Linear Programming and the Simplex Method

Similar documents
Definitions and Theorems. where x are the decision variables. c, b, and a are constant coefficients.

The Simplex algorithm: Introductory example. The Simplex algorithm: Introductory example (2)

Optimization Methods MIT 2.098/6.255/ Final exam

Optimization Methods: Linear Programming Applications Assignment Problem 1. Module 4 Lecture Notes 3. Assignment Problem

TEACHER CERTIFICATION STUDY GUIDE

IP Reference guide for integer programming formulations.

Linear Programming! References! Introduction to Algorithms.! Dasgupta, Papadimitriou, Vazirani. Algorithms.! Cormen, Leiserson, Rivest, and Stein.

Zeros of Polynomials

Inverse Matrix. A meaning that matrix B is an inverse of matrix A.

subject to A 1 x + A 2 y b x j 0, j = 1,,n 1 y j = 0 or 1, j = 1,,n 2

LINEAR PROGRAMMING. Introduction. Prototype example. Formulation of the LP problem. Excel Solution. Graphical solution

MASSACHUSETTS INSTITUTE OF TECHNOLOGY

A NEW APPROACH TO SOLVE AN UNBALANCED ASSIGNMENT PROBLEM

6.3 Testing Series With Positive Terms

Integer Linear Programming

Infinite Sequences and Series

The Method of Least Squares. To understand least squares fitting of data.

w (1) ˆx w (1) x (1) /ρ and w (2) ˆx w (2) x (2) /ρ.

TRANSPORTATION AND ASSIGNMENT PROBLEMS

Math 113 Exam 3 Practice

3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials

First, note that the LS residuals are orthogonal to the regressors. X Xb X y = 0 ( normal equations ; (k 1) ) So,

SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES

Differentiable Convex Functions

Chapter 4. Fourier Series

PROBLEM SET 5 SOLUTIONS 126 = , 37 = , 15 = , 7 = 7 1.

U8L1: Sec Equations of Lines in R 2

Question 1: The magnetic case

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

lim za n n = z lim a n n.

Summary: Congruences. j=1. 1 Here we use the Mathematica syntax for the function. In Maple worksheets, the function

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

Properties and Tests of Zeros of Polynomial Functions

Generating Functions. 1 Operations on generating functions

4.1 Sigma Notation and Riemann Sums

x c the remainder is Pc ().

Basic Iterative Methods. Basic Iterative Methods

Theorem: Let A n n. In this case that A does reduce to I, we search for A 1 as the solution matrix X to the matrix equation A X = I i.e.

Linear Regression Demystified

RADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify

Chapter 10: Power Series

Optimally Sparse SVMs

Math 61CM - Solutions to homework 3

Analysis of Algorithms. Introduction. Contents

NUMERICAL METHODS FOR SOLVING EQUATIONS

LP in Standard and Slack Forms

Scheduling under Uncertainty using MILP Sensitivity Analysis

1 Generating functions for balls in boxes

Sigma notation. 2.1 Introduction

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

(3) If you replace row i of A by its sum with a multiple of another row, then the determinant is unchanged! Expand across the i th row:

Kinetics of Complex Reactions

INFINITE SEQUENCES AND SERIES

Math 475, Problem Set #12: Answers

b i u x i U a i j u x i u x j

Direction: This test is worth 250 points. You are required to complete this test within 50 minutes.

Linearly Independent Sets, Bases. Review. Remarks. A set of vectors,,, in a vector space is said to be linearly independent if the vector equation

4.1 SIGMA NOTATION AND RIEMANN SUMS

( ) (( ) ) ANSWERS TO EXERCISES IN APPENDIX B. Section B.1 VECTORS AND SETS. Exercise B.1-1: Convex sets. are convex, , hence. and. (a) Let.

Beurling Integers: Part 2

End-of-Year Contest. ERHS Math Club. May 5, 2009

Sequences and Series of Functions

Lecture 23 Rearrangement Inequality

PAPER : IIT-JAM 2010

Mon Feb matrix inverses. Announcements: Warm-up Exercise:

( θ. sup θ Θ f X (x θ) = L. sup Pr (Λ (X) < c) = α. x : Λ (x) = sup θ H 0. sup θ Θ f X (x θ) = ) < c. NH : θ 1 = θ 2 against AH : θ 1 θ 2

Element sampling: Part 2

Sequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018

Lemma Let f(x) K[x] be a separable polynomial of degree n. Then the Galois group is a subgroup of S n, the permutations of the roots.

10-701/ Machine Learning Mid-term Exam Solution

Signals & Systems Chapter3

Algebra of Least Squares

(3) If you replace row i of A by its sum with a multiple of another row, then the determinant is unchanged! Expand across the i th row:

Ellipsoid Method for Linear Programming made simple

ARITHMETIC PROGRESSIONS

SNAP Centre Workshop. Basic Algebraic Manipulation

Lecture 1: Basic problems of coding theory

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 19 11/17/2008 LAWS OF LARGE NUMBERS II THE STRONG LAW OF LARGE NUMBERS

Chapter Vectors

The Poisson Distribution

4.3 Growth Rates of Solutions to Recurrences

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n

Best Optimal Stable Matching

CS284A: Representations and Algorithms in Molecular Biology

15.081J/6.251J Introduction to Mathematical Programming. Lecture 21: Primal Barrier Interior Point Algorithm

THE ASYMPTOTIC COMPLEXITY OF MATRIX REDUCTION OVER FINITE FIELDS

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.

CS:3330 (Prof. Pemmaraju ): Assignment #1 Solutions. (b) For n = 3, we will have 3 men and 3 women with preferences as follows: m 1 : w 3 > w 1 > w 2

Support vector machine revisited

ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory

1 Duality revisited. AM 221: Advanced Optimization Spring 2016

ROLL CUTTING PROBLEMS UNDER STOCHASTIC DEMAND

CHAPTER 10 INFINITE SEQUENCES AND SERIES

Mini Lecture 10.1 Radical Expressions and Functions. 81x d. x 4x 4

Sect 5.3 Proportions

MAT1026 Calculus II Basic Convergence Tests for Series

Introduction to Optimization, DIKU Monday 19 November David Pisinger. Duality, motivation

Polynomial Functions and Their Graphs

CHAPTER I: Vector Spaces

Transcription:

Liear Programmig ad the Simplex ethod Abstract This article is a itroductio to Liear Programmig ad usig Simplex method for solvig LP problems i primal form. What is Liear Programmig? Liear Programmig is the method of fidig a optimal solutio for a liear fuctio F of variables, whe the variables are uder some liear costraits. Here we itroduce two famous problems i the field of liear programmig: Productio Problem The productio problem ivolves determiig the rate of productio for products, so that the profit resultig from the products is maximum. Each product uses some resources, which are fiite ad provide the costraits. Here is a formal statemet of the productio problem: Products P,, P are produced from resources R,, R m. b i is the amout of resource R i available to the maufacturer. a ij is the umber of uits of R i eeded for oe uit of P j. The profit from producig oe uit of P i is kow to be c i. So the problem is to maximize f(x,, x )= uder the coditios cjxj (x j represetig the amout of P j produced) aijxj bi for all j, j m (ad every x j o-egative).

Diet Problem The diet problem is determiig the least costly amout of food to be cosumed, i order to satisfy a miimum daily requiremet of some utriets(give the amout of utriets for each uit of food ad the cost of each food uit). Primal Form I order to fid a simple algorithm for solvig LP problems, we eed a stadard form for the problem to work with. This is called the primal form ad it ca be show that every LP problem ca be trasformed to a correspodig problem i the primal form. Although we might wat to fid a miimum for the fuctio F, problems i primal form cocer fidig a maximum for F(which is called the objective fuctio). iimizig F ( ) is clearly equal to maximizig F ( ). The costraits i LP problems i primal form are oly i the form of Costraits usig or = ca be trasformed to this form. aijxj bi. All variables are o-egative i the primal form.( Although this is usually the case i LP problems, whe we eed egative variables we ca replace them with x -x, where x ad x are o-egative. Feasible Solutios Ay (x,, x ) satisfyig the costraits of the LP problem is said to be a feasible solutio for that problem. If at this solutio the objective fuctio reaches it s required maximum or miimum, the solutio is said to be optimal feasible solutio (OFS). The Solutio of a LP Problem I order to solve a LP problem, we must first fid out whether at least oe feasible solutio exist or ot. If it exists, we must either fid the OFS or show that it does ot exist (this happes whe the objective fuctio is ot bouded i the required directio).

Example of a LP problem aximize f(x, x 2 )= -2x +5x 2 if x + x 2 5, 2x 2 4, First we see that (0,0) is a feasible solutio. For all x, x 2, f (x, x 2 )= -2x +5x 2 5x 2. 0.(sice x, x 2 are o-egative). So 0 is the upper boud of the objective fuctio, ad sice f (0,2)=0, (0,2) is the OFS. Simplex ethod The Simplex algorithm for solvig LP problems was discovered i 947 by George B. Datzig. Nearly all practical LP problems are solved by a variatio of the Simplex method. We itroduce here some otios that are useful i implemetig Simplex method. Codesed Tableau The codesed tableau is a represetatio of a system of liear equatios which is i reduced-row echelo form. It is based o the cotrast betwee pivot variables ad o pivot variables. The codesed tableau below represets the system of liear equatios: p x j+ a qxkq = q= p x jm+ q= a mqxkq = m k k 2 k p a a 2 a p a 2 a 22 a 2p a m a m2 a mp b b 2 b m j j 2

j m What we have doe, basically, is to elimiate the coefficiets for the pivots,(sice they are ). The umbers j j m are the idices for pivot variables ad k k p the idices for free variable. Now we explai how to chage the role of a pivot ad free variable. Suppose we wat to chage the place of variables k q ad j i (the first a free variable ad the secod a pivot). All the elemets of the tableau should chage i this maer: k q j r. a rq /a rs /a rs. a iq - a is a rq/ a rs - a iq/ a rs b r /a rs b - a is b r / a rs k s j We ca easily see that this is a ormal row operatio, ad thus prove the method easily. Simplex Tableau Here we itroduce some chages i the primal form of LP problems that makes it fir for usig Simplex method. First we itroduce slack variables. Slack variables are itroduced to chage all the iequalities to equatios. It is based o the observatio that a b meas there is some oegative c for which a+c=b. So each costrait x + i + aijxj = j = aijxj bi ca be chaged to bi. Now the oly restrictio o the primal form to make it fit for Simplex method is that all b i are o-egative.

So we itroduce slack variables for each equatio (which is clearly pivot), ad form the followig form, which has a row added for the objective fuctio ad is called the Simplex tableau.(the idices remai the same, except for the objective row). a a 2 a p a 2 a 22 a 2p a m a m2 a mp b b 2 b p -c -c 2... -c 0 Sice all b i are o-egative, we see that (0,,0) is a feasible solutio (ad the item at the last colum, last row, which is called the corer umber gives the F(0,,0). Now it ca be show that. The pivotig process explaied above remais uchaged o the objective row 2. If we select appropriate etries i the tableau, by doig the pivotig operatio, we get a corer umber greater (or equal) to the curret oe. I other words, we get closer to the maximum. The basic solutio of a tableau is the solutio obtaied from assigig 0 to all o-basic variables( whose idex is give at the top of the tableau). The rule is to to choose a egative etry i the objective row. For every positive a ij above it, calculate a ij /b i.(this is called the _-ratio). Choose the etry with the least _-ratio (if there is a tie, choose oe of them). Pivot that etry (The purpose of the _-ratio is to keep the b i for all i o-egative, so that we ca use the method i the ext iteratio). We ca prove that the corer umber is greater or equal to the previous oe. The oly case whe it is equal is whe we have a b i =0. This leads to a basic variable beig 0 i the basic solutio, which is called degeerate case. Now we ca explai the Simplex algorithm.. See whether a egative etry exists i the objective row. If ot, the basic solutio is the OFS (the corer umber beig the maximum). 2. If every egative etry i the objective row has at least oe positive umber i it s colum, the objective fuctio is ubouded above. 3. Choose a egative etry i the objective row, choose a positive etry i it s colum which has the least _-ratio. Pivot the tableau at this etry. Go to.

It is clear that the Simplex algorithm allows a degree of freedom i choosig the etry to be pivoted, ad this leads to differet method for implemetig it. I some cases, we might be stuck with a Simplex algorithm that returs to the same tableau agai ad agai( ad thus, ever stops). The followig selectio method guaratees that we do t have to face this problem. Always choose the elemet with the smallest subscript, whe two or more choices exist. We must use this both for choosig etries i the objective row ad the positive iteger above it. Its called Blad s Termiatio Theorem. Simplex ethod i Practice 2 2 3 4-2 0 4 5-3 -2 0 3-2 4 0 4 5-3 -2 0 3 2 3 5-4 4 3 7 2 So the aswer is (4,0).