solving linear programming problems using simplex method minimization

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Solving a minimization problem using a Simplex method

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. My question is there is any condition that must be satisfied on the constraints of the optimization problem to use this method?

• linear-programming

• $\begingroup$ Would you be able to edit your question to include an example of your objective function in algebraic terms? $\endgroup$ –  danielcharters Commented Apr 10, 2020 at 7:45
• $\begingroup$ @danielcharters many thanks for ur reply. please see my edit I mean is there are any conditions that must be satisfied on the constrains inequality equations to use this method? Sorry about that. $\endgroup$ –  John adams Commented Apr 10, 2020 at 7:49
• $\begingroup$ @Johnadams, To solve the optimization problem using the simplex method, it needs to be interpreted as a standard form , in which all of the model constraints are equal. (To do that, adding slack/surplus/artificial variables.). $\endgroup$ –  A.Omidi Commented Apr 10, 2020 at 9:33
• $\begingroup$ @A.Omidi but there is no constrain on the inequality itself to use the above-mentioned method? I mean the constrain should be <= or >= to use the above-mentioned method or whatever the inequality is I can use this method? $\endgroup$ –  John adams Commented Apr 10, 2020 at 9:43
• $\begingroup$ @Johnadams, for both inequality you mentioned, $<=$ or $>=$, you could use the simplex method. In the $<=$ you need slack variables and in the $>=$ you need surplus and even artificial variables. If your problem has many variables I recommended using optimization software to do that automatically. $\endgroup$ –  A.Omidi Commented Apr 10, 2020 at 9:48

2 Answers 2

It has nothing to do even with linear programming. It's a simple mathematical fact:

$$\min \left( f \left( x \right) \right) = - \max \left( -f \left( x \right) \right)$$

which still holds when you restrict the domain of the function by the constraints (actually to a convex polyhedron in case of LP).

• 1 $\begingroup$ "Convex cone" should probably be "convex polytope" (or polyhedron), but the mathematical statement is correct. $\endgroup$ –  prubin ♦ Commented Apr 10, 2020 at 19:38

The only requirements for the constraints, that I am aware of, when using the simplex algorithm to solve a minimization (and maximization) problem is to include the slack and surplus variables where needed, and the decision variables have to be non-negative. Below is an example to illustrate how to formulate a problem to be solved using the simplex algorithm and how to include slack and surplus variables into your formulation. \begin{align}\min&\quad z = 2x_1 - 3x_2\\\text{s.t.}&\quad x_1+x_2 \leq 4\\&\quad x_1-x_2 \geq 6\\&\quad x_1,x_2 \geq 0\end{align}

The optimal solution to this would be where $ z = 2x_1-3x_2$ is the smallest, but equivalently it can be said that the optimal solution would be where $ -z = -2x_1+3x_2$ is the largest. This is done as the simplex algorithm is used to solve maximization problems, and the formulation now becomes \begin{align}\max&\quad-z = -2x_1 + 3x_2\\\text{s.t.}&\quad x_1+x_2 \leq 4\\&\quad x_1-x_2 \geq 6\\&\quad x_1,x_2 \geq 0\end{align}

We add a slack variable $s_1$ to the first constraint, which now becomes $x_1 +x_2 +s_1 = 4$ . Similarly for the second constraint, we add the surplus variable $s_2$ , and the constraint now becomes $x_1-x_2 + s_2= 6$ .

The formulation, which is now in standard form to be solved using the simplex algorithm, is as follows: \begin{align}\max&\quad-z = -2x_1 + 3x_2\\\text{s.t.}&\quad x_1 +x_2 +s_1 = 4\\&\quad x_1-x_2 + s_2= 6\\&\quad x_1,x_2 \geq 0\\&\quad s_1,s_2 \geq 0.\end{align}

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Simplex Method for Solution of L.P.P (With Examples) | Operation Research

After reading this article you will learn about:- 1. Introduction to the Simplex Method 2. Principle of Simplex Method 3. Computational Procedure 4. Flow Chart.

Introduction to the Simplex Method :

Simplex method also called simplex technique or simplex algorithm was developed by G.B. Dantzeg, An American mathematician. Simplex method is suitable for solving linear programming problems with a large number of variable. The method through an iterative process progressively approaches and ultimately reaches to the maximum or minimum values of the objective function.

Principle of Simplex Method :

It has not been possible to obtain the graphical solution to the LP problem of more than two variables. For these reasons mathematical iterative procedure known as ‘Simplex Method’ was developed. The simplex method is applicable to any problem that can be formulated in-terms of linear objective function subject to a set of linear constraints.

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The simplex method provides an algorithm which is based on the fundamental theorem of linear programming. This states that “the optimal solution to a linear programming problem if it exists, always occurs at one of the corner points of the feasible solution space.”

The simplex method provides a systematic algorithm which consist of moving from one basic feasible solution to another in a prescribed manner such that the value of the objective function is improved. The procedure of jumping from vertex to the vertex is repeated. The simplex algorithm is an iterative procedure for solving LP problems.

It consists of:

(i) Having a trial basic feasible solution to constraints equation,

(ii) Testing whether it is an optimal solution,

(iii) Improving the first trial solution by repeating the process till an optimal solution is obtained.

Computational Procedure of Simplex Method :

The computational aspect of the simplex procedure is best explained by a simple example.

Consider the linear programming problem:

Maximize z = 3x 1 + 2x 2

Subject to x 1 + x 2 , ≤ 4

x 1 – x 2 , ≤ 2

x 1 , x 2 , ≥ 4

< 2 x v x 2 > 0

The steps in simplex algorithm are as follows:

Formulation of the mathematical model:

(i) Formulate the mathematical model of given LPP.

(ii) If objective function is of minimisation type then convert it into one of maximisation by following relationship

Minimise Z = – Maximise Z*

When Z* = -Z

(iii) Ensure all b i values [all the right side constants of constraints] are positive. If not, it can be changed into positive value on multiplying both side of the constraints by-1.

In this example, all the b i (height side constants) are already positive.

(iv) Next convert the inequality constraints to equation by introducing the non-negative slack or surplus variable. The coefficients of slack or surplus variables are zero in the objective function.

In this example, the inequality constraints being ‘≤’ only slack variables s 1 and s 2 are needed.

Therefore given problem now becomes:

The first row in table indicates the coefficient c j of variables in objective function, which remain same in successive tables. These values represent cost or profit per unit of objective function of each of the variables.

The second row gives major column headings for the simple table. Column C B gives the coefficients of the current basic variables in the objective function. Column x B gives the current values of the corresponding variables in the basic.

Number a ij represent the rate at which resource (i- 1, 2- m) is consumed by each unit of an activity j (j = 1,2 … n).

The values z j represents the amount by which the value of objective function Z would be decreased or increased if one unit of given variable is added to the new solution.

It should be remembered that values of non-basic variables are always zero at each iteration.

So x 1 = x 2 = 0 here, column x B gives the values of basic variables in the first column.

So 5, = 4, s 2 = 2, here; The complete starting feasible solution can be immediately read from table 2 as s 1 = 4, s 2 , x, = 0, x 2 = 0 and the value of the objective function is zero.

Flow Chart of Simplex Method :

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Simplex Method Calculator – Two Phase Online 🥇

Simplex method calculator - free version, members-only content, do you already have a membership, get membership.

The free version of the calculator shows you each of the intermediate tableaus that are generated in each iteration of the simplex method, so you can check the results you obtained when solving the problem manually.

Advanced Functions of the simplex method online calculator – Two-Phase

Let's face it, the simplex method is characterized by being a meticulous and impractical procedure, because if you fail in an intermediate calculation you can compromise the final solution of the problem. In that sense, it is important for the student to know the step by step procedure to obtain each of the values in the iterations. Thus, in PM Calculators we have improved our application to include a complete step-by-step explanation of the calculations of the method. You can access this tool and others (such as the big m calculator and the graphical linear programming calculator ) by becoming a member of our membership .

Within the functionality that this application counts we have:

• Ability to solve exercises with up to 20 variables and 50 constraints.
• Explanation of how to determine the optimality condition.
• Explanation of the criteria to establish the feasibility condition.
• Detail of the calculations performed to obtain the vector of reduced costs, the pivot row and the other rows of the table.
• For exercises with artificial variables it becomes a two-phase method calculator .
• Explanation of the special cases such as unbounded and infeasible solutions.

You can find complete examples of how the application works in this link .

How to use the simplex method online calculator

To use our tool you must perform the following steps:

• Enter the number of variables and constraints of the problem.
• Select the type of problem: maximize or minimize .
• Enter the coefficients in the objective function and the constraints. You can enter negative numbers, fractions, and decimals (with point).
• Click on “Solve”.
• The online software will adapt the entered values ​​to the standard form of the simplex algorithm and create the first tableau .
• Depending on the sign of the constraints, the normal simplex algorithm or the two phase method is used.
• We can see step by step the iterations and tableaus of the simplex method calculator.
• In the last part will show the results of the problem.

We have considered for our application to solve problems with a maximum of 20 variables and 50 restrictions; this is because exercises with a greater number of variables would make it difficult to follow the steps using the simplex method. For problems with more variables, we recommend using other method.

Below we show some reference images of the step by step and the result of the following example:

The following problem is required to be maximized:

Objective Function Z = 3X 1 + 2X 2

Subject to the following restrictions

2X 1 + X 2 ≤ 18 2X 1 + 3X 2 ≤ 42 X 1 , X 2 ≥ 0

We enter the number of variables and constraints:

Enter the coefficients of the equations / inequalities of the problem and click on Solve:

Next you will see the step by step in obtaining the solution as well as the calculation of the vector of reduced costs:

The calculation of the values of the pivot row:

Until the final result:

Final reflection

Our free simplex minimizing and maximizing calculator is being used by thousands of students every month and has become one of the most popular online Simplex method calculators available. In addition, our full version has been helping hundreds of students study and do their homework faster and giving them more time to devote to their personal activities.

If you have questions about it or find an error in our application, we will appreciate if you can write to us on our contact page .

Solved Exercise of Minimization of 2 variables with the Big M Method Example of Linear Programming - Big M Method

Problem data.

We want to Minimize the following problem:

• X 1 + X 2 ≥ 2
• - X 1 + X 2 ≥ 1
• 0 X 1 + X 2 ≤ 3

X 1 , X 2 ≥ 0

To solve the problem, the iterations of the simplex method will be performed until the optimal solution is found.

Next, the problem will be adapted to the standard linear programming model, adding the slack, excess and/or artificial variables in each of the constraints and converting the inequalities into equalities:

• Constraint 1 : It has sign “ ≥ ” (greatest equal), therefore the excess variable S 1 will be subtracted and the artificial variable A 1 will be added. In the initial matrix, A 1 will be in the base.
• Constraint 2 : It has sign “ ≥ ” (greatest equal), therefore the excess variable S 2 will be subtracted and the artificial variable A 2 will be added. In the initial matrix, A 2 will be in the base.
• Constraint 3 : It has sign “ ≤ ” (least equal), therefore the slack variable S 3 . This variable will be placed at the base in the initial matrix.

Now we will show the problem in the standard form. We will place the coefficient 0 (zero) where appropriate to create our initial matrix:

Objective Function:

Minimize Z = X 1 - 2 X 2 + 0 S 1 + 0 S 2 + 0 S 3 + M A 1 + M A 2

Subject to:

• X 1 + X 2 - S 1 + 0 S 2 + 0 S 3 + A 1 + 0 A 2 = 2
• - X 1 + X 2 + 0 S 1 - S 2 + 0 S 3 + 0 A 1 + A 2 = 1
• 0 X 1 + X 2 + 0 S 1 + 0 S 2 + S 3 + 0 A 1 + 0 A 2 = 3

Initial Matrix

We will use the coefficients of the equations to elaborate our first table:

a ) Reduced Cost Vector Calculation (Z):

The values recorded in row Z were obtained as follows:

• Z 1 = (C b, 1 × X 1 , 1 ) + (C b, 2 × X 1 , 2 ) + (C b, 3 × X 1 , 3 ) - C j, 1 = ( M × 1 ) + ( M × -1 ) + ( 0 × 0 ) - ( 1 ) = -1
• Z 2 = (C b, 1 × X 2 , 1 ) + (C b, 2 × X 2 , 2 ) + (C b, 3 × X 2 , 3 ) - C j, 2 = ( M × 1 ) + ( M × 1 ) + ( 0 × 1 ) - ( -2 ) = 2M + 2
• Z 3 = (C b, 1 × S 1 , 1 ) + (C b, 2 × S 1 , 2 ) + (C b, 3 × S 1 , 3 ) - C j, 3 = ( M × -1 ) + ( M × 0 ) + ( 0 × 0 ) - ( 0 ) = -M
• Z 4 = (C b, 1 × S 2 , 1 ) + (C b, 2 × S 2 , 2 ) + (C b, 3 × S 2 , 3 ) - C j, 4 = ( M × 0 ) + ( M × -1 ) + ( 0 × 0 ) - ( 0 ) = -M
• Z 5 = (C b, 1 × S 3 , 1 ) + (C b, 2 × S 3 , 2 ) + (C b, 3 × S 3 , 3 ) - C j, 5 = ( M × 0 ) + ( M × 0 ) + ( 0 × 1 ) - ( 0 ) = 0
• Z 6 = (C b, 1 × A 1 , 1 ) + (C b, 2 × A 1 , 2 ) + (C b, 3 × A 1 , 3 ) - C j, 6 = ( M × 1 ) + ( M × 0 ) + ( 0 × 0 ) - ( M ) = 0
• Z 7 = (C b, 1 × A 2 , 1 ) + (C b, 2 × A 2 , 2 ) + (C b, 3 × A 2 , 3 ) - C j, 7 = ( M × 0 ) + ( M × 1 ) + ( 0 × 0 ) - ( M ) = 0
• Z 8 = (C b, 1 × R 1 ) + (C b, 2 × R 2 ) + (C b, 3 × R 3 ) = ( M × 2 ) + ( M × 1 ) + ( 0 × 3 ) = 3M

b ) Optimality Condition (Incoming Variable):

In the reduced cost vector (Z) we have positive values, so we must select the highest value for the pivot column ( minimization ).

In the vector Z (excluding the last value), we have the following numbers: [ -1, 2M + 2, -M, -M, 0, 0, 0 ] . The highest value is = 2M + 2 which corresponds to the X 2 variable. This variable will enter the base and its values in the table will form our pivot column.

Note: To check the chosen value, simply replace the variable M by a large number (e.g. M = 1000000) in the vector of reduced costs.

d ) Feasibility Condition (Outgoing Variable):

The feasibility condition will be verified by dividing the values of column R by the pivot column X 2 . To process the division, the denominator must be strictly positive (If it's negative or zero, it'll display N/A = Not applicable). The lowest value will define the variable that will exit from the base:

• Row A 1 → R 1 / X 2 , 1 = 2 / 1 = 2
• Row A 2 → R 2 / X 2 , 2 = 1 / 1 = 1 (The Lowest Value)
• Row S 3 → R 3 / X 2 , 3 = 3 / 1 = 3

The lowest value corresponds to the A 2 row. This variable will come from the base. The pivot element corresponds to the value that crosses the X 2 column and the A 2 row = 1 .

Enter the variable X 2 and the variable A 2 leaves the base. The pivot element is 1

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