simplex meathod

PRESENTATION ON SPECIAL CASES OF SIMPLEX METHOD

SPECIAL CASES IN APPLYING SIMPLEX METHOD

Several complications can occur while solving the LPP. Suchproblems are: Tie for the key row(degeneracy) Tie for the key column Unbounded problems Multiple optimal solutions Infeasible problems Redundant constraints Unrestricted Variables

TIE FOR THE KEY ROW

Degeneracy occurs when there is tie for the minimum ratio(MR) for choosing the departing variable

Basic variables

Solution variables

X1 X2 S1 S2 Min. ratio

S1 20 4 9 1 0 5

S2 10 2 7 0 1 5

Zj 0 0 0 0 0

Cj-Zj 5 3 0 0 Tie of key row

Key column

Find the coefficient of the slack variables and divide each coefficient by the corresponding positive numbers of the key column in the row, starting from left to right in order to break the tie

S1 S2

1/4= 0.25 0/4= 0

0/2= 0 1/2= 0.5

Lowest element and S1 row is a key row

X1 is replaced by S1

If the ratio do not break the tie, find the similar ratio for the coefficient of decision variables

Compare the resulting ratio, column by column Select the row which contains smallest ratio.This row

becomes the key row After resolving of this tie, simplex method is applied to obtain

the optimum solution

TIE FOR THE KEY COLUMN

This problem can arise in case of tie between identical Cj-Zj values

In such a situation selection for key column can be made arbitrarily

There is no wrong choice, although selection of one variable may result in more iteration

Regardless of which variable column is chosen the optimal solution will eventually be found

Basic variable

Solution values

X1 X2 S1 S2 Min. ratio

S1 2 3 2 1 0

S2 10 4 6 0 1

Zj 0 0 0 0 0

Cj-Zj 4 4 0 0

Tie of key column

Any one of the decision variable is selected

UNBOUNDED PROBLEM

It can be stated that a key row cannot be selected because minimum ratio column contains negative or infinity(∞) the solution is unbounded

Basic variable

Solution variable

X1 X2 S1 S2 Min. ratio

X1 7 1 0 1 0 ∞

S2 1 0 -1 -1 1 -1

Zj 35 5 0 5 0

Cj-Zj 0 4 -5 0

Unbounded solution

Key column

MULTIPLE OPTIMAL SOLUTION

If the index row indicates the value of Cj-Zj for a non basic variable to be zero, there exists an alternative optimum solution.

To find the alternative optimal solution, the non basic variable with the Cj-Zj value of zero, should be selected as an entering variable and the simplex steps continued.

Basic variable

Solution variable

X1 X2 X3 S1 S2 Min. ratio

X1 10 4 5 6 1 0

X3 12 9 8 2 0 1

Zj 56 35 34 18 2 3

Cj-Zj -33 -31 2 -2 -3X2 is not there i.e. multiple optimal solution

INFEASIBLE PROBLEMS

This condition occurs when the problem has incompatible constraints

In final simplex table, all Cj-Zj elements +ve or zero in case of minimization and –ve or zero in case of maximization

And if the basic variable include artificial variable, then LPP got an infeasible solution

Basic var. Sol. Value X1 X2 S1 S2 S3 A1X2 10 0 1 3 0 1 0

A1 20 0 0 -4 -1 -1 1

X1 40 1 0 -2 0 -1 0

Zj 190M-20M 4 3 1+4M M M-1 -M

Cj-Zj 0 0 -1-4M -M 1-M 0

Since Cj-Zj row contains all elements –ve or zero , we are having optimum solution. Since artificial variable is present as a basic variable the given problem has infeasible solution.

REDUNDANT CONSTRAINT

Consider the constraints, 3X1 + 4X2 < 7 3X1 + 4X2 < 15The second constraint is less restrictive(because both theconstraints have same co-efficient and variable) than first one,and is not required. Normally redundant constraint does not poseany problem except the computational work is unnecessarilyincreased

neglected

UNRESTRICTED VARIABLES

It is that decision variable which does not carry any value.

To solve this problem, the variable can take two values i.e. one +ve & one –ve because difference between these two same +ve and –ve value is zero

All variables become non-negative in the system and problem is solved.

Example:Max Z = 8x1 – 4x24x1 + 5x2 ≤ 20-x1 + 3x2 ≥ -23

when x1 ≥ 0 , x2 is unrestricted in signSolution :

Firstly replace the unrestricted variable x2x2 = x3 – x4

After replacing the x2 , Max Z = 8x1 – 4x3 + 4x44x1 + 5x3 – 5x4 ≤ 20x1 – 3x3 + 3x4 ≤ 23; x1,x3,x4 ≥ 0

After that slack variables are added to the constraints ,Max Z = 8x1 – 4x3 + 4x4 + 0S1 + 0S24x1 + 5x3 – 5x4 + S1 + 0S2 = 20x1 – 3x3 + 3x4 + 0S1 + S2 = 23;x1,x3,x4,S1,S2 ≥ 0

Basic variable

Solution variable

X1 X3 X4 S1 S2 Min. ratio

S1 20 4 5 -5 1 0 5

S2 23 1 -3 3 0 1 23

Zj 0 0 0 0 0 0

Cj-Zj 8 -4 4 0 0

Cj 8 -4 4 0 0

0

0

Key column

Key row

Basic variable

Solution variable

X1 X3 X4 S1 S2 Min. ratio

X1 5 1 5/4 -5/4 1/4 0 -ve

S2 18 0 -17/4 17/4* -1/4 1 72/17

Zj 40 8 10 -10 2 0

Cj-Zj 0 -14 14 -2 0

Cj 8 -4 4 0 0

8

0

Basic variable

Solution variable

X1 X3 X4 S1 S2

X1 175/17 1 0 0 3/17 5/17

X4 72/17 0 -1 1 -1/17 4/17

Zj 1688/17 8 -4 4 20/17 56/17

Cj-Zj 0 0 0 -20/17 -56/17

Since Cj-Zj ≤ 0 . The optimal solution is obtained.

Ans. x1 = 175/17x2 = x3 – x4 = 0 – 72/17 = -72/17 Z = 1688/17

THANKYOU

Presented by:-Astha Harpreet SinghShivaniShweta