...Assignment.Network Topology Paper Network topology is the arrangement of the various elements (links, nodes, etc.) of a computer network.[1][2] Essentially, it is the topological[3] structure of a network, and may be depicted physically or logically. Physical topology refers to the placement of the network's various components, including device location and cable installation, while logical topology shows how data flows within a network, regardless of its physical design. Distances between nodes, physical interconnections, transmission rates, and/or signal types may differ between two networks, yet their topologies may be identical. A good example is a local area network (LAN): Any given node in the LAN has one or more physical links to other devices in the network; graphically mapping these links results in a geometric shape that can be used to describe the physical topology of the network. Conversely, mapping the data flow between the components determines the logical topology of the network. Contents * 1 Topology * 1.1 Point-to-point * 1.2 Bus * 1.3 Star * 1.4 Ring * 1.5 Mesh * 1.6 Tree * 1.6.1 Advantages * 1.6.2 Disadvantages * 1.7 Hybrid * 1.8 Daisy chain * 2 Centralization * 3 Decentralization * 4 See also * 5 References * 6 External links Topology There are two basic categories of network topologies:[4] 1. Physical topologies 2. Logical topologies The shape of...
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... Network Topology Paper Hierarchical layout defines how the network topology is structured physically and logically. In physical topology all devices and components are placed in various locations depending on the layout. Logical topology is the flow of data throughout the layout. There are many variations of network topologies that can be implemented. The most common network topologies are star, bus, and ring. Star topology is setup with each host connected to a central hub with a point to point connection. All workstation and other devices are connected to the central hub or switch. That switch is the server and all devices connected to it are clients. All resources must be connected to a central device in order for it to be a star topology. Data that travels in the network passes through the central hub and causes it to act as a signal repeater. The advantages of a star topology are that it is easy to design and versatile in planting additional nodes. The major disadvantage of the star topology is that the hub becomes the single point of failure. Bus topology utilizes a single bus cable to connect each computer and the data from the source travels in both directions to all devices on the bus cable until it finds the specific client. Data is only accepted if the addresses matched up and if it doesn’t match, it’s simply ignored. The last computer connected to the bus needs to be terminated to prevent echoes in the network. Having a single line for bus topology makes it...
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...Network Topology Paper Assignment 1 Unit 3 Mr. Swailes The study of Network Topology recognizes 8 basic topologies * Point to point * Star * Bus * Mesh * Tree * Hybrid * Daisy chain * Ring Point to point- The simplest topology is the permanent link between two end points Star-in a local area network with a star topology. Each network host is connected to a central hub with a point-to-point connection to a central node called a hub or switch. Bus- In local area networks where Bus Topology is used, each computer is connected to a single Bus cable. A signal from the source travels in both directions to all machines connected on the Bus cable until it finds the intended recipient. If the machine address does not match the intended address for the data, the machine ignores it. However, if the data matches it is accepted. The bad part of this topology is that it has a single point of failure which is the cable and if this happens the whole network can shut down. Mesh-is a fully connected network in which each node is connected to each other. A fully connected network does not need to use switching or broadcasting. However its major disadvantage is that the number of connections grows with the number of nodes. It is extremely impractical for large networks. A two node network is technically a fully connected network. Tree-is a combination of a Bus and Star topology. It is arranged to look like a tree. The advantages are it is scalable...
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...071-326-0501 Move as a Member of a Fire Team Conditions: In a designated position (other than team leader) in a moving fire team. Standards: React immediately to the fire team leader’s example. Perform the same actions as the fire team leader does in the designated position within the formation. Performance Steps 1. Fire team formations describe the relationship of the Soldiers in the fire team to each other. Standard fire team formations are the wedge (figure 071-326-0501-1), modified wedge (figure 071-326-0501-2), diamond (figure 071-326-0501-3), and file (figure 071-326-0501-4). a. Fire team wedge (figure 071-326-0501-1). This is the basic fire team formation which— (1) Is easy to control. (2) Is flexible. (3) Allows immediate fires in all directions. (4) Offers all round local security. Figure 071-326-0501-1. Fire team wedge STP 21-1-SMCT 18 June 2009 071-326-0501 3-165 Performance Steps b. Fire team modified wedge (figure 071-326-0501-2). When rough terrain, poor visibility, or other factors reduce control of the wedge formation, the sides are closed up to (almost) a single file. When moving in less rugged terrain and control becomes easier, resume your original positions. The modified wedge is also used for extended periods when traveling on roads or trails. The modified wedge— (1) Is easier to control in reduced visibility or rough terrain than are other formations. (2) Offers less flank security than a...
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...NRIC: 3087Z SECTION: G2 NRIC: 3087Z SECTION: G2 RESEARCH QUESTION The article focused on how consumers make decisions in remote and in-store environments. Remote environments are those where products cannot be physically examined and only descriptions (both visual and verbal) are available. An example of a remote environment is an online store such as Amazon.com. In-store environments are those where real products can be handled and touched. An example of an in-store environment is the local supermarket, Cold Storage. The rationale behind the research was that consumers’ preferences might differ based on the type of information they get in the two different environments. In remote environments, consumer must rely on information provided through the sense of vision. In in-store environments, consumers can use the sense of touch to add information to their purchase decisions. The research question addressed by this study was: What is the effect of the two different types of information – information gathered through the sense of touch and information gathered through vision – on consumer decision-making and preferences? INDEPENDENT, DEPENDENT AND MODERATING VARIABLES The independent variable in the main study was the type of environment: remote or in-store. The dependent variable in this study was the number of times a product was chosen. The moderating variable in this study was the product type. A typical product can be classified into 2 general categories: geometric...
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...Unit 6: Graph Theory - Assignment Total points for Assignment: 35 points. Assignments must be submitted as a Microsoft Word document and uploaded to the Dropbox for Unit 6. All Assignments are due by Tuesday at 11:59 PM ET of the assigned Unit. NOTE: Assignment problems should not be posted to the Discussion threads. Questions on the Assignment problems should be addressed to the instructor by sending an email or by attending office hours. You must show your work on all problems. If a problem is worth 2 points and you only show the answer, then you will receive only 1 point credit. If you use a calculator or online website, give the source and tell me exactly what you provided as input. For example, if you used Excel to compute 16 * 16, state “I typed =16*16 into Excel and got 256. You may type your answer right into this document. Part I. Basic Computations 1. (4 points) The plan for a four-room house is shown below. Draw a graph that models the connecting relationships between the areas in the floor plan. [Your graph does not [Your graph does not need to be fancy. You may use any drawing software such as Visio or Creatly.com] Answer: I used viso for graph[pic] 2. a. Identify all the vertices in the above graph with odd degree. Identify the degree of each of these vertices. (2 points) Answer: The odd number of edges is a odd degree vertices are D,E,F from the graph 3,1,3 it has a odd number. So D,E,F, is odd and the rest is...
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...Food Webs Report Week 5 MTH / 221 University of Phoenix Food Webs It may be difficult to know all the factors which determine an ecological niche, and some factors may be relatively unimportant. Hence it is useful to start with the concept of competition and try to find the minimum number of dimensions necessary for a phase space in which competition can be represented by niche overlap. One approach to this question is to consider the notion of the food web of an ecological community. Definition 1 A food web of an ecological community is a directed graph with a vertex for each species in the community and a directed edge from the vertex representing species A to the vertex representing species B if and only if A preys on B We can define competition using the food web. Two species compete if and only if they have a common prey. Thus, a Hawk and Wolf compete (since Rabbit is a common prey); Rabbit and Grasshopper compete, while Deer and Toad do not compete. We use this competition relation to define a graph called the competition graph. Definition 2 The competition graph of a food web is a simple graph with a vertex for each species. Two vertices are joined by an edge if and only if the species they represent have a common prey. Definition 3 A graph is an intersection graph for a family of sets if each vertex is assigned a set in such a way that two vertices are joined by an edge if and only if the corresponding sets have non-empty...
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...MTH 333 Writing Project Abdullah Aurko MTH 333 Writing Project Topic- Connectedness: A topological property I am working on concept of connectedness of a topological space as the topic for my writing project. I wish to proceed by giving a formal definition of the concept with proof and proving that connectedness is a topological property. This will be followed up with the proofs of several interesting and essential theorems involving connectedness of a topological space and how it enables us to distinguish between different topological spaces. A topological space X is disconnected if X=A B, where A and B are disjoint, nonempty, open subsets of X. (Roseman, 1999) Definition A topological space X is connected if it is not disconnected. Examples 1. A closed interval [a,b] is connected. To show this, suppose that it was disconnected. Then there are two nonempty disjoint open sets and whose union is [a,b]. Let X be the set equal to A or B and which does not contain the point b. Let s= sup X, the supremum of X. Since X does not contain b, s must be within the interval [a,b] and thus must be within either X or [a,b]\X. If s is within X, then by the definition of supremum there is an open set (s-ε, s+ε) within X. If s is not within X, then is within [a,b]\X, which is also open, and there is an open set (s-ε, s+ε) within [a,b]\X. Either case implies that s is not the supremum. 2. The topological space X=(0,1)\{1/2) is disconnected: A=(0,1/2), B=(1/2,1) (MathWorld) Lemma The following...
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...geometry and combinatorial geometry are branches of geometry that study combinatorial properties and constructive methods of discrete geometric objects. Most questions in discrete geometry involve finite or discrete sets of basic geometric objects, such as points, lines, planes, circles, spheres, polygons, and so forth. The subject focuses on the combinatorial properties of these objects, such as how they intersect one another, or how they may be arranged to cover a larger object. Discrete geometry has large overlap with convex geometry and computational geometry, and is closely related to subjects such as finite geometry, combinatorial optimization, digital geometry, discrete differential geometry, geometric graph theory,toric geometry, and combinatorial topology. Although polyhedra and tessellations have been studied for many years by people such as Kepler and Cauchy, modern discrete geometry has its origins in the late 19th century. Early topics studied were: the density of circle packings by Thue,projective configurations by Reye and Steinitz, the geometry of numbers by Minkowski, and map colourings by Tait, Heawood, and Hadwiger. László Fejes Tóth, H.S.M. Coxeter and Paul Erdős, laid the foundations of discrete geometry. "This is an introduction to the field of discrete geometry understood as the investigation of combinatorial properties of configurations of (usually finitely many) geometric objects … . The book is written in a lively and stimulating but very precise style and contains many...
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...Sample Exam 2 - MATH 321 Problem 1. Change the order of integration and evaluate. (a) (b) 2 0 1 0 1 (x y/2 + y)2 dxdy. + y 3 x) dxdy. 1 0 0 x 0 y 1 (x2 y 1/2 Problem 2. (a) Sketch the region for the integral f (x, y, z) dzdydx. (b) Write the integral with the integration order dxdydz. THE FUNCTION f IS NOT GIVEN, SO THAT NO EVALUATION IS REQUIRED. Problem 3. Evaluate e−x −y dxdy, where B consists of points B (x, y) satisfying x2 + y 2 ≤ 1 and y ≤ 0. − Problem 4. (a) Compute the integral of f along the path → if c − f (x, y, z) = x + y + yz and →(t) = (sin t, cos t, t), 0 ≤ t ≤ 2π. c → − → − → − (b) Find the work done by the force F (x, y) = (x2 − y 2 ) i + 2xy j in moving a particle counterclockwise around the square with corners (0, 0), (a, 0), (a, a), (0, a), a > 0. Problem 5. (a) Compute the integral of z 2 over the surface of the unit sphere. → → − − → − → − − F · d S , where F (x, y, z) = (x, y, −y) and S is → (b) Calculate S the cylindrical surface defined by x2 + y 2 = 1, 0 ≤ z ≤ 1, with normal pointing out of the cylinder. → − Problem 6. Let S be an oriented surface and C a closed curve → − bounding S . Verify the equality → − → − → → − − ( × F ) · dS = F ·ds − → → − if F is a gradient field. S C 2 2 1...
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...Exercises in Classical Real Analysis Themis Mitsis Contents Chapter 1. Numbers 5 Chapter 2. Sequences, Series and Limits 11 Chapter 3. Topology 23 Chapter 4. Measure and Integration 29 3 CHAPTER 1 Numbers E 1.1. Let a, b, c, d be rational numbers and x an irrational number such that cx + d 0. Prove that (ax + b)/(cx + d) is irrational if and only if ad bc. S. Suppose that (ax + b)/(cx + d) = p/q, where p, q ∈ Z. Then (aq − cp) x = d p − bq, and so we must have d p − bq = aq − cp = 0, since x is irrational. It follows that ad = bc. Conversely, if ad = bc then (ax + b)/(cx + d) = b/d ∈ Q. E 1.2. Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be real numbers. Prove that n i=1 ai n j= 1 b j ≤ n n ak bk k=1 and that equality obtains if and only if either a1 = an or b1 = bn . S. Since {ai }n=1 and {bi }n=1 are both increasing, we have i i n n 0≤ (ai − a j )(bi − b j ) = 2n ak bk − 2 ai 1≤i, j≤n k =1 i= 1 n j=1 b j . If we have equality then the above implies (ai − a j )(bi − b j ) = 0 for all i, j. In particular (a1 − an )(b1 − bn ) = 0, and so either a1 = an or b1 = bn . E 1.3. (a) If a1 , a2 , . . . , an are all positive, then n n 1 a ≥ n2 ...
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...Chapter 1 Overview of Statistics Chapter 2 Data Collection Assignment (32 points due by 11 pm September 30th) Note: You can team up with one of your classmates to complete the assignment (not more than two in a team); if you want to work on the assignment individually, that’s also fine. If you are working in teams, then only one submission is required per team; include both the team members’ last names as part of the assignment submission file name as well as in the assignment submission document. Please provide detailed solutions to the following problems/exercises (4 problems/exercises x 8 points each): 1) Which type of data (categorical, discrete numerical, continuous numerical) is each of the following variables? a) Number of spectators at a randomly chosen US Open tennis match. Continuous numerical. b) Water consumption (in liters) by a randomly chosen US Open player during a match. Discrete numerical. c) Gender of a randomly chosen tennis player in the US Open tennis tournament. Categorical 2) Which measurement level (nominal, ordinal, interval, ratio) is each of the following variables? a) Number of annual office visits by a particular Medicare subscriber. Ordinal b) Daily caffeine consumption by a six year old child. Interval c) Type of vehicle driven by a college student. Nominal 3) Would you use a sample or a census to measure each of the following? Why? a) The number of cans of Campbell’s soup on your local supermarket’s...
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...MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1 Generally, a “solution” is something that would be acceptable if turned in in the form presented here, although the solutions given are often close to minimal in this respect. A “solution (sketch)” is too sketchy to be considered a complete solution if turned in; varying amounts of detail would need to be filled in. Problem 1.1: If r ∈ Q \ {0} and x ∈ R \ Q, prove that r + x, rx ∈ Q. Solution: We prove this by contradiction. Let r ∈ Q\{0}, and suppose that r +x ∈ Q. Then, using the field properties of both R and Q, we have x = (r + x) − r ∈ Q. Thus x ∈ Q implies r + x ∈ Q. Similarly, if rx ∈ Q, then x = (rx)/r ∈ Q. (Here, in addition to the field properties of R and Q, we use r = 0.) Thus x ∈ Q implies rx ∈ Q. Problem 1.2: Prove that there is no x ∈ Q such that x2 = 12. Solution: We prove this by contradiction. Suppose there is x ∈ Q such that x2 = 12. Write x = m in lowest terms. Then x2 = 12 implies that m2 = 12n2 . n Since 3 divides 12n2 , it follows that 3 divides m2 . Since 3 is prime (and by unique factorization in Z), it follows that 3 divides m. Therefore 32 divides m2 = 12n2 . Since 32 does not divide 12, using again unique factorization in Z and the fact that 3 is prime, it follows that 3 divides n. We have proved that 3 divides both m and n, contradicting the assumption that the fraction m is in lowest terms. n Alternate solution (Sketch): If x ∈ Q satisfies x2 = 12, then x is in Q and satisfies 2 x 2 = 3. Now prove that...
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...Recall our main theorem about vector fields. Theorem. Let R be an open region in E2 and let F be a C1 vector field on R. The following statements about F are equivalent: (1) There is a differentiable function f : R → R such that ∇f = F. (2) If C is a piecewise C1 path in R, then C F · dx depends only on the endpoints of C. (3) C F · dx = 0 for every piecewise C1 simple, closed curve in R. Furthermore, statements (1)–(3) imply (4) curl F = 0, and (4) implies (1)–(3) when R is simply connected (so all four are equivalent when R is simply connected). The purpose of this handout is to explore what happens when R is not simply connected, that is, when there exist non-gradient vector fields with zero curl. An important example Consider the vector field F = −y x2 + y2 i + x x2 + y2 j defined on R = R2\{(0, 0)}, that is, on all of R2 except the origin. Letting P = −y/(x2+y2) and Q = x/(x2 + y2), it is a simple matter to show that ∂Q ∂x = ∂P ∂y , and so curlF = 0. We can show that (1)–(3) are false for F by finding a simple, closed curve C for which C F · dx = 0. Let C be the unit circle parameterized counterclockwise by x = cost, y = sin t, 0 ≤ t ≤ 2π. We have C F · dx = C −y x2 + y2 dx + x x2 + y2 dy = 2π 0 − sin t 1 (− sin t) dt + cost 1 cost dt = 2π 0 dt = 2π. It follows from the theorem that F is not the gradient of any...
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...Math 245C Homework 1 Yunbai Cao 904066974 Apr 11, 2014 Exercise 1.10.3 Proof. Since R equipped with the usual topology F is Hausdorff, (R, F ) is Haudorff as F is stronger than F. Given any x ∈ R, for any y ∈ R\{x}, there exists By open such that By ∩ Byx = ∅. Then R\{x} = is open. Let K = [0, 1]\Q and L = { 1 }. Then K c = (−∞, 0) ∪ (1, ∞) ∪ Q ∈ F , so K 2 is closed. And L is closed by previous paragraph. Let U ⊃ K, V ⊃ L be two open neighbourhoods of K, L. Claim: U ∩ V = ∅. As F is generated by open intervals of R and {Q}, there exists 1 2 y∈R\{x} By y open and Byx x is open. Therefore {x} ∈ V ⊂ V where V is a finite intersection of open intervals (thus an open interval) and {Q}. So L ⊂ I ∩ Q ⊂ V where I is an open interval. On the other hand, as K ∩ Q = ∅, we may assume U ∈ F ( since we can always find an open subset U of U such that K ⊂ U ∈ F). As I 1 2 is an open 1 2 interval, U ∩ I = ∅ since otherwise there will be some irrational number close to in (R, F). Thus U ∩ V = ∅ as claimed. Therefore (R, F ) is not normal. that is not contained in U . Therefore U ∩ I ∩ Q = ∅ as U ∩ I ∈ F and Q is dense Exercise 1.10.6 Proof. Since X is locally compact, for any x ∈ K there exists Ux that U x is compact. Then K ⊂ cover of K, we may write K ⊂ then O = n i=1 U i x∈K Ux . Since K is compact, there n i=1 Ui with Ui ∈ {Ux : x ∈ K}. Let x open such exists a finite O= n i=1 Ui , is compact since finite union of compact sets is compact (Exercise 1.8...
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