...Runge-Kutta method (RK4) is given below. Consider the problem y = f (t, y) y(t0 ) = α Define h to be the time step size and ti = t0 + ih. Then the following formula w0 = α k1 = hf (ti , wi ) k2 = hf k3 = hf k1 h ti + , wi + 2 2 h k2 ti + , wi + 2 2 k4 = hf (ti + h, wi + k3 ) 1 wi+1 = wi + (k1 + 2k2 + 2k3 + k4 ) 6 computes an approximate solution, that is wi ≈ y(ti ). Let us look at an example: y = y − t2 + 1 y(0) = 0.5 1 The exact solution for this problem is y = t2 + 2t + 1 − 2 et , and we are interested in the value of y for 0 ≤ t ≤ 2. 1. We first solve this problem using RK4 with h = 0.5. From t = 0 to t = 2 with step size h = 0.5, it takes 4 steps: t0 = 0, t1 = 0.5, t2 = 1, t3 = 1.5, t4 = 2. Step 0 t0 = 0, w0 = 0.5. Step 1 t1 = 0.5 k1 k2 K3 K4 w1 Step 2 t2 = 1 k1 k2 K3 K4 w2 = hf (t1 , w1 ) = 0.5f (0.5, 1.425130208333333) = 1.087565104166667 = hf (t1 + h/2, w1 + k1 /2) = 0.5f (0.75, 1.968912760416667) = 1.203206380208333 = hf (t1 + h/2, w1 + k2 /2) = 0.5f (0.75, 2.0267333984375) = 1.23211669921875 = hf (t1 + h, w1 + K3 ) = 0.5f (1, 2.657246907552083) = 1.328623453776042 = w1 + (k1 + 2k2 + 2k3 + k4 )/6 = 2.639602661132812 1 = hf (t0 , w0 ) = 0.5f (0, 0.5) = 0.75 = hf (t0 + h/2, w0 + k1 /2) = 0.5f (0.25, 0.875) = 0.90625 = hf (t0 + h/2, w0 + k2 /2) = 0.5f (0.25, 0.953125) = 0.9453125 = hf (t0 + h, w0 + K3 ) = 0.5f (0.5, 1.4453125) = 1.09765625 = w0 + (k1 + 2k2 + 2k3 + k4 )/6 = 1.425130208333333 Step 3 t3 = 1.5 k1 k2 K3 K4 w3 = hf (t2 , w2 ) = 0.5f (1, 2.639602661132812)...
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...Practice 5 Problem 7 (Conditional probability, Bayes` rule). Radar detection. If an aircraft is present in a certain area, a radar correctly registers its presence with probability 0.99. If it is not present, the radar falsely registers an aircraft presence with probability 0.10. We assume that an aircraft is present with probability 0.05. What is the probability of false alarm (a false indication of aircraft presence), and the probability of missed detection (nothing registers, even though an aircraft is present)? A sequential representation of the sample space is appropriate here, as shown in Fig. 1. Figure 1: Sequential description of the sample space for the radar detection problem Solution: Let A and B be the events A={an aircraft is present}, B={the radar registers an aircraft presence}, and consider also their complements Ac={an aircraft is not present}, Bc={the radar does not register an aircraft presence}. The given probabilities are recorded along the corresponding branches of the tree describing the sample space, as shown in Fig. 1. Each event of interest corresponds to a leaf of the tree and its probability is equal to the product of the probabilities associated with the branches in a path from the root to the corresponding leaf. The desired probabilities of false alarm and missed detection are P(false alarm)=P(Ac∩B)=P(Ac)P(B|Ac)=0.95∙0.10=0.095, P(missed detection)=P(A∩Bc)=P(A)P(Bc|A)=0.05∙0.01=0.0005. Application of Bayes` rule in this...
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...Case PROBLEM The main problem that must be solved is determining whether or not Peak Selling Technologies should introduce an economy grade version of tape. SITUATION ANALYSIS Peak Selling Technologies is recognized as a leader in quality, performance, and technological innovation in the adhesive industry. Recently though small regional tape producers have found success with introducing an economy grade version of tape that is cheaper. Some members of the executive steering committee for PST thought that they should also introduce a lower quality and less expensive product into the lineup, but the President was not convinced. Peak Selling Technologies have had sales and profits grow consistently over the past five year with net sales in 2012 being $240 million with earnings before tax of almost $35 million. PST’s K2-Tape was the brand name for its pressure sensitive carton sealing tape line which was able to be used manually, by a dispenser, or automatic machine use. K2-Tape focused on the lowest overall cost to the end user through their superior performance. Their performance offered the most advanced adhesive technology, but was constantly competing against economy grade products like the ones made by Ipack. Different companies that made lower grade tape were Ipack and BOPP; Ipack tapes were lower quality and BOPP backed products that lacked K2-Tape’s strength and consistent performance, but the prices for these lower grade tapes were 25-35% below K2-Tape. K2-Tapes...
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...4/7/2014 Basic Statistics: An Overview Basic Statistics: Review Descriptive Statistics Scatter graph Measures of central tendency Mean Median, quartile, deciles, percentile Mode Weighted mean GM HM Measures of dispersion Range, IQR Semi IQR Mean deviation Standard deviation Variance Coeff of variation Inferential Statistics Populations Sampling Estimation of Parameters Point Estimation Interval Estimation Unbiased Minimum Variance Consistency Efficiency Properties of Point Estimators Statistical Inference: Hypothesis Testing T test F test Chi square test Measures of shape of the curve Moments Skewness kurtosis Probability distributions Normal Distribution T-student Distribution Chi-Square Distribution F Distribution Index Number Etc. Correlational Statistics Covariance Correlations regressions 1 4/7/2014 Some Terminology Variables are things that we measure, control, or manipulate .They may be classified as: 1. Quantitative i.e. numerical Continuous: takes fractional values ex. height in cm Discrete : takes no fractional values ex. GDP Random Variable: If the value of a variable cannot be predicted in advance Non random : If the value of a variable cannot be predicted in advance Some Terminology 2. Qualitative i.e. non numerical 1. Nominal: Items are usually categorical and may have numbers...
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...5(Non-Scenario) Conventional MC (Type A) Question Q1-K2 Description part Below you find a list of descriptions of problems that can be observed Question (stem) during testing or operation. Which is most likely a failure? Correct Answer Distracters A. The product crashed when the user selected an option in a dialog box. B. One source code file included in the build was the wrong version C. The computation algorithm used the wrong input variables. Justification for Correct Answer Partial Scoring Syllabus Ref. Learning Objective K-Level Reviewer Response Version 1.0 D. The developer misinterpreted the computational requirement for that algorithm. A is clearly a failure that can be observed during testing. B From the perspective of the configuration adding a wrong version is a fault, most probably based on a mistake by the developer / CM manager C Fault D Mistake No 1.1.2 Causes of software defects LO-1.1.5: Explain and compare the terms error, defect, fault, failure and corresponding terms mistake and bug using examples (K2) K2, Understand K3, Apply K4, Analyse Accepted with comments The ID is a different format than the other ones. However perhaps this is a better format and we should change the others instead. TM accepted changes and corrected the issues. Page 1 of 1 © International Software Testing Qualifications Board ISTQB public when released CTFL-LO-1.1.5-Q1-K2...
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...Introduction to Options Markets Chap 9 Options, Futures, and Other Derivatives, 8th Edition, John C. Hull 2011 1 Review of Option Types • A call is an option to buy • A put is an option to sell • A European option can be exercised only at the end of its life • An American option can be exercised at any time 2 Derivatives - Options Give the holder the right to buy or sell the underlying at a certain date for a certain price. (European options) • • • • • Right to buy call option Right to sell put option Payoff function: call: C(T)= max{S(T)-K, 0}, put: P(T)=max{K-S(T),0} Cash settlement Exchanges: CBOE, CBOT, Eurex, LIFFE, ... Derivatives - Options • • • • Strike K Maturity T Buy option Sell option you can buy or sell for that price date when the option expires long position (holder) short position (writer) Exercising ... ... only at maturity possible European ... at any date up to maturity possible American Option Positions • • • • Long call Long put Short call Short put 5 Long Call (Figure 9.1, Page 195) Profit from buying one European call option: option price = $5, strike price = $100, option life = 2 months 30 Profit ($) 20 10 70 0 -5 loss 80 90 100 Terminal stock price ($) 110 120 130 6 Short Call (Figure 9.3, page 197) Profit from writing (selling) one European call option: option price = $5, strike price = $100 Profit ($) 5 0 -10 -20 110 120 130 70 80 90 100 Terminal stock...
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...Technologies are a Texas based, leading manufacturer of high quality and innovative adhesive products. Their lines are heavily focused on premium industrial strength tapes, carton sealant and other packaging products. PST are steadily expanding their net sales and have experienced huge growth over the past 5 years. One of their product lines is known as K-2 tape, a pressure sealing tape of which there are nine variations. Recently Tensile, a competitor in the industry, released a new economy line tape of which PST believe may quickly become a threat to them. Now, PST are debating the consequences and outcomes of releasing an economy line tape to reduce the likelihood that Tensile’s cheaper Ecoslim and Ecoslim-wide will replace two variations of K2 tape in consumer minds. However, as this contradicts their established reputation as a premium brand, careful analysis of the industry, end customer and distributor forces, and company relationships and needs is required. PST must use this analysis to address two key challenges – retaining their current customer base, and ensuring their premium reputation remains untarnished. Customer Analysis Who are the customers? Currently, PST’s customer base largely revolves around the commercial organizational market, specifically users and distributors. The majority of end users are known to be within packaging, retail and manufacturing segments. The table below details the criteria and situation of the distributors. As they are loyal and...
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...Research on Butterfly Theorem Butterfly Theorem is one of the most appealing problems in the classic Euclidean plane geometry. The name of Butterfly Theorem is named very straightforward that the figure of Theorem just likes a butterfly. Over the last two hundreds, there are lots of research achievements about Butterfly Theorem that arouses many different mathematicians’ interests. Until now, there are more than sixty proofs of the Butterfly Theorem, including the synthetical proof, area proof, trigonometric proof, analytic proof and so on. And based on the extension and evolution of the Butterfly Theorem, people can get various interesting and beautiful results. The definition of the Butterfly Theorem is here below: “Let M be the midpoint of a chord PQ of a circle, through which two other chords AB and CD are drawn; AD cuts PQ at X and BC cuts PQ at Y. Prove that M is also the midpoint of XY.” (Bogomolny) This is the most accurate definition currently. However, Butterfly Theorem has experienced some changes and developments. The first statement of the Butterfly Theorem appeared in the early 17th century. In 1803, a Scottish mathematician, William Wallace, posed the problem of the Butterfly Theorem in the magazine The Gentlemen’s Mathematical Companion. Here is the original problem below: “If from any two points B, E, in the circumference of a circle given in magnitude and position two right lines BCA, EDA, be drawn cutting the circle in C and D, and meeting in A; and...
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...Engineer-to-Engineer Note EE-218 a Technical notes on using Analog Devices DSPs, processors and development tools Contact our technical support at dsp.support@analog.com and at dsptools.support@analog.com Or visit our on-line resources http://www.analog.com/ee-notes and http://www.analog.com/processors Writing Efficient Floating-Point FFTs for ADSP-TS201 TigerSHARC® Processors Contributed by Boris Lerner Rev 2 – March 4, 2004 Introduction So, you want to write efficient code for the ADSP-TS201 TigerSHARC® processor? Or, maybe, you have come across the optimized example floating-point FFT for this processor and would like to understand how it works and what the author had in mind when writing it. This application note tries to answer both questions by going through that FFT example and all its levels of optimization in detail. This example can be followed in developing other algorithms and code optimized for the ADSPTS201S processor. Generally, most algorithms have several levels of optimization, all of which are discussed in detail in this note. The first and most straightforward level of optimization is paralleling of instructions, as the processor architecture will allow. This is simple and boring. The second level of optimization is loop unrolling and software pipelining to achieve maximum parallelism and to avoid pipeline stalls. Although more complex than the simple parallelism of level one, this can be done in prescribed steps without good understanding...
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...Expectation Maximization in Collaborative Filtering Jonathan Baker 2010 Abstract Expectation maximization (EM) is a method of approximating maximum likelihood estimators (MLE) in models with missing data or latent variables. A straightforward application of EM is collaborative filtering (CF): using data from multiple agents to predict unreported values. In this paper, we show a simple method of applying EM to a large CF problem: predicting ratings in the Netflix Prize dataset. 1 Auxiliary Functions EM belongs to a general class1 of optimization algorithms using successive locally approximating auxiliary functions. For an objective function f : X → R, call g : X × X → R an auxiliary function if ∀x, x0 ∈ X : g(x, x0 ) ≥f (x) g(x, x) =f (x). ∞ Then for any x0 , define the sequence (xn )n=0 by xn+1 = arg min g(x, xn ) x 1 For example, the Newton Rhapson method can also be stated in terms of auxiliary functions. 1 This sequence has a non-increasing image under f . This is easy to prove: f (xn+1 ) ≤ g(xn+1 , xt ) ≤ g(xn , xn ) = f (xn ). This idea may be more clear graphically in figure (1). Each g(·, xt ) dominates f (·) but is equal at xt . It is easy to see why we might hope that the minimums of g would approach the minimum of f . Figure 1: An objective function f and the auxiliary function g on two iterations Assuming f is bounded below (and it really ought to be since we are trying to find its minimum value), then f (xn )∞ is also bounded below. We...
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...Bern University of Applied Sciences Business Option Strategies – A practical Guideline for Retail Investors Bachelor Thesis Submitted within the study program Bachelor of Science in Business Administration By XXX For the module Bachelor Thesis (BTHE) Instructor XXX Co-Supervisor XXX Submission date Friday, 16 May 2014 “Option strategies – A practical Guideline for Retail Investors” Acknowledgment The development of this paper has been observed and accompanied by instructor Professor XXX and co-instructor Professor XXX of the XXX. Both of the tutors shall be thanked gratefully at this point. Furthermore I would like to acknowledge the interviewees for their advices and information, who made it possible to complete this bachelor thesis. 2 “Option strategies – A practical Guideline for Retail Investors” Management Summary This bachelor thesis examines option strategies and what costs and fees an individual investor will be confronted with when trading such instruments in Switzerland. In theory, option strategies are an interesting and multifunctional instrument for every market direction. However, in practice, there are several difficulties which retail investors have to overcome. For example, they have only limited available assets for trading and thus the margin of the broker and the transaction costs significantly influence the potential profit (economies of scale). There are also legal restrictions like naked...
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...McGill University Fall 2014 McGill University Fall 2014 Individual Assignment: Job analysis M Pxxxx Student Number: XXXXXXX Employment ORGB 424 CHERIF ATALLAH Individual Assignment: Job analysis M Pxxx Student Number: Employment ORGB 424 CHERIF ATALLAH 08 Fall 08 Fall Contents: 1. Introduction……………………………………………………………. 3 2. Job Analysis………………………………………………………………4 3. Inventory of job duties and tasks………………………………..7 4. Analyzing the knowledge, skills and abilities.……………..11 5. Critical tasks……………………………………………………………..17 6. Selection plan……………………………………………………………19 7. Internal and external recruitment..…………………………….21 8. Critical dimensions #1……………………………………………….23 9. Critical dimensions #2……………………………………………….25 10.Conclusion………………………………………………………………...28 11. Appendix #1……………………………………………………………..29 12. References……………………………………………………………......30 Introduction The principle use of this assignment is to correlate the theories behind the staffing model and techniques. Before deciding the job requirements and specifications for hire, a manager must determine what are the tasks and duties necessary to accomplish the job efficiently. A job analysis must be done periodically due the introduction of new technologies; tasks will change and need to be adapted. Job Analysis There are three (3) key points that have to be taken into consideration when performing a job analysis. (Milkovich, Newman & al 156) 1. There...
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...The demand of the chicken is increasing throughout the year from the consumer. Poultry factory produce large amount of feather. The uncontrolled disposal used such as land filling and incineration or burial cause many problems in storage, handling, emission control and ash problem that lead to pollution in our environment. According to (reference), Five percent of the body weight of poultry is feather. Discarded feathers will also cause various infections including chlorosis, mycoplasmosis and fowl cholera. Based on the report by (reference), almost 91% of chicken feather consists of keratin, thus it would be an interesting attempt to extract this protein from the chicken feathers. Rather than being disposed of as a waste, valuable product such as protein keratin can be discovered from the chicken...
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...[pic] ENGD3016 Solid Mechanics Assignment 1: Finite Element Analysis Name: Wei Zhang ID: P14021978 Date: Dec17th 2015 Abstract 1.0 Introduction 2.0 Objectives 3.0 Matlab 4.0 Solidworks 4.1 Model of truss 1 4.2 Model of truss 2 5.0 Comparison of the two trusses 6.0 Comparison between MATLAB and SOLIDWORKS 6.1 Comparison of results 6.2 Advantages and disadvantages 7.0 Conclusion Appendix Abstract The purpose of this report is to based on two different 2D pictures of trusses to finite element analysis from two ways. First carried out the selection of materials, and choosing the size of the pipe. Two methods were used two different software for finite element analysis, solidworks and MATLAB. By comparing the simulation results with two software to make sure which is the better way. It was found that the results of MATLAB are more accurate, solidworks results more intuitive. According to the calculation results of the two methods, truss 2 is more reliable. 1.0 Introduction Finite element analysis method using a mathematical approximation of the real physical system to simulation. Also use simple and interacting elements, namely unit, you can use a limited number of unknown quantities to approximate the infinite unknowns real system.The engineering is relying on the...
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...Share PageRecommendRecent ActivityConnect via Meebo Continue Reading » Advertisement You will be redirected to the page you want to view in seconds. FIND IT: JOBS CARS HOMES APARTMENTS DATING DELAWAREONLINE TV FOODIE TV SIGNATURE SPARK 247 CLASSIFIEDS SUBSCRIBE TODAY News Communities Sports Business Entertainment Life Opinion Obituaries Help Blogs Crime & Courts Gov't & Politics Education Data Special reports Traffic Lotteries Nation/World LATEST HEADLINES They fought the law ... and they could win Aggrieved couple challenge instant ticketing and find sympathetic ears on federal appeals court- 1:03 am Suit over commercial complex set for US court in 18 months After spending some time in legal limbo, a lawsuit challenging Sussex County's denial of a...- 12:57 am Urbanization may threaten region's rich fishing industry Fish like herring, perch and striped bass don't do as well when land uses shift from rural and...- 12:48 am Wilmington Newark Hockessin/Pike Creek Chateau New Castle Middletown Kent County Sussex County Get Published Blogs High Schools UD DSU/Colleges Phillies Eagles 76ers Flyers NASCAR Gambling Buy Tickets LATEST HEADLINES Colonials capture title, tourney spot WILMINGTON — Twice, William Penn 285-pounder A.J.- 1:47 am Paul’s final shot dooms Sixers PHILADELPHIA — There wasn’t much more that Andre Iguodala could do.- 12:09 am A victory and then some for St. Elizabeth WILMINGTON — St...
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