Lower bounding and tabu search procedures for the frequency

DOI: 10.1007/s10288-004-0057-3

4OR 3: 139–161 (2005) Industry

Lower bounding and tabu search proceduresfor the frequency assignment problemwith polarization constraints

Alain Hertz1, David Schindl2, and Nicolas Zufferey2

1 Department of Mathematics and Industrial Engineering, Ecole Polytechnique de Montréal,H3C 3A7 Montréal, Canada (e-mail: [email protected])

2 Deptarment of Mathematics, Swiss Federal Institute of Technology, Lausanne, Switzerland(e-mail: {david.schindl,nicolas.zufferey}@epfl.ch)

Received: July 2003 / Revised version: October 2004

Abstract. The problem retained for the ROADEF’2001 international challenge wasa Frequency Assignment Problem with polarization constraints (FAPP). This NP-hard problem was proposed by the CELAR of the French Department of Defense,within the context of the CALMA project. Twenty seven competitors took part to thiscontest, and we present in this paper the contribution of our team that allowed us tobe selected as one of the six finalists qualified for the final round of the competition.

There is typically no solution satisfying all constraints of the FAPP. For this rea-son, some electromagnetic compatibility constraints can be progressively relaxed,and the objective is to find a feasible solution with the lowest possible level ofrelaxation. We have developed a procedure that computes a lower bound on thebest possible level of relaxation, as well as two tabu search algorithms for the FAPP,one for the frequency assignment, and one for the polarization assignment.

Key words: Frequency Assignment Problems, tabu search, lower bounds

MSC classification: 90C27, 90C35, 90C59

1 Introduction

The ever-increasing demand for communication, coupled with the limited spectraavailable, have made frequency assignment more and more difficult to accomplisheffectively. Optimization of this process has therefore become a major issue for

All correspondence to: Alain Hertz

4OR A Quarterly Journalof Operations Research

© Springer-Verlag 2005

140 A. Hertz et al.

network administration and deployment, both civil and military. One role of theCELAR (Centre d’Electronique de l’ARmement) is to study new methods for op-timizing the use of the available spectra for the French army. The ROADEF’2001

International Challenge (see Cung and Defaix 2001, or the web site Cung et al.2001 for more details) was devoted to a frequency assignment problem in Hertziantelecommunication networks. This problem was first studied within the context ofthe CALMA project (Combinatorial ALgorithms for Military Applications) of theCELAR, and further enriched so as to take polarizations into account as well as acontrolled relaxation of certain electromagnetic compatibility constraints.

The Hertzian telecommunication network consists of a set of sites in whichtransmission equipments (antennae connected to emitters or receptors) are located.An Hertzian connection joins two geographic sites; it may consist of one or morepaths. A path is a uni-directional radio-electric bond, established between antennaeat distinct sites, with a given frequency and polarization. A frequency resource isdefined as a pair (frequency, polarization) in which the components are respectivelyassociated to the carrying frequency of the transmitted signal and the wave polar-ization. To simplify the model, the polarization is supposed to be a binary variable(e.g. only vertical or horizontal).

Let T denote the set of paths.A set of available resources is defined for each patht ∈ T : it is composed on the one hand of the frequency domain, referred to as Ft ,and on the other hand of the polarization information Pt . Both Ft and Pt are finitesets of integers. The frequency assignment problem with polarization constraints(FAPP) consists in assigning a frequency ft ∈ Ft and a polarization pt ∈ Pt to eachpath t ∈ T , while satisfying the following radio-electric compatibility constraintsthat are of three different types.Let t and t ′ be two paths. The constraints of the first type impose that the absolutevalue of the difference between ft and ft ′ must be equal to or different from a givenvalue: | ft − ft ′ | = εtt ′ or | ft − ft ′ | �= εtt ′ for some given εtt ′ . In particular, ifεtt ′ = 0, then it is imposed that t and t ′ must have the same or a different frequency.The constraints of the second type impose that pt must be equal to or differentfrom pt ′ : pt = pt ′ or pt �= pt ′ . Finally, there are electromagnetic compatibilityconstraints (called CEM constraints) imposing a minimum gap between ft and ft ′ .This gap depends on the polarizations as follows:

| ft − ft ′ | ≥{

γtt ′ if pt = pt ′

δtt ′ if pt �= pt ′

where γtt ′ and δtt ′ are two given positive numbers such that γtt ′ ≥ δtt ′ . The con-straints of the first two types are considered as hard constraints and can thereforenot be violated. There are typically no solutions satisfying all constraints of theFAPP. The CEM constraints are therefore progressively relaxed as follows. Elevenrelaxation levels are defined, level zero corresponding to no relaxation, and level11 to a complete relaxation of a CEM constraint. A CEM constraint is considered as

Lower bounding and tabu search procedures for the FAPP 141

satisfied at level k (0 ≤ k ≤ 11) if

| ft − ft ′ | ≥

γ(k)

tt ′ if pt = pt ′

δ(k)

tt ′ if pt �= pt ′

where γtt ′ = γ(0)

tt ′ ≥ γ(1)

tt ′ ≥ . . . ≥ γ(11)

tt ′ = 0, δtt ′ = δ(0)

tt ′ ≥ δ(1)

tt ′ ≥ . . . ≥ δ(11)

tt ′ = 0,

and γ(k)

tt ′ ≥ δ(k)

tt ′ for all k ∈ {0, 1, . . . , 11}.

A k-feasible solution is an assignment satisfying the constraints of the first twotypes as well as those of the third type at level k. The FAPP is said k-feasible if it hasa k-feasible solution. The FAPP is a hierarchical optimization problem. Its objectivesare, by order of priority:

1. to search for the lowest relaxation level kopt for which a kopt -feasible solutionexists;

2. to minimize the number of CEM constraints not satisfied at level kopt − 1;3. to minimize the number of CEM constraints not satisfied at levels smaller than

kopt − 1.

The organizers of the challenge have generated three sets of benchmark prob-lems. A first set A of 15 instances (problems fapp01 to fapp15) has been used toselect 8 among the 27 competitors, for a second round. On the basis of the resultsobtained on the second set B of 15 instances (problems fapp16 to fapp30), six com-petitors were qualified for the final round that took place at FRANCORO III in Québec(CANADA) in May 2001. The algorithms developed by each of the six finalists werefinally tested on a set X of 10 instances (problems fapp31 to fapp40). Our teamwas classified second in the “Junior” category, while the winner was the team ofM. Vasquez and F. Trousset whose contribution is described in Vasquez (2002).

This paper describes the contribution of our team to the ROADEF’2001 challenge.We first give a mathematical formulation of the FAPP in Sect. 2. We then describe inSect. 3 a procedure that we have developed for the computation of a lower boundon the lowest possible level kopt of relaxation. This lower bounding procedure willhelp us reducing the size of some frequency domains. We then describe in Sect. 4a constructive algorithm that builds an admissible solution for the FAPP. Section 5starts with a description of a basic tabu search, and we then propose two adaptationsof this method, one for the polarization assignment (thus fixing the CEM constraints),and one for the frequency assignment. Experimental results are given in Sect. 6.

It is important to understand that the design of the procedures described in thispaper was carried out in an atmosphere of urgency. As a result, many choices hadto be made early and once for all. That was the case, for example, for the deci-sion of solving the polarization assigment and the frequency assignment problemsseparately. Also, in the context of the competition, each instance had to be solvedwithin a limit of one hour CPU-time on a PC Pentium III (500 Mhz, RAM 128Mo). This certainly explains why the competitors have all prefered local search

142 A. Hertz et al.

techniques to more complex metaheuristics such as population based methods orhybrid algorithms.

2 Mathematical formulation

The basic frequency assignment problem (FAP) is to assign a frequency ft ∈ Ft

to each path t ∈ T while satisfying interference constraints. More precisely, foreach pair (t, t ′) of paths, it is required that the value | ft − ft ′ | does not belongto a given set Stt ′ . In most models, frequencies are integer numbers and Stt ′ is theset {0, . . . , stt ′ − 1} of all positive integers that are strictly smaller than a givennumber stt ′ . In such a case, an interference constraint can simply be written as| ft − ft ′ |≥ stt ′ . It is typically impossible to satisfy all interference constraintsof a FAP . One can therefore define a penalty if | ft − ft ′ |/∈ Stt ′ . The MinimumInterference Frequency Assignment Problem (MI-FAP) is to determine a frequencyassignment with minimum total penalty. A detailed overview of solution methodsfor the FAP and the MI-FAP can be found in Aardal et al. (2003).

In the FAPP, each path t is associated not only with a frequency domain Ft ,but also with a polarization domain Pt . A solution is then an assignment of both afrequency and a polarization to each path. We now give a mathematical formulationof the FAPP. Remember first that there are only two possible values for the polariza-tion of a path. We fix these two values to 1 and -1. This means that the polarizationdomain Pt of a path t ∈ T is either equal to {−1}, {1} or {−1, 1}. Each frequencydomain Ft is a finite set of integers. The constraints linking two paths t and t ′ areof one of the following types:

| ft − ft ′ | ≥ | pt + pt ′ |2

γ(k)

tt ′ + | pt − pt ′ |2

δ(k)

tt ′ CEM(k)

| ft − ft ′ | = εtt ′ C2

| ft − ft ′ | �= εtt ′ C3

ft = ft ′ C4

ft �= ft ′ C5pt = pt ′ C6pt �= pt ′ C7

where CEM(k) represents a CEM constraint at relaxation level k. We define asolution s of the FAPP as a set of | T | couples (ft , pt ) ∈ Ft ×Pt . A solution is saidadmissible if it satisfies all constraints C2, C3, . . . , C7, while it is said k-feasible ifit is admissible and satisfies all CEM(k) constraints. To each admissible solution s,we associate a relaxation level k(s) corresponding to the smallest positive integersuch as s is k(s)-feasible. Remember that γ

(11)

tt ′ = δ(11)

tt ′ = 0, which means thatk(s) ≤ 11.


Now let � denote the total number of pairs (t, t ′) involved in a CEM constraint

(i.e., � =∣∣∣{(t, t ′) : γ

(0)

tt ′ > 0}∣∣∣. For each positive integer k ∈ {0, 1, . . . , k(s) −

1}, let V (k)(s) denote the set of CEM(k) constraints that s violates. Noticethat if a constraint belongs to a set V (k)(s), then it also belongs to the setsV (0)(s), . . . , V (k−1)(s), The objective of the FAPP, as defined by the organizersof the challenge, is to minimize

f (s) = 10 · k(s) · �2 + 10 ·∣∣∣V (k(s)−1)(s)

∣∣∣ · � +∑

i<k(s)−1

| V (i)(s) | .

Let fa(s), fb(s) and fc(s) denote the three terms of the above function f (i.e.,f (s) = fa(s) + fb(s) + fc(s)). The following property can easily be proved (seeZufferey 2002, for more details), and this shows that f respects the hierarchy ofthe objectives mentioned in the introduction.

Properties. Let s and s′ be two admissible solutions. Then

(a) fa(s) < fa(s′) ⇒ f (s) < f (s′)

(b) fa(s) = fa(s′), fb(s) < fb(s

′) ⇒ f (s) < f (s′)(c) fa(s) = fa(s

′), fb(s) = fb(s′), fc(s) < fc(s

′) ⇒ f (s) < f (s′)

3 Preprocessing

We describe in this section a preprocessing procedure which reduces the size of thefrequency domains of the paths, and computes a lower bound kinf on the best pos-sible relaxation level kopt . The procedure is a specialization of the arc-consistencyapproach for constraint programming (see for example Bessière 1994, and Mohrand Henderson 1986). Such techniques have been widely applied to the MI-FAP

(Kolen et al. 1994; Aardal et al. 2002).A frequency ft for a path t is defined as k-useless if the FAPP has no k-feasible

solution in which t has frequency ft . It would be interesting to eliminate all kopt -useless frequencies for each path t . However, kopt is not known, and we thereforecompute a lower bound kinf on kopt , and eliminate kinf -useless frequencies. Thelower bound kinf is initially set equal to 0, and is increased by one unit each time wecan prove that the FAPP is not kinf -feasible. In order to prove that the FAPP is not kinf -feasible, we associate to each path t a frequency domain, denoted Domt , whichis initially set equal to Ft , and which is then reduced by successively removingkinf -useless frequencies. If a domain Domt of a path t becomes empty, this provesthat no kinf -feasible solution exists. Otherwise, the FAPP is possibly kinf -feasible,and the reduced sets Domt may contain frequencies that are not kinf -useless.

Observe that two paths t and t ′ with Ft = Ft ′ will probably not have the samereduced sets Domt and Domt ′ . More importantly, it may happen that kinf < kopt .In such a case, we cannot guarantee that the FAPP has a kopt -feasible solution if

144 A. Hertz et al.

we restrict the frequency domains to the sets Domt instead of Ft . Indeed, kinf -useless frequencies are not necessarily kopt -useless. However, since the numberof admissible solutions is much smaller when considering the reduced sets Domt

instead of Ft , we have decided to remove kinf -useless frequencies. Notice thataccording to our experiments (see Table 1), kinf is very often equal to kopt . We nowdescribe three techniques to determine k-useless frequencies.

3.1 Reductions based on C2 and C4 constraints

The first procedure, called Reduction-C2C4 reduces the frequency domains bytaking into account C2 and C4 constraints. Both types of constraints impose that| ft − ft ′ | must be equal to a given number (εtt ′ for C2 constraints, and 0 for C4constraints). Procedure Reduction-C2C4 returns reduced frequency domains aswell as an answer which can be “NO” or “MAY BE”: the answer is “NO” if and only ifwe have proved that the FAPP has no solution satisfying all C2 and C4 constraints.The procedure is described in Fig. 1 and illustrated on the following example.

Let t and t ′ be two paths with frequency domains Ft = {1, 2, 3, 4} and Ft ′ ={3, 4, 5, 6}, respectively.Assume that | ft −ft ′ | must be equal to 1 (a C2 constraint).In that case, we easily see that these frequency domains can be reduced to Domt ={2, 3, 4} and Domt ′ = {3, 4, 5}.

Procedure Reduction-C2C4(Dom1, . . . , Dom|T |)Set continue = 1 and Reduction-C2C4(Dom1, . . . , Dom|T |) = “MAY BE”While continue = 1, do

1. set continue = 02. for all C2 or C4 constraints (t, t ′) imposing | ft − ft ′ |= α do

∀f ∈ Domt , remove f from Domt if �f ′ ∈ Domt ′ such that | ft − ft ′ |= α

3. if a frequency has been removed from a domain in step 2, then set continue = 1If ∃ t with Domt = ∅, then set Reduction-C2C4(Dom1, . . . , Dom|T |) = “NO”

Fig. 1. Reduction procedure based on C2 and C4 constraints

3.2 Reductions based on CEM constraints

In this section, we describe two reduction procedures based on CEM constraints. ACEM constraint concerning two paths t and t ′ is less restrictive when t and t ′ havedifferent polarizations. In order to determine k-useless frequencies, we thereforeassume that each CEM constraint is defined with δ

(k)

tt ′ , unless it is imposed by C6and C7 constraints, or by the polarization domains, that t and t ′ must have the samepolarization.

Consider a CEM constraint involving paths t and t ′. If there is no frequencyfor t ′ which satisfies this constraint at relaxation level k with frequency ft for t ,then ft is k-useless for t . The procedure, called Reduction1-CEM, performs this


Procedure Reduction1-CEM(k; Dom1, . . . , Dom|T |)Set continue = 1 and Reduction1-CEM(k; Dom1, . . . , Dom|T |) = “MAY BE”While continue = 1, do

1. set continue = 02. for all CEM constraints (t, t ′), do

– if C6 and C7 constraints or the polarization domains impose that t and t ′ must have the

same polarization, then set α = γ(k)

tt ′ ; else set α = δ(k)

tt ′– ∀f ∈ Domt , remove f from Domt if �f ′ ∈ Domt ′ such that | f − f ′ | ≥ α

– ∀f ′ ∈ Domt ′ , remove f ′ from Domt ′ if �f ∈ Domt such that | f − f ′ | ≥ α

3. if a frequency has been removed from a domain in step 2, then set continue = 1If ∃ t with Domt = ∅, then set Reduction1-CEM(k; Dom1, . . . , Dom|T |) = “NO”

Fig. 2. First reduction procedure based on CEM constraints

kind of reduction of the frequency domains. It returns an answer which can be“NO” or “MAY BE”: the answer is “NO” if and only if we have proved that FAPP hasno k-feasible solution. The procedure is described in Fig. 2 and illustrated on thefollowing example. Consider two paths t and t ′ with frequency domains {5, . . . , 16}and {1, . . . , 12}, respectively. Assume that it is not imposed that t and t ′ must havethe same polarization, and let δ(k)

tt ′ = 9. In that case, frequency 5 cannot be assignedto t to get a k-feasible solution, since | ft − 5 | < 9, ∀ft ∈ {1, . . . , 12}. Thesame remark holds for frequencies 6, 7, 8 and 9 for t . In a similar way, frequencies 8to 12 are k-useless for t ′. Hence, the frequency domains for t and t ′ can be reducedto {10, . . . , 16} and {1, . . . , 7}, respectively.

In the previous reductions we have considered all pairs of paths involved in aconstraint. We now show how to further reduce the frequency domains by consider-ing triplets (t, t ′, t ′′)of paths pairwise involved in CEM constraints.At least two pathsamong t, t ′ and t ′′ must have the same polarization (since -1 and 1 are the uniquepossible polarization values). Hence, we have an additional information which wasnot taken into account in procedure Reduction1-CEM. We consider four possiblesituations: pt = pt ′ �= pt ′′ , pt ′ = pt ′′ �= pt , pt = pt ′′ �= pt ′ , and pt = pt ′ = pt ′′ .For each such situation, the CEM constraints involving t, t ′ and t ′′ are preciselydefined (i.e., we know whether γ or δ should be used). Notice that some of the fourabove situations may be forbidden by C6 and C7 constraints or by the polarizationdomains. For each possible situation, the frequencies f for t , f ′ for t ′, and f ′′for t ′′ can be ordered in the six following ways: f ≤ f ′ ≤ f ′′, f ≤ f ′′ ≤ f ′,f ′ ≤ f ≤ f ′′, f ′ ≤ f ′′ ≤ f, f ′′ ≤ f ≤ f ′, f ′′ ≤ f ′ ≤ f . Hence, we have a totalof 4 ·6 = 24 cases to examine. If given a frequency f for t there are no frequenciesf ′ for t ′ and f ′′ for t ′′ which satisfy the CEM constraints at level k in at least one ofthe 24 above mentioned cases, then f is k-useless for t .

Procedure Reduction2-CEM is described in details in Fig. 3, and illustratedwith the following example. Consider three paths t, t ′ and t ′′ with frequency do-mains {1, 2, 3}, {6, . . . , 10}, and {9, . . . , 13}, respectively. Assume that it is notimposed that t, t ′, or t ′′ should have the same polarization, and consider the valuesγ

(k)

tt ′ = 16, γ(k)

tt ′′ = 12, γ(k)

t ′t ′′ = 18, δ(k)

tt ′ = 5, δ(k)

tt ′′ = 4, and δ(k)

t ′t ′′ = 3. Notice

146 A. Hertz et al.

first that procedure Reduction1-CEM is not able to reduce any of the three fre-quency domains. We now show how frequency f = 2 can be removed by procedureReduction2-CEM from the frequency domain of t . Frequencies f, f ′, and f ′′ canonly be ordered in two different ways: f ≤ f ′ ≤ f ′′ and f ≤ f ′′ ≤ f ′. Paths t ′and t ′′ cannot have the same polarization since in such a case | f ′ − f ′′ | shouldbe larger or equal to γ

(k)

t ′t ′′ = 18, which is impossible. Also, pt and pt ′ must be dif-ferent, else f ′ − f is too large. Hence the unique possible case is pt = pt ′′ �= pt ′ .Now, frequency 2 is k-useless for t since 2 + γ

(k)

tt ′′ = 14, which is too large for f ′′.In a similar way, we can see that frequency 3 is k-useless for t , and frequencies9, . . . , 12 are k-useless for t ′′. Hence the new reduced frequency domains for t, t ′and t ′′ are {1}, {6, . . . , 10} and {13}, respectively.

Consider the graphGwith vertex setT , and such that two paths t and t ′ are linkedby an edge if they are involved in a same CEM constraint. Procedure Reduction2-CEM considers triplets (t, t ′, t ′′) of paths which induce a cycle on three verticesin G. This approach can be extended to any odd cycle in G since, here again, weknow that at least two paths on such a cycle have the same polarization. No otherteam of the ROADEF’2001 challenge has implemented such reduction procedures.

3.3 Computation of the lower bound on kopt

We can now describe the procedure that computes a lower bound kinf on the bestrelaxation level kopt . This procedure uses the three reduction procedures detailedin Sects. 3.1 and 3.2. In order to determine kinf , we first repeatedly apply eachreduction procedure with k = 0, until no frequency domain Domt can be furtherreduced, or at least one of the procedures returns “NO”. In the latter case, we knowthat there is no 0-feasible solution and we restart the process with k = 1. Thisprocess is repeated by increasing k step by step, until all reduction proceduresreturn “MAY BE”, which means that no feasibility domain can be emptied. The finalvalue of k is our lower bound kinf on kopt . Moreover, each final frequency domainDomt is a subset of Ft obtained by removing kinf -useless frequencies. This processis described in Fig. 4.

The quality of the lower bound kinf , as well as the size of the reduced frequencydomains are analysed in Table 1. For each problem, we first give three valueskopt , k

′inf , and kinf , where k′

inf is the lower bound on kopt obtained without usingthe third reduction procedure. The value of kopt was given to us by the organizersof the challenge. We have also measured the average ratio |Domt ||Ft | over all paths t .To demonstrate the effectiveness of the three reduction procedures, we indicate thisaverage ratio when using only the first procedure (column labelled C2C4), the twofirst procedures (column labelled C2C4,1-CEM), and all three procedures (columnlabelled C2C4,1-CEM,2-CEM). It can be observed that, in some cases, up to 90% ofthe frequencies are kinf -useless. Notice also that when k′

inf < kinf (for exampleon fapp01 and fapp31), the above ratio may be smaller when using the two first


Procedure Reduction2-CEM(k; Dom1, . . . , Dom|T |)Set continue = 1 and Reduction2-CEM(k; Dom1, . . . , Dom|T |) = “MAY BE”While continue = 1, do

1. set continue = 02. for each CEM constraint (t, t ′), do

– if minf̂ ∈Domt

f̂ + γ(k)

tt ′ > maxf̂ ∈Domt ′

and minf̂ ∈Domt ′

f̂ + γ(k)

tt ′ > maxf̂ ∈Domt

,

introduce constraint pt �= pt ′ in the constraint set C7– if the polarization domains of pt and pt ′ impose that pt = pt ′ , but if pt �= pt ′ is a C7

constraint, then set Reduction2-CEM(k; Dom1, . . . , Dom|T |) = “NO” and STOP3. for each path t such that ∃t ′, t ′′ where t, t ′ and t ′′ are pairwise involved in a CEM constraint,

and for all f in Domt , do– set useless = 1– if C6 and C7 constraints and the sets Pt , Pt ′ , Pt" do not forbid pt =pt ′ �=pt ′′ ,

then set �tt ′ = γ(k)

tt ′ , �tt ′′ = δ(k)

tt ′′ , �t ′t ′′ = δ(k)

t ′t ′′ and apply Check-Useless– if C6 and C7 constraints and the sets Pt , Pt ′ , Pt" do not forbid pt =pt ′′ �=pt ′ ,

then set �tt ′ = δ(k)

tt ′ , �tt ′′ = γ(k)

tt ′′ , �t ′t ′′ = δ(k)

t ′t ′′ and apply Check-Useless– if C6 and C7 constraints and the sets Pt , Pt ′ , Pt" do not forbid pt ′ =pt ′′ �=pt ,

then set �tt ′ = δ(k)

tt ′ , �tt ′′ = δ(k)

tt ′′ , �t ′t ′′ = γ(k)

t ′t ′′ and apply Check-Useless– if C6 and C7 constraints and the sets Pt , Pt ′ , Pt" do not forbid pt =pt ′ =pt ′′ ,

then set �tt ′ = γ(k)

tt ′ , �tt ′′ = γ(k)

tt ′′ , �t ′t ′′ = γ(k)

t ′t ′′ and apply Check-Useless– if useless = 1, then remove f from Domt and set continue = 1

If ∃ t such that Domt = ∅, then set Reduction2-CEM(k; Dom1, . . . , Dom|T |) = “NO”

Procedure Check-Useless

If one of the six following conditions is satisfied, then set useless = 0

(a) f ≤ f ′ ≤ f ′′, f + �tt ′ ≤ maxf̂ ∈Domt ′

f̂ and f + max{�tt ′ + �t ′t ′′ , �tt ′′ } ≤ maxf̂ ∈Domt ′′

f̂

(b) f ≤ f ′′ ≤ f ′, f + �tt ′′ ≤ maxf̂ ∈Domt ′′

f̂ and f + max{�tt ′′ + �t ′t ′′ , �tt ′ } ≤ maxf̂ ∈Domt ′

f̂

(c) f ′ ≤ f ≤ f ′′, minf̂ ∈Domt ′

f̂ + �tt ′ ≤ f and f + �tt ′′ ≤ maxf̂ ∈Domt ′′

f̂

(d) f ′ ≤ f ′′ ≤ f , minf̂ ∈Domt ′

f̂ + max{�t ′t ′′ + �tt ′′ , �tt ′ } ≤ f and minf̂ ∈Domt ′′

f̂ + �tt ′′ ≤ f

(e) f ′′ ≤ f ≤ f ′, minf̂ ∈Domt ′′

f̂ + �tt ′′ ≤ f and f + �tt ′ ≤ maxf̂ ∈Domt ′

f̂

(f) f ′′ ≤ f ′ ≤ f , minf̂ ∈Domt ′′

f̂ + max{�t ′t ′′ + �tt ′ , �tt ′′ } ≤ f and minf̂ ∈Domt ′

f̂ + �tt ′ ≤ f

Fig. 3. Second reduction procedure based on CEM constraints

procedures instead of the three proposed ones. This is due to the fact that k′inf -

useless frequencies are not necessarily kinf -useless. Notice finally that our reductionprocedures are able to find the optimal level of feasibility (i.e., kinf = kopt ) for allinstances, except fapp05 and fapp39.

148 A. Hertz et al.

Procedure Lower-boundSet continue = 1 and k = −1While continue = 1, do

set continue = 0, k = k + 1, and stop = 0set Domt = Ft , ∀t ∈ T

while stop = 0, do• set stop = 1• apply Reduction-C2C4(Dom1, . . . , Dom|T |)• apply Reduction1-CEM(k; Dom1, . . . , Dom|T |)• apply Reduction2-CEM(k; Dom1, . . . , Dom|T |)• if at least one of the three procedures returns “NO”, then set continue = 1 and stop = 1• else if at least one of the three procedures returns “MAY BE” and at least one of these

procedures has modified a frequency domain, then set stop = 0Set kinf = k.

Fig. 4. Computation of a lower bound on kopt

4 Construction of an admissible solution

In order to build an admissible solution, we first assign a polarization to each path,and we then choose a frequency for each path, without taking into account the CEM

constraints. The algorithm given in Fig. 5 is a constructive procedure that assignsa polarization to each path while satisfying C6 and C7 constraints. The paths areconsidered in a random order. Let A be the subset of paths for which the polarizationhas already been fixed, and let B be the set of paths t which do not have a fixedpolarization and such that | Pt | = 1. At each iteration, if B �= ∅, we randomlychoose a path t ∈ B and we fix its polarization pt (there is only one possibility).We then determine the set It of paths (including t) for which the polarization isfixed by pt , and we propagate constaints C6 and C7 using arc-consistency. Moreprecisely, consider the graph G with vertex set T , and such that two paths t and t ′are linked by an edge if they are involved in a same C6 or C7 constraint. The setIt is the vertex set of the connected component of G containing t . For each patht ′ ∈ It , let pt→t ′ be the polarization imposed by t on t ′ when pt is assigned to t . Ifall paths t ′ ∈ It are such that pt→t ′ ∈ Pt ′ , we fix the polarization of each t ′ ∈ It topt→t ′ , and we put every such path in A; else, the procedure stops with a messageindicating that the FAPP does not have any admissible solution. Notice that whenwe fix the polarization of a path t ∈ B, we possibly reduce the polarization domainof other paths t ′ ∈ It . For example, if Pt = {1} and Pt ′ = {−1, 1}, and if a C6constraint links t to t ′, then we set Pt ′ = {1}. When (or if) B = ∅, we randomlychoose a path t /∈ A, we assign a polarization pt randomly chosen in {−1, 1} to t ,and we continue as described above by propagating C6 and C7 constraints in orderto fix the polarization of the paths in It . This random choice is possible without lossof generality because the polarization domains have been reduced when B was notempty.

The input of function Check-Pol is the current set A of paths to which apolarization has already been assigned, as well as a path t /∈ A and a polarizationpt ∈ Pt . This function returns value 0 if pt cannot be assigned to t without violatinga C6 or a C7 constraint. Otherwise, it returns value 1, and updates the set A by


Table 1. Results of the lower bounding procedure

Instance kopt k′inf

kinf C2C4 C2C4,1-CEM C2C4,1-CEM,2-CEM

fapp01 4 3 4 0.987 0.566 0.579fapp02 2 2 2 0.999 0.448 0.425fapp03 7 7 7 1 0.628 0.544fapp04 1 1 1 0.991 0.374 0.366fapp05 11 8 8 1 0.707 0.705fapp06 5 5 5 0.996 0.555 0.477fapp07 9 9 9 0.994 0.713 0.691fapp08 5 5 5 1 0.663 0.646fapp09 3 3 3 0.994 0.580 0.559fapp10 6 6 6 1 0.704 0.659fapp11 8 8 8 0.999 0.565 0.552fapp12 2 2 2 1 0.399 0.374fapp13 3 3 3 0.999 0.495 0.485fapp14 4 4 4 0.999 0.470 0.454fapp15 5 5 5 0.998 0.618 0.612fapp16 11 11 11 1 1 1fapp17 4 4 4 1 0.017 0.017fapp18 8 8 8 0.996 0.030 0.030fapp19 6 6 6 0.994 0.014 0.014fapp20 10 10 10 1 0.024 0.024fapp21 4 4 4 1 0.067 0.067fapp22 7 7 7 0.996 0.015 0.015fapp23 9 9 9 0.992 0.011 0.011fapp24 7 7 7 0.999 0.016 0.016fapp25 3 3 3 0.997 0.051 0.051fapp26 7 7 7 0.999 0.020 0.020fapp27 5 5 5 0.999 0.159 0.159fapp28 3 3 3 0.999 0.037 0.037fapp29 6 6 6 0.998 0.011 0.011fapp30 7 7 7 0.999 0.039 0.026fapp31 5 3 5 1 0.656 0.649fapp32 6 6 6 0.999 0.023 0.023fapp33 5 5 5 0.995 0.040 0.040fapp34 4 4 4 1 0.069 0.060fapp35 6 6 6 0.999 0.035 0.035fapp36 7 7 7 1 0.043 0.043fapp37 5 5 5 1 0.032 0.032fapp38 3 3 3 0.999 0.017 0.017fapp39 3 2 2 1 0.656 0.649fapp40 4 4 4 0.999 0.023 0.023

assigning polarization pt→t ′ to all t ′ ∈ It . Moreover, if t ∈ B, the polarizationdomain of each t ′ ∈ It is reduced to Pt ′ = {pt→t ′ }.

We now show how to determine a frequency assignment satisfying C2, C3, C4,and C5 constraints. Consider the interference graph G2,4 (Aardal et al. 2002) withvertex set T , and in which two vertices are linked by an edge if the correspondingpaths are linked by a C2 or a C4 constraint. For a path t , let Jt denote the vertex setof the connected component of G2,4 containing t . The algorithm described in Fig. 6is a constructive procedure that assigns a frequency to each path while satisfying allC2, C3, C4, and C5 constraints. Each path t is supposed to have a frequency domain

150 A. Hertz et al.

Procedure Init-Pol1. Set A = ∅ and determine B

2. While B �= ∅, do(a) randomly choose t ∈ B

(b) assign polarization pt to t (only one possibility)(c) if Check-Pol(A, t, pt ) = 0, then STOP: there is no admissible solution(d) set B = B − A

3. While A �= T , do(a) randomly choose t ∈ T − A

(b) randomly choose a polarization pt ∈ Pt for t

(c) if Check-Pol(A, t, pt ) = 0, then STOP: there is no admissible solution

Function Check-Pol(A, t, pt )

1. Set Check-Pol(A, t, pt ) = 12. Determine the set It of paths (including t) for which the polarizations are imposed by the one

of t , and for each t ′ ∈ It , let pt→t ′ be the polarization imposed by t to t ′ when pt is assignedto t

3. If pt→t ′ /∈ Pt ′ for some t ′ ∈ It then set Check-Pol(A, t, pt ) = 0;else set A = A ∪ It and pt ′ = pt→t ′ ∀ t ′ ∈ It ; if | Pt |= 1 then set Pt ′ = {pt→t ′ } for allt ′ ∈ It

Fig. 5. Admissible assignment of the polarizations

Domt , which can be a proper subset of Ft (see Sect. 3). The paths are consideredin a random order. Let A be the subset of paths for which a frequency has alreadybeen fixed. For a path t /∈ A, we randomly choose a frequency ft ∈ Domt , and fixthe frequency of each path in Jt using the propagation of constraints C2 and C4.More precisely, consider two adjacent vertices t and t ′ in the graph G2,4 definedabove. If t and t ′ are linked by a C4 constraint, then they must have the samefrequency, which means that when the frequency has been chosen for t , we haveno other choice for t ′. The situation is a little bit different if t and t ′ are linkedby a C2 constraint. Indeed, in such a case, given a frequency ft for t , there aretwo possibilities for the frequency ft ′ to be assigned to t ′, namely ft − εtt ′ andft + εtt ′ (if both frequencies belong to Domt ). When propagating ft in Jt , werandomly choose one of these two frequencies for ft ′ . If the choice of ft ∈ Domt

and its propagation on Jt does not lead to a partial admissible solution, we repeatthe process with another frequency for t , randomly chosen in Domt . This processmay cycle even if there are admissible solutions, but such a case never happenedin our experiments.

The input of function Check-Freq described below is the current set A ofpaths to which a frequency has already been assigned, as well as a path t /∈ A anda frequency ft ∈ Ft . This function first chooses a frequency ft ′ ∈ Domt ′ for eachpath t ′ ∈ Jt ′ , using the propagation of constraints C2 and C4. It then returns value 0if this frequency assignment violates a C3 or a C5 constraint. Otherwise, it returnsvalue 1 and updates the set A.

By combining the two procedures described in this section, we typically obtain11-feasible solutions that violate a lot of CEM(10) constraints (i.e., solutions s witha set V (10)(s) containing many constraints). In order to determine feasible solutions


Procedure Init-Freq1. Set A = ∅2. While A �= T , do

(a) randomly choose t ∈ T − A

(b) randomly choose a frequency ft ∈ Domt for t

(c) if Check-Freq(A, t, ft ) = 0, go to step 2b

Function Check-Freq(A, t, ft )1. Determine the vertex set Jt of the connected component of G2,4 containing t

2. Choose a frequency ft ′ ∈ Domt ′ for each path t ′ of Jt while satisfying constraints C2 andC4 (it is always possible, if not, procedure Lower-bound would have detected this problembefore)

3. If this frequency assignment does not violate constraints in C3 ∪ C5,then set A = A ∪ Jt and set Check-Freq(A, t, ft ) = 1;else set Check-Freq(A, t, ft ) = 0

Fig. 6. Admissible assignment of the frequencies

with a lower relaxation level, we have developed two tabu search algorithms that aredescribed in the next section: the first one tries to improve a solution by modifyingthe frequency assignment, and the second one performs changes on the polarizationassignment. Notice that these two procedures need an admissible solution as input,and they will be used alternately several times.

5 A tabu search algorithm for the FAPP

Many heuristic solution methods have been proposed for the FAP and the MI-FAP,including greedy algorithms (Zoellner and Beall 1977; Box 1978; Borndoerfer etal. 1998), local search methods (Duque-Antón et al. 1993; Knaelmann et al. 1995;Castelino et al. 1996; Al-Khaled 1998; Hao et al. 1998; Capone and Trubian 1999;Hurkens et al. 2000) and genetic algorithms (Cuppini 1994; Lai and Coghill 1996;Hao and Dorne 1996; Crisan and Muehlenbein 1998; Beckmann and Killat 1999).

We describe in this section the tabu search algorithm we have designed for theFAPP. We start with a brief description of a basic tabu search. We then propose twoadaptations of tabu search, one for the frequency assignment (called Tabu-Freq),and one for the polarization assignment (called Tabu-Pol). We end the sectionwith the final proposed algorithm which includes the lower bounding procedure,the generation of an initial admissible solution, and the two adaptations of tabusearch.

5.1 Basic tabu search

Tabu search is a local search algorithm which was originally proposed by Glover(1989), Glover (1990). Its basic version can be described as follows. Let S be theset of solutions to a combinatorial optimization problem, and let f be an objectivefunction which has to be minimized on S. A set N(s), called neighborhood of s, is

152 A. Hertz et al.

associated with each solution s ∈ S. The solutions in N(s) (also called neighborsof s) are obtained from s by performing local changes called moves. Tabu searchfirst constructs an initial solution s0 in S. Then, it successively generates solutionss1, s2, . . . in S such that si+1 ∈ N(si). When a move is performed from si to si+1,the reverse of such a move is stored in a tabu list L, and it is forbidden (with someexceptions) to use such a move for a certain number of iterations. Solution si+1is set equal to arg min

s∈N ′(si )f (s), where N ′(s) is a subset of N(s) containing all

solutions s′ which can be obtained from s by performing a move that is not in L,or such that f (s′) < f (s∗), where s∗ is the current best solution. The process isstopped after a fixed number of iterations without improvement of s∗, or when atime limit is reached. This basic version of tabu search is summarised in Fig. 7.Many variants and extensions can be found in Glover and Laguna (1997).

Initialisation1. generate an initial solution s

2. set s∗ = s, f ∗ = f (s), and L = ∅Repeat

1. determine N ′(s) = {s′ ∈ N(s) such that s′ is obtained from s by performing a move that doesnot belong to L, or such that f (s′) < f (s∗)}

2. set s′ = arg mins′′∈N ′(s)

f (s′′)

3. if f (s′) < f ∗, then set f ∗ = f (s′) and s∗ = s′4. update L and set s = s′

Until a time limit is reached or s∗ was not improved during the last q iterations (where q is a parameter)

Fig. 7. Basic tabu search

Several competitor of the ROADEF’2001 challenge have implemented a tabusearch algorithm for the FAPP. For example, Vasquez (2002) has designed a tabusearch procedure that attemps to find a k-feasible solution for a fixed level k. Ifsuch a solution is found, k is decreased by one unit. The search space S is definedas the set of all partial k-feasible assignments. More precisely, a solution s ∈ S isdefined as an assignment of a frequency ft and a polarization pt to all paths of asubset T ′ ⊆ T such that all C2, . . . , C7 and CEM(k) constraints involving paths inT ′ are satisfied. The objective is to maximize the size of T ′. A move to a neighborsolution consists in adding a frequency ft and a polarization pt to a path t /∈ T ′,and dropping those assignments which are in conflict with ft and pt . Galinier etal. (2002) have also designed a tabu search that solves the FAPP with a fixed levelk. However, they define the search space S as the set of all possible assignmentsof a frequency and a polarization to the paths. Hence, all C2, . . . , C7 and CEM(k)constraints can be violated, and these violations are penalized in the objective func-tion. A move to a neighbor solution is obtained by modifying the frequency and/orthe polarization of a single path or of two paths linked by a C2 or a C4 constraint.More details on these algorithms can be found in Vasquez (2002), and Galinier etal. (2002).


The tabu search strategy we have developed differs from those above in that wedo not solve the FAPP for a fixed level k, and that we alternately and independentlyoptimize the frequency and the polarization assignments. Details are given in thenext subsections.

5.2 Tabu search for the frequency assignment

We describe in this section a tabu search procedure that tries to improve the fre-quency assignment. The function to be minimized is the objection function f ofthe FAPP described in Sect. 2. A neighbor solution is obtained by first assigning anew frequency ft to a path t , and then repairing the neighbor solution if it is notadmissible. When modifying the frequency of a path t , we try to eliminate at leastone violation of a CEM constraint involving t at a level ≤ k∗ − 1, where k∗ denotesthe best relaxation level encountered so far. More precisely, let ns(t, f ) denote thenumber of CEM(k∗ − 1) constraints involving t that are violated if we modify thecurrent solution s by assigning frequency f to t (while the frequencies of the otherpaths are not changed). Let kt be the largest integer strictly smaller than k∗ suchthat there exists a CEM constraint involving t which is violated at level kt . Finally,let D̂omt be the subset of Domt containing all frequencies which, when assignedto t , eliminate a violation of a CEM(kt ) constraint. If D̂omt �= ∅, we randomlychoose a frequency ft ∈ D̂omt for t among those minimising ns(t, f ). Otherwise(D̂omt = ∅) we randomly select a CEM constraint violated at level kt . Let t ′ be thesecond path involved in this constraint. If there exists at least one pair (ft , ft ′) inDomt × Domt ′ which satisfies the considered constraint at level kt , we randomlychoose such a pair (ft , ft ′), and assign frequency ft to t and ft ′ to t ′ to obtain aneighbor solution. It may happen that no such pair exists. If that situation occurs,two cases are possible. If | Pt |=| Pt ′ |= 1, or if paths t and t ′ have different polar-izations, this means that there exists no kt -feasible solution. Otherwise, we changethe polarization of t or t ′, and reapply the above procedure trying to eliminate aviolation of a CEM(kt ) constraint.

It may happen that the assignment of frequency ft to t violates a C2, C3, C4 orC5 constraint. In such a case we apply procedure Check-Freq(A, t, ft ) with A =T − Jt (see Sect. 4) to check whether ft can be propagated on Jt without violatingC2, C3, C4 and C5 constraints. It may also happen that the assignment of ft ′ tot ′ (if D̂omt = ∅) creates a non-admissible solution. We then also apply procedureCheck-Freq(A, t ′, ft ′) with A = T −Jt ′ to check whether these violations can beeliminated by constraint propagation. If all violations can be eliminated, then theoutput of Check-Freq is considered as a neighbor solution. In summary, we movefrom a solution to a neighbor one by first assigning a new frequency ft to a patht so that ft eliminates at least one violation of a CEM(kt ) constraint, and creates aminimum number of violations of CEM(k∗ − 1) constraints. We then repair such aneighbor if it is not admissible.

154 A. Hertz et al.

We put in the tabu list L all paths in Jt , and we forbid to change the frequencyof these paths during | L | iterations. This may be considered as very restrictive.Indeed, when we change the frequency of a path t ′ ∈ Jt ′ from fold to fnew, onecould think about only forbidding the assignment of frequency fold to t ′ during| L | iterations. However, in our experiments, the proposed tabu list turned outto be more effective than less restrictive ones. Notice that a neighbor solution isobtained by modifying the frequency of all paths in Jt , and the move to such aneighbor is considered as tabu if at least one path in Jt belongs to L.

In order to keep control on the time needed to perform an iteration, we generateat most NF neighbors per iteration, where NF is a parameter. Let T (k) denote theset of paths involved in CEM(k) constraints that are violated in the current solutions. To generate these NF neighbor solutions, we first select min{NF , | T (k∗−1) |}paths in T (k∗−1). If | T (k∗−1) | < NF , the remaining NF − | T (k∗−1) | paths arechosen in T (k∗−2), then in T (k∗−3), and so on until NF different paths have beenfound or all violated CEM constraints have been considered. We then apply theabove mentioned neighborhood process to each one of these NF paths. In orderto accelerate the algorithm, we stop the generation of neighbor solutions when wefind a solution better than s∗ (the best solution encountered so far).

5.3 Tabu search for the polarization assignment

We now briefly describe the second tabu search procedure which tries to improve anadmissible solution by modifying the polarization assignment. If γ

(k)

tt ′ > δ(k)

tt ′ for aCEM constraint involving paths t and t ′ at level k, then this CEM constraint is easierto satisfy at level k if t and t ′ have different polarizations. Let D(k) be the numberof pairs (t, t ′) of paths having different polarizations, and such that γ

(k)

tt ′ > δ(k)

tt ′ .The proposed tabu search algorithm tries to maximize D(k∗ − 1) (i.e., this is theobjective function which has to be maximized). A neighbor solution is obtained byassigning a new polarization pt on a path t with | Pt | = 2. Such a change mayinduce violations of C6 and C7 constraints. In such a case, we apply procedureCheck-Pol(A, t, pt ) with A = T −It (see Sect. 4) to check whether the violationscan be eliminated by constraint propagation. If Check-Pol(A, t, pt ) returns value0, then no neighbor solution is generated with polarization pt for t . Otherwise, allchanges performed in Check-Pol(A, t, pt ) define a neighbor solution. We alwaysmove to the best non tabu neighbor s′ of the current solution s, and we put in thetabu list all paths which have a different polarization in s and s′.

5.4 A solution method for the FAPP

Our solution method for the FAPP is now summarized in Fig. 8. The organizers havefixed a time limit of one hour CPU-time on a PC Pentium III (500 Mhz, RAM 128Mo). The length of the tabu lists is set equal to 3, while the value of parameter NF

is an integer that is randomly chosen, at each iteration, in the interval [5, 30].


Procedure FAPP-algorithm

1. Apply procedure Lower-bound and let kinf be the obtained lower bound on kopt ;let Dom1, . . . , Dom|T | be the reduced frequency domains at feasibility level kinf

2. Apply procedures Init-Freq and Init-Pol to get an initial admissible solution s;set k∗ equal to the relaxation level k(s) of s.

3. Repeat the following process until one hour CPU is reached– Apply procedure Tabu-Freq and stop when p = 5000 iterations have been performed without

improving the best solution (according to the objective function f defined in Sect. 2), or whena (k∗ − p)-feasible solution is reached with p ≥ 1;let k denote the feasibility level of the output solution of Tabu-Freq

– If k < k∗, then set k∗ = k

– Apply Tabu-Pol and stop when p = 100 iterations have been performed without improvingthe best solution (according to the objective function D), or when a (k∗ −p)-feasible solutionis reached with p ≥ 1;let k be the feasibility level of the output solution of Tabu-Pol

– If k < k∗, then set k∗ = k

Fig. 8. The proposed solution method for the FAPP

6 Computational results

We have run the above algorithm on the 40 instances of the challenge. Rememberthat these instances are of three different types. The 15 problems of type A (fapp01 tofapp15) have between 200 and 3’000 paths, a maximum of 20’000 CEM constraints,a maximum of 2’200 other constraints, and a maximum of 250 frequencies perfrequency domain. The 15 instances of type B (problems fapp16 to fapp30) arecloser to real-life problems. The number of paths is approximately the same as fortype A instances, while the frequency domains can have up to 500 frequencies, andthe problems can have up to 31’000 CEM constraints and 13’000 other constraints.Type B instances are characterized by the fact that γ

(k)

tt ′ = δ(k)

tt ′ for each CEM

constraint, which means that the polarizations are not relevant for the satisfactionof a CEM constraint. The number of paths in the 10 instances of type X (problemsfapp31 to fapp40) ranges from 400 to 3’000. These instances have up to 30’000CEM constraints, 2’500 other constraints, and a maximum of 500 frequencies perfrequency domain.

We compare our results to those obtained by the seven other teams who haverun their algorithms on the 40 instances. The names of the teams that appear in thefollowing tables are abbreviated as follows:

– BGGS: S. Bisaillon, P. Galinier, M. Gendreau and P. Soriano, from the GERADand CRT, Montréal, Canada

– C: Y. Caseau, from e-Lab Bouygues, France– G: H. Gavranovic, from IMAG, France and the University of Sarajevo, Bosnia

and Herzegovinia– GBB: E. Gaudin, T. Benoist and E. Bourreau, from e-Lab Bouygues, France– HSZ: A. Hertz, D. Schindl and N. Zufferey, from EPF-Lausanne, Suisse and

GERAD, Montréal, Canada (our algorithm).– M: P. Michelon and 8 students, from LIA, Université d’Avignon, France

156 A. Hertz et al.

– VT: M Vasquez and F. Trousset, from LGI2P EMA-EERIE, Nîmes, France– W: B. Weinberg, from LIFL-OPAC, Université de Lille 1, France

Complete information about the 40 instances and the results obtained by the aboveeight teams can be found in Cung and Defaix (2001), or on the web site Cung et al.(2001).

In Table 2, we compare the feasibility levels reached by the eight algorithmson the 40 instances. In the three first columns, we give the name and the type ofeach instance, its number | T | of paths, and the best known feasibility level kopt

(see Table 1). The eight last columns contain the results of each algorithm. A boldnumber means that the value is the best known (i.e. it is equal to kopt ). In the lastrow, we indicate the number of best feasibility levels reached by each algorithm.We first observe that our algorithm has found the optimal feasibility level on 24instances. The VT algorithm has the best performance with 36 optimal solutions.All algorithms that produce better results than ours (i.e., BGGS,C,G,VT) have acommon feature: they all solve the FAPP with a fixed level k of feasibility. Hence,our algorithm seems to be not effective enough in the minimization of the feasibilitylevel. The 24 successful instances all have kinf = kopt (see Table 1), except fapp05and fapp39, which means that in these cases, our algorithm has proved that thereis no feasible solution with a lower feasibility level. We also notice that the 24successful instances include all type B instances.

In Table 3, we compare the eight algorithms on the 24 instances for which ouralgorithm has reached the kopt feasibility level. For each algorithm, we indicate thenumber | V (kopt−1) | of CEM constraints violated at level kopt − 1. A blank squaremeans that the algorithm has not found a kopt -feasible solution. Again, we indicatethe best results with bold numbers. Our algorithm is classified second with 17best solutions. This indicates that when we reach the optimal feasibility level kopt ,we are then very effective in the minimization of the violations of CEM(kopt − 1)constraints. Once again, we observe that the 17 successful instances include all typeB instances.

In Table 4, we further compare the eight algorithms on the 17 successful in-stances of Table 3. For each algorithm, we indicate the total number∑

k<kopt−1 | V (k) | of CEM constraints violated at a level strictly smaller thankopt − 1. Blank squares and bold numbers have the same meaning as in the pre-vious table. Our algorithm is the winner with 14 best solutions, all of them beingof type B. Putting together the results of Tables 2, 3, and 4, we conclude that ouralgorithm has found the best solution (according to the objective function definedin Sect. 2) on 14 instances. Moreover, our algorithm is strictly better than those ofthe other teams for problems fapp27 and fapp30. To summarize the results of thechallenge, we indicate in Table 5 the teams that have produced the best solutions.We observe that the BGGS algorithm has found 27 best solutions while the winnerof the challenge (i.e., the VT algorithm) has produced only 3 best solutions. The VTalgorithm was better classified than all other algorithms because it has producedsolutions with the smallest total deviation from the best known solutions. After the


Table 2. Best feasibility levels

Instance | T | kopt BGGS C G GBB HSZ M VT W

fapp01-A 200 4 4 4 4 4 4 4 4 5fapp02-A 250 2 2 2 2 2 4 11 2 7fapp03-A 300 7 7 7 7 7 7 11 7 9fapp04-A 300 1 1 1 3 12 4 2 1 6fapp05-A 350 11 11 11 11 11 11 11 11 11fapp06-A 500 5 5 5 7 7 5 6 5 7fapp07-A 600 9 9 9 10 9 9 12 9 11fapp08-A 700 5 5 5 5 10 5 11 5 10fapp09-A 800 3 3 3 4 11 5 4 3 11fapp10-A 900 6 6 6 6 11 7 11 6 9fapp11-A 1’000 8 8 8 9 11 10 11 8 11fapp12-A 1’500 2 3 7 11 11 9 11 2 11fapp13-A 2’000 3 3 7 11 11 10 11 5 11fapp14-A 2’500 4 4 8 11 11 10 11 6 11fapp15-A 3’000 5 5 7 11 11 10 11 5 11

fapp16-B 260 11 11 11 11 11 11 11 11 11fapp17-B 300 4 4 4 4 4 4 4 4 4fapp18-B 350 8 8 8 8 8 8 8 8 11fapp19-B 350 6 6 6 6 12 6 6 6 11fapp20-B 420 10 10 10 10 10 10 10 10 10fapp21-B 500 4 4 4 4 4 4 4 4 11fapp22-B 1’750 7 7 7 7 12 7 7 7 11fapp23-B 1’800 9 9 9 9 12 9 12 9 11fapp24-B 2’000 7 7 7 7 11 7 7 7 11fapp25-B 2’230 3 3 3 3 11 3 3 3 11fapp26-B 2’300 7 7 7 7 12 7 7 7 11fapp27-B 2’550 5 11 5 5 11 5 5 5 11fapp28-B 2’800 3 3 3 3 12 3 3 3 11fapp29-B 2’900 6 6 6 6 12 6 12 6 11fapp30-B 3’000 7 11 7 7 12 7 12 7 11

fapp31-X 400 5 5 5 5 12 5 5 5 11fapp32-X 550 6 10 6 6 11 11 11 6 11fapp33-X 650 5 5 5 5 11 5 5 5 11fapp34-X 750 4 4 4 4 6 5 4 4 11fapp35-X 1’500 6 7 6 6 12 10 11 6 11fapp36-X 2’000 7 7 9 8 12 10 7 7 11fapp37-X 2’250 5 11 5 8 11 10 11 5 11fapp38-X 2’500 3 11 3 3 12 10 12 9 11fapp39-X 2’750 3 3 3 3 12 3 3 11 11fapp40-X 3’000 4 11 4 8 12 10 12 4 5

Number of best 32 35 28 10 24 19 36 4

challenge, the BGGS team has implemented a filtering technique for reducing thefrequency domains. By combining this technique with their tabu search, they haveobtained 38 best solutions for the problems in Table 2, 24 for those in Table 3, and13 for those in Table 4. More details on their algorithm can be found in Galinier etal. (2002).

158 A. Hertz et al.

Table 3. Minimum number of CEM constraints violated at level kopt − 1

Instance | T | | V (kopt−1) | BGGS C G GBB HSZ M VT W

fapp01-A 200 4 4 6 14 4 5 5 14fapp03-A 300 10 10 27 16 10 13 32fapp05-A 350 1 1 892 1 1 1 97 364 495fapp06-A 500 12 12 53 24 31fapp07-A 600 22 22 132 22 26 106fapp08-A 700 16 16 53 37 19 73

fapp16-B 260 5 5 572 67 5 5 590 514 989fapp17-B 300 4 4 4 4 4 4 4 4 4fapp18-B 350 4 4 4 4 4 4 4 4fapp19-B 350 2 2 3 2 2 2 3fapp20-B 420 5 5 6 5 5 5 5 7 9fapp21-B 500 2 2 2 2 2 2 2 2fapp22-B 1’750 15 15 16 16 15 15 25fapp23-B 1’800 16 16 17 16 16 17fapp24-B 2’000 6 6 7 7 6 6 9fapp25-B 2’230 7 7 7 7 7 7 7fapp26-B 2’300 9 9 10 9 9 9 10fapp27-B 2’550 4 7 8 4 4 11fapp28-B 2’800 13 13 32 25 13 14 42fapp29-B 2’900 25 25 28 25 25 25fapp30-B 3’000 13 17 16 13 48

fapp31-X 400 4 4 161 34 21 63 117fapp33-X 650 7 7 16 7 7 7 10fapp39-X 2’750 356 356 2567 747 1066 2211

Number of best 22 4 11 9 17 11 5 1

Table 4. Minimum number of CEM constraints violated at a level < kopt − 1

Instance | T | ∑k<kopt−1

| V (k) | BGGS C G GBB HSZ M VT W

fapp05-A 350 186 372 1467 186 715fapp16-B 260 56 56 56 57fapp17-B 300 34 34 36 35 34 34 36 36 36fapp18-B 350 55 55 55 57 55 55 57 59fapp19-B 350 51 51 53 51 60fapp20-B 420 97 97 99 97 97 106fapp21-B 500 10 10 12 11 10 10 12 12fapp22-B 1’750 187 187 187 292fapp23-B 1’800 187 187 189 187fapp24-B 2’000 71 71 71 77fapp25-B 2’230 32 32 90 33 32 43 33fapp26-B 2’300 74 74 74 74 75fapp27-B 2’550 20 20 22fapp28-B 2’800 32 32 32fapp29-B 2’900 212 239 212 212 310fapp30-B 3’000 148 148

fapp33-X 650 66 66 77 69 181

Number of best 13 1 2 5 14 0 0 0


Table 5. Best teams

Instance Best teams Instance Best teams

fapp01-A BGGS fapp21-B BGGS, GBB, HSZfapp02-A BGGS fapp22-B BGGS, HSZfapp03-A GBB fapp23-B BGGS, HSZfapp04-A BGGS fapp24-B BGGS, HSZfapp05-A GBB fapp25-B BGGS, HSZfapp06-A BGGS fapp26-B BGGS, G, HSZfapp07-A GBB fapp27-B HSZfapp08-A BGGS fapp28-B BGGS, HSZfapp09-A BGGS fapp29-B G, HSZfapp10-A BGGS fapp30-B HSZfapp11-A BGGS fapp31-X BGGSfapp12-A VT fapp32-X Gfapp13-A BGGS fapp33-X BGGSfapp14-A BGGS fapp34-X BGGSfapp15-A BGGS fapp35-X Gfapp16-B BGGS, GBB fapp36-X Mfapp17-B BGGS, GBB, HSZ fapp37-X VTfapp18-B BGGS, C, GBB, HSZ fapp38-X Gfapp19-B BGGS, HSZ fapp39-X BGGSfapp20-B BGGS, GBB, HSZ fapp40-X VT

There are several reasons which can explain the success of our algorithm ontype B instances. Remember first that the reduction procedures proposed in Sect. 3assume that we are always in the most favourable situation, where the two pathsinvolved in a CEM constraint have different polarizations (i.e., δ

(k)

tt ′ can be used

instead of γ(k)

tt ′ ). We have mentioned that the polarizations are not relevant for type

B instances since γ(k)

tt ′ = δ(k)

tt ′ for each CEM constraint. Hence, these instances canbe considered as favourable cases for our lower bounding technique. The successof our algorithm on type B instances can perhaps also be explained by the bigreduction of the frequency domains (up to 90%, see Table 1), which means thatwe have drastically reduced the number of admissible solutions. Notice howeverthat for many instances of type X, our best feasibility level is much larger than kopt

(see Table 2), while the frequency domains have also been reduced a lot (see Table1). On the opposite, we have been able to produce kopt -feasible solutions for someinstances of type A (problems fapp01, fapp03, fapp05, fapp06, fapp07, and fapp08)while we have not been able to get a large reduction of the frequency domains inthese cases.

Most of the instances for which we have not been able to reach a kopt -feasiblesolution are of large size, which indicates that the performance of our algorithmdecreases with the increase of the problem size. We have however observed inSect. 3 that the quality of our lower bound kinf does not seem to depend on theproblem size. As mentioned above, the polarizations are not relevant for type B

instances, while they are for those of type A and X. We therefore think that themedium performance of our algorithm on some instances of type A or X is probably

160 A. Hertz et al.

due to the fact that most algorithms that perform better than ours on these typesof instances combine the optimization of the frequency and of the polarizationassignement in one procedure, while we solve these two problems separately. Theperformance of our algorithm can however be considered as reasonably good sincewe have always been able (except for fapp32) to find a k-feasible solution withk < 11 when such a solution exists, while this is not an easy task at all (algorithmsGBB and W have produced only 1 such solution on instances of type X).

In conclusion, our second position in the “Junior” category of a contest towhich 27 competitors have taken part gives evidence that the combination of a tabusearch with an effective reduction of the search space is an appropriate approachfor the solution of the FAPP. Algorithms that solve a series of FAPP’s with a fixedfeasibility level k (e.g., BGGS, C, G, VT) are more effective than our algorithm inproducing solution with an optimal feasibility level kopt . However, when we reachsuch a feasibility level (and this was the case for 24 instances), our algorithm is veryeffective in the minimization of the violations of CEM(k) constraints with k < kopt .Hence, the following three phase technique seems to be a promising strategy:

1. compute a lower bound kinf of the best feasibility level;2. run the VT algorithm (the winner in Table 2) until a kinf feasible solution is

reached (or the time limit is reached);3. run our HSZ algorithm starting from the output solution of Step 2, until the

time limit is reached.

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Lower bounding and tabu search procedures for the frequency