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Artificial VariableTechniques

Big M-method

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Lecture 6

Abstract If in a starting simplex tableau,

we dont have an identity submatrix (i.e. an

obvious starting BFS), then we introduce

artificial variables to have a starting BFS.

This is known as artificial variable

technique. There are two methods to find

the starting BFS and solve the problem

the Big M method and two-phase method.

In this lecture, we discuss the Big M

method.

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Suppose a constraint equation i does not

have a slack variable. i.e. there is no ith

unit vector column in the LHS of theconstraint equations. (This happens for

example when the ith constraint in the

original LPP is either or = .) Then weaugment the equation with an artificial

variable Ri to form the ith unit vector

column. However as the artificialvariable is extraneous to the given LPP,

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we use a feedback mechanism in which the

optimization process automatically attempts

to force these variables to zero level. This isachieved by giving a large penalty to the

coefficient of the artificial variable in the

objective function as follows:

Artificial variable objective coefficient

= - M in a maximization problem,= M in a minimization problem

where M is a very large positive number.

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Consider the LPP:

Minimize 212 xxz

Subject to the constraints

0,

6

93

21

21

21

xx

xx

xx

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Putting this in the standard form, the LPP is:

212 xxz


1 2 1

1 2 2

1 2 1 2

3 96

, , , 0

x x s

x x s

x x s s

Heres1,s2 are surplus variables.

Minimize

Note that we do NOT have a 2x2 identity

submatrix in the LHS.

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Introducing the artificial variables, R1,

R2, the LPP is modified as follows:

Minimize


1 2 1 1

1 2 2 2

1 2 1 2 1 2

3 9

6

, , , , , 0

x x s R

x x s R

x x s s R R

Note that we now have a 2x2 identity

submatrix in the coefficient matrix of the

constraint equations.

1 2 1 22z x x MR MR

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Now we solve the above LPP by theSimplex method.

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R2 0 1 1 0 -1 0 1 6

-2+4M -1+2M -M -M 0 0 15M

Basic z x1 x2 s1 s2 R1 R2 Sol.

z 1 -2 -1 0 0 -M -M 0R1 0 3 1 -1 0 1 0 9

R2 0 0 2/3 1/3 -1 -1/3 1 3

z 1 0 -1/3+ -2/3+ -M 2/3- 0 6+3M2M/3 M/3 4M/3

x1 0 1 1/3 -1/3 0 1/3 0 3

x1 0 1 0 -1/2 1/2 1/2 -1/2 3/2

z 1 0 0 -1/2 -1/2 1/2M 1/2-M 15/2

x2 0 0 1 1/2 -3/2 -1/2 3/2 9/2

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Note that we have got the optimal solution

to the given problem as

1 2

3 9,

2 2x x

15

2

z

It is illuminating to look at the graphical

solution also.

Optimal z = Minimum

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(3,0) (6,0)

(0.9)

(0,6)3 9

( , )2 2

(0,0)

SF

z minimum at3 9

( , )2 2

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Problem 5(a) Problem Set 3.4A Page 97

Maximize 1 2 32 3 5z x x x

Subject tothe constraints

1 2 3

1 2 3

1 2 3

7

2 5 10

, , 0

x x x

x x x

x x x

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Introducing surplus and artificial variables,

s2, R1and R2, the LPP is modified as follows:

Maximize


1 2 3 1 22 3 5z x x x MR MR

1 2 3 1

1 2 3 2 2

1 2 3 2 1 2

7

2 5 10, , , , , 0

x x x R

x x x s R

x x x s R R

Now we solve the above LPP by the Simplex

method.

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R2 0 2 -5 1 -1 0 1 10

-2-3M -3+4M 5-2M M 0 0 -17M

Basic z x1 x2 x3 s2 R1 R2 Sol.

z 1 -2 -3 5 0 M M 0R1 0 1 1 1 0 1 0 7

x1 0 1 -5/2 1/2 -1/2 0 1/2 5

z 1 0 -8 - 6 - -1 - 0 1 + 10 -7M/2 M/2 M/2 3M/2 2M

R1 0 0 7/2 1/2 1/2 1 -1/2 2

x2 0 0 1 1/7 1/7 2/7 -1/7 4/7

z 1 0 0 50/7 1/7 16/7 + -1/7 102/7

M +M

x1 0 1 0 6/7 -1/7 5/7 1/7 45/7

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The optimum (Maximum) value of

z = 102/7

and it occurs at

x1 = 45/7,x2 = 4/7,x3 = 0

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Remarks

If in any iteration, there is a tie for entering

variable between an artificial variable and

other variable (decision, surplus or slack),

we must prefer the non-artificial variable to

enter the basis.

If in any iteration, there is a tie for leaving

variable between an artificial variable andother variable (decision, surplus or slack),

we must prefer the artificial variable to

leave the basis.

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If in the final optimal tableau, an

artificial variable is present in thebasis at a non-zero level, this means

our original problem has no feasible

solution.

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Problem 4(a) Problem Set 3.4A Page 97

Maximize 1 25 6z x x


1 2

1 2

1 2

1 2

2 3 32 5

6 7 3, 0

x x

x x

x x

x x

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Introducing slack and artificial variables, s2,

s3

, and R1

, the LPP is modified as follows:

Maximize


1 2 1

1 2 2

1 2 3

1 2 1 2 3

2 3 3

2 56 7 3

, , , , 0

x x R

x x s

x x s

x x R s s

1 2 15 6z x x MR

i 1 2 1 2 3 S l

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s2 0 1 2 0 1 0 5

-5+2M -6-3M 0 0 0 -3M

Basic z x1 x2 R1 s2 s3 Sol

z 1 -5 -6 M 0 0 0R1 0 -2 3 1 0 0 3

s3 0 6 7 0 0 1 3

s2 0 -12/7 0 0 1 -2/7 29/7

z 1 1/7+ 0 0 0 6/7+ 18/7-

32M/7 3M/7 12M/7

R1 0 -32/7 0 1 0 -3/7 12/7

x2 0 6/7 1 0 0 1/7 3/7

This is the optimal tableau. As R1 is not zero, there is NO feasible

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Problem 4(d) Problem Set 3.4A Page 97

Minimize 21 64xxz


0,584

1054332

21

21

21

21

xx

xx

xx

xx

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Introducing the surplus and artificial variables,

R1

, R2

, the LPP is modified as follows:

Minimize

Subject to the constraints1 2 1

1 2 2 2

1 2 3 3

1 2 2 3 1 2 3

2 3 3

4 5 10

4 8 5

, , , , , , 0

x x R

x x s R

x x s R

x x s s R R R

1 2 1 2 34 6z x x M R M R M R

B i 1 2 2 3 R1 R2 R3 S l

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R2 0 4 5 -1 0 0 1 0 10

-4+6M -6+16M -M -M 0 0 0 18M

Basic z x1 x2 s2 s3 R1 R2 R3 Sol.

z 1 -4 -6 0 0 -M -M -M 0R1 0 -2 3 0 0 1 0 0 3

R1 0 -7/2 0 0 3/8 1 0 -3/8 9/8

z 0 -1-2M 0 -M -3/4 0 0 3/4 15/4

+M -2M +8M

R2 0 3/2 0 -1 5/8 0 1 -5/8 55/8

x2 0 1/2 1 0 -1/8 0 0 1/8 5/8

R3 0 4 8 0 -1 0 0 1 5

B i 1 2 2 3 R1 R2 R3 S l

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R2 0 3/2 0 -1 5/8 0 1 -5/8 55/8


z 1 -1-2M 0 -M -3/4 0 0 3/4 15/4

+M -2M +8MR1 0 -7/2 0 0 3/8 1 0 -3/8 9/8

s3 0 -28/3 0 0 1 8/3 0 -1 3

z 1 -8 + 0 -M 0 2 0 -M 6

22M/3 -8M/3 +5M

R2 0 22/3 0 -1 0 -5/3 1 0 5

x2 0 -2/3 1 0 0 1/3 0 0 1

x2 0 1/2 1 0 -1/8 0 0 1/8 5/8

B i 1 2 2 3 R1 R2 R3 S l

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s3 0 -28/3 0 0 1 8/3 0 -1 3

z 1 -8 + 0 -M 0 2 0 -M 6

22M/3 -8M/3 +5M

R2 0 22/3 0 -1 0 -5/3 1 0 5

x2 0 -2/3 1 0 0 1/3 0 0 1

s3 0 0 0 -14/11 1 6/11 14/11 -1

z 1 0 0 -6/11 0 2/11 12/11 -M

- M -M

x1 0 1 0 -3/22 0 -5/22 3/22 0

x2 0 0 1 -1/11 0 2/11 1/11 0

10311

126

11

15

22

16

11

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This is the optimal tableau.

The Optimal solution is:

1 215 16,22 11

x x

And Optimal z = Min z =126

11

05 simplex 3

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