Exercise 4.1 True and False Statements about Simplex x1 x2
... (a) False. In any iteration of the Simplex algorithm, the entering variable xj has negative reduced cost c̄j . When xj enters the basis, the basic variables xB are modified by xB → xB + θdjB , variable xj is modified by xj → xj + θ, and the cost changes by c′ x → c′ x + θc̄j . Therefore, if the curr ...
... (a) False. In any iteration of the Simplex algorithm, the entering variable xj has negative reduced cost c̄j . When xj enters the basis, the basic variables xB are modified by xB → xB + θdjB , variable xj is modified by xj → xj + θ, and the cost changes by c′ x → c′ x + θc̄j . Therefore, if the curr ...
NP Complexity
... • It can not contain any laterals in the same clause because are pair wise adjacent. • It can not contain lateral and its negation because there exist an edge between them. • Therefore it is easy to find an interpretation that satisfies the corresponding conjunction. Assign true to their interpretat ...
... • It can not contain any laterals in the same clause because are pair wise adjacent. • It can not contain lateral and its negation because there exist an edge between them. • Therefore it is easy to find an interpretation that satisfies the corresponding conjunction. Assign true to their interpretat ...
models solutions for the second midterm
... (4, 2) ∈ S. If we repeat this process, we can keep multiplying the first coordinate by 2. So S must contain every pair of the form (2n , 2) or (2n , 1), where n ≥ 0. But, if (2n , 2) and (2m , 2) are in S, then (3) implies that (2n , 2m ) is in S. So S is the set of all pairs of the form (2n , 2m ), ...
... (4, 2) ∈ S. If we repeat this process, we can keep multiplying the first coordinate by 2. So S must contain every pair of the form (2n , 2) or (2n , 1), where n ≥ 0. But, if (2n , 2) and (2m , 2) are in S, then (3) implies that (2n , 2m ) is in S. So S is the set of all pairs of the form (2n , 2m ), ...
PSet 1 Solutions
... (b) Give a divide-and-conquer algorithm for multiplying two polynomials of degree-bound that runs in time )* *-, . .2 . Solution: Divide the coefficients of each -degree polynomial and ! ...
... (b) Give a divide-and-conquer algorithm for multiplying two polynomials of degree-bound that runs in time )* *-, . .2 . Solution: Divide the coefficients of each -degree polynomial and ! ...
Solutions - UCR Math Dept.
... Solution: Remember that the domain of a function is the places where the function is defined. The only place where f is not defined is at x = −3, so the domain is (−∞, −3) ∪ (−3, ∞). The x and y intercepts are found the same way as problem 1. 2x − 1 y= x+3 ...
... Solution: Remember that the domain of a function is the places where the function is defined. The only place where f is not defined is at x = −3, so the domain is (−∞, −3) ∪ (−3, ∞). The x and y intercepts are found the same way as problem 1. 2x − 1 y= x+3 ...
Problems 1-3
... a) Solve the integer program graphically. If we draw an isoprofit line (e.g., 11x1 + 4x2 = 11) and move it parallelly towards the direction of increasing the objective function, the last feasible integer point it will touch will be (2, 3). Thus, the optimal integer solution is (2, 3) with value 34. ...
... a) Solve the integer program graphically. If we draw an isoprofit line (e.g., 11x1 + 4x2 = 11) and move it parallelly towards the direction of increasing the objective function, the last feasible integer point it will touch will be (2, 3). Thus, the optimal integer solution is (2, 3) with value 34. ...
Solution of Linear Programming Problems with Matlab
... If you also want to retrieve the minimal value fmin = minx (f T x), type [x,fmin]=linprog(f,A,b) If inequality and equality constraint are given, use the commands x=linprog(f,A,b,Aeq,beq) or [x,fmin]=linprog(f,A,b,Aeq,beq) Let’s solve our winemaker problem: >> f=[-12;-7];b=[10000;16000;0;0];A=[2 1;3 ...
... If you also want to retrieve the minimal value fmin = minx (f T x), type [x,fmin]=linprog(f,A,b) If inequality and equality constraint are given, use the commands x=linprog(f,A,b,Aeq,beq) or [x,fmin]=linprog(f,A,b,Aeq,beq) Let’s solve our winemaker problem: >> f=[-12;-7];b=[10000;16000;0;0];A=[2 1;3 ...
Clustering and routing models and algorithms for the design of
... given a network, which is composed of a set of access nodes with demand requirements and conduits connecting the access nodes, a set of candidate edge nodes chosen from the access nodes, and a set of facilities with different capacities and installation costs. We need to determine a partitioning of ...
... given a network, which is composed of a set of access nodes with demand requirements and conduits connecting the access nodes, a set of candidate edge nodes chosen from the access nodes, and a set of facilities with different capacities and installation costs. We need to determine a partitioning of ...
EXAM PDE 18.02.13 1. Exercise Let Ω ⊂ R 3 be
... that given a rotation matrix A (which maps bijectively both Ω onto Ω and ∂Ω onto ∂Ω ), u(Ax) is again a solution. If a solution exists, it is unique by the maximum principle, therefore u must be invarian under rotations. This implies u(x) = v(|x|) with v ∈ C 2 ((0, 1) ∩ C([0, 1]) . The Laplacian of ...
... that given a rotation matrix A (which maps bijectively both Ω onto Ω and ∂Ω onto ∂Ω ), u(Ax) is again a solution. If a solution exists, it is unique by the maximum principle, therefore u must be invarian under rotations. This implies u(x) = v(|x|) with v ∈ C 2 ((0, 1) ∩ C([0, 1]) . The Laplacian of ...
Data Representation Methods
... • A boolean formula is satisfiable iff there is at least one truth assignment to its variables for which the formula evaluates to true. • Determining whether a boolean formula in CNF is satisfiable is NP-hard. • Problem is solvable in polynomial time when no clause has more than 2 literals. • Remain ...
... • A boolean formula is satisfiable iff there is at least one truth assignment to its variables for which the formula evaluates to true. • Determining whether a boolean formula in CNF is satisfiable is NP-hard. • Problem is solvable in polynomial time when no clause has more than 2 literals. • Remain ...
PATTERNS, CONTINUED: VERIFYING FORMULAS
... A technique called Mathematic al Induction can be used to verify a formula or expression for the nth term in a pattern. This method is particularly useful when it is easy to describe the pattern recursively; that is, whenever it is easy to describe the procedure for going from the nth term to the ( ...
... A technique called Mathematic al Induction can be used to verify a formula or expression for the nth term in a pattern. This method is particularly useful when it is easy to describe the pattern recursively; that is, whenever it is easy to describe the procedure for going from the nth term to the ( ...
Study guides
... (TB is short for text book, WP is short for course web page) Norms - TB: sec 2.1; WP: items 8 / 10,11 1. What are three properties that characterize a vector norm? 2. Define some commonly used norms such as the 1, 2, p, and infinity norm. 3. Be able to prove that these are norms. 4. What are ...
... (TB is short for text book, WP is short for course web page) Norms - TB: sec 2.1; WP: items 8 / 10,11 1. What are three properties that characterize a vector norm? 2. Define some commonly used norms such as the 1, 2, p, and infinity norm. 3. Be able to prove that these are norms. 4. What are ...
Overview and History
... Fall 2013 See online syllabus (also accessible via BlueLine2): http://dave-reed.com/csc221 ...
... Fall 2013 See online syllabus (also accessible via BlueLine2): http://dave-reed.com/csc221 ...
Lecture 24
... • We have a program A that checks some candidate solution in time O(n5) • Construct a HUGE boolean formula that represents the execution of A: its variables are the candidate solution (which we don’t know) plus all memory bits • Then check if this formula is satisfiable (i.e. there exists some candi ...
... • We have a program A that checks some candidate solution in time O(n5) • Construct a HUGE boolean formula that represents the execution of A: its variables are the candidate solution (which we don’t know) plus all memory bits • Then check if this formula is satisfiable (i.e. there exists some candi ...
B. Quadratic Formula
... B. Quadratic Formula A Quadratic equation in the form ax2 + bx + c = 0 can be solved symbolically by substituting the values for a, b and c into the Quadratic Formula: ...
... B. Quadratic Formula A Quadratic equation in the form ax2 + bx + c = 0 can be solved symbolically by substituting the values for a, b and c into the Quadratic Formula: ...
Test Alg2 Chap 5N
... Questions 13 - 15, find the value of the discriminant, the number (quantity) of roots, and type of roots for each of the following problems. Do NOT find the solutions. ...
... Questions 13 - 15, find the value of the discriminant, the number (quantity) of roots, and type of roots for each of the following problems. Do NOT find the solutions. ...
Algebra 2 Name: 1.1 – More Practice Your Skills – Arithmetic
... 1.1 – More Practice Your Skills – Arithmetic Sequences ...
... 1.1 – More Practice Your Skills – Arithmetic Sequences ...
Introduction to Algorithm
... Complexity of the algorithm is O(N2) The running time of the algorithm is O(N2). Asymptotic lower-bound Asymptotic upper-bound Big-O notation Can this problem be solved in polynomial time? ...
... Complexity of the algorithm is O(N2) The running time of the algorithm is O(N2). Asymptotic lower-bound Asymptotic upper-bound Big-O notation Can this problem be solved in polynomial time? ...
Homework # 1 Solutions Problem 11, p. 4 Solve z 2 + z +1=0. Strictly
... Problem 5, p. 34 Let S be the open set consisting of all points z such that |z| < 1 or |z − 2| < 1. Explain why S is not connected. The set S consists of the points interior to two circles, one centered at 0 and one centered at 2, both of radius 1. These circles do not overlap and therefore no point ...
... Problem 5, p. 34 Let S be the open set consisting of all points z such that |z| < 1 or |z − 2| < 1. Explain why S is not connected. The set S consists of the points interior to two circles, one centered at 0 and one centered at 2, both of radius 1. These circles do not overlap and therefore no point ...
Complete Characterization of Near-Optimal Sequences for the Two
... In a two-machine flow shop scheduling problem, the set of approximate sequences ( i.e. , solutions within a factor 1+ of the optimal) can be mapped to the vertices of a permutation lattice. We introduce two approaches, based on properties derived from the analysis of permutation lattices, for charac ...
... In a two-machine flow shop scheduling problem, the set of approximate sequences ( i.e. , solutions within a factor 1+ of the optimal) can be mapped to the vertices of a permutation lattice. We introduce two approaches, based on properties derived from the analysis of permutation lattices, for charac ...
sample only Get fully solved assignment, plz drop a mail with your
... Churchman, Aackoff, and Aruoff defined operations research as “the application of scientific methods, techniques and tools to the operation of a system with optimum solutions to the problems” where 'optimum' refers to the best possible alternative. The objective of OR is to provide a scientific basi ...
... Churchman, Aackoff, and Aruoff defined operations research as “the application of scientific methods, techniques and tools to the operation of a system with optimum solutions to the problems” where 'optimum' refers to the best possible alternative. The objective of OR is to provide a scientific basi ...
Rishi B. Jethwa and Mayank Agarwal
... i) Lower Bounding Technique:- To find the lower bounds for the parallel ATSP algorithm by solving the assignment problem. ii) Upper Bounding Heuristic:- Use the solution to the assignment problem to construct a solution to the ATSP. iii) Branching rules:- Create two or more new sub-problems based on ...
... i) Lower Bounding Technique:- To find the lower bounds for the parallel ATSP algorithm by solving the assignment problem. ii) Upper Bounding Heuristic:- Use the solution to the assignment problem to construct a solution to the ATSP. iii) Branching rules:- Create two or more new sub-problems based on ...
Dynamic programming
In mathematics, computer science, economics, and bioinformatics, dynamic programming is a method for solving a complex problem by breaking it down into a collection of simpler subproblems. It is applicable to problems exhibiting the properties of overlapping subproblems and optimal substructure (described below). When applicable, the method takes far less time than other methods that don't take advantage of the subproblem overlap (like depth-first search).In order to solve a given problem using a dynamic programming approach, we need to solve different parts of the problem (subproblems), then combine the solutions of the subproblems to reach an overall solution. Often when using a more naive method, many of the subproblems are generated and solved many times. The dynamic programming approach seeks to solve each subproblem only once, thus reducing the number of computations: once the solution to a given subproblem has been computed, it is stored or ""memoized"": the next time the same solution is needed, it is simply looked up. This approach is especially useful when the number of repeating subproblems grows exponentially as a function of the size of the input.Dynamic programming algorithms are used for optimization (for example, finding the shortest path between two points, or the fastest way to multiply many matrices). A dynamic programming algorithm will examine the previously solved subproblems and will combine their solutions to give the best solution for the given problem. The alternatives are many, such as using a greedy algorithm, which picks the locally optimal choice at each branch in the road. The locally optimal choice may be a poor choice for the overall solution. While a greedy algorithm does not guarantee an optimal solution, it is often faster to calculate. Fortunately, some greedy algorithms (such as minimum spanning trees) are proven to lead to the optimal solution.For example, let's say that you have to get from point A to point B as fast as possible, in a given city, during rush hour. A dynamic programming algorithm will look at finding the shortest paths to points close to A, and use those solutions to eventually find the shortest path to B. On the other hand, a greedy algorithm will start you driving immediately and will pick the road that looks the fastest at every intersection. As you can imagine, this strategy might not lead to the fastest arrival time, since you might take some ""easy"" streets and then find yourself hopelessly stuck in a traffic jam.Sometimes, applying memoization to a naive basic recursive solution already results in a dynamic programming solution with asymptotically optimal time complexity; however, the optimal solution to some problems requires more sophisticated dynamic programming algorithms. Some of these may be recursive as well but parametrized differently from the naive solution. Others can be more complicated and cannot be implemented as a recursive function with memoization. Examples of these are the two solutions to the Egg Dropping puzzle below.