A small scale free essay sample

A little scope examination concerning the advancement of ‘self concept’ in kids, in view of the discoveries of Rosenberg (1979), utilizing 2 members of various age ranges. Dynamic This examination takes a gander at the self portrayals of two youngster members and how they create with age utilizing the discoveries of Rosenberg (1979) as a kind of perspective. Rosenberg found that the self portrayals of more youthful youngsters, physical characteristics were progressively pervasive, though more seasoned kids concentrated on their inward convictions and character properties. An open finished style of meeting was utilized to scrutinize the two participants.The answers given were then arranged into the properties found by Rosenberg; Physical (P), Character(C), Relationship(R), and Inner (I). Rosenberg’s hypothesis of ‘Locus of Self’ in which the degree to which a youngster depends on others when passing judgment on themselves, was likewise examined, with R osenberg’s discoveries that as a kid ages the more mindful they become as an asset. My discoveries concurred with Rosenberg’s, in that with age , self idea grows anyway my discoveries didn't match with Rosenberg’s convictions that self locus shifts from others to self with age, with the examination indicating no change .Introduction The examination explores the self-portrayals of two offspring of varying ages, and takes a gander at them comparable to the discoveries of Rosenberg (1979). Rosenberg accepted that more youthful youngsters utilized physical exercises and qualities to portray themselves though more established kids utilized their relationship and parts of their internal emotions to depict themselves. In Rosenberg’s study he met a choice of 8 - multi year olds, arbitrarily looked over 25 schools in Baltimore, USA.He requested that the members portray themselves utilizing a â€Å"who am I† spread out. The appropriate responses were then investigated further to incorporate sentiments of pride/disgrace, their likenesses/contrasts t o other people, and so on. He at that point ordered the appropriate responses into Physical (P), character (C), Relationships (R), and Inner qualities (I). Utilizing these classes he searched for regular examples over the age ranges. Technique Design The examination is proposed to imitate that of Rosenberg (1979), utilizing a poll (Appendix A), structured by the Open University, like that of Rosenberg.The survey will create answers to the announcement ‘Who am I’? , delivering a rundown of self descriptors about every youngster. The appropriate responses assembled would eventually be utilized and attributes into one of the four gatherings found by Rosenberg. (Informative supplement B) Following this, utilizing the finished survey, a progression of open finished inquiries dependent on the appropriate responses gave were sought after to additionally clarify the self recognitions. This gave data with respect to self idea, and its turn of events and social determination.This would help expand and wanted to discover thinking to Rosenberg’s ‘locus of self knowledge’ Participants Two kids participated. One female, one male. The female is age 8 years of age, and goes to grade school. The male is a multi year old optional school understudy. First names were utilized all through the examination. The specialists were obscure to the members. Materials ‘Who am I’ poll ( Appendix A) Pen Audio recording gear Procedure The members knew about the idea of the examination and its purposes.They were educated that there were no set in stone answers. The meetings were completed inside the school region and inside the typical tutoring hours. The multi year old was met in a structure nearby her school, where she partook in P. E consistently there. The multi year old was met in a little meeting room inverse the secretary’s office. During the meetings a sound recordist and a maker were available just as a study hall collaborator in the youthful girl’s interview.The members were solicited to contend a rundown from self descriptors headed ‘Who am I’, to which an open finished arrangement of inquiries organized around the ‘who am I’ answers, followed to help expound. The members were said thanks to for their assistance with the examination. Results The consequences of the inquiries ‘Who am I’ were ordered into one of the four of Rosenberg’s classifications. (Reference section B) and afterward the sums were included for every class. These sums were thusly, worked into rates. These rates are appeared in a diagram. Informative supplement C). The rate scores for the class physical declines from 100% to 45% over the multi year contrast. Member An, Annie scored 100% in physical characteristics. None of the participants’ answers were deciphered as scoring in the trademark or connections classes, however member B, Adam scored the dominant part, 55% of his answers in the ‘inner ‘section. These outcomes grouped would agree with Rosenberg’s discoveries that with age, ones self descriptors move away from physical ascribes to increasingly close to home beliefs.However the discoveries neglected to create proof to compare with Rosenberg’s locus of self information. The more youthful member accepted that her mom and instructors were the wellspring of data when gotten some information about her tutoring and so forth, as too did the more established member, which isn't with regards to Rosenberg’ unique discoveries that with age we can depend on our own concept of self as opposed to another legitimate individual, similar to guardians or educators. Conversation This little scope examination has not completely delivered the outcomes trusted for.The ends from the self descriptors were that physical qualities were conspicuous in the two ages yet more so in that of the more youthful member. They appeared with age a youngster changes to increasingly close to home inward accepts and wants, which to a degree would expect that Rosenberg’s discoveries were right, in spite of the fact that our example just glances at two members. The aftereffects of the locus of self information questions didn't anyway agree with that of Rosenberg as he set forward the thought that with age we change from getting information on one selves from definitive figures to accepting that we ourselves are the key.The results demonstrated that the two members still couldn't seem to influence away from picking up that data from guardians and so on. The inquiries with respect to ‘who knows you better’, were focused round proper circumstances like at school, an instructor. Member one indicated a slight comprehension, in that she replied, it depends what they said (16. 28), appearing, maybe that she figured she would know better. In any case, this was not elite to all the inquiries. Troubles in replicating the investigation were clear, as such a little example was broke down, the outcomes couldn't be over summed up.

What your dreams are telling you essays

What your fantasies are revealing to you papers Do you know what your fantasies are letting you know? On the off chance that your answer was NO youre not the only one. Numerous individuals have no clue what their fantasies are attempting to let them know. In this exploration paper I will respond to this inquiry and numerous different inquiries regarding dreams, rest and furthermore sorts of dreams. I will likewise discuss probably the most widely recognized or well known dreams and their implications. We will investigate some fantasy speculations and understandings of dreams. I have additionally done my own nearby research which I will be imparting to you the peruser in this paper. I wish to take you through the dreaming procedure bit by bit, however with the end goal for dreams to start we should initially rest. So that is the place Ill start, with the rest procedure. All dreaming begins with rest. At the point when we rest there are four phases that we experience (Carskadon 4). Stage one is the lightest period of rest. In stage one rest there is regularly visual symbolism included. Pictures show up through meandering or dream like idea which can be controlled not normal for a full dream state. In this phase of rest somebody can be stirred by low volume sounds or slight touch. Stage one rest occurs for the duration of the night. It regularly returns as a transitional condition of rest following excitement during the night. At the point when stage one reappears during rest it just most recent thirty seconds to one moment (Carskadon 5-6). Youthful grown-ups go through ten to fifteen minutes per night in stage one rest (Carskadon 6). In some cases during rest you feel your body snap or jerk, that is called Hypnic Jerks. This generally happens directly before stage two rest happens (Carskadon 6). Stage two rest may have a portion of the more slow eye development of stage one however as a rule stage two rest is liberated from any eye development. Another element of stage two is excitement. Since stage two is a more profound rest than stage one it will take more excitement to stir an individual from this phase of rest. The change from stage two rest to organize three happens mind... <!

Hilberts Third Problem (A Story of Threes)

Hilbert’s Third Problem (A Story of Threes) 1 Introduction and History 18.304 (discrete math) is one of two communication intensive (CI-M, where the M means it’s within my major) seminars I need to take as part of my math major. (The second one, which I’m taking this semester, is 18.424, information theory. The other seminar topics we get to choose from are real analysis, analysis, discrete applied mathematics, physical mathematics, theoretical computer science, logic, algebra, number theory, topology, and geometry.) The math CI-Ms are small classes, taught almost entirely by the students: in my experience so far, after a few starter lectures by the professor every student presents three lectures as a combination of slide presentations and chalktalks. Sometimes there are also p-sets; sometimes there are not. At the end of the semester we write in-depth explorations of the topics of our final presentations. In 18.304, I got to transiently experience several dozen proofs, getting closely acquainted with two of them and reading books and books and books about a third. My first topic was Dijkstra’s algorithm for finding shortest paths, my second topic was (the inifinite-ish quantity of) proofs for the infinity of primes, and my third topic was Hilbert’s third problem, which we’ll get to know much more closely in this blog post.  The first two presentations went objectively horribly. The third one went much better: I showed up early and stood in front of the bathroom mirror similing to myself for 20 seconds in a Wonder Woman power pose. On the way from the bathroom to the classroom I passed a girl from among my soon-to-be audience. We made eye contact and exchanged shy smiles and somehow that made everything better: I felt less alone, and I felt that I had an ally among the people I had been scared were going to judge me. (If you’re reading this, thank you so much, kind, mysterious friend-when-I-needed-one.) Over the course of the semester, I went from having barely any exposure to proofs beyond what I’d gotten from my computer science classes and general institute requirements to being able to follow, read, write, and (thanks to 6.046, which I took at the same time) formulate a proof. This was a huge amount of growth for me, and a big part of my goal in becoming a math major. My grandfather, who was very, very dear to me, emphasized to me that there is a lot of power at the intersection of fields. It is valuable to be able to be a translator between fields, to be an avenue for the ideas, tools, and strengths of one field to enter and innovate in another field. This has been one of my hopes for my technical education at MIT: as a 6-7/18 major, I’m split officially between three technical departmentsâ€"computer science, biology, and mathematicsâ€"and I’ve also had to take courses in physics and chemistry. My personal theory (and experience) is that computational biology, the thing I am ultimately studying, is young enough that depending on their backgrounds, people primarily approach it from the perspective of biology, computer science, or mathâ€"occasionally two of them, but very rarely all three. I’m hoping that my broader education at MIT will help me find a niche and be, as my grandfather said, a useful translator across those junctions. The rest of this blog post is my final paper, which from the very start  I wrote with the dual purpose of turning in and posting here, with an intended audience of high school students. (Feel free to skim or skip around if it helps you get a picture of the proof.) I’m hoping to write a post about Dijkstra’s algorithm someday as well, since it is one of my favorite algorithms (tied maybe with Edmonds-Karp, because flow networks are absurdly useful). I want to communicate the following things I have learned about math: Math is an incredibly diverse field, not a single path, and you may or may not have gotten a chance to actually see all the many branches in high school. Math is a human and even political field, in which a single problem can connect people across centuries. Math is not memorizationâ€"in fact, the human subject of this blog post chose math specifically for that reason. 1 Introduction and History Hilbert’s third problem, the problem of defining volume for polyhedra, is a story of both threes and infinities. We will start with some of the threes. Already in early elementary school we learn about two- and three-dimensional shapes and some of their interesting properties. We learn that a triangle is a two dimensional polygon with three edges. More generally, we learn that a polygon can be defined as a two-dimensional shape constrained to a plane, bounded by any number of straight, uncurved lines, the edges. The area contained between the edges is the single face of the polygon and the points where the edges meet are the vertices. All polygons can be cut up into a discrete number of triangles, and the area of these triangles can always be further distributed in discrete chunks to form a square of the same area as the original polygon. One method of doing this is to cut the polygon up into triangles, cut each triangle up into smaller triangles that can be reassembled into a rectangle, and cut this rectangle up and rearrange the pieces into another rectangle, this one with either its width or its height equal to the side length of the square we want to assemble. Finally, we can stack these rectangles to form a square with the same area as our original polygon, a square shared by all polygons of that area. As an example, here we chop up a two-dimensional house and rearrange it into a square of the same area. 1.1 Definition I: Equidecomposable The ability to chop up one figure and build a new one out of its pieces is the first property we are interested in: equidecomposability. Two shapes are equidecomposable if they can be divided up into congruent building blocks. For example, the house and the square that we looked at above are equidecomposable. Below is another of the many ways that they can be decomposed into congruent pieces. This relationship can be expressed mathematically. In the picture above, both the square S and the house H can be decomposed into the orange (O), purple (P), and blue (B) pieces. S = O ? P ? B = H Equidecomposability has also been called congruence by dissection and scissors congruence. 1.2 Definition II: Equicomplementable Another property that we are interested in is equicomplementability. Two shapes are equicomplementable if congruent building blocks can be added to both of them to create two equidecomposable supershapes. In addition to being equidecomposable, our house and square are also equicomplementable. The supershapes we have created are equidecomposable. We can express this relationship mathematically as well. We can add four piecesâ€"here a yellow piece Y, a pink piece P, an orange piece O, and a maroon piece Aâ€"to the square S in order to create a supershape M. We can add the same pieces to the house H to create a second supershape N. M = S ? Y ? P ? O ? A N = H ? Y ? P ? O ? A These supershapes can then be decomposed into the same pieces, pictured here as a red piece (R), a turquoise piece (T), a brown piece (B), a blue piece (L), and a green piece (G). M = R ? T ? B ? L ? G = N Equidecomposability, equicomplementability, and equality in area are intertwined for polygons in the second dimension. Throughout the 1800s there were many developments in these properties as they apply to two-dimensional polygons. Following preliminary work by William Wallace in 1807, independent proofs from Farkas Bolyai in 1832 and P. Gerwein in 1833 demonstrated that any polygon can be decomposed in such a way that its pieces can be reassembled into a square, as we illustrated earlier. This means that any pair of polygons of equal area is equidecomposable, since they can be decomposed and reassembled into the same square. In 1844, Gerling furthermore showed that it does not matter if reflections are allowed in the reassembly of the decomposed shapes. The following interesting theorems and lemmas were proved in the nineteenth century. They are outlined in the second chapter of  Boltíànskii’s 1978 Hilbert’s Third Problem: If two figures A and C are each equidecomposable with a third figure B, then A and C are also equidecomposable with each other. Every triangle is equidecomposable with some rectangle. Any two equal-area rectangles are equidecomposable. Any two equal-area polygons are equidecomposable. (This is the Bolyai-Gerwien theorem.) Any two figures that are equidecomposable are also equicomplementable. 1.3 Into the Third Dimension At the end of the nineteenth century, the question was settled in the second dimension and was expanded into the third. This is the topic that we will focus on: can the Bolyai-Gerwien theoremâ€"that any two equal-area polygons are equidecomposableâ€"be extended into the third dimension? Instead of looking at polygons, we will look at their counterparts in the third dimension, polyhedra. Polyhedra have been defined in many ways, and not all of the definitions are compatible. We will use the definition described in Cromwell’s Polyhedra. A polyhedron is the union of a finite number of polygons that has the following properties: The polygons can only meet at their edges or vertices. Each edge of each polygon is incident to exactly one other polygon. It is possible to travel from the interior of any of the polygons to the interior of any of the others without leaving the interior of the polyhedron. It is possible to travel over the polygons incident to a vertex without passing through that vertex. Within the constraints of this definition, polyhedra are diverse and varied. Below are some examples. (If the squished part of the second-to-last polyhedron were further squished into a single point, then by the third bullet point above it could no longer be a single polyhedron.) Just as any pair of polygons of equal area can be decomposed and reassembled into the same square, can any pair of polyhedra of equal volume be decomposed and reassembled into the same cube? Are polyhedra of equal volume equidecomposable? Are they equicomplementable? By the end of the nineteenth century there were several examples of equalvolume polyhedra that were both equidecomposable and equicomplementable, but there was no general solution. One simple example is prisms with the same height and equal area bases, stemming from the two-dimensional polygon result. In 1844 Gerling showed (and then Bricard proved again in 1896) that two mirror-image polyhedra are equidecomposable by cutting them up into congruent mirror-image pieces that can then be rotated into each other. There were also some specific tetrahedra equidecomposable with a cube, shown in 1896 by M.J.M. Hill. We can reduce the problem about polyhedra in general to a problem about tetrahedra. 1.4 Definition III: Tetrahedron A tetrahedron is a polyhedron with four triangular faces, six edges, and four vertices. In many current math textbooks the faces are required to be congruent. We are not going to require that any of the faces be congruent; our definition is closer to what many current math textbooks call a triangular pyramid. Just as any polygon can be cut up into triangles, any polyhedron can be cut up into tetrahedra. First, we can cut any polyhedron into a finite number of convex polyhedra. These can each then be cut up into a finite number of pyramids with polygonal bases. Because any polygon can be cut up into a finite number of triangles, each of these polygonal pyramids can be cut up into a finite number of triangular pyramids. This means that if we can prove or disprove the Bolyai-Gerwien theorem in the third dimension for tetrahedra, then we have also proved or disproved it more generally for polyhedra. Below is an example of a division of a polyhedron into triangular pyramids, based off of figures 13.12 and 13.13 in Rajwade’s 2001 Convex Polyhedra. The volume of a tetrahedron is ? the area of the base multiplied by the height; as described by Euclid, any two tetrahedra with bases of equal areas and with equal heights will also have the same volume. This definition of volume is reminiscent of the parallel result in two dimensions: the volume of a triangle is ½ the length of the base multiplied by the height. Unlike the area of a triangle, however, the volume of a tetrahedron and therefore the volume of a polyhedron is found through calculus, by dividing the three-dimensional polyhedron into infinitesimally thin two-dimensional cross sections and adding up their areas. If the Bolyai-Gerwien theorem can be expanded into the third dimension, we can define the volume of any three dimensional polyhedron the same way we define the area of a two dimensional polygon, by breaking it up into discrete building blocksâ€"tetrahedra in three dimensions and triangles in twoâ€"and reassembling the pieces into a cube or square. This would be an elementary solution, with no infinities (or calculus) required. This problem was posed by C.F. Gauss. In a letter in 1844, Gauss expressed that he wanted to see a proof that used finitely rather than infinitely many pieces. By Hilbert’s time it was not yet solved. 1.5 David Hilbert’s 23 Problems One of my favorite aspects of geometry is how seamlessly it can transition from elementary school math to the cutting edge. Modern geometry is built on fluid connections between the basic principles. This foundation was built largely by David Hilbert. The fundamentals of geometry were initially outlined by Euclid in Elements. In the nineteenth century, geometry was becoming increasingly abstract and less and less tied to the original shapes. As Hilbert said in a lecture: “One must be able to say at all timesâ€"instead of points, straight lines, and planesâ€"tables, chairs, and beer mugs.” In this context some of the unstated assumptions in Euclid’s Elements came uncovered. (As an example, one such unstated assumption was that if two lines cross, they must have a point in common.) Hlibert extended Elements, providing an axiomization of Euclidian geometry and proofs for the unchecked assumptions that stood in the way of geometry being as fully useful as algebra. David Hilbert was born in Wehla, Germany. His mother was interested in philosophy, and his father was a judge and wanted him to study law. He was homeschooled for two years and began school two years late, at age eight. The subject he went on to study was mathematics, because he did not like memorization. A turning point for Hilbert was at his first presentation, at the Technische Hochschule, where he impressed and befriended Klein, 13 years his senior. After his dissertation on invariants Hilbert went to Paris to meet the leading mathematicians upon Klein’s suggestion. In his letters back to Klein, he made comical judgements of some of the most important mathematicians of the time. In particular, he reflected that Poincar ´e, with whom Klein had a (mental breakdown inducing) rivalry, was quite shy, and that the reason he published a lot of papers was that he published even the smallest results. Hilbert continued his research in Germany, first at his hometown university and then at G ¨ottingen when he was invited over by Klein in 1894. In the beginning he continued to focus on invariant theory, a branch of abstract algebra examining how algebraic expressions change in response to change in their variables. Hilbert’s style of teaching was very different from Klein’s: unlike Klein’s prepared, perfectionist style of lecturing, Hilbert would prepare only an outline beforehand and work through the mathematics in front of his students, mistakes and all. Hilbert and Klein revolutionized the teaching of mathematics, incorporating visual aids and connecting mathematical concepts to their applications in the sciences, and elevated G ¨ottingen to a leading institution in mathematics. It was here, starting with lectures to his students, that Hilbert became interested in geometry. 1900 was the dawning of a new century and a new era of mathematics. At the second meeting of the International Congress of Mathematicians (to this day the largest math conference, meeting once every four years, and the conference at which the Fields Medal is awarded), that year in Paris, David Hilbert, then 38, presented a monograph of ten of the open problems that he considered the most important for the next century. He published all 23 later, in a report titled simply “Mathematical Problems.” Some of Hilbert’s 23 problems have been very influential in the subsequent century of mathematics, and almost all of them have been solved. (The two that have not been solved, the axiomization of physics and the foundations of geometry, are now considered less of a priority and too vague for a definitive solution, respectively.) 1.6 Hilbert’s Third Problem We are interested in the third of the 23 problems, which concerns the extension of the Bolyai-Gerwien theorem into the third dimension. Unlike Gauss, Hilbert did not believe that there was such a bridge: he asked simply for two tetrahedra that together formed a counterexample. Here we reproduce Hilbert’s third problem in its entirety, with the problem statement in bold. In two letters to Gerling, Gauss expresses his regret that certain theorems of solid geometry depend upon the method of exhaustion, i.e., in modern phraseology, upon the axiomization of continuity (or upon the axiom of Archimedes). Gauss mentions in particular the theorem of Euclid, that triangular pyramids of equal altitudes are to each other as their bases. Now the analogous problem in the plane has been solved. Gerling also succeeded in proving the equality of volume of symmetrical polyhedra by dividing them into congruent parts. Nevertheless, it seems to me probable that a general proof of this kind for the theorem of Euclid just mentioned is impossible, and it should be our task to give a rigorous proof of its impossibility. This would be obtained as soon as we succeed in exhibiting two tetrahedra of equal bases and equal altitudes which can in no way be split up into congruent tetrahedra, and which cannot be combined with congruent tetrahedra to form two polyhedra which themsel ves could be split up into congruent tetrahedra. Armed with our definitions, can find a more succinct and powerful statement of the challenge: Specify two tetrahedra of equal volume which are neither equidecomposable nor equicomplementable. Hilbert’s third problem was the first of the 23 to be solved. The first part of the problem, on equidecomposability, was solved by Hilbert’s student Max Dehn just a few months after the conference, before the full 23 problems were printed. The proof rests on a value describing a polyhedron, the Dehn invariant, which we will look at in more detail later. The Dehn invariant does not change when the polyhedron is cut apart and reassembled into a new shape: if two polyhedra are equidecomposable, then they must have same Dehn invariant (and they do). However, not all polyhedra with the same volume have the same Dehn invariant. Specifically, Dehn used the example of a regular tetrahedron and a cube of equal volume, and we will examine this case as well. Two years later Dehn showed in a second paper the second part of the problem, on equicomplementability. An incomplete and incorrect proof was published by R. Bricard four years previously in 1896. It was not cited by Hilbert, but Dehn based his proof largely on Bricard’s. Dehn’s paper was not easy to understand. It was refined by V.F. Kagan from Odessa in 1903. In the 1950s, Hadwiger, a Swiss geometer, together with his students found new properties of equidecomposability. This allowed for a more transparent presentation of Dehn’s proof. Further progress over the past century has made it even clearer and more concise. We will be presenting a very recent version of the proof, published in 2010 in Proofs from the Book. You may have noticed a theme; indeed, the story of the third of David Hilbert’s 23 problems, the quest to expand or blockade the Bolyai-Gerwien theorem from the second into the third dimension, is a story of threes. We will present the solution in three parts: three definitions, three proofs, and three examples. We have already gone through the definitions. A tetrahedron, as defined here, is a polyhedron with four (not necessarily congruent) triangular faces. Equidecomposability is a relationship between two shapes in which one can assemble one from all the pieces of another. Finally, equicomplementability is a relationship in which one can add congruent shapes to the two shapes to form two equidecomposable supershapes. Now, we will go through the three proofs, of the pearl lemma, the cone lemma, and Bricard’s condition. Afterward we will apply the result to three example tetrahedra. Two of these tetrahedra together form a counterexample, solving Hilbert’s third problem. 2 Three Proofs 2.1 Proof I: The Pearl Lemma Our first proof is by Benko. We start by defining the segments of an edge. Each edge in a decomposed shape consists of one or more segments that, placed end to end, make up the total length of that edge. In a decomposition of a polygon, the endpoints of segments are always vertices; in a decomposition of a polyhedron, the endpoints of segments can also be at the crossing of two edges. Otherwise, all non-endpoint points within any one segment belong to the same edge or edges. (In the decomposition of the square below, for example, the hypotenuse of the larger triangle is subdivided into two segments.) We examine two equidecomposable figures. They are broken up into the same pieces, which are rearranged and perhaps reflected in different ways. Imagine that we must distribute whole, indivisible tokensâ€"which we will call pearlsâ€" on all the segments in the two decompositions. The pearl lemma states that we can place the same positive number of pearls onâ€"or, in other words, assign the same positive integer toâ€"each segment in the two decompositions in such a way that each edge of a piece gets the same whole number of pearls no matter which of the two decompositions it is sitting in. Below, for example, is a correct distribution of pearls on the equidecomposable house and square we looked at earlier. For each edge of a piece, the sum of the pearls placed on the segments making up that edge in the decomposition of one figure must equal the sum of the pearls placed on the segments making up that edge in the corresponding decomposition of the other figure. In the second dimension, this simply means that the number of pearls we place on an edge in the first decomposition is equal to the number of pearls we place on that edge in the other figure’s decomposition. In the third dimension, an edge can consist of multiple segments that need not be consistent; an edge’s segments can be different, and even different in number, in the two decompositions. However, the number of pearls placed on a piece’s edge must still be the same between the two decompositions. We can express this idea as a system of linear equations. The variables to solve for are the numbers of pearls on each segment. All of the coefficients in our linear equation are positive integers; in fact, since each segment is represented once, all of the coefficients are 1. We can satisfy this system of linear equations by assigning a positive real number of pearls (which may or may not be an integer) to each segment. As an easy example, the number of pearls assigned to each segment can be equal to that segment’s length. Remember, though, that we cannot damage the pearls! We need to show that if our linear equation has positive real number solutions, then there are also positive integer solutions representing whole, unchopped pearls. This brings us to our second proof, a proof of the cone lemma. 2.2 Proof II: The Cone Lemma The cone lemma is our name for the 1903 integrality argument by Kagan, which greatly simplified Dehn’s proof. We must show that if our system of linear equations has a positive real solution, then it also has a positive integer solution. We start with a set of homogenous linear equations Ax = 0, x 0 with integer coefficients (integer values in A) and positive real solutions (positive real values in x). This is a translation of our set of linear equations in the form Ax = b, x 0 required by the pearl lemma. We need to show that if the set of real solutions to Ax = 0, x 0 is not empty, then it also contains positive integer solutions. We required that our solutions be greater than zero. However, if there are solutions that are greater than zero, then there must also be solutions greater than or equal to one, since an all-positive solution vector x can always be multiplied by some positive value to produce an equivalent vector with all values at least 1. Therefore it will suffice to show that if the set of real solutions to Ax = 0, x = 1 is not empty, then it also contains integer solutions with all values greater than or equal to 1. We will prove this using a method devised by Fourier and Motzkin. We will use Fourier-Motzkin elimination to show that there exists a lexicographically smallest real solution x to Ax = 0, x = 1 and that if the coefficient matrix A is integral, then that lexicographically smallest solution is rational. Proving that if there is a real solution, then there must exist an integer solution to Ax = 0, x = 1 is equivalent to proving the same for Ax = b, x = 1. Therefore we will show that there exists a lexicographically smallest real solution x to Ax = b, x = 1 and that if A and b are integral, then that lexicographically smallest solution is rational. (Here, the lexicographically smallest vector results from a comparison of the elements in the vectors in order. If a vector has the smallest first element, then it is the lexicographically smallest vector. If there is a tie in the first element, then we compare the second element, followed by the third, and so on.) We will use a proof by induction on N, the size of any possible solution vector x = (x1, , xN). First we consider the base case, N = 1. There is only one variable x1 to solve for, and there are only inequalities that involve x1. We simply assign to x1 the smallest value that it can take on. Now we consider solutions in multiple variables, N 1. We examine those inequalities that involve xN. Because we have declared that all elements of the solution must be greater than or equal to 1, we know that there is at least a lower bound on xN: xN = 1. There might also be additional lower bounds or upper bounds. As before, we set xN to the smallest value it could take on. Now we are looking at a smaller system, in N - 1 variables, Ax = b, x = 1. This system contains all the inequalities of the previous system in N variables except for those involving xN , which we just resolved. In addition, we add a new constraint that all upper bounds on xN are at least as large as all lower bounds on xN. We continue by looking at the inequalities involving xN-1 and setting xN-1 to the smallest value it could take on, then examining xN-2, and so on. We established that the base case system in one variable has a smallest solution. The system that preceded it is a system in two variablesâ€"we created the N = 1 system out of the N = 2 system by setting the second variable x2 to its smallest possible value. This is definitely possible to do since each variable xi has at the very least an inequality xi = 1. We know then that the system in two variables has a smallest solution as well. We can follow the same logic to find that the system in three variables has a smallest solution, and a system in four variables, and so on. If the system in N variables has a smallest solution, then the preceding system in N + 1 variables has a smallest solution as well. All of these lexicographically minimal solutions are rational, since the inequalities that we use to set the minimal value of each variable have integer coefficients. Therefore, if a system of homogeneous linear equations has a positive real solutionâ€"as does the system generated by the pearl lemma (translated so that it is a system of homogeneous linear equations)â€"then it also has a positive integer solution. We have proved the cone lemma and with it the pearl lemma. (Why is this called the cone lemma? A set of homogenous linear equations Ax = 0, x 0 with integer coefficientsâ€"integer values in Aâ€"and positive real solutionsâ€"positive real values in xâ€"is called a rational cone.) 2.3 Proof III: Bricard’s Condition Our final proof allows us to connect the pearl lemma to concrete examples and finally solve Hilbert’s third problem. Bricard’s condition was claimed but incorrectly proved by Bricard in his 1896 paper. Dehn proved it successfully, and the proof has since been refined. From here on out we will focus on the dihedral angles of three-dimensional polyhedra, the angles between faces that share an edge. A square pyramid, for example, has eight dihedral angles, one for each of its eight edges. Two of them are illustrated below A tetrahedron has six dihedral angles (one for each of its six edges) and a cube has twelve. Bricard’s condition states that, looking at two three-dimensional polyhedra that are equidecomposable or, more generally, equicomplementable, their dihedral angles must be linear combinations of each other with a difference of some multiple of p. In other words, if we define the dihedral angles of one polyhedron a1, , ar and the dihedral angles of an equidecomposable or equicomplementable polyhedron ß1, , ßs, there must be some positive integers m1, , mr and n1, , ns and an integer k such that m1  a1 + + mr  ar = n1  ÃŸ1 + + ns  ÃŸs + kp We will prove first that the property holds for any two equidecomposable polyhedra and then that it holds more generally for any two equicomplementable polyhedra. 2.3.1 Equidecomposability We start by assuming that two polyhedra are equidecomposable. As a (two-dimensional) example, we can look at the same house and square from earlier. We cut the two polyhedra up into two decompositions made up of the same pieces, and we follow the pearl lemma to assign a positive integer number of pearls to each segment in each of the two decompositions: each piece has the same total number of pearls on each of its edges in either of the two decompositions. At each pearl, we measure the dihedral angle made by the faces incident to the edge at the location of the perl. If the perl is in the interior of the figure, its dihedral angle will be p or 2p. If the pearl is on an edge of a piece but not on an edge of the figure being decomposed, then its dihedral angle will be p. We define a sum ?  1 as the sum of the dihedral angles at all of the pearls in the pieces of the first polyhedron, one angle per pearl. We also define a second sum ?  2, the sum of the dihedral angles at all of the pearls in the pieces of the second shape, again one angle per pearl. Multiple pearls can have the same dihedral angle, sometimes by being on the same edge and sometimes by being on different edges with the same dihedral angle, such as the two dihedral angles in the square pyramid illustrated earlier. In either case the dihedral angle will appear multiple times in our sum. The number of times a dihedral angle appears, and therefore its multiplier in either of the sums, must be a positive integer. Similarly, there will be an integer number of pearls with a dihedral angle of p or 2p, and therefore a nonnegative integer multiplier of p. If we define the dihedral angles of one polyhedron a1, , ar, we can represent the first sum as ?  1 = m1  a1 + + mr  ar + k1  p for some positive integers m1, , mr and a nonnegative integer k1. Similarly, if we define the dihedral angles of the second polyhedron ß1, , ßs, we can represent the second sum ?  2 = n1  ÃŸ1 + + ns  ÃŸs + k2  p for some positive integers n1, , ns and a nonnegative integer k2. We get ?  1 and ?  2 by adding the dihedral angles of each piece in the decompositions of the first polyhedron and the second polyhedron, respectively. These pieces are congruent, since the two polyhedra are equidecomposable and we are looking at decompositions that satisfy this relationship. Therefore by the perl lemma, which we used to distribute the pearls, each edge of each piece has the same dihedral angles in either of the two decompositions and the same number of pearls with each dihedral angle. In other words, ?  1 = ?  2 or, substituting in the definitions of ?  1 and ?  2, m1  a1 + + mr  ar + k1  p = n1  ÃŸ1 + + ns  ÃŸs + k2  p We define the integer k = k2 - k1 and find that our equation becomes Bricard’s condition. m1  a1 + + mr  ar = n1  ÃŸ1 + + ns  ÃŸs + kp 2.3.2 Equicomplementability Now we assume that our two polyhedra are equicomplementable. As a two dimensional example, recall that we can add congruent pieces to our house and boxâ€" â€"such that the resulting polygons are equidecomposable. We can create two superfigures from our equicomplementable polyhedra by adding congruent pieces to the two polyhedra such that the superfigures are equidecomposable. In other words, we will be able to cut apart the superfigures so that their pieces are our two original polyhedra (one original polyhedra in each superfigure) and otherwise the same pieces. These are our first two decompositions of the two superfigures and our first decomposition of each individual superfigure. In addition, we can also cut apart the superfigures into an alternative decomposition such that they share all their pieces. These form the third and fourth decompositions of our superfigures, the second decomposition of each individual superfigure. We again apply the pearl lemma to distribute the pearls on the edges of all the pieces in all four decompositions. Because we used the cone lemma to prove the perl lemma, we can impose an additional constraint: each edge of each superfigure must have the same total number of pearls in each of the two decompositions it is involved in. As before, we compute the sums of the dihedral angles at all the pearls, ?1, ?2, ?1, and ?2. ?1 and ?2 are the first decomposition of each superfigure. ?1 includes our first original polyhedron, ?2 includes our second original polyhedron, and each other piece of ?1 is congruent to a piece in ?2 and vice versa. ?1 and ?2 are the alternative decompositions of the two superfigures. Each piece in ?1 has a congruent counterpart in ?2 and vice versa. First, we notice that ?1 and ?2 are decompositions into the same pieces. As in our proof for equidecomposable polyhedra, this means that ?1 and ?2 must be identical: ?1 = ?2. Next, we notice that ?1 and ?1 are decompositions of the same polyhedron, our first superfigure. Recall that we restricted our placement of pearls so that each edge of the superfigure needed to have the same number of pearls. Recall also that pearls that are placed inside of the superfigure or on the inside of one of its faces, rather than on one of its edges, yield dihedral angles of p or 2p. Though the two decompositions of our first superfigure might have different numbers of pearls, the edges of the superfigure itself must have the same numbers of pearls. This means that ?1 and ?1 can differ by an integer multiple of p, which we here call l1.                    ?1 = ?1 + l1  p By the same logic, with an integer l2,                    ?2 = ?2 + l2  p Since we deduced above that ?1 = ?2, we can restate these two equations as a relationship between ?1 and ?2, with l  =  l2 -  l1.                    ?2 = ?1 + lp We now have a statement describing the relationship between the original decompositions of the two superfigures, ?1 and ?2â€"we are almost done! Remember that the difference between ?1 and ?2 is that the first contains our first polyhedron and the second contains our second polyhedron. Each other piece in ?1 has a congruent counterpart in ?2 and vice versa. Since they are identical, we can subtract the contributions of these congruent pieces from each side of the equation. This leaves the first polyhedron’s contributions to its superfigure and the second polyhedron’s contributions to its superfigure. As before, they will differ by an integer multiple of pâ€"lp. m1  a1 + + mr  ar = n1  ÃŸ1 + + ns  ÃŸs +  lp This is Bricard’s condition as described in the opening of this proof, replacing the integer k with the integer l. We have now shown that Bricard’s condition holds for both equidecomposable and equicomplementable polyhedra. 2.3.3 The Dehn Invariant We do not use it in this version of the proof, but we will pause for a moment to formally define the Dehn invariant, which was used by Dehn in his original proof. You may notice a few parallels to Bricard’s conditionâ€"indeed, though it was published earlier, Bricard’s condition is a consequence of the relationship between the Dehn invariant and equidecomposability. We define in radians the dihedral angles of a polyhedron P: a1, a2, , ap. We also define the lengths l1,  l2, ,  lp of the corresponding edges in P. The Dehn invariant f(P) of the polyhedron P with respect to the function f is the sum: f(P) =  l1  f(a1) +  l2  f(a2) + +  lp  f(ap) f is an additive functionâ€"every linear dependence in the input values also exists  in the output values. We also add the additional constraint that f(p) = 0. In his original proof, Dehn showed that the Dehn invariant does not change when a polyhedron is cut apart and reassembled into a new shape: if two polyhedra P and Q are equidecomposable, then they have same Dehn invariant. In other words, for every additive function f that is defined for P’s and Q’s dihedral angles and for which f(p) = 0, f(P) = f(Q). Dehn found, however, that not all polyhedra with the same volume have the same Dehn invariant, demonstrating that not all polyhedra with the same volume are equidecomposable and thereby solving Hilbert’s third problem. This proof is more complicated than the three-part proof we presented here; it is described in detail in the corresponding chapter of the third edition of Proofs from the Book (otherwise we refer to the fourth edition). 3 Three Examples Finally, we can apply Bricard’s condition to the dihedral angles of three example tetrahedra and identify a counterexample, our solution to Hilbert’s third problem. 3.1 Example I: A Regular Tetrahedron Our first example is the same example used by Dehn in his first paper. Here we examine a regular tetrahedron, a tetrahedron in which each of its four faces is an equilateral triangle. All dihedral angles in a regular tetrahedron are arccos ?. In contrast, all dihedral angles in a cube are p/2. We will show that the regular tetrahedron cannot be equidecomposable nor equicomplementable with any cube. Because of the work we have already done, this is very easy to do! If a regular tetrahedron were equidecomposable or equicomplementable with a cube, then by Bricard’s condition, there must be some positive integers m1 and n1 and some integer k such that m1 arccos ? = n1p/2 + kp We can rearrange this equation to solve for k. k = 1/p (m1 arccos ? - n1 p/2) k = m1 1/p arccos ? - ½ n1 1/p arccos ? is irrational (we will not examine the proof here, but it is in chapter 7 of Proofs from the Book). k, then, must also be irrationalâ€"it cannot be an integer. Bricard’s condition does not hold for these two figures. This means that a regular tetrahedron cannot be equidecomposable nor equicomplementable with any cube, even a cube of the same volume. This example alone demonstrates that two polyhedra of equal volume are not necessarily equidecomposable or equicomplementable with each other. However, though we have gotten at the spirit of Hilbert’s third problem we have not solved it exactly. We will look at two more examples to produce two tetrahedra of equal volume that are neither equidecomposable nor equicomplementable. 3.2 Example II: A Tetrahedron with a Vertex Incident to Three Orthogonal Edges Our next example very closely parallels the logic we saw above. Now we look at a tetrahedron with three orthogonal edges of equal length u that share a vertex. Three of this tetrahedron’s six dihedral angles (the turquoise angles) are right angles (size p/2). The other three (the red angles) have size arccos v?. Like the regular tetrahedron in our previous example, this tetrahedron cannot be equidecomposable nor equicomplementable with a cube. If it were, then by Bricard’s condition there must be some positive integers m1, m2, and n1 and some integer k such that m1 arccos ? + m2 p/2 = n1 p/2 + kp As before, we can rearrange this equation to solve for k. k = 1/p (m1 arccos ? + (m2 - n1) p/2) k = m1 1/p arccos v? + ½ (m2 - n1) 1/p arccos v? is irrational (this proof is also covered in chapter 7 of Proofs from the Book). As in the first example, k must be irrational as well and therefore cannot be an integer. Bricard’s condition does not hold for this relationship. This means that a tetrahedron with a vertex incident to three orthogonal edges also cannot be equidecomposable or equicomplementable with a cube. 3.3 Example III: A Tetrahedron with an Orthoscheme Finally we look at a tetrahedron with a three-edge orthoscheme: a sequence of three edges that are mutually orthogonal and in this case of identical length u, the same length as each of the three orthogonal edges in the previous example. The dihedral angles in this tetrahedron are p/2 (the three turquoise angles), p/4 (the two orange angles), and p/6 (the blue angle). The dihedral angles of this tetrahedron are all rational multiples of p/2, the size of any dihedral angle of a cube. In fact, a cube can be decomposed into six congruent tetrahedrons with an orthoscheme. Notice that this tetrahedron has the same base and height as the tetrahedron in the second example, and therefore the same volume. However, it cannot be eqidecomposable nor equicomplementable with a tetrahedron with those dihedral angles (or with the regular tetrahedron in the first example). If it were, then by Bricard’s condition there must be some positive integers m1, m2, m3, n1, and n2, and some integer k such that m1 p/2 + m2 p/4 + m3 p/6 = n1 arccos v? + n2 p/2 + kp As in the previous two examples, we can rearrange this equation to solve for k. k = 1/p (m1 p/2 + m2 p/4 + m3 p/6 - n1 arccos v? - n2 p/2) k = ½ m1 + ¼ m2 + ? m3 - ½ n2 - n1 1/p arccos v? Recall from the second example that 1/p arccos v? is irrational. k, then, cannot be an integer and Bricard’s condition cannot hold for this relationship. A tetrahedron with an orthoscheme of three identical-length edges cannot be equidecomposable nor equicomplementable with a tetrahedron with a vertex incident to three orthogonal edges of that same length. We have found two tetrahedra with the same base and height (and therefore the same volume) that are neither equidecomposable nor equicomplementable. In them we have found a solution to Hilbert’s third problem. 4 Open Problems The theory of the equidecomposabililty and equicomplementability of polyhedra, revived by Hilbert’s third problem, is largely solved. When we look at polytopes in higher dimensions, however, some gaps do remain. Though we did not use it in this proof, we mentioned earlier the Dehn invariant and its relationship with the equidecomposability of polyhedra: if two polyhedra are equidecomposable, then they necessarily have the same Dehn invariant. In 1965, Sydler proved that this condition is not only necessary, but also sufficient for equidecomposability: if two polyhedra have the same Dehn invariant, then they are necessarily equidecomposable. This result has not yet been provedâ€"and its validity is unknownâ€"for spheres and hyperbolic spaces of at least three dimensions, and in general for dimensions of at least five. If we only allow translations while reconstructing the second polytope from the decomposition of the first, then this problem is open in four or more dimensions. In addition, the problem of finding the minimum number of allowed motions necessary for equidecomposability is also unsolved in four or more dimensions. The consequences of restricting the motions in equidecomposability (to translations, to translations and central inversions, or to all motions that preserve orientation) and the existing proofs in lower dimensions are explored in depth in Boltíànskii’s Hilbert’s Third Problem. 5 Conclusion Hilbert’s third problem is one example of the necessity and beauty of a rigorous mathematical proof. If the Bolyai-Gerwien theorem could have been expanded into the third dimension, then we could define the volume of any three dimensional polyhedron using the discrete methods from the second dimension. Instead, progressing from area in the second dimension to volume in the third dimension requires breaking the shape into continuous, infinitely small building blocks and the use of calculus, an entire other toolbox and an entire other field: we share the third dimension with unavoidable infinities. 6 References The chapter “Hilbert’s Third Problem: Decomposing Polyhedra” (pages 53-61) from the 4th edition of Proofs from the Book  by Martin Aigner and Günter M. Ziegler (published in 2010 by Springer in Berlin). That same chapter (pages 45-52) from the 3rd edition of Proofs from the Book (published in 2004)â€"this one is based on Dehn’s original proof. “A New Approach to Hilbert’s Third Problem” by David Benko, pages 665-76 of volume 114 of The American Mathematical Monthly  (2007). Hilbert’s Third Problem by V.G. Boltíànskii (published in 1978 by V.H. Winston Sons in Washington, D.C.). Polyhedra: One of the Most Charming Chapters of Geometry  by Peter R. Cromwell (published in 1997 by Cambridge University Press in New York). Scissors Congruences, Group Homology and Characteristic Classes by Johan L. Dupont  (published in 2001 by World Scientific in Singapore). Revolutions of Geometry by Michael O’Leary (published in 2010 by John Wiley Sons in Hoboken, NJ). Convex Polyhedra with Regularity Conditions and Hilbert’s Third Problem by A.R. Rajwade (published in 2001 by the Hindustan Book Agency in New Delhi). Hilbert’s Third Problem: Scissors Congruence by Chih-Han Sah (published in 1979 by Fearon Pitman in San Francisco).                             If you’re hungry for more college-level math, I recommend starting with Proofs from the Book, the source of this proof and the textbook we used in 18.304â€"they just came out with their fifth edition and you can read it for free online. Proofs is divided into five sections, each one containing a sample of a field of mathematics:  number theory,  geometry,  analysis,  combinatorics, and  graph theory. Together these cover a lot of what we learn in course 18  and, in the case of combinatorics and graph theory, course 6-3 (or the course 6 part of 6-7, in my case).

Absence of Evidence, or Evidence of Absence; A paper on...

Absence of Evidence, or Evidence of Absence? A paper on Animal Consciousness nbsp;nbsp;nbsp;nbsp;nbsp;Consciousness is a difficult term to grasp; so much so, that many scientists will not even attempt to define the term, much less search for it’s evidence. Most however, do agree that consciousness must include certain aspects; specifically cognition, self-awareness, memory, and abstract thought. nbsp;nbsp;nbsp;nbsp;nbsp;Lesley J. Rogers describes consciousness as, â€Å"related to awareness, intelligence, and complex cognition, as well as language. Consciousness may be manifested in self-awareness, awareness of others, intentional behavior, including intentional communication, deception of others, and in the†¦show more content†¦nbsp;nbsp;nbsp;nbsp;nbsp;In Dr. Pepperberg’s book, The Alex Studies, she taught the parrot to be able to recognize different objects by color, shape, and material. He was even able to eventually distinguish between concepts such as â€Å"bigger,† â€Å"smaller,† â€Å"same,† â€Å"different,† â€Å"over,† and â€Å"under.† When asked to identify objects, Alex correctly identified, on first try, 80% of all objects presented in over 200 tests (45). He was also able to correctly pair different labels together to fit a certain object; for example color and material. After only two years of traini ng, Alex was able to communicate with contextual and conceptual use of human speech. He could identify, request, and refuse a set of objects for play or food (50). Dr. Pepperberg also took precautions to ensure that she had not allowed for any â€Å"cues† to tip off Alex to a correct answer, as in the case of â€Å"Clever Hans.† nbsp;nbsp;nbsp;nbsp;nbsp;These animals also demonstrated memory, another of the aspects of consciousness. Many behaviorists believe that animals act only on instinct, or on conditioned responses to stimuli. Others, like Lesley J. 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It s Only A Game - Isn t It - 1635 Words

It’s Only a Game - Isn’t It? A small group of people sat around a ouija board expecting nothing to happen. A few nervous giggles were exchanged amongst each other. The laughter quickly stopped when the planchette stopped moving and the room became cold; the air turned thick and heavy. Something entered the space, a feeling of distress surrounded the group. They became paranoid, and felt like there was something looking at them from the dark corners. After what felt like hours trying to end the session, the planchette still refused to move, almost as if it was frozen on the board. As their impatience quickly grew, they forced the planchette to goodbye. The feeling of despair had vanished, but the feeling of being watched remained as everyone left the room that night. 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Cheat Sheet MDM Risk analysis Free Essays

Decisions based on them are dangerous! A single point only ever tells us what the average of two cases is, never what happens between the two cases! Poor understanding of downside risk poor understanding of upside opportunity 2) Scenario analysis: Define your scenarios; best-worst-base There are a range of results! Check if risk makes a difference 3) Use distributions for the uncertainties to describe key risk drivers Choose distribution based on historical data or expert opinion Distribution is important for the simulation; based on the given distribution, the simulator ill be more/less likely to pick numbers in specific ranges Uniform: same probability of all numbers in a given range Triangle: point within the range is much more likely than the other points Normal: you know the middle point but it could be off by X in either direction 4) Run (at)Risk (Monte-Carlo simulation) Define distributions (step 3) Define output cell fir which to simulate results Things to look out for Mean of objective variable (usually NAP) Compare results with scenario results (atria’s will give better indication of the range than the scenarios! Look at full range of outcomes Look at standard deviation and at confidence range Look at downside risk and upside potential. What is % of being above/below specific n umber? What is breakable probability? What is the distribution like? Perform Monte-Carlo simulation to Evaluate different possible outcomes Determine expected result, range of results, probability of results (e. G. We will write a custom essay sample on Cheat Sheet MDM Risk analysis or any similar topic only for you Order Now Probability of break-even), downside risk, etc.. Advantages: avoid the Flaw of Averages, understand the risk, test your intuition 5) Sensitivity analysis Purpose Examine sensitivity of results when model parameters are varied Observe change in results due to change in assumptions Identify main uncertainty drivers / key risk drivers Methodology What-if analysis (simple changing of numbers to see what happens) One-way two-way sensitivity analysis Tornado diagrams One-way two-way sensitivity analysis Use one-way sensitivity analysis (data table) to check how changes to a variable effect the output variable. Use Goal Seek to find breakable point of that variable. Use two-way sensitivity analysis (data table) to check for changes in two different variables at the same time Tornado diagram Check for impact of each variable / parameter, sorted in order of magnitude Shows you on which variables you should focus most, where the most important risks lie! Some Excel info points: Simulation settings: EXAMPLE QUESTIONS ON RISK ANALYSIS 1 . In what type of decision context could risk analysis be useful and why may it be dangerous to rely on single point forecasts? What techniques can you use to overcome the problems of such forecasts? How do you decide what technique is most appropriate to use? Every business decision entails risk dangerous! A single point only ever tells us what the average Of two cases is, never what happens between the two cases! Example answer for this part: These numbers are based on the average scenario which is not necessarily representative of the true value (argue why could over- or underestimate). Furthermore, they do not tell us anything about the risk. Technique: scenario analysis or simulation 2. Explain in your own words how Monte Carlo Simulation could be useful to a decision maker Evaluate different possible outcomes Averages, understand the risk, test your intuition 3. Explain how the simulation process works to produce results that are useful to a decision maker Example answer: This is different from the E,250 that Carolina’s predecessor estimated because the original estimate was made using only single-value estimates for each of the variables. However, by using a Monte Carlo simulation that allows for a range of possible values (with a triangular distribution to account for the higher likelihood of the values Of 5% and 20% for economy and business, respectively). This means that, based on 1 ,OHO iterations of possible combinations for each of the variables as per the arranging definition of the potential values for each variable under each iteration, the mean of the cost is E 10,277. 4. A friend of yours has just learned about simulation methods and has asked you to conduct a complicated risk analysis to help her making a choice. She said she would be happy to let you solve the problem and then recommend what action she should take. Explain why she needs to be involved in the analysis and modeling process and what kind of information you need from her. Risk analysis requires information about the characteristics of a particular uncertainty (e. G. Shape of probability striation function, range of likely values etc) 5. A simulation model has produced the following three risk profiles displayed below. What advice would you give to the decision maker on the basis of this output? Choice depends on risk attitude, personal wealth, importance of project success and cost of investment alternative. Alternative C has the highest associated payoff. However, range of possible payoffs is quite large. The steeper the shape of the probability distribution function, the smaller the range of possible expected payoffs (look at standard deviation of outcomes). Consider 5% confidence interval of most likely payoffs. Alternative A has quite a big confidence interval with relatively flat slope at the edges. Look at intersection of B and C and argue which one is less risky. 6. Your boss has asked you to work up a simulation model to examine the uncertainty regarding the success or failure of five different investment projects. He provides probabilities for the success of each project individually (numbers given). Because the projects are run by people in different segments of their investment market, you both agree that it would be reasonable to believe that, given these probabilities, he outcomes of the projects are independent. He points out, however, that he really is not fully confident in these probabilities and that they could be off by as much as 0. 05 in either direction on any given probability. (a) How can you incorporate this uncertainty about the probabilities in the simulation model? Use normal distributions for each project with Sd= 0. 05 (b) Now suppose he changes probability to include ranges. How can you update your simulation model to take this additional information into account? Update probability distributions – triangle, discrete, uniform, normal Example answer: He should use historical data and his expert judgment to estimate the distribution of inputs. He should apply a normal distribution if the different values are independent of each other. Example for normal distribution argument: However, since the number of high quality applications is the sum of the individual decisions â€Å"whether or not to apply/’ of a substantial amount of high caliber young professionals, and since this decision is taken by each potential applicant to a large extend independently of each other, the normal distribution with mean 630 seems reasonable. Moreover, given the potential range of high quality applications is between 51 0 and 750, a standard deviation of 60 seems reasonable; that is, the range of 240 students corresponds to 4 standard deviations. Since the proportion of offers accepted is again the sum of many individual decisions, the normal distribution with mean 58% and standard deviation of 2% might be reasonable. 7. Interpret the following risk analysis result tables ask at: Minimum, expected, maximum, P(loss) = x % (downside risk), P( X) = Y% (upside potential) 8. Interpret sensitivity analysis Describe how output variable is sensitive to given assumptions/parameters. Describe how output variable minimizes and maximizes with the different scenarios; what is the upside potential and downside risk Example answer: The total cost decreases by El ,800 for each 5% increase in the business class no-show rate from 15% to 20% (at which point it is minimized), but then increases by E,700 per percentage point increase from 20% to 30%. The rate Of increase is consistent regardless of the rate of economy no-show. (could include more insights!!! ) The two-way sensitivity table and the accompanying chart show us that in the lower ranges of the possible no-show rates, the total cost is sensitive to both variables in fairly similar proportion, until the optimum combination (I. E. The minimized cost) is reached at 5% economy and 20% business. After this inflection point, the total cost becomes much more sensitive to changes in the business class no-show rate. 9. Describe, compare and explain the shape of a distribution. Risk profile: probability of making a loss vs. a profit Minimum versus maximum Variance Size of 90% confidence interval around the mean Expected return mean average) Include arguments why distributions might differ with different scenarios 1 0) Make recommendation based on the results. Will usually be trade-off between high risk for higher return on average and lower risk for lower return on average Include risk profiles, probabilities, maximum and minimum numbers†¦ Example answer: The policy that we have recommended is better than the others, because it has the lowest average total cost. Furthermore, the 95% confidence interval has the narrowest range of possible values, as well as the lowest probability that costs will exceed El 7,000. However, even though our recommended policy is better overall, it is not necessarily going to be the best on each individual flight. However, this doses t matter since the average cost is the single most important criterion when choosing a policy because you have 365 * 4 flights per year. One additional insight you could generate is the simulated cost difference between the current and suggested policies. The new policy is worse than the original policy 6% of the times. 1 1) What can be further done to improve profitability and manage the risks involved? How to cite Cheat Sheet MDM Risk analysis, Papers

Needs of training free essay sample

Examine the need and benefits of Training All organizations must manage four resources: money, equipment, information, and people. Investments in better equipment may speed up production or reduce waste. Investments in training and development of employees can make them more productive or more effective in their jobs, directly contributing to the bottom line. The singular function of training is to produce change. It is the upgrading of a person’s skill or the addition of a new skill, which in turn can bring about the desired change an agency is seeking. It is important to understand that training in and of itself cannot motivate a work force The purpose of training and management development programs is to improve employee capabilities and organizational capabilities. When the organization invests in improving the knowledge and skills of its employees, the investment is returned in the form of more productive and effective employees. Training and development programs may be focused on individual performance or team performance. We will write a custom essay sample on Needs of training or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page The creation and implementation of training and management development programs should be based on training and management development needs identified by a training needs analysis so that the time and money invested in training and management development is linked to the mission or core business of the organization Value and importance of Training According to Edwin B Flippo, â€Å"training is the act of increasing the knowledge and skills of an employee for doing a particular job† From the above definition, it can be observed that training is the act of increasing the knowledge and skills of an employee. It imparts specific skills for specific purposes. It is mainly job oriented. Training is needed to be given to both old and new employees throughout their tenure in the organization. †¢to foster growth and development †¢to provide opportunities for employees to accept greater challenges †¢to aid employees in contributing to the achievement of department goals and the agency’s mission and vision †¢to build employee self-confidence and commitment †¢to produce a measurable change in performance †¢to bring about the desired changes that can solve a variety of problems Types of Training Programmes Induction or Orientation Training: This training programme is meant to induct a new employee into the new social setting of this work setting. The new employee is introduced to his job situations and informed about the rules, working conditions, privileges and activities. Most of the information is likely to be embodied in a handbook which is distributed to an employee and in case of rank and file workers, the orientation may consist only of brief explanations by a member of Personnel department or the supervisor under whom the employee will work In order to maximize the effectiveness of training and development, organizations must constantly assess their employees’ current training and development needs and identify training and development needs to prepare employees for their next position. This requires that organizations recognize that different employees will have different needs and that these needs will change over time as these workers continue in their careers The different types of training programs can be broadly classified as Job Training: The purpose of job training is to increase the knowledge of workers about the jobes within which they are concerned so that their efficiency and skill of performance are improved. In job training workers learn correct methods of handling machines and equipments, avoiding accidents, removing bottlenecks,etc Promotional Training: Many companies have adopted a policy of filing some of the vacancies at higher levels by promoting existing employees. When existing employees are promoted in the Organization, they are required to shoulder new responsibilities. For this, they require training so that they may not experience any difficulty to shoulder the responsibilities of the new position to which they have been promoted Refresher Training: At the time of initial appointment, employees are formally training for their jobs, but with the passage of time, they may forget some of the methods which were taught to them or some of the methods or all of them may have become out dated because of technological development. Hence, refresher training is arranged for existing employees in order to enable them to revive and improve their knowledge.