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Lecture 7 Lecture 7 Linearization of Linearization of

a Quadratic Assignment Problem a Quadratic Assignment Problem Indicator VariablesIndicator Variables

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OutlineOutline

Linearization of a Quadratic Assignment Linearization of a Quadratic Assignment Problem Problem

Indicator VariablesIndicator Variables

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Linearization of Linearization of a Quadratic Assignment Problema Quadratic Assignment Problem

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ExampleExample1, if { , , } is assigned to { , },0, . .ij

i a b c j AB Co w

1, ;

1, ;

{0,1}, { , , }, { , , }.

i ij

j ij

ij

j

i

i a b c i A B C

, , , 1, ,

1min ,2

nijkl ij kl

i j k li k j l

c

. .s t

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Linearization of Linearization of the Quadratic Assignment Problem the Quadratic Assignment Problem

non-linear objective function with terms non-linear objective function with terms such as such as ijijklkl

to linearize the non-linear term to linearize the non-linear term ijijklkl

let let ijklijkl = = ij ij klkl

need to ensure that need to ensure that ijklijkl = 1 = 1 ij ij klkl = 1 = 1

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Linearization of Linearization of the Quadratic Assignment Problem the Quadratic Assignment Problem

ijklijkl = 1 = 1 ij ij klkl = 1 = 1 ijklijkl = 1 = 1 ij ij = 1 and = 1 and klkl = 1 = 1

two parts two parts ijklijkl = 1 = 1 ij ij = 1 and = 1 and klkl = 1 = 1

ij ij = 1 and = 1 and klkl = 1 = 1 ijklijkl = 1= 1

trickstricks ijkl ijkl ijij and and ijkl ijkl kl kl

ijij + + klkl 1 + 1 + ijklijkl

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Linearization of Linearization of the Quadratic Assignment Problem the Quadratic Assignment Problem

drawback of the methoddrawback of the method addition of one variable and three constraints addition of one variable and three constraints

for one cross-product for one cross-product nn((nn1)/2 cross products for 1)/2 cross products for nnijij’s ’s

any method to add less variables or less any method to add less variables or less constraintsconstraints

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Linearization of Linearization of the Quadratic Assignment Problem the Quadratic Assignment Problem

three cross-products 2three cross-products 21122+3+311331144 of of

four variables four variables 11, , 22, , 33, , 44

the previous method: 3 new variables and 9 the previous method: 3 new variables and 9 new constraints new constraints

another method with 1 new variable and 3 another method with 1 new variable and 3 new constraintsnew constraints

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Linearization of Linearization of the Quadratic Assignment Problem the Quadratic Assignment Problem

let let ww = 2 = 21122+3+311331144 = = 11(2(222+3+33344))

hope to have: hope to have: 11 = 1 = 1 ww = 2 = 222+3+33344, and , and 11

= 0 = 0 ww = 0 = 0 possible values of possible values of ww {-1, 0, 1, 2, 3, 4, 5} {-1, 0, 1, 2, 3, 4, 5}

11 = 0 = 0 ww = 0: = 0: ww 5 511

how to model how to model 11 = 1 = 1 ww = 2 = 222+3+33344??

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Linearization of Linearization of the Quadratic Assignment Problem the Quadratic Assignment Problem

to model to model 11 = 1 = 1 ww = 2 = 222+3+33344

ww = 2 = 222+3+33344

ww 2 222+3+33344 and and ww 2 222+3+33344

11 = 1 = 1 ww 2 222+3+33344 and and ww 2 222+3+33344 when when 11 = 1: two valid constraints = 1: two valid constraints

ww 2 222+3+33344 and and ww 2 222+3+33344

when when 11 = 0: two redundant constraints = 0: two redundant constraints

ww 2 222+3+33344+5+55511 and and ww 2 222+3+33344+5+51155 mind that when mind that when 11 = 0, = 0, ww = 0 by = 0 by ww 5 511

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Indicator VariablesIndicator Variables

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To Model the Setup CostTo Model the Setup Cost

cost to produce cost to produce xx items, items, xx {0, 1, 2, …} {0, 1, 2, …}cc((xx) = 0 if ) = 0 if xx = 0, and = 0, and cc((xx) = ) = ff + + ax ax if if xx > >

00 suppose there exists suppose there exists = 1 = 1 xx > 0, i.e., > 0, i.e., = 1 = 1

for for xx > 0 and > 0 and = 0 for = 0 for xx = 0 = 0   then then cc((xx) = ) = ff + + ax ax Question: how to construct such a Question: how to construct such a ??

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To Model the Setup CostTo Model the Setup Cost

= 1 x > 0 means that = 1 x > 0 and x > 0 = 1 let max{0, let max{0, xx} < } < MM and and > 0 such that > 0 such that MM is practically infinity is practically infinity

and and < 1 < 1 to model to model = 1 x > 0

xx to model to model x > 0 = 1

xx MM to model to model = 1 x > 0

xx and and xx MM

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To Model the Setup CostTo Model the Setup Cost

cost to produce cost to produce xx items, items, xx {0, 1, 2, …} {0, 1, 2, …} cc((xx) = 0 if ) = 0 if xx = 0, and = 0, and cc((xx) = 10 + 4) = 10 + 4xx, if , if xx > 0 > 0

if there exists if there exists = 1 = 1 xx > 0, then > 0, then cc((xx) = ) = 1010+4+4xx

question: how to define such a question: how to define such a ? ? xx > 0 > 0 = 1 and = 1 and = 1 = 1 xx > 0 > 0

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To Model the Setup CostTo Model the Setup Cost

xx > 0 > 0 = 1 = 1 let let MM be a large positive number be a large positive number xx MM

= 1 = 1 xx > 0 > 0 (( = 1 = 1 xx > 0) > 0) ( (xx = 0 = 0 = 0) = 0) let let be a small positive number be a small positive number xx

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To Model the Setup CostTo Model the Setup Cost

cost to produce cost to produce xx items, items, xx {0, 1, 2, …} {0, 1, 2, …} cc((xx) = 0 if ) = 0 if xx = 0, and = 0, and cc((xx) = 10 + 4) = 10 + 4xx, if , if x > x > 0 0

then then cc((xx) = 10) = 10 + 4 + 4xx, where , where x,x, and and

xx MM (plus some other constraints for (plus some other constraints for xx) )

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Simplifying the Formulation Simplifying the Formulation by Problem Structure by Problem Structure

minimizationminimization min min cc((xx) = 10) = 10 + 4 + 4xx, where , where xx MM (plus (plus

some other constraints for some other constraints for xx) ) question: is it possible to omit question: is it possible to omit xx??

function of the constraint: function of the constraint: xx = 0 = 0 = 0 = 0 for minimization, even without the constraint, for minimization, even without the constraint,

will be forced to 0 at minimum if will be forced to 0 at minimum if xx = 0 = 0

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An Indicator An Indicator to Represent a to Represent a -Constraint-Constraint

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ContextContext

two types of goods, type 1 and type 2 weight of each piece

type 1: 2 ton

type 2: 5 ton

capacity of a truck: 9 ton

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Model Model = 1 = 1 -Constraint-Constraint let be the number of trucks decided by the manager if is set to 1, i.e., the manager has decided to use 1

truck, x1 and x2 satisfy certain relationship = 1 2x1 + 5x2 9 how to model the relationship between and the

constraint 2x1 + 5x2 9? need to have something

giving 2x1 + 5x2 9 if = 1 Having no effect if = 0

2x1 + 5x2 + M M + 9

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Model Model -Constraint -Constraint = 1 = 1

let be the number of trucks if the amount of goods 9 ton, only one vehicle is

needed 2x1 + 5x2 9 = 1 how to model the relationship between and the

constraint 2x1 + 5x2 9? need to have something

when 2x1 + 5x2 9, = 1

Having no effect if 2x1 + 5x2 > 9

9 2x1 5x2 + M 2x1 + 5x2 + M 9+

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Model Model = 1 = 1 -Constraint-Constraint

= the number or trucks used = the number or trucks used = 1 = 1 iffiff the amount of good is no more than 9 the amount of good is no more than 9

tonton combining the two casescombining the two cases

2x1 + 5x2 + M M + 9 and 2x1 + 5x2 + M 9+

consider 2x1 + 5x2 + M 9+

= 0 2x1 + 5x2 + M 9+ = 0 2x1 + 5x2 > 9

2x1 + 5x2 9 = 1

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An Indicator An Indicator to Represent a to Represent a -Constraint-Constraint

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Model Model = 1 = 1 -Constraint-Constraint let be the indicator that the manager orders wasting no

capacity If the manager orders wasting no capacity, the truck is loaded at

least to full capacity = 1 2x1 + 5x2 9 how to model the relationship between and the constraint 2x1 +

5x2 9? need to have something

if = 1, 2x1 + 5x2 9 if = 0, the manager has said nothing, the truck may or may not loaded to

capacity

2x1 + 5x2 + M M + 9

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Model Model -Constraint -Constraint = 1 = 1

now suppose that there is no wasted capacity only if the manager has said so

2x1 + 5x2 9 = 1 how to model the relationship between and

the constraint 2x1 + 5x2 9? need to have something

if 2x1 + 5x2 9, = 1 if 2x1 + 5x2 < 9, there is no constraint on

2x1 + 5x2 ≤ M + 9 -

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Model Model -Constraint -Constraint = 1 = 1

combining the previous two cases combining the previous two cases 2x1 + 5x2 + M M + 9 and 2x1 + 5x2 ≤ M + 9 -

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An Indicator An Indicator to Represent a =-Constraintto Represent a =-Constraint

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=-Constraint=-Constraint

an equality constraint an equality constraint an an - constraint and an - constraint and an --

constraint constraint

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Model Model = 1 = 1 =-Constraint =-Constraint

= 1 2x1 + 5x2 = 9

= 1 (2x1 + 5x2 9 and 2x1 + 5x2 9)

from the previous results

2x1 + 5x2 + M M + 9, and

2x1 + 5x2 + M M + 9 

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Model =-Constraint Model =-Constraint = 1 = 1

2x1 + 5x2 = 9 = 1 = 0 2x1 + 5x2 9

2x1 + 5x2 9: either 2x1 + 5x2 < 9 or 2x1 + 5x2 > 9, but not both

from previous results: 2x1 + 5x2 + M1 9 + , 2x1 + 5x2 + M2 + 9

1 + 2 = + 1

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Model =-Constraint Model =-Constraint = 1 = 1

to model 2x1 + 5x2 = 9 iff = 1 2x1 + 5x2 + M M + 9, 2x1 + 5x2 + M M + 9, 2x1 + 5x2 + M1 9 + , 2x1 + 5x2 M2 + 9, 1 + 2 = + 1.