...Competitive Algorithms for VWAP and Limit Order Trading Sham M. Kakade Michael Kearns Computer and Information Science University of Pennsylvania Computer and Information Science University of Pennsylvania kakade@linc.cis.upenn.edu mkearns@cis.upenn.edu Yishay Mansour Luis E. Ortiz Computer Science Tel Aviv University Computer and Information Science University of Pennsylvania mansour@post.tau.ac.il leortiz@linc.cis.upenn.edu ABSTRACT We introduce new online models for two important aspects of modern financial markets: Volume Weighted Average Price trading and limit order books. We provide an extensive study of competitive algorithms in these models and relate them to earlier online algorithms for stock trading. Categories and Subject Descriptors F.2 [Analysis of Algorithms and Problem Complexity]: Miscellaneous; J.4 [Social and Behavioral Sciences]: Economics General Terms Algorithms, Economics Keywords Online Trading, Competitive Analysis, VWAP 1. INTRODUCTION While popular images of Wall Street often depict swashbuckling traders boldly making large gambles on just their market intuitions, the vast majority of trading is actually considerably more technical and constrained. The constraints often derive from a complex combination of business, regulatory and institutional issues, and result in certain kinds of “standard” trading strategies or criteria that invite algorithmic analysis. One of the most common...
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...Case Analysis – Specialty Toys 1. Senior sales forecaster predicted and expected demand of 20,000 units with .95 probability that demand would be between 10,000 and 30,000 units. P (10,000 < x< 30,000) = .95 (30,000-20,000)/1.96 = 5,102 X = the demand of Weather Teddy Mean µ = 20,000 Standard Deviation α = 5,102 The normal distribution of the demand for the Weather Teddy is represented in the graph below. This is based off of the forecast of previous selling history of other similar toys. The forecast shows that the demand for the Weather Teddy will be at 20,000 units, but with a probability of .95 of selling anywhere between 10k to 20k units. Therefore, with the information given the standard deviation of this forecast is at 5,102. 2. With various order quantities suggested by members of the management team, it would be wise to compute the probability of a stock-out for each of the order quantities suggested. The probability of a stock-out is the inverse of the probability that the quantity sold is less than or equal to the amount purchased by the company. 1st Formula used: z = (x - µ) / σ The probability of a stock out is calculated by subtracting the probability found in the chart from 1. Suggested Quantity to Order | Probability of a Stock-Out | 15,000 | 83.65% | 18,000 | 65.17% | 24,000 | 21.77% | 28,000 | 5.82% | 3. Based on the various order quantities suggested by members of the management team, a simple profit analysis should...
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...A Comparison of Solution Procedures for the Flow Shop Scheduling Problem with Late Work Criterion Jacek Błażewicz *) Erwin Pesch 1) Małgorzata Sterna 2) Frank Werner 3) *) Institute of Computing Science, Poznań University of Technology Piotrowo 3A, 60-965 Poznań, Poland phone: +48 (61) 8790 790 fax: +48 (61) 8771 525 blazewic@sol.put.poznan.pl Institute of Information Systems, FB 5 - Faculty of Economics, University of Siegen Hölderlinstrasse 3, 57068 Siegen, Germany pesch@fb5.uni-siegen.de Institute of Computing Science, Poznań University of Technology Piotrowo 3A, 60-965 Poznań, Poland Malgorzata.Sterna@cs.put.poznan.pl Faculty of Mathematics, Otto-von-Guericke-University PSF 4120, 39016 Magdeburg, Germany Frank.Werner@mathematik.uni-magdeburg.de 1) 2) 3) 1 A Comparison of Solution Procedures for the Flow Shop Scheduling Problem with Late Work Criterion Abstract In this paper, we analyze different solution procedures for the two-machine flow shop scheduling problem with a common due date and the weighted late work criterion, i.e. for problem F2 | dj = d | Yw, which is known to be binary NP-hard. In computational experiments, we compare the practical efficiency of a dynamic programming approach, an enumerative method and a heuristic list scheduling procedure. Test results show that each solution method has its advantages and none of them can be rejected from the consideration a priori. Keywords: flow shop, late work, dynamic programming, enumerative...
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...Perceptual Map Situation Analysis The Situation Analysis will help your company understand current market conditions and how the industry will evolve over the next eight years. The analysis can be done as a group or you can assign parts to individuals and then report back to the rest of the company. An online version of the Situation Analysis is available in the Getting Started area. (customers want better performing products) and for size is -0.7 (customers want smaller products). At the end of Round 1 the center of the Traditional segment will have a performance of 5.7 and a size of 14.3. 5.0 + 0.7 = 5.7 and 15.0 - 0.7 = 14.3 Table 2 displays the segment center locations at the end of each round. Print the Perceptual Map Form in the Industry Conditions Report then use Table 2 to find the location of each segment center for Rounds 1 through 8. Mark the approximate locations on the form (see the example in Figure 1). Remember, the locations in Table 2 are the centers of the segment circles, not product positions. Product positions are reported on page 4 of The Capstone Courier. The exercises require two reports: The Industry Conditions Report and The Capstone Courier, which are available from the website’s Reports link. The Courier is also available from the Capstone Spreadsheet’s Reports menu bar. The Courier available at the start of Round 1 displays the results for Round 0, when all companies are equal. If you access the report from the website, use the Round 0...
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...STEP 1: 123 * 6 = 8 → 3 → 7 in linked list. STEP 2: 123 * 50= 0 → 5 → 1 → 6 in linked list. STEP 3: ADD STEP 1 AND STEP 2 8 → 8 → 8 → 6 in linked list. STEP 4: 123 * 400= 0 → 0 → 2 → 9 → 4 in linked list. STEP 5: ADD STEP 3 AND STEP 4 8 → 8 → 0 → 6 → 5 in linked list. Exponentiation: The initial step is to arrange in decreasing order of exponents and then perform the action. The other method is by using Θ(log n) time algorithm based on binary representation. 1.5. Representation of one’s and two’s complement 1’s complement: + 4= 0100 - 4= 1011 (taking 1’s complement of +4 is -4 (i.e.) inverting the bit). So, 1’s complement is used to represent both positive and negative integers. 2’s complement: Converting the value to 1’s complement and then adding 1 to that complement. - 4= (1’s complement of 4) + 1 = 1011+1 = 1100. It is used in implementation of ADT, which performs arithmetic operation. So it is qualified under data structure. 1.9. Does every problem have an algorithm? Every problem doesn’t have an algorithm. For example if I were sorting the records stored in an array, the searching becomes difficult for the largest element. The value of an...
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...UCCC2063 Analysis Algorithm Assignment Lecturer / Tutor : Mr. Wong Chee Siang Course: CS Name | ID | Cheng Shong Wei | 1103771 | Er Siang Chiang | 1203122 | Ng Yuan Wen | 1201733 | Question 1 : Fibonacci a. Iterative Fibonacci 1. Code: #include <iostream> #include <time.h> using namespace std; int main() { //loop until 80th Fibonacci long long count =10; while (count<=10000000000) { //variables time_t start,end; unsigned long long a = 1, b = 1; //prompt start = clock(); //start timer for(unsigned long long n = 3; n <= count; ++n) //calculated by iterative { unsigned long long fib = a + b; a = b; b = fib; if (n==count) cout<<"Fibonacci Number: "<<fib<<endl; } //end timer show Runtime end=clock(); double diff ((double)(end - start)); double time = diff / CLOCKS_PER_SEC; cout<<"Runtime ("<< count <<"th) : "<< time <<" seconds"<<endl; cout<<"------------------------------------------------------"<<endl; count = count *10; } system ("pause"); return 0; } 2. Analysis: n=3count1=count-3+1 T(n) =count-2 =O(count) =O(n) 3. Table n-th Fibonacci no. | Time 1(sec) | Time 2(sec) | Time 3(sec) | Average(sec) | 10 | 0.001 | 0.001 | 0.001 | 0.001 | 100 | 0.001 | 0.001 | 0.001 | 0.001 | 1’000 | 0.002 | 0.001 | 0.001 | 0.0013 | 10’000 | 0.003 | 0.002 | 0.001 | 0.0020 | 100’000 |...
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...Faculty of Computer Studies M180 Data Structures and Algorithms in Java Mid-Term Assessment (MTA) Fall – 2013 Saturday 17, November 2012 Number of MTA Pages: (including this cover sheet) ( 3 ) Time Allowed: ( 2 ) Hour Instructions: 1- Write all your answers on the same exam paper. 2- Electronic devices (especially calculators) are not allowed. Question Type Max. Mark Part 1: Multiple Choice Questions 10 Part 2: Short questions 20 Part 3: Coding questions 30 Total 60 Student Mark M180 Midterm-Ex am Fall 2012-2013 PART 1: ALL QUESTIONS ARE REQUIRED [10 Marks] Question 1: Choose the correct answer: (10 marks, one mark each) 1) a) b) c) d) One of the following methods return the top of the stack without applying any modifications: pop( ) push( ) peek( ) isEmpty( ) 2) a) b) c) d) Two main measures for the efficiency of an algorithm are: Processor and memory Time and space Complexity and capacity Data and space 3) a) b) c) d) A doubly nested “for loop” typically takes time in: Θ(n2) Θ(n) Θ(log n) Θ(log n2) 4) a) b) c) d) Arrays are best data structures: For relatively permanent collections of data For the size of the structure and the data in the structure are constantly changing For both of above situation For none of above situation 5) The elements of an array are stored successively in memory cells because: a) The architecture of computer memory does not allow...
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...BOOKS ABOUT TASHTEGO CASE ANALYSIS Rdiana.com TASHTEGO CASE ANALYSIS Updated: 02/26/2015 DISCLAIMER: RDIANA.COM uses the following Tashtego Case Analysis book available for free PDF download which is also related with TASHTEGO CASE ANALYSIS Tashtego Case Analysis can be easily downloaded from our library. Don’t you believe? It is completely free. You just have to register on our site – click on the link below and answer simple questions. It will provide you for free access to Tashtego Case Analysis and other eBooks. We ask you to pass a registration because of hard hackers’ attacks that knock out of service our library and prevent our users from downloading Tashtego Case Analysis as well as other books when it is necessary. When pass the registration, you can be sure of free and unlimited access to Tashtego Case Analysis and lots of other PDF data. Files can be downloaded on your device when you want. Therefore, if you still need Tashtego Case Analysis and cannot download it from other sites, register on our site and get a free access to a rich collection of eBooks right now. Save your time and efforts. PDF FILE: TASHTEGO CASE ANALYSIS Rdiana.com BOOKS ABOUT TASHTEGO CASE ANALYSIS PAGE: 2 TASHTEGO CASE ANALYSIS RR.DVI INSTITUT NATIONAL DE RECHERCHE EN INFORMATIQUE ET AUTOMATIQUE An Average-case Analysis of the Gaussian Algorithm for Lattice Reduction Herve Daude , Philippe Flajolet , Brigitte Vallee N 2798 Fevrier 1996 PROGRAMME 2...
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...COMPUTATIONAL COMPLEXITY OPERATIONS RESEARCH Pooja Punjabi Manash Hazarika INDIAN INSTITUTE OF MANAGEMENT, KOZHIKODE COMPUTATIONAL COMPLEXITY Computational Complexity is a measure of the computational time taken by a particular algorithm. In a scenario where there are multiple algorithms available for a particular problem, the effectiveness of any particular algorithm is gauged on the basis of the time constraint. This is done by breaking the algorithm into its basic steps and then taking a count of each of them. Hence greater is the number of steps, greater is the complexity. Now for example, if we take two 5 bit binary numbers and XOR them, the number of steps taken is 5 and if the same process is repeated for a 100 bit binary number, the number of steps goes up to 100. The algorithm employed in either case is the same; the complexity is given by the size of the numbers. When we say size of a number n, it is defined as the number of binary bits which are required to denote ‘n’ in base 2. For example, 5 in base 10 when expressed in binary takes the form 101, thereby giving n = 3. Similarly 20 is given by 101002 which makes n = 5. Now if we XOR any two numbers each of size n=b, the number of steps taken will be ‘b’. Hence we can say that XORing those numbers has computational complexity of order ‘b’ which is denoted by O(b). This can be applied to even simpler applications like addition wherein if we are to add two n digit numbers, the minimum number of steps taken (ignoring...
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...1 Big-Oh Notation Let f and g be functions from positive numbers to positive numbers. f (n) is O(g(n)) if there are positive constants C and k such that: f (n) ≤ C g(n) whenever n > k f (n) is O(g(n)) ≡ ∃C ∃k ∀n (n > k → f (n) ≤ C g(n)) To prove big-Oh, choose values for C and k and prove n > k implies f (n) ≤ C g(n). Standard Method to Prove Big-Oh 1. Choose k = 1. 2. Assuming n > 1, find/derive a C such that f (n) C g(n) ≤ =C g(n) g(n) This shows that n > 1 implies f (n) ≤ C g(n). Keep in mind: • n > 1 implies 1 < n, n < n2, n2 < n3, . . . • “Increase” numerator to “simplify” fraction. 2 Proving Big-Oh: Example 1 Show that f (n) = n2 + 2n + 1 is O(n2). Choose k = 1. Assuming n > 1, then f (n) n2 + 2n + 1 n2 + 2n2 + n2 = < =4 2 2 g(n) n n Choose C = 4. Note that 2n < 2n2 and 1 < n2. Thus, n2 + 2n + 1 is O(n2) because n2 + 2n + 1 ≤ 4n2 whenever n > 1. Proving Big-Oh: Example 2 Show that f (n) = 3n + 7 is O(n). Choose k = 1. Assuming n > 1, then f (n) 3n + 7 3n + 7n 10n = < = = 10 g(n) n n n Choose C = 10. Note that 7 < 7n. Thus, 3n + 7 is O(n) because 3n + 7 ≤ 10n whenever n > 1. 3 Proving Big-Oh: Example 3 Show that f (n) = (n + 1)3 is O(n3). Choose k = 1. Assuming n > 1, then f (n) (n + 1)3 (n + n)3 8n3 = < = 3 =8 3 3 g(n) n n n Choose C = 8. Note that n + 1 < n + n and (n+n)3 = (2n)3 = 8n3. Thus, (n+1)3 is O(n3) because (n + 1)3 ≤ 8n3 whenever n > 1. Proving Big-Oh: Example 4 Show that f (n) = Σ i is O(n2). i=1 n Choose k = 1. Assuming n > 1, then f (n)...
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...DATA STRUCTURES ANALYSIS OF ALGORITHMS Definition:- The method of solving a problem is known as an algorithm.It is a sequence of instructions that act on some input data to produce some output in a finite number of steps. Properties:- a) Input:An algorithm must receive some input data supplied externally. b)Output:An algorithm must produce atleast one output as the result. c)Finiteness:The algorithm must terminate after a finite number of steps. d)Definiteness:The steps to be performed in the algorithm must be clear and unambiguous. e)Effectiveness:One must be able to perform the steps in the algorithm without applying any intelligence. All algorithms basically fall under two broad categories-Iterative and Recursive algorithms. Iterative Algorithms typically use loops and conditional statements. Recursive Algorithms use a divide and Conquer strategy. As per this,the recursive algorithm breaks down a large problem into small pieces and then applies the algorithm to each of these smal pieces. Determining which algorithm is efficient than the other involves analysis of algorithms. While analyzing,time required to execute it determined .’time’ represents the number of operations that are carried out while executing the algorithm. While analyzing iterative algorithms we need to determine how many times the loop is executed. To analyze a recursive algorithm one needs to determine amount of work done for three things: ...
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...An Iterated Dynasearch Algorithm for the Single-Machine Total Weighted Tardiness Scheduling Problem Faculty of Mathematical Studies, University of Southampton, Southampton, SO17 1BJ, UK Faculty of Mathematical Studies, University of Southampton, Southampton, SO17 1BJ, UK Department of Decision and Information Sciences, Rotterdam School of Management, Erasmus University, P.O. Box 1738, 3000 DR Rotterdam, The Netherlands Richard.Congram@paconsulting.com • C.N.Potts@maths.soton.ac.uk • S.Velde@fac.fbk.eur.nl Richard K. Congram • Chris N. Potts • Steef L. van de Velde T his paper introduces a new neighborhood search technique, called dynasearch, that uses dynamic programming to search an exponential size neighborhood in polynomial time. While traditional local search algorithms make a single move at each iteration, dynasearch allows a series of moves to be performed. The aim is for the lookahead capabilities of dynasearch to prevent the search from being attracted to poor local optima. We evaluate dynasearch by applying it to the problem of scheduling jobs on a single machine to minimize the total weighted tardiness of the jobs. Dynasearch is more effective than traditional first-improve or best-improve descent in our computational tests. Furthermore, this superiority is much greater for starting solutions close to previous local minima. Computational results also show that an iterated dynasearch algorithm in which descents are performed a few random moves away from previous...
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...Data Structure and Algorithms UNIT – I PART- A Questions & Answers 1. What is an algorithm? An algorithm is a finite set of instructions that , if followed, accomplishes a particular task. 2. What are the characteristics of an algorithm? Input – Zero or more quantities Output – At least one quantity Definiteness – each instruction is clear and unambiguous Finiteness – terminate after a finite number of steps Efficiency – easily understandable 3. Define Space Complexity The Space complexity of an algorithm is the amount of memory it needs to run to completion 4. Define Time Complexity Time complexity of an algorithm is the amount of computer time it needs to run to completion 5. What are asymptotic notations? The notations that enables us to make meaningful statements about the time and space complexity of a program is called asymptotic notations. 6. What are the various asymptotic notations used to define the efficiency of an algorithm? Big ‘Oh’ (O) Omega (Ω) Theta (θ) Little ‘oh’ (o) Little Omega 7. What is information? Information is a recorded or communicated material that has some meaning associated with symbolic representation. 8. What are the aspects to be considered to ensure information transfer between source and destination? Syntactic – Physical form of information Semantic – Meaning Pragmatic – Action taken as a result of interpretation of information 9. State the Markov algorithm It takes the input string X and through a number...
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...CPS 230 DESIGN AND ANALYSIS OF ALGORITHMS Fall 2008 Instructor: Herbert Edelsbrunner Teaching Assistant: Zhiqiang Gu CPS 230 Fall Semester of 2008 Table of Contents 1 I 2 3 4 5 Introduction D ESIGN T ECHNIQUES Divide-and-Conquer Prune-and-Search Dynamic Programming Greedy Algorithms First Homework Assignment S EARCHING 3 4 5 8 11 14 17 18 19 22 26 29 33 34 35 38 41 44 IV 13 14 15 16 G RAPH A LGORITHMS Graph Search Shortest Paths Minimum Spanning Trees Union-Find Fourth Homework Assignment T OPOLOGICAL A LGORITHMS 17 18 19 Geometric Graphs Surfaces Homology Fifth Homework Assignment G EOMETRIC A LGORITHMS 20 21 22 Plane-Sweep Delaunay Triangulations Alpha Shapes Sixth Homework Assignment NP-C OMPLETENESS 23 24 25 Easy and Hard Problems NP-Complete Problems Approximation Algorithms Seventh Homework Assignment 45 46 50 53 56 60 61 62 65 68 72 73 74 77 81 84 85 86 89 92 95 V II 6 7 8 9 Binary Search Trees Red-Black Trees Amortized Analysis Splay Trees Second Homework Assignment P RIORITIZING VI III 10 11 12 Heaps and Heapsort Fibonacci Heaps Solving Recurrence Relations Third Homework Assignment VII 2 1 Introduction Meetings. We meet twice a week, on Tuesdays and Thursdays, from 1:15 to 2:30pm, in room D106 LSRC. Communication. The course material will be delivered in the two weekly lectures. A written record of the lectures will be available on the web, usually a day after the lecture. The web also contains other information...
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...ASSIGNMENT - 3 Answers 1 Answer 2 Answer 3 4. For what types of workloads does SJF deliver the same turnaround times as FIFO? ANSWER: - In the above case if the jobs are same in size or the jobs periodically applied (i.e. first shortest job then later second shortest job and continues) then the turnaround time of SJF is same as FIFO. 5. For what types of workloads and quantum lengths does SJF deliver the same response times as RR? ANSWER: - The response time delivered by SJF is equal to the response time of RR but this happens only when all the jobs arrived are at the point when the planning quantum of RR is bigger than the bigger occupation to be administrations in order of increasing size. 6. What happens to response time with SJF as job lengths increase? Can you use the simulator to demonstrate the trend? ANSWER: - If you suppose the length of the job increases then average response time varies (increases).If every job is sorted in increasing job order than the last job response time will be equal to the sum of current job and previous (n-1) jobs. In this way if the size of the job increases the response time will also increases for all larger jobs. 7. What happens to response time with RR as quantum lengths increase? Can you write an equation that gives the worst-case response time, given N jobs? ANSWER: - In the case of RR, the response time increases as the quantum lengths increases. This happens because the waiting time of a process for its turn to...
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