1. 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) OR

2. Every feasible solution of (1) can be extended, in the unique way as given above, into a feasible solution of (2). Every feasible solution of (2) can be restricted, by deleting slack variables, into a feasible solution of (1). The correspondence between feasible solutions of (1) and (2) carries optimal solutions of (1) onto optimal solutions of (2), and vice versa.
So solve (2) instead of (1) and disregard the values of slack variables to obtain an optimal solution to the original problem. OR

3. Surplus variable (잉여변수) :
𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 ≥ 𝑏 𝑖  ( 𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 ) − 𝑥 𝑛+𝑖 = 𝑏 𝑖 , 𝑥 𝑛+𝑖 ≥0
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 Chap 8. 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. OR

4. 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).
(zeroth equation)
(first equation)
(second equation)
(third equation)
(nonnegativity constraints)
(Note that, unlike the text, we place the objective function at the top. Such presentation style is used more widely and we follow that convention) OR

5. 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. OR

6. 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. OR

7. Initial feasible solution
To find a better solution, find a nonbasic variable having positive coefficient in zeroth equation (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 𝑥 1 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 𝑥 1 ? OR

8. x1  (5/2) most binding (called minimum ratio test), get new solution
(continued)
x1  (5/2) most binding (called minimum 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 basic feasible solution 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

9. 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 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. OR

10. OR

11. OR

12. Equivalent to performing row operations

13. 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 = 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. OR

14. Perform substitutions (elementary row operations)
Next iteration:
Select 𝑥 3 as the nonbasic variable to increase the value (called entering nonbasic variable).
𝑥 6 becomes 0 (changes status from basic variable to nonbasic variable, called leaving basic variable)
Perform substitutions (elementary row operations) OR

15. New solution is
It is optimal since any feasible solution must have nonnegative values and 𝑧=13−3 𝑥 2 − 𝑥 4 − 𝑥 6 implies that z  13 for any nonnegative feasible solution.
Hence if the coefficients of the nonbasic variables in zeroth equation are all non-positive, current solution is optimal (note that it is a sufficient condition for optimality but not a necessary condition) OR

16. 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 𝑥 1 = 𝑥 0 +𝑡𝑑, 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 +𝑡𝑑≥0. OR

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

18. 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 zeroth equation)
x3=0
x6=0 A
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.
x1=0 d
x4=0 B OR

19. 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.)
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)
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. Basic solutions obtained from a dictionary during simplex iterations are always feasible. OR

20. 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.)
Two view points:
(1) The zeroth equation may be considered as part of the system of equations. In that case, z variable is regarded as basic variable. It always remains basic during the simplex iterations.
(2) On the other hand, zeroth equation may be regarded as a separate equation which is used to read off the objective function values and other equations and nonnegativity describes the feasible solution set of LP. Note that the set of feasible solutions does not change although we perform simplex iterations.
Both viewpoints are useful. OR

21. The set of basic variables is called a basis (기저) of the basic solution. (note that the set of columns in matrix 𝐴 corresponding to basic variables spans the subspace generated by the columns of matrix 𝐴 (which is 𝑅 𝑚 ).)
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

22. Remarks Maximum number of b.f.s. in augmented form is 𝑚+𝑛 𝑛 .
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, details later in Chapter 7). OR

23. Obtaining all optimal solutions
For any feasible solution 𝑥 =( 𝑥 1 , 𝑥 2 , …, 𝑥 6 ), if any of 𝑥 2 , 𝑥 4 , 𝑥 6 (nonbasic) is greater than 0, then the objective value is less than 13. (from zeroth equation)
Hence we must have 𝑥 2 = 𝑥 4 = 𝑥 6 =0 for any optimal solution. Then the current optimal solution is the only one solution which satisfies the constraints, which means it is a unique optimal solution.
If all coefficients in the zeroth equation are < 0, it gives a sufficient condition for the uniqueness of the current optimal solution. OR

24. Any feasible solution with 𝑥 3 =0 is an optimal solution.
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 OR

25. Tableau format OR

26. Tableau format only maintains coefficients in the equations.
It is convenient to carry out simplex iterations in the tableau by performing elementary row operations. OR