Lecture 11: Post-Optimal Analysis. September 23, 2009

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1 Lecture : Post-Optimal Analysis September 23, 2009

2 Today Lecture Dual-Simplex Algorithm Post-Optimal Analysis Chapters 4.4 and 4.5. IE 30/GE 330

3 Lecture Dual Simplex Method The dual simplex method will be crucial in the post-optimal analysis It used when at the current basic solution, we have The z-coefficients (reduced costs) satisfy optimality condition But the basic solution is infeasible Technical detail: all constraints in the problem have to be converted to IE 30/GE 330 2

4 Lecture minimize z = 3x + 2x 2 + x 3 subject to 3x + x 2 + x 3 3 3x + 3x 2 + x 3 6 x + x 2 + x 3 3 x, x 2, x 3 0 Example 4.4- minimize z = 3x + 2x 2 + x 3 subject to 3x x 2 x 3 3 3x 3x 2 x 3 6 x + x 2 + x 3 3 x, x 2, x 3 0 Introducing slack variables x 4, x 5, x 6, we generate the initial simplex table using the slacks as basic variables Basic x x 2 x 3 x 4 x 5 x 6 RHS z x x x A basic variable with the most negative value (infeasible) has to leave In this case x 5 is leaving the basis IE 30/GE 330 3

5 Lecture The entering variable is determining by the absolute value ratio test for the z-row entries and the x 5 -row entries, but only on the negative values: (here the entries in x 2 and x 3 column) minimum of 2 3, The minimum is attained in x 2 -column so x 2 enters; -3 is the pivot element (in x 5 -row and x 2 -column) We carry out the regular pivoting operations, and obtain Basic x x 2 x 3 x 4 x 5 x 6 RHS z x x x IE 30/GE 330 4

6 Lecture In the next iteration x 4 leaves and x 2 enters and the pivoting element is 2 3 Again, we carry out the regular pivoting operations, and obtain The table is optimal - why? Basic x x 2 x 3 x 4 x 5 x 6 RHS z x x x IE 30/GE 330 5

7 Lecture Dual in Sensitivity and Post-Optimal Analysis Sensitivity analysis without the use of the dual problem Increase or decrease of available resources LP Model: Change in the right-hand side - D i Increase or decrease in product prices/production costs LP Model: Change in the objective coefficients - d i Using dual simplex method we can determine an optimal solution for the perturbed problem from the optimal table of the original problem Using the dual problem/duality we can analyze additional changes Introducing new operations (resources) LP Model: New constraints in the original (primal) problem Introducing new products (activities) LP Model: New variables in the original (primal) problem IE 30/GE 330 6

8 Lecture We discuss this changes in the light of duality Changes affecting feasibility LP Model: right-hand side change or a new constraint How to recover optimal if the perturbation causes the change in basic optimal solution? Changes affecting optimality LP Model: objective coefficient or new variable How to find new optimal? IE 30/GE 330 7

9 Lecture Changes affecting feasibility TOYCO Revenue Maximization (Primal) Problem maximize z = 3x + 2x 2 + 5x 3 subject to x + 2x 2 + x (machine ) 3x + 2x (machine 2) x + 4x (machine 3) x, x 2, x 3 0 Current optimal: z = 350 x = 0, x 2 = 00, x 3 = 230 Suppose TOYCO wants to increase its operations to 602, 644, and 588. New TOYCO Constraints subject to x + 2x 2 + x x + 2x x + 4x x, x 2, x 3 0 How will this change affect the TOYCO S optimal solution and revenue? IE 30/GE 330 8

10 Lecture Optimal Table Basic x x 2 x 3 x 4 x 5 x 6 RHS z x x x The changes in the operations affect the right-hand side result in Perturbed Table Basic x x 2 x 3 x 4 x 5 x 6 RHS z ? x ? 4 3 x ? 2 2 x ? IE 30/GE 330 9

11 Lecture Use column rule to determine the RHS data except for z-row new solution = inverse optimal new right-hand side New solution x 2 x 3 x 6 = New solution is feasible, hence stays optimal. = Optimal operating point x = 0, x 2 = 322 and x 3 = 28 What is the new optimal value? IE 30/GE 330 0

12 Lecture New optimal value: substitute the new solution in the objective z = 3x + 2x 2 + 5x 3 = = 890 IE 30/GE 330

13 Lecture Alternative TOYCO plan TOYCO Revenue Maximization (Primal) Problem maximize z = 3x + 2x 2 + 5x 3 resource slack subject to x + 2x 2 + x (machine ) x 4 3x + 2x (machine 2) x 5 x + 4x (machine 3) x 6 x, x 2, x 3 0 Current optimal: z = 350 x = 0, x 2 = 00, x 3 = 230 As an alternative TOYCO considers shifting 20 units of time from machine 3 to machine, because at the current optimal there is a slack of 20 units Optimal Table Basic x x 2 x 3 x 4 x 5 x 6 RHS z x x x IE 30/GE 330 2

14 Lecture New TOYCO Constraints subject to x + 2x 2 + x x + 2x x + 4x x, x 2, x 3 0 How will this change affect the TOYCO S optimal solution and cost? IE 30/GE 330 3

15 Lecture The changes in the operations affect the right-hand side result in Perturbed Table Basic x x 2 x 3 x 4 x 5 x 6 RHS z ? x ? 4 3 x ? 2 2 x ? Use column rule to determine the RHS data except for z-row new solution = inverse optimal new right-hand side New solution x 2 x 3 x 6 = New solution is not feasible = IE 30/GE 330 4

16 Lecture Corresponding operating point x = 0, x 2 = 0 and x 3 = 230 Corresponding cost: substitute the above values in the objective z = 3x + 2x 2 + 5x 3 = = 370 The table corresponding to these changes Basic x x 2 x 3 x 4 x 5 x 6 RHS z x x x What shall we do to find the new optimal solution? IE 30/GE 330 5

17 Lecture Apply the dual simplex method Basic x x 2 x 3 x 4 x 5 x 6 RHS z x x x leaving the basis Which nonbasic variable will enter the basis? IE 30/GE 330 6

18 Lecture x 4 enters since the only negative entry in x 6 -row is at that place Basic x x 2 x 3 x 4 x 5 x 6 RHS z x x x leaving the basis Use 2 as pivoting element and perform the standard basis exchange Basic x x 2 x 3 x 4 x 5 x 6 RHS 5 z x x x The new optimal value is worse than the old optimal value! The revenue is reduced by 20. There is a slack of 20 on machine!!! IE 30/GE 330 7