More adaptive robust stable routingMateusz Zotkiewicz

TELECOM & Management SudParis9, rue Charles Fourier, 91011 Evry, France

Institute of Telecommunications, Warsaw University of TechnologyNowowiejska 15/19, 00-665 Warszawa, Poland

Email: mateusz.zotkiewicz@it-sudparis.eu

Walid Ben-AmeurTELECOM & Management SudParis

9, rue Charles Fourier, 91011 Evry, FranceEmail: walid.benameur@it-sudparis.eu

Abstract—In the paper we deal with the problem of optimalpartitioning of a traffic demand polytope using a hyperplane. Inthe considered model all possible demand matrices belong to apolytope. The polytope can be divided into parts, and differentrouting schemes can be considered while dealing with trafficmatrices from different parts of the polytope. The model canbe applied to all networks that support unrestricted routing ofbifurcated flows, e.g., MPLS networks or optical networks. Inthe paper we present an algorithm that solves one of the mostpractical versions of the considered problem, i.e., reservationvectors on both sides of the hyperplane have to be the same.Moreover, we present another (faster) algorithm that solves amore restricted version of the problem. Finally, we presentnumerical results proving the applicability of the introducedalgorithms.

I. INTRODUCTION

Modern telecommunication networks deal with traffic gen-erated by variety of different applications utilized by a largenumbers of users. It makes prediction of traffic patternsincreasingly difficult. The introduction of new services andthe mobility of customers make the task harder.

Some models have been proposed in the past to address theuncertainty. The first approach consists in building a trafficmatrix based on the worst case situation for each trafficcomponent. Routing is then computed based on it. While thisapproach is simple, it can provide expensive solutions.

A second approach is based on probability information.Probability models are chosen to model the traffic variations.Then, one may look for a routing optimizing a probabilis-tic criterium: throughput expectation, mean delays, blockingprobability, etc. The solution obtained in this way is goodon average but can be very bad in some cases. This kind ofapproach is generally called stochastic programming.

Another general approach called robust optimization (see[5]) is based on a finite number of scenarios. It consists incomputing a solution that is compatible with all scenarios.In the networking context, we want to determine a routingscheme such that each traffic matrix belonging to a givenfinite set of traffic matrices can be carried through the network.Robust optimization can be viewed as an alternative, and moretractable, way to solve the stochastic programming problems.

Contract grant sponsor: Polish Ministry of Science and Higher Education;Contract grant numbers: N517 397334 and 280/N-DFG/2008/0

Another robustness model was proposed in [9] (and inde-pendently in [8]). It assumes that the traffic generated in eachnode is limited. Limitations on the incoming traffic can alsobe considered. The traffic matrix is then any matrix satisfyingthis kind of constraints. This model is called Hose model.

A more general polyhedral model is considered in [2]–[4].The model assumes that each possible traffic matrix belongsto a polytope. In [4] a polynomial time algorithm is proposedto compute a routing scheme that solves this problem. Therouting is robust (it is compatible with all matrices) and stable(it does not change when the matrix changes).

A completely different approach to deal with uncertaintyconsists in allowing the network to change routing in adynamic way each time there is a change in terms of trafficmatrix. This problem was intensively studied in the context ofswitched networks. Different rules can be applied to connectcalls depending on the current situation, e.g., alternate dynamicrouting, sequential routing, trunk reservation (see [1] andreferences therein). While this routing has many benefits, it isgenerally difficult to implement. Moreover, it is not optimal,because the rules that are used are fixed in advance.

For more detailed discussion of different approaches to theproblem of uncertainty proposed in the past readers can consulta recent survey that can be found in [6].

In [4] the following question was raised: given a trafficpolytope, is it theoretically easy to compute a fully dynamicrouting (routing scheme that depends on the current trafficmatrix only). Recently it was proven in [7] that this problemis difficult. Moreover, it is clearly not easy to implement thiskind of routing.

Therefore, it is worth to consider something between robuststable routing and fully dynamic routing. Instead of computinga robust solution, it is possible to partition the uncertainty setinto some subsets, and compute a robust routing for each ofthem. This class of problems was considered in [2]. In thispaper we present algorithms that solve problems introducedthere. Moreover, we present numerical results proving theapplicability of our algorithms.

We deal with a class of problems where it is assumed thata direction of a hyperplane dividing the uncertainty set isknown but its position is subject to optimization. We makethis assumption because some hyperplane’s directions can beimplemented easier than others, e.g., an amount of traffic

This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings.978-1-4244-4148-8/09/$25.00 ©2009

generated between a pair of nodes or an amount of trafficgenerated by a specific node. In both situations a positionof a threshold that triggers changes in routing is subject tooptimization. In other words, we provide two routings: thefirst is used when the amount of traffic, which was set asa direction of the hyperplane, is smaller than the threshold,otherwise the second routing is used. Note that when werestrict possible directions to a finite (and tractable) set we canuse our algorithm to obtain an optimal solution in reasonabletime (by calling it once for each possible direction).

Another possibility is to consider time as a direction of thehyperplane. In this case time is represented by an additionaldimension of the uncertainty set. Our goal is to optimize themoment when routing should be changed.

The paper is organized as follows. In Section II we presentthe notation and formulations of two basic problems. InSection III the presented models are enhanced by introducingthe partitioning of the uncertainty set. In Section IV we presentan algorithm that solves the introduced problems. Another(faster) algorithm is presented in Section V. Numerical resultscan be found in Section VI and conclusions in Section VII.

II. NOTATION AND PROBLEM FORMULATIONS

We consider a directed graph G = (V,A), where V is theset of nodes and A is the set of arcs. The graph represents abackbone and the arcs depict the unidirectional transmissionlinks. For each arc a ∈ A an installed capacity ca ∈ IR+ anda routing cost wa ∈ IR+ for one unit of traffic are given.

Let t = (tij)i,j∈V be a vector of IR|V|(|V|−1) that specifiestraffic demands (or capacity requirements) between all pairsof nodes of V . This vector will be called a traffic matrix. Thetraffic matrix is supposed to be variable and can be any pointof the traffic demand polytope D. D is generally defined bysome linear constraints involving the variables tij , for i, j ∈ V .However, it can be also defined by a set of traffic matrices(convex hull of those matrices) or by an oracle.

To express routing problems as mathematical programs,we introduce the following notation.

P(i, j): finite set of acyclic paths of G from i to j (i, j ∈ V).xij

p : proportion of the traffic demand from i to j (i, j ∈V) carried through a path p ∈ P(i, j). Note that 0 ≤xij

p ≤ 1. For a current traffic matrix t ∈ D the trafficcarried through p is then given by tij · xij

p .xij

a : proportion of traffic from i to j flowing through anarc a ∈ A.

fa : maximum amount of traffic carried on an arc a ∈ A.It depends on the polytope D and the routing pattern.It can be considered as the minimum capacity thathas to be reserved on a link a.

w(D) : routing cost.

A. Robust-Routing-Problem

The problem of computing the minimum cost robust stablerouting of an uncertainty domain D, denoted by Robust-

Routing-Problem, can be formulated as follows:

Minimize:

wRR(D) =∑

a∈Awafa

Subject to:∑

p∈P(i,j)

xijp ≥ 1 ∀i, j ∈ V, (1a)

∑

p∈P(i,j),p�a

xijp ≤ xij

a ∀i, j ∈ V, ∀a ∈ A, (1b)

∑

i,j∈Vxij

a tij ≤ fa ∀a ∈ A, ∀t ∈ D, (1c)

fa ≤ ca ∀a ∈ A, (1d)

xijp ≥ 0 ∀p ∈ P(i, j), ∀i, j ∈ V, (1e)

xija ≥ 0 ∀a ∈ A, ∀i, j ∈ V. (1f)

The objective is to minimize the total routing cost. Inequali-ties (1a) express the fact that the traffic demand between everypair of nodes may be split among many paths. Every variablexij

a is defined by (1b). For a given matrix t the traffic flowingthrough a link a is given by the left-hand side of (1c). Thus, thecapacity fa that should be reserved on an arc a must be higherthan the traffic carried in all the situations. In other words,inequalities (1c) must be valid for each traffic matrix t ∈ D.Inequality (1d) indicates that fa is lower than the capacityof a link a. Inequalities (1e) and (1f) force the variables tobe nonnegative. Note that the variables xij

a can be eliminatedfrom the formulation. However, we keep them for the sake ofclearness and simplicity.

Presented in this way, Robust-Routing-Problem seems to bedifficult. First, the number of paths of P(i, j) can be very high.Second, inequalities (1c) are nonlinear. Finally, they have tobe satisfied for each t ∈ D, and D is generally an infinite set.

Fortunately, the problem was proven to be easily solvableusing an algorithm based on constraint generation (see [3],[4]). Considering the current solution of the relaxation ofRobust-Routing-Problem (only a finite set of traffic matrices istaken into account instead of the whole polytope D), we onlyhave to check wether there is an arc a and a traffic matrixt ∈ D such that (1d) is violated. This can be easily done bythe following linear program:

maxt∈D

∑

i∈V

∑

j∈V\{i}xij

a tij .

If the maximum is larger than fa we add a violated inequality.Routing paths can also be generated in an iterative way by

solving shortest path problems where arc weights are given bythe values of the dual variables of (1b).

B. No-Sharing-Problem

Another set of problems emerges when we impose harsherrules on capacities’ reservation and forbid different relationsto share resources. We denote the core problem of this set asNo-Sharing-Problem and formulate it as follows:

This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings.978-1-4244-4148-8/09/$25.00 ©2009

Minimize:

wNS(D) =∑

a∈Awafa

Subject to:∑

p∈P(i,j)

xijp ≥ 1 ∀i, j ∈ V, (2a)

∑

p∈P(i,j),p�a

xijp ≤ xij

a ∀i, j ∈ V, ∀a ∈ A, (2b)

xija tij ≤ f ij

a ∀i, j ∈ V, ∀a ∈ A, ∀t ∈ D, (2c)∑

i,j∈Vf ij

a ≤ fa ∀a ∈ A, (2d)

fa ≤ ca ∀a ∈ A, (2e)

xijp ≥ 0 ∀p ∈ P(i, j), ∀i, j ∈ V, (2f)

xija ≥ 0 ∀a ∈ A, ∀i, j ∈ V. (2g)

In the formulation variables f ija denote the amount of

capacity reserved on a link a by traffic generated betweennodes i and j. In fact, they [like xij

a – here and in (1)] areobsolete and can be eliminated from the formulation. However,we keep them for the sake of clearness and simplicity.

Note that the basic No-Sharing-Problem is equivalent toreserving tmax

ij = maxt∈D tij of flow for each relation.Therefore, it is easily solvable by means of an appropriatelinear program facilitated by a path generation mechanism.However, it gets more complicated when we allow D to bepartitioned.

III. PARTITIONING PROBLEMS

Given a traffic demand polytope D and a normal vectorα that defines a direction of a hyperplane α.x = β (β isa variable subject to optimization) we can partition D intotwo subsets L(D,β) = D ∩ {t, α.t ≤ β} and R(D,β) =D ∩ {t, α.t ≥ β}, called a left hand side polytope (LHSP)and a right hand side polytope (RHSP), respectively. Then,we can consider a robust routing for each subset. Said anotherway, instead of having only one routing scheme, we will havetwo schemes: if the current traffic matrix belongs to the first(second) subset it uses the first (second) routing scheme.

Depending on restrictions imposed on the reservation vec-tors two different strategies can be considered. The reservationvectors on opposite sides of the hyperplane can be eitherallowed to be different or required to be identical. The formercase models a situation when we are the owners of the networkand our task is to minimize the usage of resources. The lattercorresponds to a situation when we rent capacities from otheroperators and we cannot renegotiate those renting agreementseach time we change the routing.

A. The reservation vectors can be different

We denote the problem by Different-Reservation-Problem,and for Robust-Routing-Problem we formulate it as follows:

wDR/RR(D) = minβ

max{wRR(L(D,β)), wRR(R(D,β))

}.

This version can be approximately solved within an arbitraryprecision using a polynomial algorithm presented in [2].Knowing that wRR(L(D,β)) is a nondecreasing function ofβ and wRR(R(D,β)) is a nonincreasing function of β we canfind β� such that wRR(L(D,β�)) = wRR(R(D,β�)) using abinary search.

In order to formulate Different-Reservation-Problem for No-Sharing-Problem we have to substitute wNS for wRR. Notethat this variant can be solved using the same algorithm.

B. The reservation vectors are identical

Here we assume that the reservation vectors correspond-ing to the two uncertainty subsets should be the same. Wewill call it Identical-Reservation-Problem (for Robust-Routing-Problem or No-Sharing-Problem). A version that correspondsto Robust-Routing-Problem can be formulated as follows:

Minimize:

wIR/RR(D) =∑

a∈Awafa

Subject to:∑

p∈P(i,j)

xijp ≥ 1 ∀i, j ∈ V, (3a)

∑

p∈P(i,j),p�a

xijp ≤ xij

a ∀i, j ∈ V, ∀a ∈ A, (3b)

∑

i,j∈Vxij

a tij ≤ fa ∀a ∈ A, ∀t ∈ L(D,β), (3c)

∑

p∈P(i,j)

xijp ≥ 1 ∀i, j ∈ V, (3d)

∑

p∈P(i,j),p�a

xijp ≤ xij

a ∀i, j ∈ V, ∀a ∈ A, (3e)

∑

i,j∈Vxij

a tij ≤ fa ∀a ∈ A, ∀t ∈ R(D,β), (3f)

fa ≤ ca ∀a ∈ A, (3g)

xijp , xij

p ≥ 0 ∀p ∈ P(i, j), ∀i, j ∈ V, (3h)

xija , xij

a ≥ 0 ∀a ∈ A, ∀i, j ∈ V. (3i)

In the formulation non-overlined variables correspond toLHSP, and overlined variables correspond to RHSP.

While dealing with a version of Identical-Reservation-Problem for No-Sharing-Problem we have to replace (3c) and(3f) with:

xija tij ≤ f ij

a ∀i, j ∈ V, ∀a ∈ A, ∀t ∈ L(D,β), (4a)

xija tij ≤ f ij

a ∀i, j ∈ V, ∀a ∈ A, ∀t ∈ R(D,β), (4b)

and add the following two sets of constraints:∑

i,j∈Vf ij

a ≤ fa ∀a ∈ A, (5a)

∑

i,j∈Vf ij

a ≤ fa ∀a ∈ A. (5b)

This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings.978-1-4244-4148-8/09/$25.00 ©2009

The way this family of problems (for both Robust-Routing-Problem and No-Sharing-Problem) can be approached is themajor novelty of our paper.

IV. DOUBLE BINARY SEARCH ALGORITHM

In this section we present an algorithm that solves Identical-Reservation-Problem. Although it can successfully deal withboth versions of the problem (for Robust-Routing-Problem andNo-Sharing-Problem), for the sake of simplicity we will con-centrate on the former one.

Consider a simpler problem that consists in answeringa question if there exists a feasible solution to Identical-Reservation-Problem whose cost is smaller or equal to agiven value. We will refer to this simplified problem asIdentical-Reservation-Limit. Obviously, knowing the way ofsolving Identical-Reservation-Limit we can solve Identical-Reservation-Problem relatively fast using a binary search.

First we have to know upper and lower bounds for thecost wIR/RR(D,β) denoted by wmin and wmax, respectively.In our implementation we set wmin to the optimal objectiveof Different-Reservation-Problem, and wmax to the optimalobjective of Identical-Reservation-Problem with β set to theoptimal value obtained while computing the lower bound(Identical-Reservation-Problem for fixed β was solved in [2]).

The question arises how to solve Identical-Reservation-Limit. In this section we present an algorithm that can do thisfor both Robust-Routing-Problem and No-Sharing-Problem.Although it is not polynomial, it performs well and can solveeven medium size problems (see the numerical results inSection VI). The key ideas of the approach can be seen inAlgorithm 1.

Algorithm 1 Identical-Reservation-Limit(wlim)β2 = βmax

while β2 > βmin doβ1 = findMax(β2, wlim)if β1 = β2 then

return YESelse

β2 = β1

end ifend whilereturn NO

The algorithm returns YES, if a solution to Identical-Reservation-Problem, whose cost is smaller or equal to wlim,exists. Otherwise, it returns NO. It uses constants βmin andβmax that define an interval of possible values of β. It alsouses the function findMax. The basic idea of the method is tolimit the interval of possible positions of the hyperplane. Atthe begin the hyperplane can be anywhere in [βmin, βmax].It means that we cannot judge if any t ∈ D has to belongeither to LHSP or to RHSP (we assume that the interval issufficiently large). Then, in the loop, we limit the intervalof possible positions of the hyperplane. More precisely, weextend the number of matrices that have to belong to RHSP.

Having RHSP defined for a particular loop we can, us-ing findMax method, calculate the maximal size of LHSP(RHSP is given) which satisfies the cost limit. If the wholepolytope D can be divided between LHSP and RHSP theproblem has been solved. If some matrices cannot be addedto LHSP for a given RHSP, it means that they have to belongto RHSP. We add them to RHSP and repeat the loop.

Let us now introduce a new problems that will facilitateexplanation of findMax method. Assume that values β1 andβ2 are given. Two-Known-Planes-Problem consists in findingthe optimal routing when LHSP is defined as L(D,β1) andRHSP is defined as R(D,β2). Note that β1 and β2 are givenso the problem can be easily solved using techniques presentedin [3], [4] and briefly described in Section II.

Method findMax returns the maximal value of β1 thatallows Two-Known-Planes-Problem to be solved for a givenβ2 and within a given cost limit wlim. If such a value does notexist it returns −∞. Note that β1 is a variable subject to opti-mization and β2 is constant. In such a situation the objectivefunction of Two-Known-Planes-Problem is nondecreasing forβ1 as we cannot decrease a cost of a solution by adding newtraffic matrices to the problem. Therefore, the method can use(and it does) a binary search.

V. LIMITED NUMBER OF EXTREME POINTS

In Section IV we presented an algorithm that solvesIdentical-Reservation-Problem for both Robust-Routing-Problem and No-Sharing-Problem. In this section we takeadvantage of a special feature of No-Sharing-Problem, andpresent an algorithm that can solve this version of Identical-Reservation-Problem in polynomial time when D is a convexhull of a given set of traffic matrices.

Assume that the polytope D and hyperplanes α.t = β1

and α.t = β2 (β1 < β2) are given, and the middlepolytope Dmid = R(D,β1) ∩ L(D,β2) is not empty butsimultaneously does not contain any extreme points of D.We define the left and right polytopes as Dleft = L(D,β1)and Dright = R(D,β2), respectively. We use the followingnotation: Dβ = D ∩ {t, α.t = β}.

Lemma 5.1: For any β ∈ (β1, β2) all points t ∈ Dβ can beexpressed as t = λt′ +(1−λ)t′′, where λ = β2−β

β2−β1, t′ ∈ Dβ1

and t′′ ∈ Dβ2 .Proof: Take any point t ∈ Dβ . As it belongs to D, it

can be expressed as a convex combination of extreme pointsof D. Dmid does not contain any extreme points of D, so allof them have to belong either to Dleft or to Dright (Dleft ∪Dmid ∪Dright = D). It means that we can calculate a convexcombination of only those extreme points that belong to Dleft

and obtain a point in Dleft (together with its weight). Thesame operation can be performed on Dright. The consideredmatrix t will be a convex combination of those two points.Obviously, those two points can be connected by a line that iscontained in D. What is more, this line also contains one pointfrom Dβ1 (denoted by tβ1) and one point from Dβ2 (denotedby tβ2). Certainly, t = λtβ1 + (1 − λ)tβ2 , and we can taket′ = tβ1 and t′′ = tβ2 .

This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings.978-1-4244-4148-8/09/$25.00 ©2009

When we think about β2 as a variable we can write thefollowing lemma.

Lemma 5.2: For any β ∈ (β1, β2) all points t ∈ Dmid ∩{t, α.t ≤ β} can be expressed as t = λt′ + (1 − λ)t′′, whereλ = β2−β

β2−β1, t′ ∈ Dβ1 and t′′ ∈ Dmid ∩ {t, α.t ≤ β2}.

Proof: Take any point t ∈ Dmid ∩ {t, α.t ≤ β}. Notethat t ∈ Dβ′ , where β1 ≤ β′ ≤ β. Now let us take β′

2 =β1 + β′−β1

β−β1(β2 − β1). Obviously, β1 ≤ β′

2 ≤ β2, so there areno extreme points of D between β and β′

2. What is more,β′2−β′

β′2−β1

= λ. Therefore, we can use Lemma 5.1, and constructeach matrix t ∈ Dmid ∩ {t, α.t ≤ β} from matrices t′ ∈ Dβ1

and t′′ ∈ Dmid ∩ {t, α.t ≤ β2}, using the formula t = λt′ +(1 − λ)t′′.

Note that Dleft = L(D,β1) and Dleft ⊂ L(D,β2).Therefore, the previous lemma can be extended to also coverpoints that belong to Dleft.

Corollary 5.3: For any β ∈ (β1, β2) all points t ∈ L(D,β)can be expressed as t = λt′ + (1 − λ)t′′, where λ = β2−β

β2−β1,

t′ ∈ L(D,β1) and t′′ ∈ L(D,β2).Obviously, it works also in the opposite direction.Corollary 5.4: For any β ∈ (β1, β2) all points t ∈ R(D,β)

can be expressed as t = λt′ + (1 − λ)t′′, where λ = β2−ββ2−β1

,t′ ∈ R(D,β1) and t′′ ∈ R(D,β2).

Both Corollaries 5.3 and 5.4 can be rewritten taking intoaccount tmax

ij . By tmaxij (A) we denote maxt∈A tij , where A

is a traffic demand polytope.Corollary 5.5: For any β ∈ (β1, β2) maximal values of

tmaxij = tmax

ij (L(D,β)) satisfies an inequality:

tmaxij ≤ λtmax

ij′ + (1 − λ)tmax

ij′′,

where λ = β2−ββ2−β1

, tmaxij

′ = tmaxij (L(D,β1)) and tmax

ij′′ =

tmaxij (L(D,β2)). Moreover, maximal values of tmax

ij =tmaxij (R(D,β)) satisfies an inequality:

tmaxij ≤ λtmax

ij

′ + (1 − λ)tmaxij

′′,

where λ = β2−ββ2−β1

, tmaxij

′ = tmaxij (R(D,β1)) and tmax

ij

′′ =tmaxij (R(D,β2)).

Let us now assume that the optimal solutions of Identical-Reservation-Problem (for No-Sharing-Problem) for given β1

and β2 are provided, and we want to construct a solution forβ ∈ (β1, β2). Variables of the solution for β will be denotedas described in Section II. Variables of the solution for β1 willbe additionally followed by a prim, e.g., f ′

a. Variables of thesolution for β2 will be followed by a double prim, e.g., f ′′

a .Let us now construct routing to the solution for β in the

following way:

xijp = (λxij

p

′tmaxij

′ + (1 − λ)xijp

′′tmaxij

′′)/tmaxij (6a)

xijp = (λxij

p

′tmaxij

′ + (1 − λ)xijp

′′tmaxij

′′)/tmaxij (6b)

Lemma 5.6: The routing constructed using (6) is feasible,i.e., satisfies (3a) and (3d).

Proof: Let us focus on the first equality. We use thefirst inequality from Corollary 5.5, substitute u for λtmax

ij′

and substitute v for (1 − λ)tmaxij

′′. Summing the obtained

inequalities for all p ∈ P(i, j) we receive the followingformula:

∑

p∈P(i,j)

xijp ≥ u

u + v

∑

p∈P(i,j)

xijp

′+

v

u + v

∑

p∈P(i,j)

xijp

′′.

Variables xijp

′and xij

p′′

satisfy (3a), so xijp also has to satisfy

it. Similar proof can be used to show that xijp satisfy (3d).

Knowing that the routing expressed by (6) is feasible, andusing (4) in order to substitute f ij

a for xijp tij we obtain the

following corollary.Corollary 5.7: If routing of the solution for β satisfies (6)

then f ija ≤ λf ij

a′+(1−λ)f ij

a′′

and f ija ≤ λf ij

a

′+(1−λ)f ij

a

′′,

for all a ∈ A and all i, j ∈ V .Note that according to (3) (modified for No-Sharing-

Problem) the cost function wIR/NS(D,β) depends on fa

only. What is more, according to (5) fa depends on f ija only.

Therefore, from Corollary 5.7 we can deduce that:Corollary 5.8: Function wIR/NS(D,β) is convex for β ∈

[β1, β2], if Dmid = R(D,β1)∩L(D,β2) does not contain anyextreme point of D.

Let us now return to the algorithm. We can easily calculatea list of increasing values of β that describe all hyperplaneswhich contain extreme points of D (note that the polytopeis a convex hull of a given set of demand matrices). Wecan formally define the list as B = {β1, β2, . . . , βn} suchthat βi < βi+1, for i = 1, 2, . . . , n − 1, and each extremepoint of D belongs to

⋃ni=1 Dβi

. Knowing from Corollary 5.8that function wIR/NS(D,β) is convex for β ∈ [βi, βi+1] andi = 1, 2, . . . , n−1, we can calculate optimal w

�IR/NSβ∈[βi,βi+1]

(D),for i = 1, 2, . . . , n − 1, using any known polynomial timealgorithm that can find the minimum of a convex function(e.g., golden ratio or Fibonacci search). Obviously, optimalw�IR/NS(D) = mini=1,2,...,n−1 w

�IR/NSβ∈[βi,βi+1]

(D).Proposition 5.9: Identical-Reservation-Problem for No-

Sharing-Problem can be solved in polynomial time, if D isdefined as a convex hull of a given set of traffic matrices.

Proof: The problem can be solved using the presentedalgorithm. The algorithm is polynomial as it invokes a poly-nomial method finding the minimum of a convex function.The method is invoked a polynomial number of times.

VI. NUMERICAL RESULTS

We tested our algorithms on an Intel 2.4 GHz CPU with 3.25GB RAM, using a linear programming solver CPLEX 11.0[10]. We built our example cases using real world networksavailable in SNDlib [11]. We used altanta and francetopologies. Both were tested using two different sets of activenodes (we will refer to them as VLANs) and three differenttraffic demand polytopes defined for each of the VLANs. Thefirst polytope for each VLAN satisfied restrictions of Hosemodel presented in [8], i.e., traffic can originate and terminatein active nodes only and both outgoing and incoming traffic arelimited by given bounds. Another two traffic demand polytopeswere convex hulls of a number (3 or 10) of extreme points ofthe first polytope.

This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings.978-1-4244-4148-8/09/$25.00 ©2009

TABLE IPERFORMANCE OF THE PRESENTED ALGORITHMS

Robust-Routing-Problem No-Sharing-Problemnetwork Different-Reservation Identical-Reservation Different-Reservation Identical-Reservation

topology |V| |A| dim. D variant gain[%] time[s] gain[%] time[s] gain[%] time[s] gain[%] time[s]3 pts. 19.2 0.3 13.2 2.1 56.8 0.3 48.9 0.6 / 0.3

12 10 pts. 29.7 0.3 22.1 2.6 54.4 0.3 39.5 3.0 / 0.3atlanta 15 44 Hose 26.9 0.4 22.0 2.7 54.4 0.3 39.5 3.0

3 pts. 41.6 1.2 39.0 10.5 47.1 0.6 41.4 1.6 / 0.556 10 pts. 14.2 6.1 9.2 525.4 22.8 0.8 18.4 2.5 / 0.9

Hose 3.5 80.0 1.7 3785.5 10.8 0.6 9.2 1.43 pts. 29.1 1.0 19.7 6.4 45.6 0.8 40.3 1.5 / 0.8

20 10 pts. 4.2 1.7 1.6 12.2 34.7 1.1 28.7 2.2 / 1.3france 25 90 Hose 1.4 3.3 0.2 48.0 28.0 1.1 24.3 2.8

3 pts. 28.6 5.6 24.8 52.4 46.1 3.4 37.4 8.3 / 3.456 10 pts. 16.0 20.5 5.5 3115.1 20.5 3.4 17.2 6.9 / 4.0

Hose 4.6 7542.2 1.1-4.6 9000.0 5.0 2.8 4.7 4.6

Our test cases represent congested network configurationsand our task is to minimize additional costs incurred by thecongestion. Therefore, we introduced additional uncapacitatedlinks between all pairs of nodes. Routing costs of theseadditional links were set to 1, while routing costs of linksfrom the original network were set to 0. We can considerthose additional links as possibilities to rent capacity fromother operators. Each Hose model polytope was scaled in sucha way that the minimal congestion on the most heavily utilizedlink for the polytope built on its 3 extreme points was 1.1. Foreach topology and each polytope we used up to five differentdirections α of the hyperplane.

In Table I we present networks used in our experiments,possible gains when different strategies of dividing a trafficdemand polytope are implemented (we present results obtainedfor the best case out of 10 randomly generated test cases), andrunning times of the algorithms that are capable of solvingthe considered scenarios. The first five columns describenetworks used in the simulations: network – name of thetopology in SNDlib, |V| – number of nodes, |A| – numberof directed links, dim. D – number of tij > 0 (dimensionof a traffic demand polytope), variant – way traffic demandpolytopes were created (either a full Hose model polytope, ora convex hull built on 3 or 10 extreme points of the full Hosemodel polytope). The following eight columns present resultsobtained for four different variants of the problem. By gainwe understand the percentage of rented capacity that can besaved by using the partitioning of the traffic demand polytope.

Both versions of Different-Reservation-Problem were solvedby means of the method of Section III-A. Identical-Reservation-Problem for Robust-Routing-Problem was solvedusing the double binary search algorithm of Section IV, whileIdentical-Reservation-Problem for No-Sharing-Problem wassolved using both the double binary search method and thealgorithm of Section V. That is why the execution times forthis version of the problem are presented as a/b, where a istime consumed by the double binary search algorithm and bis time consumed by the algorithm of Section V.

The 9000-second time limit has been hit once. Identical-Reservation-Problem for Robust-Routing-Problem has not

been solved to optimality for the last test case. Therefore, inthis case the gain is presented as a − b, where a is the gainof the best solution found and b is the upper bound for it. Inother words, the optimal gain belongs to [a, b].

VII. CONCLUSIONS

In the paper we presented two algorithms: the doublebinary search algorithm that solves Identical-Reservation-Problem, and another (faster) algorithm that solves Identical-Reservation-Problem for No-Sharing-Problem when the trafficdemand polytope is given by a set of its extreme points. Thealgorithms were thoroughly tested and proven to be applicableto networks of practical sizes. We also showed possible savingswhile migrating from the robust stable routing in a directionof the dynamic routing.

The presented algorithms can be easily extended to solvea problem when the traffic demand polytope is divided bymany hyperplanes of the same direction. However, a moregeneral problem that consists in providing an optimal set ofhyperplanes of unknown directions still remains a challenge.

REFERENCES

[1] G. R. Ash. Dynamic routing in telecommunications networks. McGraw-Hill Professional, New York, USA, 1997.

[2] W. Ben-Ameur. Between fully dynamic routing and robust stablerouting. In Proc. DRCN, 2007.

[3] W. Ben-Ameur and H. Kerivin. New economical virtual private net-works. Commun. ACM, 46(6):69–73, 2003.

[4] W. Ben-Ameur and H. Kerivin. Routing of uncertain traffic demands.Optim. Eng., 6(3):283–313, 2005.

[5] A. Ben-Tal and A. Nemirovski. Robust optimization methodology andapplications. Math. Program., 92(3):453–480, May 2002.

[6] C. Chekuri. Routing and network design with robustness to changingor uncertain traffic demands. SIGACT News, 38(3):106–129, 2007.

[7] C. Chekuri, F. B. Shepherd, G. Oriolo, and M. G. Scutella. Hardnessof robust network design. Networks, 50(1):50–54, August 2007.

[8] N. G. Duffield, P. Goyal, A. Greenberg, P. Mishra, K. K. Ramakrishnan,and J. E. Van Der Merwe. A flexible model for resource managementin virtual private networks. In Proc. ACM SIGCOMM, 1999.

[9] J. A. Fingerhut, S. Suri, and J. S. Turner. Designing least-costnonblocking broadband networks. J. Algorithm, 24(2):287–309, 1997.

[10] ILOG. CPLEX 11.0 User’s Manual. ILOG, 2007. http://www.lingnan.net/lab/uploadfile/200864184419679.pdf.

[11] S. Orlowski, M. Pioro, A. Tomaszewski, and R. Wessaly. SNDlib1.0–Survivable Network Design Library. In Proc. INOC, 2007.http://sndlib.zib.de.

This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings.978-1-4244-4148-8/09/$25.00 ©2009