Lecture : Post-Optimal Analysis September 23, 2009
Today Lecture Dual-Simplex Algorithm Post-Optimal Analysis Chapters 4.4 and 4.5. IE 30/GE 330
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
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 3 2 0 0 0 0 x 4 3 0 0 3 x 5 3 3 0 0 6 x 6 0 0 3 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
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 5 0 0 2 0 4 2 3 x 4 4 0 2 0 3 3 x 2 0 0 2 3 3 2 x 6 2 0 0 3 3 IE 30/GE 330 4
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 3 0 0 9 0 2 2 2 x 3 6 0 3 3 0 2 2 2 x 2 3 0 3 0 2 2 2 x 6 2 0 0 0 0 IE 30/GE 330 5
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
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
Lecture Changes affecting feasibility TOYCO Revenue Maximization (Primal) Problem maximize z = 3x + 2x 2 + 5x 3 subject to x + 2x 2 + x 3 430 (machine ) 3x + 2x 3 460 (machine 2) x + 4x 2 420 (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 3 603 3x + 2x 3 644 x + 4x 2 588 x, x 2, x 3 0 How will this change affect the TOYCO S optimal solution and revenue? IE 30/GE 330 8
Lecture Optimal Table Basic x x 2 x 3 x 4 x 5 x 6 RHS z 4 0 0 2 0 350 x 2 0 4 2 0 00 4 3 x 3 0 0 0 230 2 2 x 6 2 0 0 2 20 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 4 0 0 2 0? x 2 0 4 2 0? 4 3 x 3 0 0 0? 2 2 x 6 2 0 0 2? IE 30/GE 330 9
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 = 0 2 4 0 0 2 2 602 644 588 New solution is feasible, hence stays optimal. = 40 322 28 Optimal operating point x = 0, x 2 = 322 and x 3 = 28 What is the new optimal value? IE 30/GE 330 0
Lecture New optimal value: substitute the new solution in the objective z = 3x + 2x 2 + 5x 3 = 3 0 + 2 40 + 5 322 = 890 IE 30/GE 330
Lecture Alternative TOYCO plan TOYCO Revenue Maximization (Primal) Problem maximize z = 3x + 2x 2 + 5x 3 resource slack subject to x + 2x 2 + x 3 430 (machine ) x 4 3x + 2x 3 460 (machine 2) x 5 x + 4x 2 420 (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 4 0 0 2 0 350 x 2 0 4 2 0 00 4 3 x 3 0 0 0 230 2 2 x 6 2 0 0 2 20 IE 30/GE 330 2
Lecture New TOYCO Constraints subject to x + 2x 2 + x 3 450 3x + 2x 3 460 x + 4x 2 400 x, x 2, x 3 0 How will this change affect the TOYCO S optimal solution and cost? IE 30/GE 330 3
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 4 0 0 2 0? x 2 0 4 2 0? 4 3 x 3 0 0 0? 2 2 x 6 2 0 0 2? 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. 0 2 4 0 0 2 2 450 460 400 = 0 230 40 IE 30/GE 330 4
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 = 3 0 + 2 0 + 5 230 = 370 The table corresponding to these changes Basic x x 2 x 3 x 4 x 5 x 6 RHS z 4 0 0 2 0 370 x 2 0 4 2 0 0 4 3 x 3 0 0 0 230 2 2 x 6 2 0 0 2 40 What shall we do to find the new optimal solution? IE 30/GE 330 5
Lecture Apply the dual simplex method Basic x x 2 x 3 x 4 x 5 x 6 RHS z 4 0 0 2 0 370 x 2 0 4 2 0 0 4 3 x 3 0 0 0 230 2 2 x 6 2 0 0 2 40 leaving the basis Which nonbasic variable will enter the basis? IE 30/GE 330 6
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 4 0 0 2 0 370 x 2 0 4 2 0 0 4 3 x 3 0 0 0 230 2 2 x 6 2 0 0 2 40 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 5 0 0 0 350 2 2 x 2 0 0 0 00 4 4 3 x 3 0 0 0 230 2 2 x 4 0 0 2 20 2 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