2. OR-1 20131 Simplex method (algebraic interpretation) Add slack variables( 여유변수 ) to each constraint to convert them to equations. (We may refer it as an augmented LP) (1) (2)

3. OR-1 20132

4. 3  Remark: If LP includes equations in the constraints, we need to replace each equation with two inequalities to express the problem in standard form as we have seen earlier. Then we may add slack or surplus variables to convert them to equations. However, this procedure will increase the number of constraints and variables. Equations in an LP can be handled directly without changing them to inequalities. Detailed method will be explained in Chap8. General LP Problems. For the time being, we assume that we follow the standard procedure to replace each equation with two inequalities to obtain a standard form.

5. OR-1 20134 Changes in the solution space when slack is added  x2x2 x1x1 x1x1 x3x3 x2x2 1 1 1 1 1 Solution set is still 2-dimensional

6. OR-1 20135  Next let Then find solution to the following system which maximizes z (tableau form) In the text, dictionary form is used, i.e. each dependent variable (including z) (called basic variable) is expressed as linear combinations of indep. var. (called nonbasic variable). (Note that, unlike the text, we place the objective function in the first row. Such presentation style is used more widely and we follow that convention)

7. OR-1 20136  From previous lectures, we know that if the polyhedron P has at least one extreme point and the LP over P has a finite optimal value, the LP has an extreme point optimal solution. Also an extreme point of P for our problem is a basic feasible solution algebraically. We obtain a basic solution by setting x 1 = x 2 = x 3 = 0 and finding the values of x 4, x 5, and x 6, which can be read directly from the dictionary. (also z values can be read.) If all values of x 4, x 5, and x 6 are nonnegative, we obtain a basic feasible solution.

8. OR-1 20137  Now, we look for another basic feasible solution (extreme point of the polyhedron) which gives a better objective value than the current solution. Such solution can be examined by setting 7 – 4 = 3 variables at 0 (called nonbasic variables) and solve the equations for the remaining 4 variables (called basic variables). Here z may be regarded as a basic variable and it remains basic at any time during the simplex iterations.

9. OR-1 20138

10. 9 (continued) x 1  (5/2) most binding (called ratio test), get new solution x 1 = (5/2), x 2, x 3 = 0, x 4 = 0, x 5 = 1, x 6 = (1/2), z = 25/2 This is a new basic feasible solution since x 4 now can be treated as a nonbasic variable (has value 0) and x 1 is basic. (We need a little bit of caution here in saying that the new solution is a basic feasible solution since we must be able to obtain it by setting x 2, x 3, and x 4 at 0 and obtain a unique solution after solving the remaining system of equations)

11. OR-1 201310  Change the dictionary so that the new solution can be directly read off x 1 : 0  (5/2), x 4 : 5  0 So change the role of x 1 and x 4. x 4 becomes independent (nonbasic) variable and x 1 becomes dependent (basic) variable. Why could we find a basic feasible solution easily? 1) all independent(nonbasic) variables appear at the right of equality (have value 0) 2) each dependent (basic) variable appears in only one equation 3) each equation has exactly one basic variable appearing ( z variable may be interpreted as a basic variable, but usually it can be treated separately since it always remains basic and it is irrelevant to the description of the feasible solutions) So change the dictionary so that it satisfies the above properties.

12. OR-1 201311

13. OR-1 201312 

14. OR-1 201313 Equivalent to performing row operations

15. OR-1 201314

16. OR-1 201315

17. OR-1 201316

18. OR-1 201317  Moving directions in R n in the example x 1 = (5/2), x 2, x 3 = 0, x 4 = 0, x 5 = 1, x 6 = (1/2), z = 25/2

19. OR-1 201318 Geometric meaning of an iteration  Notation x1x1 x3x3 x2x2 x 1 =0 x 2 =0 x 3 =0

20. OR-1 201319  Our example : assume x 2 does not exist. It makes the polyhedron 2 dimensional since we have 5 variables and 3 equations (except nonnegativity and objective row) x 1 =0 x 4 =0 x 3 =0 x 6 =0 d A B

21. OR-1 201320 Terminology

22. OR-1 201321

23. OR-1 201322

24. OR-1 201323 Remarks  The basic feasible solution to the augmented form is an extreme point of the corresponding polyhedron. Also it corresponds to the extreme point of the polyhedron for standard LP (after ignoring the slack variables). (If the given LP is not in standard form, we should be careful in saying the equivalence, especially when free variables exist.)  Simplex method searches the extreme points of the polyhedron during the iterations.  Note that we used (though without proof) the equivalence of the extreme points (geometric definition) and the basic feasible solution (algebraic definition) for augmented form LP.

25. OR-1 201324

26. OR-1 201325 Obtaining all optimal solutions

27. OR-1 201326  Another example

28. OR-1 201327 Tableau format 

29. OR-1 201328  Tableau format only maintains coefficients in the equations. It is convenient to carry out simplex iterations in the tableau.