the simplex method. standard linear programming problem standard maximization problem 1. all...

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)

Maximize the objective function:P(x,y) = 5x + 4ySubject to:x + y ≤ 202x + y ≤ 35-3x + y ≤ 12x ≥ 0y ≥ 0

Solution

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Sixth

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 - 3ySubject to:5x + 4y ≤ 32x + 2y ≤ 10x ≥ 0y ≥ 0

SolutionLetq(x) = - p(x) = - ( -2x -3y) = 2x + 3yTo minimize p is to maximize q. Thus, we solve the

following standard maximization linear programming problem:

Maximize the objective function:q(x) = 2x + 3ySubject to:5x + 4y ≤ 32x + 2y ≤ 10x ≥ 0y ≥ 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

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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


Minimize the objective function:p(x,y) = 6x + 8ySubject to:40x + 10y ≥ 240010x + 15y ≥ 21005x + 15y ≥ 1500x ≥ 0y ≥ 0

Minimize the objective function: p(x,y) = 6x + 8ySubject to:40x + 10y ≥ 2400, 10x + 15y ≥ 2100 , 5x + 15y ≥ 1500, x ≥ 0 and y ≥ 0We 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 240010 15 21005 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

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Minimize the objective function:p(x,y) = x + 2ySubject to:-2x + y ≥ 1- x + y ≥ 2x ≥ 0y ≥ 0

Minimize the objective function: p(x,y) = x + 2ySubject 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

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Homework

1. Using the simplex method, maximize: p = x + (6/5)y subject to:2x + y ≤ 180 , x + 3y ≤ 300 , x ≥ 0 , y ≥ 0Solution: p(48,84) = 148.8

2. Minimize: p(x,y) = - 5x - 4y Subject to: x + y ≤ 20 , 2x + y ≤ 35 , -3x + y ≤ 12 , x ≥ 0y ≥ 0Solution: 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 ≥ 0Solution: p(20,16) = 92Maximize the objective function:

the simplex method. standard linear programming problem standard maximization problem 1. all...

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