Simplex Method

MSci331—Week 3~4

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Simplex Algorithm

• Consider the following LP, solve using Simplex:

2

1 2

1 2

1 2

1 2

1 2

3 2

. . 2 100

80

40

, 0

MAX Z x x

s t x x

x x

x x

x x

Step 1: Preparing the LP

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Step 2: Express the LP in a tableau form

Z X1 X2 S1 S2 S3 RHS RatioRow 0 1 -3 -2 0 0 0 0 --Row 1 - 2 1 1 0 0 100Row 2 - 1 1 0 1 0 80Row 3 - 1 0 0 0 1 40

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Step 3: Obtain the initial basic feasible solution (if available)

Z X1 X2 S1 S2 S3 RHS RatioRow 0 1 -3 -2 0 0 0 0 --Row 1 - 2 1 1 0 0 100Row 2 - 1 1 0 1 0 80Row 3 - 1 0 0 0 1 40

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a) Set n-m variables equal to 0 These n-m variables the NBV

b) Check if the remaining m variables satisfy the condition of BV

= If yes, the initial feasible basic solution (bfs) is readily a available

= else, carry on some ERO to obtain the initial bfs

Step 4: Apply the Simplex Algorithm

Z X1 X2 S1 S2 S3 RHS RatioRow 0 1 -3 -2 0 0 0 0 --Row 1 - 2 1 1 0 0 100 100/2Row 2 - 1 1 0 1 0 80 80/1Row 3 - 1 0 0 0 1 40 40/1

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a) Is the initial bfs optimal? (Will bringing a NBV improve the value of Z?)

b) If yes, which variable from the set of NBV to bring into the set of BV? - The entering NBV defines the pivot column

c) Which variable from the set of BV has to become NBV?

- The exiting BV defines the pivot rowPivot cell

Summary of Simplex Algorithm for Papa Louis

m≠0

Set: n-m=0

1

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BFS (intial)

BFS (1)

BFS (2)

BFS (3)

The optimal solution is x1=20, x2=60 The optimal value is Z=180

The BFS at optimality x1=20, x2=60, s3=20

Geometric Interpretation of Simplex Algorithm

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Class activity

• Consider the following LP:

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This is a maximizing LP, in normal form. So an initial BFS exists.

Class activity

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Class activity

11

Z x1 x2 s1 s2 s3 s4 RHS

1 -3 -3 0 0 0 0 100 -----1 1 1 0 0 0 41 2 0 1 0 0 62 -3 0 0 1 0 2-1 2 0 0 0 1 4

Class activity

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Z x1 x2 s1 s2 s3 s4 RHS

1 -3 -3 0 0 0 0 100 -----1 1 1 0 0 0 41 2 0 1 0 0 62 -3 0 0 1 0 2-1 2 0 0 0 1 4

4/1

6/1

2/2*

---

Make this coefficient equal 1 and pivot all other rows relative to it

Class activity

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Z x1 x2 s1 s2 s3 s4 RHS

1 -3 -3 0 0 0 0 100 -----1 1 1 0 0 0 41 2 0 1 0 0 61 -3/2 0 0 1/2 0 1-1 2 0 0 0 1 4

0 -7.5 0 0 3/2 0 103

0 2.5 1 0 -1/2 0 3

0 3.5 0 1 -1/2 0 5

0 1/2 0 0 1/2 1 5

Class activity

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Z x1 x2 s1 s2 s3 s4 RHS

1 -3 -3 0 0 0 0 100 -----1 1 1 0 0 0 41 2 0 1 0 0 61 -3/2 0 0 1/2 0 1-1 2 0 0 0 1 4

0 -7.5 0 0 3/2 0 103

0 2.5 1 0 -1/2 0 3

0 3.5 0 1 -1/2 0 5

0 1/2 0 0 1/2 1 5

3/2.5*

5/3.5

---

5/0.5

Make this coefficient equal 1 and pivot all other rows relative to it

Class activity

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Z x1 x2 s1 s2 s3 s4 RHS

1 -----0 -7.5 0 0 3/2 0 103

0 1 2/5 0 -1/5 0 6/5

0 3.5 0 1 -1/2 0 5

0 0 3 0 0 0 112

0 0 -1.4 1 1/5 0 0.8

1 -3/2 0 0 1/2 0 11 0 3/5 0 0.8 0 2.8

0 1/2 0 0 1/2 1 50 0 -1/5 0 0.6 1 4.4

Example: LP model with Minimization Objective

• Solve the following LP model:

• Initial Tableau

1 2

1 2

1 2

1 2

2 3

. .

4

6

, 0

Min z x x

s t

x x

x x

x x

Row Basic Variable Z x1 x2 s1 s2 RHS Ratio

test 0 -1 2 -3 0 0 0 1 s1 0 1 1 1 0 4 2 s2 0 1 -1 0 1 6

1 2

1 2

1 2

1 2

2 3

. .

4

6

, 0

MAX z x x

s t

x x

x x

x x

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Example: LP model with Minimization Objective

• Iteration 0

• Iteration 1

• Optimality test:

Row Basic Variable Z x1 x2 s1 s2 RHS Ratio

test 0 -1 2 -3 0 0 0 1 s1 0 1 1 1 0 4 4 2 s2 0 1 -1 0 1 6 -

Row Basic Variable Z x1 x2 s1 s2 RHS Ratio

test 0 -1 5 0 3 0 12 1 x2 0 1 1 1 0 4 2 s2 0 2 0 1 1 10

1 112 5 3z x s

17

4040

3535

3030

2525

2020

1515

1010

55

5 10 15 20 25 30 35 40 5 10 15 20 25 30 35 40

> Constraint

1 2

1 2

1 2

1

1 2

Max 2 3

. .

5

3 35

20

, 0

Z x x

s t

x x

x x

x

x x

1 2

1 2

1 2

1 2

1

1 2

Max 2 3

. .

5

3 35

20

3.5

, 0

5 35x

Z x x

s t

x x

x x

x

x

x

x

xx11

xx22

Constraint 1

Constraint 2

Constraint 3

Z

New feasible regionConstraint 418

Equality Constraint

1 2

1 2

1 2

1

1 2

Max 2 3

. .

5

3 35

20

, 0

Z x x

s t

x x

x x

x

x x

1 2

1 2

1 2

1

1 2

Max 2 3

. .

5

3 35

20

, 0

Z x x

s t

x x

x x

x

x x

4040

3535

3030

2525

2020

1515

1010

55

5 10 15 20 25 30 35 40 5 10 15 20 25 30 35 40 xx11

xx22

Constraint 1

Constraint 2

Constraint 3

Z

New feasible region19

The Problem of Finding an Initial Feasible BV

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

Max 2 3 4

. . 30

2 3 60

2 20

, , 0

Z x x x

s t x x x

x x x

x x x

x x x

An LP Model

Standard Form

1 2 3 1 2

1

2

1

1 2 3

1 2 3

1 2 3

1 2 23

Max 2 3 4 0 0

. . 30

2 3 60

2 20

, , , , 0

Z x x x

s t x x x

x x x

x x x

x x

s e

s

e

s ex

Cannot find an initial basic variable that is feasible.

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Example: Solve Using the Big M Method

1 2 1

1 2 1

1 2 1

1 2 1

1 2 1

MAX 3

. . 5 25

3 3 4 2 12

0

, , , 0

x x y z

s t x x y z

x x y z

x x y z

x x y z

1 2 1

1 2 1

1 2 1

1

31 2 1

1 2 1 1 3

MAX 3

. . 5 25

3 3 4 2 12

0

, , , , , 0

x x y z

s t x x y z

x x y z

x x y z

x x y

e

e

s

z s

Write in standard form

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Example: Solve Using the Big M Method

1 2 1

1 2 1

1 2 1

1

31 2 1

1 2 1 1 3

MAX 3

. . 5 25

3 3 4 2 12

0

, , , , , 0

x x y z

s t x x y z

x x y z

x x y z

x x y

e

e

s

z s

1 2 1 1 3 2 3

1 2 1

1 2 1

1 2 1

1 2 1 1 3 2

2

1

3

3

3

MAX 3 0 0

. . 5 25

3 3 4 2 12

0

, , , , , , , 0

x x y z s e Ma Ma

s t x x y z

x x y z

x x y z

x x y z s e a

ae

a

a

s

Adding artificial variables

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Example: Solve Using the Big M Method

1 2 1 1 3 2 3

1 2 1

1 2 1

1 2 1

1 2 1 1 3 2

2

1

3

3

3

MAX 3 0 0

. . 5 25

3 3 4 2 12

0

, , , , , , , 0

x x y z s e Ma Ma

s t x x y z

x x y z

x x y z

x x y z s e a

ae

a

a

s

Put in tableau form

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 -3 1 1 1 0 M 0 M 0

1s 1 0 1 -1 1 5 1 0 0 0 25

2a 2 0 3 -3 -4 2 0 1 0 0 12

3a 3 0 1 -1 -1 -1 0 0 -1 1 0

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Example: Solve Using the Big M Method

Eliminating a2 from row 0 by operations: new Row 0 = old Row 0 -M*old Row 2

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 -3-3M 1+3M 1+4M 1-2M 0 0 0 M -12M

1s 1 0 1 -1 1 5 1 0 0 0 25

2a 2 0 3 -3 -4 2 0 1 0 0 12

3a 3 0 1 -1 -1 -1 0 0 -1 1 0

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 -3 1 1 1 0 M 0 M 0

1s 1 0 1 -1 1 5 1 0 0 0 25

2a 2 0 3 -3 -4 2 0 1 0 0 12

3a 3 0 1 -1 -1 -1 0 0 -1 1 0

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Example: Solve Using the Big M Method

Eliminating a3 from the new row 0 by operations: new Row=old Row-M*old Row 3

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 -3-3M 1+3M 1+4M 1-2M 0 0 0 M -12M

1s 1 0 1 -1 1 5 1 0 0 0 25

2a 2 0 3 -3 -4 2 0 1 0 0 12

3a 3 0 1 -1 -1 -1 0 0 -1 1 0

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 -3-4M 1+4M 1+5M 1-M 0 0 M 0 -12M

1s 1 0 1 -1 1 5 1 0 0 0 25

2a 2 0 3 -3 -4 2 0 1 0 0 12

3a 3 0 1 -1 -1 -1 0 0 -1 1 0

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Example: Solve Using the Big M Method

The initial basic variables are s1=25, a2=12, and a3=0. Now ready to proceed for the simplex algorithm.

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 -3-4M 1+4M 1+5M 1-M 0 0 M 0 -12M

1s 1 0 1 -1 1 5 1 0 0 0 25

2a 2 0 3 -3 -4 2 0 1 0 0 12

3a 3 0 1 -1 -1 -1 0 0 -1 1 0

The initial Tableau

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 -3-4M 1+4M 1+5M 1-M 0 0 M 0 -12M

1s 1 0 1 -1 1 5 1 0 0 0 25 25

2a 2 0 3 -3 -4 2 0 1 0 0 12 4

3a 3 0 1 -1 -1 -1 0 0 -1 1 0 0

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Example: Solve Using the Big M Method

Using EROs change the column of x1 into a unity vector.

Iteration 1

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 -3-4M 1+4M 1+5M 1-M 0 0 M 0 -12M

1s 1 0 1 -1 1 5 1 0 0 0 25 25

2a 2 0 3 -3 -4 2 0 1 0 0 12 4

3a 3 0 1 -1 -1 -1 0 0 -1 1 0 0

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 0 -2 -2+M -2-5M 0 0 -3-3M -3+4M -12M

1s 1 0 0 0 2 7 1 0 1 -1 25 3.57

2a 2 0 0 0 -1 5 0 1 3 -3 12 2.4

1x 3 0 1 -1 -1 -1 0 0 -1 1 0 --

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Example: Solve Using the Big M Method

Using EROs change the column of z into a unity vector.

Iteration 2

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 0 -2 -2+M -2-5M 0 0 -3-3M -3+4M -12M

1s 1 0 0 0 2 7 1 0 1 -1 25 3.57

2a 2 0 0 0 -1 5 0 1 3 -3 12 2.4

1x 3 0 1 -1 -1 -1 0 0 -1 1 0 --

Basic Row/Eq. no. Coefficient of RHS MRT

W 1x 2x 1y z 1s 2a 3e 3a

0 1 0 -2 -12/5 0 0 (2+5M)/5 -9/5 (-21/5)+M 4.8

1s 1 0 0 0 17/5 0 1 -7/5 -16/5 16/5 8.2 2.35 z 2 0 0 0 -1/5 1 0 1/5 3/5 -3/5 2.4 --

1x 3 0 1 -1 -6/5 0 0 1/5 --2/5 2/5 2.4 --

Students to try more iterations. The solution is infeasible. See the attached solution. 28

Special case 1: Alternative Optima

1 2

1 2

1 2

1 2

0.5

. . 2 4

2 3

, 0

Max z x x

s t x x

x x

x x

.

29See Notes on this slide (below) for more information

Special case 1: Alternative Optima1 2

1 2

1 2

1 2

0.5

. . 2 4

2 3

, 0

Max z x x

s t x x

x x

x x

30

Special case 2: Unbounded LPs1 2

1 2

1 2

MAX 2

. . 1

2 2

1, 2 0

z x x

s t x x

x x

x x

01

s1

x1

31See Notes on this slide (below) for more information

Special Case 3: Degeneracy

32

Special Case 5: Degeneracy

Iteration 0

Iteration 1

Iteration 2

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Special Case 5: Degeneracy Degeneracy reveals from practical standpoint that the model has at least one redundant constraint.

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