2. OR-1 20091  Now, we look for other basic feasible solutions which gives better objective values than the current solution. Such solutions 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.

3. OR-1 20092  Initial feasible solution To find a better solution, find a nonbasic variable having positive coefficient in z row (say x 1 ) and increase the value of the chosen nonbasic variable while other nonbasic variables remain at 0. We need to obtain a solution that satisfies the equations. Since x 1 increases and other nonbasic variables remain at 0, the value of basic variables must change so that the new solution satisfies the equations and nonnegativity. How much can we increase x 1 ?

4. OR-1 20093 (continued) x 1  (5/2) most binding (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 b.f.s 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)

5. OR-1 20094  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 is 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.

6. OR-1 20095

7. 6 

8. 7 Equivalent to performing row operations

9. OR-1 20098  Note that the previous solution x 1 = x 2 = x 3 = 0, x 4 = 5, x 5 = 11, x 6 = 8, z = 0 and the new solution x 1 = (5/2), x 2, x 3 = 0, x 4 = 0, x 5 = 1, x 6 = (1/2), z = 25/2 satisfies the updated system of equations. Only difference is that the new solution can be read off directly from the new dictionary. We update the dictionary to read a new basic solution directly, but the set of solutions is not changed.

10. OR-1 20099  Next iteration: Select x 3 as the nonbasic bariable to increase the value (called entering nonbasic variable). x 6 becomes 0 (changes status to nonbasic variable from basic variable) Perform substitutions (elementary row operations)

11. OR-1 200910 New solution is It is optimal since any feasible solution must have nonnegative values and implies that z  13 for any nonnegative feasible solution Hence if the coefficients of the nonbasic variables in z- row are all non-positive, current solution is optimal (note that it is a sufficient condition for optimality but not a necessary condition)

12. OR-1 200911  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 Then we obtained x 1 = x 0 +t d, where d = (1, 0, 0, -2, -4, -3) and t = 5/2 Note that the d vector can be found from the dictionary.( the column for x 1 ) We make t as large as possible while x 0 +t d  0.

13. OR-1 200912 Geometric meaning of an iteration  Notation x1x1 x3x3 x2x2 x 1 =0 x 2 =0 x 3 =0

14. OR-1 200913  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 obj row) x 1 =0 x 4 =0 x 3 =0 x 6 =0 d A B We move from A, which is an extreme point defined by 3 eq. and x 1 =x 3 = 0 to B defined by the 3 eq. and x 3 = x 4 = 0.

15. OR-1 200914 Terminology Assume that we have max c’x, Ax = b, x  0, where A is m  (n + m) and full row rank.  A solution x * is called a basic solution ( 기저해 ) if it can be obtained by setting n of the variables equal to 0 and then solving for the remaining m variables, where the columns of the A matrix corresponding to the m variables are linearly independent. (Hence provides a unique solution.)  In the text, basic solution is defined as the solution which can be obtained by setting the right-hand side variables (independent var.) at zero in the dictionary. This is the same definition as the one given above. But the text does not make clear distinction between basic solution and basic feasible solution.

16. OR-1 200915  For a basic solution x *, the n variables which are set to 0 are called nonbasic variables ( 비기저변수 ) (independent var.) and the remaining m variables are called basic variables ( 기저변수 ) (dependent var.)  The z-row may be considered as part of system of equations. In that case, z var. is regarded as basic variable. It always remains basic during the simplex iterations. On the other hand, z-row may be regarded as a separate equation which is used to read off objective function values and other equations and nonnegativity describes the solution set. Both viewpoints are useful.  A solution x * is called a basic feasible solution ( 기저가능해 ) if it is a basic solution and satisfies x  0. (feasible solution to the augmented LP)

17. OR-1 200916  The set of basic variables are called basis ( 기저 ) of the basic solution. (note that the set of basic variables spans the subspace generated by the columns of A matrix.)  In a simplex iteration, the nonbasic variable which becomes basic in that iteration is called entering (nonbasic) variable ( 도입변수 ) and the basic variable which becomes nonbasic is called leaving (basic) variable ( 탈락변수 )  Minimum ratio test ( 최소비율검사 ) : test to determine the leaving basic variable  Pivoting : computational process of constructing the new dictionary (elementary row operations)

18. OR-1 200917 Remarks  For standard LP, the basic feasible solution to the augmented form corresponds to the extreme point of the feasible set of points. (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 in its 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.

19. OR-1 200918  Maximum number of b.f.s. in augmented form is  In the simplex method, one nonbasic variable becomes basic and one basic variable becomes nonbasic in each iteration (except the z variable, it always remains basic.)  In real implementations, we do not update entire dictionary ( or tableau). We maintain information about the current basis. Then entire tableau can be constructed from that information and the simplex iteration can be performed (called revised simplex method).

20. OR-1 200919 Obtaining all optimal solutions If all coefficients in the z- row are < 0, it gives a sufficient condition for the uniqueness of the current optimal solution.

21. OR-1 200920  Another example Any feasible solution with x 3 = 0 is an optimal solution. The set of feasible solutions with x 3 = 0 is given by

22. OR-1 200921 Tableau format 

23. OR-1 200922  Tableau format only maintains coefficients in the equations. It is convenient to carry out a simplex iteration in the tableau.