Assignment-2014s2
awaiting reviewer selection与assignment
awaiting reviewer selection与assignment Awaiting reviewer selection and assignment is a crucial phase in the publication process of any research paper or academic article. This stage marks the transition from the initial submission of the manuscript to the formal evaluation by experts in the relevant field. It involves the careful selection of reviewers who possess the necessary expertise and qualifications to provide meaningful and constructive feedback on the work.等待审稿人选择和分配是任何研究论文或学术文章发表过程中的关键阶段。
这一阶段标志着从手稿的初步提交到相关领域专家正式评审的过渡。
它涉及仔细选择具备必要专业知识和资格的审稿人,以便他们能够提供有意义且建设性的反馈。
During this period, the editors of the journal or publishing house assess the manuscript, ensuring it meets the basic criteria for publication, such as its relevance to the journal's scope, the quality of the research, and the adherence to academic standards and ethical guidelines. Once the manuscript is deemed suitable for review, the next step is to identify potential reviewers.在此期间,期刊或出版机构的编辑会评估手稿,确保其满足出版的基本标准,例如与期刊范围的相关性、研究质量以及对学术标准和道德规范的遵守。
Assignment+1_2014_
Assignment 11.Give a descriptive definition for each of the following terms.(1)Feature size,(2)Flexible block,(3)Datapath library,(4)Base array,(5)Primitive cell,(6)Prediffused array,(7)Floorplaning,(8)Placement,(9)Wire-load model,(10)Routing model.Solution:(1) Feature size:特征尺寸,指某种工艺下最小晶体管的沟道长度,约为最小晶体管长度的一半。
(2) Flexible block:可变的单元块,是CBIC中的标准单元区,包含了各种标准单元行,设计时可以改变其形状大小。
(3) Datapath library:数据通路(数据总线)库,用来编译生成数据通路逻辑的库,一般包括加法器、减法器、乘法器和简单的算术逻辑单元。
它使得数据通路的版图设计方法更快速且密度更高。
(4) Base array:基本阵列,门阵列上预先确定的晶体管图案即为基本阵列,它是由最小单元重复排列组成的。
(5) Primitive cell:基元,指的是上述基本阵列中的最小单元。
(6) Prediffused array:预扩散阵列,在基于门阵列的ASIC中,可以将已完成扩散并形成晶体管的硅圆片储备待用,所以有时把门阵列成为预扩散阵列。
(7) Floorplaning:布图规划,在芯片上排列网表的模块。
(8) Placement:布局,确定功能模块中具体单元的摆放位置。
(9) Wire-load model:连线-负载模型,为了在实际布线完成前估计出引线的寄生电容,就需要对给定大小的电路模块中线网的电容进行估算。
这常要采用查找表的方式,也成为连线-负载模型。
(10) Routing model:布线模型,它可以用来告诉自动布线工具在单元上何处进行布线,以及连接到单元的位置和类型。
Assignment 1S
Let is the proportion of the fine-dining restaurants that have instituted policies restricting the use of cell-phones defective items .
= 0.20
P (1 )
3. A company is trying to decide which of the two brands of tyres to buy for its trucks. It would like to adopt brand G unless there is some evidence that brand F is better, in which case it will adopt the latter brand. An experiment in which 16 tyres from each brand are used is conducted. The results are as follows: Jenama F/Brand F Jenama G/Brand G
ESEE 2201 STATISTICS II SEM 2, 2013/2014 - ASSIGNMENT I DUE BY WEDNESDAY, 16 APRIL, 2014
Students are allowed to work as a group (no more than three (3) students per group) on the assignment. 1. The lifetimes of light bulbs produced by a particular manufacturer have mean 1,200 hours and standard deviation 400 hours. The population distribution is normal. Suppose that you purchase 16 bulbs, which can be regarded as a random sample from the manufact urer’s output. a) What is the mean of the sample mean lifetime? Explain. Mean of the sample mean lifetime is 1,200 hours. If a population is normal with mean μ and standard deviation σ, the sampling distribution of the mean is also normally distributed with mean . b) What is the standard error of the sample mean? Explain. Standard error of the sample mean: x
assignment2
Chap24/11P, 33,44P; Chap25/8P,9PProblem 1 (Chap 24/11P) A point charge q is placed at one corner of a cube of edge a . What is the flux through each of the cube faces? [Solution]Note that when a point charge is placed at one corner of a cube, the field lines from the charge is tangential to the three faces meeting at the point charge (See the above diagram). So the flux through the these faces must be zero.To find the flux through the other faces, we can build a large cube composed of 8 identical smaller cubes and place the charge at the center of the large one (See the above diagram). According to the Gauss ’ law, 0encE q εΦ=, the flux flowing out of the large cube is 0E qεΦ=.Note that the faces of the large cube is composed of 24 faces of the smaller cube, and the fluxes through each smaller face have the same value due to the symmetry, namely24each small face q εΦ=. This is just the flux through each of the other three faces of a cube when a point charge q is placed at its corner.[Problem 2] Chap24/25P. Charge is distributed uniformly throughout the volume of an infinitely long cylinder of radius R . (a) Show that, at a distance r from the cylinder axis (for r R <),2r E ρε=where ρis the volume charge density. (b)Write an expression for E when r R > .[Solution] Due to the cylindrical symmetry of the charge distribution, we choose a circular cylinder of radius r and height h , coaxial with the charged cylinder, as the Gaussian surface (asshown in the diagram). Two end caps are included to ensure the Gaussian surface is closed. Then the electric field lines will flow throught out of the cylindrical part of the Gaussian surface uniformly and perpendicularly no matter r R < or r R > if the charge is positive.:ρhAccordingly, the electric flux through the Gaussian surface is given by(2)E E rh πΦ=where E is the field magnitude at the cylindrical surface of radius r and 2rh π is the area of this side surface.On the other hand, the net charge enclosed by the Gaussian surface is given by22(), for (), for encr h r R q R h r R ρπρπ⎧<⎪=⎨>⎪⎩Applying the Gauss ’s law 0encE q εΦ=, we have the magnitude of the electric field at a distancer from the cylinder axis as20, for 2, for 2rr R E R r R rρερε⎧<⎪⎪=⎨⎪>⎪⎩The direction of the field is radially outward.[Problem 3] Chap24/33. A planar slab of thickness d and volume charge densityρ.[Solution](1) The magnitude of the electric field inside the slab, at a distance x from the central plane of theslab.The charge distribution has a planar symmetry, so we choose a closed cylinder perpendicular to the slab as the Gaussian surface. The end caps of area A are located at a distance x from the central plane of the slab, above and below respectively.From the symmetry, we know that the electric field Eat any point on the end caps must havethe same magnitude and be perpendicular to the caps in an either outward or inward direction.( In the above diagram, the electric field lines are assumed to flow out of the end caps. ) So the flux through the end caps is determined by12cap cap EA Φ=Φ=For the cylindrical part, the electric field is parallel to the surface at any point so that there is no flux through it, namely,0cyl Φ=The total flux through the Gauss surface is given by122cly cap cap EA Φ=Φ+Φ+Φ=And the charge enclosed by this Gauss surface is2enc xA ρρ=⋅According the the Gauss ’ law 0/enc ρεΦ= , we have022/EA xA ρε=So the magnitude of the electric field at a distance x<d/2 from the central plane is given by0/E x ρε=(2) The magnitude of the electric field outside the slab, at a distance x from the central plane ofthe slab.Similarly, we choose a closed cylinder as the Gaussian surface. The distance from the end caps to the central plane now satisfies x >d /2, as shown in the above diagram. The total flux through the Gaussian surface is still122cly cap cap EA Φ=Φ+Φ+Φ=,But the enclosed charge is given by enc Ad ρρ=. Then the Gauss ’ law 0/enc ρεΦ= reduces to02/EA Ad ρε=So that the magnitude of electric field outside the slab is always0/2E d ρε=,independent of the distance x from the central plane.[Problem 4] Chap.24/44P The figure shows a spherical shell of charge with uniform volme charge density ρ. Plot E due to the shell for distances r from the center of t he shell ranging from zero to 30cm. Assume that 631.010C/m ρ-=⨯, 10cm a = and 20cm b =.[Solution ] According to the spherical symmetry of the charge distribution, we choose a concentric sphere of radius r as the Gaussian surface. The net charge enclosed by this surface is133enc23330, for ( S in the diagam)4(), for ( S34(), for ( S in the diagam)3r a r a q a r b b a r b πρπρ⎧⎪<⎪-⎪=⋅≤<⎨⎪⎪-⋅≥⎪⎩Since the electric field Eat any point on the Gaussiansurface must have the same magnitude and be perpendicular to the surface in any case, the electric flux Φ through the Gaussian surface is always given by24E r πΦ=⋅.Then the Gauss ’ law 0/enc ρεΦ=reduces to 24encq E r πε⋅=, so the magnitude of the electricfield inside the sphere at a distance r from the centre is332033200, for(), for 3(), for 3r a r a E a r b r b a r b r ρερε⎧⎪<⎪⎪-⎪=⋅≤<⎨⎪⎪-⎪⋅≥⎪⎩. Substituting 10cm a =, 20cm b = and 31C/m ρμ= we have3220, for 10cm (0.001), for 10cm 20cm , for 20cm r A r E r r Br r ⎧<⎪⎪-⎪=≤<⎨⎪⎪≥⎪⎩ with -1-137664(N C m )A ≈⋅⋅ and -12263.65(N C m )B ≈⋅⋅.From these expressions we can get the value of the field magnitude at any positions. For example,3976(N/C)E ≈ at 10cm r =, 6591(N/C)E ≈ at 20cm r =, and 2929(N/C)E ≈ at30cm r =. Also we can find30.002(1)0dE A dr r=+> and2240.006(1)0d E A dr r =-< for 10cm 20cm r ≤<. The dependence of E on the distance r is given in the following figure.0.000.050.100.150.200.250.30-100001000200030004000500060007000E =3976N/CE =6591N/CE (N /C )r (m)E =2929N/CChap 25/ 8PA nonconducting sphere of radius R . The electric field inside the sphere 30()4qr E r Rπε=. TakeV =0 at the center. V (r )=? Potential difference between the surface and the center? Which point has a higher potential if q is positive?[Solution](a) Suppose point P located inside the sphere at a distance r from the center O . Set V=0 at the center, the potential at P is given by()P OV r E ds =-⋅⎰Since Eis directed radially, we choose a pathalong the field lines from O to P as shown in the diagram. The differential displacement dsof this path is parallel to the electric field(here we have assumed q is positive) and has length dr ’, so the integral is evaluated as22''333'0'0000'()(')''4428r rr rr r qr q r qr V r E r dr dr R R Rπεπεπε=====-=-=-=-⎰⎰If the charge q is negative, it is easy to find that this result still holds.(b) On the surface, the potential is0()8q V R Rπε=-This is also the difference in potential between the point on the surface and the center.(c) If q is positive, the potential on the surface is negative, which means the potential at the center is higher. We can also get this conclusion according to the positive work done by the electric field when a positive charge is moved from O to P .Chap 25/ 9PThe same sphere as that in 8P, but take V =0 at infinity. [Solution]Since the potential is set as zero at infinity, thepotential at point P is now given byinfinity()PV r E ds =-⋅⎰According the formula of the field due to auniformly-charged sphere, the electric field Eat adistance r ’ from the center is directed radially and has magnitude2300', for ' (inside the sphere)4('), for ' (outside the sphere)4'qr r R R E r q r R r πεπε⎧<⎪⎪=⎨⎪>⎪⎩We choose a path along the field lines from infinity to point P so that the differential displacement dslocated at r ’ is antiparallel to the field and has length -dr ’(Note here '0dr < since 'r decreases from ∞ to point P). The potential at P can be calculated as2''''3''''00''223300000''22233000'()(')'(')'''44''()4'84883(3)888r Rr rr Rr rr r Rr r R r R r rr r Rq qr V r E r dr E r dr dr dr R r qqr qqr qr R R R Rq qr q R r R R Rπεπεπεπεπεπεπεπεπεπε=====∞==∞====∞==-⋅-⋅=-⋅-=+-=--=-=-⎰⎰⎰⎰The difference between the center and the surface is given by0003()(0)488q q q V r R V r RRRπεπεπε=-==-=-The value of the potential at P is different from the result in 8P since different potential reference is chosen, while the potential difference is independent of the location of the zero-potential point.V =。
Eaton AT-S+ 2自动测试系统用户指南说明书
© 2014 Eaton, All Rights Reserved.
5
Planning example for AT-S+
3rd floor
Planning example for AT-S+ with only one riser
UVA
AT-S+ SU1
DLS
2nd floor
RV-E30
UVA
© 2014 Eaton, All Rights Reserved.
Modules – Control unit CU S+
Control unit CU S+
• Graphic display • Foil keyboard • Two supplementary signal contacts • Access to servicing menu via special service SD card • Displaced SD card slot • Downwards-compatible to ST-S control unit • New LED / button designations • Web connection not directly accessible • Energy buffer
© 2014 Eaton, All Rights Reserved.
10
Modules – Control unit CU S+
Technical data CU S+
Mechanic Dimensions (WxHxD) Installation Degree of protection Ambient temperature Relative humidity Allowed degree of pollution Input voltage (Mains) Permissible mains frequency Power consumption Dimensions (WxHxD) Relayouputsoops S1S2 / S3S4
纹理物体缺陷的视觉检测算法研究--优秀毕业论文
摘 要
在竞争激烈的工业自动化生产过程中,机器视觉对产品质量的把关起着举足 轻重的作用,机器视觉在缺陷检测技术方面的应用也逐渐普遍起来。与常规的检 测技术相比,自动化的视觉检测系统更加经济、快捷、高效与 安全。纹理物体在 工业生产中广泛存在,像用于半导体装配和封装底板和发光二极管,现代 化电子 系统中的印制电路板,以及纺织行业中的布匹和织物等都可认为是含有纹理特征 的物体。本论文主要致力于纹理物体的缺陷检测技术研究,为纹理物体的自动化 检测提供高效而可靠的检测算法。 纹理是描述图像内容的重要特征,纹理分析也已经被成功的应用与纹理分割 和纹理分类当中。本研究提出了一种基于纹理分析技术和参考比较方式的缺陷检 测算法。这种算法能容忍物体变形引起的图像配准误差,对纹理的影响也具有鲁 棒性。本算法旨在为检测出的缺陷区域提供丰富而重要的物理意义,如缺陷区域 的大小、形状、亮度对比度及空间分布等。同时,在参考图像可行的情况下,本 算法可用于同质纹理物体和非同质纹理物体的检测,对非纹理物体 的检测也可取 得不错的效果。 在整个检测过程中,我们采用了可调控金字塔的纹理分析和重构技术。与传 统的小波纹理分析技术不同,我们在小波域中加入处理物体变形和纹理影响的容 忍度控制算法,来实现容忍物体变形和对纹理影响鲁棒的目的。最后可调控金字 塔的重构保证了缺陷区域物理意义恢复的准确性。实验阶段,我们检测了一系列 具有实际应用价值的图像。实验结果表明 本文提出的纹理物体缺陷检测算法具有 高效性和易于实现性。 关键字: 缺陷检测;纹理;物体变形;可调控金字塔;重构
Keywords: defect detection, texture, object distortion, steerable pyramid, reconstruction
II
ssa编译原理
SSA(Static Single Assignment)是一种中间表示形式,用于编译器的优化和分析。
它基于一个简单的原则:每个变量在程序中只能被赋值一次。
这种形式可以方便地进行数据流分析、寄存器分配和优化等操作。
SSA形式的特点是每个变量都有唯一的定义点,而且所有使用该变量的地方都引用同一定义点。
这样就消除了变量的重定义和修改,使得程序的数据流更加清晰和可控。
编译器通常通过以下步骤将代码转换为SSA形式:1. 静态单赋值化(Static Single Assignment)在这一步中,编译器会遍历源代码,为每个变量插入额外的赋值语句,以确保每个变量只被赋值一次。
如果存在变量的重新赋值,就会生成一个新的变量来代表不同的版本。
例如,对于语句`x = x + 1`,编译器会将其转换为`x1 = x + 1`,其中`x1`是一个新的变量。
2. φ函数插入(Phi-function Insertion)在SSA形式中,当一个变量有多个前驱定义点时,需要使用φ函数(Phi-function)来选择正确的定义。
φ函数会根据控制流的不同路径选择合适的定义。
编译器会在需要的地方插入φ函数,以确保使用变量时能够根据不同的路径选择正确的定义。
3. 优化和分析一旦代码转换为SSA形式,编译器可以进行各种优化和分析,例如常量传播、复写传播、活跃变量分析等。
这些操作基于SSA形式的特性,可以更准确地分析数据流和控制流,从而进行更有效的优化。
4. 寄存器分配在SSA形式中,每个变量只有一个定义点,因此寄存器分配变得更加简单。
编译器可以根据变量的使用情况和生命周期,选择适当的寄存器进行分配。
总之,SSA形式是一种对程序进行中间表示的方法,通过消除变量的重定义和修改,使得编译器可以更准确地进行优化和分析。
它在编译器领域扮演着重要的角色,并广泛应用于现代编译器的优化阶段。
linear sum assignment 算法 -回复
linear sum assignment 算法-回复什么是linear sum assignment 算法?Linear sum assignment 算法,也称为匈牙利算法或者Kuhn–Munkres 算法,是一种解决线性和分配问题(Linear Sum Assignment Problem)的优化算法。
线性和分配问题是指给定一个n x n的代价矩阵,任务是将其中的n个任务分配给n个执行者,并使得总的分配代价最小化。
这个问题在实际生活中有很多应用,比如任务分配、人员调度以及机器分配等。
接下来,我们来了解linear sum assignment算法的实现过程。
1. 初始化代价矩阵首先,我们需要创建一个n x n的代价矩阵,其中的元素表示每个任务分配给每个执行者的代价。
如果该任务与该执行者不兼容,可以设置一个较大的代价(比如无穷大)。
这个代价矩阵可以根据具体问题进行定义和计算。
2. 行减最小值接下来,我们需要对代价矩阵进行行减最小值的操作。
对于每一行,我们找到该行的最小值,并将该行的所有元素都减去最小值。
这个操作能够确保每一行的最小元素变为0,并保持相对代价的大小顺序。
3. 列减最小值类似于行减最小值的操作,我们需要对代价矩阵进行列减最小值的操作。
对于每一列,我们找到该列的最小值,并将该列的所有元素都减去最小值。
这个操作能够确保每一列的最小元素变为0,并保持相对代价的大小顺序。
4. 找出最优解接下来,我们需要找出代价矩阵的最优解。
我们可以通过递归搜索的方法来找到最优解。
具体来说,我们从代价矩阵的左上角开始,沿着边缘上代价为0的路径,一直搜索到右下角。
如果找到一条路径,使得每一行和每一列都仅有一个代价为0的元素,我们就找到了一个可能的最优解。
5. 确认最优解在找到可能的最优解后,我们需要对其进行确认。
这一步是通过检查每一行和每一列中是否仅有一个代价为0的元素来实现的。
如果满足这个条件,则可以确认这个解是最优解。
二次指派问题 匈牙利算法
二次指派问题匈牙利算法
匈牙利算法(匈牙利算法)是一个经典的用于解决二分图最大匹配问题的算法。
它的基本思想是通过交替路径的方法来寻找增广路径,从而不断增加匹配的数量。
以下是匈牙利算法的步骤:
1. 初始化一个空的匹配,即所有顶点都不属于任何一个匹配;
2. 从左侧的一个未匹配顶点开始,尝试为其查找增广路径;
3. 在查找增广路径时,可以使用深度优先搜索(DFS)或广度优先搜索(BFS)等方法;
4. 如果找到增广路径,则将路径上的边进行反转,即原来是匹配边的变成非匹配边,原来是非匹配边的变成匹配边;
5. 如果没有找到增广路径,则说明已经得到了最大匹配,算法终止;
6. 重复步骤2至5,直到所有的顶点都被匹配。
匈牙利算法的时间复杂度为O(V*E),其中V是顶点的数量,E是边的数量。
在实际应用中,通过使用一些优化技巧,可以达到更高的效率。
需要注意的是,匈牙利算法只能解决二分图最大匹配问题,即图的边集可以分割成两个互不相交的子集,将每个子集的顶点与另一个子集的顶点相连。
对于一般的图,可能需要使用其他算法来解决最大匹配问题。
2014年新课标高考英语信息卷2含解析
泄露天机—2014高考信息卷(二)考试时间120分钟,满分150分。
第一部分:听力第一节(共5小题;每小题1.5分,满分7.5分)1. What is the weather like?A. Rainy.B. Sunny.C. Cloudy.2. Where are the speakers?A. In the store.B. At the school.C. At home.3. When must the cameras be returned?A. Before 8:00 am.B. Before 8:00 pm.C. By the next day.4. What can we know about the man according to the conversation?A. He doesn’t like high-speed travel.B. He is interested in high-speed travel.C. He asks the woman to travel by train.5. What did the man tell the woman to do?A. To see him this week.B. To call him on Tuesday.C. To go home and wait for his call.第二节(共15小题,每小题1.5分,满分22. 5分)听第6段材料,回答第6至8题。
6. What are the speakers talking about?A. About a piece of cake.B. About what a child is able to do.C. About the opinion of a driver’s test.7. What can we learn from the conversation?A. The man felt the test was easy.B. A child could have passed the test.C. The man wanted to have a piece of cake.8. What is the most probable relationship between the speakers?A. Friends.B. Teacher and student.C. Strangers.听第7段材料,回答第9至11题。
Assignment2
It is known that x A sin n t represents the complete solution to this system equation. (a) What are the physical meanings of the parameters n and ? Show that the
MECH 364
Assignment 2
Problems from the book: 2.2, 2.4, 2.6, 2.7, 2.8, 2.13, 2.16, 2.17, 2.22, 2.24 2.2 Consider the undamped, simple oscillator given by
Torsional Spring k
Mass m1
Mass m2 y
meq y keq
l1 l2
Figure P2.8
(a) A lumped-parameter mechanical system (b) An equivalent system.
First consider the lumped-parameter system shown in Figure P2.8(a). A light, yet rigid, beam is hinged at one end using a frictionless pivot and restrained using a torsional spring having torsional stiffness k, at the hinged end. Two point masses m1 and m2 are attached at distances l1 and l2, respectively, from the hinged end. Under static equilibrium conditions, the beam remains in a horizontal configuration. Suppose that the beam is excited by a small initial push and left to vibrate with angular displacement . As a result the mass m2 undergoes a lateral-displacement (up and down) vibration y. An equivalent vibratory system is shown in Figure P2.8(b). Here meq is an equivalent mass assumed to be present at the location of m2 and restrained by an equivalent linear spring there with stiffness keq. (a) Explain why gravity effects do not enter into the equations of motion of Figure P2.8, assuming that and y are measured from the static equilibrium configuration. (b) (c) Obtain expressions for meq and keq in terms of the parameters of the original system shown in Figure P2.8(a). What is the natural frequency of the system in Figure P2.8(a)? What is the natural frequency of the system in Figure P2.8(b)? Comment on these results. 2.13 The inverted pendulum may be used as a simple model in the stability studies of inherently unstable systems such as rockets. Consider an inverted pendulum of point mass m and length l, which is restrained at its pivot (assumed to be smooth) by a torsional spring of stiffness k. This arrangement is sketched in Figure P2.13.
Supporting-sentences
And he also shouted at his boss and at last he lost his job
第2页,共21页。
Bob was having trouble getting to sleep at night.
第5页,共21页。
Ⅱ Learning aims:
Get to know the types and functions of supporting sentences by reading and analyzing.
2. Get to know the strategies that can be used to write supporting sentences.
He went to see his doctor
who gave him some extra-strong sleeping pills. He slept so well after taking the pills.
He went work with a light heart. When his boss saw Bob in the office, he shouted at him angrily,
He went to see his doctor,
i“annyoodrudhesirshtodouoglcdetothrraisdvaeoidfathgheoeohtdraodmubaoloem.deenvtearlTyddhiasaeyna”ks,se.tthoethdeoctrtoerattmolednt of the Hheimw.ent to work with a light heart. doctor, he BHuetbBeogbanwtaossliantge.songs and talk with his colleagues.
ce2014基准函数
ce2014基准函数CEC 2014是一个经典的基准函数集合,用于评估和比较进化算法和优化算法的性能。
该函数集合由30个优化问题组成,涵盖了不同的优化特性,包括单峰优化、多峰优化和基准函数中的连续优化。
以下是CEC 2014基准函数的列表:1. Sphere Function2. Schwefel's Problem 1.23. Rosenbrock's Function4. Rastrigin's Function5. Griewank's Function6. Ackley's Function7. Weierstrass Function8. FEF8F29. FEF8F810. FEF8F611. Composition Function 112. Composition Function 213. Composition Function 314. Composition Function 415. Composition Function 516. Composition Function 617. Composition Function 718. Composition Function 819. Composition Function 920. Composition Function 1021. Composition Function 1122. Composition Function 1223. Composition Function 1324. Composition Function 1425. Composition Function 1526. Composition Function 1627. Composition Function 1728. Composition Function 1829. Composition Function 1930. Composition Function 20这些基准函数的具体数学表达式和性质,请参考相关文献或CEC 2014官方网站。
稀疏恢复和傅里叶采样
Accepted by . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Leslie A. Kolodziejski Chair, Department Committee on Graduate Students
2
Sparse Recovery and Fourier Sampling by Eric Price
Submitted to the Department of Electrical Engineering and Computer Science on August 26, 2013, in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Computer Science
Author . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Department of Electrical Engineering and Computer Science August 26, 2013
Certified by . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Piotr Indyk Professor Thesis Supervisor
434203574_1_Assignment1
Assignment1Due date:2014-10-17Question1:Search in the Appendices of the English textbook(or google)for the data of Youngs moduli(E),thermal expansion coefficients(α),and melting temperatures(T m)of the following materials,and then make a comparison and an explanation:Steel1040,Al2O3(99.9%pure),and PE(HDPE)Question2:The net potential energy between two adjacent ions E N may be represented by the expressionE N=−Ar+Br n(1)Calculate the bonding energy E0in terms of the parameters A,B,and n. Question3:The net potential energy between two adjacent ions E N may be represented by the expressionE N=−Cr+Dexp(−rρ)(2)In which r is the interionic separation and C,D,andρare constants whose values depend on the specific material.(a)Derive and expression for the bonding energy E0in terms of the equilibrium interionic separation r0and the constants D andρ(b)Derive another expression for E0in terms of r0,C andρQuestion4:Compute the percentage ionic character of the interatomic bond for each of the following compounds:MgO,CdS and FeO.Question5:If the atomic radius of aluminum is0.143nm,calculate the volume of its unit cell in cubic meters.Question6:Show that the atomic packing factor of HCP is0.74.Question7:Calculate the radius of an iridium atom,given that Ir has an FCC crystal structure,a density of22.4g/cm3,and an atomic weight of192.2g/mol.Alloy Atomic Weight(g/mol)Density(g/cm3)Atomic Radius(nm)A43.1 6.400.122B184.412.300.146C91.69.600.137Question8:Below are listed the atomic weight,density,and atomic radius for three hypothetical alloys.For each determine whether its crystal structure is FCC, BCC,or simple cubic and then justify your determination.*Bonus1:As we have learned,most metals possess three lattices,i.e.,BCC, FCC,and HCP.Why no metal holds simple cubic structure(exceptα-Po)?。
MATS自动转换开关说明书
MATSAutomatic transfer switches Intelligent design High reliabilityMATS Automatic Transfer SwitchContentsAutomatic transfer switchesProduct description Design graphical and operation theory Product guide Technical parameter Controller description State indication Delay setup Second terminal wiring description Controller technical data Demission (234671*********)Product type●CB class ;●PC class 。
MATS automatic transfer switchesMATS series automatic transfer switches is based on Eaton's MCCB MCB and disconnect switches. Mechanical andelectrical interlocking are build-in and make two source no paralleling. Three types of controllers can adapt complex applications.●Model intelligent and compact design;●Low noise and safe transfer with the engine drive mechanism;●Low consumption in running time;●Mechanical and electricalinterlocking make no paralleling in transfer process;●Undervoltage 、lack phase 、 Loss voltage sensor;automatic pick up;●CB class is based on breaker which has line protectionfunction; PC class is based on disconnect switch; ●Auto / Man;●Online time set;●Fire interlocking and LED indicating are standard;●Can not transfer when the emergency is error;●Condition indication.Product descriptionMATS automatic transfer switches2M ats Automatic Transfer Switches /electrical ContentsDescription PageMATS Automatic Transfer SwitchDesign graphical and operation theoryStandardGB 14048.11-2002《Low voltageswitches and automatic transferswitches》Operating environment● Temperature: -25℃~700C● Altitude: <2000m● Humidity: <95%● Pollution: 3● IP20Design graphicalOperation theory● The automatic transfer switchis made of controller and acouple of switches that makethe transfer between normaland emergency;● Auto and Manual model can beprogrammable on thefaceplate;● Normal and emergency sensorwhen choose the auto model.The power source will transfer to emergency when normal iserror and emergency is right;● When normal is ready, thepower source will transfer tonormal achieve pick upfunction. When choosemanual, the controller will turn off, use the handle to transferthe power source. This modelis used to repairing andabnormal states;● MATS Automatic TransferSwitch is build-infire-interlocking function. Whenfire-interlocking signal iseffective, the controller willopen both normal andemergency to cut the powersource. After single has beenremoved, you must re-plug itand reset the controller;● Normal and emergency stateindication;● MATSB is used for Utility-Generator application.3Mats Automatic Transfer Switches /electricalCB class PC class4MATS Automatic Transfer SwitchProduct guideMats Automatic Transfer Switches /electrical CB class product guideMCB based●Breaker: PL9, 6kA;●Trip curve: C type or D type;●Amperes: 0.5, 1, 2, 3, 4, 6, 10, 13, 16, 20, 25, 32, 40, 50, 63A;●Controller: A type only.NoteMCCB based●Controller: A, B, C;●Break: LZM;●Inm: 1: 160A, 2: 250A, 3: 630A;●Break capability: C-36kA, N-50kA, S-70kA;●Protective unite: A: Thermomagnetic trip;AE: Electronic tripping device;●Amperes: 1: 20, 25, 32, 40, 50, 63, 80, 100, 125, 160A ;2: 160, 200, 250A ;3: 320, 400, 500, 630A 。
2014组合测试解析
两两组合掩盖测试举例
两两组合掩盖测试举例
一个网络信息系统如表所示,需要进展兼容性测试,假设需要进展完全的 测试,那么该系统需要 81 个测试用例,但是假设承受两两组合测试技术 进展测试,只需要 9 个测试用例。
由于每个参数都有不同的取值,使用贪心算法来选择适 宜的参数值添加到相应的测试用例中,当照旧有未被掩盖的 组合时,需要添加新的测试用例来掩盖这些漏掉的组合。
贪心算法
IPO 算法
美国北卡罗莱纳大学计算机系的 Y. Lei
和 提出一种基于参数挨次的渐进扩展的两两
组合掩盖测试数据生成方法。与 AETG 算法不
IPO 算法举例
IPO 算法举例
元启发式搜寻算法
贪心算法的根本思想是从小到大构造矩阵,直到全部的掩盖条件都得到 满足.另外一种查找最优的(或者较优的)掩盖数组的方法是,利用一个 已有的数组,通过适宜的变换得到一个更优的掩盖矩阵.这样,通过逐 次变换得到较优的矩阵.为了避开陷入局部最优,算法承受了一些启发 式的策略,通过多样化或变异的方法跳出局部最优点.
组合测试其根本思想是从统计学的角度对测试试验进展设计。该方法设计的 测试用例以参数及参数之间的交互为掩盖标准,虽然不能够进展完全的测试, 但是利用该方法进展的测试得到的结果能够在确定程度上反映系统的内部规 律,具有确定的代表性。
这种方法对由系统输入、外部大事等相互作用而产生的故障系统,具有较强 的检错力气。
类似算法有:CAST,TCG,DDA,PSSA等。
基于one-test-at-a-time策略的算法
设某个被测系统有3个因素〔A、B、C〕,因素A有2个值〔A1和A2〕,因 素B有2个值〔B1和B2〕,因素C有3个值〔C1、C2、C3〕。
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ECON3206/5206, Financial Econometrics, 2014 S2 Group Assignment (10% + 2% + 5% in Total)1.This group assignment has three components. Of the total 17%, 10% is for the assignmentitself, 2% for the teamwork peer review, and 5% for an individually written piece entitled Reflection on Teamwork. These are detailed as the points/pages below. Please read thisdocument thoroughly (pages 0-6).2.This assignment must be completed in a group of 3 or 4 students. The members of a groupcome from the same tutorial class. If you are unaware of what group you are in, please email your tutor.3.Each group must submit one hard copy of the assignment and one online (soft) copy.4.The online submission deadline is 11:59pm, Monday, 20/October/2014. Each groupMUST submit the online copy to Turnitin on Moodle. A link to the assignment submission will be available prior to this date. Only one member of the group should make thesubmission – that person must be nominated on the Group Cover Sheet.5.tutorial .6.For both online and hard-copy submission, the assignment Cover Sheet must be properlyfilled, which includes the class number, group number, names and student numbers of the group members.7. A late-submission penalty of 20% will apply for each 24 hours late. This applies to both thehard copy and the soft copy submissions.8.For hard-copy submission, DO NOT use plastic sheets or binders. Simply staple the pagestogether. Do not submit loose-leaf sheets of paper.9.Each student MUST complete a confidential online self and peer assessment by11.59 pm,Friday, 24/October/2014. No marks will be awarded for this component, which is worth 2%, if the self and peer assessment is not completed by the deadline.10.Each student MUST submit a soft copy and a hard copy of the written piece Reflection onTeamwork (no more than 500 words) with the same deadlines as in Item 4 and 5. The soft copy must be submitted online to Turnitin (a separate link will be available). The hard copy must be submitted to your tutor at the beginning of your Week 12 tutorial.11.All submissions will be checked for plagiarism. The University regards plagiarism as a formof academic misconduct, and has very strict rules regarding plagiarism. Where it can beestablished that individual students are responsible for the plagiarism, those individualstudents will be penalised. However, where it is judged that the plagiarism should have been clear to the other group members, the penalty will apply to all members of the group.THE UNIVERSITY OF NEW SOUTH WALESSchool of EconomicsECON3206/5206, Financial Econometrics, 2014 S2Group AssignmentGroup Cover SheetCheck-list1.Choose one member to submit the assignment: one soft-copy to be submitted online and onehard-copy to be submitted to your tutor.2.Class number, group number, all names and student numbers of the group must be filled in onthis Cover Sheet.3.Each group member must complete the online self and peer assessment. Marks on teamworkwill be zero if this assessment is not completed by the deadline. Details about the teamworkassessment will be announced shortly (see Teamwork Assessment section at the end of thisdocument).Class: ________ Group: __RequirementsYou need to use EViews to complete this assignment, see Project Description on Page 2. The relevant topics mainly include descriptive statistics (Slides-01), ARMA models (Slides-03, 04), and ARCH/GARCH models (Slides-06, 07, 08). The examples in the lecture slides and tutorial exercises are useful and may give you hints for completing the assignment. In your analysis, you should exploit the models, techniques and principles covered in the lectures and tutorials. Your answers should be justified either statistically or economically.This is a group assignment. Groups have been assigned (see the spreadsheets in “assignment Groups”). All members in a group should make their best efforts to cooperate toward the completion and perfection of the assignment. The assignment should be clearly typed. Answers, results, tables and charts should be presented in the relevant places in the assignment, which should be arranged according to the order of the questions. The length of the assignment should not exceed 9 pages (exclusive of the cover page but inclusive of all text, tables and charts).SubmissionEach group must submit one hardcopy AND one electronic copy of their assignment. The electronic copy must be submitted to the course website by 11:59pm, Monday, 20 October 2014. Instructions will soon be available on the website.Each group must also submit one hardcopy of the assignment to their tutor at the beginning of the tutorial time in Week 12. The names, IDs and signatures of the group members must be on the cover page. The assignment should be properly stapled. Do not use plastic sheets or binders.All electronic copies will be checked for plagiarism on the Turnitin software. Please also see notes on Plagiarism in the course outline.DataThis assignment is based on the data in the Excel file DJdata-Assign.xls. There are three columns in the file: Date, Price and iRate, whereiRate is the interest rate on the 3-month US treasury bill (in annualised percentages),Price is the Dow Jones Composite Average (DJCA) index, andDate is the date when the Price and iRate were recorded.The observations are weekly, recorded at the end of each Wednesday. The series span from 1954-01-13 to 2012-06-27 with a total of 3051 observations.Project DescriptionSuppose that you work for an investment firm now, which is 2012-06-27. The firm intends toinvest in the US market and is interested in a portfolio that exactly resembles the compositionof the DJCA index. As a quantitative analyst in the firm, you are assigned the task of analysing the DJCA return series, defined by r =100×ln (P /P ), where P is the Price series. In particular, the risk management division of the firm has posted the followingquestions to you.1. [2 marks] What are the main characteristics of the DJCA returns? What are the maincharacteristics of the excess returns?12. [3 marks] Can you build an adequate model for the DJCA return series, whichcaptures the dependence structure in the return series?3. [1 mark] Return volatilities tend to be “persistent”. To what degree is the DJCA returnvolatility persistent?4. [1 mark] According to folklores, negative returns tend to cause more volatility thanpositive ones. Does this hold for the DJCA return series?5. [1 mark] By the risk return trade-off logic, investing in a portfolio at a more volatiletime should be rewarded by a higher expected return. Does this hold for the DJCAreturns?6. [2 marks] At the end of the sample, 2012-06-27, assuming the DJCA portfolio isvalued at $1m, what is the 1-week ahead value at risk at the 99% level?Further, your supervisor points out that there is a structural break in the volatility level withthe break point at the 821th observation (1969-10-01).You need to impress the risk management division and your supervisor by carryingout an in-depth analysis and writing a nicely-presented report that thoroughly answers thequestions. Of course, you know exactly how to find your answers and how to justify youranswers, because you did a fantastic course called Financial Econometrics back in yourUNSW days.1Here, “excess return” = “DJCA return”–“risk free interest rate”, the latter is approximated by the 3-month treasury bill rate. Pay attention to the fact that the treasury bill rate is in annual percentages.HintsRegarding VaRIn question 6, to evaluate the 1-week ahead VaR, you need y | =E(y |Ω ) and σ =Var(y |Ω ), where T =3051 or 2012-06-27, from your preferred model (see Tutorial 9, sample answers and Slides-08 page 16). Once the model is estimated, these two objects can be calculated from the estimation results.For example, suppose my model (which is not necessarily your model!) is the AR(1)-GARCHy =c +ϕ y +ε , ε |Ω ∼N(0,σ ), σ =α +α ε +β σ . Then, from EViews estimation results, I can see all parameter estimates (c,ϕ ,α ,α ,β ) as well as the residual ε , the standardised residual v , and the conditional variance σ for allt ≤T . Therefore, I will be able to compute y | =c +ϕ y and σ =α +α ε +β σ as required.Teamwork Peer Review (2 marks)The Group Assignment provides an environment for students to cooperate with one another to bring out the best of a group. Students in a group are expected to make effective and positive contributions to the project. More specifically, students are expected to (and likely expect their team members to)a)contribute positively to task planning and coordination;b)help to raise the team morale;c)attend meetings punctually and be well prepared;d)communicate/discuss with team members constructively;e)share ideas generously and be a ble t o b uild o n o ther’s i deas;f)resolve disagreements with team members respectfully;g)complete allocated tasks with quality and in time;h)provide support and help to team members;i)demonstrate good knowledge and skills for the project.At the completion of this project, each student is required to give the following confidential assessment for each teammate (including herself/himself) online:Please r ate t his p erson’s o verall c ontribution t o t he p roject.(0 = nil; 1 = very low; 2 = low; 3 = satisfactory; 4 = high; 5 = very high)Your assessment should be based on the criteria in the list a)-i) above. Each student, who completed this assessment, will receive a Teamwork mark that is a summary of the self and peer assessment. The details for this online assessment will be announced shortly.Reflection on Teamwork (Max 500 words)This is an individual written reflective piece on your experience of working in a team. It is to be submitted separately from the group assignment, with the same deadlines as the assignment.This task helps to develop important skills which employers look for in graduates, including the ability to honestly reflect on and analyse your own performance in order to improve your practice.In your reflection, you should discuss your group learning process, including your own contribution, the effectiveness of the group as a team, and what you have learned about teamwork as a result of completing the project.The following questions should help direct your reflection (you can, but do not have to, use these as your headings).YouHow d id y ou c ontribute t o t he t eamwork a nd t o t he g roup’s l earning?What were your strengths (and weaknesses) as a collaborative team member?What did you learn about yourself as a team player?The teamWhat were the strengths of your team? What worked well?What issues/problems (if any) did your team encounter? How did the team address them?How could this particular teamwork experience have been improved?Your learning about teamworkWhat did you learn about teamwork?What would you do differently or better next time you work in a team?。