Chapter 4. Solving Linear Programs the Simplex Method

8/14/2019 Chapter 4. Solving Linear Programs the Simplex Method

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Chapter 4. Solving Linear ProgramsChapter 4. Solving Linear ProgramsThe Simplex MethodThe Simplex Method

LP Model: The Augmented (Canonical) FormLP Model: The Augmented (Canonical) Formmax 3 x1 + 5 x2s.t. x1 = 4

2 x2 = 12

3 x1 + 2 x2 = 18x1, x2, >= 0

Variable.Slack :

0X,,X, ,, 0,0

=X++++

X++++ X++++

s.t.

+++max

xxxx

bxaxaxa

bxaxaxa bxaxaxa

xcxcxc

i+n

m+n 1+nn21

mm+nnmn2m21m1

22+nn2n222121

11+nn1n212111

nn2211

=

=

+x3+x4

+x5x3, x4, x5


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x1 + x3 = 4

2 x2 + x4 = 123 x1 + 2 x2 + x5 = 18

Augmented Solution : A solution for the originalvariables that has been augmented by the correspondingvalues of the slack variables.

In the standard form: (3, 2)

In the augmented form: (3, 2, 1, 8, 5 )

How does the augmented corner solution look like?


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Rank of MatrixRank of Matrix A A

.0 0

0 0

...

...

......

...

...

...

and

if onlyandif rank fullhas

mn

2n

n1

m2

22

12

m1

21

11

==

==

=

y y A

x Ax A

a

aa

a

aa

a

aa

A

T


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Basis of MatrixBasis of Matrix A A

(B, N). A

a

aa

a

aa

a

aa

B

A A B A

B A

mm

mm

mnnm

=

=

>

mm

2m

m1

m2

22

12

m1

21

11

:Example

.of columnsselectedarecolumnsitsand

of matrix-subrank -fullais if of basis

aisThen,. andrank fullhavematrixLet

.

...

...

......

...

...

...

,


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Basic Solution of Basic Solution of Ax=b Ax=b

. and 1

is, basistoingcorrespondsolution, basicThe

as decomposed

becanequationthe,of basisselectedaFor

.equationmatrixheConsider t

0

.

==

=+

=

N B

N B

xb B x

Bb Nx Bx

A B

b Ax


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A Basis Solution of the Augmented SystemA Basis Solution of the Augmented System

.0

0 and

8

12

4

. and

1

0

0

0

1

0

0

0

1

),,(Select.

1

0

0

002

102

010

3

0

1

N

1

B

2

1

N

5

4

3

B

543

=

==

=

==

=

=

xb B x

x

x x

x

x

x

x

A A A B A

x1 + x3 = 4

2 x2 + x4 = 123 x1 + 2 x2 + x5 = 18


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s Basic Solution: An augmented corner-point solution.

Basic variables are x 1, x 2 and x 5In the standard form: (4, 6)

In the augmented form: ( 4, 6 , 0, 0, -6) s Basic Feasible Solution: An augmented feasible corner-

point solution.

Basic variables are x 2, x 3 and x 5In the standard form: ( 0, 6)

In the augmented form: ( 0, 6, 4, 0, 6)


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Basic Solution of the Augmented Form

Basic VariableBasic Variable: A basic variable solution has at leastA basic variable solution has at least nn variables, original or variables, original or

slack, whose value are zero. These variables are called non-slack, whose value are zero. These variables are called nonbasic variables, and the others are called basic variables. basic variables, and the others are called basic variables.

( 4, 6, 0, 0, -6)( 0, 6, 4, 0, 6 )

A basic solution can be obtained directly form the augmentedA basic solution can be obtained directly form the augmentedform by settingform by setting nn ( non-basic ) variables equal to zero and( non-basic ) variables equal to zero andsolvingsolving mm equations for equations for mm ( basic ) variables( basic ) variables

The coefficient matrix of theThe coefficient matrix of the mm equations is called basis.equations is called basis. If the values of theIf the values of the mm basic variables are nonnegative, then it is basic variables are nonnegative, then it is

a basic feasible solution corresponding to a feasible corner-a basic feasible solution corresponding to a feasible corner- point in the original form. point in the original form.


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Adjacent Basic Solutions

Two basic solutions are adjacent if all but one of their Two basic solutions are adjacent if all but one of their basic variables are the same. Two adjacent basic basic variables are the same. Two adjacent basicsolutions represent two neighbor corner-point in thesolutions represent two neighbor corner-point in theoriginal form.original form.

( 4, 6, 0, 0, -6 ) and ( 4, 3, 0, 6, 0 )( 4, 6, 0, 0, -6 ) and ( 4, 3, 0, 6, 0 )

Two basic feasible solutions are adjacent if all but oneTwo basic feasible solutions are adjacent if all but oneof their basic variables are the same. Two adjacentof their basic variables are the same. Two adjacent

basic feasible solutions represent two feasible neighbor basic feasible solutions represent two feasible neighbor corner-point in the original form.corner-point in the original form.

( 4, 0, 0, 12, 6 ) and ( 4, 3, 0, 6, 0 )( 4, 0, 0, 12, 6 ) and ( 4, 3, 0, 6, 0 )


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Algebraic Outline of the Simplex Method

Start from a basic feasible solution

Is there a better adjacent basic feasible solution exists?

Move to a better adjacent basic feasible solution

The movement involves converting one non-basic variable into a basicThe movement involves converting one non-basic variable into a basicvariable ( called entering basic variable ) and simultaneously converting avariable ( called entering basic variable ) and simultaneously converting a

basic variable into a non-basic variable ( called the leaving basic variable ), basic variable into a non-basic variable ( called the leaving basic variable ),and then compute the new basic feasible solution.and then compute the new basic feasible solution.

NO

YES


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Z = 3x1 + 5x2

x1 + x3 =42x2 + x4 = 12

3x1 + 2x2 + x5 = 18

1. What is the criterion for selecting the entering basic1. What is the criterion for selecting the entering basicvariable or to show that the current basic feasible solutionvariable or to show that the current basic feasible solutionis optimal?is optimal?

The candidate for the entering basic variables are the n currentnonbasic variables. The one chosen would be changed from nonbasic

into basic, so its value would be increased from zero to some positivenumber and the others would be kept at zero. Since we require animprovement, the rate of change in the objective function must be a

positive one.


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Z = 3x1 + 5 x2

x1 + x3 =4

2x2 + x4 = 123x1 + 2 x2 + x5 = 18

1. What is the criterion for selecting the entering basic1. What is the criterion for selecting the entering basicvariable or to show that the current basic feasible solutionvariable or to show that the current basic feasible solutionis optimal?is optimal?

Express the objective function in terms of only nonbasic variables andalso the basic variables in terms of the nonbasic variables---calledCanonical Form . Then, the coefficient of the entering variable must be

positive. If the coefficients are all nonpositive, then no better adjacent basic feasible solution exists and the current one is optimal.


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2. How to select the leaving basic variable?2. How to select the leaving basic variable?Z = 5Z = 5 X2X2

+ X3+ X3 = 4= 4

22X2X2 + X 4 = 12+ X 4 = 12

22X2X2 + X5 = 18+ X5 = 18

X3X3 = 4= 4

X4 = 12 - 2X4 = 12 - 2 X2X2

X5 = 18 - 2X5 = 18 - 2 X2X2


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2. How to select the leaving basic variable?2. How to select the leaving basic variable?Z = 5Z = 5 X2X2

+ X3+ X3 = 4= 4

22X2X2 + X 4 = 12+ X 4 = 12

22X2X2 + X5 = 18+ X5 = 18

X3X3 = 4= 4

X4X4 = 12 - 2= 12 - 2 X2X2

X5 = 18 - 2X5 = 18 - 2 X2X2

3. How to compute the new basic feasible solution?3. How to compute the new basic feasible solution?

Z = 3X1Z = 3X1 - (5/2) X4 +30- (5/2) X4 +30

X1 + X3 = 4X1 + X3 = 4

X2 + ( 1/2 ) X4 = 6X2 + ( 1/2 ) X4 = 6

X1X1 - X4 + X5 = 6- X4 + X5 = 6


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Iteration 2 for the example:

Z = 3Z = 3 X1X1 + 30+ 30

X1X1 + X3+ X3 = 4= 4X2X2 = 6= 6

33X1X1 + X5 = 6+ X5 = 6

Z =Z = + 30 + 3X1+ 30 + 3X1

X3X3 = 4 -= 4 - X1X1

X2X2 = 6= 6

X5X5 = 6 -= 6 - 3X13X1

Z =Z = - (5/2 ) X4 - X5 + 36- (5/2 ) X4 - X5 + 36

X3 + ( 1/3 ) X4 - ( 1/3 ) X5 = 2X3 + ( 1/3 ) X4 - ( 1/3 ) X5 = 2

X2X2 + ( 1/2 ) X4+ ( 1/2 ) X4 = 6= 6

X1X1 - ( 1/3 ) X4 + ( 1/3 ) X5 =2- ( 1/3 ) X4 + ( 1/3 ) X5 =2


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Summary of the Simplex MethodSummary of the Simplex Method

1.1. Select a basic feasible solutionSelect a basic feasible solution

2.2. Express the basic variables in terms of the nonbasic variables, andExpress the basic variables in terms of the nonbasic variables, andexpress the objective function in terms of only nonbasic variablesexpress the objective function in terms of only nonbasic variables.

If all objective coefficients are non-positive, then stop: the currentIf all objective coefficients are non-positive, then stop: the current basic feasible solution is optimal. basic feasible solution is optimal.Otherwise, select the entering variable such that its coefficient is theOtherwise, select the entering variable such that its coefficient is thelargest, and among basic variables select the leaving variable suchlargest, and among basic variables select the leaving variable suchthat it becomes zero first when increasing the entering variable andthat it becomes zero first when increasing the entering variable andkeeping the other nonbasic variables at zero. If no basic variablekeeping the other nonbasic variables at zero. If no basic variable

becomes zero, then stop: the objective function is unbounded. becomes zero, then stop: the objective function is unbounded.

3.3. Goto Step 2.Goto Step 2.


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The Original Simplex Method: Tabular FormThe Original Simplex Method: Tabular Forms Gaussian Elimination

X1 + X3 = 4X2 - X3 = 1

3X1 + 2X2 - 4X3 = -5

1 0 1 40 2 -1 1

3 2 -4 -5Row 3 - 3 x Row 1:

1 0 1 40 2 -1 10 2 -7 -17

( Row 2 ) / 2:1 0 1 40 1 -1/2 1/20 2 -7 -17


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Row 3 - 2 x (Row 2):1 0 1 40 1 -1/2 1/2

0 0 -6 -18

(Row 3 ) / (-6 ):1 0 1 40 1 -1/2 1/20 0 1 3

Row 2 + ( Row 3 ) / 2:1 0 1 40 1 0 20 0 1 3

Row 1 - Row 3:1 0 0 10 1 0 20 0 1 3


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Variable.Slack :

0X,,X, ,, 0,0

=X++++

X++++

X++++

s.t.

+++max

x

xxx bxaxaxa

bxaxaxa

bxaxaxa

xcxcxc

i+n

m+n 1+nn21

mm+nnmn2m21m1

22+nn2n222121

11+nn1n212111

nn2211

=

=

max 3 x1+ 5 x2

s.t. x1 + x3 = 42 x2 + x4 = 12

3 x1 + 2 x2 + x5 = 18x1, x2, x3, x4, x5 >= 0


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=X++++

X++++

X++++

0= - - -

bxaxaxa

bxaxaxa

bxaxaxaxcxcxc-z

mm+nnm n2m 21m1

22+nn2n222121

11+nn1n212111

nn2211

=

=

z - 3 x1 -5 x2 = 0x1 + x3 = 4

2 x2 + x4 = 123 x1 + 2 x2 + x5 = 18

x1, x2, x3, x4, x5 >= 0


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Step 0. Initial TableStep 0. Initial Table

Basic Right-HandVar.. Side

x x x x m+n 1+n21 ...

baa x

baa x

baa x

cc

mmmmn

n

n

1 0

0 0 0 0

1 1

0 0 0 - - Z

21

222212

112111

21

+

+

+

Basic Variable RHS

Z -3 -5 0 0 0 0

3 1 0 1 0 0 4

4 0 2 0 1 0 12

5 3 2 0 0 1 18


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s Step 1. Optimality TestStep 1. Optimality TestThe current basic feasible solution is optimal if and only if everyThe current basic feasible solution is optimal if and only if everycoefficient in Row 0 is nonnegative. If it is, stop; otherwise, go tocoefficient in Row 0 is nonnegative. If it is, stop; otherwise, go toStep 2.Step 2.

s Step 2. Update the Basic Feasible SolutionStep 2. Update the Basic Feasible Solution

1.1. Determine the variable by selecting the variable with the lowestDetermine the variable by selecting the variable with the lowestcoefficient in Row 0. The corresponding column is called the pivotcoefficient in Row 0. The corresponding column is called the pivotcolumn.column.

Basic VariableBasic Variable RHSRHS

ZZ -3-3 -5-5 00 00 00 00

33 11 00 11 00 00 44

44 00 22 00 11 00 1212

55 33 22 00 00 11 1818


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Step 2. ( Continued )Step 2. ( Continued )

2.2. Determine the leaving variable byDetermine the leaving variable bya)a) picking out each coefficient in the pivot column that is strictly picking out each coefficient in the pivot column that is strictly

positive, ( if no positive coefficient exists, then stop; the positive, ( if no positive coefficient exists, then stop; the problem is unbounded. ) problem is unbounded. )

b) b) dividing each of these coefficients into RHS for the samedividing each of these coefficients into RHS for the samerow,row,

c)c) identifying the row that has the smallest of these ratios, andidentifying the row that has the smallest of these ratios, and

d)d) selecting the basic variable for this row, which is called theselecting the basic variable for this row, which is called the pivot rowpivot row.

Basic Variable RHSBasic Variable RHS RatioRatio

ZZ -3-3 -5-5 00 00 00 00

33 11 00 11 00 00 44

44 00 22 00 11 00 12 6 = 12/212 6 = 12/2

55 33 22 00 00 11 16 9 = 18/216 9 = 18/2


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Step 2. ( Continued )Step 2. ( Continued )

3.3. The intersection of the pivot column and pivot row is the pivotThe intersection of the pivot column and pivot row is the pivotnumber. Replace the leaving basic variable by the entering basicnumber. Replace the leaving basic variable by the entering basicvariable in Column 0. Perform Gaussian elimination for the pivotvariable in Column 0. Perform Gaussian elimination for the pivotnumber. Goto Step 1.number. Goto Step 1.


ZZ -3-3 -5-5 00 00 00 00

33 11 00 11 00 00 44

22 00 22 00 11 00 1212

55 33 22 00 00 11 1818

( Row 2 ) / 2 :( Row 2 ) / 2 : Basic VariableBasic Variable RHSRHSZZ -3-3 -5-5 00 00 00 00

33 11 00 11 00 00 44

44 00 11 0 1/20 1/2 00 66

55 33 22 00 00 11 1818


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Row 3 - 2 x Row 2


Z -3 -5 0 0 0 03 1 0 1 0 0 42 0 1 0 1/2 0 65 3 0 0 -1 1 6

Row 0 + 5 x Row 2

Basic VariableBasic Variable RHSRHSZ -3 0 0 5/2 0 30

3 1 0 1 0 0 42 0 1 0 1/2 0 65 3 0 0 -1 1 6


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Iteration 2:

Basic Variable RHS Ratio

Z -3 0 0 5/2 0 303 1 0 1 0 0 4 42 0 1 0 1/2 0 65 3 0 0 -1 1 6 2

Basic Variable RHS

Z -3 0 0 5/2 0 303 1 0 1 0 0 4

2 0 1 0 1/2 0 6

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

Basic Variable RHSZ 0 0 0 3/2 1 36

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

2 0 1 0 1/2 0 6

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


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Initial BFS for General FormInitial BFS for General Form

Artificial Variable for Equality ConstrainsArtificial Variable for Equality Constrains

maximizemaximize 3x1 + 5x23x1 + 5x2subject tosubject to x1 = 4x1 = 42x2 = 0

maximizemaximize 3x1 + 5x23x1 + 5x2subject tosubject to x1 + x3 = 4x1 + x3 = 4

2x2 + x4 = 122x2 + x4 = 123x1 + 2x2 + x5 = 183x1 + 2x2 + x5 = 18

x1, x2, x3, x4, x5 >= 0x1, x2, x3, x4, x5 >= 0


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maximizemaximize 3x1 + 5x2 - Mx33x1 + 5x2 - Mx3subject tosubject to x1 + x3 = 4x1 + x3 = 4

2x2 + x4 = 122x2 + x4 = 123x1 + 2x2 + x5 = 183x1 + 2x2 + x5 = 18

x1, x2, x3, x4, x5 >= 0x1, x2, x3, x4, x5 >= 0

Basic VariableBasic Variable RHSRHSZZ -3-3 -5-5 100100 00 00 00

33

11

00

11

00

00

44

44 00 22 00 11 00 121255 33 22 00 00 11 1818


ZZ -103-103 -105-105 00 00 00 -400-40033 11 0 10 1 00 00 4444 00 22 00 11 00 12125 35 3 22 00 00 11 1818


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The BigThe Big M M MethodMethod

1.1. Change every > = constraint to = constraint to


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The Two-Phase MethodThe Two-Phase Method

1.1. Change every >= constraint to


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Tie BreakingTie Breaking

Degeneracy: The number of zero-value variables (including slacks) isalways greater than or equal to the number of original variables. In otherwords, the number of positive-value variables is always less than or equalto the number of inequality constraints. When they are not equal, we havea degenerate corner or basic feasible solution.

Left-Right and Top-Bottom Rule

Max 2x1 - 3x2 + 5x3s.t. 8x1 + 5x2 + 4x3


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Simplex Method:Basic Variable RHS Ratio

Z -2 3 -5 0 0 0 04 8 -5 4 1 0 0 24 65 1 -7 0 0 1 0 176 5 3 10 0 0 1 60 6


Z 8 -13/4 0 5/4 0 0 303 2 -5/4 1 1/4 0 0 6

5 1 -7 0 0 1 0 17

6 -15 31/2 0 -5/2 0 1 0 0


Z 301/62 0 0 45/62 0 13/62 30

3 49/62 0 1 3/62 0 5/62 6

5 -179/31 0 0 -35/31 1 14/31 17

2 -30/31 1 0 -5/31 0 2/31 0


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Min - (3/4)x4 +20x5 - (1/2)x6 + 6x7

s.t. x1 + (1/4)x4 - 8x5 - x6 + 9x7 = 0x2 + (1/2)x4 - 12x5 - (1/2)x6 + 3x7 = 0

x3 + x6 = 1x1, x2, x3, x4, x5, x6, x7 >= 0

Cycling:Cycling:

Degeneracy can cause the phenomenon called cycling, theinfinite repetition of a finite sequence of basic feasiblesolutions producing non change in the value of the objectiveand thereby preventing the finite termination of theSimplex method.


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Methods Handling DegeneracyMethods Handling Degeneracys

Random ChoiceRandom Choice Chose the row randomly from among all thoseChose the row randomly from among all thosethat yield the minimum ratiothat yield the minimum ratio

s Perturbation of the RHSPerturbation of the RHS b b ii ++ ii, i=1,,m, i=1,,m

s Lexicographic OrderingLexicographic Orderings Double Least Index RuleDouble Least Index Rule

Least indexed variable among all eligibleLeast indexed variable among all eligibleentering variablesentering variables

Least indexed variable among all rows that yieldLeast indexed variable among all rows that yieldthe minimum ratiothe minimum ratio

Chapter 4. Solving Linear Programs the Simplex Method

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Transcript of Chapter 4. Solving Linear Programs the Simplex Method