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Linear Programming, 1 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0 Standard form.

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Presentation on theme: "Linear Programming, 1 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0 Standard form."— Presentation transcript:

1 Linear Programming, 1 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0 Standard form

2 Linear Programming, 2 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0 n +1 Variables

3 Linear Programming, 3 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0 Objective function

4 Linear Programming, 4 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0 n Objective function coefficients (data)

5 Linear Programming, 5 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0 m Constraints

6 Linear Programming, 6 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0 m*n technical coefficients (data)

7 Linear Programming, 7 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0 m resource limits (data)

8 Linear Programming Example Max 2*X1 + 3*X2 = z s.t. X1 + 2*X2  10 X1, X2  0

9 Feasibility Region Max 2*X1 + 3*X2 = z s.t. X1 + 2*X2  10 X1, X2  0 X2 X X2  5 – 0,5*X1 X1,X2  0

10 Feasibility Region Convex Set X2 X1 10 5

11 Objective Function Isoclines - 1 Max 2*X1 + 3*X2 = z s.t. X1 + 2*X2  10 X1, X2  0 X2 X X2 = z/3 – 2/3*X1 z=6 X2=2–2/3*X1

12 Objective Function Isoclines - 2 Max 2*X1 + 3*X2 = z s.t. X1 + 2*X2  10 X1, X2  0 X2 X X2 = z/3 – 2/3 * X1 z=15 z=6 z=0

13 Graphical Solution Max 2*X1 + 3*X2 = z s.t. X1 + 2*X2  10 X1, X2  0 X2 X X2 = z/3 – 2/3 * X1 z=20

14 Graphical Solution X2 X z=20 Is at extreme point!

15 Other Extreme Points X2 X1 10 5

16 Solving Arithmetically Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n  b 1 … a m1 *X 1 +…+ a mn *X n  b m X 1, X n  0

17 Convert Inequalities Max c 1 *X 1 +…+c n *X n +0*S 1 +…+0*S m = z s.t. a 11 *X 1 +…+a 1n *X n +1*S 1 +…+0*S m = b 1 … a m1 *X 1 +…+a mn *X n +0*S 1 +…+1*S m = b m X 1, X 2, S 1, S m  0 Max c 1 *X 1 +…+c n *X n = z s.t. a 11 *X 1 +…+a 1n *X n  b 1 … a m1 *X 1 +…+a mn *X n  b m X 1, X 2  0

18 Arithmetic Problem, 1 Number of unknown variables: n+m Number of equations (constraints): m  Can only solve m equations for m unknown variables

19 Arithmetic Problem, 2 From graphical illustration, we know Solution occurs at extreme point

20 1 Constraint  1 Non-Zero Variable Max Objective s.t. X1 + 2*X2 + S = 10 X1, X2, S  0 X2 X1 10 5

21 2 Constraints X2 X Max Objective s.t. X1 + 2*X2 + S1 = 10 2*X1 + X2 + S2 = 10 X1, X2, S1, S2  0

22 Max Objective s.t. X1 + 2*X2 + S1 = 10 2*X1 + X2 + S2 = 10 X1, X2, S1, S2  0 X2 X Constraints  2 Non-Zero Variables

23 Arithmetic Problem, 3 Extreme points have only m non-zero variables and n-m zero variables Thus, if we knew which n-m variables are zero, we could solve for the remaining m variables using our m equation system New task: Which variables should remain zero?

24 Arithmetic Solution 1.Pick n-m variables and set to zero 2.Solve mxm equation system 3.Compute optimality indicators 4.If optimal, done; otherwise

25 Optimal solution is extreme point Number of non-zero variables = number of equations Non-zero (positive) variables are called basic variables All remaining variables are called non- basic variables Non-basic variables are zero.

26 Decomposed LP in Matrix Notation Max C B X B + C NB X NB s.t. B X B + A NB X NB = b X B, X NB  0

27 Solution of LP in Matrix Notation BX B =b-AX NB BX B =b X B =B -1 b

28 Solver tasks Determine which variables form basis Invert coefficient matrix to calculate optimal variable values

29 If optimal solution is an extreme point, why not simply calculate all extreme points and choose the one with highest objective function value?

30 Number of Possible Extreme Points n.. Number of Variables m.. Number of Equations

31 Calculating all extreme points?


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