Dual simplexmethod
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The document describes the steps of the dual simplex method for solving a maximization linear programming problem. It begins with ensuring all reduced costs in the simplex tableau are nonnegative before attempting the method. The key steps are: (1) check if the right-hand sides are nonnegative and stop if so, (2) pick an exiting variable if a right-hand side is negative, (3) use the minimum ratio test to select an entering variable, and (4) pivot and return to step 1. The example problem demonstrates applying these steps to solve a maximization problem using the dual simplex method.
Dual simplexmethod
- 1. Dual Simplex Method Assume we have a maximization problem. Step (0): Correction! We need all reduced costs (i.e., not the original cT vector but c T = c T B −1 A − c T ) in the simplex tableau to be nonnegative before we can even attempt B to use the method. Example (Corrected from class on 10/14) max -2x1 - 2x2 - x3 s.t. x1 +3x2 +3x3 +x4 = 12 -2x1 +10x2 +x3 -x5 = 21 xj ≥ 0 Here is the problem in tableau form with the basic variables having +1 coefficients when isolated in their appropriate rows: Basis X1 X2 X3 X4 X5 RHS X4 1 3 3 1 0 12 X5 2 -10 -1 0 1 -21 Z 2 2 1 0 0 0 We see that we do have nonnegative z-row coefficients. Step (1): If RHS >= 0 for all rows, STOP. The current basis is optimal. For the example problem, the basis is primal infeasible since -21 < 0. Thus, it cannot be optimal (despite a nonnegative z-row!) Step (2): Pick an EXITING variable. Any basic variable with RHS < 0 is a candidate to exit. Check to make sure we don’t have an unbounded LP. If all entries in the ROW of the EXITING variable are nonnegative STOP. The optimal dual cost is UNBOUNDED (so the primal problem is infeasible.) X5 is selected to leave the basis. The is no indication of unboundedness. Step (3): Use Minimum Ratio Test to determine ENTERING variable. Form the ratios z − row coefficient exiting row coefficient for all columns where the exiting row coefficient is negative. For the example problem we have min{2/|-10|, 1/|-1|} = 2/10 so X2 enters the basis. 1
- 2. Step (4): Perform the pivot operation and go to Step (1). Basis X1 X2 X3 X4 X5 RHS X4 1.6 0 2.7 1 0.3 5.7 X2 -0.2 1 0.1 0 -0.1 2.1 Z 2.4 0 0.8 0 0.2 -4.2 Step (1): Observe that the Dual Simplex Method always maintains a nonnegative z-row. Now that all RHS coefficients are nonnegative, the method stops with the current basis as optimal. X* = (0, 2.1, 0, 5.7, 0) Z* = -4.2 2