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1 THE REVISED SIMPLEX METHOD CONTENTS Linear Program in the Matrix Notation Basic Feasible Solution in Matrix Notation Revised Simplex Method in Matrix.

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Presentation on theme: "1 THE REVISED SIMPLEX METHOD CONTENTS Linear Program in the Matrix Notation Basic Feasible Solution in Matrix Notation Revised Simplex Method in Matrix."— Presentation transcript:

1 1 THE REVISED SIMPLEX METHOD CONTENTS Linear Program in the Matrix Notation Basic Feasible Solution in Matrix Notation Revised Simplex Method in Matrix Notation Reference: Chapter 5 in Bazaraa, Jarvis and Sherali.

2 2 Matrix Notation Maximize  j=1,n c j x j subject to  j=1,n a ij x j  b i for all i = 1, 2, …, m x j  0 for all j = 1, 2, …, n Add the slack variables: x n+i = b i -  j=1,n a ij x j for all i = 1, 2, …, m Problem in the matrix notation: Maximize cx subject to Ax = b x  0

3 3 Matrix Notation Maximize cx subject to Ax = b x  0 where

4 4 The Primal Simplex Method B : The set of indices corresponding to basic variables N : The set of indices corresponding to nonbasic variables The linear program in the matrix form: Maximize z = c B x B + c N x N subject to Bx B + N x N = b x B  0, x N  0

5 5 The Primal Simplex Method (contd.) Constraint Matrix: Bx B + N x N = b or Bx B = b - Nx N Let x B define a basis, then x B = B -1 b - B -1 Nx N where B is an invertible mxm matrix (that is, whose columns are linearly independent). Objective Function: z = c B x B + c N x N z = c B (B -1 b - B -1 Nx N ) + c N x N z = c B B -1 b + (c N - c B B -1 N) x N

6 6 The Primal Simplex Method (contd.) Basis in the Matrix Notation: z = c B B -1 b + (c N - c B B -1 N) x N x B = B -1 b - B -1 Nx N Basic Solution associated with this Basis: x N = 0 x B = B -1 b

7 7 Computing Simplex Multipliers in Matrix Notation The simplex multipliers  must be such that z j – c j : = c j -  i=1,m a ij  i = 0 for each basic variable x j Alternatively, c j =  i=1,m a ij  i for each basic variable x j or c B =  B or  = c B B -1

8 8 Summary of Formulas w.r.t. a Basis Suppose that B is a set of basic variables at some iteration. Let B denote the associated basis (columns of A in the basis). Then we can obtain the simplex tableau for this iteration using the following formulas: x j column in the constraints = B -1 a j Right-hand side of the tableau = B -1 b The z j – c j = c B B -1 a j - c j Right-hand side of row 0 = c B B -1 b Simplex multipliers  = c B B -1

9 9 Example of the Revised Simplex Method Maximize 4x 1 + 3x 2 subject to x 1 - x 2  1 2x 1 - x 2  3 x 2  5 x 1, x 2  0 B = {3, 4, 5} and N = {1, 2} N = b

10 10 Iteration 1 Step 1. The current primal solution x B : x B = = B -1 b == Step 2. The current dual solution  : y = c B B -1 = = d j = c j – z j = c j –  i=1,m a ij y i = [4 3] Entering variable = x 1

11 11 Iteration 1 (contd.) Step 3. Determine. = B -1 A 1 = = Step 4. Perform the minimum ratio test. b’ = = Minimum ratio = 1/1 = 1 Leaving variable = x 3 B =

12 12 Iteration 2 B = {1, 4, 5} and N = {3, 2} N =

13 13 Iteration 2 (contd.) Step 1. The current primal solution x B : x B = = B -1 b == Step 2. Determine the simplex multipliers y and for the nonbasic variables: y = c B B -1 = [4 0 0] B -1 Entering variable = x 2 d j = c j – z j = c j –  i=1,m a ij y i = [-4 7], z 3 – c 3 = – 4, z 2 – c 2 = – 7 =

14 14 Iteration 2 (contd.) Step 3. Determine A’ 2. A’ 2 = B -1 A 2 = = Step 4. Perform the minimum ratio test. b’ = A’ 2 = Minimum ratio = 1/1 = 1 Leaving variable = x 4 B =

15 15 Revised Simplex Method Step 1. Obtain the initial primal feasible basis B. Determine the corresponding feasible solution = B -1 b Step 2. Obtain the corresponding simplex multipliers  = c B B -1. Check the optimality of the current BFS. If the current basis is optimal, then STOP. Step 3. If the current BFS is not optimal, identify the entering variable x k (that is, z k – c k =  i=1,m a ik  i - c k > 0). Step 4. Obtain the column = B -1 a k and perform the minimum ratio test to determine the leaving variable x l. Step 5. Update the basis B (or B -1 ) and go to Step 2. Time per iteration = m 2 + time needed to find an entering variable.

16 16 (Original) Simplex Method Step 1. Obtain the initial feasible basis. Step 2. Check the optimality of the current basis (that is, z j - c j  0 for each j  N). If optimal, STOP. Step 3. If the current basis is not optimal, identify the entering variable x k (that is, z k - c k > 0). Step 4. Perform the minimum ratio test to determine the leaving variable x l. Step 5. Perform a pivot operation to update the basis and go to Step 2. Time per iteration: proportional to nm


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