Operations Research- Lecture 3 - Algebraic Simplex Method_2

8/11/2019 Operations Research- Lecture 3 - Algebraic Simplex Method_2

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Lecture 3 The Simplex Method for LP

Principles of SimplexGeometric Interpretation and Key Concepts

The Algebraic Simplex MethodSlack variablesAugmented Solution, Basic SolutionBasic/Non-basic VariablesCanonical System of EquationsOptimality Test

Learning outcomes:understand the geometric concepts behind theSimplex method; relate the geometric and algebraic interpretationsof the Simplex method; apply the algebraic Simplex method step bystep to solve small LP problems.

Operations Research


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Principles of Simplex

Developed by G. Dantzig in 1947, this is a very efficient

algebraic procedureto solve very large LP problems.

The underlying principles of the Simplex method arebased on the geometric concepts:

constraint boundaries corner-point solutions

corner-point feasible (CPF) solutions

corner-point infeasible solutions

adjacent CPF solutions edges of the feasible region

a CPF solution is optimal if it does not have adjacent CPFsolutions that are better


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3

0

1

2

3

4

5

6

1 2 3 4 5 6

x1

x2

Geometric Concepts of Simplex

(4)

(5)

(2)

(3)

exampleproblemATLAS

(6)0,0

(5)2

(4)1

(3)62

(2)2446:Subject to

(1)45:Maximise

21

2

21

21

21

21

xx

x

xx

xx

xx

xxZ


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4

0

1

2

3

4

5

6

1 2 3 4 5 6

x1

x2

(4)

(5)

(2)

(3)

The Simplex Procedure

1. Initialisation

2. Optimality Test

3. If Optimal, Stop

4. If not Optimal, performiteration: move to betteradjacent CPF solutionand then go to step 2.

(6)0,0

(5)2

(4)1(3)62

(2)2446

:Subject to

(1)45:Maximise

21

2

21

21

21

21

xx

x

xx

xx

xx

xxZ


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Key Concepts of the Simplex Procedure

It focuses only on CPF solutions (assuming there is at least

one). In each iteration, another CPF solution is explored.

If possible, the origin (xi= 0 i) being a CPF solution, isselected as the initial point.

The procedure moves to adjacent CPF solutions by movingalong the edges of the feasible region.

Better adjacent CPF solutions are identified by positive rateof improvementon the objective value Z.

If the rate of improvement along the edges emanating fromthe current CPF solution are all negative, then the currentCPF solution is optimal.


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Importance of the Simplex Method

Simplex is the standard methodfor solving LP problems

Even modern algorithms to solve large IP problems arebased on successive LP relaxations solved using Simplex

Variants of the Simplex method are tailored for specifictypes of problems

The Primal Simplexmethod is simply known as Simplex

The Dual Simplexmethod is efficient for re-optimizationafter the model has changed (change of coefficients, changeof right-hand side values, addition/deletion of constraints)

The Revised Simplexmethod is useful as part of atechnique called Column Generationfor solving very largeLP problems


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Special Cases in the Simplex Method

The form of the Simplex method explained in this lecture

works for a given format of the LP model: Maximisation objective (starting from solution xi=0 i

is straightforward)

Inequalities in the constraints are of type (adding

slack variables makes sense) The right-hand side values biin the constraints are

positive

Non-negativity constraints in all decision variables.

To deal with variations of the above LP model formal,some algebraic manipulationis required to prepare theaugmented model.


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Unbounded Search Space

No leaving basic variable can be identified in Step 2 of agiven iteration. This will happen if there is no bound onthe increase of the entering basic variable.

Multiple Optimal Solutions

The Simplex method stops immediately after the firstoptimal solution is found. Multiple optimal solutionsexist if at least one of the non-basic variables has acoefficient of zero in the objective function equation.

Additional optimal solutions can be obtained byperforming additional iterations of the Simplex method,each time choosing a non-basic variable with zerocoefficient as the entering basic variable.


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The Algebraic Simplex Method

Convert constraints to equations

Introduce 1 slack variable in each functional constraint

to obtain equalities:0and24462446 332121 xxxxxx

0and6262 442121 xxxxxx

0and11 552121 xxxxxx

0and22 6622 xxxx

Example 5.1Apply the algebraic Simplex method to solve the

ATLAS maximisation problem.

(6)0,0

(5)2

(4)1

(3)62

(2)2446:Subject to

(1)45:Maximise

21

2

21

21

21

21

xx

x

xx

xx

xx

xxZ


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The result is an augmented model

(6)6...1for0

(5)2

(4)1

(3)62

(2)2446:Subject to

(1)045:Maximise

62

521

421

321

21

ix

xx

xxx

xxx

xxx

xxZ

i

0

1

2

3

4

5

6

1 2 3 4 5 6

x1

x2

(4)

(5)

(2)

(3)

For a given solution, thevalue of the slack variableindicates if the solution is: On the constraint

boundary On the feasible region On the infeasible region


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Augmented solutions and basic solutions

0

1

2

3

4

5

6

1 2 3 4 5 6

x1

x2

(4)

(5)

(2)

(3)

For a given solution, theaugmented solutionisobtained by calculating

the value of the slackvariables. A basic solutionis an augmented corner-point solution.

(6)6...1for0

(5)2

(4)1

(3)62

(2)2446:Subject to

(1)045:Maximise

62

521

421

321

21

ix

xx

xxx

xxx

xxx

xxZ

i


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Algebraic procedure from the augmented model

From the set of variables in the augmented model, someare designated as basic variablesand the others are

designated as non-basic variablesduring the solutionprocedure.

The non-basic variables are set to zero and the basic

variables are calculated by solving the equations(augmented constraints).

Initialisation

0

so,variablesbasic-nontheasdesignatedareand

21

21

xx

xx

6543

21

,,,:basic

,:basic-non

xxxx

xx


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Solving for the basic variables from the augmented model:

(6)6...1for0

2(5)2

1(4)1

6(3)62

24(2)2446

:Subject to

(1)045:Maximise

662

5521

4421

3321

21

ix

xxx

xxxx

xxxx

xxxx

xxZ

i

The initial basic feasible (BF) solution: (0,0,24,6,1,2)

Optimality Test

The initial BF solution (0,0,24,6,1,2) is not optimal

6543

21

,,,:basic

,:basic-non

xxxx

xx

(1)45 21 xxZ


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Iteration

1. Determine direction of movement.

Rate of improvement 5 (corresponding to x1) is better.

The non-basic variable x1is now called the entering basicvariable.

2(5)2

boundsnoso1(4)1

6so6(3)62

4so624(2)2446

662

115521

114421

113321

xxx

xxxxxx

xxxxxx

xxxxxx

2. Determine when to stop the movement.

Increase x1as much as possible but maintain the other non-basic variables on zero (x2= 0), maintain the satisfaction ofthe equations, and maintain variables non-negative.

(1)45 21 xxZ


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(cont. Step 2)

Then, x1can be increased up to 4 as indicated by theaugmented constraint (2).

The basic variable x3becomes the leaving basic variable.

(5)2

(4)1

(3)62

(2)2446

(1)045

62

521

421

321

21

xx

xxx

xxx

xxx

xxZ

6541

32

,,,:basic

,:basic-non

xxxx

xx3. Solving for the new BF solution

Initial BF solution: (0,0,24,6,1,2)

New BF solution: (4,0,0,?,?,?)


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(cont. Step 3)

The coefficient pattern of the leaving basic variable x3(0,1,0,0,0) is reproduced for the entering variable x1by

means of algebraic manipulation.

(5)2

(4)1

(3)62

(2)2446

(1)045

62

521

421

321

21

xx

xxx

xxx

xxx

xxZ(1)02

6

5

3

232 xxZ

Given the new non-basic variables

x2=0 and x3=0 then:Z=20, x1=4, x4=2, x5=5, x6=2

The new BF solution: (4,0,0,2,5,2)6541

32

,,,:basic

,:basic-non

xxxx

xx

(2)46

1

3

2321 xxx

(3)261

34 432 xxx

(5)262 xx

(4)56

1

3

5532 xxx


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(1)6

5

3

220 32 xxZ

Optimality Test

The new BF solution (4,0,0,2,5,2) is not optimal.

Iteration 2

1. Determine direction of movement.

Rate of improvement 2/3 (corresponding to x2) is better.

The non-basic variable x2is now the entering basic variable.

2. Determine when to stop the movement.

Increase x2as much as possible but maintain the other non-basic variables on zero (x3= 0), maintain the satisfaction ofthe equations, and maintain variables non-negative.


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(cont. Step 2)

2so2(5)2

3so3

55(4)5

6

1

3

5

4

6so

3

42(3)2

6

1

3

4

6so3

24(2)4

6

1

3

2

22662

225532

224432

221321

xxxxx

xxxxxx

xxxxxx

xxxxxx

Then, x2can be increased up to 6/4=1.5 as indicated by theaugmented constraint (3).

The basic variable x4becomes the leaving basic variable.

6521

43

,,,:basic

,:basic-non

xxxx

xx


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(cont. Step 3)

Since the new non-basic variables x3=0 and x4=0 then:Z=21, x

1

=3, x2

=1.5, x5

=5/2, x6

=1/2

The new BF solution: (3,1.5,0,0,2.5,0.5)

Optimality Test

The new BF solution (3,1.5,0,0,2.5,0.5) is optimal becausein the new objective function both x3and x4have negativecoefficient, so increasing any of them would lead to a worseadjacent BF solution.

(1)21

4321 43 xxZ

6521

43

,,,:basic

,:basic-non

xxxx

xx


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Example 5.2Apply the algebraic Simplex method to solve the WENBUmaximisation problem.

Augmented Model

0,0

(4)60075

(3)150025(2)80001:Subject to

(1)1025:Maximise

21

21

1

2

21

xx

xx

x

x

xxZ

0,,,,0

60075)4(

150025(3)

80001)2(

01025)1(

54321

521

41

32

21

xxxxx

xxx

xx

xx

xxZ


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Example 5.2 (cont.)

Initialisation

543

21

,,:basic,:basic-non

xxx

xx

60060075)4(

1500150025(3)

80080001)2(01025)1(

5521

441

332

21

xxxx

xxx

xxx

xxZ

(1)1024 21 xxZ

Optimality Test

New BF solution (0,0,800,1500,600) with Z = 0 is not optimal.


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Example 5.2 (cont.)

Iteration

120560060075)4(

60251500150025(3)onboundno80080001)2(

01025)1(

115521

11441

1332

21

xxxxxx

xxxxx

xxxx

xxZ

x1can increase up to 60 as given by (3)Leaving basic variable: x

4New BF solution: (60,0,?,0,?)531

42

,,:basic

,:basic-non

xxx

xx

60075)4(150025(3)

80001)2(

01025)1(

521

41

32

21

xxx

xx

xx

xxZ

60251

41 xx

150010 42 xxZ

80010 32 xx

3005

17 542 xxx

New BF solution (60,0,800,0,300) Z = 1500

Entering basic variable: x1


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Example 5.2 (cont.)

(1)101500 42 xxZ

Optimality Test

New BF solution (60,0,800,0,300) with Z = 1500 is not optimal.

531

42

,,:basic

,:basic-non

xxx

xx

Iteration 2

7

30073003005

17)4(

onboundno6060251(3)

801080080001)2(

150010)1(

225542

2141

22332

42

xxxxxx

xxxx

xxxxx

xxZ

Entering basic variable: x2

x2can increase up to 300/7 42.86 as given by (4)

Leaving basic variable: x5New BF solution: (?,300/7,?,0,0)321

54

,,:basic

,:basic-non

xxx

xx

b i


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Example 5.2 (cont.)

3005

17)4(

6025

1(3)

80001)2(

150010)1(

552

41

32

42

xxx

xx

xx

xxZ

6025

141 xx

713500

710

3525

54 xxZ

72600

710

3510

543 xxx

7300

71

351

542 xxx

New BF solution (60,300/7,2600/7,0,0) Z = 13500/7

321

54

,,:basic

,:basic-non

xxx

xx

(1)7

10

35

25

7

1350054 xxZ

Optimality Test

New BF solution (60,42.86,371.42,0,0) with Z = 1928.57 is optimal


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Example 5.4For the following LP model and with reference to theSimplex method in algebraic form, write the augmented model, findthe initial (basic feasible) BF solution and determine if this initialsolution is optimal or not. Then, continue the process and work

through the steps of the Simplex method (in algebraic form) to solvethis model. Please label clearly each of the steps and calculationscarried out.

27

)4(0,0

(3)303(2)303:Subject to

(1):Maximise

21

21

21

21

xx

xx

xx

xxZ

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Operations Research- Lecture 3 - Algebraic Simplex Method_2

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