L17_Dual simplex method for solving the primal

# L17_Dual simplex method for solving the primal - Dual...

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Dual simplex method for solving the primal

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In this lecture we describe the important Dual Simplex method and illustrate the by doing one or two problems.
Dual Simplex Method Suppose a “basic solution” satisfies the optimality conditions but not feasible, we apply dual simplex algorithm. In regular Simplex method, we start with a Basic Feasible solution (which is not optimal) and move towards optimality always retaining feasibility. In the dual simplex method, the exact opposite occurs. We start with an (more than) “optimal” solution (which is not feasible) and move towards feasibility always retaining optimality conditions. The algorithm ends once we obtain feasibility.

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1. The objective function must satisfy  the  optimality conditions of the regular Simplex  method. 2. All the constraints must be of the   type  . To start the dual Simplex method, the following three conditions are to be met: ( Note : As in the simplex method, we must have an identity matrix in the constraint matrix; however, the RHS constants b i need NOT be 0.) 3. All variables should be   0.
In any iteration, we first decide the “leaving” variable and then decide which variable should enter (the basis). Dual Feasibility Condition (=Condition for a variable to leave (the basis)) The leaving variable, x i , is the basic variable having the most negative value (i.e. in the Simplex tableau, the corresponding constraint row has the most negative RHS). Ties are broken arbitrarily. If all the basic variables are 0, the algorithm ends and we have obtained the optimal solution.

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The entering variable, x j , is determined from among the non-basic variables as the one for which the ratio < - 0 : ij ij j j a a c z is least, where z j - c j is the coefficient of x j in the z-row, a ij is the coefficient of x j in the leaving variable row. Ties are broken arbitrarily. If no variable can enter the basis, the problem has no feasible solution. (Why?) Dual Optimality Condition (=Condition for a non-basic variable x j to enter the basis)
Problem 2(a) Problem Set 4.4A Page 141 Solve by Dual Simplex method Minimize 2 1 3 2 x x z + = subject to 1 2 1 2 1 2 2 2 30 2 10 , 0 x x x x x x + ≤ + ≥

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Putting it in the form “all constraints of the type and adding slack variables, the problem becomes Minimize 2 1 3 2 x x z + = subject to Since all the objective coefficients are 0 and the problem is minimization, optimality is met. We start the solution by writing the Simplex tableau. 1 2 1 1 2 2 1 2 1 2 2 2 30 2 10 , , , 0 x x s x x s x x s s + + = - - + = -
s2 0 -1 -2 0 1 -10 z 1 -2 -3 0 0 0 Basic z x1 x2 s1 s2 Sol s1 0 2 2 1 0 30 x2 0 1/2 1 0 -1/2 5 z 1 -1/2 0 0 -3/2 15 s1 0 1 0 1 1 20 This is the optimal tableau.

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## This note was uploaded on 02/23/2011 for the course MATH 112 taught by Professor Ritadubey during the Spring '11 term at Amity University.

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L17_Dual simplex method for solving the primal - Dual...

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