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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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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

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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

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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)

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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

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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)

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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)

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

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

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Feasibility Region Convex Set X2 X1 10 5

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Objective Function Isoclines - 1 Max 2*X1 + 3*X2 = z s.t. X1 + 2*X2 10 X1, X2 0 X2 X1 10 5 X2 = z/3 – 2/3*X1 z=6 X2=2–2/3*X1

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Objective Function Isoclines - 2 Max 2*X1 + 3*X2 = z s.t. X1 + 2*X2 10 X1, X2 0 X2 X1 10 5 X2 = z/3 – 2/3 * X1 z=15 z=6 z=0

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Graphical Solution Max 2*X1 + 3*X2 = z s.t. X1 + 2*X2 10 X1, X2 0 X2 X1 10 5 X2 = z/3 – 2/3 * X1 z=20

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Graphical Solution X2 X1 10 5 z=20 Is at extreme point!

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Other Extreme Points X2 X1 10 5

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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

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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

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

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

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

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2 Constraints X2 X1 10 5 5 Max Objective s.t. X1 + 2*X2 + S1 = 10 2*X1 + X2 + S2 = 10 X1, X2, S1, S2 0

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Max Objective s.t. X1 + 2*X2 + S1 = 10 2*X1 + X2 + S2 = 10 X1, X2, S1, S2 0 X2 X1 10 5 5 2 Constraints 2 Non-Zero Variables

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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?

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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

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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.

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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

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

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

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

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

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Calculating all extreme points?

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