L5_10 - 1/1 ESI 6314 Deterministic Methods in Operations Research Lecture Notes 5 University of Florida Department of Industrial and Systems

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1/1 ESI 6314 Deterministic Methods in Operations Research Lecture Notes 5 University of Florida Department of Industrial and Systems Engineering / REEF 2/1 Converting an LP to standard form LP in standard form has only equality and nonnegativity constraints. Inequality constraints are converted into equality constraints by introducing a new variable in the left-hand side. In j-th ≤ constraint, we add a slack variable s j : x 1 + 2 x 2 + x 3 ≤ 5 ↓ x 1 + 2 x 2 + x 3 + s j = 5 ⇔ s j = 5- x 1- 2 x 2- x 3 In j-th ≥ constraint, we subtract an excess variable e j : x 1 + 2 x 2 + x 3 ≥ 5 ↓ x 1 + 2 x 2 + x 3- e j = 5 ⇔ e j =- 5 + x 1 + 2 x 2 + x 3 3/1 Example Consider the following LP: maximize 5 x 1 + 5 x 2 + 3 x 3 subject to x 1 + 3 x 2 + x 3 ≤ 3- x 1 + 3 x 3 ≤ 2 2 x 1- x 2 + 2 x 3 ≤ 4 2 x 1 + 3 x 2- x 3 ≤ 2 x 1 , x 2 , x 3 ≥ . m z = 5 x 1 + 5 x 2 + 3 x 3 s 1 = 3- x 1- 3 x 2- x 3 s 2 = 2 + x 1- 3 x 3 s 3 = 4- 2 x 1 + x 2- 2 x 3 s 4 = 2- 2 x 1- 3 x 2 + x 3 Here s 1 , s 2 , s 3 , s 4 are slack variables. 4/1 Example The standard form of this LP: maximize 5 x 1 + 5 x 2 + 3 x 3 subject to x 1 + 3 x 2 + x 3 + s 1 = 3- x 1 + 3 x 3 + s 2 = 2 2 x 1- x 2 + 2 x 3 + s 3 = 4 2 x 1 + 3 x 2- x 3 + s 4 = 2 x 1 , x 2 , x 3 ≥ s 1 , s 2 , s 3 , s 4 ≥ 5/1 Example: A Basic Feasible Solution z = 5 x 1 + 5 x 2 + 3 x 3 s 1 = 3- x 1- 3 x 2- x 3 s 2 = 2 + x 1- 3 x 3 s 3 = 4- 2 x 1 + x 2- 2 x 3 s 4 = 2- 2 x 1- 3 x 2 + x 3 To get a feasible solution, set all variables in the rhs (which we will call non-basic variables ) to 0: x 1 = x 2 = x 3 = 0 ⇒ s 1 = 3 , s 2 = 2 , s 3 = 4 , s 4 = 2; z = 0 . Basic variables : s 1 , s 2 , s 3 , s 4 . Non-basic variables : x 1 , x 2 , x 3 . The corresponding solution is a basic feasible solution (bfs). 6/1 Definitions Definition: A basic solution to a linear system Ax = b ( m equations, n variables, n ≥ m ) is obtained by setting n- m variables equal to 0 and solving for the values of the remaining m variables. This assumes that setting the n- m variables equal to 0 yields unique values for the remaining m variables, or, equivalently, the columns for the remaining m variables are linearly independent. Definition: A basic feasible solution is any basic solution in which all variables are non-negative. 7/1 Important Fact Theorem: A point in the feasible region of an LP is an extreme point if and only if it is a basic feasible solution to the LP. 8/1 Iterative improvement: pivot variable (column) z = 5 x 1 + 5 x 2 + 3 x 3 s 1 = 3- x 1- 3 x 2- x 3 s 2 = 2 + x 1- 3 x 3 s 3 = 4- 2 x 1 + x 2- 2 x 3 s 4 = 2- 2 x 1- 3 x 2 + x 3 To increase the value of z , we can try to increase the value of one of the non-basic variables with a positive (and as large as possible) coefficient in the objective....
View Full Document

This note was uploaded on 12/14/2010 for the course ESI 6314 taught by Professor Vladimirlboginski during the Fall '09 term at University of Florida.

Page1 / 32

L5_10 - 1/1 ESI 6314 Deterministic Methods in Operations Research Lecture Notes 5 University of Florida Department of Industrial and Systems

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online