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A relaxed projection method for finite-dimensional equilibrium problemsS. Scheimberg a & P.S.M. Santos ba PESC/COPPE, IM, UFRJ , Rio de Janeiro 21941-972 , Brazilb Department of Mathematics , UFPI, Campus Ministro PetronioPortela, Bl – 04, Teresina 64049-610 , BrazilPublished online: 15 Nov 2010.

To cite this article: S. Scheimberg & P.S.M. Santos (2011) A relaxed projection method for finite-dimensional equilibrium problems, Optimization: A Journal of Mathematical Programming andOperations Research, 60:8-9, 1193-1208, DOI: 10.1080/02331934.2010.527974

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OptimizationVol. 60, Nos. 8–9, August–September 2011, 1193–1208

A relaxed projection method for finite-dimensional

equilibrium problems

S. Scheimberga and P.S.M. Santosb*

aPESC/COPPE, IM, UFRJ, Rio de Janeiro 21941-972, Brazil;bDepartment of Mathematics, UFPI, Campus Ministro Petronio Portela,

Bl – 04, Teresina 64049-610, Brazil

(Received 2 March 2010; final version received 23 September 2010)

In this article we consider an equilibrium problem for a differentiablebifunction which is not necessarily monotone. We present an implemen-table projection method. At each iteration, only one inexact projectiononto a simple approximation of the constraint set is performed, such as apolyhedron, which renders it numerically attractive. The algorithm canidentify, in practice, a subsequence that converges to a solution underreasonable assumptions. Some numerical results are reported showing theperformance of our algorithm.

Keywords: projection method; equilibrium problem; weak cocoercivity

1. Introduction

Let C be a nonempty closed convex subset of Rn and let f :Rn

�Rn! (�1,þ1] be a

function such that f(x, x)¼ 0 for all x2C and C�C is contained in the domain of f.We consider the following Equilibrium problem (denoted by EP( f,C) in the sequel):

EP f,Cð ÞFind x� 2C such that

f ðx�, yÞ � 0 8y2C:

�ð1Þ

In this article, we assume that function f(x, �) :Rn! (�1,þ1] is convex and

differentiable at x, for all x2E, where C�E and E�E� dom( f ). Let fy(x, x) :¼fy(x, �)(x) for x2E.

The equilibrium problem provides a unified framework for a wide class ofproblems. Convex optimization, variational problems, nonlinear complementarityproblems, Nash equilibrium problems, fixed point problems and vector optimiza-tion problems can be regarded as particular cases of the equilibrium problem.

Results on the existence of solutions for equilibrium problems have beenextensively studied (see, e.g. [2,3,12,26] and the references therein).

Numerical algorithms have been proposed for solving the equilibrium problemin finite- or infinite-dimensional spaces using the auxiliary problem principle,

*Corresponding author. Email: psergio@ufpi.edu.br

ISSN 0233–1934 print/ISSN 1029–4945 online

� 2011 Taylor & Francis

http://dx.doi.org/10.1080/02331934.2010.527974

http://www.tandfonline.com

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the proximal point method and combinations of these methods with projections (see,e.g. [5,13,15,18,22,23] and the references therein). Methods for solving the equilib-rium problem with a differentiability condition on f(x, �) were given, for example in[17,29] and the references therein. This condition is also considered in [24] for thegeneralized Nash equilibrium problem.

The goal of this article is to present an implementable and tractable algorithm tosolve the class of equilibrium problems considered here. At each iteration wecompute only one inexact projection onto a convenient set, like a polyhedron, whichapproximates the original set C. We can handle inner, outer or other kinds ofapproximations of the original set. We emphasize the well-known advantage ofprojection methods from a computational point of view. We prove the existence of afeasible cluster point of the generated sequence by the algorithm if EP( f,C) issolvable and a weak cocoercivity condition is verified. It is easy to identify,in practice, a subsequence converging to such a cluster point. We prove that thecluster point is a solution if certain continuity assumptions hold for f(x, �) orfy(x, �)(x). Furthermore, we show that the whole sequence converges to a solution byassuming the cocoercivity condition with respect to the solution set. This article isorganized as follows: in Section 2 we recall some useful tools that will be used in thesequel. In Section 3 we define the algorithm and study its convergence. Finally, inSection 4, some computational experiments are reported.

2. Preliminaries

In this section, we present some basic concepts, properties and notations that we willconsider in the development of our work.

Given a symmetric positive definite matrix D of order n, we recall that theD-projection of a point x onto a nonempty closed convex set C�R

n, PDC :Rn! R

n,is defined by

PDCðxÞ ¼ argmin

y2C

1

2kx� yk2D

� �, ð2Þ

where uk k2D¼ hu, uiD ¼ hu,Dui for all u2Rn and h�, �i denotes the Euclidean inner

product.The D-projection operator has the following useful properties which can be

found in [25].

LEMMA 2.1 Let x2Rn. Then, the following assertions hold:

(1) The point px is the D-projection of x onto C, PDCðxÞ, if, and only if, it holds:

hx� px, y� pxiD � 0 8y2C:

(2) The D-projection operator onto C is nonexpansive in relation to the D-norm, i.e.

kPDCðxÞ � PD

Cð yÞkD � kx� ykD 8x, y2Rn:

The following lemma will be used in our convergence analysis.

LEMMA 2.2 Let D be a symmetric positive definite matrix of order n, x, y2Rn and

�2 [0, 1] . Then, it holds:

k�xþ ð1� �Þ yk2D ¼ �kxk2D þ ð1� �Þk yk

2D � �ð1� �Þkx� yk2D: ð3Þ

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The following characterization of the equilibrium problem in terms of theD-projection operator will be used in the convergence analysis.

LEMMA 2.3 Assume that f(x, �) :Rn! (�1,þ1] is convex and differentiable at x for

all x2C. Then, a point x2C is a solution of EP( f,C) if, and only if,

x ¼ PDC ½x� �D

�1fyðx, xÞ, �4 0: ð4Þ

Proof The point x is a solution of EP( f,C) if and only if x verifies

x2 argminx2C

f ðx, xÞ: ð5Þ

Furthermore, since f ðx, �Þ is convex and differentiable at x, this last relation isequivalent to affirm that x satisfies

h fyðx, xÞ, x� xi � 0 8x2C: ð6Þ

In other words, for every x2C, x verifies:

hx� ½x� �D�1fyðx, xÞ,x� xiD ¼ �h fyðx, xÞ, x� xi � 0: ð7Þ

Hence, from Lemma 2.1, we conclude that (4) and (5) are equivalent. g

Notice that the solution set of the equilibrium problem (1) is equal to the solutionset of the variational inequality problem (6) [13,14].

We also recall the definition of �D-distance between sets introduced in [25] whichis based on the definition of �-distance of functions given in [1]. We denote byNCCS(E ), the family of nonempty closed convex subsets of E.

Definition 2.4 Let C1,C22NCCS(E ) and � � 0. The �D-distance between C1 and C2

is given by

d�DðC1,C2Þ :¼ supkxkD��

kPDC1ðxÞ � PD

C2ðxÞkD: ð8Þ

Let us observe that d�D is a pseudo-distance.

We close this section with the following property used to obtain the convergenceof the whole sequence to a solution (see, e.g. [4]).

PROPOSITION 2.5 (Quasi-Fejer convergence theorem) Let SRn be a nonempty set

and let {xk}Rn be a sequence such that for every x� 2S there is a summable sequence

{�k}Rþ verifying

kxkþ1 � x�k2D � kxk � x�k2D þ �k ð9Þ

for all k2N. Then, the sequence {xk} is bounded. Furthermore, if a cluster point x of{xk} belongs to S, then the whole sequence converges to x.

3. The relaxed projection method for equilibrium problems

Let D be a symmetric positive definite matrix of order n and let {�k} and {�k} besequences of nonnegative parameters such that:

R1. {�k} [0, 1] ,P�k(1� �k)¼þ1 and {�k} [a, b], 05a� b.

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We will consider the following notation:

Tðx,�Þ :¼ PDC ½x� �D

�1fyðx, xÞ;Tkðx,�Þ :¼ PD

Ck½x� �D�1fyðx, xÞ.

3.1. Algorithm RPMEP

Initialization: Choose x02E and C12NCCS(E ). Set k¼ 0.Iterative step: Let xk2E and Ckþ12NCCS(E ).

Obtain an inexact evaluation of Tkþ1(xk, �k), denoted by Tkþ1(x

k, �k)þwk. Set

xkþ1 :¼ ð1� �kÞxk þ �k½Tkþ1ðx

k,�kÞ þ wk

¼ð1� �kÞxk þ �k PD

Ckþ1½xk � �kD

�1fyðxk, xkÞ þ wk

n o: ð10Þ

We observe that (10) corresponds to an inexact Mann iterate [6]. When Ck¼C,for all k2N, the algorithm RPMEP becomes an instance of the method given in [28]for finding a fixed point of a nonexpansive operator.

At each iteration k, an estimate of the projection onto the set C or onto asimple approximation Ck of C (like a polyhedron) is computed. The inclusion of anerror wk in (10) gives a more realistic model for the implementation of the algorithm(see, e.g. [6]).

The method allows us to consider inner (Ck�C), outer (C�Ck) or another kindof approach. Approximations of the constraint set have also been considered in[11,16,21,25] and the references therein.

3.2. Convergence analysis

From now on we denote by {xk}, the sequence generated by Algorithm RPMEP.The following result used to derive convergence properties of {xk} is a direct

consequence of Definition 2.4.

LEMMA 3.1 Let x2Rn, �40 and �40 . If kx� �D�1fy(x, x)kD5�, then it holds:

kTkþ1ðx,�Þ � Tðx,�ÞkD � d�D ðCkþ1,CÞ 8k2N: ð11Þ

We consider first the following additional assumptions in order to obtaincontractive properties.

A1 The solution set of EP( f,C), denoted by S( f,C), is nonempty.A2 The differential operator fy is cocoercive on E with respect to x� 2S( f,C) with

modulus �40, i.e.

h fyðx, xÞ � fyðx�, x�Þ, x� x�i � �k fyðx, xÞ � fyðx

�, x�Þk2 8x2E: ð12Þ

Let us observe that Assumption A1 is a usual requirement for EP( f,C) (see, e.g.[15,18,22]). We also note that Assumption A2 is weaker than the classicalcocoercivity condition, which is a common requirement for methods using onlyone projection to solve variational inequalities (see, e.g. [9,20,28] and the referencestherein).

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The next nonexpansive property related to a solution of EP( f,C) will be used in

the convergence analysis.

PROPOSITION 3.2 Assume that A1 and A2 are verified. Let �m(D) be the minimum

eigenvalue of D. Then, for every x2E, �2 [0, 2��m(D)] and k2N it hold

kTkðx,�Þ � Tkðx�,�ÞkD � kx� x�kD ð13Þ

and

kTðx,�Þ � Tðx�,�ÞkD � kx� x�kD: ð14Þ

Proof By combining the hypotheses, the definition of Tk and Lemma 2.1, it results

kTkðx,�Þ � Tkðx�,�Þk2D ¼ kP

DCk½x� �D�1fyðx, xÞ � PD

Ck½x� � �D�1fyðx

�,x�Þk2D

� kx� x� � �D�1ð fyðx,xÞ � fyðx�, x�ÞÞk2D

¼ kx� x�k2D � 2�hx� x�, fyðx, xÞ � fyðx�, x�Þi

þ �2 fyðx, xÞ � fyðx�, x�Þ,D�1½ fyðx, xÞ � fyðx

�, x�Þ� �

� kx� x�k2D þ ð�2kD�1k � 2��Þk fyðx, xÞ � fyðx

�, x�Þk2: ð15Þ

Thus, we conclude that (13) is verified, by taking �2 [0, 2��m(D)]. The same

argument is used to prove (14). g

From now on we assume condition R1 with b¼ 2��m(D).The next proposition allows us to obtain the boundedness of the sequence and,

therefore, the existence of cluster points.

THEOREM 3.3 Assume that Assumptions A1, A2 and R1 are fulfilled. If

R2Pþ1

k¼0 �k½d�DðCkþ1,CÞ þ kwkkD5þ1 8� � 0,

then:

(i) The sequence {xk}k2N is bounded;(ii) The sequence {xk}k2N has cluster points.

Proof (i) Let x� 2S( f,C) verifying A2. Then by Lemma 2.3 we have

x� ¼ Tðx�,�kÞ 8k2N: ð16Þ

Taking x�¼ (1� �k)x�þ �kT(x

�, �k) and using the definition of xkþ1 we get

kxkþ1 � x�kD ¼ kð1� �kÞðxk � x�Þ þ �k Tkþ1ðx

k,�kÞ � Tðx�,�kÞ þ wk� �

kD

� ð1� �kÞkxk � x�kD þ �kkTkþ1ðx

k,�kÞ � Tkþ1ðx�,�kÞkD

þ �k kTkþ1ðx�,�kÞ � Tðx�,�kÞkD þ kw

kkD�

: ð17Þ

Thus, using this inequality and (13), with �¼�k, we deduce that

kxkþ1 � x�kD � kxk � x�kD þ �k kTkþ1ðx

�,�kÞ � Tðx�,�kÞkD þ kwkkD

� �: ð18Þ

Now, by R1 we have that {�k} [a, b]. Therefore, we can take �� such that

kx�� �kD�1fy(x

�, x�)kD� �� for every k2N. Therefore, by Lemma 3.1 and (18)

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we obtain

kxkþ1 � x�kD � kxk � x�kD þ �k½d��DðCkþ1,CÞ þ kw

kkD

� kx0 � x�kD þXkj¼0

�j½d��DðCjþ1,CÞ þ kw

jkD

� kx0 � x�kD þXþ1j¼0

�j½d��DðCjþ1,CÞ þ kw

jkD: ð19Þ

Hence, due to Assumption R2 we conclude that the sequence {kxkþ1� x�k} is

bounded. Therefore, {xk}k2N is also bounded.Part (ii) is obtained from (i). g

Let us note that, for Ck¼C for all k2N, the condition R2 becomesP�kkw

kk5þ1 which is considered in [28].

The following corollary is a straightforward consequence of Theorem 3.3,

Proposition 3.2 and A2.

COROLLARY 3.4 Under the assumptions of Theorem 3.3 it holds that all the sequences

{fy(xk, xk)}, {kxk��k fy(x

k,xk)k} and {T(xk,�k)} are bounded.

The next technical result is used to derive an important property involving the

residual sequence {kxk�T(xk, �k)k}.

PROPOSITION 3.5 Assume that conditions A1, A2, R1 and R2 are verified. Then,

for all k2N it holds

kxkþ1 � x�k2D � kxk � x�k2D � �kð1� �kÞkx

k � Tðxk,�kÞk2D þ �k ð20Þ

where {�k}Rþ andPþ1

k¼0 �k 5þ1.

Proof Let x� 2S( f,C) verifying condition A2. By Lemma 2.3 we can take

x� ¼ ð1� �kÞx� þ �kTðx

�,�kÞ: ð21Þ

Now, define

ek ¼ Tkþ1ðxk,�kÞ � Tðxk,�kÞ þ wk: ð22Þ

According to (21), (22) and (10) we have

kxkþ1 � x�k2D ¼ kð1� �kÞðxk � x� þ �ke

kÞ

þ �k Tðxk,�kÞ � Tðx�,�kÞ þ �kek

� �k2D: ð23Þ

Using (3) and (21) in the right-hand side of the above equality, we get

kxkþ1 � x�k2D ¼ ð1� �kÞkxk � x� þ �ke

kk2D

þ �kkTðxk,�kÞ � Tðx�,�kÞ þ �ke

kk2D

� �kð1� �kÞkxk � Tðxk,�kÞk

2D: ð24Þ

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Hence, by the Cauchy–Schwarz inequality it follows

kxkþ1 � x�k2D � ð1� �kÞ kxk � x�k2D þ 2kxk � x�kDk�ke

kkD þ k�kekk2D

� �þ �k kTðx

k,�kÞ � Tðx�,�kÞk2D þ 2kTðxk,�kÞ � Tðx�,�kÞkDk�ke

kkD�

þ k�kekk2D

�� �kð1� �kÞkx

k � Tðxk,�kÞk2D: ð25Þ

Therefore, by Proposition 3.2 we obtain

kxkþ1 � x�k2D � kxk � x�k2D þ 2kxk � x�kDk�ke

kkD þ k�kekk2D

� �kð1� �kÞkxk � Tðxk,�kÞk

2D: ð26Þ

We claim that the second and third terms of the right-hand side of this inequality are

bounded. Indeed, in view of Theorem 3.3 we can take an upper bound L of thesequence {kxk� x�kD}. In order to obtain the boundedness of {kekkD} we first apply

Corollary 3.4 to get an upper bound �4 0 of the sequence {kxk��k fy(xk,xk)kD}.

Then, using Lemma 3.1 we obtain

kekkD � d�DðCkþ1,CÞ þ kwkkD: ð27Þ

Hence, in view of (26) and (27) we get

kxkþ1 � x�k2D � kxk � x�k2D � �kð1� �kÞkx

k � Tðxk,�kÞk2D þ �k,

where

�k :¼ 2L�k d�DðCkþ1,CÞ þ kwkkD

� þ��k d�D ðCkþ1,CÞ þ kw

kkD� �2

:

According to R2 we have thatP1

k¼0 �k½d�DðCkþ1,CÞ þ kwkkD51. Therefore,

it follows thatP1

k¼0½�kd�DðCkþ1,CÞ þ kwkkD

2 51. So,Pþ1

k¼0 �k 5þ1. g

The result below is important for the convergence analysis.

THEOREM 3.6 Assume that conditions A1, A2, R1, and R2 are verified. Then,

lim infk!þ1

kxk � Tðxk,�kÞkD ¼ 0: ð28Þ

Proof By Proposition 3.5, we have that there exists a summable sequence {�k} suchthat for all k2N we get

�kð1� �kÞkxk � Tðxk,�kÞk

2D � kx

k � x�k2D � kxkþ1 � x�k2D þ �k:

It results in the following equation,

Xmk¼0

�kð1� �kÞkxk � Tðxk,�kÞk

2D � kx

0 � x�k2D � kxmþ1 � x�k2D þ

Xmk¼0

�k

� kx0 � x�k2D þXþ1k¼0

�k: ð29Þ

Hence, we get

X1k¼0

�kð1� �kÞkxk � Tðxk,�kÞk

2D 5þ1: ð30Þ

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Consider the conditions on the parameter �k in R1 so (30) implies

lim infk!þ1

kxk � Tðxk,�kÞk2D ¼ 0:

g

COROLLARY 3.7 Assume that the hypotheses of Theorem 3.6 are fulfilled. Then, there

is a subsequence fxkjg of {xk} and a point x2C such that

(i) limj!þ1 xkj ¼ x;(ii) limj!þ1 kx

kj � Tðxkj ,�kj Þk ¼ 0:

Proof By considering Theorems 3.3 and 3.6 and Corollary 3.4 we get that there is a

subsequence fxkjg of {xk} converging to a point x and there is a point p such that

limj!þ1

kxkj � Tðxkj ,�kj Þk ¼ 0 ð31Þ

and

limj!þ1

Tðxkj ,�kjÞ ¼ p: ð32Þ

Hence,

0 � kxkj � pk � kxkj � Tðxkj ,�kj Þk þ kTðxkj ,�kj Þ � pk:

Therefore, combining (31) and (32), and taking limits in the above inequalities

give us

limj!þ1

xkj ¼ x ¼ p:

Since Tðxkj ,�kj Þ 2C for all j2N and C is closed, it follows that x2C. g

Let us note that the subsequence fxkjg verifying (31) is easy to be identified from a

numerical point of view.Now, we prove that x obtained in Corollary 3.7, which is an accumulation point

of {xk}, is a solution of the equilibrium problem when a continuous assumption on

fy(�, �)(x) :¼ fy(x, �)(x) or on f is considered. Next we have the first result.

THEOREM 3.8 Assume that conditions A1,A2,R1, and R2 are satisfied. In addition,

assume that the following condition holds:

A3 The operator fy(�, �) is continuous on E.Then, the point x obtained in Corollary 3.7 is a solution of EP( f,C).

Proof By condition R1 we have that {�k} is bounded. Therefore, we can assume,

without loss of generality, that f�kjg is convergent, that is, there exists �4 0 such that

limj!þ1

�kj ¼ �: ð33Þ

From Lemma 2.1 and (33) we obtain that:

limj!þ1

PDCðx

kj � �kjD�1fyðx

kj , xkj ÞÞ ¼ PDCðx� �D

�1fyðx, xÞÞ:

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Hence, using (31) it results

0 ¼ limj!þ1

kxkj � PDCðx

kj � �kjD�1fyðx

kj , xkjÞÞkD ¼ kx� PDCðx� �D

�1fyðx, xÞÞkD:

ð34Þ

So, we get

x ¼ PDCðx� �D

�1fyðx, xÞÞ:

The conclusion follows from Lemma 2.3. g

Let us note that Assumption A3 is considered, for example, in [29] in the

framework of monotone equilibrium problems.Now, we establish our second result.

THEOREM 3.9 Assume that conditions A1, A2, R1, and R2 are fulfilled and that

(x, x)¼ 0 for all x2E. If the following condition holds:

A4 f(�, u) is upper semicontinuous for all u2C,then, the point x obtained in Corollary 3.7 is a solution of EP( f,C).

Proof Let fxkjg be the subsequence converging to x. Take an arbitrary u2C.

By Lemma 2.1 and the Cauchy–Schwarz inequality it follows that

0 � hTðxkj ,�kjÞ � ½xkj � �kjD

�1fyðxkj , xkjÞ, u� Tðxkj ,�kj ÞiD

¼ �kjh fyðxkj , xkj Þ, u� xkji þ �kjh fyðx

kj , xkj Þ, xkj � Tðxkj ,�kjÞi

þ hTðxkj ,�kjÞ � xkj , u� Tðxkj ,�kjÞiD

� �kjh fyðxkj , xkjÞ, u� xkji þ

��kjk fyðx

kj , xkj Þk:

þ ku� Tðxkj ,�kj ÞkkDk�kxkj � Tðxkj ,�kj Þk: ð35Þ

On the other hand, by the convexity of f ðxkj , �Þ, we have

f ðxkj , uÞ � f ðxkj , xkjÞ þ h fyðxkj , xkjÞ, u� xkji: ð36Þ

Since �kj 4 a4 0, (35) and (36) imply that

f ðxkj , xkjÞ � f ðxkj , uÞ

þ k fyðxkj , xkjÞk þ

ku� Tðxkj ,�kjÞkkDk

a

" #kxkj � Tðxkj ,�kj Þk: ð37Þ

Now, Corollary 3.4 implies that

k fyðxkj , xkj Þk þ

ku� Tðxkj ,�kjÞk

akDk

( )

is bounded. By taking lim inf in (37) and recalling that f ðxkj , xkjÞ ¼ 0 for all j2N,

we get

lim infkj!þ1

f ðxkj , uÞ � 0: ð38Þ

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We conclude from Assumption A4 and (38) that

f ðx, uÞ � lim supj!þ1

f ðxkj , uÞ

� lim infj!þ1

f ðxkj , uÞ

� 0:

Hence, x is a solution of EP( f,C). g

For inner approximations of the constraint set C, the above theorem holds byconsidering E¼C.

Note that condition A4 is also used in [13,14], for instance.We say that a sequence {xk} is a generalized asymptotically solving sequence for

EP( f,C) when there is a subsequence verifying (38) and {xk}E . Let us recall that{xk} is an asymptotically solving sequence for EP( f,C) when (38) is verified for thewhole sequence [13,15].

Finally, we obtain the convergence of the whole sequence under the cocoercivitycondition of fy(x, x) with respect to the solution set of EP( f,C).

THEOREM 3.10 Assume that conditions A1, R1 and R2 are fulfilled and A2 is verifiedwith respect to each solution of EP( f,C). If there is an accumulation point of {xk}which is a solution of EP( f,C), then, the whole sequence {xk} converges to a solution ofthe equilibrium problem.

Proof Let x� 2S( f,C). Then, by hypotheses, A2 holds with respect to x�.Therefore, by Proposition 3.5, there is a sequence {�k}Rþ,

Pþ1k¼0 �k 5þ1 such

that

kxkþ1 � x�k2D � kxk � x�k2D þ �k:

Hence, by Proposition 2.5 we reach our goal. g

4. Numerical results

In this section we illustrate the algorithm RPMEP. Some comparisons are alsoreported.

The algorithm has been coded in MATLAB 7.8 with the stopping conditionkxk�Tk(x

k, �k)kD510�8, on a 1GB RAM Pentium IV Dual Core desktop. InExamples 4.1 and 4.4, we choose D as in [7], i.e. D is taken to be the diagonal matrixwhose elements are equal to the diagonal elements of the linear part of the operatorfy(�, �). In the remaining examples, we consider D as the identity.

Example 4.1 Consider a two-player game with the following payoff functions onthe region C¼ {(x1, x2)2R

2 :� 10� x1, x2� 10},

�jðx1, x2Þ :¼ðx1 þ x2Þ

2

4þðx1 � x2Þ

2

9, j ¼ 1, 2: ð39Þ

This problem is solved in [19] by considering the bifunction

f ðx, yÞ :¼ ½�1ð y1, x2Þ � �1ðx1, x2Þ þ ½�2ðx1, y2Þ � �2ðx1, x2Þ:

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The Nash equilibrium point for this game is x�¼ (0, 0). Let us observe that the

derivative of f(x, �) with respect to y, at x, is

fyðx, xÞ ¼1318

518

518

1318

!x1

x2

�

Hence, we have that fy(�, �) is cocoercive with modulus �¼ 1.

We consider the starting iterate x0¼ (10, 5) given in [19], Ck¼C¼E, �k ¼k

kþ1,

D ¼ diagð1318 ,1318Þ and �k¼ 2�m(D)�. In Table 1, we show the behaviour of RPMEP in

relation to the relaxation algorithm (RA) given in [19].

Example 4.2 Consider the river basin pollution problem given in [19] which consists

of three players with payoff functions:

�jðxÞ ¼ ujx2j þ 0:01xjðx1 þ x2 þ x3Þ � vjxj, j ¼ 1, 2, 3,

where u¼ (0.01, 0.05, 0.01) and v¼ (2.90, 2.88, 2.85) and the constraints are given

by x1, x2, x3� 0, 3.25x1þ 1.25x2þ 4.125x3� 100 and 2.291x1þ 1.5625x2þ

2.8125x3� 100. In this case, f ðx, yÞ ¼P3

j¼1½�jð yj j xÞ � �jðxÞ, where (y1 jx)¼

(y1, x2, x3), . . . , (y3 j x)¼ (x1, x2, y3). So, we get

fyðx, xÞ ¼0:04 0:01 0:010:01 0:12 0:010:01 0:01 0:04

0@

1A x1

x2x3

0@

1A � 2:90

2:882:85

0@

1A

is cocoercive with modulus �¼ 8.1466. We take Ck¼C¼E, �k ¼ ð1:99�1kÞ�,

�k ¼k

kþ1 and D¼ I and the starting iterate x0¼ (0, 0, 0) like in [19]. The Nash

equilibrium point of this game is x�¼ (21.149, 16.028, 2,722).

Table 2 gives the results obtained by RPMEP algorithm and by the RA

used in [19].

Example 4.3 Consider three equilibrium problems given in [22], where C ¼ Rnþ

and the bifunction is of the form

f ðx, yÞ ¼ hPxþQyþ q, y� xi: ð40Þ

Table 1. A two-player game.

RA RPMEP

Iteration (k) xk1 xk2 xk1 xk2

1 1.2849 �1.4668 1.3888 �1.38882 1.5639 �0.4942 0.5658 �0.56583 0.1901 �0.2169 0.1886 �0.18864 0.0834 �0.0731 0.0544 �0.05445 0.0281 �0.0321 0.0141 �0.01416 0.0123 �0.0108 0.0033 �0.00337 0.0082 �0.0078 0.0007 �0.0007

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The matrix Q and the vector q are defined by

Q ¼

1:6 1 0 0 0

1 1:6 0 0 0

0 0 1:5 1 0

0 0 1 1:5 0

0 0 0 0 2

26666664

37777775

and q ¼

�1

�2

�1

2

�1

26666664

37777775

for the first two instances and by

Q ¼

2:3550 1:6364 1:8430 2:1540 0:7586

1:6364 1:6620 1:5323 1:4876 0:2901

1:8430 1:5323 2:4317 2:2961 1:0964

2:1540 1:4876 2:2961 2:8473 1:2273

0:7586 0:2901 1:0964 1:2273 0:8085

26666664

37777775

and q ¼

�1

�1

0

0

0

26666664

37777775

for the last instance.The matrix P is given by

P ¼

3:1 2 0 0 0

2 3:6 0 0 0

0 0 3:5 2 0

0 0 2 3:3 0

0 0 0 0 3

26666664

37777775,

3:1 2 0 0 0

2 3:6 0 0 0

0 0 3:5 2 0

0 0 2 3:3 0

0 0 0 0 2

26666664

37777775

and 10I,

respectively, where I denotes the identity matrix of order 5.It follows that fy(x, x)¼ (PþQ)xþ q is cocoercive with modulus �, equal to

0.1256, 0.1256 and 0.0545, respectively.

We consider the starting point given in [22], �k¼ 2�, Ck¼C¼E, �k ¼k

kþ1

and D¼ I.

Table 2. A river basin pollution problem.

RA RPMEP

Iteration (k) xk1 xk2 xk3 xk1 xk2 xk3

1 19.35 17.19 3.79 6.1256 9.4720 4.42462 20.71 16.11 3.043 11.6052 17.0027 5.90613 20.88 16.07 2.924 15.4625 16.6220 6.01274 21.08 16.03 2.776 17.6922 16.1206 5.21605 21.10 15.03 2.757 19.0110 16.0847 4.3561... ..

. ...

10 21.14 16.03 2.730 20.9895 16.0301 2.8475... ..

. ...

20 21.14 16.03 2.729 21.1443 16.0278 2.7263

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In Table 3, we compare our algorithm with the interior proximal extragradient

(IPE) algorithm given in [22].Now, we apply our algorithm to the following optimization problem in order to

illustrate the use of simple approximations of the constraint set.

Example 4.4 Consider the well-known Rosen–Suzuki optimization problem and

its reformulation as an equilibrium problem. The bifunction is given by

f(x, y)¼�(y)��(x) with

�ðxÞ ¼ x21 þ x22 þ 2x23 þ x24 � 5x1 � 5x2 � 21x3 þ 7x4:

Hence, fy(x, x)¼ (2x1�5, 2x2�5, 4x3�21, 2x4þ 7)T.The constraint set is defined by C¼ {x2R

2 : gi(x)� 0, i¼ 1, 2, 3}, where

g1ðxÞ ¼ x21 þ x22 þ x23 þ x24 þ x1 � x2 þ x3 � x4 � 8,

g2ðxÞ ¼ x21 þ 2x22 þ x23 þ 2x24 � x1 � x4 � 10,

g3ðxÞ ¼ 2x21 þ x22 þ x23 þ 2x1 � x2 � x4 � 5:

The optimal point is x�¼ (0, 1, 2,�1). Let us note that fy is cocoercive with

modulus �¼ 0.25.

We consider x0¼ (5,�5, 5,�5) =2C, �k ¼k

kþ1 and D¼ diag(2, 2, 4, 2). We consider

the outer approximations of the constraint set given in [11]. We observe that the

method converges to the solution of the problem. However, the condition R2 is not

satisfied. We take C0¼E¼R4. Given xk and Ck�R

4, we define Ckþ1 as

Ckþ1 :¼\j2 Ik

fx2R4: giðx

jÞ þ hrgiðxjÞ, x� x ji � 0, i ¼ 1, 2, 3g,

where Ik {1, 2, . . . , k}. In Table 4, we show our results with different forms to

compute �k.

Table 3. A general equilibrium problem.

Problem 1 Problem 2 Problem 3

Algorithm IPE RPMEP IPE RPMEP IPE RPMEP

Iterations 19 12 20 12 40 16cpu (s) 1.078 0.0856 1.296 0.0764 10.875 0.1159

Table 4. A reformulation of Rosen–Suzuki optimizationproblem.

�k Iteration (k) cpu (s) kxk� x�kD

0.15(k2� 1)/k2 37 0.0488 6.531D� 100.25(k2� 1)/k2 20 0.0248 2.835D� 100.35(k2� 1)/k2 24 0.0344 6.338D� 100.45(k2� 1)/k2 100 0.2049 7.881D� 11

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In the two examples below, we test a set of complementarity problems given in[8,10] and [27] defined by

x � 0, FðxÞ � 0, hFðxÞ,xi � 0, ð41Þ

with F(x)¼S(x)þMxþ q, where S(x) and Mxþ q are the nonlinear part and thelinear part of F(x), respectively. Hence, the bifunction f and the constraint set aregiven by

f ðx, yÞ ¼ hFðxÞ, y� xi and C ¼ Rnþ: ð42Þ

Notice that, depending on the data, this problem could not be necessarily cocoercive.We compare the behaviour of our algorithm with the projection methods givenin [8,10] and [27].

Example 4.5 Consider the linear complementarity problem given in [27] with

M ¼

1 2 . . . . . . 2

0 1 2 . . . ...

..

. . .. . .

. . .. ..

.

..

. . .. . .

.2

0 . . . . . . 0 1

2666666664

3777777775, q ¼ ð�1, � 1, . . . , � 1ÞT:

In this case, we have S(x)¼ 0.

Following [27], we consider x0 generated uniformly in (0, 1), �¼ 0.9 and the stopcriterion kxk�T(xk, �k)k� 10�6. We take �k ¼

kkþ1 and �k¼ 2�, D¼ I.

In Table 5, we compare the numerical results of RPMEP with those algorithmsgiven in [8,27] which compute at least two projections per iteration.

Example 4.6 Consider the nonlinear complementarity problem (41) given in [10,27].The linear part of F is of the form M¼ATAþB, where A is an n� n matrix whoseentries are randomly generated in the interval (�5, 5) and the skew symmetric matrixB is generated in the same way. The vector q is generated from a uniform distribution

Table 5. A linear complementarity problem.

RPMEP Algorithm [8] Algorithm [27]

Dimension (n) Iteration (k) cpu (s) Iteration (k) cpu (s) Iteration (k) cpu (s)

500 17 0.0388 64 0.8412 14 0.2403600 18 0.0569 47 0.8813 12 0.3805700 17 0.0760 71 1.6323 15 0.4807800 17 0.0982 65 2.0129 15 0.6610900 16 0.1204 57 2.2532 15 0.82121000 17 0.1570 87 4.0859 15 1.02152000 17 0.6181 75 14.5209 14 4.05583000 17 1.3442 79 37.5840 16 10.0044

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in the interval (�500, 500). The components of the nonlinear operator S are definedby Sj(x)¼ sj (xj) where sj is a random variable in (0, 1).

We consider x0¼ 0, �k ¼k

kþ1, D¼ I and the stepsize sequence, {�k}¼ {�,�/2, �/2,�/3,�/3,�/3, . . .} proposed in [30]. We take �¼ 10�3, 2� 10�3, 6� 10�3 for n¼ 100,200, 500, respectively.

In Table 6, we compare our method with the forward–backward splittingalgorithm (FBA12) and the extra-gradient algorithm (EGA12) presented in [10].

Acknowledgements

The authors would like to thank the anonymous referees for their constructive comments andsuggestions which helped us to improve the presentation of this article. Scheimberg waspartially supported by CNPq, FAPERJ. Santos was Partially supported by CNPq, FAPEPI.

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