# Simplex Method Minimization Examples

Simplex Method Section 4 Maximization and Minimization with Problem Constraints Introduction to the Big M Method In this section, we will present a generalized version of the si l th d th t ill l b th i i ti dimplex method that will solve both maximization and minimization problems with any combination of ≤, ≥, = constraints 2 Example.

• There is a method of solving a minimization problem using the simplex method where you just need to multiply the objective function by -ve sign and then solve it using the simplex method. All you need to do is to multiply the max value found again by -ve sign to get the required max value of the original minimization problem.
• Minimization example Find solution using Simplex(BigM) method MIN Z = x1 + x2 subject to 2x1 + 4x2 = 4 x1 + 7x2 = 7 and x1,x2 = 0 Solution: Problem is.
• Mathematically speaking, in order to use the “flipped” simplex method to solve a linear programming problem, we need the standard minimization problem: an objective function, and one or more constraints of the form, a1x1 + a2x2 + a3x3 + anxn ge V.

## Linear Programming Simplex Method Examples

### 4.2 The Simplex Method: Standard Minimization Problems

#### Learning Objectives

• Use the Simplex Method to solve standard minimization problems.

#### Notes

This section is an optional read. This material will not appear on the exam.

We can also use the Simplex Method to solve some minimization problems, but only in very specific circumstances. The simplest case is where we have what looks like a standard maximization problem, but instead we are asked to minimize the objective function.

 Minimize \$C=-2x-3y\$ subject to \$leftlbracebegin{align}5x+4y&leq 32x+2y&leq 10xgeq 0,&~ygeq 0end{align}right.\$

We notice that minimizing \$C\$ is the same as maximizing \$P=-C\$. Thus the solution to the minimization problem can be found by solving the standard maximization problem below with the techniques learned in Section 4.1.

 Maximize \$P=2x+3y\$ subject to \$leftlbracebegin{align}5x+4y&leq 32x+2y&leq 10xgeq 0,&~ygeq 0end{align}right.\$

The solution to this example is left as an exercise. The other important class of minimization problems we encounter are called standard minimization problems. Definition. A standard minimization problem is a linear programming problem in which we seek to minimize an objective function \$C=c_1x_1+ldots+c_nx_n\$ subject to the constraints\$\$begin{align}a_{11}x_1 + ldots + a_{1n}x_n & geq b_1& vdotsa_{m1}x_1 + ldots + a_{mn}x_n & geq b_mend{align}\$\$where all \$x_igeq 0\$. Onenote 2016 download for mac.

Definition. Each maximization problem in linear programming is associated with a counterpart minimization problem, and vice versa. For the purposes of identification, the given problem will be referred to as the primal problem, and the counterpart to this problem is called the dual problem.

Example. Consider the following standard minimization problem.

 Minimize \$C=4x+2y\$ subject to \$leftlbracebegin{align}5x+y&geq 55x+3y&geq 10xgeq 0,&~ygeq 0end{align}right.\$

## Simplex Method Minimization Examples Pdf

We first write down the following tableau for the given primal problem:\$\$begin{array}{cc c}x & y &hline5 & 1 & 55 & 3 & 10hline4 & 2end{array}\$\$Note that this is not a typical Simplex tableau. There are no slack variables, and we don't rewrite the objective function for the last row. Next we transpose.\$\$begin{array}{cc c}u & v &hline5 & 5 & 41 & 3 & 2hline5 & 10end{array}\$\$Then we construct the dual problem:

## Simplex Method Minimization Examples

 Maximize \$P=5u+10v\$ subject to \$leftlbracebegin{align}5u+5v&leq 4u+3v&leq 2ugeq 0,&~vgeq 0end{align}right.\$

## Simplex Method Minimization Problem

Then we set up the Simplex tableau and proceed as the Section 4.1, using \$x\$ and \$y\$ as our slack variables.\$\$begin{array}{ccccc c}u & v & x & y & P & text{Constant}hline5 & 5 & 1 & 0 & 0 & 41 & 3 & 0 & 1 & 0 & 2hline-5 & -10 & 0 & 0 & 1 & 0end{array}\$\$Solving this tableau is left to the reader as an exercise. The final tableau looks like this.\$\$begin{array}{ccccc c}u & v & x & y & P & text{Constant}hline1 & 0 & 3/10 & -1/2 & 0 & 1/50 & 1 & -1/10 & 1/2 & 0 & 3/5hline0 & 0 & 1/2 & 5/2 & 1 & 7end{array}\$\$Interpreting this in the usual manner, the solution of the dual problem is \$u=1/5\$, \$v=3/5\$, and \$P=7\$. The solution to the primal problem appears under the respective slack variables in the last row of the final tableau: \$x=1/2\$ and \$5/2\$.