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In the standard simplex table The coefficients of the basic variables inthe z-row should be zero. • Th r iv iin in h n r in m rix 1 must be identity coef ficients. • In case it is not we have to use elementary row operations to keep it in the standard simplex form before weapply thesimplex method

Big-M Two Phase Methods

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Page 1: Big-M Two Phase Methods

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In the standard simplex table

bull The coefficients of the basic variables in the z-rowshould be zero

bull Th r iv i i n in h n r in m rix

1

must be identity coefficients

bull In case it is not we have to use elementary row

operations to keep it in the standard simplex formbefore we apply the simplex method

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Contents

bull Artificial Variables Techniques

bull Big M Method

bull Two Phase Method

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If in a starting simplex tableau we donrsquot have an

identity sub matrix (ie 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

so ve t e prob em ndash

(1) Big M Method

(2) Two-Phase Method

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Big M Method

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Suppose that ith constraint equation doesnt have aslack variable ie

ith constraint in the original LPP is either ge or = type

Then an artificial variable Ri is added to form the ithunit vector column

Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal

zero in the optimum solution

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Artificial variable objective coefficient

= - M (Sum of artificial variables) in a maximization

problem

= sum o t e art c a var a es n a m n m zat onproblem

where M is a very large positive number

Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin

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Example 1

Min z = 4x1 + 3x2

Subject to

2x1 + x2 ge 10x1 + x2 ge 6

-3x + 2x le 6

x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the

constraints

z ndash 4x1 ndash 3x2 = 0

2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6

-3x1 + 2x2 + S3 = 6

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Addition of artificial variables to obtain a starting BFS

Artificial variables

R1 and R2 added tothe constraints

2x1 + x2 ndash S1 +R1 = 10

x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6

These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to

them in the objective function

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z = 4x1 + 3x2 + MR1 + MR2

2x1 + x2 ndash s1 +R1 = 10

x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6

Objective function

constraints

Where M is a very large penalty assigned onartificial variables R1 and R2

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Tabular form

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Since the z column does not change in the iterations it may be

deleted

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983122983089

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The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

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Apply the usual simplex method for solvingminimization problem

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983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

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102

61

---

Select the most positive element of the z row

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983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

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512

112

2172

983089983087983090

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Final iteration

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Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

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Phase - II

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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Standard simplex table

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Applying the usual simplex method for solving aminimization problem

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Minratio

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983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

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At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

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Is this standard simplex table

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z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

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Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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Contents

bull Artificial Variables Techniques

bull Big M Method

bull Two Phase Method

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If in a starting simplex tableau we donrsquot have an

identity sub matrix (ie 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

so ve t e prob em ndash

(1) Big M Method

(2) Two-Phase Method

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Big M Method

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Suppose that ith constraint equation doesnt have aslack variable ie

ith constraint in the original LPP is either ge or = type

Then an artificial variable Ri is added to form the ithunit vector column

Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal

zero in the optimum solution

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Artificial variable objective coefficient

= - M (Sum of artificial variables) in a maximization

problem

= sum o t e art c a var a es n a m n m zat onproblem

where M is a very large positive number

Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin

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Example 1

Min z = 4x1 + 3x2

Subject to

2x1 + x2 ge 10x1 + x2 ge 6

-3x + 2x le 6

x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the

constraints

z ndash 4x1 ndash 3x2 = 0

2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6

-3x1 + 2x2 + S3 = 6

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Addition of artificial variables to obtain a starting BFS

Artificial variables

R1 and R2 added tothe constraints

2x1 + x2 ndash S1 +R1 = 10

x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6

These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to

them in the objective function

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z = 4x1 + 3x2 + MR1 + MR2

2x1 + x2 ndash s1 +R1 = 10

x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6

Objective function

constraints

Where M is a very large penalty assigned onartificial variables R1 and R2

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983106983137983155983145983139 983162 983160983089

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983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Tabular form

983122983090

983088 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

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Since the z column does not change in the iterations it may be

deleted

983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

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983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

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983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983106983137983155

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983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

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983123983090

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983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

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983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983155983090

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983122983090

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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

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983155983090

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983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 3: Big-M Two Phase Methods

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If in a starting simplex tableau we donrsquot have an

identity sub matrix (ie 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

so ve t e prob em ndash

(1) Big M Method

(2) Two-Phase Method

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Big M Method

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Suppose that ith constraint equation doesnt have aslack variable ie

ith constraint in the original LPP is either ge or = type

Then an artificial variable Ri is added to form the ithunit vector column

Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal

zero in the optimum solution

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Artificial variable objective coefficient

= - M (Sum of artificial variables) in a maximization

problem

= sum o t e art c a var a es n a m n m zat onproblem

where M is a very large positive number

Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin

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Example 1

Min z = 4x1 + 3x2

Subject to

2x1 + x2 ge 10x1 + x2 ge 6

-3x + 2x le 6

x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the

constraints

z ndash 4x1 ndash 3x2 = 0

2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6

-3x1 + 2x2 + S3 = 6

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Addition of artificial variables to obtain a starting BFS

Artificial variables

R1 and R2 added tothe constraints

2x1 + x2 ndash S1 +R1 = 10

x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6

These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to

them in the objective function

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z = 4x1 + 3x2 + MR1 + MR2

2x1 + x2 ndash s1 +R1 = 10

x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6

Objective function

constraints

Where M is a very large penalty assigned onartificial variables R1 and R2

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983106983137983155983145983139 983162 983160983089

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983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Tabular form

983122983090

983088 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983088 983085983091 983090 983088 983088 983088 983088 983089 983094

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Since the z column does not change in the iterations it may be

deleted

983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983106983137983155

983145983139

983160983089

983160983090

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983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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Big M Method

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Suppose that ith constraint equation doesnt have aslack variable ie

ith constraint in the original LPP is either ge or = type

Then an artificial variable Ri is added to form the ithunit vector column

Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal

zero in the optimum solution

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Artificial variable objective coefficient

= - M (Sum of artificial variables) in a maximization

problem

= sum o t e art c a var a es n a m n m zat onproblem

where M is a very large positive number

Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin

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Example 1

Min z = 4x1 + 3x2

Subject to

2x1 + x2 ge 10x1 + x2 ge 6

-3x + 2x le 6

x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the

constraints

z ndash 4x1 ndash 3x2 = 0

2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6

-3x1 + 2x2 + S3 = 6

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Addition of artificial variables to obtain a starting BFS

Artificial variables

R1 and R2 added tothe constraints

2x1 + x2 ndash S1 +R1 = 10

x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6

These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to

them in the objective function

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z = 4x1 + 3x2 + MR1 + MR2

2x1 + x2 ndash s1 +R1 = 10

x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6

Objective function

constraints

Where M is a very large penalty assigned onartificial variables R1 and R2

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983106983137983155983145983139 983162 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Tabular form

983122983090

983088 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983088 983085983091 983090 983088 983088 983088 983088 983089 983094

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Since the z column does not change in the iterations it may be

deleted

983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983106983137983155

983145983139

983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 5: Big-M Two Phase Methods

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Suppose that ith constraint equation doesnt have aslack variable ie

ith constraint in the original LPP is either ge or = type

Then an artificial variable Ri is added to form the ithunit vector column

Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal

zero in the optimum solution

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Artificial variable objective coefficient

= - M (Sum of artificial variables) in a maximization

problem

= sum o t e art c a var a es n a m n m zat onproblem

where M is a very large positive number

Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin

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Example 1

Min z = 4x1 + 3x2

Subject to

2x1 + x2 ge 10x1 + x2 ge 6

-3x + 2x le 6

x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the

constraints

z ndash 4x1 ndash 3x2 = 0

2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6

-3x1 + 2x2 + S3 = 6

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Addition of artificial variables to obtain a starting BFS

Artificial variables

R1 and R2 added tothe constraints

2x1 + x2 ndash S1 +R1 = 10

x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6

These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to

them in the objective function

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z = 4x1 + 3x2 + MR1 + MR2

2x1 + x2 ndash s1 +R1 = 10

x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6

Objective function

constraints

Where M is a very large penalty assigned onartificial variables R1 and R2

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983106983137983155983145983139 983162 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Tabular form

983122983090

983088 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983088 983085983091 983090 983088 983088 983088 983088 983089 983094

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Since the z column does not change in the iterations it may be

deleted

983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983106983137983155

983145983139

983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

21

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

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983155983090

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983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

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983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

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983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

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Tutorial - 4

48

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 6: Big-M Two Phase Methods

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Artificial variable objective coefficient

= - M (Sum of artificial variables) in a maximization

problem

= sum o t e art c a var a es n a m n m zat onproblem

where M is a very large positive number

Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin

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Example 1

Min z = 4x1 + 3x2

Subject to

2x1 + x2 ge 10x1 + x2 ge 6

-3x + 2x le 6

x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the

constraints

z ndash 4x1 ndash 3x2 = 0

2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6

-3x1 + 2x2 + S3 = 6

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Addition of artificial variables to obtain a starting BFS

Artificial variables

R1 and R2 added tothe constraints

2x1 + x2 ndash S1 +R1 = 10

x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6

These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to

them in the objective function

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z = 4x1 + 3x2 + MR1 + MR2

2x1 + x2 ndash s1 +R1 = 10

x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6

Objective function

constraints

Where M is a very large penalty assigned onartificial variables R1 and R2

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983106983137983155983145983139 983162 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Tabular form

983122983090

983088 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983088 983085983091 983090 983088 983088 983088 983088 983089 983094

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Since the z column does not change in the iterations it may be

deleted

983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983106983137983155

983145983139

983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

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983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

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983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

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Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

48

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 7: Big-M Two Phase Methods

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Example 1

Min z = 4x1 + 3x2

Subject to

2x1 + x2 ge 10x1 + x2 ge 6

-3x + 2x le 6

x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the

constraints

z ndash 4x1 ndash 3x2 = 0

2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6

-3x1 + 2x2 + S3 = 6

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Addition of artificial variables to obtain a starting BFS

Artificial variables

R1 and R2 added tothe constraints

2x1 + x2 ndash S1 +R1 = 10

x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6

These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to

them in the objective function

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z = 4x1 + 3x2 + MR1 + MR2

2x1 + x2 ndash s1 +R1 = 10

x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6

Objective function

constraints

Where M is a very large penalty assigned onartificial variables R1 and R2

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983106983137983155983145983139 983162 983160983089

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983155983151983148983157983156983145983151983150

983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Tabular form

983122983090

983088 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983088 983085983091 983090 983088 983088 983088 983088 983089 983094

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Since the z column does not change in the iterations it may be

deleted

983106983137983155983145983139 983160983089

983160983090

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983155983151983148983157983156983145983151983150

983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

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983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

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983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

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983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

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983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

48

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 8: Big-M Two Phase Methods

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Addition of artificial variables to obtain a starting BFS

Artificial variables

R1 and R2 added tothe constraints

2x1 + x2 ndash S1 +R1 = 10

x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6

These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to

them in the objective function

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z = 4x1 + 3x2 + MR1 + MR2

2x1 + x2 ndash s1 +R1 = 10

x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6

Objective function

constraints

Where M is a very large penalty assigned onartificial variables R1 and R2

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983106983137983155983145983139 983162 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Tabular form

983122983090

983088 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983088 983085983091 983090 983088 983088 983088 983088 983089 983094

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Since the z column does not change in the iterations it may be

deleted

983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983106983137983155

983145983139

983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

22

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

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983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

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983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

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983155983090

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983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

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983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

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Applying the usual simplex method for solving aminimization problem

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983085

Minratio

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983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

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983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

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983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

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983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

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Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 9: Big-M Two Phase Methods

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z = 4x1 + 3x2 + MR1 + MR2

2x1 + x2 ndash s1 +R1 = 10

x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6

Objective function

constraints

Where M is a very large penalty assigned onartificial variables R1 and R2

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983106983137983155983145983139 983162 983160983089

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983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Tabular form

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Since the z column does not change in the iterations it may be

deleted

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983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

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The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

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983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

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983123983090

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983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

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102

61

---

Select the most positive element of the z row

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983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

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983123983090

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983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

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Phase - II

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983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

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form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

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Applying the usual simplex method for solving aminimization problem

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983085

Minratio

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983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

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983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

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At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

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983160983091 983155983089

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983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

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Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 10: Big-M Two Phase Methods

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983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Tabular form

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Since the z column does not change in the iterations it may be

deleted

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983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

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The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

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102

61

---

Select the most positive element of the z row

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983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

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512

112

2172

983089983087983090

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983122983090

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983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

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Final iteration

983160983090

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983123983091

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Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

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983155983090

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983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

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983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

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983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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7172019 Big-M Two Phase Methods

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

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Tutorial - 4

48

7172019 Big-M Two Phase Methods

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 11: Big-M Two Phase Methods

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Since the z column does not change in the iterations it may be

deleted

983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088

983122983089

983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

The initial BFS as obtained from the table is

R1 = 10 R2 = 6 and z = 0 Is this tablecorrect

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983106983137983155

983145983139

983160983089

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983123983090

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983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

18

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

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983155983090

983122983089

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983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

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983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

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983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 12: Big-M Two Phase Methods

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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as

z = 4x1 + 3x2 + MR1 + MR2

Wh hi in n i n r

This is because of the fact that the elementsof the basic variable have a non zero

coefficient in the z-row

The starting table now becomes

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983106983137983155

983145983139

983160983089

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983123983090

983122983089

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983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

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983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

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983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983106983137983155983145983139 983160983089

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983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

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Tutorial - 4

48

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 13: Big-M Two Phase Methods

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983106983137983155983145983139 983160983089

983160983090

983123983089

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983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094

Apply the usual simplex method for solvingminimization problem

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983106983137983155983145983139 983160983089

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983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

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102

61

---

Select the most positive element of the z row

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983106983137983155

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983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

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983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

22

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

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983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983155983090

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983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983160983090

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983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

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Tutorial - 4

48

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117

983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088

Selecting pivot row and pivot column for a minimization problem

Minratio

983122983090

983089 983089 983088 983085983089 983088 983089 983088 983094

983123983091

983085983091 983090 983088 983088 983088 983088 983089 983094

102

61

---

Select the most positive element of the z row

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983106983137983155

983145983139

983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

18

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

21

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

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Tutorial - 4

48

7172019 Big-M Two Phase Methods

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 15: Big-M Two Phase Methods

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983106983137983155

983145983139

983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983085983090983083983117

983090

983085983092983085983117

983090

983085983117 983092 991251 983091983117

983090

983088 983088 983090983088 983083983117

Minratio

Second iteration

983160983089

983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093

983122983090

983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089

983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089

512

112

2172

983089983087983090

15

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

18

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

7172019 Big-M Two Phase Methods

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

20

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

21

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Tutorial - 3

22

7172019 Big-M Two Phase Methods

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

26

7172019 Big-M Two Phase Methods

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

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Tutorial - 4

48

7172019 Big-M Two Phase Methods

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 16: Big-M Two Phase Methods

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983106983137983155983145983139 983160983089

983160983090

983123983089

983123983090

983122983089

983122983090

983123983091

983155983151983148983157983156983145983151983150

983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090

983160983089

983089 983088 983085983089 983088 983089 983088 983088 983092

Final iteration

983160983090

983088 983089 983089 983085983089 983088 983089 983088 983090

983123983091

983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092

Solution x1 = 4 x2 = 2 z = 22

16

7172019 Big-M Two Phase Methods

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

7172019 Big-M Two Phase Methods

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

18

7172019 Big-M Two Phase Methods

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

7172019 Big-M Two Phase Methods

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

20

7172019 Big-M Two Phase Methods

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

21

7172019 Big-M Two Phase Methods

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Tutorial - 3

22

7172019 Big-M Two Phase Methods

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

26

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2751

Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

7172019 Big-M Two Phase Methods

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

28

7172019 Big-M Two Phase Methods

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3051

Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 17: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

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Consider the following LPP

Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

17

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 1851

Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

18

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 1951

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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

7172019 Big-M Two Phase Methods

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

20

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2151

If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

21

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2251

Tutorial - 3

22

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2351

(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2551

(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2651

TWO PHASE METHOD

26

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2751

Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2851

Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

7172019 Big-M Two Phase Methods

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

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Tutorial - 4

48

7172019 Big-M Two Phase Methods

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 18: Big-M Two Phase Methods

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Introducing surplus and artificial variables s2 R1 and R2 the

LPP is modified as follows

Maximize

Subject to

1 2 3 1 2

2 3 5 z x x x M R M R= + minus minus minus

1 2 3 1 7 x x x R+ + + =

Now we solve the above LPP by the Simplex method

1 2 3 2 2

1 2 3 2 1 2

2 5 1 0 0

x x x s R

x x x s R R

minus + minus + =

ge

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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7172019 Big-M Two Phase Methods

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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7172019 Big-M Two Phase Methods

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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7172019 Big-M Two Phase Methods

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 19: Big-M Two Phase Methods

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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 0

R1 0 1 1 1 0 1 0 7

z 1 0 -8 - 6 - -1 - 0 1 + 10 -

x1 0 1 -52 12 -12 0 12 5

7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2

x2 0 0 1 17 17 27 -17 47

z 1 0 0 507 17 167 + -17 1027M +M

x1 0 1 0 67 -17 57 17 457

7172019 Big-M Two Phase Methods

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The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

20

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2151

If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

21

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2251

Tutorial - 3

22

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2351

(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2451

(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2551

(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2651

TWO PHASE METHOD

26

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2751

Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2851

Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

28

7172019 Big-M Two Phase Methods

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3051

Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3251

Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 20: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2051

The optimum (Maximum) value of

z = 1027

and it occurs at

x1 = 457 x2 = 47 x3 = 0

20

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2151

If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

21

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2251

Tutorial - 3

22

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2351

(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2451

(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2551

(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2651

TWO PHASE METHOD

26

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2751

Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2851

Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

48

7172019 Big-M Two Phase Methods

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable

(decision surplus or slack) we must prefer the

artificial variable to leave the basis

into the basis

If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our

original problem has no feasible solution

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Tutorial - 3

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(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 22: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2251

Tutorial - 3

22

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2351

(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2451

(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2551

(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2651

TWO PHASE METHOD

26

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2751

Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2851

Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

28

7172019 Big-M Two Phase Methods

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

7172019 Big-M Two Phase Methods

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3251

Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 23: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2351

(1) Solve the following LPPs using Simplex method

(i) Max

subject to

1 21 2 1 5 z x x= minus minus

1 2

1 2

1 2

4 3 1 22 5 1 0

0

x x x x

x x

+ le

+ le

ge

(ii) Max

subject to

23

1 22 3 z x x= minus

1 2

1 2

1 2

1 2

2

2 2

2

0 u n r e s t r i c e d

x x

x x

x x

x x

minus + le

minus le

minus minus le

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2451

(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2551

(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2651

TWO PHASE METHOD

26

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2751

Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2851

Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

28

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

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Tutorial - 4

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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(iii) Max

subject to

1 2 33 2 5 z x x x= + +

1 2 3

1 3

1 2

1 2 3

2 4 3 0

3 2 4 6 0

4 4 2 0

0

x x x

x x

x x

x x x

+ + le

minus minus ge minus

+ le

ge

(iv) Minsubject to

24

1 2 36 2 6 z x x x= minus minus minus

1 2 3

1 2 3

1 2 3

1 2 3

2 3 1 4

4 4 1 0 4 6

2 2 4 3 7

2 1 3

x x x

x x x

x x x

x x x

minus + le

minus + + le

+ minus le

ge ge ge

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

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TWO PHASE METHOD

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

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s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

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Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

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z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 25: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

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(2) Solve the following LPPs using Big-M method

(i) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 23 3

4

x x

x x

x

+ ge

+ le

le

(ii) Min

subject to

1 2 0 x x ge

25

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 5

2 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2651

TWO PHASE METHOD

26

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2751

Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2851

Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

28

7172019 Big-M Two Phase Methods

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

7172019 Big-M Two Phase Methods

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3151

s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3251

Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 26: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2651

TWO PHASE METHOD

26

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2751

Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2851

Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

28

7172019 Big-M Two Phase Methods

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

7172019 Big-M Two Phase Methods

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3151

s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

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Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

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r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 27: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

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Why use two phase method when Big M - methodis already there

One major disadvantage of Big M - method is that itis computationally inefficient

Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result

27

7172019 Big-M Two Phase Methods

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

28

7172019 Big-M Two Phase Methods

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

7172019 Big-M Two Phase Methods

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3151

s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

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Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

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Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

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Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 28: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

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Phase - I

983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150

983152983154983151983138983148983141983149983081983086

983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155

983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155

983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141

983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086

28

7172019 Big-M Two Phase Methods

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Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

7172019 Big-M Two Phase Methods

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Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3151

s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3251

Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

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Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

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983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 29: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 2951

Phase - II

983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141

983085

983152983154983151983138983148983141983149983086

29

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3051

Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3151

s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3251

Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 30: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3051

Example

Min z = 4x1 + 6x2 + 5x3

Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28

x1 x2 x3 ge 0

30

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3151

s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3251

Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 31: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3151

s1 s2 ndash surplus variableR1 R2 ndash artificial variables

Min z = 4x1 + 6x2 + 5x3

Subject to

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Artificial variables are added to get an identity sub matrix

31

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3251

Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 32: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3251

Phase I (construct an auxiliary LPP)

construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero

Min r = R1 + R2New objective function

2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0

Constraints

32

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 33: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3351

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983122983089

983090 983092 983091 983085983089 983088 983089 983088 983091983090

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

form a new r row (similar to the manner in which new z- row is

formed in Big M method)

ThusNew r-row = r-row + R1 row + R2 row

33

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 34: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3451

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Standard simplex table

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096

34

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 35: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3551

Applying the usual simplex method for solving aminimization problem

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088

983085

Minratio

983089

983122983090

983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092

35

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 36: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3651

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983122983089

983122983090

983155983151983148983157983156983145983151983150

983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089

983122983089

983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090

983160983091

983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095

36

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 37: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3751

983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150

983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088

983160983090

983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093

At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable

37

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 38: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3851

Phase II

We will use the BFS obtained in Phase-I as a starting

BFS of the given problem and use simplex method tofind the optimal solution

38

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 39: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 3951

Form a simplex table

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085983092 983085983094 983085983093 983088 983088 983088

983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Is this standard simplex table

39

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

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r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 40: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4051

983106983137983155983145983139 983160983089

983160983090

983160983091 983155983089

983155983090

983155983151983148983157983156983145983151983150

983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093

z ndashrow = z row + 6(x2 row) + 5(x3 row)

983090

983160983091

983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093

Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point

40

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 41: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4151

Solve the following LPP using Two-Phase method

Minimize

Sub ect to

321 352 x x x z ++=

1 2 3

1 2 3

1 2 3

2 4 5 0

0

x x x

x x x

x x x

minus +

+ + =

ge

41

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 42: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4251

Phase I

Minimize

Subject to

1 2r R R= +

1 2 3 1 12 20 x x x s Rminus + minus + =

=1 2 3 2

1 2 3 1 1 2 0 x x x s R R ge

42

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 43: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4351

r 1 0 0 0 0 -1 -1 0

Basic r x1 x2 x3 s1 R1 R2 Sol

3 2 2 -1 0 0 70

R1 0 1 -2 1 -1 1 0 20

R2 0 2 4 1 0 0 1 50

x1 0 1 -2 1 -1 1 0 20

r 1 0 8 -1 2 -3 0 10

R2 0 0 8 -1 2 -2 1 10

43

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 44: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4451

r 1 0 8 -1 2 -3 0 10

Basic r x1 x2 x3 s1 R1 R2 Sol

x1 0 1 -2 1 -1 1 0 20

R2 0 0 8 -1 2 -2 1 10

x1 0 1 0 34 -12 12 14 452

r 1 0 0 0 0 -1 -1 0

x2 0 0 1 -18 14 -14 18 54

44

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 45: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4551

This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal

solution of Phase I

Phase II

subject to the same constraints as given in the original

problem

321

45

Basic z x1 x2 x3 s1 R1 R2 Sol

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 46: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4651

z 1 -2 -5 -3 0 0

1 2 3 1 1 2

x1 0 1 0 34 -12 452

x2 0 0 1 -18 14 54

- -

0 0 -178 14 2054

x1 0 1 2 12 0 25

s1 0 0 4 -12 1 5

Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the

optimal z = Min z = 5046

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4751

In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4851

Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5151

(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 47: Big-M Two Phase Methods

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In Phase I we always minimize the sum of the

artificial variables

If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em

has no feasible solution

47

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Tutorial - 4

48

7172019 Big-M Two Phase Methods

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Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 48: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

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Tutorial - 4

48

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 49: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 4951

Solve the following LPPs using Two-Phase Method

(i) Maximize

Subject to

1 2 32 3 5 z x x x= + minus

1 2 3 7 x x x+ + =

1 2 3

1 2 3

2 5 10

0

x x x

x x x

minus + ge

ge

49

7172019 Big-M Two Phase Methods

httpslidepdfcomreaderfullbig-m-two-phase-methods 5051

(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 50: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

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(ii) Maximize

Subject to

321 422 x x x z ++=

1 2 32 2 x x x+ + le

1 2 3

1 2 3 0

x x x

x x x

+ +

ge

50

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge

Page 51: Big-M Two Phase Methods

7172019 Big-M Two Phase Methods

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(iii) Max

subject to

1 23 z x x= minus +

1 2

1 2

1

2 2

3 34

0

x x

x x

x

x x

+ ge

+ le

le

ge

(iv) Min

subject to

51

1 2 39 2 z x x x= + +

1 2 3

1 2 3

1 2 3

4 2 52 3 4

0

x x x

x x x

x x x

+ + ge

+ + ge

ge