Lecture 12 - Midterm 2 Review

Lecture 12 - Midterm 2 Review - 540:311DETERMINISTICMODELSIN

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540:311 DETERMINISTIC MODELS IN OPERATIONS RESEARCH Lecture 12: Midterm 2 Review Class Mee±ng: Thu Apr 7 th 10:20‐11:40am Prof. W. Art Chaovalitwongse
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Topics Big‐M Two Phase Graphical SensiEvity Analysis Computer SensiEvity Analysis Matrix NotaEon Matrix SensiEvity Analysis 100% Rule ( maybe )
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4.12 – The Big M Method The simplex method algorithm requires a starEng bfs. Previous problems have found starEng bfs by using the slack variables as our basic variables. If an LP have ≥ or = constraints, however, a starEng bfs may not be readily apparent. Idea: Introduce arEFcial variables to start the simplex with a bfs Use the simplex method and perform ero’s unEl the opEmal soluEon found, infeasible, or unbounded If opEmal soluEon is found and the arEFcial variable(s) is a BVs, the problem is infeasible.
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4.12 – The Big M Method Letting x1 = number of ounces of orange soda in a bottle of Oranj x2 = number of ounces of orange juice in a bottle of Oranj min z = 2x 1 + 3x 2 st 0.5x 1 + 0.25x 2 4 (sugar constraint) x 1 + 3x 2 20 (Vitamin C constraint) x 1 + x 2 = 10 (10 oz in 1 bottle of Oranj) x 1 , x 2 , > 0 Row 1: z - 2x 1 - 3x 2 = 0 Row 2: 0.5x 1 + 0.25x 2 + s 1 = 4 Row 3: x 1 + 3x 2 - e 2 = 20 Row 4: x 1 + x 2 = 10 s 1 could be used for BVs but row 2 would violate sign restrictions and row 3 no readily apparent basic variable.
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4.12 – The Big M Method In order to use the simplex method, a bfs is needed -> artificial variables are created. In the optimal solution, all artificial variables must be set equal to zero . For a min LP , a term Ma i is added to the objective function for each artificial variable a i . For a max LP, the term –Ma i is added to the objective function for each a i . M represents some very large number. Row 1: z - 2x 1 - 3x 2 = 0 Row 2: 0.5x 1 + 0.25x 2 + s 1 = 4 Row 3: x 1 + 3x 2 - e 2 + a 2 = 20 Row 4: x 1 + x 2 + a 3 = 10 z - 2x 1 - 3x 2 -Ma 2 - Ma 3 = 0 0.5x 1 + 0.25x 2 + s 1 = 4 x 1 + 3x 2 - e 2 + a 2 = 20 x 1 + x 2 + a 3 = 10 It is extremely costly for an artificial variable to be positive. The optimal solution should force a 2 = a 3 =0.
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4.12 – The Big M Method 1. Modify the constraints so that the rhs of each constraint is nonnegative . 2. Convert each inequality constraint to standard form (add a slack variable for constraints, add an excess variable for constraints) . 3. For each or = constraint, add artificial variables. 4. Add (for each artificial variable) Ma i to min problem objective functions or -Ma i to max problem objective functions. 5. Since each artificial variable will be in the starting basis, all artificial variables must be eliminated from row 0 before beginning the simplex . Row z x1 x2 s1 e2 a2 a3 rhs 0 1.00 -2.00 -3.00 -M -M 0.00 1 0.50 0.25 1.00 4.00 2 1.00 3.00 -1.00 1.00 20.00 3 1.00 1.00 1.00 10.00
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4.12 – The Big M Method Pivot 1 z x1 x2 s1 e2 a2 a3 rhs ratio ero 0 1.00 2m - 2 4M -3 -M 30M Row 0 + M(Row 2) + M(Row 3) 1 0.50 0.25 1.00 4.00 16 2 1.00 3 -1.00 1.00 20.00 6.67 3 1.00 1.00 1.00 10.00 10 ero 1 z x1 x2 s1 e2 a2 a3 rhs ero 0 1.00 2m - 2 4M -3 -M 30M 1 0.50 0.25 1.00 4.00 2 0.33 1 -0.33 0.33 6.67 Row 2 divided by 3 3 1.00 1.00 1.00 10.00 ero 2 z x1 x2 s1 e2 a2 a3 rhs ero 0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/3 Row 0 - (4M-3)*(Row 2) 1 0.50 0.25 1.00 4.00 2 0.33 1 -0.33 0.33 6.67 3 1.00 1.00 1.00 10.00 ero 3 z x1 x2 s1 e2 a2 a3 rhs ero 0 1.00 (2M-3)/3 (M-3)/3
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This note was uploaded on 09/20/2011 for the course ENG 300 taught by Professor Albin during the Fall '11 term at Rutgers.

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Lecture 12 - Midterm 2 Review - 540:311DETERMINISTICMODELSIN

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