Connection and costs distribution problems

Size: px
Start display at page:

Download "Connection and costs distribution problems"

Transcription

1 MaMaEuSch Management Mathematics for European Schools mamaeusch Connection and costs distribution problems Federico Perea Justo Puerto MaMaEuSch Management Mathematics for European Schools CP DE - COMENIUS - C21 University of Seville This project has been carried out with the partial support of the European Union in the framework of the Sokrates programme. The content does not necessarily reflect the position of the European Union, nor does it involve any responsibility on the part of the European Union. 0

2 1 Previous concepts 1.1 What is a graph? In this work we describe some situations which can be represented by graphs. In order to investigate such situations we do not need to study graphs in detail. We start by introducing the idea of a graph through some examples Examples Route Maps The diagram in figure 1 is a map of what will be the Sevilla Underground. Like all maps, it does not represent every feature of the city in question, but only those of relevance to the people who use it. In the case of that map, the exact geographical locations of the stations are unimportant. What is important, however, is the way in which the various stations are interconnected, so that a passenger can plan a route from one station to another. The map is simply a diagrammatic way of indicating how the stations are interconnected. Architectural Floor Plans The plan of the lower floor of a house is represented in figure 2. For small plans like this, such a diagram is very convenient for showing which rooms have mutual access, but for large plans a less cumbersome representation is useful. One such representation is to draw the rooms as small solid circles as shown in figure 3. Such diagrams are known to architects as circulation diagrams, because of their use in analyzing the movements of people in large buildings. In particular, they have been used in the designing of airports, and in planning the layout of supermarkets. Such diagrams are useful in representing the connection between the various rooms, but they do not give us any information about the size or shape of the rooms The definition of a graph The common feature in all the preceding examples is that in each case we have a system of objects which are interrelated in some way. In the first example the objects are stations interconnected by rails and in the second example they are rooms with mutual access. In each case we can draw a diagram in which the objects are represented by points and the interconnections are represented by lines. Such a diagram is called a graph. The points representing the objects are called vertices (also called nodes or points), and the lines representing the 1

3 Figure 1: Future Sevilla Underground. interconnections are called edges (also called arcs or just lines). For example, the circulation diagram of the house is a graph with seven vertices (corresponding to the playroom, kitchen, hall, etc.) and ten edges (corresponding to the interconnections between these rooms). We can formalize these ideas as follows: Definition 1.1 A graph is a diagram consisting of points, called vertices, joined together by lines, called edges; each edge joins exactly two vertices. We also need to know two concepts: cycle and connection. We say that a graph G is connected if there is a path in G between any given pair of vertices. Both the example on the underground and the example on the lower floor of a house give us connected graphs. The graph shown in figure 4 is disconnected, that is, not connected. A path that connects a vertex to itself is called a cycle. In figure 3 the path 2

4 Figure 2: Lower floor. living room - sitting room - study - living room is a cycle, because it joins the living room to itself. Those concepts lead us to an entity we are using in this work: a tree network. A connected graph with contains no cycles is called a tree. Some examples are shown in figure 5. If G is a connected graph, then a spanning tree in G is a tree which includes every vertex of G. For example let us consider the graph shown in figure 6. The number of spanning trees in a graph can be very large (bounded above by 2 n 2, where n is the number of nodes that the network has). In figure 7 we present three possible spanning trees for the graph shown in figure Cooperative situations In many situations we find several agents that, in case they join their efforts, can either get a higher profit or have lower costs after their actions (for instance business) are taken. Those situations are called cooperative situations, because it is allowed for the agents to cooperate. Let us see one example Example Three friends, Justin, Paula and Malcom, are looking into the possibilities of starting a care center for elderly people from 60 to 80 years. After a thorough investigation of the local regulations and market, they come to the following assessment of the situation. They need 3

5 Figure 3: Graph. one nurse per four people from 71 to 80, and one nurse per 10 people from 60 to 70. For each person from 71 to 80 they are required to have 8 square meters indoors inside and 4 outdoors. For a person from 60 to 70 the requirements are 5 and 6 respectively. After calculating all costs they realize that they can make a net profit of 200 Euros per month per elderly man/woman from 71 to 80 and a net profit of 150 Euros per month per elderly man/woman from 60 to 70. Justin knows 9 people whom he can hire as nurses. He has the possibility of renting 260 square meters inside and 200 outside. Therefore he has to solve a linear programming problem to calculate the maximum profit that he can make on his own while complying with all regulations. If we call x 1 the number of elderly people from 71 to 80 and x 2 those that are from 60 to 70, the problem to solve in order to calculate the maximum profit that Justin can have is: 1 max 200x x 2 1 s.t. x x x 1 + 5x x 1 + 6x x 1 0, x For further details on linear programming see: H.W.Hamacher, E.Korn, R.Korn, S.Schwarze Production Planning and Linear Optimization in MaMaEuSch web site mamaeusch/allgemein e/frames2 e.html (2004). 4

6 Figure 4: Disconnected Graph. After doing that he can get a profit of 7000 Euros on his own by having 20 elderly people of each group. Paula knows 5 people whom she can hire as nurses and Malcom knows 14. Paula has the possibility of renting 120 square meters indoors and 200 outdoors, and Malcom can rent 590 square meters inside and 400 outside. When solving the linear programming problems for Paula and Malcom, we have that Paula can have a maximum profit of 3600 Euros if acting on her own, and Malcom can earn Euros. They can also decide to combine forces. We assume that there is no overlap between the people that they can hire as nurses. If Justin and Paula decide to work together they will have a profit of Euros. If Justin and Malcom worked together they would get a profit of Euros. Paula and Malcom would have a profit of And, if all three of them worked together they would get a profit of Euros. That situation can be summarized in the following table: S v(s) S v(s) S v(s) S v(s) {1} 7000 {3} {1,3} {1,2,3} {2} 3600 {1,2} {2,3} where we denote Justin by number 1, Paula by number 2 and Malcom by number 3, and v(s) is the maximum profit that the group S can get by acting on its own. 5

7 Figure 5: Some trees. Their total profit is maximized if they combine forces. In this case they will have to figure out a way to divide the profit of Euros. 2 Connection and costs distribution problems In many daily situations we face up to optimization problems. It normally happens that, when several agents (people, companies,etc.) join their efforts in order to carry out an action in common which will serve all of them, the cost generated is lower than the cost that would be generated without acting allied. It is clear that they will join efforts, because they will spend less money acting together and they will get the same service. But then a new problem arises: how to allocate the costs generated by the common action. Let us imagine that several farmers are going to build a wire fence to isolate their farmlands. Since those fences delimit two land properties, they are used by more than one farmer. Therefore it seems logical to install the fences between them, and it seems illogical that they build the fences individually. But the time to pay approaches and, consequently, problems 6

8 Figure 6: This graph is not a tree. arises. In many similar situations it is agreed to pay the whole total cost in equal shares (from now on, this rule will be called proportional rule), and it is even thought to be the fairest way. Through some examples, we will see that the proportional rule can be unfair, taking into account the problems that the word fairness can bring on. 3 On cost allocation in connection problems The first situation we deal with belongs to the class of minimum spanning tree problems. In these problems we have several users (located in one node of a graph) who want to benefit by a product coming from a fixed source (could be water, electricity, petrol, etc.). We want to connect every user to the source. The network connection is not fixed, that is, the nodes can be connected in any possible way. Example 3.1 Let us consider a group of villages, each one of them needs to be connected to a reservoir directly or through other villages. Each possible connection has an associated cost and the problem is how to connect all the villages to the reservoir so that the total expense generated is as low as possible. The network with the minimum cost associated is called minimum spanning tree. Nevertheless, the construction of the minimum spanning tree is only one part of the problem. Besides minimizing the total expenses, a problem about the cost allocation must be considered, 7

9 Figure 7: Some trees. that is, it must be decided how much each village should pay for the network. In the following examples we will propose several allocation rules and we will discuss the idea of fair allocation. Formally, a minimum spanning tree problem is a 3-upla T = (N,, t), where N = {1,..., n} is the set of nodes (villages, in the example above), is the source (the reservoir) and t : E N R + is the nonnegative costs function that assigns a cost to each arc (how much it would cost building a pipe). E S is defined as the subset of all arcs between pairs of nodes of S N = N, so (S, E S ) is the complete graph over S. E S = {{i, j} i, j S, i j}. Since the connection costs are nonnegative, it is obvious that a graph at minimum cost connecting every node to the source is a tree itself, and that explains the name of the problem minimum spanning tree. Given a minimum spanning tree problem T = (N,, t), the Bird allocation β R (T ) is built by assigning to each point i N the cost of the first arc in the path from the node i to the source. The computation of this allocation can be taken from Prim s algorithm, which 8

10 builds a minimum spanning tree, starting from the fixed root, by adding the arcs at lowest price, without making loops (paths that connect a point to itself). Let us see an example. In the case shown in figure 8, we want to connect three nodes; A, B and C to a source (S) located in the red point at minimum cost. Figure 8: Example of Prim algorithm. We first join the source to its nearest node, in this example it is B. After that we find which is the nearest node to B or the source, and it results A. A is nearer to B than to the source, so we connect A with B. To finish we have to connect C to the graph. To get the minimum spanning tree we have to do it at minimum cost, that is to say, we connect C to its nearest node between those that are already in the graph. In the example, the nearest node to C is the source, so we build the arc joining C to the source. The result is the fourth picture of figure 8, the minimum spanning tree of the nodes shown in the first picture. Now we have to face the other problem: how to allocate the expenses generated after building the minimum spanning tree. To do that we follow Bird s Rule. Let us see how the rule works (Bird s Rule). Input: a minimum spanning tree problem (N,, t). Output: a set of arcs R E N and the corresponding Bird allocation β R (T ). 1. Choose the source as the root. 9

11 2. Initialize R =. 3. Find an arc at minimum cost e = {i, j} E N \ R incident in, or in any of the nodes of an arc of R, in such a way that the inclusion of e in R does not create any loop. 4. One of the nodes i, j, let us say j, was connected previously to the source and the other one i is a node that was not connected to the source yet. Let us assign the cost β R i (T ) = t(e) to the node i. 5. We join e to R. 6. If there are some nodes that have not been connected to the root yet in the graph (N, R), go back to the step number 3. The next example helps us to understand these rule. Example 3.2 Let us consider the minimum spanning tree problem T, with N = {1, 2, 3} as shown in the figure 9, where the numbers over the arcs represent the costs associated to each arc. Figure 9: Minimum spanning tree problem T. When we apply the algorithm to this problem, the first arc we join to R could be {, 1} or {, 3} (these are the arcs at minimum cost starting from the source). Let us suppose that we choose the first one, then the cost β R 1 (T ) = 10. Later, following the Bird s algorithm, we add {1, 2} to R, and we get β R 2 (T ) = 6. To finish, we add {2, 3} to R, and it results β R 3 (T ) = 5. 10

12 This leads us to the following cost allocation: (10, 6, 5). On the other hand, if we started from {, 3}, we would finally obtain the cost allocation β R (T ) = (6, 5, 10). Both of these minimum spanning trees are represented in figure 10. Figure 10: Two minimum spanning trees. That is to say, in the first solution we have that, according to the Bird s rule, the node 1 has to pay 10 units, the second node 6 and the node number 3 has to pay 5. Analogously, in the second solution the node have to pay 6, 5 and 10 units respectively. 4 Application Spain is divided into 17 different regions, and one of them is called Andalusia. It is divided into 8 provinces, and in each one of them there is a capital. The geographical map of this region, where the eight capitals are denoted by a black dot, is shown in figure 11. We present the following cases as application of minimum spanning tree: 4.1 Case 1: connection without source Let us suppose that a telecommunication company wants to connect the eight capitals by a optical fibre cable. The distances between the cities are given in the following table: 11

13 Figure 11: Andalusia. Almería Cádiz Córdoba Granada Huelva Jaén Málaga Sevilla Almería Cádiz Córdoba Granada Huelva Jaén Málaga Sevilla Build the minimum spanning tree for this case and allocate how much each capital has to pay for its construction. 12

14 4.2 Case 2: connection with source Let us assume that the eight capitals must be connected to each other and to an energy source located in the city of Málaga. Build the minimum spanning tree for this case and decide how much each city has to pay taking into account that the city of Málaga should not pay anything for the network. 4.3 Solutions In the first case we do not have any source, so the first arc will be one that joins the nearest two cities to one another, in this case Sevilla and Huelva. After that we will join to the network the nearest city to Sevilla or Huelva choosing that city between the six ones that are not in the network yet. In our case Cádiz is the nearest to one of those two, and it is closer to Sevilla than to Huelva. So, the following arc to be in the network is Sevilla-Cádiz. We proceed this way with all the other cities and we get the minimum spanning tree made up by: Sevilla, Huelva, 94 Sevilla, Cádiz, 125 Sevilla, Córdoba, 138 Jaén, Córdoba, 104 Jaén, Granada, 99 Málaga, Granada, 129 Granada, Almería, 166 total distance, 855 For the case when we have a fixed source, the solution is the same and the only thing that changes is the order in which the arcs appear: Málaga, Granada, 129 Jaén, Granada, 99 Jaén, Córdoba, 104 Sevilla, Córdoba, 138 Sevilla, Huelva, 94 Sevilla, Cádiz, 125 Granada, Almería, 166 total distance,

15 The connection at minimum cost is shown in the figure 12. Figure 12: Minimum spanning tree. 4.4 Costs allocation In the example 1, if we suppose that each Km of the network costs Euros and the overall cost of the network must be paid by the cities, how much should each city pay? In the second example, if we suppose that Málaga paid the construction of the source that will supply the other cities and, because of that, it should not pay anything for the network, how much should each one of the other cities pay? 4.5 Allocation after the connection without source In the first case, since we do not have any source, we can not use Bird s rule. Many rules to allocate the expenses can be proposed and we will apply the next ones: Proportional allocation: we divide the total expense( Euros) between the number of cities connected by the network (8). Each city pays the same, ( Euros). 14

16 Each city pays half of the price of the arcs that finish or begin in it, for instance, the connection between Sevilla-Huelva is paid by both cities in equal shares. 4.6 Allocation after the connection with source In the second case we do have a fixed source, Málaga. That is, in this case we can apply Bird s rule. Before then let us see how the proportional allocation works and the problems that it causes, and finally we will propose a third way of allocating costs. Proportional allocation. The total expense of the network amounted to , and because of that each of the seven cities connected to Málaga has to pay = Euros. Is that the 7 fairest way? Let us see what happens if Granada and Jaén decide to make their own network on their own, that is, without joining the rest. Applying the algorithm to build a minimum spanning tree as before, we would conclude that the shortest connection between both cities and Málaga has a length of Km, that is to say, a total expense of Euros. If those two cities had to accept the proportional rule, they would pay between them Euros. Logically, these two cities do not agree with sharing the total expense in equal shares, since they can build a network by themselves that gives them the same service and at a lower price. This is an example where the proportional allocation does not make everyone happy. If when making an allocation we find subgroups of agents (cities in our case) that can connect themselves and pay less money, that allocation would probably be defeated. In Game Theory, those allocation that hold that there is no subgroup of players (in our case the players are the cities) that can spend less money by acting on their own, are called Core allocations. Let us see now the Bird s rule. Each city pays: Granada Euros. Jaén Euros. Córdoba Euros. Sevilla Euros. 15

17 Huelva Euros. Cádiz Euros. Almería Euros. (Each city pays for the arc that connects it to the network). One can check that after this allocation there is no subgroup of cities that can spend less money by acting on their own, that is, we would say in Game Theory that this allocation belongs to the core. It is well known that in every minimum spanning tree problem the Bird s rule gives us an allocation belonging to the core. As a conclusion we present another kind of allocation used in connection problems. An allocation in which each city pays the proportional part of the arcs that it uses (if an arc is used by several cities, they pay the cost of that arc between all of them in equal shares). Let us see how this allocation works: The arc Málaga-Granada is used by all the cities, so they divide its cost ( Euros) in seven parts. The arc Granada-Almería is only used by Almería, therefore this city bears the cost of it ( Euros). Jaén, Córdoba, Sevilla, Cádiz and Huelva use the arc Granada-Jaén, so each one of them pays Euros. The arc Jaén-Córdoba is used to connect to the source by Córdoba, Sevilla, Cádiz and Huelva. Therefore each of these cities should pay Euros. The connection Córdoba-Sevilla is used by Sevilla, Cádiz and Huelva. Because of that they have to pay / 3 Euros each one. The connection Sevilla-Cádiz is only used by Cádiz, so this city bears the cost of that arc ( Euros). Finally, the arc Sevilla-Huelva is only used by Huelva and therefore it has to pay the Euros for its construction. In conclusion, each city has to pay: Granada: Almería: = Jaén: =

18 Córdoba: = Sevilla: = Cádiz: = Huelva: = Although this allocation seems to be fair it does not belong to the core, since if Sevilla, Cádiz and Huelva decide to make their own network to connect them to Málaga, the costs would amount to (Sevilla-Málaga) (Sevilla-Huelva) (Sevilla-Cádiz), what means a total expense of Euros, a lower price than the Euros that they have to pay after the allocation described above. That is, this allocation does not satisfy the proposed fairness condition and, therefore, it will never be taken. 5 How to pay a new elevator in a resident s association? Nowadays there still exist buildings which do not have any elevator. At a given point in time their neighbours think that this is the time to install it. In those cases the controversy of how to pay the elevator appears. The way of allocating the common costs motivate that some neighbours decide to act on their own without taking into account the remainder. This may happen when there is a group of neighbours that can spend less money if they build an elevator only for their own use than the money they would spend if they had to pay their share for the construction of the elevator for the whole community. Let us see the following case: In a five storey block, with one apartment in each floor, an elevator is going to be installed. The company in charge of building the elevator has a fixed price depending on the number of floors the elevator has. Due to technical problems the cost of each floor is increased with its height, that is, it is more expensive to build the part of the elevator between the sixth and the seventh floor than the one between the first and the second. The prices of the company are shown in the following table: Number of storeys Cost of the elevator

19 The draft of the elevator is shown in figure 13. Figure 13: Elevator draft. In the resident s association it was proposed the following cost allocation: each one pays proportionally, that is, Euros. Is this allocation fair? How would you allocate the expenses among all the neighbours? 5.1 Sharing expenses Let us see the three rules to share costs that we saw before in our new situation: 1. Proportional allocation says that each neighbour has to pay Euros. But let us thing a bit, if the neighbour of the first floor wanted to build an elevator only for his own use, he would have to pay less (10000 Euros), so he will not want to be part of the joint construction of the elevator. 2. Bird s rule. When using that rule to share costs no subgroup of neighbours will be able to build an elevator for themselves at less cost. In this case, Bird s rule proposes the following cost allocation: The neighbour of the first floor pays Euros (the cost of building an elevator from the hall to the first floor). The neighbour of the second floor pays (the cost of building the elevator from the first floor to the second) and so on. With this allocation rule the neighbour of the third floor pays Euros, has to pay the one living 18

20 in the fourth floor and the neighbour of the fifth floor pays (Recall that those allocations, so that there is no subgroup of agents that can spend less money by acting by themselves, are known in Cooperative Game Theory as core allocations.) 3. Another rule that can be proposed is the next: each neighbour pays for the number of floors that he has to pass through in order to get to his floor, that is, the neighbour of the third floor has to pay for the first part of the elevator, the second and the third. In this case each neighbour pays: The first part costs Euros and it is paid among all the neighbours, because all of them use it ( = 2000 Euros). The part between the first and the second storey is paid by the neighbours from the second floor to the fifth ( = 2750 Euros). The third part costs and is paid by the neighbours of the third, fourth and fifth floor ( = 4000 Euros). The part between the third and the fourth storey is Euros and it has to be paid by those neighbours who use it, that is the ones living in the fourth and the fifth floor, ( = 6500 Euros). The last part, between the fourth and the fifth is paid only by the neighbour of the fifth floor, ( = Euros). So the payoff vector is: (2000, 4750, 8750, 15250, 29250), where the i th neighbour pays the i th component of the vector above. Although that is a stable allocation belonging the Core of the corresponding cooperative game, it seems to be excessive that the neighbour of the fifth floor pays 15 times what the neighbour of the first storey pays. What do you think about the concept of fairness in allocation of expenses? 19

Population and sample. Sampling techniques

Population and sample. Sampling techniques MaMaEuSch Management Mathematics for European Schools http://www.mathematik.unikl.de/ mamaeusch Population and sample. Sampling techniques Paula Lagares Barreiro Justo Puerto Albandoz MaMaEuSch Management

More information

A simple analysis of the TV game WHO WANTS TO BE A MILLIONAIRE? R

A simple analysis of the TV game WHO WANTS TO BE A MILLIONAIRE? R A simple analysis of the TV game WHO WANTS TO BE A MILLIONAIRE? R Federico Perea Justo Puerto MaMaEuSch Management Mathematics for European Schools 94342 - CP - 1-2001 - DE - COMENIUS - C21 University

More information

1. Sorting (assuming sorting into ascending order) a) BUBBLE SORT

1. Sorting (assuming sorting into ascending order) a) BUBBLE SORT DECISION 1 Revision Notes 1. Sorting (assuming sorting into ascending order) a) BUBBLE SORT Make sure you show comparisons clearly and label each pass First Pass 8 4 3 6 1 4 8 3 6 1 4 3 8 6 1 4 3 6 8 1

More information

Chapter 6: Graph Theory

Chapter 6: Graph Theory Chapter 6: Graph Theory Graph theory deals with routing and network problems and if it is possible to find a best route, whether that means the least expensive, least amount of time or the least distance.

More information

Decision Mathematics D1 Advanced/Advanced Subsidiary. Tuesday 5 June 2007 Afternoon Time: 1 hour 30 minutes

Decision Mathematics D1 Advanced/Advanced Subsidiary. Tuesday 5 June 2007 Afternoon Time: 1 hour 30 minutes Paper Reference(s) 6689/01 Edexcel GCE Decision Mathematics D1 Advanced/Advanced Subsidiary Tuesday 5 June 2007 Afternoon Time: 1 hour 30 minutes Materials required for examination Nil Items included with

More information

Analysis of Algorithms, I

Analysis of Algorithms, I Analysis of Algorithms, I CSOR W4231.002 Eleni Drinea Computer Science Department Columbia University Thursday, February 26, 2015 Outline 1 Recap 2 Representing graphs 3 Breadth-first search (BFS) 4 Applications

More information

Flow and Activity Analysis

Flow and Activity Analysis Facility Location, Layout, and Flow and Activity Analysis Primary activity relationships Organizational relationships» Span of control and reporting hierarchy Flow relationships» Flow of materials, people,

More information

CSE 326, Data Structures. Sample Final Exam. Problem Max Points Score 1 14 (2x7) 2 18 (3x6) 3 4 4 7 5 9 6 16 7 8 8 4 9 8 10 4 Total 92.

CSE 326, Data Structures. Sample Final Exam. Problem Max Points Score 1 14 (2x7) 2 18 (3x6) 3 4 4 7 5 9 6 16 7 8 8 4 9 8 10 4 Total 92. Name: Email ID: CSE 326, Data Structures Section: Sample Final Exam Instructions: The exam is closed book, closed notes. Unless otherwise stated, N denotes the number of elements in the data structure

More information

Project and Production Management Prof. Arun Kanda Department of Mechanical Engineering Indian Institute of Technology, Delhi

Project and Production Management Prof. Arun Kanda Department of Mechanical Engineering Indian Institute of Technology, Delhi Project and Production Management Prof. Arun Kanda Department of Mechanical Engineering Indian Institute of Technology, Delhi Lecture - 15 Limited Resource Allocation Today we are going to be talking about

More information

MATHEMATICAL THOUGHT AND PRACTICE. Chapter 7: The Mathematics of Networks The Cost of Being Connected

MATHEMATICAL THOUGHT AND PRACTICE. Chapter 7: The Mathematics of Networks The Cost of Being Connected MATHEMATICAL THOUGHT AND PRACTICE Chapter 7: The Mathematics of Networks The Cost of Being Connected Network A network is a graph that is connected. In this context the term is most commonly used when

More information

Lecture 3. Linear Programming. 3B1B Optimization Michaelmas 2015 A. Zisserman. Extreme solutions. Simplex method. Interior point method

Lecture 3. Linear Programming. 3B1B Optimization Michaelmas 2015 A. Zisserman. Extreme solutions. Simplex method. Interior point method Lecture 3 3B1B Optimization Michaelmas 2015 A. Zisserman Linear Programming Extreme solutions Simplex method Interior point method Integer programming and relaxation The Optimization Tree Linear Programming

More information

Unit 6 Coordinate Geometry

Unit 6 Coordinate Geometry Mathematics I Frameworks Student Edition Unit 6 Coordinate Geometry 2 nd Edition Table of Contents Introduction:... 3 Video Game Learning Task... 6 New York Learning Task... 11 Quadrilaterals Revisited

More information

An approach to Descriptive Statistics through real situations

An approach to Descriptive Statistics through real situations MaMaEuSch Management Mathematics for European Schools http://www.mathematik.unikl.de/ mamaeusch An approach to Descriptive Statistics through real situations Paula Lagares Barreiro 1 Federico Perea Rojas-Marcos

More information

Outline. NP-completeness. When is a problem easy? When is a problem hard? Today. Euler Circuits

Outline. NP-completeness. When is a problem easy? When is a problem hard? Today. Euler Circuits Outline NP-completeness Examples of Easy vs. Hard problems Euler circuit vs. Hamiltonian circuit Shortest Path vs. Longest Path 2-pairs sum vs. general Subset Sum Reducing one problem to another Clique

More information

Lecture 16 : Relations and Functions DRAFT

Lecture 16 : Relations and Functions DRAFT CS/Math 240: Introduction to Discrete Mathematics 3/29/2011 Lecture 16 : Relations and Functions Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT In Lecture 3, we described a correspondence

More information

Computational Geometry. Lecture 1: Introduction and Convex Hulls

Computational Geometry. Lecture 1: Introduction and Convex Hulls Lecture 1: Introduction and convex hulls 1 Geometry: points, lines,... Plane (two-dimensional), R 2 Space (three-dimensional), R 3 Space (higher-dimensional), R d A point in the plane, 3-dimensional space,

More information

Lecture 17 : Equivalence and Order Relations DRAFT

Lecture 17 : Equivalence and Order Relations DRAFT CS/Math 240: Introduction to Discrete Mathematics 3/31/2011 Lecture 17 : Equivalence and Order Relations Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT Last lecture we introduced the notion

More information

12 Abstract Data Types

12 Abstract Data Types 12 Abstract Data Types 12.1 Source: Foundations of Computer Science Cengage Learning Objectives After studying this chapter, the student should be able to: Define the concept of an abstract data type (ADT).

More information

VISUALIZING HIERARCHICAL DATA. Graham Wills SPSS Inc., http://willsfamily.org/gwills

VISUALIZING HIERARCHICAL DATA. Graham Wills SPSS Inc., http://willsfamily.org/gwills VISUALIZING HIERARCHICAL DATA Graham Wills SPSS Inc., http://willsfamily.org/gwills SYNONYMS Hierarchical Graph Layout, Visualizing Trees, Tree Drawing, Information Visualization on Hierarchies; Hierarchical

More information

Seminar. Path planning using Voronoi diagrams and B-Splines. Stefano Martina stefano.martina@stud.unifi.it

Seminar. Path planning using Voronoi diagrams and B-Splines. Stefano Martina stefano.martina@stud.unifi.it Seminar Path planning using Voronoi diagrams and B-Splines Stefano Martina stefano.martina@stud.unifi.it 23 may 2016 This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International

More information

MATHEMATICS Unit Decision 1

MATHEMATICS Unit Decision 1 General Certificate of Education January 2008 Advanced Subsidiary Examination MATHEMATICS Unit Decision 1 MD01 Tuesday 15 January 2008 9.00 am to 10.30 am For this paper you must have: an 8-page answer

More information

2.3 Scheduling jobs on identical parallel machines

2.3 Scheduling jobs on identical parallel machines 2.3 Scheduling jobs on identical parallel machines There are jobs to be processed, and there are identical machines (running in parallel) to which each job may be assigned Each job = 1,,, must be processed

More information

IE 680 Special Topics in Production Systems: Networks, Routing and Logistics*

IE 680 Special Topics in Production Systems: Networks, Routing and Logistics* IE 680 Special Topics in Production Systems: Networks, Routing and Logistics* Rakesh Nagi Department of Industrial Engineering University at Buffalo (SUNY) *Lecture notes from Network Flows by Ahuja, Magnanti

More information

Session 7 Bivariate Data and Analysis

Session 7 Bivariate Data and Analysis Session 7 Bivariate Data and Analysis Key Terms for This Session Previously Introduced mean standard deviation New in This Session association bivariate analysis contingency table co-variation least squares

More information

Computational Approach for Assessment of Critical Infrastructure in Network Systems

Computational Approach for Assessment of Critical Infrastructure in Network Systems Computational Approach for Assessment of Critical Infrastructure in Network Systems EMIL KELEVEDJIEV 1 Institute of Mathematics and Informatics Bulgarian Academy of Sciences ABSTRACT. Methods of computational

More information

Lecture 15 An Arithmetic Circuit Lowerbound and Flows in Graphs

Lecture 15 An Arithmetic Circuit Lowerbound and Flows in Graphs CSE599s: Extremal Combinatorics November 21, 2011 Lecture 15 An Arithmetic Circuit Lowerbound and Flows in Graphs Lecturer: Anup Rao 1 An Arithmetic Circuit Lower Bound An arithmetic circuit is just like

More information

A permutation can also be represented by describing its cycles. What do you suppose is meant by this?

A permutation can also be represented by describing its cycles. What do you suppose is meant by this? Shuffling, Cycles, and Matrices Warm up problem. Eight people stand in a line. From left to right their positions are numbered,,,... 8. The eight people then change places according to THE RULE which directs

More information

Scheduling Home Health Care with Separating Benders Cuts in Decision Diagrams

Scheduling Home Health Care with Separating Benders Cuts in Decision Diagrams Scheduling Home Health Care with Separating Benders Cuts in Decision Diagrams André Ciré University of Toronto John Hooker Carnegie Mellon University INFORMS 2014 Home Health Care Home health care delivery

More information

Switches and Bridges. COS 461: Computer Networks Spring 2009 (MW 1:30-2:50 in COS 105) Guest Lecture Jennifer Rexford 1

Switches and Bridges. COS 461: Computer Networks Spring 2009 (MW 1:30-2:50 in COS 105) Guest Lecture Jennifer Rexford 1 Switches and Bridges COS 461: Computer Networks Spring 2009 (MW 1:30-2:50 in COS 105) Guest Lecture Jennifer Rexford 1 Goals of Todayʼs Lecture Devices that shuttle data at different layers Repeaters and

More information

Spring 2007 Math 510 Hints for practice problems

Spring 2007 Math 510 Hints for practice problems Spring 2007 Math 510 Hints for practice problems Section 1 Imagine a prison consisting of 4 cells arranged like the squares of an -chessboard There are doors between all adjacent cells A prisoner in one

More information

The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 72.

The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 72. ADVANCED SUBSIDIARY GCE UNIT 4736/01 MATHEMATICS Decision Mathematics 1 THURSDAY 14 JUNE 2007 Afternoon Additional Materials: Answer Booklet (8 pages) List of Formulae (MF1) Time: 1 hour 30 minutes INSTRUCTIONS

More information

Natural Neighbour Interpolation

Natural Neighbour Interpolation Natural Neighbour Interpolation DThe Natural Neighbour method is a geometric estimation technique that uses natural neighbourhood regions generated around each point in the data set. The method is particularly

More information

6.3 Conditional Probability and Independence

6.3 Conditional Probability and Independence 222 CHAPTER 6. PROBABILITY 6.3 Conditional Probability and Independence Conditional Probability Two cubical dice each have a triangle painted on one side, a circle painted on two sides and a square painted

More information

Social Media Mining. Graph Essentials

Social Media Mining. Graph Essentials Graph Essentials Graph Basics Measures Graph and Essentials Metrics 2 2 Nodes and Edges A network is a graph nodes, actors, or vertices (plural of vertex) Connections, edges or ties Edge Node Measures

More information

Why? A central concept in Computer Science. Algorithms are ubiquitous.

Why? A central concept in Computer Science. Algorithms are ubiquitous. Analysis of Algorithms: A Brief Introduction Why? A central concept in Computer Science. Algorithms are ubiquitous. Using the Internet (sending email, transferring files, use of search engines, online

More information

MD5-26 Stacking Blocks Pages 115 116

MD5-26 Stacking Blocks Pages 115 116 MD5-26 Stacking Blocks Pages 115 116 STANDARDS 5.MD.C.4 Goals Students will find the number of cubes in a rectangular stack and develop the formula length width height for the number of cubes in a stack.

More information

A Non-Linear Schema Theorem for Genetic Algorithms

A Non-Linear Schema Theorem for Genetic Algorithms A Non-Linear Schema Theorem for Genetic Algorithms William A Greene Computer Science Department University of New Orleans New Orleans, LA 70148 bill@csunoedu 504-280-6755 Abstract We generalize Holland

More information

Approximation Algorithms

Approximation Algorithms Approximation Algorithms or: How I Learned to Stop Worrying and Deal with NP-Completeness Ong Jit Sheng, Jonathan (A0073924B) March, 2012 Overview Key Results (I) General techniques: Greedy algorithms

More information

2 SYSTEM DESCRIPTION TECHNIQUES

2 SYSTEM DESCRIPTION TECHNIQUES 2 SYSTEM DESCRIPTION TECHNIQUES 2.1 INTRODUCTION Graphical representation of any process is always better and more meaningful than its representation in words. Moreover, it is very difficult to arrange

More information

Unit 7 Quadratic Relations of the Form y = ax 2 + bx + c

Unit 7 Quadratic Relations of the Form y = ax 2 + bx + c Unit 7 Quadratic Relations of the Form y = ax 2 + bx + c Lesson Outline BIG PICTURE Students will: manipulate algebraic expressions, as needed to understand quadratic relations; identify characteristics

More information

Chapter 13: Binary and Mixed-Integer Programming

Chapter 13: Binary and Mixed-Integer Programming Chapter 3: Binary and Mixed-Integer Programming The general branch and bound approach described in the previous chapter can be customized for special situations. This chapter addresses two special situations:

More information

Decision Mathematics 1 TUESDAY 22 JANUARY 2008

Decision Mathematics 1 TUESDAY 22 JANUARY 2008 ADVANCED SUBSIDIARY GCE 4736/01 MATHEMATICS Decision Mathematics 1 TUESDAY 22 JANUARY 2008 Additional materials: Answer Booklet (8 pages) Graph paper Insert for Questions 3 and 4 List of Formulae (MF1)

More information

Network Flow I. Lecture 16. 16.1 Overview. 16.2 The Network Flow Problem

Network Flow I. Lecture 16. 16.1 Overview. 16.2 The Network Flow Problem Lecture 6 Network Flow I 6. Overview In these next two lectures we are going to talk about an important algorithmic problem called the Network Flow Problem. Network flow is important because it can be

More information

princeton univ. F 13 cos 521: Advanced Algorithm Design Lecture 6: Provable Approximation via Linear Programming Lecturer: Sanjeev Arora

princeton univ. F 13 cos 521: Advanced Algorithm Design Lecture 6: Provable Approximation via Linear Programming Lecturer: Sanjeev Arora princeton univ. F 13 cos 521: Advanced Algorithm Design Lecture 6: Provable Approximation via Linear Programming Lecturer: Sanjeev Arora Scribe: One of the running themes in this course is the notion of

More information

Maximum Range Explained range Figure 1 Figure 1: Trajectory Plot for Angled-Launched Projectiles Table 1

Maximum Range Explained range Figure 1 Figure 1: Trajectory Plot for Angled-Launched Projectiles Table 1 Maximum Range Explained A projectile is an airborne object that is under the sole influence of gravity. As it rises and falls, air resistance has a negligible effect. The distance traveled horizontally

More information

Discuss the size of the instance for the minimum spanning tree problem.

Discuss the size of the instance for the minimum spanning tree problem. 3.1 Algorithm complexity The algorithms A, B are given. The former has complexity O(n 2 ), the latter O(2 n ), where n is the size of the instance. Let n A 0 be the size of the largest instance that can

More information

Problem Set 7 Solutions

Problem Set 7 Solutions 8 8 Introduction to Algorithms May 7, 2004 Massachusetts Institute of Technology 6.046J/18.410J Professors Erik Demaine and Shafi Goldwasser Handout 25 Problem Set 7 Solutions This problem set is due in

More information

Task: ASC Ascending Paths

Task: ASC Ascending Paths Task: ASC Ascending Paths You are visiting the Royal Botanical Gardens. In the gardens there are n intersections connected by m roads. The intersections are numbered from 1 to n. Each road connects two

More information

8.5 PETRI NETS. Figure A computer program. Figure 8.5.2

8.5 PETRI NETS. Figure A computer program. Figure 8.5.2 8.5 PETRI NETS Consider the computer program shown in Figure 8.5.1. Normally, the instructions would be processed sequentially first, A = 1, then B = 2, and so on. However, notice that there is no logical

More information

Operations and Supply Chain Management Prof. G. Srinivasan Department of Management Studies Indian Institute of Technology Madras

Operations and Supply Chain Management Prof. G. Srinivasan Department of Management Studies Indian Institute of Technology Madras Operations and Supply Chain Management Prof. G. Srinivasan Department of Management Studies Indian Institute of Technology Madras Lecture - 41 Value of Information In this lecture, we look at the Value

More information

Exam 2 Study Guide and Review Problems

Exam 2 Study Guide and Review Problems Exam 2 Study Guide and Review Problems Exam 2 covers chapters 4, 5, and 6. You are allowed to bring one 3x5 note card, front and back, and your graphing calculator. Study tips: Do the review problems below.

More information

Problem of the Month: Once Upon a Time

Problem of the Month: Once Upon a Time Problem of the Month: The Problems of the Month (POM) are used in a variety of ways to promote problem solving and to foster the first standard of mathematical practice from the Common Core State Standards:

More information

Random Map Generator v1.0 User s Guide

Random Map Generator v1.0 User s Guide Random Map Generator v1.0 User s Guide Jonathan Teutenberg 2003 1 Map Generation Overview...4 1.1 Command Line...4 1.2 Operation Flow...4 2 Map Initialisation...5 2.1 Initialisation Parameters...5 -w xxxxxxx...5

More information

Induction. Margaret M. Fleck. 10 October These notes cover mathematical induction and recursive definition

Induction. Margaret M. Fleck. 10 October These notes cover mathematical induction and recursive definition Induction Margaret M. Fleck 10 October 011 These notes cover mathematical induction and recursive definition 1 Introduction to induction At the start of the term, we saw the following formula for computing

More information

Operations and Supply Chain Management Prof. G. Srinivasan Department of Management Studies Indian Institute of Technology, Madras

Operations and Supply Chain Management Prof. G. Srinivasan Department of Management Studies Indian Institute of Technology, Madras Operations and Supply Chain Management Prof. G. Srinivasan Department of Management Studies Indian Institute of Technology, Madras Lecture - 36 Location Problems In this lecture, we continue the discussion

More information

Why Vedic Mathematics?

Why Vedic Mathematics? Why Vedic Mathematics? Many Indian Secondary School students consider Mathematics a very difficult subject. Some students encounter difficulty with basic arithmetical operations. Some students feel it

More information

Lecture 3: Linear Programming Relaxations and Rounding

Lecture 3: Linear Programming Relaxations and Rounding Lecture 3: Linear Programming Relaxations and Rounding 1 Approximation Algorithms and Linear Relaxations For the time being, suppose we have a minimization problem. Many times, the problem at hand can

More information

Problem of the Month: Fair Games

Problem of the Month: Fair Games Problem of the Month: The Problems of the Month (POM) are used in a variety of ways to promote problem solving and to foster the first standard of mathematical practice from the Common Core State Standards:

More information

Graph Theory Problems and Solutions

Graph Theory Problems and Solutions raph Theory Problems and Solutions Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles November, 005 Problems. Prove that the sum of the degrees of the vertices of any finite graph is

More information

Data Structure [Question Bank]

Data Structure [Question Bank] Unit I (Analysis of Algorithms) 1. What are algorithms and how they are useful? 2. Describe the factor on best algorithms depends on? 3. Differentiate: Correct & Incorrect Algorithms? 4. Write short note:

More information

Bar Graphs and Dot Plots

Bar Graphs and Dot Plots CONDENSED L E S S O N 1.1 Bar Graphs and Dot Plots In this lesson you will interpret and create a variety of graphs find some summary values for a data set draw conclusions about a data set based on graphs

More information

Current California Math Standards Balanced Equations

Current California Math Standards Balanced Equations Balanced Equations Current California Math Standards Balanced Equations Grade Three Number Sense 1.0 Students understand the place value of whole numbers: 1.1 Count, read, and write whole numbers to 10,000.

More information

Lecture 1: Course overview, circuits, and formulas

Lecture 1: Course overview, circuits, and formulas Lecture 1: Course overview, circuits, and formulas Topics in Complexity Theory and Pseudorandomness (Spring 2013) Rutgers University Swastik Kopparty Scribes: John Kim, Ben Lund 1 Course Information Swastik

More information

TYPES OF NUMBERS. Example 2. Example 1. Problems. Answers

TYPES OF NUMBERS. Example 2. Example 1. Problems. Answers TYPES OF NUMBERS When two or more integers are multiplied together, each number is a factor of the product. Nonnegative integers that have exactly two factors, namely, one and itself, are called prime

More information

Advice Complexity of Online Coloring for Paths

Advice Complexity of Online Coloring for Paths Advice Complexity of Online Coloring for Paths Michal Forišek 1, Lucia Keller 2, and Monika Steinová 2 1 Comenius University, Bratislava, Slovakia 2 ETH Zürich, Switzerland LATA 2012, A Coruña, Spain Definition

More information

Midterm Practice Problems

Midterm Practice Problems 6.042/8.062J Mathematics for Computer Science October 2, 200 Tom Leighton, Marten van Dijk, and Brooke Cowan Midterm Practice Problems Problem. [0 points] In problem set you showed that the nand operator

More information

1. The Fly In The Ointment

1. The Fly In The Ointment Arithmetic Revisited Lesson 5: Decimal Fractions or Place Value Extended Part 5: Dividing Decimal Fractions, Part 2. The Fly In The Ointment The meaning of, say, ƒ 2 doesn't depend on whether we represent

More information

Number Models for Area

Number Models for Area Number Models for Area Objectives To guide children as they develop the concept of area by measuring with identical squares; and to demonstrate how to calculate the area of rectangles using number models.

More information

Social Media Mining. Network Measures

Social Media Mining. Network Measures Klout Measures and Metrics 22 Why Do We Need Measures? Who are the central figures (influential individuals) in the network? What interaction patterns are common in friends? Who are the like-minded users

More information

3 Some Integer Functions

3 Some Integer Functions 3 Some Integer Functions A Pair of Fundamental Integer Functions The integer function that is the heart of this section is the modulo function. However, before getting to it, let us look at some very simple

More information

Reading 13 : Finite State Automata and Regular Expressions

Reading 13 : Finite State Automata and Regular Expressions CS/Math 24: Introduction to Discrete Mathematics Fall 25 Reading 3 : Finite State Automata and Regular Expressions Instructors: Beck Hasti, Gautam Prakriya In this reading we study a mathematical model

More information

Explore architectural design and act as architects to create a floor plan of a redesigned classroom.

Explore architectural design and act as architects to create a floor plan of a redesigned classroom. ARCHITECTURAL DESIGN AT A GLANCE Explore architectural design and act as architects to create a floor plan of a redesigned classroom. OBJECTIVES: Students will: Use prior knowledge to discuss functions

More information

MATHEMATICS Unit Decision 1

MATHEMATICS Unit Decision 1 General Certificate of Education June 2007 Advanced Subsidiary Examination MATHEMATICS Unit Decision 1 MD01 Thursday 7 June 2007 9.00 am to 10.30 am For this paper you must have: an 8-page answer book

More information

Solution Guide for Chapter 6: The Geometry of Right Triangles

Solution Guide for Chapter 6: The Geometry of Right Triangles Solution Guide for Chapter 6: The Geometry of Right Triangles 6. THE THEOREM OF PYTHAGORAS E-. Another demonstration: (a) Each triangle has area ( ). ab, so the sum of the areas of the triangles is 4 ab

More information

V. Adamchik 1. Graph Theory. Victor Adamchik. Fall of 2005

V. Adamchik 1. Graph Theory. Victor Adamchik. Fall of 2005 V. Adamchik 1 Graph Theory Victor Adamchik Fall of 2005 Plan 1. Basic Vocabulary 2. Regular graph 3. Connectivity 4. Representing Graphs Introduction A.Aho and J.Ulman acknowledge that Fundamentally, computer

More information

INTRODUCTION TO MATHEMATICAL MODELLING

INTRODUCTION TO MATHEMATICAL MODELLING 306 MATHEMATICS APPENDIX 2 INTRODUCTION TO MATHEMATICAL MODELLING A2.1 Introduction Right from your earlier classes, you have been solving problems related to the real-world around you. For example, you

More information

7 Relations and Functions

7 Relations and Functions 7 Relations and Functions In this section, we introduce the concept of relations and functions. Relations A relation R from a set A to a set B is a set of ordered pairs (a, b), where a is a member of A,

More information

Session 6 Number Theory

Session 6 Number Theory Key Terms in This Session Session 6 Number Theory Previously Introduced counting numbers factor factor tree prime number New in This Session composite number greatest common factor least common multiple

More information

Technology Solution Guide. Deploying Omnitron PoE Media Converters with Aruba Access Points and AirMesh Routers

Technology Solution Guide. Deploying Omnitron PoE Media Converters with Aruba Access Points and AirMesh Routers Technology Solution Guide Deploying Omnitron PoE Media Converters with Aruba Access Points and AirMesh Routers WARRANTY DISCLAIMER THE FOLLOWING DOCUMENT, AND THE INFORMATION CONTAINED HEREIN IS PROVIDED

More information

Fry Instant Word List

Fry Instant Word List First 100 Instant Words the had out than of by many first and words then water a but them been to not these called in what so who is all some oil you were her sit that we would now it when make find he

More information

SOLVING DATABASE QUERIES USING ORTHOGONAL RANGE SEARCHING

SOLVING DATABASE QUERIES USING ORTHOGONAL RANGE SEARCHING SOLVING DATABASE QUERIES USING ORTHOGONAL RANGE SEARCHING CPS234: Computational Geometry Prof. Pankaj Agarwal Prepared by: Khalid Al Issa (khalid@cs.duke.edu) Fall 2005 BACKGROUND : RANGE SEARCHING, a

More information

Optimization in ICT and Physical Systems

Optimization in ICT and Physical Systems 27. OKTOBER 2010 in ICT and Physical Systems @ Aarhus University, Course outline, formal stuff Prerequisite Lectures Homework Textbook, Homepage and CampusNet, http://kurser.iha.dk/ee-ict-master/tiopti/

More information

Components: Interconnect Page 1 of 18

Components: Interconnect Page 1 of 18 Components: Interconnect Page 1 of 18 PE to PE interconnect: The most expensive supercomputer component Possible implementations: FULL INTERCONNECTION: The ideal Usually not attainable Each PE has a direct

More information

A couple of Max Min Examples. 3.6 # 2) Find the maximum possible area of a rectangle of perimeter 200m.

A couple of Max Min Examples. 3.6 # 2) Find the maximum possible area of a rectangle of perimeter 200m. A couple of Max Min Examples 3.6 # 2) Find the maximum possible area of a rectangle of perimeter 200m. Solution : As is the case with all of these problems, we need to find a function to minimize or maximize

More information

136 CHAPTER 4. INDUCTION, GRAPHS AND TREES

136 CHAPTER 4. INDUCTION, GRAPHS AND TREES 136 TER 4. INDUCTION, GRHS ND TREES 4.3 Graphs In this chapter we introduce a fundamental structural idea of discrete mathematics, that of a graph. Many situations in the applications of discrete mathematics

More information

Topology-based network security

Topology-based network security Topology-based network security Tiit Pikma Supervised by Vitaly Skachek Research Seminar in Cryptography University of Tartu, Spring 2013 1 Introduction In both wired and wireless networks, there is the

More information

Cash $331 $9,095. Receivables 4,019 2,342. Prepaid expenses 2,729 456. Prepaid catalog costs 2,794 1,375. Deferred catalog costs 7,020 3,347

Cash $331 $9,095. Receivables 4,019 2,342. Prepaid expenses 2,729 456. Prepaid catalog costs 2,794 1,375. Deferred catalog costs 7,020 3,347 Coldwater Creek Coldwater Creek, located in Idaho, operates a direct-mail catalog business. Items marketed through these catalogs include women's and men's apparel, jewelry, and household items. The beginning

More information

Module 5. Broadcast Communication Networks. Version 2 CSE IIT, Kharagpur

Module 5. Broadcast Communication Networks. Version 2 CSE IIT, Kharagpur Module 5 Broadcast Communication Networks Lesson 1 Network Topology Specific Instructional Objectives At the end of this lesson, the students will be able to: Specify what is meant by network topology

More information

Fry Instant Words High Frequency Words

Fry Instant Words High Frequency Words Fry Instant Words High Frequency Words The Fry list of 600 words are the most frequently used words for reading and writing. The words are listed in rank order. First Hundred Group 1 Group 2 Group 3 Group

More information

Graphical Integration Exercises Part Four: Reverse Graphical Integration

Graphical Integration Exercises Part Four: Reverse Graphical Integration D-4603 1 Graphical Integration Exercises Part Four: Reverse Graphical Integration Prepared for the MIT System Dynamics in Education Project Under the Supervision of Dr. Jay W. Forrester by Laughton Stanley

More information

Ampere's Law. Introduction. times the current enclosed in that loop: Ampere's Law states that the line integral of B and dl over a closed path is 0

Ampere's Law. Introduction. times the current enclosed in that loop: Ampere's Law states that the line integral of B and dl over a closed path is 0 1 Ampere's Law Purpose: To investigate Ampere's Law by measuring how magnetic field varies over a closed path; to examine how magnetic field depends upon current. Apparatus: Solenoid and path integral

More information

International Indian School, Riyadh SA1 Worksheet 2015-2016 Class: VI Mathematics

International Indian School, Riyadh SA1 Worksheet 2015-2016 Class: VI Mathematics International Indian School, Riyadh SA1 Worksheet 2015-2016 Class: VI Mathematics CH KNOWING OUR NUMBERS I Fill In the blanks 1. 1km = mm 2. 1 gram = milligrams 3. The roman numeral M stands for the number

More information

Decision Making under Uncertainty

Decision Making under Uncertainty 6.825 Techniques in Artificial Intelligence Decision Making under Uncertainty How to make one decision in the face of uncertainty Lecture 19 1 In the next two lectures, we ll look at the question of how

More information

Development of FIDO - A Network Design Tool for Fibre Cable Layout

Development of FIDO - A Network Design Tool for Fibre Cable Layout Development of FIDO - A Network Design Tool for Fibre Cable Layout Andrew J Mason Andy B Philpott Operations Research Group University of Auckland New Zealand a.mason@auckland.ac.nz a.philpott@auckland.ac.nz

More information

System Interconnect Architectures. Goals and Analysis. Network Properties and Routing. Terminology - 2. Terminology - 1

System Interconnect Architectures. Goals and Analysis. Network Properties and Routing. Terminology - 2. Terminology - 1 System Interconnect Architectures CSCI 8150 Advanced Computer Architecture Hwang, Chapter 2 Program and Network Properties 2.4 System Interconnect Architectures Direct networks for static connections Indirect

More information

Using Proportions to Solve Percent Problems I

Using Proportions to Solve Percent Problems I RP7-1 Using Proportions to Solve Percent Problems I Pages 46 48 Standards: 7.RP.A. Goals: Students will write equivalent statements for proportions by keeping track of the part and the whole, and by solving

More information

FACTORS, PRIME NUMBERS, H.C.F. AND L.C.M.

FACTORS, PRIME NUMBERS, H.C.F. AND L.C.M. Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 1 of 16 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Higher Tier FACTORS, PRIME NUMBERS, H.C.F. AND L.C.M. Version:

More information

CALCULATING PERIMETER. WHAT IS PERIMETER? Perimeter is the total length or distance around a figure.

CALCULATING PERIMETER. WHAT IS PERIMETER? Perimeter is the total length or distance around a figure. CALCULATING PERIMETER WHAT IS PERIMETER? Perimeter is the total length or distance around a figure. HOW DO WE CALCULATE PERIMETER? The formula one can use to calculate perimeter depends on the type of

More information

GEOMETRIC SEQUENCES AND SERIES

GEOMETRIC SEQUENCES AND SERIES 4.4 Geometric Sequences and Series (4 7) 757 of a novel and every day thereafter increase their daily reading by two pages. If his students follow this suggestion, then how many pages will they read during

More information

Examples of Functions

Examples of Functions Examples of Functions In this document is provided examples of a variety of functions. The purpose is to convince the beginning student that functions are something quite different than polynomial equations.

More information