MAXIMIZATIONBY

DR. ARSHAD ZAHEER

Simplex Method

Steps of Simplex Method

1. Express the problem in equation form2. Write the inequalities in form of equalities3. Determine an initial feasible solution4. Write initial tableau / iteration5. Determine the criteria for optimality6. Select an entering variable using the

optimality condition7. Select a leaving variable using the

feasibility condition8. Write the optimal solution

Steps of Simplex Method (Details)

There are following steps related to solving a Maximization problem by simplex method

1. Express the problem in equation form In the case the given problem is in the form

of statements we need to translate it in equations

2. Write the inequalities in form of equalities

In this step we encounter normally three kinds of problem

Type Less than equal to form (≤)

Example 4X1 + 3X2 < 240

Solution Introduce slack variable to solve this kind of problemSlack variables are always added

New Equation

4X1 + 3X2 + S1= 240

Steps of Simplex Method (Details)

Type Equal to form (=)

Example 4X1 + 3X2 = 240

Solution Introduce artificial variable to solve this kind of problemArtificial variables are always added

New Equation

4X1 + 3X2 + A1= 240

Steps of Simplex Method (Details)

Type Greater than Equal to form (≥)

Example 4X1 + 3X2 ≥ 240

Solution Introduce surplus variable and these are always subtractedIntroduce artificial variable to solve this kind of problem and these are always added

New Equation

4X1 + 3X2 - S1+ A1= 240

Steps of Simplex Method (Details)

3. Determine an initial feasible solutionTo determine an initial feasible solution see how

much equations and variables are there. This will help in deciding how much variables would be given arbitrary values

Arbitrary values = # of Variables -- # of EquationsSuppose we have 6 variables and 4 equations;Arbitrary values = 6 – 4

= 2So the two variables will be given arbitrary values

equal to zeroAll the variables and the RHSs of equations should be

non negative

Steps of Simplex Method (Details)

4. Write initial tableau / IterationThe values from the initial solution will be put

into the initial tableau;

Basic RHS

f

All the non zero variables from initial solution

Non basic Variables having zero values called non basic variable

The RHS of every equation will came here

The coefficient of constraints will be written hereThe coefficient of objective function

Steps of Simplex Method (Details)

5. Determine the criteria for optimality

Solution of current tableau is said to be optimal if all the objective function co-efficient becomes Non Negative / Positive

Steps of Simplex Method (Details)

6. Select an entering variable using the optimality condition

The entering variable in a maximization probe is non basic variable with most negative co efficient in the f- row. The optimum is reached at the iteration where all the f- row coefficients are non negative

Steps of Simplex Method (Details)

7. Select a leaving variable using the feasibility condition

In case of maximization the leaving variable is the basic variable associated with the smallest non negative ratio with the strictly positive denominator. There is no need to take ratios for negative and zero because they can not be entered into basic variable column.

Steps of Simplex Method (Details)

8. Write the optimal solutionWhen the solution criteria is satisfied, the

values of the decision variables will be taken from tableau from corresponding RHS.

Put these values in the original maximization function to calculate optimal point.

Steps of Simplex Method (Details)

Illustration No 1

Illustration

Maximize: f=5x1+ 4x2 (Objective Function)

Subject to: (Constraints)

6x1+ 4x2 ≤ 24 (1)

x1+ 2x2 ≤ 6 (2)

-x1+ x2 ≤ 1 (3)

x2 ≤ 2 (4)

(Non-Negative Constraints)

x1, x2 ≥ 0

Rewriting Constraints equations in homogeneous order

Step 1.Rewriting Constraints equations in homogeneous order

6x1+ 4x2 ≤ 24 (1)

x1+ 2x2 ≤ 6 (2)

-x1+ x2 ≤ 1 (3) 0x1+ x2 ≤ 2 (4)

x1, x2 ≥ 0

Inequalities Constraints in Equation Form

Write inequalities constraints in equation form by adding slack variable “S”.

e.g. 6x1+ 4x2+ S = 24 (S is slack variable)Let S1, S2, S3 and S4 be the slack variables for first, second, third and fourth constraints respectively.

6x1+ 4x2+ S1 = 24 ………….(5)

x1+ 2x2+ S2 = 6 ………….(6)

-x1+ x2 + S3 = 1 ………….(7)

0x1+ x2+ S4 = 2 ………….(8)

x1, x2,S1, S2,S3, S4 ≥ 0 f-5X1 -4X2 =0

Initial feasible solution

How many variable will be given Arbitrary values =

# of Variables -- # of Equations

6 - 4 = 2

Let x1= 0, x2 = 0Putting above values in objective function (f=5x1+ 4x2) and equation 5-8,

f = 0S1 = 24

S2 = 6S3 = 1S4 = 2x1, x2,S1, S2,S3, S4 ≥ 0

Basic x1 x2 S1 S2 S3 S4 RHS RatioS1 6 4 1 0 0 0 24 24/6 = 4 (min)S2 1 2 0 1 0 0 6 6/1=6

S3 -1 1 0 0 1 0 1

Pivot Column coefficient is negative so ratio is not possible

S4 0 1 0 0 0 1 2

Pivot Column coefficient is zero so ratio is not possible

f -5 -4 0 0 0 0 0  

Pivot Column Pivot Row

Pivot NoInitial Tableau / Iteration

Some cross checking measures

Basic x1 x2 S1 S2 S3 S4 RHSS1 6 4 1 0 0 0 24S2 1 2 0 1 0 0 6S3 -1 1 0 0 1 0 1S4 0 1 0 0 0 1 2f -5 -4 0 0 0 0 0

The basic variables are equal to 1 under their respective columns and all the other values in columns will be zero, so if you don’t have zero or one on these position it means you have done some thing wrong.

Calculation for next Iteration

New Pivot Row = 1

X Old Pivot RowPivot No.

New Pivot Row =

1X [6 4 1 0 0 0 24]

6= [1 2/3 1/6 0 0 0 4]

New Row = Old Row – Pivot Column Coefficient x New Pivot RowNew S2 Row = [1 2 0 1 0 0

6]- (1)[1 2/3 1/6 0 0 0

4]= [0 4/3 -1/6 1 0 0

2]

Calculation

New S3 Row =[-1 1 0 0 1 0 1]-(-1) [1 2/3 1/6 0 0 0 4]= [0 5/3 1/6 0 1 0 5]

New S4 Row = [0 1 0 0 0 1 2] - (0) [1 2/3 1/6 0 0 0 4]

= [0 1 0 0 0 1 2]New f Row = [-5 -4 0 0 0 0 0]

-(-5) [1 2/3 1/6 0 0 0 4]= [0 -2/3 5/6 0 0 0 20]

Calculation

Basic x1 x2 S1 S2 S3 S4 RHS Ratio

x1 1 2/3 1/6 0 0 0 4 4/(2/3)=6

S2 0 4/3 -1/6 1 0 0 2 2/(4/3) = 1.5 (min)

S3 0 5/3 1/6 0 1 0 5 5/(5/3) = 3

S4 0 1 0 0 0 1 2 2/1=2

f 0 -2/3 5/6 0 0 0 20

Still the solution is not optimal as there is -2/3 as negative coefficient of objective function so we again repeat the whole process

Calculation

New Pivot Row= 3/4[04/3 -1/6 1 0 0 2]

[0 1 -1/8 3/4 0 0

3/2]

New Pivot Row = 1

X Old Pivot RowPivot No.

New Pivot

Row =

1

X [0 4/3 -1/6 1 0 0 2]

4/3

Calculation

New Row = Old Row – Pivot Column Coefficient x New Pivot Row

New x1 Row = [1 2/3 1/6 0 0 0 4]

- (2/3)[0 1 -1/8 3/4 0 0 3/2]= [1 0 1/4 -1/2 0 0 3]

New S3 Row =[0 5/3 1/6 0 1 0 5] - 5/3[01 -1/8 3/4 0 0 3/2]

= [0 0 3/8 -5/4 1 0 5/2]New S4 Row = [0 1 0 0 0 1 2]

- (1) [0 1 -1/8 3/4 0 03/2]

= [0 0 1/8 -3/4 0 1 1/2]New f Row = [0 -2/3 5/6 0 0 0 20]

- (-2/3) [0 1 -1/8 3/4 0 0 3/2] = [0 0 3/4 1/2 0 0

21]

Criteria for optimality

Solution of current tableau is said to be optimal if all the objective function co-efficient becomes Non Negative / Positive

Criteria for optimality satisfied

Basics

x1 x2 S1 S2 S3 S4 RHS

x1 1 0 1/4 -1/2 0 0 3x2 0 1 -1/8 3/4 0 0 3/2S3 0 0 3/8 -5/4 1 0 5/2S4 0 0 1/8 -3/4 0 1 ½f 0 0 3/4 1/2 0 0 21

Criteria for Optimum Solution because all coefficients (of objective function) are non-negative or zero

Optimal Solution

X1 = 3X2 = 3/2f = 21

Cross checking of maximization pointput values of X1 and X2 from above solution

into original objective functionf=5x1+ 4x2

=5 (3) + 4(3/2) =21

Stay Blessed