L4-Lp Standard Form

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Math 407A: Linear Optimization

Lecture 4: LP Standard Form 1

1Author: James Burke, University of Washington
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LPs in Standard Form

Minimization maximization

Linear equations to linear inequalities

Lower and upper bounded variables

Interval variable bounds

Free variable

Two Step Process to Standard Form
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We say that an LP is in standard form

if its matrix representationhas the form
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max cTx

s.t. Ax b

0

x
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max cTx It must be a maximization problem.

s.t. Ax b

0

x
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We say that an LP is in standard form if its matrix representationhas the form


s.t. Ax b Only inequalities of the correct direction.

0

x
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We say that an LP is in standard form if its matrix representationhas the form


s.t. Ax b Only inequalities of the correct direction.

0 x

All variables must be non-negative.
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Every LP can be Transformed to Standard Form
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minimization maximization
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To transform a minimization problem to a maximizationproblem multiply the objective function by 1.
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linear inequalities
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linear inequalities

If an LP has an inequality constraint of the form

ai1x1+ ai2x2+ +ainxn bi,

it can be transformed to one in standard form by multiplying

the inequality through by 1 to get

ai1x1 ai2x2 ainxn bi.
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linear equations
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linear equationsThe linear equation

ai1xi+ +ainxn= bi

can be written as two linear inequalities

ai1x1+ +ainxn bi

and

ai1x1+ +ainxn bi.

The second of these inequalities can be transformed tostandard form by multiplying through by 1.
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variables with lower bounds
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If a variable xihas lower bound liwhich is not zero (li xi),one obtains a non-negative variable wiwith the substitution

xi =wi+li.

In this case, the bound li

xi

is equivalent to the bound0 wi.
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xi =wi+li.

In this case, the bound li

xi


variables with upper bounds

f S
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xi =wi+li.

In this case, the bound li xi


variables with upper bounds

If a variable xihas an upper bound ui (xi ui) one obtains anon-negative variable wiwith the substitution

xi=ui wi.

In this case, the bound xi ui is equivalent to the bound0 wi.

E LP b T f d S d d F
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variables with interval bounds

E LP b T f d S d d F
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variables with interval boundsAn interval bound of the form li xi uican be transformedinto one non-negativity constraint and one linear inequalityconstraint in standard form by making the substitution

xi =wi+li.

E LP b T f d t St d d F
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variables with interval boundsAn interval bound of the form li xi uican be transformedinto one non-negativity constraint and one linear inequalityconstraint in standard form by making the substitution

xi =wi+li.

In this case, the bounds li xi uiare equivalent to theconstraints

0 wi and wi ui li.

E LP b T f d t St d d F
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free variables
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Transformation to Standard Form


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Put the following LP into standard form.

minimize 3x1 x2subject to x

1 + 6x

2 x

3 + x

4 3

7x2 + x4 = 5x3 + x4 2

1 x2, x3 5, 2 x4 2.

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The hardest part of the translation to standard form, or at least

the part most susceptible to error, is the replacement of existingvariables with non-negative variables.

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Reduce errors by doing the transformation in two steps.

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Step 1: Make all of the changes that do not involve a variablesubstitution.

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Step 1: Make all of the changes that do not involve a variablesubstitution.

Step 2: Make all of the variable substitutions.

Must be a maximization problem


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minimize 3x1 x2subject to x1 + 6x2 x3 + x4 3

7x2 + x4 = 5x3 + x4 2

1 x2,

x3 5,

2 x4 2.

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7x2 + x4 = 5x3 + x4 2

1 x2,

x3 5,

2 x4 2.

Minimization =maximization

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7x2 + x4 = 5x3 + x4 2

1 x2,

x3 5,

2 x4 2.

Minimization =maximization

maximize 3x1+ x2.

Inequalities must go the right way.


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q g g y

x1+ 6x2 x3+ x4 3

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q g g y

x1+ 6x2 x3+ x4 3

becomes



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q g g y

x1+ 6x2 x3+ x4 3

becomes

x1 6x2+ x3 x4 3.

Equalities are replaced by 2 inequalities.
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q p y q

7x2+ x4= 5

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7x2+ x4= 5

becomes

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7x2+ x4= 5

becomes

7x2+ x4 5

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7x2+ x4= 5

becomes

7x2+ x4 5

and

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7x2+ x4= 5

becomes

7x2+ x4 5

and

7x2 x4 5

Grouping upper bounds with the linear inequalities.
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The double bound 2 x4 2 indicates that we should group theupper bound with the linear inequalities.

Grouping upper bounds with the linear inequalities.
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The double bound 2 x4 2 indicates that we should group theupper bound with the linear inequalities.

x4 2

Step 1: Transformation to Standard Form
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Combining all of these changes gives the LP

maximize 3x1 + x2subject to x1 6x2 + x3 x4 3

7x2 + x4 5 7x2 x4 5x3 + x4 2

x4 2

1 x2, x3 5, 2 x4.

Step 2: Variable ReplacementTh i bl i f l i b
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The variable x1 is free, so we replace it by

Step 2: Variable ReplacementTh i bl i f l i b
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x1=z+1 z

1 with 0 z+1 , 0 z

1 .

Step 2: Variable ReplacementTh i bl i f l it b
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x1=z+1 z

1 with 0 z+1 , 0 z

1 .

x2 has a non-zero lower bound (1 x2) so we replace it by

Step 2: Variable ReplacementTh i bl i f l it b
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x1=z+1 z

1 with 0 z+1 , 0 z

1 .


z2=x2+ 1 or x2=z2 1 with 0 z2.

Step 2: Variable ReplacementThe variable x is free so we replace it by
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x1=z+1 z

1 with 0 z+1 , 0 z

1 .


z2=x2+ 1 or x2=z2 1 with 0 z2.

x3 is bounded above (x3 5), so we replace it by

Step 2: Variable ReplacementThe variable x is free so we replace it by
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x1=z+1 z

1 with 0 z+1 , 0 z

1 .


z2=x2+ 1 or x2=z2 1 with 0 z2.


z3= 5 x3 or x3= 5 z3 with 0 z3.

Step 2: Variable ReplacementThe variable x1 is free so we replace it by


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x1=z+1 z

1 with 0 z+1 , 0 z

1 .


z2=x2+ 1 or x2=z2 1 with 0 z2.


z3= 5 x3 or x3= 5 z3 with 0 z3.

x4 is bounded below (2 x4), so we replace it by

Step 2: Variable ReplacementThe variable x1 is free so we replace it by
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x1=z+1 z

1 with 0 z+1 , 0 z

1 .


z2=x2+ 1 or x2=z2 1 with 0 z2.


z3= 5 x3 or x3= 5 z3 with 0 z3.

x4 is bounded below (2 x4), so we replace it by

z4=x4+ 2 or x4=z4 2 with 0 z4.

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Substituting x1=z+1 z

1 into

maximize 3x1 + x2subject to x1 6x2 + x3 x4 3

7x2 + x4 5

7x2 x4 5x3 + x4 2

x4 2

1 x2, x3 5, 2 x4.

gives

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maximize 3z+1 + 3z

1 + x2subject to z+1 z

1 6x2 + x3 x4 37x2 + x4 5

7x2 x4 5x3 + x4 2

x4 2

0 z+1 , 0 z

1 , 1 x2, x3 5, 2 x4.

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Substituting x2=z2 1 into

maximize 3z+1 + 3z

1 + x2subject to z+1 z

1 6x2 + x3 x4 37x2 + x4 5

7x2 x4 5x3 + x4 2

x4 2

0 z+1 , 0 z

1 , 1 x2, x3 5, 2 x4.

gives

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maximize 3z+1 + 3z

1 + z2 1subject to z+1 z

1 6z2 + x3 x4 3 6 = 37z2 + x4 5 + 7 = 12

7z2 x4 5 7 = 12x3 + x4 2

x4 2

0 z+1 , 0 z

1 , 0 z2, x3 5, 2 x4.

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maximize 3z+1 + 3z

1 + z2subject to z+1 z

1 6z2 + x3 x4 37z2 + x4 12

7z2 x4 12x3 + x4 2

x4 2

0 z+1 , 0 z

1 , 0 z2, x3 5, 2 x4.

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Substituting x3= 5 z3 into

maximize 3z+1 + 3z


1 6z2 + x3 x4 37z2 + x4 12

7z2 x4 12x3 + x4 2

x4 2

0 z+1 , 0 z

1 , 0 z2, x3 5, 2 x4.

gives

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maximize 3z+1 + 3z


1 6z2 z3 x4 3 5 = 87z2 + x4 12

7z2 x4 12 z3 + x4 2 5 = 3

x4 2

0 z+1 , 0 z

1 , 0 z2, 0 z3, 2 x4.



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Substituting x4=z4 2 into

maximize 3z+1 + 3z


1 6z2 z3 x4 87z2 + x4 12

7z2 x4 12 z3 + x4 3

x4 2

0 z+1 , 0 z

1 , 0 z2, 0 z3, 2 x4.

gives

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maximize 3z+1 + 3z


1 6z2 z3 z4 8 2 = 17z2 + z4 12 + 2 = 1

7z2 z4 12 2 = 1 z3 + z4 3 + 2 =

z4 2 + 2 =

0 z+1 , 0 z

1 , 0 z2, 0 z3, 0 z4.

which is in standard form.

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After making these substitutions, we get the following LP instandard form:

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After making these substitutions, we get the following LP instandard form:

maximize 3z+1 + 3z


1 6z2 z3 z4 107z2 + z4 14

7z2 z4 14 z3 + z4 1

z4 4

0 z+1

, z

1 , z

2, z

3, z

4.

Transformation to Standard Form: Practice


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Transform the following LP to an LP in standard form.

minimize x1 12x2 + 2x3

subject to 5x1 x2 + 3x3 = 152x1 + x2 20x3 300 x2 , 1 x3 4

L4-Lp Standard Form

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