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

solving linear programming problems using simplex method minimization

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:

solving linear programming problems using simplex method minimization

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.

solving linear programming problems using simplex method minimization

Flow Chart of Simplex Method :

solving linear programming problems using simplex method minimization

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9.3: Minimization By The Simplex Method

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  • Page ID 32490

  • Rupinder Sekhon and Roberta Bloom
  • De Anza College

Learning Objectives

In this section, you will learn to solve linear programming minimization problems using the simplex method.

  • Identify and set up a linear program in standard minimization form
  • Formulate a dual problem in standard maximization form
  • Use the simplex method to solve the dual maximization problem
  • Identify the optimal solution to the original minimization problem from the optimal simplex tableau.

In this section, we will solve the standard linear programming minimization problems using the simplex method. Once again, we remind the reader that in the standard minimization problems all constraints are of the form \(ax + by ≥ c\).

The procedure to solve these problems was developed by Dr. John Von Neuman. It involves solving an associated problem called the dual problem . To every minimization problem there corresponds a dual problem. The solution of the dual problem is used to find the solution of the original problem. The dual problem is a maximization problem, which we learned to solve in the last section. We first solve the dual problem by the simplex method.

From the final simplex tableau, we then extract the solution to the original minimization problem.

Before we go any further, however, we first learn to convert a minimization problem into its corresponding maximization problem called its dual .

Example \(\PageIndex{1}\)

Convert the following minimization problem into its dual.

\[\begin{array}{ll} \textbf { Minimize } & \mathrm{Z}=12 \mathrm{x}_{1}+16 \mathrm{x}_{2} \\ \textbf { Subject to: } & \mathrm{x}_{1}+2 \mathrm{x}_{2} \geq 40 \\ & \mathrm{x}_{1}+\mathrm{x}_2 \geq 30 \\ & \mathrm{x}_{1} \geq 0 ; \mathrm{x}_{2} \geq 0 \end{array} \nonumber \]

To achieve our goal, we first express our problem as the following matrix.

\[\begin{array}{cc|c} 1 & 2 & 40 \\ 1 & 1 & 30 \\ \hline 12 & 16 & 0 \end{array} \nonumber \]

Observe that this table looks like an initial simplex tableau without the slack variables. Next, we write a matrix whose columns are the rows of this matrix, and the rows are the columns. Such a matrix is called a transpose of the original matrix. We get:

\[\begin{array}{cc|c} 1 & 1 & 12 \\ 2 & 1 & 16 \\ \hline 40 & 30 & 0 \end{array} \nonumber \]

The following maximization problem associated with the above matrix is called its dual.

\[\begin{array}{ll} \textbf { Maximize } & \mathrm{Z}=40 \mathrm{y}_{1}+30 \mathrm{y}_{2} \\ \textbf { Subject to: } & \mathrm{y}_{1}+\mathrm{y}_{2} \leq 12 \\ & 2 \mathrm{y}_1+\mathrm{y}_2 \leq 16 \\ & \mathrm{y}_{1} \geq 0 ; \mathrm{y}_{2} \geq 0 \end{array} \nonumber \]

Note that we have chosen the variables as y's, instead of x's, to distinguish the two problems.

Example \(\PageIndex{2}\)

Solve graphically both the minimization problem and its dual maximization problem.

Our minimization problem is as follows.

\[\begin{array}{ll} \textbf { Minimize } & \mathrm{Z}=12 \mathrm{x}_1+16 \mathrm{x}_2 \\ \textbf { Subject to: } & \mathrm{x}_{1}+2 \mathrm{x}_{2} \geq 40 \\ & \mathrm{x}_{1}+\mathrm{x}_{2} \geq 30 \\ & \mathrm{x}_{1} \geq 0 ; \mathrm{x}_{2} \geq 0 \end{array} \nonumber \]

We now graph the inequalities:

imageedit_3_7200556551.png

We have plotted the graph, shaded the feasibility region, and labeled the corner points. The corner point (20, 10) gives the lowest value for the objective function and that value is 400.

Now its dual is:

\[\begin{array}{ll} \textbf { Maximize } & \mathrm{Z}=40 \mathrm{y}_1+30 \mathrm{y}_{2} \\ \textbf { Subject to: } & \mathrm{y}_{1}+\mathrm{y}_{2} \leq 12 \\ & 2 \mathrm{y}_1+\mathrm{y} 2 \leq 16 \\ & \mathrm{y}_{1} \geq 0 ; \mathrm{y}_{2} \geq 0 \end{array}\nonumber \]

We graph the inequalities:

imageedit_6_8472935815.png

Again, we have plotted the graph, shaded the feasibility region, and labeled the corner points. The corner point (4, 8) gives the highest value for the objective function, with a value of 400.

The reader may recognize that Example \(\PageIndex{2}\) above is the same as Example 3.1.1, in section 3.1. It is also the same problem as Example 4.1.1 in section 4.1, where we solved it by the simplex method.

We observe that the minimum value of the minimization problem is the same as the maximum value of the maximization problem; in Example \(\PageIndex{2}\) the minimum and maximum are both 400. This is not a coincident. We state the duality principle.

The Duality Principle

The objective function of the minimization problem reaches its minimum if and only if the objective function of its dual reaches its maximum. And when they do, they are equal.

Our next goal is to extract the solution for our minimization problem from the corresponding dual. To do this, we solve the dual by the simplex method.

Example \(\PageIndex{3}\)

Find the solution to the minimization problem in Example \(\PageIndex{1}\) by solving its dual using the simplex method. We rewrite our problem.

\[\begin{array}{ll} \textbf { Minimize } & \mathrm{Z}=12 \mathrm{x}_{1}+16 \mathrm{x}_{2} \\ \textbf { Subject to: } & \mathrm{x}_{1}+2 \mathrm{x}_{2} \geq 40 \\ & \mathrm{x}_{1}+\mathrm{x}_{2} \geq 30 \\ & \mathrm{x}_{1} \geq 0 ; \mathrm{x}_{2} \geq 0 \end{array} \nonumber \]

\[\begin{array}{ll} \textbf { Maximize } & \mathrm{Z}=40 \mathrm{y}_{1}+30 \mathrm{y}_{2} \\ \textbf { Subject to: } & \mathrm{y}_{1}+\mathrm{y}_{2} \leq 12 \\ & 2 \mathrm{y}_{1}+\mathrm{y}_{2} \leq 16 \\ & \mathrm{y}_{1} \geq 0 ; \mathrm{y}_{2} \geq 0 \end{array} \nonumber \]

Recall that we solved the above problem by the simplex method in Example 4.1.1, section 4.1. Therefore, we only show the initial and final simplex tableau.

The initial simplex tableau is

\[\begin{array}{ccccc|c} \mathrm{y}_1 & \mathrm{y}_2 & \mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{Z} & \mathrm{C} \\ 1 & 1 & 1 & 0 & 0 & 12 \\ 2 & 1 & 0 & 1 & 0 & 16 \\ \hline-40 & -30 & 0 & 0 & 1 & 0 \end{array}\nonumber \]

Observe an important change. Here our main variables are \(\mathrm{y}_1\) and \(\mathrm{y}_2\) and the slack variables are \(\mathrm{x}_1 and \mathrm{x}_2\).

The final simplex tableau reads as follows:

\[\begin{array}{ccccc|c} \mathrm{y}_1 & \mathrm{y}_2 & \mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{Z} & \\ 0 & 1 & 2 & -1 & 0 & 8 \\ 1 & 0 & -1 & 1 & 0 & 4 \\ \hline 0 & 0 & 20 & 10 & 1 & 400 \end{array} \nonumber \]

A closer look at this table reveals that the \(\mathrm{x}_1\) and \(\mathrm{x}_2\) values along with the minimum value for the minimization problem can be obtained from the last row of the final tableau. We have highlighted these values by the arrows.

\[\begin{array}{ccccc|c} \mathrm{y}_1 & \mathrm{y}_2 & \mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{Z} & \\ 0 & 1 & 2 & -1 & 0 & 8 \\ 1 & 0 & -1 & 1 & 0 & 4 \\ \hline 0 & 0 & 20 & 10 & 1 & 400 \\ & & \uparrow & \uparrow & & \uparrow \end{array} \nonumber \]

We restate the solution as follows:

The minimization problem has a minimum value of 400 at the corner point (20, 10)

We now summarize our discussion.

Minimization by the Simplex Method

  • Set up the problem.
  • Write a matrix whose rows represent each constraint with the objective function as its bottom row.
  • Write the transpose of this matrix by interchanging the rows and columns.
  • Now write the dual problem associated with the transpose.
  • Solve the dual problem by the simplex method learned in section 4.1.
  • The optimal solution is found in the bottom row of the final matrix in the columns corresponding to the slack variables, and the minimum value of the objective function is the same as the maximum value of the dual.

Two Phase Method: Linear Programming

In Two Phase Method , the whole procedure of solving a linear programming problem (LPP) involving artificial variables is divided into two phases.

In phase I, we form a new objective function by assigning zero to every original variable (including slack and surplus variables) and -1 to each of the artificial variables. Then we try to eliminate the artificial varibles from the basis. The solution at the end of phase I serves as a basic feasible solution for phase II. In phase II, the original objective function is introduced and the usual simplex algorithm is used to find an optimal solution. The following are examples of Two Phase Method .

Example 1, Two Phase Method: Example 2

Two Phase Method: Minimization Example 1

Minimize z = -3x 1 + x 2 - 2x 3

x 1 + 3x 2 + x 3 ≤ 5 2x 1 – x 2 + x 3 ≥ 2 4x 1 + 3x 2 - 2x 3 = 5

x 1 , x 2 , x 3 ≥ 0

If the objective function is in minimization form, then convert it into maximization form.

Changing the sense of the optimization

Any linear minimization problem can be viewed as an equivalent linear maximization problem, and vice versa. Specifically:

If z is the optimal value of the left-hand expression, then -z is the optimal value of the right-hand expression.

Maximize z = 3x 1 – x 2 + 2x 3

Converting inequalities to equalities

x 1 + 3x 2 + x 3 + x 4 = 5 2x 1 – x 2 + x 3 – x 5 = 2 4x 1 + 3x 2 - 2x 3 = 5

x 1 , x 2 , x 3 , x 4 , x 5 ≥ 0

Where: x 4 is a slack variable x 5 is a surplus variable

The surplus variable x 5 represents the extra units.

Now, if we let x 1 , x 2 and x 3 equal to zero in the initial solution, we will have x 4 = 5 and x 5 = -2, which is not possible because a surplus variable cannot be negative. Therefore, we need artificial variables .

x 1 + 3x 2 + x 3 + x 4 = 5 2x 1 – x 2 + x 3 – x 5 + A 1 = 2 4x 1 + 3x 2 - 2x 3 + A 2 = 5

x 1 , x 2 , x 3 , x 4 , x 5 , A 1 , A 2 ≥ 0

Where A 1 and A 2 are artificial variables.

Phase 1 of Two Phase Method

In this phase, we remove the artificial variables and find an initial feasible solution of the original problem. Now the objective function can be expressed as

Maximize 0x 1 + 0x 2 + 0x 3 + 0x 4 + 0x 5 + (–A 1 ) + (–A 2 )

Initial basic feasible solution

The intial basic feasible solution is obtained by setting x 1 = x 2 = x 3 = x 5 = 0

Then we shall have A 1 = 2 , A 2 = 5, x 4 = 5

Two Phase Method: Table 1

On small screens, scroll horizontally to view full calculation

Key column = x 1 column Minimum (5/1, 2/2, 5/4) = 1 Key row = A 1 row Pivot element = 2 A1 departs and x 1 enters.

A2 departs and x 2 enters. Here, Phase 1 terminates because both the artificial variables have been removed from the basis.

Phase 2 of Two Phase Method

The basic feasible solution at the end of Phase 1 computation is used as the initial basic feasible solution of the problem. The original objective function is introduced in Phase 2 computation and the usual simplex procedure is used to solve the problem.

Use horizontal scrollbar to view full calculation

"The most useful virtue is patience" - John Dewey

Two Phase Method: Final Optimal Table

An optimal policy is x 1 = 5/2, x 2 = 0, x 3 = 5/2. The associated optimal value of the objective function is z = 3 X (5/2) – 0 + 2 X (5/2) = 25/2.

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solving linear programming problems using simplex method minimization

  • max Z = 3x1 + 5x2 + 4x3 subject to 2x1 + 3x2 2x2 + 5x3 3x1 + 2x2 + 4x3 and x1,x2,x3 >= 0
  • max Z = 5x1 + 10x2 + 8x3 subject to 3x1 + 5x2 + 2x3 4x1 + 4x2 + 4x3 2x1 + 4x2 + 5x3 and x1,x2,x3 >= 0
  • max Z = 4x1 + 3x2 subject to 2x1 + x2 x1 + x2 x1 x2 and x1,x2 >= 0
  • =,>=`4,7`');">min Z = x1 + x2 subject to 2x1 + 4x2 >= 4 x1 + 7x2 >= 7 and x1,x2 >= 0
  • =,>=`80,60`');">min Z = 600x1 + 500x2 subject to 2x1 + x2 >= 80 x1 + 2x2 >= 60 and x1,x2 >= 0
  • =`12,10,10`');">min Z = 5x1 + 3x2 subject to 2x1 + 4x2 2x1 + 2x2 = 10 5x1 + 2x2 >= 10 and x1,x2 >= 0
  • max Z = x1 + 2x2 + 3x3 - x4 subject to x1 + 2x2 + 3x3 = 15 2x1 + x2 + 5x3 = 20 x1 + 2x2 + x3 + x4 = 10 and x1,x2,x3,x4 >= 0
  • max Z = 3x1 + 9x2 subject to x1 + 4x2 x1 + 2x2 and x1,x2 >= 0
  • max Z = 3x1 + 2x2 + x3 subject to 2x1 + 5x2 + x3 = 12 3x1 + 4x2 = 11 and x2,x3 >= 0 and x1 unrestricted in sign
  • max Z = 3x1 + 3x2 + 2x3 + x4 subject to 2x1 + 2x2 + 5x3 + x4 = 12 3x1 + 3x2 + 4x3 = 11 and x1,x2,x3,x4 >= 0
  • =`30,24,3`');">max Z = 6x1 + 4x2 subject to 2x1 + 3x2 3x1 + 2x2 x1 + x2 >= 3 and x1,x2 >= 0
  • =`6,10,1`');">max Z = 3x1 + 5x2 subject to x1 - 2x2 x1 x2 >= 1 and x1,x2 >= 0
  • =`5,8`');">max Z = 6x1 + 4x2 subject to x1 + x2 x2 >= 8 and x1,x2 >= 0
  • =,>=`-5,8`');">max Z = 6x1 + 4x2 subject to -x1 - x2 >= -5 x2 >= 8 and x1,x2 >= 0

solving linear programming problems using simplex method minimization

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4.2.1: Maximization By The Simplex Method (Exercises)

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  • Page ID 37871

  • Rupinder Sekhon and Roberta Bloom
  • De Anza College

SECTION 4.2 PROBLEM SET: MAXIMIZATION BY THE SIMPLEX METHOD

Solve the following linear programming problems using the simplex method.

1) \[\begin{array}{ll} \text { Maximize } & \mathrm{z}=\mathrm{x}_{1}+2 \mathrm{x}_{2}+3 \mathrm{x}_{3} \\ \text { subject to } & \mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_3 \leq 12 \\ & 2 \mathrm{x}_{1}+\mathrm{x}_{2}+3 \mathrm{x}_{3} \leq 18 \\ & \mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3} \geq 0 \end{array} \nonumber \]

2) \[\begin{array}{ll} \text { Maximize } \quad z= & x_{1}+2 x_{2}+x_{3} \\ \text { subject to } & x_{1}+x_{2} \leq 3 \\ & x_{2}+x_{3} \leq 4 \\ & x_{1}+x_{3} \leq 5 \\ & x_{1}, x_{2}, x_{3} \geq 0 \end{array} \nonumber \]

3) A farmer has 100 acres of land on which she plans to grow wheat and corn. Each acre of wheat requires 4 hours of labor and $20 of capital, and each acre of corn requires 16 hours of labor and $40 of capital. The farmer has at most 800 hours of labor and $2400 of capital available. If the profit from an acre of wheat is $80 and from an acre of corn is $100, how many acres of each crop should she plant to maximize her profit?

4) A factory manufactures chairs, tables and bookcases each requiring the use of three operations: Cutting, Assembly, and Finishing. The first operation can be used at most 600 hours; the second at most 500 hours; and the third at most 300 hours. A chair requires 1 hour of cutting, 1 hour of assembly, and 1 hour of finishing; a table needs 1 hour of cutting, 2 hours of assembly, and 1 hour of finishing; and a bookcase requires 3 hours of cutting, 1 hour of assembly, and 1 hour of finishing. If the profit is $20 per unit for a chair, $30 for a table, and $25 for a bookcase, how many units of each should be manufactured to maximize profit?

5). The Acme Apple company sells its Pippin, Macintosh, and Fuji apples in mixes. Box I contains 4 apples of each kind; Box II contains 6 Pippin, 3 Macintosh, and 3 Fuji; and Box III contains no Pippin, 8 Macintosh and 4 Fuji apples. At the end of the season, the company has altogether 2800 Pippin, 2200 Macintosh, and 2300 Fuji apples left. Determine the maximum number of boxes that the company can make.

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