the simplex method
DESCRIPTION
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). - PowerPoint PPT PresentationTRANSCRIPT
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
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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
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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
Illustrating Example (3)
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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Illustrating Example (4)
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
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- 2x' - y' ' + x = 1
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- x' - 2y' + q = 0
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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: