刚才做的事情产生作用没有，控制台输入： mvnhelp:system 如果没有任何问题，执行完该命令之后，在D:\maven\repository下面就会多出很多文件，这些文件就是maven从中央仓库下载到本地仓库的文件，maven已经开始为我们工作了。Maven的安装配置和本地仓库地址的配置就已经结束了。 国内较快的maven镜像 国内连接maven官方的仓库更新依赖库，网速一般很慢，收集一些国内快速的maven仓库镜像以备用。 ====================国内OSChina提供的镜像，非常不错===================
Maven源配置教程
Maven安装及基本使用 1、安装maven首先到maven官网下载maven安装包, Maven下载地址, 下载apache-maven-3.3.9-bin.zip文件 > 由于当前系统为windows, 所以下载apache-maven-3.3.9-bin.zip文件; 如果您使用的系统为unix类系统请下载apache-maven-3.3.9-bin.tar.gz文件 > > 注: 由于maven是依赖于jdk的, 所以在安装maven前需要提前准备好jdk环境。jdk的安装方式不是本文的重点, 所以本文不进行展开说明了。 右键解压apache-maven-3.3.9-bin.zip文件到当前目录中即可 >如果您使用的为unix环境, 以CentOS为例, 使用以下命令进行解压 tar -zxvf apache-maven-3.3.9-bin.tar.gz 到此maven已经安装完成。 为了之后使用方便,所以需要将maven添加到环境变量中, 配置方式: 右键 "我的电脑" -> 选择 "属性" -> 选择 "高级系统设置" -> 选择 "高级" 标签页 -> 选择 "环境变量" 在新打开的窗口中的"系统变量"中新建变量名为"MAVEN_HOME", 变量值为 maven解压后的路径, 例如: E:-maven-3.3.9 然后编辑环境变量PATH, 在PATH的环境变量值的尾部添加";%MAVEN_HOME%", 一路确认后即添加成功。注: windows中多个环境变量的值需要用";"分隔 unix系统中环境变量的配置方式, 以CentOS为例: 使用以下方式为当前用户添加maven的环境变量打开~/.bashrc文件并在文件末尾添加以下内容 vim ~/.bashrc MAVEN_HOME=/opt/apache-maven-3.3.9 PATH=$PATH:$MAVEN_HOME/bin export MAVEN_HOME PATH 保存并退出, 并使用下面的命令使新添加的环境变量立即生效
data mining 练习题
1.Below is a table representing eight transactions and five items: Beer, Coke, Pepsi, Milk, and Juice. The items are represented by their first letters; e.g., "M" = milk. which of the pairs are frequent itemsets? 2.Here is a table with seven transactions and six items, A through F. An "x" indicates that the item is in the transaction. A B C D E F x x x x x x x x x x x x x x x x x x x x x x x x x x x x Assume that the support threshold is 2. Find all the closed frequent itemsets. Answer: There are many ways to find the frequent itemsets, but the amount of data is small, so we'll just list the results. Among the pairs, all but AF are frequent. The counts are: AC, CE: 5AE, CD: 4AD, BC, BD, BE, BF, DE: 3AB, CF, DF, EF: 2 Here are the counts of the frequent triples: ACE: 4ACD, BCE, CDE: 3ABC, ABE, ADE, BCD, BDE, BDF, BCF, BEF, CEF: 2 There are four quadruples that are frequent, all with counts of 2: BCEF, BCDE, ACDE, and ABCE. There are no frequent sets of five items.
(仓库管理)设置Maven组件库仓库
（仓库管理）设置Maven组件库仓库
设置Maven仓库 1. 介绍 Maven是Java开发者中流行的构建工具，Maven的好处之一是可以帮助减少构建应用程序时所依赖的软件构件的副本，Maven建议的方法是将所有软件构件存储于一个叫做repository的远程仓库中。 Maven会从https://www.wendangku.net/doc/a414654818.html,中的公用仓库中同步构件，这个公用仓库下载缓慢、不稳定，并且不包含一些构件的最新版本，而且不能上传团队私有的构件。通过设置内部Maven仓库，团队可以更有好地利用Maven仓库的优势并且克服使用ibiblio上的仓库时的缺点。 本文着眼于Maven仓库工具应提供的一些功能，列出了选择Maven仓库工具的标准，说明了使用Artifactory设置Maven仓库的相关步骤。在Linux和Windows 中设置Maven仓库的过程是相同的，少数不同点在文中已经用高亮显示，本文列出了与这个Maven仓库的使用有关的POM文件的范例，用截图来向用户阐述如何设置Maven仓库，Maven和artifactory配置范例也有列出。 1.1 背景知识 本文假定读者熟悉以下的概念和技术： ●Maven 2 ●J2EE web 服务器部署过程(e.g. 在Tomcat中部署Web应用程序) ●XML 1.2 Maven仓库的目的 Maven仓库的目的是作为团队内所使用的所有软件构件的内部私有构件仓库，将Maven构件（jar和pom）存储到一个专门的Maven仓库比将它们存储到版本控制系统中更有优势，这是因为： ●构件（jar）是二进制文件，不属于版本控制系统，版本控制系统在处理文本文件方面比较好 ●保持较小的版本控制数据库 ●Checkout、update和其他版本控制的操作可以更快 1.3 建立内部私有仓库的优势： ●减少可能的版本冲突 ●减少首次构建时需要的手动干涉
eclipse maven 项目没有maven依赖仓库问题解决
eclipse maven 项目没有maven依赖仓库问题解决 之前是没有Maven Dependencies 这个文件夹仓库 解决办法：
武汉工程大学文科基金项目
所属学科及学科代码： 项目编号： 武汉工程大学文科基金项目 申请书 项目名称： 项目负责人： 联系电话： 依托学院部门： 申请日期： 武汉工程大学科技处制 2007年9月
简表填写要求 一、简表内容将输入计算机，必须认真填写，采用国家公布的标准简化汉 字。简表中学科（专业）代码按GB/T13745-92“学科分类与代码”表填写。 二、部分栏目填写要求： 项目名称——应确切反映研究内容，最多不超过25个汉字（包括标点符号）。 学科名称——申请项目所属的第二级或三级学科。 申请金额——以万元为单位，用阿拉伯数字表示，注意小数点。 起止年月——起始时间从申请的次年元月算起。 项目组其他主要成员——指在项目组内对学术思想、技术路线的制定理论分析及对项目的完成起主要作用的人员。
一、项目信息简表
二、选题：本课题国内外研究现状述评；选题的意义。 三、内容：本课题研究的基本思路和方法；主要观点。 四、预期价值：本课题理论创新程度或实际应用价值。 五、研究基础：课题负责人已有相关成果；主要参考文献。 六、完成项目的条件和保证：包括申请者和项目组主要成员业务简历、项目申请人和主要成员承担过的科研课题以及发表的论文；科研成果的社会评价；完成本课题的研究能力和时间保证；资料设备；科研手段。 （请分5部分逐项填写）。
七、经费预算
六、项目负责人承诺 我确认本申请书及附件内容真实、准确。如果获得资助，我将严格按照学校有关项目管理办法的规定，认真履行项目负责人职责，积极组织开展研究工作，合理安排研究经费，按时报送有关材料并接受检查。若申请书失实或在项目执行过程中违反有关科研项目管理办法规定，本人将承担全部责任。 负责人签字： 年月日 七、所在学院意见 负责人签字：学院盖章： 年月日 八、科技处审核 已经按照项目申报要求对项目申请人的资格及项目申请书内容进行了审核。项目如获资助，科技处将根据项目申请书内容，落实项目研究所需经费及其它条件；以保证项目按时顺利完成。 科技处盖章 年月日
Oracle的数据挖掘工具--Data mining suite
Oracle Data Mining 能够让公司构建高级商务智能应用，这些应用可以挖掘公司数据库中的数据、发现新的内在信息，并把这些信息集成到业务应用中。Oracle Data Mining 为分类、预测和关联嵌入了数据挖掘功能。所有的模式构建和标记(scoring) 功能都可以通过基于Java 的API 访问。 Oracle的数据挖掘工具--Data mining suite 数据挖掘(Data Mining)是决策分析技术的一个更高层次，数据挖掘技术采用人工智能的决策分析方法，按照用户既定的业务目标，对数据 仓库中浩如烟海的数据进行探索，揭示隐藏其中的规律，并进一步将其 模型化。 从业务问题的定义到分析结果的实施，Data mining suite提供完整的工具： 1.确定业务问题 定义的业务目标和相关的数据及文件存到集中管理的项目夹 (Project Folder)中。Data mining suite为项目中的每一步自 动生成文档方便项目跟踪。这些工作可使用可视化的工作流对象 完成。 2.为挖掘准备数据 Data mining suite可以访问企业网络上的所有数据，这些数据 类型包括： - 数据仓库 - 关系型数据库 - 文本文件 - 多维数据库 Data mining suite提供Import 向导程序简化数据装载过程。 Find Missing Values向导可以快速确定和解决数据不全的问题。 Data mining suite提供一套完整的数学、统计学、字符、比较、 逻辑运算函数用于数据的抽样、过滤和转换过程。Data mining suite还提供关键字段向导帮助确定最重要的包含信息的字段， 这有助于提高后续模型分析的精确性和计算速度。 3.建立多种模型 Data mining suite目前的版本支持如下数据挖掘模型： -Neural Networks(Net) -Classification and regression decision trees(Tree) -Momory-based reasoning(Match) -Bayesian learning(Bayes) -Clustering(Cluster) 这些数据挖掘模型相结合，可以使客户从数据仓库中获取最大程
202X美国留学文科专业申请建议.doc
202X美国留学文科专业申请建议 现在，去美国留学，文科专业比较容易申请一些，但是，美国的文科专业众多，该如何申请呢，下面来说说美国留学文科专业申请建议。 1、文科专业非常庞杂，常见专业有：语言类，新闻和传播，政治学，社会学，人类学，历史学，经济学，法学，教育学，心理学，建筑学，城市规划和景观设计，艺术类等。 2、若是英语专业，可以申请教育学，比如教育心理，TESOL，早期教育等；文学类，比如比较文学等；传媒类，公关，广告等；或者政治学、社会学、历史学。具体申请什么专业根据申请人的背景经历进行确定。中文和日语专业的可以申请东亚研究，有东亚研究设置的专业都是TOP50的学校，所以申请难度也是很大的。 3、美国的顶尖大学综合排名前30的学校里设置传播学院的并不多。因为传播学是后兴起的专业才有七、八十年的历史，很多老牌的学校他们排斥这样的新兴学科，所以他们不设置这样的学院，例如哈佛，耶鲁，牛津，剑桥等根本没有传播学校。有些学校虽然有传播学院或者有相关的专业但是规模一般也很小，有些学生甚至只招生本校的本科生。例如：斯坦福大学，传播学一直没有权威的专业排名，比较出名的学校有：哥伦比亚大学，纽约大学，但是这两个牛校都只设置新闻学院，并没有传播学研究。而且这两个学校的新闻类专业是属于那种超级牛人才能申请到的，一般在国内每年招生只有1-2个，甚至没有。总的来说新闻类专业申请难度很大。并且对申请人的英语及GRE成绩要求非常高，并且申请人最好有在央视，新华社，这样的背景才会比较有利。 4、政治学，社会学，人类学和历史，一般硕士的申请很难拿到奖学金，博士奖学金设置会很多，但是同时对学生的研究兴趣有要求，并希望看到学生对未来的职业发展规划。
高中理科申请书
篇一：高一文理分科申请表 高一文理分科申请表 明： 1、传 媒含播音与主持艺术、广播电视编导、服饰艺术与表演、影视表演、空中乘务等专业；美术含 绘画、书法艺术和书法教育、设计等专业；音乐含声乐、舞蹈、器乐、理论作曲、指挥等专业。 2、本 申请表经学生、家长、班主任签名后，不能再更改，学校以此为依据重新分班。 上梅 中学教务处 2013年12月28日 篇二：文理分科申请书 请书 敬的老师： 我 是，现在高（）班，本人喜欢科技制作、科普读物、科技发明和生命科学等相关 知识与能力锻炼，对理科的兴趣大于对文科的兴趣。经过慎重考虑，并与家长商量，我决定申 请就读（文科、理科、美术、体育、音乐、舞蹈、传媒）班，敬请批准。 谢谢！ （学 生本人签名）年月日 意上述申请。 （家 长签名）年月日 篇三：转班申请书 转班申请书尊敬的杨老师： 您好！ 我想转到理科高二（16）班.因为经过假期的思考，我发现自己对文科没有很大的兴趣，而且
我也了解到自己在文科方面很难得到提高。对此，我恳请杨老师能让我转到理科高二（16） 班，我真切希望自己能转理科高二（16）班，我总结了我转班的原因，有以下几点： 一、 我觉得自己对文科的兴趣没有了之前的那种热情，而且我觉得自己在理科方面还有待提高，我 对它也很感兴趣。人们都说兴趣是学习最好的老师，有了兴趣就是成功的一半。而现在我对文 科已经没有了那种热情，又怎么会对学习文科尽心尽力、认真努力的去做呢？所以我真心的想 转到理科。 二、 我的文科成绩不是很好，而且从小我就贪玩把英语落下了很多。然而英语在文科当中可以占很 大的优势，我的英语很差，读文科没有优势，所以我希望自己能转到理科。 三、 学习文科需要很好的记忆力，但我这个人很赖，不喜欢背诵，而文科又需要背诵和 忆。 四、 根据社会的需求和我个人的发展空间，我想理科更适合我。 亡羊 补牢，为时不晚，希望杨老师能给我一次机会。 此致 敬礼 申请 人： 011年7月31日 篇四：申请书 尊敬的政府领导： 我叫 xxx，19xx年xx月xx日生，系xxx居民，配偶19xx年生，也是xxx人，我们都无固定 职业，现有家庭成员x人，家庭月收入xxx元。我们也曾经多次想买房子，但就我们这点收 入想买完全属于我们自己的房子那简直是奢望。所以本人和家人至今也一直居无定所，但为了 生活，又不得不在城区内四处奔波打工，并租房居住。由于本人家庭生活的实际困难和无住房 的实际情况，现想申请政府廉租住房一套，望领导给予批准为盼！ 谢谢！ 请人：