MIE 379 Deterministic Operations Research Today s Goals

1. MIE 379 Deterministic Operations ResearchToday�s Goals Understand the Revised Simplex Method. Read Ch. 5.1-5.2 Midterm Elab 305 6-8p.m. Oct 9. Review Session Tuesday Oct. 7 6p.m. Elab 325 Project: Initial solution due Oct. 16 Start working on alternative solutions First draft due Oct. 31 (new date)

2. Homework 4.3-5 Max 5x1+3x2+4x3 s.t. 2x1+x2+x3 = 20 3x1+x2+2x3 = 30 x1,x2,x3 = 0 How many basic variables are there in this problem? If we know that x2 and x3 are non-zero, how can we find the optimal solution directly? How can this help us use the simplex method more efficiently, assuming we start from the usual initial solution? What if I told you that x3 was the only non-zero decision variable in a solution. How could you find that solution?

3. Linear ProgrammingDescription Max Z = c1x1 + � + cnxn s.t. a11x1 + � + a1nxn = b1 am1x1 + � + amnxn = bm x1 = 0, �, xn = 0

4. Revised Simplex Uses Matrix notation A shortcut.

5. Revised Simplex Methodre-write problem in matrix form Max Z = cx s.t. Ax = b and x = 0 c = [c1,c2,�,cn] contribution of each xi to Z

6. Revised Simplex

7. Revised Simplex:Initial Tableau

8. Revised Simplex Method Put Wyndor Glass problem in Matrix form Fill in values for each Matrix or vector: c, A, x, b, I. xs Convince yourself that this really represents the initial tableau.

9. Revised Simplex:Initial Tableau

10. Revised Simplex:Initial Tableau

11. Revised Simplex Recall that in any basic solution m variables are basic (positive) and n variables are non-basic (equal to zero) Ignore all the non-basic variables. Let xB be the vector of basic variables Let B be the Matrix of coefficients of basic variables. Then it is true that BxB=b

12. Initial Basis

13. Revised Simplex Consider the following constraint set, in augmented form. x3 and x4 are slacks. x1 + 2x2 + x3 = 5 2x1 + 3x2 + x4 = 8 If x1 and x3 are basic, then we know that x2 and x4 each equal 0, so we can rewrite the constraint set as x1 + x3 = 5 2x1 = 8

14. Revised Simplex Consider the following constraint set, in augmented form. x3 and x4 are slacks. x1 + 2x2 + x3 = 5 2x1 + 3x2 + x4 = 8 If x1 and x3 are basic, then we know that x2 and x4 each equal 0, so we can rewrite the constraint set as x1 + x3 = 5 2x1 = 8

15. Revised Simplex In any iteration it is true that BxB = b We can solve for xB B-1BxB = B-1b xB = B-1b So, if I tell you the basis, you can tell me the value of the basic variables.

16. Revised Simplex Method Example The basic variables are x1, x3, x4. What are their values? xB = B-1b

17. Revised Simplex Method Example The basic variables are x3, x2, x5. What are their values?

18. Revised Simplex Method Let cB be the objective coefficients of the basic variables. Then cBxB = Z Z = cBxB = cBB-1b

19. Revised Simplex Method Example The basic variables are x3, x2, x1. What are their values? What is the value of Z?

20. Revised Simplex Method Example x1 is basic, x2 is non-basic, and the third constraint is binding. What are the basic variables? What is the value of x1? Is this a feasible solution?

21. Revised Simplex Let xB be the vector of basic variables Let B be the Matrix of coefficients of basic variables. cB = coefficients in objective row of basic variables.

22. Revised simplex: solving for a basic feasible solution If we know the basic variables, how can we find the RHS of the tableau? Now, we can use this knowledge to fill in the rest of the tableau.

23. Revised Simplex

24. Revised Simplex

28. Fundamental Insight

29. Example Max 5x1+3x2+4x3 s.t. 2x1+x2+x3 = 20 3x1+x2+2x3 = 30 x1,x2,x3 = 0 The optimal basis is x2 and x3 Fill in the final tableau

30. Revised Simplex Method Find entering basic variable Calc top row Find leaving basic variable Calc rhs and column under entering basic variable New Solution Calc new B-1

31. Revised Simplex Initial Tableau

32. Revised Simplex Initial Tableau

33. Revised Simplex Initial Tableau

34. Revised Simplex: Check for optimality

35. Revised Simplex

36. Revised Simplex

37. Revised Simplex

38. Revised simplex method Old basis xB=[s1, x2, s3] Entering basic variable x1 Leaving basic variable s3 New basis xB=[s1, x2, x1] Now continue the simplex method, using this basis

39. Revised Simplex Initial Tableau Optimal basis is x1, x2, s1. Fill in final tableau.