Dual simplex method for solving the primal

In this lecture we describe the important Dual Simplex method and illustrate the method 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.

To start the dual Simplex method, the following three conditions are to be met: The objective function must satisfy the optimality conditions of the regular Simplex method. All the constraints must be of the type . 3. All variables should be  0. (Note: As in the simplex method, we must have an identity matrix in the constraint matrix; however, the RHS constants bi need NOT 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, xi, 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.

Dual Optimality Condition (=Condition for a non-basic variable xj to enter the basis) The entering variable, xj , is determined from among the non-basic variables as the one for which the ratio is least, where zj - cj is the coefficient of xj in the z-row, aij is the coefficient of xj in the leaving variable row. Ties are broken arbitrarily. If no variable can enter the basis, the problem has no feasible solution. (Why?)

Problem 2(a) Problem Set 4.4A Page 141 Solve by Dual Simplex method Minimize subject to

Putting it in the form “all constraints of the type  ” and adding slack variables, the problem becomes Minimize 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.

This is the optimal tableau. Basic z x1 x2 s1 s2 Sol z 1 -2 -3 0 0 0 s1 0 2 2 1 0 30 s2 0 -1 -2 0 1 -10 z 1 -1/2 0 0 -3/2 15 s1 0 1 0 1 1 20 x2 0 1/2 1 0 -1/2 5 This is the optimal tableau. Optimal sol: x1= 0, x2=5; Optimum Value: z =15

(0,15) 2x1+2x2=30 2x1+3x2=45 2x1+3x2=30 (0,5) 2x1+3x2=20 (0,0) (10,0) (15,0) 2x1+3x2=15 x1+2x2=10

Problem Solve by Dual Simplex method Minimize subject to

Putting it in the form “all constraints of the type  ” and adding slack variables, the problem becomes Minimize 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.

This is the optimal tableau. Basic z x1 x2 s1 s2 Sol z 1 -2 -3 0 0 0 s1 0 - 2 -2 1 0 -30 s2 0 -1 -2 0 1 -10 z 1 0 -1 -1 0 30 x1 0 1 1 -1/2 0 15 s2 0 0 -1 -1/2 1 5 This is the optimal tableau. Optimal sol: x1=15, x2=0; Optimum Value: z =30

Problem 2(c) Problem Set 4.4A Page 142 Solve by Dual Simplex method Minimize subject to Note this ! That is subject to

Adding slack variables, the problem becomes Minimize 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.

Optimal Tableau Basic z x1 x2 s1 s2 s3 Sol s1 0 1 1 1 0 0 1

Problem (From Hillier & Lieberman) Solve by Dual Simplex method Maximize subject to

Adding slack variables, the problem becomes Maximize subject to Since all the objective coefficients are ≤ 0 and the problem is maximization, optimality is met. We start the solution by writing the Simplex tableau.

Optimal Tableau Basic z x1 x2 s1 s2 s3 Sol s1 0 1 1 1 0 0 8

Problem (From Hillier & Lieberman) Consider the following LPP: Maximize subject to

Write the dual of the above problem. Solve the problem by Regular Simplex method. Solve the dual by the dual Simplex method. Compare the resulting sequence of basic solutions with the complementary basic solutions obtained in part (b).

Solution: (a) The Dual of the given LPP is: Minimize subject to (b) We now solve the given LPP by regular Simplex method.

Optimal Tableau Basic z x1 x2 s1 s2 s3 Sol s1 0 3 1 1 0 0 12

(c) We now solve the dual by Dual Simplex method. Putting it in the form “all constraints of the type  ” and adding slack variables, the problem becomes Minimize subject to

Optimal Tableau Basic w y1 y2 y3 t1 t2 Sol t1 0 -3 -1 -5 1 0 - 3

We now enumerate the basic solutions of the primal & the complementary basic solutions Itera-tion No. Primal Basic Solution Fea-sible? z= w Compl. Basic Solution 0 (0,0, 12, 6,27) Y 0 (0,0, 0,-3, -2) N 1 (4,0, 0, 2, 7) Y 12 (1,0, 0, 0, -1) N 2 (3,3, 0, 0,3) Y 15 (1/2,3/2, 0, 0, 0) Y Thus we observe that dual Simplex method solves the dual on the primal tableau.

Problem 3 Problem Set 4.4A Page 142 Dual Simplex with Artificial Constraints Consider the LPP Maximize subject to

Adding the surplus variables s1 and s2 to the first and second constraints and the slack variable s3 to the third constraint, the LPP becomes Maximize subject to

The starting Basic Solution consisting of surplus s1 and s2 and slack s3 is infeasible as s1 = -4 and s2 = -3. However the dual Simplex method is NOT applicable as the optimality condition is not met. We will solve the problem by augmenting the artificial constraint where M is large enough NOT to eliminate any feasible points of the original solution space. Using the new constraint row as the pivot row and taking x1 as the entering variable as it has the most negative coefficient in the z-row will give an all-optimal objective function row. Next we carry out the dual simplex method. The working follows.

We now allow s4 to leave and x1 to enter the basis Basic z x1 x2 x3 s1 s2 s3 s4 Sol z 1 -2 1 -1 0 0 0 0 0 s1 0 -2 -3 5 1 0 0 0 -4 s2 0 1 -9 1 0 1 0 0 -3 s3 0 4 6 3 0 0 1 0 8 s4 0 1 0 1 0 0 0 1 M We now allow s4 to leave and x1 to enter the basis z 1 0 1 1 0 0 0 2 2M s1 0 0 -3 7 1 0 0 2 -4+2M s2 0 0 -9 0 0 1 0 -1 -3- M s3 0 0 6 -1 0 0 1 -4 8-4M x1 0 1 0 1 0 0 0 1 M Now dual simplex method starts: s3 leaves;s4 enters the basis

This is the optimal tableau. Basic z x1 x2 x3 s1 s2 s3 s4 Sol z 1 0 4 1/2 0 0 1/2 0 4 s1 0 0 0 13/2 1 0 1/2 0 0 s2 0 0 -21/2 1/4 0 1 -1/4 0 -5 s4 0 0 -3/2 1/4 0 0 -1/4 1 -2+M x1 0 1 3/2 3/4 0 0 1/4 0 2 z 1 0 0 25/42 0 8/21 17/42 0 44/21 s1 0 0 0 13/2 1 0 1/2 0 0 x2 0 0 1 -1/42 0 -2/21 1/42 0 10/21 s4 0 0 0 3/14 0 -1/7 -3/14 1 -9/7+M x1 0 1 0 11/14 0 1/7 3/14 0 9/7 This is the optimal tableau.

Problem 4(b) Problem Set 4.4A Page 142 Dual Simplex with Artificial Constraints Consider the LPP Maximize subject to

Adding the slack variable s1 to the first constraint and the surplus variables s2 and s3 to the second and third constraints, the LPP becomes Maximize subject to

The starting Basic Solution consisting of slack s1 and surplus s2 and s3 is infeasible as s2 = - 4 and s3 = - 3. However the dual Simplex method is NOT applicable as the optimality condition is not met. We will solve this by augmenting the artificial constraint where M is large enough NOT to “eliminate” any feasible points of the original solution space. Using the new constraint row as the pivot row and taking x1 as the entering variable as it has the negative coefficient in the z-row will give an all-optimal objective function row. Next we carry out the dual simplex method. The working follows.

We now allow x1 to enter and s4 to leave the basis z 1 0 3 0 0 0 1 M Basic z x1 x2 s1 s2 s3 s4 Sol z 1 -1 3 0 0 0 0 0 s1 0 1 -1 1 0 0 0 2 s2 0 -1 -1 0 1 0 0 -4 s3 0 -2 2 0 0 1 0 -3 s4 0 1 0 0 0 0 1 M We now allow x1 to enter and s4 to leave the basis z 1 0 3 0 0 0 1 M s1 0 0 -1 1 0 0 -1 2 - M s2 0 0 -1 0 1 0 1 -4+ M s3 0 0 2 0 0 1 2 -3+2M x1 0 1 0 0 0 0 1 M Now dual simplex method starts: s1 leaves;s4 enters the basis

Basic z x1 x2 s1 s2 s3 s4 Sol s1 0 0 - 1 1 0 0 - 1 2 - M z 1 0 3 0 0 0 1 M s1 0 0 - 1 1 0 0 - 1 2 - M s2 0 0 -1 0 1 0 1 -4+ M s3 0 0 2 0 0 1 2 -3+2M x1 0 1 0 0 0 0 1 M z 1 0 2 1 0 0 0 2 s4 0 0 1 -1 0 0 1 M - 2 s2 0 0 - 2 1 1 0 0 -2 s3 0 0 0 2 0 1 0 1 x1 0 1 - 1 1 0 0 0 2

This is the optimal tableau. Basic z x1 x2 s1 s2 s3 s4 Sol z 1 0 2 1 0 0 0 2 s4 0 0 1 -1 0 0 1 M - 2 s2 0 0 -2 1 1 0 0 -2 s3 0 0 0 2 0 1 0 1 x1 0 1 -1 1 0 0 0 2 z 1 0 0 2 1 0 0 0 s4 0 0 0 -1/2 1/2 0 1 M - 3 x2 0 0 1 -1/2 -1/2 0 0 1 s3 0 0 0 2 0 1 0 1 x1 0 1 0 1/2 -1/2 0 0 3 This is the optimal tableau.

Generalized Simplex Algorithm The (primal) Simplex algorithm starts with a feasible solution which is not optimal and then moves towards optimality always retaining the feasibility. The dual Simplex algorithm starts with (better than) optimal solution which is not feasible and moves towards feasibility always retaining the optimality. What if the starting solution is neither feasible nor optimal? We have to use either (primal) Simplex Algorithm with Artificial variables or Dual Simplex Algorithm with Artificial constraints.

In both these algorithms we look for corner point solutions (feasible or not). The generalized Simplex algorithm exploits this fact and without using artificial variables or constraints moves from one corner point solution to another till optimality is obtained (or the criterion that the problem is unbounded or infeasible is detected). We illustrate with two examples. We now redo Problem 4(b) of Problem Set 4.4A Page 142 without adding an artificial constraint. The starting Simplex tableau is neither feasible nor optimal.

Basic z x1 x2 s1 s2 s3 Sol s1 0 1 -1 1 0 0 2 s2 0 -1 -1 0 1 0 -4 We now allow s2 to leave the basis as it has the most –ve value and x1 to enter as it satisfies the minimum ratio test. z 1 0 4 0 -1 0 4 s1 0 0 -2 1 1 0 -2 x1 0 1 1 0 -1 0 4 s3 0 0 4 0 -2 1 5 We now allow s1 to leave the basis as it has the –ve value and x2 to enter as it has the only –ve coefficient in the leaving row.

This is the optimal tableau Basic z x1 x2 s1 s2 s3 Sol z 1 0 0 2 1 0 0 x2 0 0 1 -1/2 -1/2 0 1 x1 0 1 0 1/2 -1/2 0 3 s3 0 0 0 2 0 1 1 This is the optimal tableau

Problem 4(c) Problem Set 4.4A Page 143 Consider the LPP Minimize subject to

Adding the slack variable s2 to the second constraint and the surplus variables s1 and s3 to the first and third constraints, the LPP becomes Minimize subject to

The new s2 row shows that the problem has no feasible solution. Basic z x1 x2 s1 s2 s3 Sol z 1 1 -1 0 0 0 0 s1 0 -1 4 1 0 0 -5 s2 0 1 -3 0 1 0 1 s3 0 -2 5 0 0 1 -1 We now allow s1 to leave the basis as it has the most –ve value and x1 to enter as it has the only –ve coefft in the leaving row. z 1 0 3 1 0 0 5 x1 0 1 -4 -1 0 0 5 s2 0 0 1 1 1 0 -4 s3 0 0 -3 -2 0 1 9 The new s2 row shows that the problem has no feasible solution.

Problem 4(b) Problem Set 4.4A Page 142 Consider the LPP Maximize subject to

Adding the slack variable s1 to the first constraint and the surplus variables s2 and s3 to the second and third constraints, the LPP becomes Maximize subject to We did this by generalized Simplex algorithm earlier. We redo it in another way.

We now allow x1 to enter and s1 to leave the basis Basic z x1 x2 s1 s2 s3 s4 Sol z 1 -1 3 0 0 0 0 0 s1 0 1 -1 1 0 0 0 2 s2 0 -1 -1 0 1 0 0 -4 s3 0 -2 2 0 0 1 0 -3 We now allow x1 to enter and s1 to leave the basis z 1 0 2 1 0 0 0 2 x1 0 1 -1 1 0 0 0 2 s2 0 0 -2 1 1 0 0 -2 s3 0 0 0 2 0 1 0 1 Now dual simplex method starts: s2 leaves;x2 enters the basis

This is the optimal tableau Max z = 0; x1 = 3; x2 = 1. Basic z x1 x2 s1 s2 s3 s4 Sol z 1 0 0 2 1 0 0 0 x1 0 1 0 1/2 -1/2 0 0 3 x2 0 0 1 -1/2 -1/2 0 0 1 s3 0 0 0 2 0 1 0 1 This is the optimal tableau Max z = 0; x1 = 3; x2 = 1.

z = -3 SF (3,1) Optimal Solution