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OR-1 20151 Simplex method (algebraic interpretation) Add slack variables( 여유변수 ) to each constraint to convert them to equations. (We may refer it as.

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Presentation on theme: "OR-1 20151 Simplex method (algebraic interpretation) Add slack variables( 여유변수 ) to each constraint to convert them to equations. (We may refer it as."— Presentation transcript:

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2 OR 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   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.

4 OR

5 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). (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) (zeroth equation) (first equation) (second equation) (third equation) (nonnegativity constraints)

6 OR  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.

7 OR  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.

8 OR

9 8 (continued) x 1  (5/2) most binding (called minimum 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)

10 OR  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.

11 OR

12 OR 

13 OR Equivalent to performing row operations

14 OR

15 OR

16 OR

17 OR  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

18 OR Geometric meaning of an iteration  Notation x1x1 x3x3 x2x2 x 1 =0 x 2 =0 x 3 =0

19 OR  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

20 OR Terminology

21 OR

22 OR

23 OR Remarks

24 OR Obtaining all optimal solutions

25 OR  Another example

26 OR Tableau format 

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


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