Mute ur call

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex method when some constraints are not constraintsWe employ a mathematical trick to jumpstart the problem by adding artificial variables to the equations.

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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex method when some constraints are not constraints (cont.)Example:

Max 16x1+15x2+20x3-18x4ST2x1 + x2 + 3x3 3000[1]3x1 + 4x2 + 5x3 60x4 2400[2]x4 32[3]X2 200[4]X1 + x2 + x3 800[5]X1 x2 x3 =0[6]Xj 0 for all J

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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex method when some constraints are not constraints (cont.)Example:

Max 16x1+15x2+20x3-18x4ST2x1 + x2 + 3x3 3000[1]3x1 + 4x2 + 5x3 60x4 2400[2]x4 32[3]X2 200[4]X1 + x2 + x3 800[5]X1 x2 x3 =0[6]Xj 0 for all J

*We add artificial : R4, R5, R6, respectively to the fourth, fifth, and sixth equations.We assign a very large negative objective function coefficient , -M , ( +M for minimization problem) to each artificial variable

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex method when some constraints are not constraints (cont.)The solutionMax 16x1+15x2+20x3-18x4ST2x1 + x2 + 3x3 + s1= 3000[1]3x1 + 4x2 + 5x3 60x4 + s2 = 2400[2]x4 + s3 = 32[3]X2 s4 + R4 = 200[4]X1 + x2 + x3 s5 + R5 = 800[5]X1 x2 x3 + R6= 0[6]Xj 0 , Sj 0, Rj 0 for all J *MR4 MR5 MR6The simplex algorithm can then be used to solve this problem

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variablesExample # 1:

MAX 2x1+ 5x2STX1 4x1 + 4x2 323x1+ 2x2 = 24*

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variables (cont.)The Solution By adding the appropriate slack, surplus, and artificial variables, we obtain the following:

MAX 2x1 + 5x2 MR1 MR3STX1 s1 + R1 =4X1 + 4x2 + s2 = 323x1 + 2x2 + R3= 24

X1,x2,s1,s2,R1,R3 0

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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variables (cont.)

The initial table :

Make z consistent; (R1, R3) in z-row coefficient (+M,+M) it must be zero; By apply:

New z-row = old z-row + ( -M * R1 row M * R3 row)

New z-row = old z-row + ( M * R1 row +M * R3 row)*MAX objective functionMIN objective function

BasisX1X2S1S2R1R3RHSR110-10104S214010032R332000124Z-2-500+ M+ M0

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variables (cont.)Starting table:*

BasisX1X2S1S2R1R3RHSR110-10104S214010032R332000124Z-2-4M-5-2M+M0- M- M-28M

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variables (cont.)To determine Entering Variable; We should look to the largest negative number in z-row.

*Largest negative number

BasisX1X2S1S2R1R3RHSR110-10104S214010032R332000124Z-2-4M-5-2M+M0- M- M-28M

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variables (cont.)Calculate the ratio; then, determine the smallest positive number as Leaving Variable

Pivot element = ( 1, 0, -1, 0, 1, 0, 4)/ (1) ( 1, 0, -1, 0, 1, 0, 4)*Leaving Variable

BasisX1X2S1S2R1R3RHSRatioR110-101044S21401003232R3320001248Z-2-4M-5-2M+M0- M- M-28M

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variables (cont.)First iteration

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BasisX1X2S1S2R1R3RHSRatioX110-10104.S20411-102828R30230-31124Z0-5-2M-2-3M02+3M-M8-12M

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variables (cont.)Second iteration*

BasisX1X2S1S2R1R3RHSRatioX112/30001/3812S2010/3010-1/3247.2S102/310-11/346Z0-11/30002/3+16

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variables (cont.)Third iteration

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BasisX1X2S1S2R1R3RHSRatioX110-10104S200-515-24X2013/20-3/21/26Z0011/30-11/25/238

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Maximization] when there are artificial variables (cont.)*

pointsClassificationReasonX1=0, X2=0Not FeasibleR1, R3 both Positive (4, 24)X1=4, X2=0Not FeasibleR3 positive= 12X1=8, X2=0Feasible but not optimalX2 is negativeX1=4, X2=6Feasible and optimal All x1,X2 0

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Minimization] when there are artificial variablesExample # 2:

Min 4x1 + x2ST3x1+ x2 = 34x1 + 3x2 6X1+ 2x2 4X1, x2 0

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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Minimization] when there are artificial variables (cont.)The Solution By adding the appropriate slack, surplus, and artificial variables, we obtain the following:

Min 4x1 + x2 + MR1 + MR2ST3x1+ x2 + R1= 34x1 + 3x2 s1 + R2 = 6X1+ 2x2 + s2 = 4X1, x2 , s1, s2, R1, R2 0

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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Minimization] when there are artificial variables (cont.)The initial table:

New z-row = old z-row +( M * R1 row +M * R3 row)

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BasisX1X2S1R1R2S2RHSR13101003R243-10106S21200014Z-4-10-M-M00

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Minimization] when there are artificial variables (cont.)Starting table:

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BasisX1X2S1R1R2S2RHSR13101003R243-10106S21200014Z-4+7M-1+4M-M0009M

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Minimization] when there are artificial variables (cont.)First iteration*

BasisX1X2S1R1R2S2RHSX111/301/3001R205/3-1-4/3102S205/30-1/3013Z0(1+5M)/3-M(4-7M)/3004+2M

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Minimization] when there are artificial variables (cont.)Second iteration

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BasisX1X2S1R1R2S2RHSX1101/53/5-1/503/5X201-3/5-4/53/506/5S20011-111Z001/58/5 - M-1/5 -M018/5

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Solving For the optimal solution of [Minimization] when there are artificial variables (cont.)Third iteration

Optimal solution : x1= 2/5, x2= 9/5, z= 17/5

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BasisX1X2S1R1R2S2RHSX11002/50-1/52/5X2010-1/503/59/5s10011-111Z0007/5 M-M-1/517/5

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

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

DegeneracyAlternative optimaUnbounded solutionNonexisting ( infeasible ) solution *

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

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.

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. *

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)Example:Max 3x1 + 9x2STX1 + 4x2 8X1 + 2x2 4X1, x2 0

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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)The solution:The constraints:X1 + 4x2 + s1= 8X1 + 2x2 + s2= 4X1, x2 ,s1,s2 0

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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)*

BasisX1X2S1S2RHSs114108s212014Z-3-9000

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)*

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

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)

Same objective no change and improve ( cycle)It is possible to have no improve and no termination for computation. *

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

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)Alternative optima

If the z-row value for one or more nonbasic variables is 0 in the optimal tubule, alternate optimal solution is exist.*

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)Example:

Max 2x1+ 4x2STX1 + 2x2 5X1 + x2 4X1, x2 0*

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)The solution

Max 2x1+ 4x2STX1 + 2x2 + s1= 5X1 + x2 + s2 = 4X1, x2, s1, s2 0

*

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)*

BasisX1X2S1S2RHSs112104s211015Z-2-4000

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

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

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

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

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

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

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

*

BasisX1X2S1S2RHSx2011-11x110-123Z002010

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)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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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)ExampleMax 2x1+ x2STX1 x2 102x1 40X1, x20

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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)The solution

Max 2x1+ x2STX1 x2 +s1= 102x1 +s2= 40X1, x2,s1,s20

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Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)

All value if x2( nonbasic variable) either zero or negative.So, solution space is unbounded*

BasisX1X2S1S2RHSx21-11010x1200140Z-2-1000

Hamdy A. Taha, Operations Research: An introduction, Prentice Hall*

Simplex Algorithm Special cases (cont.)Infeasible solutionR coefficient at end 0

This situation can never occur if all the constraints are of the type with nonnegative RHS*