*Four Special cases in Simplexhttp://studygalaxy.com/

Simplex Algorithm Special casesThere are four special cases arise in the use of the simplex method.

DegeneracyAlternative optimaUnbounded solutionNonexisting ( infeasible ) solution *

Simplex Algorithm Special cases (cont.)Degeneracy ( no improve in objective)

It typically occurs in a simplex iteration when in the minimum ratio test more than one basic variable determine 0, hence two or more variables go to 0, whereas only one of them will be leaving the basis. Model has at least one Redundant Constraint.This is in itself not a problem, but making simplex iterations from a degenerate solution may give rise to cycling, meaning that after a certain number of iterations without improvement in objective value the method may turn back to the point where it started. *

Degenearcy Special cases (cont.)Example:Max 3x1 + 9x2STx1 + 4x2 8X1 + 2x2 4X1, x2 0

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Simplex Algorithm Special cases (cont.)The solution:The constraints:X1 + 4X2 + s1= 8X1 + 2X2 + s2= 4X1, X2 ,s1,s2 0

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Simplex Algorithm Special cases (cont.)Iteration 0*Here X1 and X2 tie for leaving variable( with same minimum ratio 2)

BasisX1X2S1S2RHSs114108s212014Z-3-9000

Simplex Algorithm Special cases (cont.)Iteration 1*Here basic variable S2 is 0 resulting in a degenerate basic solution.

BasisX1X2S1S2RHSX21/411/402s20-1/210Z-3/402/4018

Simplex Algorithm Special cases (cont.)

Iteration 2

Same objective no change and improve ( cycle)Is it possible to stop computation at Iteration1?*

BasisX1X2S1S2RHSX201-1/22X110-120Z003/23/218

It is possible to have no improve and no termination for computation.

Maximize Z= 0.75x1-20x2+0.5x3-6x4

Such that0.25x1-8x2-x3+9x4 0 0.5x1-12x2-0.5x3+3x4 0 x3 1x1,x2,x3, x4 0

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Temporarily DegenerateMaximize Z= 3X1 +2X2

Such that4X1-X2 84X1+3X2 124X1+X28X1,X2 0*

Maximize Z= 3X1 +2X2

Such that

4X1-X2+S1= 8

4X1+3X2+S2 =12

4X1+X2+S3=8

X1,X2 ,S1,S2,S30

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Alternative optima

If the z-row value for one or more nonbasic variables is 0 in the optimal tubule, alternate optimal solution exists.When the objective function is parallel to a binding constraint objective function will assume same optimal value.We have infinite number of such points*

Example:

Max 2x1+ 4x2

ST

x1 + 2x2 5

x1 + x2 4

x1, x2 0*

The solution

Max 2x1+ 4x2

ST

x1 + 2x2 + s1= 5

x1 + x2 + s2 = 4

x1, x2, s1, s2 0

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Iteration0

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BasisX1X2S1S2RHSs112104s211015Z-2-4000

Simplex Algorithm Special cases (cont.)Optimal solution is 10 when x2=5/2, x1=0.

How do we know from this tubule that alternative optima exist ?*

BasisX1X2S1S2RHSx21/211/205/2s21/20-1/213/2Z002010

Simplex Algorithm Special cases (cont.)By looking at z-row coefficient of the nonbasic variable.

The coefficient for x1 is 0, which indicates that x1 can enter the basic solution without changing the value of z.*

BasisX1X2S1S2RHSx21/211/205/2s21/20-1/213/2Z002010

Simplex Algorithm Special cases (cont.)The second alternative optima is:

The new optimal solution is 10 when x1=3, x2=1

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BasisX1X2S1S2RHSx2011-11x110-123Z002010

Any point on BC represents an alternate optimum with z=10*

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In practice alternate optima are useful as they allow us to choose from many solutions experiencing deterioration in the objective value.

In a product-mix problem C(3,1) would be more appealing.*

*3.Unbounded solution:

It occurs when nonbasic variables are zero or negative in all constraints coefficient (max) and variable coefficient in objective is negative

ExampleMax 2x1+ x2

ST

x1 x2 10

2x1 40

x1, x20

SolutionMax 2x1+ x2

ST

x1 x2 +s1 =10

2x1+s2 =40

x1, x2,s1,s2 0

Simplex Algorithm Special cases (cont.)

All value if x2( nonbasic variable) either zero or negative.

So, solution space is unbounded*

BasisX1X2S1S2RHSx21-11010x1200140Z-2-1000

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We can conclude that after 3rd iteration if we increase the value of x2 ,value of z will increase correspondingly up to infinite. Hence we can conclude that solution is unbounded.

Example:Maximize Z=20 x1+10 x2S.T.3x1-3x2+5x3 50x1+x3 10x1-x2+4x3 20x1,x2,x3 0

Solution:Maximize Z=20x1+10x2

3x1-3x2+5x3+s1 = 50

x1+x3+s2 = 10

x1-x2+4x3+s3 = 20

x1,x2,x3,s1,s2,s3 0

Basisx1x2x3s1s2s3RHSS13-3510050S210101010S31-1400120Z-20-10-10000

Infeasible Solution:This situation can never occur if all the constraints are of the type with non-negative RHS because slack provide feasible solution.High penalty is provided with R in objective function to reduce them to 0 at optimum.Artificial variable R coefficient at end 0 if solution is infeasible.

Example:Maximize Z=3x1+2x2

S.T.

2x1+x2 2

3x1+4x2 12

x1,x2=0

Solution (Graphical):Pseudo optimal solution

Solution (Algebric) : M=1000Maximize Z=3x1+2x2+0S1+0S2+MR

S.T.

2x1+x2+S1 = 2

3x1+4x2-S2+R = 12

x1,x2,S1,S2,R 0

Solution (Algebric) : M=1000

Pseudo optimumR 0SOLUTION INFEASIBLE

REFERENCESOperations Research an introduction, seventh edition ,Hamdy A. TahaTORA

THANK YOU

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