Continuous Facility Location Problems · 2015. 12. 14. · Chrystal{Peirce (Sylvester)'s Algorithm z Construct a large circle (with center x ) which covers all the points and which

Continuous Facility Location Problems

PreDoc Course on Operations Research V��ctor Blanco

Sevilla, Noviembre 2015 Universidad de Granada

Introduction

Single-facility Location Problems

Multi-facility Location Problems

New Approaches

An Extension: Location under Refraction

Location Theory

Given a set of demand points, the goal of classical location prob-lems is to �nd one or several points for placing new facilities suchthat they optimize one or several possibly constrained objectivefunctions.

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Is LT interesting for the society?

z An expanding market: It will require the addition of more capacity ata certain geographic point, either in an existent facility or in a newone.

z Introduction of new products or services.z A contracting demand, or changes in the location of the demand: It

may require the shut down and/or relocation of operations.

z Obsolescence of a manufacturing facility due to the appearance of newtechnologies. It means the creation of a new modern plant somewhereelse.

z The pressure of the competence. To increase the level of service, it canforce the company to increase capacity of certain plants or relocatesome of them.

z Change in other resources, like labor conditions or subcontractedcomponents, or change in the political or economic environment in acertain region.

z Mergers and acquisitions. Some facilities may appear as redundants,or bad located with respect to others.

What can/need to be considered in LT?

z Proximity to Customersz Business Climatez Total Costsz Infraestructurez Quality of Laborz Suppliersz Other Facilitiesz Political Risksz Government Barriersz Trading Blocksz Environmental Regulationz Host Communityz Competitive Advantage

Discrete vs. Continuous Location

DISCRETE CONTINUOUS

Facilities To be chosen from an specified finite set To be chosen from a continuousspace.

Costs/Dist Given Part of the decision problem.

Forbidden Regions Filter Potential Facilites (Preprocess) To be modeled.

Input Data Matrix of distances Coordinates of demand points.

Mathematical Programming Framework

We are given:

z A set of demand points (clients) A = fa1; : : : ;ang.z The number of facilities to be located: p.z A potential set of facilities X (= X1 � � � � �Xp).z A function that measures the cost of locating any set of

(x1; : : : ; xp) 2 X : fA(x1; : : : ; xp).

min(x1;:::;xp)2X

fA(x1; : : : ; xp)

X � Rd (CONTINUOUS)

Weber Problem: The Torricelli Point

a

b

c

Pierre Fermat (1601-1665): minx2R2

kx � ak2 + kx � bk2 + kx � ck2


a

b

c

x

Pierre Fermat (1601-1665): minx2R2

kx � ak2 + kx � bk2 + kx � ck2


a

b

c

x

x 0b0

kx � ak+ kx � bk+ kx � ck = kx 0 � xk+ kx 0 � b0k+ kx � ckx is the unique point which x 0 is in the shortest path (straight line)from b0 to c!!! The same for all vertices!

Weber Problem (Simpson, 1705)

Given:

z A set of demand points fa1; : : : ;ang � Rd .z a norm k � k in Rd .

minx2Rd

nXi=1

kx � aik

Weighted Weber problem:

minx2Rd

nXi=1

wikx � aik

for some weights w1; : : : ;wn .

The `1-norm case

minx2Rd

nXi=1

wikx � aik1 =nX

i=1

dXk=1

wi jxk � aik j

Linear Programming:

minx2Rd

nXi=1

wi

dXk=1

zik

s :t :zik � xk � aik ;8i = 1; : : : ;n ; k = 1; : : : ;d ;zik � �xk + aik ;8i = 1; : : : ;n ; k = 1; : : : ;d ;zik � 0:

The `2-norm case: R2

min(x ;y)2R2

f (x ; y) =nX

i=1

wip(x � ai )2 + (y � bi )2

z @f@x =nX

i=1

wi (x � ai )p(x � ai )2 + (y � bi )2

= 0.

z @f@y =nX

i=1

wi (y � bi )p(x � ai )2 + (y � bi )2

= 0.

x =

Xi

wiaik(x � ai ; y � bi )k2X

i

wik(x � ai ; y � bi )k2

y =

Xi

wibik(x � ai ; y � bi )k2X

i

wik(x � ai ; y � bi )k2

Weiszfeld's Algorithm (1937)

Let x0 2 Rd an initial feasible solution.

while xk+1 6= xk do

xk+1 T (xk ) =

8>>>>>><>>>>>>:

nXi=1

wiaikxk � aik

nXi=1

wikxk � aik

if xk 6= ai , 8i ,

ai if xk = ai .

end

Example

a1 = (1; 0), a2 = (0; 1), a3 = (1; 1).wi = 1, i = 1; 2; 3.

1 x0 = (0; 0)! f (x1) = 1:984069.

2 x1 =

(1; 0)

k(1; 0)k+

(0; 1)

k(0; 1)k+

(1; 1)

k(1; 1)k1

k(1; 0)k+

1

k(0; 1)k+

1

k(1; 1)k

= (0:6306019374; 0:6306019374)!

f (x2) = 1:945387.

3 x2 = (0:7057929076; 0:7057929076)! f (x2) = 1:936498.

4 x3 = (0:7394342516; 0:7394342516)! f (x3) = 1:933659.

5 x4 = (0:7577268212; 0:7577268212)! f (x4) = 1:932604.

6 x5 = (0:7686160612; 0:7686160612)! f (x5) = 1:932178.

7 x6 = (0:7754326925; 0:7754326925)! f (x6) = 1:931996.

8 x7 = (0:7798314131; 0:7798314131)! f (x7) = 1:931917.

9 x8 = (0:7827248172; 0:7827248172)! f (x8) = 1:931881.

10 x9 = (0:7846518796; 0:7846518796)! f (x9) = 1:931865.

11 ...

Example

a1

a2a3 xk f (xk )

(0; 0) 1:984069

(0:6306019374; 0:6306019374) 1:936498

(0:7057929076; 0:7057929076) 1:936498

(0:7394342516; 0:7394342516) 1:933659

(0:7577268212; 0:7577268212) 1:932604

(0:7686160612; 0:7686160612) 1:932178

(0:7754326925; 0:7754326925) 1:931996

(0:7798314131; 0:7798314131) 1:931917

(0:7827248172; 0:7827248172) 1:931881

(0:7846518796; 0:7846518796) 1:931865

(0:7859459291; 0:7859459291) 1:931858

(0:7868196823; 0:7868196823) 1:931855

(0:7874118293; 0:7874118293) 1:931853

(0:7878141326; 0:7878141326) 1:931852

(0:7880879197; 0:7880879197) 1:931852

Weiszfeld's algorithm

Weisfeld's converges to the optimal solution (Kuhn 1973; Katz1974) for Euclidean Distances in O(n) (if the optimum is notone of the demand points).

Accelerations:

z x (�) = xk + �(xk+1 � xk ), � 2 [1; dd�1 ] (Chen, 1984).z Update �'s (Drezner, 1992).

For any `� -norm:

T(xk )j =

8>>>>><>>>>>:

nXi=1

wi jxkj � aij j��2aij

kxk � aik��1�

nXi=1

wi jxkj � aij j��2

kxk � aik��1�

if xk 6= ai , 8i ,

aij if xkj = aij 8j .

The algorithm converges if the generated sequence is regular : Thenonregular sequences have measure zero in the solution space(Brimberg & Chen; 1998).

Weiszfeld's and beyond

z For `� -norms (� 2 [1; 2]): (Brimberg & Love; 1992).z For `� -norms (� > 2): (Rodr��guez-Ch��a & Valero; 2013).

aproximate v 2 R+ bypv2 + " .

z `� -norms: Perturbations (Morris & Verdini; 2001):

Approximate

nXi=1

kx � aik� by

"nX

i=1

�(xk � aik )2 + �

� p2

# 1p

:

z Banach Spaces. (Eckhardt, 1980; Puerto & Rodr��guez-Ch��a, 1999,2006)

z Sphere. (Zhang, 2003). minx2R3nX

i=1

wi cos�1(a ti x ).

z Regional Demands.z Demand sets.z Radial distances.

The center problem

\It is required to �nd the least circle which shall containa given system of points in the plane (Sylvester 1857)"

minx2R2

maxi=1;:::;n

kx � aik

a1

a2 a3

a4

a5

\If a circle is drawn through three points, then two cases arise. If thethree points do not lie on the same semicircle, no smaller circle than thisone can be drawn that contain the three points. If the points do lie in thesame semicircle, it is obvious that a circle described upon the linejoining the outer two as a diameter will be smaller than the circlepassing through all three and will contain them all. (Sylvester 1860)"

Chrystal{Peirce (Sylvester)'s Algorithm

z Construct a large circle (with center x ) which covers all the points andwhich passes through ai and aj .

z Find ak such that the angle � = aiakaj is minimum.

if � is obtuse thenx � = x and f � = 1

2kai � aj k.

elseCompute the center of the circle passing through ai ; aj and ak : x .if The triangle formed by those points is not obtuse then

x � = x .else

Drop the point with the obtuse angle and Repeat

a1

a2 a3

a4

a5

x0

�

x1

x� = (1:16666; 1:16666), r = 1:178511

Elzinga{Hearn's Algorithm

z Pick any two demand points ai and aj .z Construct a circle based on the segment connecting ai and aj .

if The circle covers all points thenSTOP

elseAdd a point outside the circle ak .

if The triangle with the three points at its vertices is obtuse thenDrop the obtuse vertex and Repeat.

if The circle passing through the three points covers all points thenSTOP.

if There is a point outside the circle: a` thenAdd it as a fourth point. Discard one of then:

z Keep a` and it farthest point, ai .z Extend the diameter of the current circle through ai de�ning two half planes.z Select the point which is not on the same half plane as ak .

a1

a2 a3

a4

a5

x0

x1

x� = (1:16666; 1:16666), r = 1:178511

Considerations

Euclidean Unweighted Center Problems on the Plane:

z Silvester{Chrystal's needs to construct a initial circle passingthrough two points and covering the rest. Then, for theiterations, it needs to select the point with minimum angle (lawof cosines...). Complexity at most O(n3) (Chrystal; 1985).

z Elzinga{Hearn's need to compute the circle enclosing an acutetriangle (Drezner{Wesolowsky; 1980). Then, �nding a pointoutside the circle. Complexity at least O(n2) (Drezner & Sheda,1987).

z (Drezner, 2011) tested for up to 10000 points and determinedthat \empirically" Erzinga{Hearn's is more e�cient for largerproblems, although similar for small instances.

Improvements

z (Shamos & Hoey, 1975): O(n log(n)) using Voronoi diagrams.z (Megiddo, 1983): O(n) linear programming on the plane.z (Elzinga & Hearn, 1975): Extension to dimension d .z `1-norm:

minx2Rd

t

s :t :t �dX

k=1

zik ;8i = 1; : : : ;d ;

zik � xk � aik ;8i = 1; : : : ;n ; k = 1; : : : ;d ;zik � �xk + aik ;8i = 1; : : : ;n ; k = 1; : : : ;d ;zik � 0:

z Weighted Case: Euclidean Case on the plane O(n) (Dyer, 1986;Megiddo, 1983) and O(3d+2

2

n) for dimension d .

Ordered Median Objective

minx2Rd

nXi=1

kx � aik minx2Rd

maxi=1;:::;n

kx � aik

The median solutions:

z Concern with the spatiale�ciency.

z Remote and low-populationdensity areas arediscriminated in terms ofaccessibility to facilities, ascompared with centrallysituated and high-populationdensity areas.

The center solutions:

z Promote spatial equity.z But, may cause a large

increase in the total distance(losing spatial e�ciency).


z Denote zi (x ) = kx � aik.z For a given x , sort zi : z(1) � z(2) � � � � � z(n).z For center problems we would like to minimize z(1).z For median problems we would like to minimize the sum of z(i)

(which equals the sum of zi ).

z What if we wish to minimize the sum of the k -largest z 's:z(1) + z(2) + � � �+ z(k).

z What if we wish obtain the solution with minimum range?z(1) � z(n).

These are Ordered median functions!!!

OM�(z1; : : : ; zn) =nX

i=1

�iz(i)

Particular choices of � allow to model many problems!!


a1

a2

y

x

OM�(x ) = �1kx � a1k+ �2kx � a2k.OM�(y) = �1ky � a2k+ �2ky � a1k

Ordered Median Location Problems

minx2Rd

OM�(w1kx � a1k; : : : ;wnkx � ank)

Special Interesting Cases:

z Weber Problem � = (1; : : : ; 1).z Center Problem � = (1; 0; : : : ; 0).

z k -centrum Problem � = (kz }| {

1; : : : ; 1; 0; : : : ; 0).

} Introduced by Slater (1978) for discrete facility location.

} For k = 1, Center, for k = n , Median.

} (Rodr��guez-Ch��a, Espejo & Drezner; 2010): Gradient descentmethod for the planar Euclidean case.

} (Ogryczak & Tamir, 2003):

min kt +

nXi=1

qi

s :t :qi � w1kx � a1k � t ; 8i = 1; : : : ;n ;

qi � 0; 8i = 1; : : : ;n ;

x 2 Rd :

(Ogryczak & Tamir; 2003)

Let �k (z ) =

kXi=1

z(i) and h(t) =

nXi=1

(k(t � zi )� + (n � k)(zi � t)+).

z h is piecewise linear and convex, and z(k) is a minimum of h .z h(z(k)) = n

Pki=1

z(i) � kPn

i=1z(i) = n�k (z )� k

Pni=1

zi .

z �k (z ) =1

n

�kPn

i=1zi +mint2R h(t)

�.

z De�ning qi = (zi � t)+ and pi = (zi � t)�:

�k (z ) =min1

n

nX

i=1

(kpi + (n � k)qi + kzi )

!s:t : zi � t = qi � pi ;8i = 1; : : : ; n;

qi ; pi � 0;8i = 1; : : : ; n:

z Since pi = qi � yi + t :

�k (z ) = mint2R

kt +

nXi=1

qi

!

Ordered Median Objectives

min kt +nX

i=1

qi

s :t :qi � w1kx � a1k � t ;8i = 1; : : : ;n ;qi � 0;8i = 1; : : : ;n ;x 2 Rd :

z For polyhedral norms : A Linear Program.z For `1-norm: d + 1 variables and 2dn constraints, for �xed d ,

solved in O(n) (Megiddo, 1984; Zemel, 1984).

Ordered Median Problems

The result in (Ogryczak & Tamir, 2003) extends to monotone �:�1 � �2 � � � � � �n (assuming �n+1 = 0):

��(z ) =

nXi=1

�iz(i)

= (�1 � �2)z(1) + (�2 � �3)(z(1) + z(2)) + (�3 � �4)(z(1) + z(2) + z(3))

+ � � �+ (�n�1 � �n)(z(1) + � � �+ z(n�1)) + �n(z(1) + � � �+ z(n))

=

nXk=1

(�k � �k+1)�k (z )

mint1;:::;tn2R

nXi=1

(�k � �k+1) ktk +

nXi=1

qik

!

s :tqik � zi � tk ;8i ; k = 1; : : : ;n ;qik � 0;8i ; k = 1; : : : ;n :

General Ordered Median Objectives

z a.k.a. Ordered Weighted Averaging in Multicriteria Analysis (Yager,1988).

z Introduced in Location Theory (in network location) by (Nickel &Puerto; 1999) extending median, center and cent-dian criteria.

z For the continuous case:

} (Puerto & Fern�andez; 1995,2000): Geometric characterization ofthe solutions.

} (Puerto, Rodr��guez-Ch��a & Fern�andez-Palac��n; 1997):Semiobnoxious location problems.

} (Rodr��guez-Ch��a, Nickel, Puerto & Fern�andez; 2000): Polyhedralgauges on the plane - Polynomially bounded algorithm based oniterating on sorted bisectors and solving LP's.

} (Drezner & Nickel, 2008): Euclidean planar case (BTST).

} (Espejo, Rodr��guez-Ch��a & Valero, 2009): Approximated gradientdescent method for the convex case for � 2 [1; 2].

} (Blanco, ElHaj, Puerto; 2013): A general SDP-relaxation.

} (Blanco, Puerto, ElHaj; 2014, 2015): SOCP & SDP exactformulations for single/multi-facility problems.

Multifacility Problems

We are given:

z A set of demand points (clients) A = fa1; : : : ;ang.z The number of facilities to be located: p.z A potential set of facilities X (= X1 � � � � �Xp).z A function that measures the cost of locating any set of

(x1; : : : ; xp) 2 X : fA(x1; : : : ; xp).

min(x1;:::;xp)2X

fA(x1; : : : ; xp)

Multiple-allocation Weber Problem

Several facilities (each producing a di�erent product) are to belocated in order to minimize the sum of the weighted distancesbetween all facilities and all users as well as between facilities.


wij : weight between demand point ai and the facility xj .�ij : weight between facilities xj and xj 0 .

minx1;:::;xp2Rd

nXi=1

pXj=1

wij kxj � aik+p�1Xj=1

pXj 0=j+1

�jj 0 kxj � xj 0k


z The multiple-allocation Weber problem is strictly convex if thepoints are not collinear and w ; � � 0.

z For linear distances (block norms): LP formulation (Ward &Wendell, 1985).

z For Euclidean distances:

} Practically e�cient: (Calamai & Conn, 1980).

} Poly-time: (Xue, Rosen & Pardalos, 1996).

Single-allocation Weber Problem

Several facilities (producing the same product) are to be locatedin order to minimize the sum of the weighted distances betweenall users as to their closest facility.


wj : weight for facility xj .

minx1;:::;xp2Rd

nXi=1

pXj=1

wj minjfkxj � aikg

minx1;:::;xp2Rd

nXi=1

pXj=1

zijwj kxj � aik

s :t :Xi

zij = 1;8j = 1; : : : ;p;

zij 2 f0; 1g;8i = 1; : : : ;n ; j = 1; 1; : : : ;p:


z The objective function is neither convex or concave (Cooper, 1967).z Eilon, Watson{Ghandi{Christo�des, 1971) found a 50-point data set

such that for 5 facilities has 61 local minima (the worst deviated 41%from the best).

z With the Euclidean norm, the optimal locations of new facilities are inthe convex hull of existing facilities. (Francis & Cabot, 1972).

z With any norm on the plane, at least one optimal location of each newfacility, belongs to the convex hull of existing facilities (Hansen,Perreur & Thisse, 1980).

z When a mixed norm problem on the plane involves `� -norms (� � 1)one optimal location of each new facility belongs to the octagonal hullof existing facilities (Hansen, Perreur & Thisse, 1980).

z Optimal locations for all the new facilities can be found in the metrichull (intersection of all metric balls containing them) of existingfacilities (Michelot, 1987).

z NP-hard (Megiddo, 1984) { Enumeration of Voronoi partitions of thecustomer set.


z Some heuristics (for the planar Euclidean case)} Iterate on the location{allocation phases until no improvement is

made: (Cooper, 1964).

} Local search (Love & Juel, 1982), (Brimberg, Drezner,Mladenovic & Salhi, 2014).

} Tabu search (Brimberg & Mladenovic, 1996).

} p-median based approach (Hansen, Maldenovic & Taillard, 1998).

z Exact methods (on the plane):

} Euclidean distance: B-&-b partitioning the space (Kuenne &Soland, 1972), descents algorithms (p = 2) (Ostresh; 1973, 1975),separating hyperplanes (Drezner; 1984), B-&-b + covering(Rosing, 1992) ...

} Rectilinear distances: (Love & Morris, 1975)

} Collinear points (Love, 1976).

} DC programming for p = 2 (Chen, Hansen, GJaumard & Tuy,1998).

} BSSS - Column Generation: (Krau, 1997)?

B., Puerto, ElHaj. Revisiting several problems and algo-rithms in continuous location with `� norms. COA2013.

Single-Facility Convex Ordered Median Problems

z A set of demand points fa1; : : : ;ang � Rd .z For each demand point ai a weight !i .z a norm k � k� in Rd (� � 1; � 2 Q): kxk� = (

Pdj=1 jxj j� )1=� .

z A set of weights �1 � � � � � �n � 0

minx2Rd

nXi=1

�i!(i)kx � a(i)k

where !(1)kx � a(1)k� � : : : � !(n)kx � a(n)k� .

The formulation

Let us denote zi = !ikx � aik:minx2Rd

nXi=1

�i!(i)kx � a(i)k = minx2Rd

nXi=1

�iz(i).

Then, if Pn is the set of permutations of f1; : : : ;ng:nX

i=1

�iz(i) = max�2Pn

nXi=1

�iz�(i)

(Hardy, Littlewood & P�olya; 1934):Let � be other permutation of the indices of z , then there exists i , jsuch that z�(i) < z�(j ):�iz�(j ) + �j z�(i) � (�iz�(i) + �j z�(j )) = (�i � �j )(z�(j ) � z�(i)) � 0Then, we may exchange zi by zj ... after a �nite number of steps...

The formulation

How to represent the feasible set Pn?pik =

�1 if zi goes in position k ,0 otherwise.

, so:

nXi=1

pik = 1;8k = 1; : : : ;n ;nX

i=1

pik = 1;8k = 1; : : : ;n ;

The formulation

nXi=1

�i z(i) = max

nXi=1

nXk=1

�k zipik

s:t

nXi=1

pik = 1; 8k = 1; :::; n;

nXk=1

pik = 1; 8i = 1; :::; n;

pik 2 f0; 1g:

For given z 0s, the problem is TU (is an assignment problem), so equivalent to:

nXi=1

�i z(i) = max

nXi=1

nXk=1

�k zipik

s:t

nXi=1

pik = 1; 8k = 1; :::; n;

nXk=1

pik = 1; 8i = 1; :::; n;

pik � 0:

So, its solution coincides with the one of its dual:

min

nXk=1

vk +

nXi=1

wi

s:t vi + wk � �k zi ; 8i ; k = 1; :::; n;

The formulation

Hence, to compute minx2RdnX

i=1

�i!(i)kx � a(i)k, we have:

min

nXk=1

vk +

nXi=1

wi

s :t vi + wk � �k zi ; 8i ; k = 1; :::;n ;zi � !ikx � aik� ; i = 1; :::;n :

The formulation

!ik�x � aik� � �zi () !i

0@ dX

j=1

j�xj � aij j rs1A

sr

� �zsr

i �z1�

i

() !i

0@ dX

j=1

j�xj � aij j rs �zrs(� r�s

r)

i

1A

sr

� �zsr

i

() !rs

i

dXj=1

j�xj � aij j rs �z�r�ss

i � �zi

which holds if and only if 9ui 2 Rd , uij � 0; 8j = 1; :::; d such that

j�xj � aij j rs �z�r�ss

i � uij ; satisfying !rs

i

dXj=1

uij � �zi ;

equivalently j�xj � aij jr � u sij �z r�si ; !rs

i

Pdj=1 uij � �zi :

The formulation

So, minx2RdnX

i=1

�i!(i)kx � a(i)k is reformulated as:

minnX

k=1

vk +nX

i=1

wi

s :tvi + wk � �kzi ; 8i ; k = 1; :::;n ;yij � xj + aij � 0; i = 1; : : : ;n ; j = 1; :::; d ;yij + xj � aij � 0; i = 1; : : : ;n ; j = 1; :::; d ;yrij � u sij z r�si ; i = 1; : : : ;n ; j = 1; :::; d ; ;

!rs

i

dXj=1

uij � zi ; i = 1; : : : ;n ;

uij � 0; i = 1; : : : ;n ; j = 1; : : : ;d :

The `1-norm case

If � = 1:

minnX

k=1

vk +nX

i=1

wi

s :t vi + wk � �kzi ;8i ; k = 1; :::;n ;

zi � !idX

j=1

uij ;i = 1; :::;n ;

xj � aij � uij i = 1; :::;n ; j = 1; : : : ;d ;�xj + aij � uij i = 1; :::;n ; j = 1; : : : ;d :

A General Model

In general for � = rs :

min

nXk=1

vk +

nXi=1

wi

s :tvi + wk � �kzi ; 8i ; k = 1; :::;n ;yij � xj + aij � 0; i = 1; : : : ;n ; j = 1; :::; d ;yij + xj � aij � 0; i = 1; : : : ;n ; j = 1; :::; d ;yrij � u sij z r�si ; i = 1; : : : ;n ; j = 1; :::; d ; ;

!rs

i

dXj=1

uij � zi ; i = 1; : : : ;n ;

uij � 0; i = 1; : : : ;n ; j = 1; : : : ;d :

How to handle with the constraints x r � u s t r�s??

Handling x r � u st r�s

For � = 2 (r = 2; s = 1): x 2 � u t (SECOND ORDER CONE CONSTRAINT)For � = 3 (r = 3; s = 1): x 3 � u t2 (??)Let w =

pux : w2 � ux and x 4 � ut2x = w2t2 ) x 2 � wt .

Actually, if both constraints hold:x 4 � w2t2 � uxt2 ) x 3 � ut2.so the constraint x 3 � u t2 is equivalent to:

w2 � ux ;x 2 � wt ;w � 0:

SECOND ORDER CONE CONSTRAINTS!!

Handling x r � u st r�s

Let � = rs > 1; � 6= 2 be such that r ; s 2 N n f0g and gcd(r ; s) = 1.Let x ; u and t be non negative and satisfying

x r � u s t r�s : (1)

Let k = blog2(r)c and � = bin(s), � = bin(r � s) and

= bin(2k � r) 2 f0; 1gk .Then, if there exists w such that either:

Handling NonLinear Constraints

1. (x ; t ;u ;w) is a solution of the following system, if�i + �i + i = 1, for all 0 < i � k � 1.8<

:w21 � u�0t�0x 0 ;w2i+1 � wiu�i t�i x i ; i = 1; : : : ; k � 2x 2 � wk�1u�k�1t�k�1x k�1 ;

Handling NonLinear Constraints

2. Let c = #fi : �i + �i + i = 3; i = 2; :::; k � 2g, (x ; t; u; w) is a solution of the following system, ifthere exist ij and il(j), j = 1; : : : ; c such that:

1. 0 < i1 < i2 < : : : < ic � k � 2,2. ij < il(j) < ij+1,

3. �ij+ �ij

+ ij= 3, �il(j)

+ �il(j)+ il(j)

= 0 and �h + �h + h = 2 for h = ij + 1; : : : ; il(j)�1:8>>>>>>>>>>>><>>>>>>>>>>>>:

w21

� u�0 t�0 x0 ;

w2i+1

� wi u�i t�i xi ; i 2 f1; : : : ; i1 � 1g

� � � �� for each j = 1; : : : ; c ��

w2�(j)

� ut;

w2�(j)+1

� w�(j)�1x

w2�(j)+2�s

� w�(j)+2(s�1)aij+s

w2�(j)+2�s+1

� w�(j)+2s�1bij+s

o;

s = 1; : : : ; il(j) � ij � 1 and

aij+s+ bij+s

= u�ij+s t

�ij+s x

ij+s ;

w2�(j)+2(il(j)�ij )

� w�(j)+2(il(j)�ij�1)w�(j)+2(il(j)�ij�1)+1

;

nif m � 1 ��(j ) + 2(il(j) � ij � 1)

;

w2�(j)+2(il(j)�ij )+s

� w�(j)+2(il(j)�ij )+s�1u�il(j)+s t

�il(j)+s x

il(j)+s ;

nfor all s = 1; : : : ;

ij+1 � il(j) � 1��

x2 � wmd:

Conversely, if (x ; t; u; w) is a solution of one of those systems then (x ; t; u) verifies xr � us tr�s .

Example

Let us consider � = 10000070001 which in turns means that r = 105 and

s = 70001.

x 100000 � u70001t29999

In this case k = log2(105) = 17 and

z bin(70001) = (1; 0; 0; 0; 1; 1; 1; 0; 1; 0; 0; 0; 1; 0; 0; 0; 1).z bin(29999) = (1; 1; 1; 1; 0; 1; 0; 0; 1; 0; 1; 0; 1; 1; 1; 0; 0).z bin(31072) = (0; 0; 0; 0; 0; 1; 1; 0; 1; 0; 0; 1; 1; 1; 1; 0; 0).

According to the requirement of 2.:

i1 = 5 i2 = 8 i3 = 12il(1) = 7 il(2) = 9 il(3) = 15�(1) = 6 �(2) = 11 �(3) = 16

,

the total number of inequalities is m = 1 + 2� 6 + 9 = 22:

Example

level 1 level 2 level 3 level 4 level 5

w21� ut w2

2� w1t w

23� w2t w

24� w3t w

25� w4t

Bloc i1level 6 level 7 level 8

w26� ut w2

8� w6u w

210

� w8w9w27� w5x w

29� w7x

Bloc i2level 8 level 9 level 10

w210

� w8w9 w211

� ut w213

� w11w12w212

� w10x

level 11 level 12

w214

� w13t w215

� w14x

Bloc i3level 13 level 14 level 15 level 16

w216

� ut w218

� w16t w220

� w18t w222

� w20w21w217

� w15x w219

� w17x w221

� w19x

level 17

x2 � w22u

SDP Representation

a2 � bc ,

�b + c 0 2a0 b + c b � c2a b � c b + c

�� 0; b + c � 0,

�

2ab � c

�

2

� b + c;

For any set of lambda weights satisfying �1 � ::: � �n and� = rs such that r ; s 2 N n f0g, r > s and gcd(r ; s) = 1,the OM location problem can be represented as a semide�niteprogramming problem with n2 + n(2d + 1) linear constraintsand at most 4nd log r positive semide�nite constraints.

Let " > 0 be a prespeci�ed accuracy and (X 0;S0) be a feasibleprimal-dual pair of initial solutions. An optimal primal-dualpair (X ;S) satisfying X � S � " can be obtained in at mostO(� log X

0�S0

" ) iterations and the complexity of each iterationis bounded above by O(��3; �2�2; �2) being � = 3n +2nd(1+log r) and � = p, the dimension of the dual matrix variable Sp .

Single-facility constrained location problems

Consider the restricted problem:

minx2K�Rd

nXi=1

�i!�(i)kx � a�(i)k� : (2)

Assume that any of the following conditions holds:

1 gi (x) are concave for i = 1; : : : ; ` and �P

`

i=1�ir

2gi (x) � 0 for each

dual pair (x ; �) of the problem of minimizing any linear functional ctxon K (Positive De�nite Lagrange Hessian (PDLH)).

2 gi (x) are sos-concave on K for i = 1; : : : ; ` or gi (x) are concave on K andstrictly concave on the boundary of K where they vanish, i.e.@K \ @fx 2 Rd : gi (x) = 0g, for all i = 1; : : : ; `.

3 gi (x) are strictly quasi-concave on K for i = 1; : : : ; `.

Then, there exists a constructive �nite dimension embedding,which only depends on � and gi , i = 1; : : : ; `, such that (2) is asemide�nite problem.

Experiments: Weber

d

2 3 10Time(Ave) Gap(Ave) Time(Ave) Gap(Ave) Time(Ave) Gap(Ave)

� n SDP SOCP SDP SOCP SDP SOCP SDP SOCP SDP SOCP SDP SOCP

1.5

10 0.20 0.06 10�8 < 10�8 0.28 0.06 10�8 < 10�8 0.86 0.09< 10�8 < 10�8

100 1.71 0.27 10�8 < 10�8 3.16 0.40 10�8 < 10�8 10.89 8.77 10�8 < 10�8

500 10.78 15.34 10�8 < 10�8 15.84 22.32 10�8 < 10�8 51.23 1883.8< 10�8 < 10�8

1000 21.22 128.53 10�8 < 10�8 30.67179.72 10�8 < 10�8 103.1724170.12 10�8 < 10�8

5000 103.5057013.83 10�8 < 10�8 178.50 NaN 10�8 NaN 566.64 NaN 10�8 NaN

10000210.22 NaN 10�8 NaN455.05 NaN 10�8 NaN1330.36 NaN 10�8 NaN

2

10 0.06 0.04< 10�8 < 10�8 0.12 0.04< 10�8 < 10�8 0.46 0.04< 10�8 < 10�8

100 0.40 0.05< 10�8 < 10�8 0.99 0.07< 10�8 < 10�8 4.63 0.04< 10�8 < 10�8

500 1.50 0.12< 10�8 < 10�8 5.83 0.14< 10�8 < 10�8 21.63 0.14< 10�8 < 10�8

1000 3.27 0.42< 10�8 < 10�8 11.08 0.48< 10�8 < 10�8 44.04 0.36< 10�8 < 10�8

5000 17.06 19.90< 10�8 < 10�8 58.16 22.09< 10�8 < 10�8 218.82 15.59< 10�8 < 10�8

10000 33.19 137.81< 10�8 < 10�8 118.91162.98< 10�8 < 10�8 455.30 107.06< 10�8 < 10�8

3

10 0.19 0.06 10�8 < 10�8 0.31 0.14 10�8 10�8 0.99 0.54 10�8 5�10�8

100 1.88 1.77 10�8 2�10�6 3.60 3.66 10�8 2�10�5 12.71 43.54 10�8 9�10�8

500 10.82 24.44 10�8 8�10�4 17.87124.97 10�8 8�10�4 57.91 3362.08 10�8 5�10�5

1000 21.73 55.03 10�8 7�10�4 33.99279.74 10�8 2�10�3 118.11 NaN 10�8 NaN

5000 110.87 NaN 10�8 NaN181.17 NaN 10�8 NaN 646.46 NaN 10�8 NaN


3.5

10 0.33 0.12 10�8 < 10�8 0.47 0.16 10�8 2�10�8 1.75 0.43 10�8 < 10�8

100 3.87 3.31 10�8 4�10�6 5.44 5.44 10�8 9�10�8 19.71 124.31 10�8 2�10�7

500 18.62 49.23 10�8 3�10�3 26.99242.89 10�8 3�10�3 92.6416092.68 10�8 10�3

1000 37.06 190.64 10�8 2�10�3 51.50799.74 10�8 2�10�3 192.7721409.00 10�8 7�10�4

5000 280.27 NaN 10�8 NaN304.94 NaN 10�8 NaN1178.17 NaN 10�8 NaN


Experiments: Center

d



1.5

10 0.22 0.05 10�6 < 10�8 0.39 0.21 < 10�8 4�10�5 1.02 0.10 10�4 2�10�2

100 2.13 0.20 < 10�8 < 10�8 8.75 0.53 < 10�8 4�10�5 43.28 9.032�10�5 2�10�2

500 12.59 8.20 < 10�8 < 10�8 63.16 38.33 < 10�8 9�10�5 237.991435.458�10�6 8�10�3

1000 27.11 56.80 < 10�8 < 10�8 115.35315.47 < 10�8 5�10�5 327.04 NaN6�10�6 NaN

5000 150.3426786.61 10�8 < 10�8 357.49 NaN < 10�8 NaN1231.78 NaN5�10�6 NaN

10000371.39 NaN < 10�8 NaN1297.23 NaN < 10�8 NaN2762.51 NaN7�10�6 NaN

2

10 0.11 0.035�10�8 5�10�6 0.17 0.04 10�8 6�10�5 0.44 0.042�10�5 10�2

100 1.34 0.05 10�7 7�10�6 2.09 0.05 10�8 4�10�6 7.68 0.062�10�4 8�10�3

500 8.80 0.098�10�8 6�10�6 13.69 0.10 10�8 10�5 50.25 0.14 10�5 5�10�3

1000 20.85 0.115�10�7 5�10�6 30.27 0.16 10�8 10�5 115.64 0.26 10�5 6�10�3

5000 119.90 0.68 10�6 4�10�6 212.21 0.78 10�8 2�10�4 912.44 1.374�10�3 5�10�3

10000287.13 1.313�10�6 10�5 467.08 1.39 10�8 8�10�5 1510.41 2.488�10�3 5�10�3

3

10 0.23 0.072�10�7 < 10�8 0.37 0.23 10�8 2�10�5 1.08 0.09 10�4 7�10�3

100 2.32 0.28 < 10�8 < 10�8 7.48 0.48 < 10�8 3�10�7 37.66 9.58 10�5 10�2

500 14.47 14.78 10�8 < 10�8 52.27 34.852�10�7 4�10�6 209.461446.296�10�6 5�10�3

1000 28.93 110.24 10�8 < 10�8 119.96297.476�10�7 4�10�5 293.48 NaN2�10�5 NaN

5000 160.9619057.62 10�8 < 10�8 456.11 NaN 10�4 NaN1223.41 NaN2�10�4 NaN

10000434.79 NaN < 10�8 NaN6829.70 NaN3�10�5 NaN2663.15 NaN3�10�4 NaN

3.5

10 0.33 0.06 < 10�8 10�8 0.56 0.076�10�6 5�10�5 1.80 0.175�10�5 10�2

100 4.47 0.38 10�7 < 10�8 13.72 0.834�10�7 2�10�5 68.91 19.602�10�5 6�10�3

500 21.93 19.682�10�8 < 10�8 90.65 58.542�10�7 9�10�6 373.213658.092�10�6 6�10�3

1000 44.82 246.902�10�8 < 10�8 179.48963.052�10�5 6�10�7 603.41 NaN 10�5 NaN

5000 244.04 NaN2�10�8 NaN 551.16 NaN2�10�6 NaN2279.93 NaN2�10�4 NaN

10000510.25 NaN2�10�8 NaN2618.87 NaN2�10�5 NaN4814.26 NaN6�10�5 NaN

Experiments: 0:5-centrum

d



1.5

10 0.29 0.11 10�8< 10�8 0.42 0.10 < 10�8 10�8 1.01 0.132�10�4 2�10�4

100 4.67 0.53 < 10�8< 10�8 8.01 1.18 < 10�8 < 10�8 30.45 17.073�10�8 5�10�8

500 39.92 32.40 10�8< 10�8 49.26 96.39 < 10�8 < 10�8 205.37 2794.702�10�8 10�8

1000 68.11 214.23 10�8< 10�8 109.52 659.33 < 10�8 < 10�8 437.95 53531.382�10�8 < 10�8

5000 476.95 NaN < 10�8 NaN 668.08 NaN < 10�8 NaN 4738.70 NaN2�10�8 NaN

100001242.57 NaN 10�8 NaN2016.01 NaN < 10�8 NaN15348.57 NaN2�10�8 NaN

2

10 0.13 0.07 < 10�8< 10�8 0.18 0.05 < 10�8 < 10�8 0.45 0.058�10�5 4�10�4

100 1.36 0.13 < 10�8< 10�8 2.03 0.10 < 10�8 < 10�8 7.56 0.10 10�8 < 10�8

500 8.40 0.68 < 10�8< 10�8 15.63 0.48 < 10�8 < 10�8 53.87 0.47 < 10�8 < 10�8

1000 22.38 3.71 < 10�8< 10�8 30.90 2.13 < 10�8 < 10�8 108.22 1.99 < 10�8 < 10�8

5000 128.18 397.93 < 10�8< 10�8 195.56 293.61 < 10�8 < 10�8 815.34 228.04 < 10�8 < 10�8

10000 337.173112.30 < 10�8< 10�8 460.982129.80 < 10�8 < 10�8 2373.23 2545.92 < 10�8 < 10�8

3

10 0.30 0.10 10�8< 10�8 0.42 0.11 < 10�8 2�10�8 1.09 0.133�10�5 4�10�7

100 5.65 0.66 10�8< 10�8 10.20 1.46 < 10�8 < 10�8 40.08 21.772�10�7 < 10�8

500 50.36 41.43 10�8< 10�8 72.35 125.78 < 10�8 < 10�8 225.13 2331.944�10�8 < 10�8

1000 100.17 418.97 10�8< 10�8 145.24 774.01 < 10�8 < 10�8 463.74 NaN4�10�8 NaN

5000 582.84 NaN2�10�8 NaN 894.95 NaN < 10�8 NaN 4067.13 NaN2�10�8 NaN

100001715.00 NaN 10�8 NaN2565.21 NaN2�10�8 NaN13649.88 NaN5�10�8 NaN

3.5

10 0.44 0.116�10�8< 10�8 0.60 0.12 < 10�8 < 10�8 1.81 0.262�10�4 2�10�6

100 10.90 1.58 10�8< 10�8 16.97 3.54 < 10�8 < 10�8 60.45 61.054�10�8 10�8

500 80.28 133.692�10�8< 10�8 124.50 347.442�10�8 < 10�8 379.20 6691.804�10�8 < 10�8

1000 171.62 869.752�10�8< 10�8 252.792367.652�10�8 10�8 852.59211506.044�10�8 < 10�8

5000 1033.28 NaN2�10�8 NaN1700.71 NaN2�10�8 NaN 8510.86 NaN6�10�8 NaN


Experiments: Random �'s

d



1.5

10 0.24 0.19 < 10�8 < 10�8 0.40 0.09 < 10�8 < 10�8 1.19 0.143�10�6 10�5

100 4.03 1.33 10�8 < 10�8 6.73 1.88 10�8 < 10�8 22.22 18.46 < 10�8 < 10�8

500 159.42 75.98 10�8 < 10�8 190.77141.67 10�8 < 10�8 380.99 1322.44 < 10�8 3�10�8

10001270.76 NaN 10�7 NaN1730.61 NaN2�10�8 NaN 2379.03 NaN < 10�8 NaN

2

10 0.14 0.043�10�8 10�8 0.18 0.06 10�8 10�8 0.62 0.063�10�5 2�10�6

100 5.11 0.313�10�6 10�8 6.94 0.372�10�8 10�8 17.07 0.293�10�8 < 10�8

500 427.54 12.883�10�6 10�8 443.47 15.293�10�6 < 10�8 1092.70 11.355�10�7 < 10�8

10002079.97 100.513�10�6 10�8 7702.03127.812�10�6 < 10�8 9235.59 81.987�10�7 < 10�8

3

10 0.51 0.105�10�8 2�10�8 0.72 0.16 10�7 6�10�8 1.89 0.284�10�6 3�10�6

100 64.14 3.495�10�6 10�6 58.10 5.03 10�6 5�10�7 152.45 46.06 10�6 2�10�7

500 1532.17 272.562�10�4 10�4 2269.39413.243�10�5 7�10�5 7950.09 4100.409�10�6 2�10�5

10004546.732080.303�10�4 10�4 5678.17 NaN9�10�5 NaN18011.7929734.013�10�5 10�5

3.5

10 0.63 0.16 10�6 10�8 1.48 0.19 10�5 4�10�8 7.05 0.342�10�6 2�10�5

100 33.32 10.01 10�6 4�10�6 302.76 15.039�10�5 3�10�6 596.20 163.092�10�4 7�10�8

500 1555.08 438.345�10�6 3�10�5 2774.06817.914�10�4 4�10�4 8705.0019654.02 �10�4 2�10�4


Range & Non Convex Constraints

d3

� n Time(Ave) Gap(Ave)

2

10 0.46 0.00001623100 9.45 0.00457982500 80.56 0.000302631000 204.96 0.00094492

K = f(x1; x2) 2 R2 : x 21 � 2x 22 � 2x 23 � 0;�2x 21 + 5x 22 + 4x 23 � 0g.

Comparisons

Algorithm

n Problem DN09 ERV09 New

100Weber 0.47 N/A 0.05

k-centrum 0.39 1.76 0.13random 7.13 0.79 0.31

500Weber 7.28 N/A 0.12

k-centrum 3.99 5.63 0.68random 85.56 4.69 12.88

1000Weber 27.69 N/A 0.42

k-centrum 15.04 25.32 3.71random 340.2 17.17 100.51

R2 R3� �

2 3 2 3

n Problem BHP12 New BHP12 New BHP12 New BHP12 New

100Weber 3.55 0.05 5.21 1.77 4.79 0.07 7.32 3.60Center 30.83 0.05 34.07 0.28 48.51 0.05 57.85 0.48

k-centrum 37.58 0.13 34.41 0.66 52.52 0.10 53.87 1.46

500Weber 17.74 0.12 27.46 10.82 25.32 0.14 37.22 17.87Center 305.36 0.09 299.41 14.47 566.29 0.10 600.27 34.85

k-centrum 285.02 0.68 291.8 41.43 452.85 0.48 449.46 72.35

1000Weber 39.82 0.42 58.32 21.73 56.86 0.48 84.06 33.99Center 736.25 0.11 864.93 28.93 1494.76 0.16 1606.89 119.96

k-centrum 666.2 3.71 729.3 100.17 1149.9 2.13 1280.1 145.24

Blanco, Puerto, ElHaj. Continuous multifacility ordered me-dian location problems . EJOR2015.

Multiple-allocation OM Location Problems

Let us consider a set of demand points fa1; a2; : : : ;ang � Rd . Wewant to locate p new facilities X = fx1; x2; : : : ; xpg which minimizethe following expression:

f NI� (x1; x2; : : : ; xp) =

nXi=1

pXj=1

�ijd(i)(xj )+

p�1Xj=1

pXj 0=j+1

�jj 0kxj �xj 0k� ; (3)

where for any x 2 Rd , di (x ) = !ikai � xk� and d(i)(x ) is the i -thelement in the permutation of (d1(x ); : : : ;dn (x )) such thatd(1)(x ) � d(2)(x ) � : : : � d(i)(x ) � : : : � d(n)(x ). In this model, it isassumed that:

�1j � �2j � : : : � �nj � 0; 8j = 1; : : : ;p: (4)�jj 0 � 0 for any j ; j 0 = 1; : : : ;p and, as mention above, d(i)(xj ) is theexpression, which appears at the i -th position in the ordered versionof the list

LNIj := (w1kxj � a1k� ; : : : ;wnkxj � ank� ) for j = 1; 2; : : : ;p: (5)


�NI� := minxff NI� (x ) : x = (x1; : : : ; xp); xj 2 Rd ; 8j = 1; : : : ;pg;

(LOCOMF�NI)Theorem

Assume that � 62 f1;+1g, the demand points in A are notcollinear and for all i = 1; : : : ;n there exists at least onej 2 f1; : : : ;pg such that �ij 6= 0. Then the optimal solution ofProblem (LOCOMF�NI) is unique.

Collocation

A = f(0; 0); (1; 1); (2; 2); (10; 10)g , `2-norm and consider the followingweights wi = 1 for all i = 1; : : : ; 4, �12 = 0 and

�11 = �21 = �31 = �41 = 1;

�12 = 1; �22 = �32 = �42 = 0:

f � = 16p2 which is attained by x �1 2 f(x ; x ) : x 2 [1; 2]g, x �2 = (5; 5).


�NI� = min

nXi=1

pXj=1

vij +

nX`=1

pXj=1

w`j +

p�1Xj=1

pXj 0=j+1

tjj 0 (NIMFOMP�)

s.t. vij + w`j � �`j uij ; 8i ; ` = 1; : : : ; n; j = 1; : : : ; p (6)

yijk � xjk + aik � 0; 8i = 1; : : : ; n; j = 1; : : : ; p; k = 1; : : : ; d (7)

yijk + xjk � aik � 0; 8i = 1; : : : ; n; j = 1; : : : ; p; k = 1; : : : ; d; (8)

yrijk � &sijku

r�sij

; 8i = 1; : : : ; n; j = 1; : : : ; p; k = 1; : : : ; d; (9)

!rsi

dXk=1

&ijk � uij ;8i = 1; : : : ; n; j = 1; : : : ; p (10)

zjj 0k � xjk + xj 0k � 0; 8j ; j0 = 1; : : : ; p; k = 1; : : : ; d; (11)

zjj 0k + xjk � xj 0k � 0; 8j ; j0 = 1; : : : ; p; k = 1; : : : ; d; (12)

z rjj 0k � �

s

jj 0k tr�sjj 0

; 8j ; j 0 = 1; : : : ; p; k = 1; : : : ; d; (13)

�rs

jj 0

dXk=1

�jj 0k � tjj 0 ; 8j ; j0 = 1; : : : ; p; (14)

&ijk � 0; 8i = 1; : : : ; n; j = 1; : : : ; p; k = 1; : : : ; d; (15)

�jj 0k � 0; 8j ; j0 = 1; : : : ; p; k = 1; : : : ; d; (16)

vij 2 R;w`j 2 R; tjj 0 � 0; 8i = 1; : : : ; n; j ; j0 = 1; : : : ; p (17)

yijk � 0; zjj 0k � 0; 8i = 1; : : : ; n; j ; j0 = 1; : : : ; p; k = 1; : : : ; d: (18)

Moreover, (NIMFOMP�) satis�es Slater condition and it can be represented as a

second order cone program with (np + p2)(2d + 1) + p2 linear inequalities and at most

4(p2d + npd) log r second order cone inequalities.

Example

A = f(9:46; 9:36); (8:93; 7:00); (2:20; 1:12); (1:33; 8:89)g (A is subset of size 4of the 50-cities data set from (Eilon, Watson-Gandy & Christo�des), and(randomly generated-) lambda weights:�11 = 147:31, �21 = 24:44, �31 = 24:16, �41 = 10:77, �12 = 119:08),�22 = 0:56, �32 = 0:00, �42 = 0:00, �12 = 0:56 and wi = 1 for alli = 1; : : : ;n .

minx1;x22R2

147:31d(1)(x1) + 24:44d(2)(x1) + 24:16d(3)(x1) + 10:77d(4)(x1) + 119:08d(1)(x2)+

+ 0:56d(2)(x2) + 0:00d(3)(x2) + 0:00d(4)(x2) + 0:56kx1 � x2k2

x �1 = (5:24; 6:41) and x�2 = (5:61; 5:44), with objective value

f � = 1704:55.

Experiments

Gurobi 5.6 executed in PC with an Intel Core i7 processor at 2x 2.40GHz and 4 GB of RAM.(Eilon, Watson-Gandy & Christo�des; 1971) Dataset { 50 cities.Random � and � weights.

�

p 1.5 2 32 2.5095 2.1157 3.74705 12.7794 6.5161 9.813010 29.1873 10.5726 19.545515 49.4854 19.1129 40.450630 148.7449 40.5635 85.5676

Single allocation multifacility location problems

Let us consider a set of demand points fa1; a2; : : : ;ang � Rd . Wewant to locate p new facilities X = fx1; x2; : : : ; xpg.ti : Rpd 7! R, ti (x1; : : : ; xp) := min

jkxj � aik.

Each client ai will be allocated to its closest facility:

~fi (x ) : Rpd 7! Rx = (x1; : : : ; xp) 7! ti := min

j=1::pfkxj � aikg:

�� := minxf

nXi=1

�i~f(i)(x ) : x = (x1; : : : ; xp); xj 2 K; 8j = 1; : : : ;pg;

(LOCOMF)where:

z K � Rd satis�es the Archimedean property. Without loss ofgenerality we shall assume that we know M > 0 such thatkxj k2 �M , for all j = 1; : : : ;p.

z � := rs � 1, r ; s 2 N with gcd(r ; s) = 1.z �` � 0 for all ` = 1; : : : ;n :

Example

A = f (0; 0), (0; 1), (1; 1), (1; 0)g. �ij = 1 8i ; j .

x �1 2 f0g � [0; 1] and x �2 2 f1g � [0; 1]

MISOCO Formulation

min

nX`=1

�`�`

s.t. ti � �` +UBi (1� wi`);8i = 1; : : : ;n ; ` = 1; : : : ;n ;

�` � �`+1;8` = 1; : : : ;n � 1;

uij � ti +UBi (1� zij );8 i = 1; : : : ;n ; j = 1; : : : ; p;

vijk � xjk + aik � 0; 8i = 1; : : : ;n ; j = 1; : : : ; p; k = 1; : : : ; d ;

vijk + xjk � aik � 0; 8i = 1; : : : ;n ; j = 1; : : : ; p; k = 1; ::; d ;

v rijk � �sijku

r�sij ; 8i = 1; : : : ;n ; j = 1; : : : ; p; k = 1; : : : ; d ;

dXk=1

�ijk � uij ;8i = 1; : : : ;n ; j = 1; : : : ; p;

pXj=1

zij = 1;8i = 1; : : : ;n ;

nXi=1

wi` = 1;8` = 1; : : : ;n ;

nXl=1

wil = 1;8i = 1; : : : ;n ;

wi` 2 f0; 1g; 8 i ; ` = 1; : : : ;n ;

zij 2 f0; 1g; 8 i = 1; : : : ;n ; j = 1; : : : ; p;

�`; ti 2 R+; 8i ; ` = 1; : : : ;n ;

vijk ; �ijk ; uij 2 R+; 8i = 1; : : : ;n ; j = 1; : : : ; p; k = 1; : : : ; d ;xj 2 K; 8j = 1; : : : ; p:

MISOCO Formulation

Let x be a feasible solution of the location problem thenthere exists a solution (x ; z ;u ; v ; �;w ; t ; �) of the above prob-lem such that their objective values are equal. Conversely, if(x ; z ;u ; v ; �;w ; t ; �) is a feasible solution for the above problemthen x is a feasible solution for the location problem. Further-more, if K satis�es Slater condition then the feasible region ofthe continuous relaxation of the above problem also satis�esSlater condition and �� = �̂�.

Example

p = 3, � = 75 ,� = (2:25; 1:70; 1:14; 1:11; 1:06; 1:03; 1:01; 1:01; 1:00; 1:00).

x �1 = (6:19; 1:58), x�2 = (5:00; 9:36), and x

�3 = (1:44; 6:55), with

optimal objective value f � = 30:1460

Constrained Case

If any of the following conditions hold:

1 gi (x ) are concave for i = 1; : : : ;m and�Pmi=1 �ir2gi (x ) � 0 for each dual pair (x ; �) of theproblem of minimizing any linear functional ctx on K(Positive De�nite Lagrange Hessian (PDLH)).

2 gi (x ) are sos-concave on K for i = 1; : : : ;m or gi (x ) areconcave on K and strictly concave on the boundary of Kwhere they vanish, i.e. @K \ @fx 2 Rd : gi (x ) = 0g, for alli = 1; : : : ;m .

3 gi (x ) are strictly quasi-concave on K for i = 1; : : : ;m .

Then, there exists a constructive �nite dimension embedding,which only depends on � and gi , i = 1; : : : ;m , such that theproblem is mixed-integer SDP representable.

Experiments

p-median p-center p-25-centrump � CPUTime f � CPUTime f � CPUTime f �

21.5 22.31 150.955 1.03 4.9452 10.08 100.84742 1.13 135.5222 0.28 4.8209 0.38 95.08923 23.68 130.8560 13.51 4.7880 139.03 89.0238

51.5 55.28 78.6074 3.73 2.8831 33.09 53.49952 12.49 72.2369 5.37 2.6610 7.61 49.69323 125.10 68.1791 2.87 2.5094 18.23 46.9844

101.5 5.36 45.0525 2.66 1.6929 68.36 30.71372 2.31 41.6851 5.3 1.6113 17.93 28.90173 4.76 39.7222 55.76 1.5950 225.64 27.5376

151.5 6.70 30.0543 9.44 1.1139 49.92 22.41652 43.91 27.6282 0.62 1.0717 11.26 20.65363 150.99 26.6047 50.08 1.0530 244.59 20.8544

301.5 14.45 9.9488 74.43 1.0080 202.54 9.08062 4.81 8.7963 1.53 0.9192 5.29 8.52163 198.78 8.6995 57.37 0.8508 287.90 8.0016

Blanco, Puerto, Ponce Continuous location under the e�ectof refraction. Submitted. http://arxiv.org/abs/1404.3068

http://arxiv.org/abs/1404.3068

Refraction

Change in direction of propagation of any wave as a result of itstraveling at di�erent speeds at di�erent points along the wavefront.

Applications: Transportation Systems connecting urban and ruralareas; natural barriers or borders, ...

Euclidean Planar Snell's Law

A

H

B

a

b

�b

�a

!A; !B refraction indices.

minx2H

!Aka � xk+ !Bkx � bk

+

!A sin �a = !B sin �b

SP between points separated by a hyperplane

H = fx 2 Rd : �tx = �g.

a 2 HA = fx 2 Rd : �tx � �g (pA = rAsA -norm)

b 2 HB = fx 2 Rd : �tx > �g (pB = rBsB -norm)

Lemma

If 1 < pA; pB < +1, the length dpApB (a ; b) of the shortest weightedpath between a and b is

dpApB (a ; b) = !akx � � akpA + !bkx � � bkpB ;

where x � = (x �1 ; : : : ; x�d )

t , �tx � = � must satisfy the followingconditions:

SP between points separated by a hyperplane

1 For all j such that �j = 0:

!a

�jx �j � aj j

kx � � akpA

�pA�1sg(x �j �aj )+!b

�jx �j � bj j

kx � � bkpB

�pB�1sg(x �j �bj ) = 0:

2 For all i ; j such that �i�j 6= 0.

!a

�jx �i � ai j

kx � � akpA

�pA�1sg(x �i � ai )

�i+ !b

�jx �i � bi j

kx � � bkpB

�pB�1sg(x �i � bi )

�i=

!a

�jx �j � aj j

kx � � akpA

�pA�1 sg(x �j � aj )�j

+ !b

�jx �j � bj j

kx � � bkpB

�pB�1 sg(x �j � bj )�j

:

Generalized Snell's Law

sinpA aj :=j�jaj � �j x �j jka � x �kpA

; j = 1; : : : ;d :

x ��tb��

b

b�ta��

a

a

�t z = �


Corollary (Snell's-like result)

The point x � in H must satisfy:


!a

�jx �j � aj j

kx � � akpA

�pA�1sg(x �j �aj )+!b

�jx �j � bj j

kx � � bkpB

�pB�1sg(x �j �bj ) = 0:

2 For all i ; j ; �i�j 6= 0.

!a

�sinpA aij�i j

�pA�1sg(x �i � ai )

�i+ !b

�sinpB bij�i j

�pB�1sg(x �i � bi )

�i=

!a

�sinpA ajj�j j

�pA�1 sg(x �j � aj )�j

+ !b

�sinpB bjj�j j

�pB�1 sg(x �j � bj )�j

;


Corollary (Snell's Law)

If d = 2, pA = pB = 2 the point x� satis�es:

!A sin �A = !B sin �B ;

where �A and �B are:

1 if �1 � �2, the angles between the vectors a � x � and(��2; �1)t , and b � x � and (�2;��1)t .

2 if �1 > �2, the angles between the vectors a � x � and(�2;��1)t , and b � x � and (��2; �1)t .

Location under Refraction

Given H = fx 2 Rd : �tx = �g, A � HA;B � HB :

f � := infx2Rd

Xa2A

!a dpA;pB (x ; a) +Xb2B

!b dpA;pB (x ; b) (P)

where dpA;pB (x ; y) is the length of the SP between x ; y 2 Rd .

z Parlar, 1994. (Planar and heuristics).z Carrizosa & Rodr��guez-Ch��a, 1997. (Rapid transit lines induced by

networks on the plane).

z Brimberg, Kakhki & Wesolowsky, 2003, 2005 (Planar and `1-`2 norms,bounded regions).

z Zaferanieh, Taghizadeh, Brimberg & Wesolowsky, 2008. (Planar andBSSS based method).

z Fathali & Zaferanieh, 2011. (Planar and block norms).

Our Goal: Exact approach for solving (P) for any d and any `p-norms.

Shortest paths

HB

HA

H

x

a

HB

HA

H

xb

Shortest paths

For x 2 Rd , then we assume that the shortest path length between xand a 2 HA or b 2 HB is:

dpA;pB (x ; a) =

( kx � akpA if x 2 HA,miny2H

ky � akpA + kx � ykpB if x 2 HB ,

and

dpA;pB (x ; b) =

( kx � bkpB if x 2 HB ,miny2H

ky � bkpB + kx � ykpA if x 2 HA.

Theorem

Assume that minfjAj; jB jg > 2. If the points in A or B are notcollinear and pA < +1, pB > 1 then Problem (P) always has aunique optimal solution.

Formulation

minXa2A

!aZa +Xb2B

!bZb

s.t. za � Za � Ma (1� ); 8a 2 A;

�a + ua � Za � Ma ; 8a 2 A;

zb � Zb � Mb ; 8b 2 B ;

�b + ub � Zb � Mb (1� ); 8b 2 B ;

za � kx � akpA ; 8a 2 A;

�a � kx � yakpB ; 8a 2 A;

ua � ka � yakpA ; 8a 2 A;

zb � kx � bkpB ; 8b 2 B ;

�b � kx � ybkpA ; 8b 2 B ;

ub � kb � ybkpB ; 8b 2 B ;

�tx � � � M (1� );

�tx � � � �M;

�tya = �; 8a 2 A;

�tyb = �; 8b 2 B ;

Za ; za ; �a ; ua � 0; 8a 2 A;

Zb ; zb ; �B ; uB � 0; 8b 2 B ;

ya ; yb 2 Rd ; 8a 2 A; b 2 B ;

2 f0; 1g:

Divide et impera

Theorem

Let x � 2 Rd be the optimal solution of (P). Then, x � is thesolution of one of the following two problems:

minx2HA

f � (PA) minx2HB

f � (PB )

Divide et impera

PA

minXa2A

!a za +Xb2B

!b�b +Xb2B

!bub

s.t.

za � kx � akpA ; 8a 2 A;

�b � kx � ybkpA ; 8b 2 B ;

ub � kb � ybkpB ; 8b 2 B ;

�tyb = �; 8b 2 B ;

�tx � �;

za � 0; 8a 2 A;

�b ; ub � 0; 8b 2 B ;

x ; yb 2 Rd :

PB

minXb2B

!a zb +Xa2A

!a�a +Xa2A

!aua

s.t.

zb � kx � bkpB ; 8b 2 B ;

�a � kx � yakpB ; 8a 2 A;

ua � ka � yakpA ;8a 2 A;

�tya = �;8a 2 A;

�tx � �;

zb � 0; 8b 2 B ;

�a ; ua � 0; 8a 2 A;

x ; ya 2 Rd :

NLP Formulation

Lemma

Z � kX �Y kp, for any p = rs with r ; s 2 N n f0g, r > s andgcd(r ; s) = 1, and X ;Y variables in Rd , can be equivalentlywritten as the following set of constraints:

Qk +Xk �Yk � 0; k = 1; : : : ;d ;Qk �Xk +Yk � 0; k = 1; : : : ;d ;

Qrk � �skZ r�s ;k = 1; : : : ;d ;dX

k=1

�k � Z ;

�k � 0;k = 1; : : : ;d :

NLP Formulation

Theorem

Let k � kpi be a `pi -norm with pi = risi > 1, ri ; si 2 N n f0g, andgcd(ri ; si ) = 1 for i 2 fA;Bg. Then, solving (PA) is equivalent to

minXa2A

!a za +Xb2B

!j �j +Xb2B

!bub

s.t. x 2 HA; y 2 H ;

tak � xk + ak � 0;

tak + xk � ak � 0;

vbk + xk � ybk � 0;

vbk � xk + ybk � 0;

gbk � ybk + bk � 0;

gbk + ybk � bk � 0;

trAak� �

sAakz rA�sAa ;

vrAbk� �

sAbk�rA�sAb

;

grBbk�

sBbk

urB�sBb

;

dXk=1

�ak � za ;

dXk=1

�bk � �b ;

dXk=1

bk � ub ;

�ak ; tak ; �bk ; vbk ; bk ; gbk � 0;

za ; �b ; ub � 0;

x ; yb 2 Rd :

SOCP Formulation

Corollary

Problem (PA) can be represented as a semide�nite programmingproblem with:

z jAj(2d + 1) + jB j(4d + 3) + 1 linear constraints, andz at most 4d(jAj log rA + jB j log rA + jB j log rB ) positive

semide�nite constraints.

Constrained Case

TheoremLet K := fx 2 Rd : gj (x) � 0; j = 1; : : : ; lg be a basic closed, compact semialgebraic set withnonempty interior, and consider the restricted problem:

minx2K

Xa2A

!a d(x ; a) +Xb2B

!b d(x ; b): (19)

Assume that K satis�es the Archimedean property and further that any of the followingconditions hold:

1 gi (x) are concave for i = 1; : : : ; l and �P

l

i=1�ir

2gi (x) � 0 for each dual pair

(x ; �) of the problem of minimizing any linear functional ctx on K (Positive De�niteLagrange Hessian (PDLH)).

2 gi (x) are sos-concave on K for i = 1; : : : ; l or gi (x) are concave on K and strictly

concave on the boundary of K where they vanish, i.e. @K \ @fx 2 Rd : gi (x) = 0g, forall i = 1; : : : ; l .

3 gi (x) are strictly quasi-concave on K for i = 1; : : : ; l .

Then, there exists a constructive �nite dimension embedding, which only depends on pA, pBand gi , i = 1; : : : ; l , such that the solution of (19) can be obtained by solving twosemide�nite programming problems.

Hyperplane Endowed with a third norm...

B

A

b

a

y�x �

�A

�B

dt (a; b) =

�ka � bkpi if a; b 2 Hi ; i 2 fA;Bg;

minx ;y2H

kx � akpA + kx � ykpH + ky � bkpB if a 2 HA, b 2 HB ,

(DT)

Snell's like result

Assume that k � kpA ; k � kpB ; k � kpH are `p-norms with 1 < p < +1. Letx �; y� 2 Rd , �tx � = �ty� = �. Then, x � and y� de�ne the shortestweighted path between a and b when traversing the hyperplane isallowed if and only if the following conditions are satis�ed:


!a

hjx�j � aj j

kx� � akpA

ipA�1sg(x�j � aj ) + !H

hjx�j � y

�j j

kx� � y�kpH

ipH�1sg(x�j � y

�j ) = 0;

!b

hjy�j � bj j

ky� � bkpB

ipB�1sg(y�j � bj )� !H

hjx�j � y

�j j

kx� � y�kpH

ipH�1sg(x�j � y

�j ) = 0:

Snell's like result

2 For all i ; j , such that �i�j 6= 0:

!a

hsin aij�i j

ipA�1 sg(x�i � ai )�i

+ !H

hjx�i � y

�i j

kx� � y�kpH

ipH�1 sg(x�i � y�i )�i

=

!a

hsin aj

j�j j

ipA�1 sg(x�j � aj )�j

+ !H

hjx�j � y

�j j

kx� � y�kpH

ipH�1 sg(x�j � y�j )�j

;

and

!a

hsin bij�i j

ipB�1 sg(y�i � bi )�i

� !H

hjx�i � y

�i j

kx� � y�kpH

ipH�1 sg(x�i � y�i )�i

=

!a

hsin bj

j�j j

ipB�1 sg(y�j � bj )�j

� !H

hjx�j � y

�j j

kx� � y�kpH

ipH�1 sg(x�j � y�j )�j

:

Snell's like result

Corollary

If d = 2, pA = pB = pH = 2 and H = f(x1; x2) 2 R2 : x2 = 0g, thepoints x �, y� satisfy one of the following conditions:

1 !a sin �a = !b sin �b = !Hjy�1 j

kx��y�kpHand x � 6= y�, or

2 !a sin �a = !b sin �b and x� = y�,

where �a is the angle between the vectors a � x � and (0;�1) and�b the angle between b � y� and (0; 1).

Location if the hyperplane is endowed with third norm

minx2Rd

Xa2A

!adt (x ; a) +Xb2B

!b dt (x ; b): (PT)

Location if the hyperplane is endowed with third norm

Theorem

Assume that minfjAj; jB jg > 2. If the points in A or B are notcollinear and pB > 1 or pA < +1 then Problem (PT) always hasa unique optimal solution.

Proposition

Let A;B � Rd and H = fx 2 Rd : �tx = �g. Then, ifpA � pB � pH , Problem (PT) reduces to Problem (P).

Theorem

(PT) can also be formulated as a SOCP Problem.

Experiments: Comparisons

SOCP coded in Gurobi 5.6 (PC with an Intel Core i7 processor at 2x2.40 GHz, and 4GB RAM). Barrier convergence tol. QCP: 10�10.(*) Parlar '94 (**) Zaferanieh, Taghizadeh, Brimberg &Wesolowsky, '08. `pH =

14`1

jA [ B j H CPUTime f � CPUTime(�);(��) f (�);(��)

4 (*) y = x 0.037041 26.951942 49.62 26.95195818 (*) y = 1:5x 0.057064 112.350633 35.54 112.35070230 (**) y = 0:5x 0.056049 301.378686 8.25 301.49136130 (**) y = x 0.076050 265.971645 15.31 265.97331530 (**) y = 1:5x 0.074053 257.814199 16.94 257.81424750 (**) y = 0:5x 0.107079 1126.392248 35.00 1127.38231350 (**) y = x 0.116091 966.377027 30.61 966.37761550 (**) y = 1:5x 0.095062 939.487369 29.44 939.487629

With a third norm for the hyperplane:

jA [ B j H CPUTime f � x�

4 (*) y = x 0.0000 20.5307 (0:000000; 0:000001)18 (*) y = 1:5x 0.0000 108.3362 (8:811381; 7:119336)30 (**) y = 0:5x 0.0156 254.7805 (6:000000; 3:000000)30 (**) y = x 0.0000 230.7513 (5:234851; 5:234838)30 (**) y = 1:5x 0.0156 244.4072 (5:153294; 5:102873)50 (**) y = 0:5x 0.0156 917.1736 (11:923664; 5:961832)50 (**) y = x 0.0156 808.2990 (10:000020; 9:999995)50 (**) y = 1:5x 0.0156 892.4482 (10:521522; 9:571467)

Eilon, Watson-Gandy & Christo�des data set

jA [ Bj = 50 H = fy = 1:5xg (jAj = 15) H = fy = xg (jAj = 18) H = fy = 0:5xg (jAj = 39)pA pB pH CPUTime f

� CPUTime f � CPUTime f �

1.5 1 0.0000 230.8447 0.0313 212.9341 0.0156 200.6406

21 0.0158 227.9991 0.0156 202.6576 0.0000 185.95251.5 0.0313 194.1881 0.0313 189.0401 0.0156 182.1283

31 0.0313 223.8203 0.0469 194.1612 0.0156 174.04441.5 0.0156 192.0466 0.0469 180.9279 0.0313 170.31992 0.0156 178.2223 0.0312 174.8964 0.0313 168.5066

1

1 0.0000 219.8367 0.0000 182.1900 0.0000 161.20331.5 0.0313 188.7783 0.0156 168.9589 0.0000 157.21462 0.0156 175.4420 0.0156 163.6797 0.0000 155.61243 0.0156 164.5924 0.0156 159.3740 0.0156 154.3965

1 1

1.5 0.0156 237.4732 0.0156 224.9178 0.0000 236.13002 0.0000 237.3162 0.0156 218.9480 0.0000 235.46893 0.0156 236.3904 0.0156 213.5591 0.0156 234.98071 0.0000 233.7967 0.0156 204.3500 0.0000 234.7300

1.5

12 0.0156 230.8165 0.0313 206.9512 0.0469 200.55143 0.0625 228.5484 0.0938 201.5863 0.0156 200.30681 0.0313 225.9387 0.0156 192.4722 0.0156 200.1428

1.52 0.0313 196.5559 0.0469 193.3584 0.0313 196.48643 0.0469 196.5561 0.0469 188.3989 0.0313 196.30081 0.0156 196.5431 0.0469 179.3396 0.0313 196.1787

2

13 0.0156 225.7539 0.0313 197.2805 0.0156 185.95011 0.0156 223.1421 0.0156 188.1506 0.0156 185.9133

1.53 0.0469 194.1881 0.0469 184.0770 0.0313 182.12711 0.0156 194.1881 0.0313 175.0117 0.0158 182.0955

23 0.0156 180.1096 0.0156 178.0624 0.0156 180.10971 0.0156 180.1097 0.0156 169.7842 0.0156 180.0857

3

1

1

0.0313 221.2011 0.0156 184.9957 0.0313 174.04421.5 0.0313 192.0466 0.0313 171.8455 0.0313 170.31992 0.0156 178.2223 0.0313 166.6027 0.0156 168.50663 0.0312 166.8362 0.0469 162.3214 0.0313 166.8361

Experiments: Larger Instances

jA [ Bj = 5000 jA [ Bj = 10000 jA [ Bj = 50000pA pB pH d = 2 d = 3 d = 5 d = 2 d = 3 d = 5 d = 2 d = 3 d = 5

1.5 1 3.2034 5.4599 10.1520 7.4852 9.2511 19.0804 40.9418 74.9246 115.2941

21 1.5939 2.2502 7.6415 5.1255 8.2040 14.0078 21.8708 25.9411 59.77861.5 3.9692 6.0632 4.5474 8.1728 14.0797 23.8067 55.2635 83.8310 154.2883

31 3.9222 5.1412 6.9852 6.8132 9.4927 20.6114 42.9964 61.4724 116.46651.5 5.4850 10.0950 13.4449 14.3149 21.0337 34.0574 91.9616 106.6900 206.69972 7.9385 9.8603 10.1802 14.2672 17.7362 38.0629 95.3150 135.0647 180.6230

1

1 0.3125 0.6940 9.4607 0.8750 1.6096 6.3288 6.0945 25.7856 89.77721.5 1.2346 2.2502 8.6333 5.6724 4.9605 9.1259 18.8410 32.5503 54.03102 0.8908 1.2188 15.9704 1.9534 2.7346 7.9853 18.8615 17.2053 40.54643 3.4691 2.7346 12.0584 9.5637 6.7195 9.5323 71.7654 70.1868 49.5907

1 1

1.5 18.9396 28.7109 15.6735 37.5415 80.9833 401.8414 596.6057 878.6363 3171.62352 13.7043 24.4318 13.2359 29.2056 68.3894 372.3283 354.3334 721.5562 3166.15113 17.5702 25.1258 3.8570 39.3008 93.4990 415.0733 541.8219 1014.1090 3945.82341 4.9695 11.7517 3.1101 13.7673 26.7468 96.7260 133.7586 632.9736 2492.2830

1.5

12 5.2506 8.2509 4.6457 13.7986 16.0956 37.3793 105.4177 103.2694 273.08663 6.2975 11.9545 4.0473 13.2135 24.9720 57.8267 96.9583 128.9880 326.76601 3.6722 5.5632 4.1409 7.0632 13.1580 31.0345 46.1239 81.3482 118.2435

1.52 12.9546 15.8455 3.7347 23.3466 29.3155 46.6898 138.6629 200.2891 385.13073 13.5232 14.9234 4.5473 22.2837 33.9099 53.9483 171.0538 175.6803 697.50711 12.0022 11.5482 3.9533 21.8464 22.1743 37.0102 111.1779 144.5975 241.2852

2

13 3.5316 7.6883 125.3288 9.8294 11.5794 41.0986 61.4067 62.9410 158.66351 1.7034 3.3288 145.9833 3.5629 7.7041 15.4610 22.8465 38.9976 98.4269

1.53 5.6255 9.3605 105.3967 13.4234 19.0805 45.4697 71.1114 101.3439 269.33031 5.1256 5.4850 137.3159 7.6791 16.5075 24.8255 63.0027 85.4602 134.8291

23 6.6725 9.4387 132.3028 12.1731 20.4003 39.2473 79.9453 121.0863 220.78751 4.6879 5.4607 153.6319 9.4696 14.5639 22.6620 68.1690 63.1358 118.4005

3

1

1

3.7357 6.5511 17.7052 7.8602 10.1575 34.1457 37.1292 48.5630 140.35461.5 7.7665 10.4455 17.7145 15.2061 26.2626 37.2546 84.7931 119.5438 235.11772 7.6569 10.6885 17.4306 16.5483 23.6745 44.5896 99.2611 227.0411 219.49033 9.8843 10.0948 19.1583 19.2838 21.8153 43.0209 129.5420 153.3979 243.4983

`Easy' Extensions

z Norms for each demand points: Each point provided with twonorms k � kAa and k � kBb .

z Critical Reection angle principle: Shortest paths between pointsin the same halfspace are allowed to \traverse and reect".

HA`1

HB

`1

(4; 3)

(8; 7)

HA`1

HB

`1

(4; 3)

(8; 7)

length = 8. length = 6.

Extensions

minXa2A

!a za +Xb2B

!b�b +Xb2B

!bub

s.t. z1a � kx � akpA ; 8a 2 A;

z2a � kx � y1a kpA ; 8a 2 A;

z3a � ky1a � y

2a kpB ; 8a 2 A;

z4a � ky2a � akpA ; 8a 2 A;

�b � kx � ybkpA ; 8b 2 B ;

ub � kyb � bkpB 8b 2 B ; (PEXTA )

za � z1a +Ma (�a � 1); 8a 2 A;

za � z2a + z

3a + z

a4 �Ma�a ; 8a 2 A;

�tx � �;

�ty ja = �; 8j = 1; 2;

�tyb = �; 8a 2 A;

�a 2 f0; 1g; 8a 2 A;

z ka � 0; 8a 2 A; k = 1; 2; 3; 4

�b ; ub � 0; 8b 2 B ;

x ; y1a ; y2a ; yb 2 R

d :

minXa2A

!a za +Xb2B

!b�b +Xb2B

!bub

s.t. z1b � kx � bkpB ; 8b 2 B ;

z2b � kx � y1b kpB ; 8b 2 B ;

z3b � ky1b � y

2b kpA ; 8b 2 B ;

z4b � ky2b � bkpB ; 8b 2 B ;

�a � kx � yakpA ; 8a 2 A;

ua � kya � akpB 8a 2 A; (PEXTB )

zb � z1b +Mb(�b � 1); 8b 2 B ;

zb � z2b + z

3b + z

a4 �Mb�b ; 8b 2 B ;

�tx � �;

�ty j = �; 8j = 1; 2;

�tya = �; 8a 2 A;

�b 2 f0; 1g; 8b 2 B ;

z kb � 0; 8a 2 A; k = 1; 2; 3; 4

�b ; ub � 0; 8b 2 B ;

x ; y1b ; y2b ; ya 2 R

d :

Extensions

`pH =14`1, `pA = `pB = `1

HA

HB

x �

x 0

HA

HB

x �

x 0

f � = 128:00f 0 = 132:9166.

d((6; 1); x �) = 6:3333d1((6; 1); x

�) = 6:66666.

Extensions

pA = pB = 1 and k � kH = 14`1 , H = f(x ; y) : y = �1 xg

�1 N x0T

f 0T

CPUTimeT x�Re

f �Re

CPUTimeRe Improvement

0.5

4 (5; 2:5) 16.75 0.0000 (5; 2:5) 16.75 0.0156 0.00%18 (9; 4:5) 97.75 0.0000 (9; 4:5) 89.50 0.0313 9.22%30 (6; 3) 266.50 0.0000 (6; 3) 251.00 0.0313 6.18%

50 (*) (12; 6) 959.75 0.0000 (11; 5:5) 911.50 100.1884 5.29%50 (**) (5:89; 2:945) 201.55 0.0000 (5:89; 2:945) 189.91 100.1563 6.13%

1

4 (5; 6) 24.17 0.0000 (4; 6) 23.67 0.0156 2.11%18 (9; 8) 132.92 0.0000 (3:3333; 5) 128.00 0.0156 3.84%30 (5; 5) 299.75 0.0000 (2:6667; 4) 269.75 0.0625 11.12%

50 (*) (11; 10) 1076.58 0.0156 (5:3333; 8) 1009.25 100.4547 6.67%50 (**) (3:7133; 5:570) 206.37 0.0156 (3:5; 5:250) 195.52 100.0241 5.55%

1.5

4 (0; 0) 22.50 0.0000 (5; 5) 22.50 0.0156 0.00%18 (8; 8) 123.00 0.0000 (8; 8) 105.50 0.0781 16.59%30 (5; 5) 265.25 0.0000 (5; 5) 251.25 1.2971 5.57%

50 (*) (1; 10) 927.75 0.0000 (1; 10) 873.50 100.0066 6.21%50 (**) (5; 5) 177.52 0.0000 (5:57; 5:57) 170.40 11.6622 4.18%

(*) Zaferanieh, Taghizadeh Kakhki, Brimberg, J. & Wesolowsky, '08.

(**) Eilon, Watson-Gandy & Christo�des, '71.

More?

What about locating other objects?

z Segments (Imai, Lee & yang; 1992), (Agarwal, Efrat, Sharir &Toledo, 1993), (Petersen, 1997).

z Rectilinear trajectories (D��az-Ba~nez & Mesa, 1996).z Rectangular facilities: (Carrizosa, Mu~noz-M�arquez, Puerto;

1998).

z Lines & Hyperplanes: (Schobel, 1999).z Point + Segment (Service and Rapid Transit Line): (Espejo &

Rodr��guez-Ch��a, 2011).

z Polyhedral Structures.z Minisum Hyperspheres: (Korner, 2011).z Competitive/Hu� Location models. (Drezner, 1995).z Any dimension? Any norm? other criteria? other convex bodies?

new approaches?

IntroductionSingle-facility Location ProblemsMulti-facility Location ProblemsNew ApproachesAn Extension: Location under Refraction

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