1. Simplex method (algebraic interpretation)
Add slack variables(여유변수) to each constraint to convert them to equations.
(1)
(2) OR

2. (Surplus variable (잉여변수) : a’x  b  a’x – xs = b, xs  0 )
Hence we have a 1-1 mapping which maps each feasible point in (1) to a feasible point in (2) uniquely (and conversely) and the objective values are the same for the points.
So solve (2) instead of (1) and disregard the value of slack variables to obtain an optimal solution to the original problem.
(Surplus variable (잉여변수) : a’x  b  a’x – xs = b, xs  0 ) OR

3. Remark: If LP includes equations, we need to convert each equation to 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 convert equations to inequalities. OR

4. Changes in the solution space when slack is addedx2 x3 1 1 x1 1 x1 1 1 x2
Solution set is still 2-dimensional OR

5. Next let
Then find solution to the following system which maximizes z (tableau form)
In the text, dictionary form 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) OR

6. 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 x1 = x2 = x3 = 0 and finding the values of x4, x5, and x6 , which can be read directly from the dictionary. (also z values can be read.) If all values of x4, x5, and x6 are nonnegative, we obtain a basic feasible solution.
The equation for z may be regarded as part of the systems of equations, or we may think of it as a separate equation used to evaluate the objective value for the given solution. OR

7. 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. OR

8. Initial feasible solution
To find a better solution, find a nonbasic variable having positive coefficient in z row (say x1) 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 x1 increases and other nonbasic variables remain at 0, the values of basic variables must change so that the new solution satisfies the equations and nonnegativity. How much can we increase x1? OR

9. x1  (5/2) most binding (ratio test), get new solution
(continued)
x1  (5/2) most binding (ratio test), get new solution
x1 = (5/2), x2, x3 = 0, x4 = 0, x5 = 1, x6 = (1/2), z = 25/2
This is a new b.f.s since x4 now can be treated as a nonbasic variable (has value 0) and x1 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 x2, x3, and x4 at 0 and obtain a unique solution after solving the remaining system of equations) OR

10. Why could we find a basic feasible solution easily?
Change the dictionary so that the new solution can be directly read off
x1 : 0 (5/2), x4 : 5  0
So change the role of x1 and x x4 becomes independent (nonbasic) variable and x1 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. OR

11. OR

12. OR

13. Equivalent to performing row operations

14. Note that the previous solution
x1 = x2 = x3 = 0, x4 = 5, x5 = 11, x6 = 8, z = 0
and the new solution
x1 = (5/2), x2, x3 = 0, x4 = 0, x5 = 1, x6 = (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. OR

15. x6 becomes 0 (changes status to nonbasic variable from basic variable)
Next iteration:
Select x3 as the nonbasic bariable to increase the value (called entering nonbasic variable).
x6 becomes 0 (changes status to nonbasic variable from basic variable)
Perform substitutions (elementary row operations) OR

16. implies that z  13 for any nonnegative feasible solution
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) OR

17. x1 = (5/2), x2, x3 = 0, x4 = 0, x5 = 1, x6 = (1/2), z = 25/2
Moving directions in Rn in the example
x1 = (5/2), x2, x3 = 0, x4 = 0, x5 = 1, x6 = (1/2), z = 25/2
Then we obtained x1 = x0 +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 x1)
We make t as large as possible while x0 +t d  0. OR

18. Geometric meaning of an iteration
Notation
x1=0
x2=0 x3
x3=0 x1 x2 OR

19. Our example : assume x2 does not exist
Our example : assume x2 does not exist. It makes the polyhedron 2 dimensional since we have 5 variables and 3 equations (except nonnegativity and obj row)
x3=0
x6=0 A
We move from A, which is
an extreme point defined by 3 eq. and x1=x3= 0 to B defined by the 3 eq. and x3 = x4 = 0.
x1=0 d
x4=0 B OR

20. 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. OR

21. Both viewpoints are useful.
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) OR

22. The set of basic variables are called basis (기저) of the basic solution
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) OR

23. 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. OR

24. 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). OR

25. 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. OR

26. Any feasible solution with x3 = 0 is an optimal solution.
Another example
Any feasible solution with x3 = 0 is an optimal solution.
The set of feasible solutions with x3 = 0 is given by OR

27. Tableau format OR

28. Tableau format only maintains coefficients in the equations.
It is convenient to carry out a simplex iteration in the tableau. OR