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Simplex Method Meeting 5

Simplex Method Meeting 5. Course : D0744 - Deterministic Optimization Year : 2009. What to learn?. Artificial variables Big-M method. The Facts. To start, we need a canonical form

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Simplex Method Meeting 5

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  1. Simplex MethodMeeting 5 Course : D0744 - Deterministic Optimization Year : 2009

  2. What to learn? Artificial variables Big-M method

  3. The Facts To start, we need a canonical form If we have a  constraint with a nonnegative right-hand side, it will contain an obvious basic variable (which?) after introducing a slack var. If we have an equality constraint, it contains no obvious basic variable If we have a  constraint with a nonnegative right-hand side, it contains no obvious basic variable even after introducing a surplus var.

  4. 2x + 3y  5  2x + 3y + s = 5, s  0 (s basic) 2x + 3y = 5  ??????? Infeasible if x=y=0! 2x + 3y  5  2x + 3y -s = 5, s  0 (??????) Infeasible if x=y=0! ?????????????????? Compare!

  5. One Equality??? 2x + 3y = 5  2x + 3y + a = 5, a = 0 (I) (s basic, but it should be 0!) How do we force a = 0? This is of course not feasible if x=y=0, as 0+0+0 5!

  6. 2x + 3y = 5  2x + 3y + a = 5, a = 0 (I) (a basic, but it should be 0!) How do we force a = 0? This is of course not feasible if x=y=0, as 0+0+0 5 Idea: solve a first problem with Min {a | constraint (I) + a  0 + other constraints }! One Equality???

  7. Artificial Variables Notice: In an equality constraint, the extra variable is called an artificial variable. For instance, in 2x + 3y + a = 5, a = 0 (I) a is an artificial variable.

  8. One Inequality  ??? 2x + 3y  5  2x + 3y - s = 5, s  0 (I) s could be the basic variable, but it should be  0 and for x=y=0, it is -5 ! How do we force s  0? ?

  9. 2x + 3y  5  2x + 3y - s = 5, s  0 (I) s could be the basic variable, but it should be  0 and it is -5 for x=y=0! How do we force s  0? By making it 0! how?

  10. 2x + 3y  5  2x + 3y - s = 5, s  0 (I) s could be basic, but it should be  0 and it is -5 for x=y=0! How do we force s  0? By making it 0! But we have to start with a canonical form… so treat is as an equality constraint! 2x + 3y - s+ a = 5, s  0, a  0 and Min a

  11. Artificial Variables Notice: In a  inequality constraint, the extra variable is called an artificial variable. For instance, in 2x + 3y – s+ a = 5, s  0, a  0 (I) a is an artificial variable. In a sense, we allow temporarily a small amount of cheating, but in the end we cannot allow it!

  12. What if we have many such = and  constraints? 7x - 3y – s1+ a1 = 6, s1,a1  0 (I) 2x + 3y + a2 = 5, a2  0 (II) a1 and a2 are artificial variables, s1 is a surplus variable. One minimizes their sum: Min {a1+a2 | a1, a2  0, (I), (II), other constraints} i.e., one minimizes the total amount of cheating!

  13. Then What? We have two objectives: Get a “feasible” canonical form Maximize our original problem Two methods: big M method phase 1, then phase 2

  14. Big-M Method Combine both objectives : (1) Min iai (2) Max j cjxj into a single one: (3) Max – M iai + j cjxj where M is a large number, larger than anything subtracted from it. If one minimizes j cjxj then the combined objective function is Min M iai + j cjxj

  15. The Big M Method The simplex method algorithm requires a starting bfs. Previous problems have found starting bfs by using the slack variables as our basic variables. If an LP have ≥ or = constraints, however, a starting bfs may not be readily apparent. In such a case, the Big M method may be used to solve the problem. Consider the following problem.

  16. Example Bevco manufactures an orange-flavored soft drink called Oranj by combining orange soda and orange juice. Each orange soda contains 0.5 oz of sugar and 1 mg of vitamin C. Each ounce of orange juice contains 0.25 oz of sugar and 3 mg of vitamin C. It costs Bevco 2¢ to produce an ounce of orange soda and 3¢ to produce an ounce of orange juice. Bevco’s marketing department has decided that each 10-oz bottle of Oranj must contain at least 30 mg of vitamin C and at most 4 oz of sugar. Use linear programming to determine how Bevco can meet the marketing department’s requirements at minimum cost.

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