Simplex Method in OR

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Simplex Method

It is a method to translate the geometric definitions of

extreme points (as in graphical Method) into an algebraicdefinitions.

It requires that each of the constraints be put in a specialstandard form, which normally results in a set of

simultaneous equations in which the number of variablesexceeds the number of equations.

The extreme points of graphical solution space can beidentified algebraically by the Basic Solution of the systemof equations.

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Standard LP Form:

LP model include constraints of all types (≤, ≥, =).To develop a general solution method the LP must be put in

common format.

The properties of the standard form are as follows:

1. All the constraints are equations.

2. All variable are non negative

3. The objective function may be maximization or minimization



There are two types of the variables.

1. Slack Variable

It represents unused amount of resource.For Constraints with the (≤), the right hand side represents thelimit on the availability of resource, whereas the left hand siderepresents the usage of this limited resource by differentactivities.

2. Surplus Variable

For Constraints with the (≥), normally set the minimum specification



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Constraints

A constraint of the type (≤, ≥) can be converted to an equation by adding a slack variable to (subtracting surplus variable from) the

left side of the constraint, e.g.

x1 + 2x2 ≤ 6 add a slack s1 ≥ 0 to the left side of the constraint

x1 + 2x2 + s1 = 6

3x1 + 2x2 – 3x3 ≥ 5, subtract a surplus s2 ≥ 0 from the left side ofthe constraint

3x1 + 2x2 – 3x3 – s2 = 5

The right side of an equation can always be made nonnegative bymultiplying both sides by -1.

2 x1 +3x2 -7x3 = -5; is mathematically equal to ,

-2 x1 -3x2 +7x3 =5



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Basic Solutions

Consider the following system of equations:

2x1 + x2 + 4x3 + x4 =2 & x1 + 2x2 + 2x3 + x4 =3In which we have; m=2 and n=4

A basic solution will be associated withn - m = 4-2 =2 zero variables.

The given set of equations can have n! /m! (n-m)! =4!/2! (4-2)! = 6 possible basic solutions.

The n-m variables, that are set equal to zero, are non basic

variables. The remaining m variables are basic variables.

If all solution values are nonnegative, the basic solution is said tobe feasible.

If any of the value is negative, the basic solution is said to beinfeasible



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Example 1

Maximize z = 3x1 + 2x2

subject to-x1 + 2x2 4

3x1 + 2x2 14x1 – x2 3

x1, x2 0

Solution:

Convert every inequality constraint in to an equality constraint,so that the problem may be written in the standard form, by

adding the slack variable to each constraint.



-x1 + 2x2 + s1 = 4

3x1 + 2x2 + s2 = 14

x1 – x2 + s3 = 3

x1, x2, s1, s2, s3 0, Where s1, s2 and s3 are slack variables.

Since slack variables represent unused resources, their contribution in the

objective function is zero.Including these slack variables in the objective function, we get

Maximize z = 3x1 + 2x2 + 0 s1 + 0 s2 + 0 s3



Initial basic feasible solution

Now we assume that nothing can be produced. Therefore, the valuesof the decision variables are zero.x1 = 0, x2 = 0, z = 0

When we are not producing anything, obviously we are left withunused capacitys1 = 4, s2 = 14, s3 = 3

We note that the current solution has three variables (slack

variables s1, s2 and s3) with non-zero solution values and twovariables (decision variables x1 and x2) with zero values.

Variables with non-zero values are called basic variables.Variables with zero values are called non-basic variables.



Trial and Error

n=5, m=3, therefore basic variables =3 and non basic variables

=2.

Possible basic solutions = n! /m! (n-m)! = 5!/3! (5-3)!= 10 possible basic solutions.

-x1 + 2x2 + s1 = 4

3x1 + 2x2 + s2 = 14

x1 – x2 + s3 = 3

S. No. Non basic

variable

Basic Variable Z value

1 x1 , x2 s1 =4, s2 =14, s3 =3 0

2 x1 , s1 x2 =2, s2 =10, s3 =5 43 x1 , s2 x2 =7, s1 =-10, s3 =10 I.S

z = 3x1 + 2x2



S. No. Non basic

variable


4 x1 , s

3 x

2 =-3, s

1 =10, s

2 =20 I.S

5 x2 , s1 x1 =-4, s2 =2, s3 =7 I.S

6 x2 , s2 x1 =14/3, s1 =26/3, s3 =-5/3 I.S

z = 3x1 + 2x2

S. No. Non basic

variable


7 x2 , s3 x1 =3, s1 =7, s2 =5 9

8 s1, s2 x1 =-3.5, x2 =0.25, s3 =6.75 I.S

9 s1, s3 x1 =10, x2 =7, s2 =-30 I.S

10 s2 , s3 x1 =4, x2 =1, s1 =6 14

The largest profit of Rs. 14 is obtained, when 1 unit of x2 and 4

units of x1 are produced. It also indicates that 6 units are stillunutilized, as shown by slack variable s1 =6

Simplex Method in OR

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