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

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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....
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## 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.

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

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