Four Special Cases In Simplex 3g261i

Four Special cases in Simplex

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Simplex Algorithm – Special cases 

There are four special cases arise in the use of the simplex method. 1. 2. 3. 4.

Degeneracy Alternative optima Unbounded solution Nonexisting ( infeasible ) solution

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Simplex Algorithm – Special cases (cont.) 1.

Degeneracy ( no improve in objective)

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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.

 

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Degenearcy – Special cases (cont.) Example: Max 3x1 + 9x2

ST x1 + 4x2 ≤ 8 X1 + 2x2 ≤ 4 X1, x2 ≥ 0

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

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Simplex Algorithm – Special cases (cont.) 

Iteration “0” Entering Variable

Leaving Variable

Basis

X1

X2

S1

S2

RHS

s1

1

4

1

0

8

s2

1

2

0

1

4

Z

-3

-9

0

0

0

Here X1 and X2 tie for leaving variable( with same minimum ratio 2)

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Simplex Algorithm – Special cases (cont.) 

Iteration 1 Entering Variable Leaving Variable Basis

X1

X2

S1

S2

RHS

X2

1/4

1

1/4

0

2

s2

½

0

-1/2

1

0

Z

-3/4

0

2/4

0

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Here basic variable S2 is 0 resulting in a degenerate basic solution.

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Simplex Algorithm – Special cases (cont.) 

 

Iteration 2 Basis

X1

X2

S1

S2

RHS

X2

0

1

½

-1/2

2

X1

1

0

-1

2

0

Z

0

0

3/2

3/2

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Same objective

Same objective no change and improve ( cycle) Is it possible to stop computation at Iteration1? 8

It is possible to have no improve and no termination for computation. Maximize Z= 0.75x1-20x2+0.5x3-6x4 Such that 0.25x1-8x2-x3+9x4 ≤ 0 0.5x1-12x2-0.5x3+3x4 ≤ 0 x3 ≤ 1 x1,x2,x3, x4 ≥0 10

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Temporarily Degenerate 

Maximize Z= 3X1 +2X2

Such that

4X1-X2≤ 8 4X1+3X2 ≤12 4X1+X2≤8 X1,X2 ≥0 13

Maximize Z= 3X1 +2X2 Such that 4X1-X2+S1= 8

4X1+3X2+S2 =12 4X1+X2+S3=8 X1,X2 ,S1,S2,S3≥0

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Alternative optima 

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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

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Example: Max 2x1+ 4x2

ST x1 + 2x2 ≤ 5

x1 + x 2 ≤ 4 x1, x2 ≥0

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The solution Max 2x1+ 4x2

ST x1 + 2x2 + s1= 5

x1 + x 2 + s2 = 4 x1, x2, s1, s2 ≥0

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

Iteration”0” Entering Variable Leaving Variable Basis

X1

X2

S1

S2

RHS

s1

1

2

1

0

4

s2

1

1

0

1

5

Z

-2

-4

0

0

0

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Simplex Algorithm – Special cases (cont.) 



Optimal solution is 10 when x2=5/2, x1=0.

Basis

X1

X2

S1

S2

RHS

x2

1/2

1

1/2

0

5/2

s2

1/2

0

-1/2

1

3/2

Z

0

0

2

0

10

How do we know from this tubule that alternative optima exist ? 24

Simplex Algorithm – Special cases (cont.) 

By looking at z-row coefficient of the nonbasic variable. Entering Variable



Leaving Variable

Basis

X1

X2

S1

S2

RHS

x2

1/2

1

1/2

0

5/2

s2

1/2

0

-1/2

1

3/2

Z

0

0

2

0

10

The coefficient for x1 is 0, which indicates that x1 can enter the basic solution without changing the value of z. 25

Simplex Algorithm – Special cases (cont.) 



The second alternative optima is: Basis

X1

X2

S1

S2

RHS

x2

0

1

1

-1

1

x1

1

0

-1

2

3

Z

0

0

2

0

10

The new optimal solution is 10 when x1=3, x2=1 26

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.

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3.Unbounded solution:

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

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Example Max 2x1+ x2 ST x1 – x2 ≤10 2x1 ≤ 40 x1, x2≥0

Solution Max 2x1+ x2 ST x1 – x2 +s1 =10

2x1+s2 =40 x1, x2,s1,s2 ≥0

Simplex Algorithm – Special cases (cont.) Basis

X1

X2

S1

S2

RHS

x2

1

-1

1

0

10

x1

2

0

0

1

40

Z

-2

-1

0

0

0



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



So, solution space is unbounded

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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 x2 S.T. 3x1-3x2+5x3 ≤ 50 x1+x3 ≤ 10 x1-x2+4x3 ≤ 20 x1,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

Basis x1

x2

x3

s1

s2

s3

RHS

S1

3

-3

5

1

0

0

50

S2

1

0

1

0

1

0

10

S3

1

-1

4

0

0

1

20

Z

-20

-10

-1

0

0

0

0

Infeasible Solution: 





This situation can never occur if all the constraints are of the type “≤” with nonnegative 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=1000 Maximize 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

R≠0 Pseudo optimum

SOLUTION INFEASIBLE

REFERENCES 



Operations Research an introduction, seventh edition ,Hamdy A. Taha TORA

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