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The Simplex Method. Standard Linear Programming Problem. Standard Maximization Problem 1. All variables are nonnegative . 2. All the constraints (the conditions) can be expressed as inequalities of the form: ax + by ≤ c, where c is a positive constant. Illustrating Example (1).
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Standard Linear Programming Problem Standard Maximization Problem 1. All variables are nonnegative. 2. All the constraints (the conditions) can be expressed as inequalities of the form: ax + by ≤ c, where c is a positive constant
Illustrating Example (1) Maximize the objective function: P(x,y) = 5x + 4y Subject to: x + y ≤ 20 2x + y ≤ 35 -3x + y ≤ 12 x ≥ 0 y ≥ 0
What about when all of the constraints (the inequalities) are of the type “≤ positive constant”But we want to minimize the objective function instead of maximizing.
Minimization with “≤” constraintsIllustrating Example (2) Minimize the objective function: p(x,y) = -2x - 3y Subject to: 5x + 4y ≤ 32 x + 2y ≤ 10 x ≥ 0 y ≥ 0
Solution Let q(x) = - p(x) = - ( -2x -3y) = 2x + 3y To minimize p is to maximize q. Thus, we solve the following standard maximization linear programming problem: Maximize the objective function: q(x) = 2x + 3y Subject to: 5x + 4y ≤ 32 x + 2y ≤ 10 x ≥ 0 y ≥ 0
Rewriting the inequalities as equations, by introducing the “slack” variables u and v and the formula of the objective function as done in example (1). 5x + 4y ≤ 32 , x + 2y ≤ 10 and q = 2x +3y Are transformed to: 5x + 4y + u = 32 x + 2y + v = 10 - 2x - 3y + q = 0
Standard Linear Programming Problem Standard Minimization Problem 1. All variables are nonnegative. 2. All the constraints (the conditions) can be expressed as inequalities of the form: ax + by ≥ c, where c is a positive constant
Solving The Standard Minimization Problem We use the fundamental theorem of Duality
Illustrating Example (3) Minimize the objective function: p(x,y) = 6x + 8y Subject to: 40x + 10y ≥ 2400 10x + 15y ≥ 2100 5x + 15y ≥ 1500 x ≥ 0 y ≥ 0
Minimize the objective function: p(x,y) = 6x + 8y Subject to: 40x + 10y ≥ 2400, 10x + 15y ≥ 2100 , 5x + 15y ≥ 1500, x ≥ 0 and y ≥ 0 We will refer to the above given problem by the primal (original) problem First: We construct the following table, which we will refer to by the “primal” table: x y constant --------------------------------- 40 10 2400 10 15 2100 5 15 1500 --------------------------------- 6 8 Second: We construct a dual (twin) table from interchanging the rows and columns in the primal table: x' y' z' constant ----------------------------------------------------------- 40 10 5 6 10 15 15 8 --------------------------------------------------------- 2400 2100 1500 Third: We interpret the “dual table” as a standard maximization problem, which will refer to as the “dual problem” or “twin problem” of the “primal problem” or the “original problem” Miaximoze the objective function: q( x ' , y ' , z ' ) = 2400x' + 2100y' + 1500z' Subject to: 40x' + 10y' + 5z' ≤ 6, 10x' + 15y' + 15z' ≤ 8, x' ≥ 0 and y' ≥ 0, z' ≥ 0
Fourth: We apply the simplex method explained in example (1) to solve this problem Maximize the objective function: q(x,y,z) = 2400x' + 2100y' + 1500z' Subject to: 40x' + 10y' + 5z' ≤ 6, 10x' + 15y' + 15z' ≤ 8, x' ≥ 0 and y' ≥ 0, z' ≥ 0 4.a.Rewriting the inequalities and the formula of the objective function, with the slack variables being the same x and y (in that order) of the original (minimization) problem : 40x' + 10y' + 5z' + x = 6 10x' + 15y' + 15z' + y = 8 - 2400x' - 2100y' - 1500z‘ + q = 0 4.b. We construct the simplex table for this problem
Illustrating Example (4) Minimize the objective function: p(x,y) = x + 2y Subject to: -2x + y ≥ 1 - x + y ≥ 2 x ≥ 0 y ≥ 0
Minimize the objective function: p(x,y) = x + 2y Subject to: -2x + y ≥ 1, - x + y ≥ 2 We will refer to the above given problem by the primal (original) problem First: We construct the following table, which we will refer to by the “primal” table: x y constant --------------------------------- -2 1 1 -1 1 2 --------------------------------- 1 2 Second: We construct a dual (twin) table from interchanging the rows and columns in the primal table: x' y' constant ------------------------------------------- -2 -1 1 1 1 2 ----------------------------------------- 1 2 Third: We interpret the “dual table” as a standard maximization problem, which will refer to as the “dual problem” or “twin problem” of the “primal problem” or the “original problem” Maximize the objective function: q( x ' , y ‘ ) = x' + 2y' Subject to: -2x' - y' ≤ 1, x' + y' ≤ 2, x' ≥ 0 and y' ≥ 0
Fourth: We apply the simplex method explained in example (1) to solve this problem Maximize the objective function: q( x ' , y ‘ ) = x' + 2y' Subject to: - 2x' - y' ≤ 1, x' + y' ≤ 2, x' ≥ 0 and y' ≥ 0 4.a.Rewriting the inequalities and the formula of the objective function, with the slack variables being the same x and y (in that order) of the original (minimization) problem : - 2x' - y'' + x = 1 x' + y' + y = 2 - x' - 2y' + q = 0 4.b. We construct the simplex table for this problem
Homework 1. Using the simplex method, maximize: p = x + (6/5)y subject to: 2x + y ≤ 180 , x + 3y ≤ 300 , x ≥ 0 , y ≥ 0 Solution: p(48,84) = 148.8 2. Minimize: p(x,y) = - 5x - 4y Subject to: x + y ≤ 20 , 2x + y ≤ 35 , -3x + y ≤ 12 , x ≥ 0 y ≥ 0 Solution: p(15,5) = - 95 3. Using the dual theorem, minimize: p = 3x + 2y subject to: 8x + y ≥ 80 , 8x + 5y ≥ 240 , x + 5y ≥ 100, x ≥ 0 , y ≥ 0 Solution: p(20,16) = 92 Maximize the objective function:
