J Glob Optim
DOI 10.1007/s10898-011-9814-y
A new class of hybrid extragradient algorithms
for solving quasi-equilibrium problems
Jean Jacques Strodiot · Thi Thu Van Nguyen ·
Van Hien Nguyen
Received: 24 May 2011 / Accepted: 1 November 2011
© Springer Science+Business Media, LLC. 2011
Abstract Generalized Nash equilibrium problems are important examples of quasi-equilibrium problems. The aim of this paper is to study a general class of algorithms for solving
such problems. The method is a hybrid extragradient method whose second step consists in
finding a descent direction for the distance function to the solution set. This is done thanks to
a linesearch. Two descent directions are studied and for each one several steplengths are proposed to obtain the next iterate. A general convergence theorem applicable to each algorithm
of the class is presented. It is obtained under weak assumptions: the pseudomonotonicity of
the equilibrium function and the continuity of the multivalued mapping defining the constraint set of the quasi-equilibrium problem. Finally some preliminary numerical results are
displayed to show the behavior of each algorithm of the class on generalized Nash equilibrium
problems.
Keywords Quasi-equilibrium problems · Quasi-variational inequalities ·
Hybrid extragradient methods · Generalized Nash equilibrium problems
Mathematics Subject Classification (2000)
49J40 · 65K10 · 91A12
This research was supported by the Institute for Computational Science and Technology at Ho Chi Minh
City, Vietnam (ICST HCMC).
J. J. Strodiot · T. T. V. Nguyen · V. H. Nguyen
Institute for Computational Science and Technology, Ho Chi Minh City, Vietnam
T. T. V. Nguyen
e-mail: [email protected]
J. J. Strodiot (B) · V. H. Nguyen
Department of Mathematics, University of Namur, Namur, Belgium
e-mail: [email protected]
V. H. Nguyen
e-mail: [email protected]
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J Glob Optim
1 Introduction
The aim of this paper is first to present a general double-projection method as well as two
of its concrete realizations for solving a quasi-equilibrium problem in the sense of Blum and
Oettli [2], and then to illustrate numerically these methods on some test problems and in particular, on some generalized Nash equilibrium problems (GNEPs)[4]. Let X be a nonempty
closed convex subset of IRn , let K be a multivalued mapping from X into itself such that for
every x ∈ X, K(x) is a nonempty closed convex subset of X and let f : X × X → IR be a
mapping such that for every x ∈ X, f (x, x) = 0 and f (x, ·) is a convex function on X. The
quasi-equilibrium problem QE(K; f ) consists in finding x ∗ ∈ K(x ∗ ) such that
f (x ∗ , y) ≥ 0 for all y ∈ K(x ∗ ).
We denote the solution set of QE(K; f ) by K ∗ .
When f (x, y) = F (x), y−x where F : X → IRn , problem QE(K; f ) can be expressed
as: Find x ∗ ∈ K(x ∗ ) such that
F (x ∗ ), y − x ∗ ≥ 0 for all y ∈ K(x ∗ ).
This problem is called a quasi-variational inequality problem and is denoted QV I (K; F ).
When K(x) = K for every x ∈ X, problems QE(K; f ) and QV I (K; F ) are denoted
E(K; f ) and V I (K; F ), and called equilibrium problem and variational inequality problem,
respectively.
Problems QE(K; f ) and QV I (K; F ) have a number of important applications, for example, in economics, engineering, and operations research (see, for example, [3,7,11,19]). In
particular the GNEP can be modeled as a problem QE(K; f ) and when the data are differentiable as a problem QV I (K; f ). More precisely, assume there are N players each controlling
the variables xi ∈ IRni . We denote by x the overall vector of all variables: x = (x1 , . . . , xN ),
while we use the notation x−i = (x1 , . . . , xi−1 , xi+1 , . . . , xN ) to denote the vector of all players’ variables except that of player i. The aim of player i, given the other players’ strategies,
is to choose an xi ∈ Ki ⊂ IRni that solves the problem
Minimize fi (xi , x−i )
subject to xi ∈ Ki ,
where fi : IRn → IR and n = N
i=1 ni . We make the blanket assumption that the objective functions fi are, as a function of xi alone, convex, while the sets Ki are all nonempty
closed and convex. A point x is feasible if xi ∈ Ki for all players i. A solution of the Nash
equilibrium problem (NEP) is a feasible point x ∗ such that
∗
∗
fi (xi∗ , x−i
) ≤ fi (xi , x−i
) ∀xi ∈ Ki
holds for each player i = 1, . . . , N . So, a Nash equilibrium is a feasible strategy profile x ∗
with the property that no single player can benefit from a unilateral deviation from xi∗ as long
as the other players do not change their own strategies.
The GNEP extends the classical NEP by assuming that each player’s strategy set can
depend on the rival players’ strategies x−i . So we will write Ki (x−i ) to indicate that we
might have a different closed convex set Ki for each different x−i . Analogously to the NEP
case, the aim of each player i is, given x−i , to choose a strategy xi ∈ Ki (x−i ) that solves the
problem
Minimize fi (xi , x−i )
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subject to xi ∈ Ki (x−i ).
J Glob Optim
∗ ) such that,
A generalized Nash equilibrium (GNE) is a tuple of strategies x ∗ = (x1∗ , . . . , xN
for all i = 1, . . . , N ,
∗
∗
∗
fi (xi∗ , x−i
) ≤ fi (xi , x−i
) ∀xi ∈ Ki (x−i
).
The requirement that the feasible sets depend on the variables of players’ rivals is natural in
many applications (for example, the case in which the players share some common resource).
A GNEP is termed with shared constraints when the feasible sets Ki (x−i ) are defined by
Ki (x−i ) = {yi ∈ Ki | (yi , x−i ) ∈ Q},
where Ki is the nonempty closed and convex set of individual constraints of player i and Q
is a nonempty closed convex set in IRn representing the set of shared constraints, common
for all the players.
When the functions fi (·, x−i ) are convex and differentiable, we can define
N
F (x) = ∇xi fi (x) i=1
∗
and we see that x ∗ is a GNE if x ∗ ∈ K(x ∗ ) = N
i=1 Ki (x−i ) and
F (x ∗ )T (y − x ∗ ) ≥ 0 ∀y ∈ K(x ∗ ).
This problem is the quasi-variational inequality Problem QV I P (K; F ) (see, for instance,
[1,4,5,8,16–18,22,25,28,31] and the references quoted therein).
When the functions fi (·, x−i ) are convex and nondifferentiable, we define the NikaidoIsoda function
Ψ : IRn × IRn → IR
Ψ (x, y) =
N
fi (xi , x−i ) − fi (yi , x−i )
i=1
and we see that x ∗ is a GNE if x ∗ ∈ K(x ∗ ) =
N
i=1
∗ ) and
Ki (x−i
f (x ∗ , y) ≥ 0 ∀y ∈ K(x ∗ )
where f (x, y) = −Ψ (x, y) [2]. This problem is the quasi-equilibrium problem QE(K; f ).
We are interested in this work to develop efficient numerical methods for solving problems
QE(K; f ) and QV I (K; F ) and thus, in particular, for solving GNEPs.
Recently Fukushima [6], Taji [23] and Kubota et al. [13] proposed a class of gap functions for solving problem QV I (K; F ) (see also [15]). Unfortunately, contrary to the case of
variational inequalities, their gap functions were only directionally differentiable when the
function F is differentiable and the constraint set is given by convex differentiable inequalities. Furthermore, the gap functions being nonconvex, supplementary conditions must be
imposed to obtain that any stationary point of these functions is a global minimum. Such
conditions are known for problem V I (K; F ) but, to our knowledge, unknown for problem
QV I (K; F ). So in this paper we prefer to consider projection and extragradient methods
rather than methods based on gap functions for solving problems QEP (K; f ). More precisely, our aim is to generalize a class of double-projection methods for solving problem
V I (K; F ) to the case of equilibrium problems with a moving constraint set K(x).
After recalling in Sect. 2 several extragradient-type methods for solving variational
inequality problems and as well as their main properties, we propose in the next section
(Sect. 3) a very general algorithm inspired by these ideas for solving problem QE(K, f ).
The strategy is to reduce at each step the distance to the solution set. After that, conditions
are given on the data to force the convergence of this very general algorithm. In particular the
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function f is supposed to be pseudomonotone and continuous and the multivalued function
K to be continuous. Let us mention that a similar general algorithm has been proposed in
[30], Section 4 for solving problem V I (K; F ).
In Sect. 4 we introduce two linesearches that give rise to two implementable algorithms
denoted Algorithms 1 and 2, respectively. Their convergence is established under mild
assumptions. In the next two sections (Sects. 5 and 6) we consider two variants of each
of the previous algorithms. The variants of Algorithm 1 are related to the classical extragradient method. When K(x) = K for every x ∈ X, one of them coincides with an algorithm
due to Tran et al. [24]. The variants of Algorithm 2 are related to Noor’s modified extragradient method [14]. When f (x, y) = F (x), y − x, one of them coincides with an algorithm
proposed in [31] for solving problem QV I (K; F ). Finally the last section of the paper is
devoted to preliminary numerical results.
2 Preliminaries
Let us consider for a moment the case of a variational inequality with a fixed constraint
set K(x) = K for all x ∈ X. It is well-known that this problem can be reformulated as a
fixed-point equation:
x − PK (x − λF (x)) = 0
(1)
where PK denotes the orthogonal projection from IRn onto K, and λ > 0 is a constant. The
corresponding fixed point algorithm: x k+1 = PK (x k − λF (x k )) is convergent to a solution
of problem V I (K; F ) under strong assumptions: F is Lipschitz continuous and strongly
monotone. To avoid that, the following modified fixed point equations have been introduced:
x − PK (x − λF (x̄)) = 0 or x − PK (x̄ − λF (x̄)) = 0
(2)
where x̄ = PK (x − λF (x)). When F is Lipschitz continuous, it can be proven (see [12,29])
that if x satisfies (2), then x satisfies (1) (and thus is a solution of problem V I P (K; F )) provided that the Lipschitz constant L is such that λ < 1/L and λ < min{3, 1/L2 }, respectively.
The first equation of (2) gives rise to the classical extragradient method [12] and its variants
[9,10]: Given x k ∈ K, x k+1 is obtained after two projections as follows:
k
y
= PK (x k − λF (x k ))
x k+1 = PK (x k − λF (y k )).
The second equation of (2) gives rise to the Noor’s modified extragradient iteration [14]:
k
y
= PK (x k − λF (x k ))
x k+1 = PK (y k − λF (y k )).
These two methods generate sequences converging to a solution of problem V I (K; F ) under
the assumption that F is pseudomonotone and Lipschitz continuous with a condition on the
Lipschitz constant.
A well-known strategy [26,27] to avoid the use of the Lipschitz constant is first to define
y k = PK (x k − λF (x k )) and then to find a direction d k such that the inequality
d k , x k − x ∗ ≥ μx k − y k 2 with μ > 0
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holds for any solution x ∗ of problem V I P (K; F ). When y k = x k , the direction −d k is a
descent direction at x k for the distance function to K ∗ : 21 x − x ∗ 2 (x ∗ ∈ K ∗ ). For the
k
classical extragradient method an example of such a direction is given by d k = F α(zk ) where
zk = (1 − αk )x k + αk y k and αk = α mk with mk being the smallest nonnegative integer m
satisfying the inequality
F (x k − α m (x k − y k )), x k − y k ≥ c x k − y k 2
(4)
and α, c ∈ (0, 1). In fact, this vector zk gives rise to a hyperplane
H k = {x ∈ IRn | F (zk ), x − zk = 0}
(5)
which strictly separates x k from the solution set K ∗ of problem V I P (K; F ). The direction
d k satisfies (3) with μ = c and the next iterate x k+1 is given by
x k+1 = PK (x k − βk d k )
where βk > 0 is chosen such that x k+1 − x ∗ 2 < x k − x ∗ 2 for every solution x ∗ ∈ K ∗
(see [9,26] for more details). For example, the steplength βk can be chosen in such a way
that x k − βk d k be the orthogonal projection of x k onto H k . It is easy to see that this step is
given by
βk = αk
F (zk ), x k − zk .
F (zk )2
There is a similar strategy to avoid the use of the Lipschitz constant for the modified extragradient method. The parameter μ, the direction d k , the inequality (4) defining the linesearch,
and the hyperplane H k become
μ = 1 − c, d k = x k − y k +
F (zk )
,
αk
F (x k ) − F (x k − α m (x k − y k )), x k − y k ≤ c x k − y k 2 ,
(6)
Gk = {x ∈ IRn | d k , x k − x = x k − y k , x k − y k − F (x k ) + F (zk )}.
(7)
= PK
− βk
where, for example, the steplength βk
The next iterate is given by
can be chosen in such a way that x k − βk d k be the orthogonal projection of x k onto Gk .
It is easy to see that this step is given by
x k+1
βk =
(x k
dk)
x k − y k , x k − y k − F (x k ) + F (zk )
d k 2
(see [27,30] for more details).
The general algorithm and its six implementable versions proposed in this paper for solving problem QE(K; f ) are inspired by these ideas. First y k is computed as the solution of
the convex minimization problem
1
min f (x k , y) + y − x k 2 .
2
y∈K(x k )
This solution y k coincides with PK(x k ) (x k − F (x k )) when f (x, y) = F (x), y − x. In a second step, a direction d k is supposed to be computed satisfying (3) for any solution x ∗ ∈ K ∗
and in a third step a line search is performed along the opposite of this direction to reduce
the distance of x k to the solution set.
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In order to prove the convergence of our algorithms we assume that f is defined and finite
on X × where ⊂ IRn is an open convex set containing X.
Another important assumption used in the convergence proof of our algorithms is the
continuity of the multivalued mapping K. Let us recall that
– K is upper semicontinuous (u.s.c) at x̄ ∈ X if
⎫
{x k } ⊂ X and x k → x̄ (k → ∞) ⎬
y k ∈ K(x k )
⎭
y k → ȳ (k → ∞)
⇒
ȳ ∈ K(x̄).
– K is lower semicontinuous (l.s.c) at x̄ ∈ X if x k ∈ X and x k → x̄, then for any ȳ ∈ K(x̄),
there exists a sequence {y k } with y k ∈ K(x k ), such that y k → ȳ (k → ∞).
– K is continuous on X if K is u.s.c. and l.s.c. at every point of X.
3 A general Algorithm
From now on the following assumption is supposed to be satisfied for problem (QEP):
Assumption (A)
(a) f (x, ·) is convex on for all x ∈ X, f is continuous on X × , and f (x, x) = 0 for
all x ∈ X.
(b) K is continuous on X and K(x) is a nonempty closed convex subset of X for all x ∈ X;
(c) x ∈ K(x) for all x ∈ X;
(d) S ∗ := {x ∈ S | f (x, y) ≥ 0, ∀y ∈ T } is nonempty, where S = ∩x∈X K(x) and
T = ∪x∈X K(x).
(e) f is pseudomonotone on X with respect to S ∗ , i.e.,
∀x ∗ ∈ S ∗ ∀y ∈ X f (y, x ∗ ) ≤ 0.
Assumption (d) has been used by Zhang et al. [31] to prove the convergence of their method
for solving problem QV I (K; F ). Since S ∗ is contained in the solution set of QE(K; f ),
this assumption can be seen as a generalization to problem QE(K; f ) of the nonemptyness
of the solution set of problem E(K; f ). Indeed, when K(x) = K for every x ∈ X, the sets
S, T , and K coincide and the set S ∗ is the solution set of EP (K; f ). However, for problem
QE(K; f ), this assumption is not easy to check. Let us also mention that assumption (c) is
true for the GNEP when there is a shared constraint between the players.
Our general algorithm can be expressed as follows:
General Algorithm
Let x 0 ∈ X, let μ ∈ (0, 1) and let γ ∈ (0, 2). Set k = 0.
Step 1 Compute y k = arg miny∈K(x k ) [f (x k , y) + 21 y − x k 2 ]. If y k = x k , Stop.
Step 2 Otherwise find d k such that d k , x k − x ∗ ≥ μx k − y k 2 > 0 for every x ∗ ∈ S ∗ .
Step 3 Compute x k (βk ) = PK(x k ) (x k − βk d k ) where βk is such that
x k (βk ) − x ∗ 2 ≤ x k − x ∗ 2 − γ (2 − γ ) μ2
x k − y k 4
d k 2
for every x ∗ ∈ S ∗ . Set x k+1 = x k (βk ), k := k + 1 and go to Step 1.
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Comments
(1) When y k = x k , the vector −d k is a descent direction at x k for the function 21 x − x ∗ 2
for every x ∗ ∈ S ∗ . In particular, d k = 0 for every k.
(2) An example of βk satisfying (8) is given by
βk = γ μ
x k − y k 2
.
d k 2
Indeed for any x ∗ ∈ S ∗ , we have x ∗ ∈ K(x k ) and consequently, using the nonexpansiveness of the orthogonal projection and the property of d k , we obtain
x k (βk ) − x ∗ 2 ≤ x k − x ∗ 2 − 2βk d k , x k − x ∗ + βk2 d k 2
(9)
≤ x k − x ∗ 2 − 2μβk x k − y k 2 + βk2 d k 2 .
= x k − x ∗ 2 − γ (2 − γ ) μ2
x k − y k 4
.
d k 2
(3) When f (x, y) = F (x), y − x for every x, y ∈ X, Step 1 becomes: Compute y k =
PK(x k ) (x k − F (x k )).
First we give a characterization of y k computed from x k at Step 1 of the General Algorithm.
Proposition 1 For every y ∈ K(x k ), we have
f (x k , y) ≥ f (x k , y k ) + x k − y k , y − y k .
In particular, f (x k , y k ) + x k − y k 2 ≤ 0.
Proof The vector y k being a solution of a convex minimization problem, the optimality
conditions imply that there exists s k ∈ ∂f (x k , y k ) such that
0 ∈ s k + y k − x k + NK(x k ) (y k )
where NK(x k ) (y k ) denotes the normal cone to K(x k ) at y k . Hence, by definition of this cone,
we obtain that
∀y ∈ K(x k ) x k − y k − s k , y − y k ≤ 0.
(10)
On the other hand, since s k ∈ ∂f (x k , y k ), we can write
∀y ∈ K(x k ) f (x k , y) ≥ f (x k , y k ) + s k , y − y k .
(11)
Combining (10) and (11) and taking y = x k , we obtain the desired result because x k ∈ K(x k )
by Assumption (A)(c).
Now we justify the stopping criterion: y k = x k .
Proposition 2 If y k = x k , then x k is a solution of problem QE(K; f ).
Proof Since y k = x k and x k ∈ K(x k ), it follows from Proposition 1 that
∀y ∈ K(x k ) f (x k , y) ≥ f (x k , x k ) + x k − x k , y − x k = 0,
i.e., that x k is a solution of QE(K; f ).
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Next we assume that x k = y k for every k and we prove that the sequence {x k } generated
by the General Algorithm is bounded.
Proposition 3 The sequence {x k } is bounded.
Proof Since, by construction (see Step 3 of the General Algorithm), {x k −x ∗ } is a decreasing sequence, we have for every k,
x k ≤ x k − x ∗ + x ∗ ≤ x 0 − x ∗ + x ∗ ,
and thus {x k } is bounded.
To prove the boundedness of the sequence {y k }, we need the next lemma.
Lemma 1 x k − y k ≤ g for every g ∈ ∂f (x k , x k ).
Proof Let g ∈ ∂f (x k , x k ). Then
f (x k , y k ) ≥ f (x k , x k ) + g, y k − x k = g, y k − x k .
Using successively Proposition 1, the previous inequality and the Cauchy-Schwarz inequality,
we obtain
x k − y k 2 ≤ −f (x k , y k ) ≤ −g, y k − x k ≤ g x k − y k ,
and thus x k − y k ≤ g.
The sequence {x k } being bounded, let x̄ be one of its limit points. Then there exists a
subsequence {x kj } converging to x̄. Thanks to Lemma 1 we can prove that the corresponding
sequence {y kj } is also bounded.
Proposition 4 The sequence {y kj } is bounded.
Proof By Lemma 1 it is sufficient to prove that there exists M > 0 such that
g ≤ M
for every g ∈ ∂f (x kj , x kj ) and all j.
Since x̄ ∈ , {x kj } ⊂ , f (x̄, ·) is finite on and since the sequence of convex functions
{f (x kj , ·)} converges pointwise on to the convex function f (x̄, ·), it follows from [20]
Theorem 24.5 that there exists j0 such that for all j ≥ j0 ,
∂f (x kj , x kj ) ⊆ ∂f (x̄, x̄) + B
where B denotes the closed Euclidean unit ball of IRn . Since B and ∂f (x̄, x̄) are bounded,
there exists M > 0 such that
g ≤ M
for every g ∈ ∂f (x kj , x kj ) and all j ≥ j0 .
Hence the sequence {y kj } is bounded.
Proposition 5 Let x̄ be a limit point of {x k }. Assume that x kj → x̄ and that x kj −y kj → 0
as j → ∞. Then x̄ is a solution of problem QE(K; f ).
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Proof By assumption y kj = y kj − x kj + x kj → x̄. Since y kj ∈ K(x kj ) for every j and
since K is u.s.c. on X, we obtain that x̄ ∈ K(x̄).
Now let ȳ ∈ K(x̄). We have to prove that f (x̄, ȳ) ≥ 0. Since K is l.s.c. on X, there exists
a sequence {ȳ kj } such that
ȳ kj ∈ K(x kj ) for every j and ȳ kj → ȳ.
So, for every j , we have, by definition of y kj , that
1
1
f (x kj , y kj ) + x kj − y kj 2 ≤ f (x kj , ȳ kj ) + x kj − ȳ kj 2 .
2
2
Taking the limit as j → ∞ and remembering that f is continuous, we obtain
1
1
0 = f (x̄, x̄) + x̄ − x̄2 ≤ f (x̄, ȳ) + x̄ − ȳ2 .
2
2
(12)
But this implies that f (x̄, ȳ) ≥ 0. Indeed, the inequality (12) means that x̄ is a solution of
the convex minimization problem
1
2
min f (x̄, y) + y − x̄ .
y∈K(x̄)
2
Hence 0 ∈ ∂f (x̄, x̄) + NK(x̄) (x̄), that is, x̄ is a solution of min f (x̄, y) subject to y ∈ K(x̄).
Consequently f (x̄, ȳ) ≥ 0.
Finally we obtain the convergence of the whole sequence {x k } to a solution of problem
QE(K; f ) when the function f is strictly monotone at a limit point of {x k }. Let us recall
that f is strictly monotone at x ∈ X if for all y ∈ X, y = x, we have
f (x, y) + f (y, x) < 0.
Proposition 6 If, in addition to the assumptions of Proposition 5, the function f is strictly
monotone at x̄, then the whole sequence {x k } converges to x̄ as k → ∞. Furthermore x̄ is a
solution of problem QE(K; f ).
Proof Let x kj → x̄. By Proposition 5, x̄ is a solution of problem QE(K; f ).
(1) First we prove that x̄ ∈ S ∗ . Let x ∗ ∈ S ∗ . Then x ∗ ∈ ∩x∈X K(x) and f (x ∗ , y) ≥ 0 for
every y ∈ K(x) and every x ∈ X. Since x̄ ∈ K(x̄), we have f (x ∗ , x̄) ≥ 0.
On the other hand f (x̄, x ∗ ) = 0. Indeed, by Assumption (A)(e), we have that f (x̄, x ∗ ) ≤
0 and, since x̄ ∈ K ∗ and x ∗ ∈ K (x̄), we have that f (x̄, x ∗ ) ≥ 0. Consequently x ∗ = x̄.
Indeed, if x ∗ = x̄, we deduce from the strict monotonicity of f at x̄ that
f (x ∗ , x̄) = f (x ∗ , x̄) + f (x̄, x ∗ ) < 0,
which contradicts f (x ∗ , x̄) ≥ 0. Hence x̄ = x ∗ ∈ S ∗ .
(2) Next we prove that x k → x̄. Since x̄ = x ∗ ∈ S ∗ , it follows from Step 3 of the General
Algorithm that the sequence {x k − x̄} is nonincreasing and consequently converges
to some a ≥ 0. Since x kj → x̄, we deduce that the whole sequence x k − x̄ → 0,
that is, x k → x̄ as k → ∞.
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The convergence of the General Algorithm is obtained under the assumption that x k −
→ 0 as k → ∞. Thanks to the inequality (8) the sequence {x k − x ∗ } is nonincreasing
and converges to some a ≥ 0, which implies that
yk x k − y k 4
→ 0 as k → ∞.
d k 2
Consequently x k − y k → 0 when the sequence {d k } is bounded. It remains to examine
the case when the sequence {d k } is unbounded.
4 A class of double-projection methods
In this section we give two examples of directions d k such that for all k,
d k , x k − x ∗ ≥ μx k − y k 2 > 0 for some μ ∈ (0, 1) and for every x ∗ ∈ S ∗ .
They are based on a linesearch performed between y k and x k . This linesearch has the property
that when the steplengths tend to zero, then the sequence {d k } is unbounded and x k −y k →
0. More precisely, Step 2 of the General Algorithm is replaced by one of the following linesearch procedures:
Linesearch 1: Given x k , y k as before and α, c ∈ (0, 1)
Find m the smallest nonnegative integer such that
f (zk,m , x k ) − f (zk,m , y k ) ≥ c x k − y k 2
where zk,m := (1 − α m )x k + α m y k .
k
Set αk = α m and zk = zk,m . Set d k = αg k where g k ∈ ∂f (zk , x k ).
When K(x) = K for every x ∈ X, this linesearch has been used by Tran et al. [24] for
solving problem E(K; f ). When f (x, y) = F (x), y − x for every x, y ∈ X, the inequality
k
satisfied by the linesearch coincides with (4) and the direction d k becomes equal to F α(zk ) .
Linesearch 2: Given x k , y k as before and α, c ∈ (0, 1)
Find m the smallest nonnegative integer such that
f (zk,m , x k ) − f (zk,m , y k ) + f (x k , y k ) ≥ −c x k − y k 2
where zk,m := (1 − α m )x k + α m y k .
k
Set αk = α m and zk = zk,m . Set d k = x k − y k + αg k where g k ∈ ∂f (zk , x k ).
When f (x, y) = F (x), y − x for every x, y ∈ X, the inequality satisfied by this
k
linesearch coincides with (6) and the direction d k becomes equal to x k − y k + F α(zk ) .
This linesearch has been used by Zhang et al. [31] for solving problem QV I (K; F ).
Let us also mention that Linesearch 2 is closely related to Linesearch 1. Indeed, from
Proposition 1, we have that
f (x k , y k ) + x k − y k 2 ≤ 0.
So the vector zk,m obtained from Linesearch 2 satisfies
f (zk,m , x k ) − f (zk,m , y k ) ≥ −f (x k , y k ) − cx k − y k 2 ≥ (1 − c) x k − y k 2 ,
that is, the inequality defining Linesearch 1 with c replaced by 1 − c.
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J Glob Optim
Finally, let us also observe that when f (x, y) = F (x), y − x, only one projection is
used in our algorithms to generate zk . This differs from many algorithms for solving problem V I (K; F ) where a projection is required at each trial point PK(x k ) (x k − αk F (x k )) of a
backtracking procedure to obtain zk . This can lead to expensive computation (see [30] and
the references quoted therein).
First we prove that the two linesearches are finite when y k = x k .
Proposition 7 Assume y k = x k . Then the two Linesearches 1 and 2 give αk and zk after
finitely many iterations.
Proof Suppose that the linesearches are not finite. Then for all m ∈ IN, we have (for Linesearch 1):
f (zk,m , x k ) − f (zk,m , y k ) < c x k − y k 2 .
Since f (·, x) is continuous for all x ∈ , and zk,m → x k as m → ∞, we obtain
−f (x k , y k ) = f (x k , x k ) − f (x k , y k ) ≤ c x k − y k 2 .
On the other hand, by Proposition 1, we have that
f (x k , y k ) + x k − y k 2 ≤ 0.
Combining these last two inequalities yields
x k − y k 2 ≤ c x k − y k 2 .
Since c ∈ (0, 1), we deduce that y k = x k , which contradicts the assumption y k = x k .
Similarly, for Linesearch 2, we have for all m ∈ IN, that
f (zk,m , x k ) − f (zk,m , y k ) + f (x k , y k ) < −c x k − y k 2 .
Taking the limit as m → ∞ and noting that f (·, x) is continuous for all x ∈ , we obtain
0 = f (x k , x k ) − f (x k , y k ) + f (x k , y k ) ≤ −c x k − y k 2 .
But then y k = x k , which contradicts the assumption y k = x k .
Now our aim is to prove that the direction d k obtained from each linesearch satisfies the
property: d k , x k − x ∗ ≥ μx k − y k 2 for some μ ∈ (0, 1) and for every x ∗ ∈ S ∗ .
Proposition 8 For the two linesearches and for every g k ∈ ∂f (zk , x k ) and x ∗ ∈ S ∗ , we have
k
g
k
∗
, x − x ≥ f (zk , x k ) − f (zk , y k ).
αk
Proof Let g k ∈ ∂f (zk , x k ) and x ∗ ∈ S ∗ . Then
f (zk , x ∗ ) ≥ f (zk , x k ) + g k , x ∗ − x k .
Since f is pseudomonotone on X with respect to S ∗ , we have f (zk , x ∗ ) ≤ 0. So
g k , x k − x ∗ ≥ f (zk , x k ).
(13)
Since f (zk , ·) is convex, we can write, using the definition of zk ,
f (zk , zk ) ≤ (1 − αk )f (zk , x k ) + αk f (zk , y k ),
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J Glob Optim
that is,
f (zk , x k ) ≥ αk [f (zk , x k ) − f (zk , y k )].
Combining (13) and (14) yields the announced result.
(14)
It follows immediately from Proposition 8 that if Linesearch 1 is used, then the required
k
property on d k = αg k :
d k , x k − x ∗ ≥ μ x k − y k 2
is satisfied for μ = c. In particular, when Linesearch 1 is used (that is, when y k = x k ), the
direction d k = 0 whatever g k ∈ ∂f (zk , x k ).
To get a similar property but for Linesearch 2, we need to prove the following proposition.
Proposition 9 For every k and every x ∗ ∈ S ∗ , we have
x k − y k , x k − x ∗ ≥ f (x k , y k ) + x k − y k 2 .
Proof Let x ∗ ∈ S ∗ . Since f is pseudomonotone on X with respect to S ∗ , and since x ∗ ∈
K(x k ), we can write, using Proposition 1, that
0 ≥ f (x k , x ∗ ) ≥ f (x k , y k ) + x k − y k , x ∗ − y k .
But then we have
x k − y k , y k − x k + x k − y k , x k − x ∗ ≥ f (x k , y k ),
that is, the thesis.
From Propositions 8 and 9, we deduce immediately that the required property on d k is
k
satisfied for the second linesearch with μ = 1 − c. Indeed for d k = x k − y k + αg k we have
d k , x k − x ∗ = x k − y k +
gk
k
αk , x
− x∗
≥ f (zk , x k ) − f (zk , y k ) + f (x k , y k ) + x k − y k 2
≥ (1 − c) x k − y k 2 .
In particular, when Linesearch 2 is used, the direction d k = 0 whatever g k ∈ ∂f (zk , x k ).
From now on we denote by Algorithms 1 and 2 the General Algorithm with Step 2 replaced
by Linesearches 1 and 2, respectively.
Algorithm 1
Let x 0 ∈ X, let α, c ∈ (0, 1), and let γ ∈ (0, 2). Set k = 0.
Step 1 Compute y k = arg miny∈K(x k ) [f (x k , y) + 21 y − x k 2 ]. If y k = x k , Stop.
Step 2 Otherwise find m the smallest nonnegative integer such that
f (zk,m , x k ) − f (zk,m , y k ) ≥ c x k − y k 2
where zk,m := (1 − α m )x k + α m y k .
Set αk = α m and zk = zk,m .
Step 3 Compute g k ∈ ∂f (zk , x k ) and x k+1 = PK(x k ) (x k − βk d k ) where
dk =
123
gk
αk
and βk = γ c
x k −y k 2
.
d k 2
Set k := k + 1 and go to Step 1.
J Glob Optim
Algorithm 2
Let x 0 ∈ X, let α, c ∈ (0, 1), and let γ ∈ (0, 2). Set k = 0.
Step 1 Compute y k = arg miny∈K(x k ) [f (x k , y) + 21 y − x k 2 ]. If y k = x k , Stop.
Step 2 Otherwise find m the smallest nonnegative integer such that
f (zk,m , x k ) − f (zk,m , y k ) + f (x k , y k ) ≥ −c x k − y k 2
where zk,m := (1 − α m )x k + α m y k .
Set αk = α m and zk = zk,m .
Step 3 Compute g k ∈ ∂f (zk , x k ) and x k+1 = PK(x k ) (x k − βk d k ) where
dk = xk − yk +
gk
αk
and βk = γ (1 − c) xd−y
k 2 . Set k := k + 1 and go to Step 1.
k
k 2
Now for Algorithms 1 and 2 we have the following boundedness properties:
Proposition 10 Let x̄ be a limit point of {x k } and assume that x kj → x̄. Then the sequences
{y kj }, {zkj }, and {g kj } are bounded.
Proof Since the sequences {x kj } and {y kj } are bounded (see Propositions 3 and 4), it follows
that the sequence {zkj } is also bounded because zkj belongs to the segment [x kj ; y kj ] for
every j . So a subsequence of {zkj }, again denoted {zkj }, converges to some z̄ ∈ X. Since
x̄ ∈ X ⊆ , {x kj } ⊆ X ⊆ , x kj → x̄, and the sequence of convex functions {f (zkj , ·)}
converges pointwise to the convex function f (z̄, ·), it follows from Theorem 24.5 of [20]
that there exists j0 such that for all j ≥ j0 ,
∂f (zkj , x kj ) ⊆ ∂f (z̄, x̄) + B
where B denotes the closed Euclidean unit ball of IRn . Since B and ∂f (z̄, x̄) are bounded,
the sequence {g kj } is also bounded.
In order to apply Proposition 5, we need to prove the next result.
Proposition 11 Let x kj → x̄. Then x kj − y kj → 0 as j → ∞.
Proof We examine two cases:
(1) inf j αkj > 0: The sequence {d kj } is bounded because the sequences {x kj }, {y kj }, and
{g kj } are bounded (Proposition 10) and
d kj =
g kj
αkj
(Linesearch 1) or d kj = x kj − y kj +
g kj
αkj
(Linesearch 2).
Since, from (8), xd−y
→ 0, we deduce that x kj − y kj → 0.
k 2
(2) inf j αkj = 0: Then αkj → 0 (for a subsequence). But this implies that αkj < 1 for j
k
k 4
large enough and that the linesearch conditions are not satisfied for
αkj
αkj k
x kj +
y j.
z̄kj = 1 −
α
α
αkj
α
. Let us denote
It is immediate that z̄kj → x̄. Now let us examine separately the two linesearches.
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J Glob Optim
(a) If the second linesearch is used, we have
f (z̄kj , x kj ) − f (z̄kj , y kj ) + f (x kj , y kj ) < −c x kj − y kj 2 .
Let ȳ be a limit point of the bounded sequence {y kj }. Then, f being continuous, the
left-hand side of the previous inequality tends to
f (x̄, x̄) − f (x̄, ȳ) + f (x̄, ȳ) = 0.
Consequently, we obtain that x kj − y kj → 0.
(b) If the first linesearch is used, we have
f (z̄kj , x kj ) − f (z̄kj , y kj ) < c x kj − y kj 2 .
By definition of y kj we also have
x kj − y kj 2 ≤ −f (x kj , y kj ).
Let ȳ be a limit point of {y kj }. Then combining the two inequalities and taking the limit
as j → ∞, we obtain
f (x̄, x̄) − f (x̄, ȳ) ≤ −cf (x̄, ȳ),
which implies that f (x̄, ȳ) ≥ 0. So −f (x kj , y kj )→−f (x̄, ȳ) ≤ 0 and x kj −y kj →0.
Finally using successively Propositions 5, 6 and 11, we obtain the following convergence
result for Algorithms 1 and 2.
Theorem 1 Any limit point of the sequence {x k } generated by Algorithm 1 or Algorithm 2
is a solution of problem QE(K; f ). If f is strictly monotone at a limit point x̄ of {x k }, then
x k → x̄ as k → ∞.
When f (x, y) = F (x), y−x for all x, y ∈ X, our Algorithm 2 coincides with Algorithm
2 in [31]. So, as a by-product we find again the convergence results obtained in Theorems
5.1 and 5.2 of [31]. In the next section we consider other choices for βk .
5 Variants of Algorithm 1
In Algorithm 1 the next iterate was defined as x k+1 = x k (βk ) where for every β > 0, x k (β) =
k
PK(x k ) (x k − βd k ). The direction was equal to d k = αg k with g k ∈ ∂f (zk , x k ) and αk obtained
by using the first linesearch. Furthermore the step βk was chosen such that the following
inequality holds:
x k (βk ) − x ∗ 2 ≤ x k − x ∗ 2 − γ (2 − γ ) c2
x k − y k 4
.
d k 2
(15)
An example of such a step is βk = γ c xd−y
k 2 .
In this section we show that it is possible to choose other steps βk while keeping true
the inequality (15). So the convergence results obtained in the previous section remain valid
for these variants. These steps will give rise to better decreases on the distance between the
iterates and the set S ∗ .
k
123
k 2
J Glob Optim
By definition of g k ∈ ∂f (zk , x k ), we have
g k , x k − x ∗ ≥ f (zk , x k ) − f (zk , x ∗ )
where x ∗ is any element in S ∗ . Since f is pseudomonotone on X with respect to S ∗ , we
obtain that f (zk , x ∗ ) ≤ 0 and thus that
d k , x k − x ∗ = gk k
f (zk , x k )
, x − x∗ ≥
.
αk
αk
(16)
Using this inequality in (9) and taking
βk = γ
f (zk , x k )
αk ,
g k 2
(17)
we deduce that
x k+1 − x ∗ 2 ≤ x k − x ∗ 2 − 2
β2
βk
f (zk , x k ) + k2 g k 2
αk
αk
≤ x k − x ∗ 2 − γ (2 − γ )
f (zk , x k )2
.
g k 2
Now from the convexity of f (zk , ·) and the first linesearch, we obtain that
f (zk , x k ) ≥ αk (f (zk , x k ) − f (zk , y k )) ≥ αk cx k − y k 2 .
(18)
So if we use the new step βk given in (17), we can conclude that
x k+1 − x ∗ 2 ≤ x k − x ∗ 2 − γ (2 − γ ) αk2 c2
x k − y k 4
,
g k 2
and thus that inequality (15) holds.
Replacing βk by its new value in Step 3 of the General Algorithm, we obtain
f (zk , x k )
k
α
d
= PK(x k ) [x k − γ σk g k ]
x k+1 = PK(x k ) x k − γ
k
g k 2
where σk =
f (zk ,x k )
.
g k 2
Consequently, in that case, Algorithm 1 becomes
Algorithm 1a
Let x 0 ∈ X, let α, c ∈ (0, 1), and let γ ∈ (0, 2). Set k = 0.
Step 1 Compute y k = arg miny∈K(x k ) [f (x k , y) + 21 y − x k 2 ]. If y k = x k , Stop.
Step 2 Otherwise find m the smallest nonnegative integer such that
f (zk,m , x k ) − f (zk,m , y k ) ≥ c x k − y k 2
where zk,m := (1 − α m )x k + α m y k .
Set αk = α m and zk = zk,m .
Step 3 Compute g k ∈ ∂f (zk , x k ) and x k+1 = PK(x k ) (x k − γ σk g k ) where
σk =
f (zk ,x k )
.
g k 2
Set k := k + 1 and go to Step 1.
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J Glob Optim
Comments
(1) When K(x) = K for every x ∈ X, Algorithm 1a coincides with Algorithm 2a of Tran
et al. [24]. Moreover, when f (x, y) = F (x), y − x for every x, y ∈ X, Algorithm 1a
corresponds to an extragradient algorithm due to Iusem and al. [9].
(2) Consider the hyperplane H k defined by
H k = x ∈ IRn | g k , x k − x = f (zk , x k )
and observe that when f (x, y) = F (x), y − x for every x, y ∈ X, the hyperplane H k
coincides with the one defined in (5). Now this hyperplane H k separates x k from S ∗ .
Indeed, from (18), it follows that f (zk , x k ) > 0 = g k , x k − x k and from (16), that
g k , x k − x ∗ ≥ f (zk , x k ) for every x ∗ ∈ S ∗ . Furthermore g k is the normal vector to
H k and since x k − σk g k ∈ H k , we can say that x k − σk g k is the orthogonal projection
of x k onto H k . So when γ = 1, the new iterate x k+1 is obtained by projecting onto
K(x k ) the projection of x k onto H k .
Since the set S ∗ ⊆ K ∗ is contained in K(x k ) ∩ H+k where
H+k = x ∈ IRn | g k , x k − x ≥ f (zk , x k ) ,
a variant of Algorithm 1a consists in replacing in Step 3 the iterate x k+1 = PK(x k ) (x k −
γ σk g k ) by x k+1 = PK(x k )∩H k (x k − γ σk g k ). Using the nonexpansiveness of PK(x k )∩H k
+
+
instead of the one of PK(x k ) , we immediately obtain that the inequality (15) holds. So the
convergence of the sequence {x k } is preserved for this variant. In the sequel we denote by
Algorithm 1b the Algorithm 1a where Step 3 has been replaced by
Step 3 Compute g k ∈ ∂f (zk , x k ) and x k+1 = PK(x k )∩H k (x k − γ σk g k ) where
+
σk =
f (zk ,x k )
g k 2
and H+k = {x ∈ IRn | g k , x k − x ≥ f (zk , x k )}.
Set k := k + 1 and go to Step 1.
When γ = 1 we have that x k − γ σk g k = PH k (x k ) and we can use the proof of Lemma 2.2
+
in [21] to show that
x k+1 = PK(x k )∩H k (x k ).
+
When γ = 1, it is also possible to give a geometric interpretation of Step 3 in Algorithms 1a
and 1b. In that purpose we recall the following property of the orthogonal projection onto a
convex set.
Lemma 2 [26] Let C be a nonempty closed convex subset in IRn . Then for any x ∈ IRn and
z ∈ C,
PC (x) − z2 ≤ x − z2 − PC (x) − x2 .
Using Lemma 2 with C = K(x k ), x = x k − βd k and z = x ∗ , we can write
x k (β) − x ∗ 2 ≤ x k − βd k − x ∗ 2 − x k (β) − x k + βd k 2 .
Developing the first term of the right-hand side of this inequality and using successively (9),
(16), and (17), we obtain
x k (β) − x ∗ 2 ≤ x k − x ∗ 2 + ϕ k (β)
123
(19)
J Glob Optim
where
ϕ k (β) = ϕk (β) − x k (β) − x k +
β k 2
g αk
with
ϕk (β) = −2
β
β2
f (zk , x k ) + 2 g k 2 .
αk
αk
Since ϕ k (β) ≤ ϕk (β), we have, in particular, that
x k (β) − x ∗ 2 ≤ x k − x ∗ 2 + ϕk (β).
(20)
f (zk ,x k )
αk = σk αk minimizes the right-hand side of (20). Since
g k 2
k
σk αk d ) = PK(x k ) (x k − σk g k ), it follows that the new iterate x k+1
It is easy to check that β1 =
x k (β1 ) = PK(x k ) (x k −
in Algorithm 1a is given by
x k+1 = x k (β1 ).
Now if we minimize the right-hand side of (19), it can be shown exactly as in [26] that the
function ϕ k (β) is convex and admits a minimum for a steplength β2 ≥ β1 . Computing an
explicit value for β2 seems difficult but it is possible, using a proof similar to the one of
Lemma 3.2 in [26], to show that
x k (β2 ) = PK(x k )∩H k (x k − β1 d k ) = PK(x k )∩H k (x k − σk g k ).
+
Hence the new iterate
x k+1
+
in Algorithm 1b is given by
x k+1 = x k (β2 ).
To end this section we summarize Algorithm 1b.
Algorithm 1b
Let x 0 ∈ X, let α, c ∈ (0, 1), and let γ ∈ (0, 2). Set k = 0.
Step 1 Compute y k = arg miny∈K(x k ) [f (x k , y) + 21 y − x k 2 ]. If y k = x k , Stop.
Step 2 Otherwise find m the smallest nonnegative integer such that
f (zk,m , x k ) − f (zk,m , y k ) ≥ c x k − y k 2
where zk,m := (1 − α m )x k + α m y k .
Set αk = α m and zk = zk,m .
Step 3 Compute g k ∈ ∂f (zk , x k ) and x k+1 = PK(x k )∩H k (x k − γ σk g k ) where
+
σk =
f (zk ,x k )
g k 2
and H+k = {x ∈ IRn | g k , x k − x ≥ f (zk , x k )}.
Set k := k + 1 and go to Step 1.
When γ = 1, we have seen that x k+1 = PK(x k )∩H k (x k ). Consequently, when K(x) = K
+
for every x ∈ X and f (x, y) = F (x), y −x, we find again the projection method introduced
by Solodov and al. for solving variational inequality problems [21] (see also Algorithm 3.1
in [26]).
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J Glob Optim
6 Variants of Algorithm 2
In Algorithm 2 the next iterate was defined as x k+1 = x k (βk ) where for every β > 0, x k (β) =
k
PK(x k ) (x k − βd k ). The direction was equal to d k = x k − y k + αg k with g k ∈ ∂f (zk , x k ) and
αk obtained by using the second linesearch. Furthermore the step βk was chosen such that
the following inequality holds:
x k (βk ) − x ∗ 2 ≤ x k − x ∗ 2 − γ (2 − γ ) (1 − c)2
x k − y k 4
.
d k 2
(21)
An example of such a step is βk = γ (1 − c) xd−y
k 2 .
In this section we show that it is possible to choose other steps βk while keeping true
the inequality (21). So the convergence results obtained in Section 3 remain valid for these
variants. These steps will give rise to better decreases on the distance between the iterates
and the set S ∗ .
From the definition of d k , (16) and Proposition 9, we obtain that
k
d k , x k − x ∗ ≥
k 2
f (zk , x k )
+ f (x k , y k ) + x k − y k 2 .
αk
(22)
Using this inequality in (9) and taking
βk = γ σk where σk =
f (zk ,x k )
αk
+ f (x k , y k ) + x k − y k 2
d k 2
,
we deduce that
x k+1 − x ∗ 2 ≤ x k − x ∗ 2 − 2βk σk d k 2 + βk2 d k 2
≤ x k − x ∗ 2 − γ (2 − γ ) σk2 d k 2 .
Now we observe that
σk2 d k 2 ≥ (1 − c)2
x k − y k 4
> 0.
d k 2
(23)
Indeed, using successively the definition of σk , the second linesearch and the convexity of
f (zk , ·) (see the first inequality in (18)), we obtain
σk d k 2 =
≥
f (zk , x k )
+ f (x k , y k ) + x k − y k 2
αk
f (zk , x k )
+ f (zk , y k ) − f (zk , x k ) + (1 − c)x k − y k 2
αk
≥ (1 − c)x k − y k 2 .
(24)
Taking the square of both members of (24) we deduce the inequality (23) as well as the
estimation
x k+1 − x ∗ 2 ≤ x k − x ∗ 2 − γ (2 − γ ) (1 − c)2
x k − y k 4
.
d k 2
So inequality (21) is satisfied and the convergence of Algorithm 2 is preserved when the
stepsize βk is equal to γ σk .
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J Glob Optim
With this stepsize βk , Algorithm 2 becomes
Algorithm 2a
Let x 0 ∈ X, let α, c ∈ (0, 1), and let γ ∈ (0, 2). Set k = 0.
Step 1 Compute y k = arg miny∈K(x k ) [f (x k , y) + 21 y − x k 2 ]. If y k = x k , Stop.
Step 2 Otherwise find m the smallest nonnegative integer such that
f (zk,m , x k ) − f (zk,m , y k ) + f (x k , y k ) ≥ −c x k − y k 2
where zk,m := (1 − α m )x k + α m y k .
Set αk = α m and zk = zk,m .
Step 3 Compute g k ∈ ∂f (zk , x k ) and x k+1 = PK(x k ) (x k − γ σk d k ) where
gk
and σk =
d =x −y +
αk
k
k
k
f (zk ,x k )
αk
+ f (x k , y k ) + x k − y k 2
d k 2
.
Set k := k + 1 and go to Step 1.
Comments
(1) When for every x, y ∈ X, K(x) = K and f (x, y) = F (x), y − x, Algorithm 2a
corresponds to Algorithm NVE-1 due to Wang et al. [27].
(2) Consider the hyperplane Gk defined by
f (zk , x k )
k
n
k
k
k
k
k
k 2
+ f (x , y ) + x − y G = x ∈ IR | d , x − x =
αk
and observe that when f (x, y) = F (x), y − x for every x, y ∈ X, this hyperplane
coincides with the one defined in (7). Now using (22) and (24), it is easy to prove that
Gk separates x k from S ∗ and that x k − σk d k is the orthogonal projection of x k onto
Gk . So when γ = 1, the new iterate x k+1 is obtained by projecting onto K(x k ) the
projection of x k onto Gk .
Since the set S ∗ ⊆ K ∗ is contained in K(x k ) ∩ Gk+ where
f (zk , x k )
+ f (x k , y k ) + x k − y k 2 ,
Gk+ = x ∈ IRn | d k , x k − x ≥
αk
a variant of Algorithm 2a consists in replacing in Step 3 the iterate x k+1 = PK(x k ) (x k −
γ σk d k ) by x k+1 = PK(x k )∩Gk (x k − γ σk d k ). Using the nonexpansiveness of PK(x k )∩Gk
+
+
instead of the one of PK(x k ) , we immediately obtain that the inequality (21) holds. So the
convergence of the sequence {x k } is preserved for this variant. In the sequel we denote by
Algorithm 2b the Algorithm 2a where Step 3 has been replaced by
Step 3 Compute g k ∈ ∂f (zk , x k ) and x k+1 = PK(x k )∩Gk (x k − γ σk d k ) where
+
dk = xk − yk +
gk
and σk =
αk
f (zk ,x k )
αk
+ f (x k , y k ) + x k − y k 2
d k 2
.
Set k := k + 1 and go to Step 1.
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J Glob Optim
When γ = 1 we have that x k − γ σk g k = PGk (x k ) and we can use the proof of Lemma 2.2
+
in [21] to show that
x k+1 = PK(x k )∩Gk (x k ).
+
When γ = 1, it is also possible to give a geometric interpretation of Step 3 in Algorithms 2a
and 2b. Using (9), (22) and Lemma 2 as in the previous section, we obtain for any x ∗ ∈ S ∗
x k (β) − x ∗ 2 ≤ x k − x ∗ 2 − ϕ k (β)
(25)
where
ϕ k (β) = ϕk (β) − x k (β) − x k + βd k 2
with
ϕk (β) = −2β σk d k 2 + β 2 d k 2 .
Since ϕ k (β) ≤ ϕk (β), we have, in particular, that
x k (β) − x ∗ 2 ≤ x k − x ∗ 2 + ϕk (β).
(26)
It is easy to check that β1 = σk minimizes the right-hand side of (26). Since x k (β1 ) =
PK(x k ) (x k − β1 d k ) = PK(x k ) (x k − σk d k ), it follows that the iterate x k+1 in Algorithm 2a is
given by
x k+1 = x k (β1 ).
Now if we minimize the right-hand side of (25), it can be shown exactly as in [27] that the
function ϕ k (β) is convex and admits a minimum for a steplength β2 ≥ β1 . Computing an
explicit value for β2 seems difficult but it is possible, using a proof similar to the one of
Lemma 3.2 in [27], to show that
x k (β2 ) = PK(x k )∩Gk (x k − β1 d k ) = PK(x k )∩Gk (x k − σk d k )
+
+
Hence the iterate x k+1 in Algorithm 2b is given by
x k+1 = x k (β2 ).
Since ϕ k (β2 ) ≤ ϕ k (β1 ) ≤ ϕk (β1 ), Algorithm 2b should be faster than Algorithm 2a. To end
this section we give a description of Algorithm 2b.
Algorithm 2b
Let x 0 ∈ X, let α, c ∈ (0, 1), and let γ ∈ (0, 2). Set k = 0.
Step 1 Compute y k = arg miny∈K(x k ) [f (x k , y) + 21 y − x k 2 ]. If y k = x k , Stop.
Step 2 Otherwise find m the smallest nonnegative integer such that
f (zk,m , x k ) − f (zk,m , y k ) + f (x k , y k ) ≥ −c x k − y k 2
where zk,m := (1 − α m )x k + α m y k .
Set αk = α m and zk = zk,m .
Step 3 Compute g k ∈ ∂f (zk , x k ) and x k+1 = PK(x k )∩Gk (x k − γ σk d k ) where
+
dk = xk − yk +
123
gk
,
αk
σk =
f (zk ,x k )
αk
+ f (x k , y k ) + x k − y k 2
d k 2
J Glob Optim
and
Gk+ = {x ∈ IRn | d k , x k − x ≥
f (zk , x k )
+ f (x k , y k ) + x k − y k 2 }.
αk
Set k := k + 1 and go to Step 1.
When γ = 1, and for every x, y ∈ X, K(x) = K and f (x, y) = F (x), y − x, we find
again the projection method NVE-2 introduced by Wang et al. [27] for solving variational
inequality problems.
7 Numerical results
In order to compare the numerical behavior of the six algorithms presented in the previous sections we have implemented them in MATLAB and we have first solved an academic
quasi-equilibrium problem of size n = 5 and then three examples of generalized Nash games.
Example 1 The bifunction f of the quasi-equilibrium problem comes from the Cournot-Nash
equilibrium model considered in [24]. It is defined for each x, y ∈ IR5 , by
f (x, y) = P x + Qy + q, y − x
where q ∈ IR5 , and P and Q are two square matrices of order 5 such that Q is symmetric
positive semidefinite and Q − P is negative semidefinite. In our tests the vector q and the
matrices P and Q are chosen as follows:
⎤
⎡
⎤
⎡
⎤
⎡
1.6 1
0
0 0
3.1 2
0
0 0
1
⎢ 1 1.6 0
⎢ 2 3.6 0
⎢ −2 ⎥
0 0⎥
0 0⎥
⎥
⎢
⎥
⎢
⎥
⎢
⎢
⎥
⎢
⎥
⎢
0 1.5 1 0 ⎥
0 3.5 2 0 ⎥ ; Q = ⎢ 0
q = ⎢ −1 ⎥ ; P = ⎢ 0
⎥.
⎣ 0
⎣ 0
⎣ 2⎦
0
1 1.5 0 ⎦
0
2 3.3 0 ⎦
0
0
0
0 2
0
0
0
0 3
−1
Instead of the fixed constraint set used in [24] we consider the moving set K(x) =
1≤i≤5 Ki (x) where for each x ∈ IR5 and each i, the set Ki (x) is defined by
⎧
⎫
⎨
⎬
xj ≥ −1 .
Ki (x) = yi ∈ IR | yi +
⎩
⎭
1≤j ≤5,j =i
The numerical results obtained for this example are listed in Table 1 for Algorithms 1, 1a,
and 1b and in Table 2 for Algorithms 2, 2a, and 2b. Two different starting points are chosen
and for each test the number of iterations and the CPU time needed to get a solution are
reported.
Table 1 The results of Example 1 for Algorithms 1, 1a, and 1b
Starting point
Number of iterations
CPU (s)
Algorithm 1
Algorithm 1a
Algorithm 1b
Algorithm 1
Algorithm 1a
Algorithm 1b
(1, 3, 1, 1, 2)T
27
20
20
0.64
0.48
0.53
(0, 0, 0, 0, 0)T
20
15
15
0.59
0.54
0.31
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J Glob Optim
Table 2 The results of Example 1 for Algorithms 2, 2a, and 2b
Starting point
Number of iterations
CPU (s)
Algorithm 2
Algorithm 2a
Algorithm 2b
Algorithm 2
Algorithm 2a
Algorithm 2b
(1, 3, 1, 1, 2)T
105
41
41
3.04
0.96
1.20
(0, 0, 0, 0, 0)T
59
32
32
1.42
0.65
0.53
Algorithm 1a
Algorithm 1b
Table 3 The results of Example 2 for Algorithms 1, 1a, and 1b
Starting point
Number of iterations
CPU (s)
Algorithm 1
Algorithm 1a
Algorithm 1b
289
145
128
9.23
4.78
2.42
3
2
2
0.26
0.20
0.15
(10, 10)T
255
120
120
8.43
3.70
2.57
(0, 10)T
259
193
195
8.20
6.10
3.76
(5, 5)T
254
119
119
8.21
3.81
2.40
(0, 0)T
(10, 0)T
Algorithm 1
In a second step we consider three examples of generalized Nash games. They are described
in the paper by Zhang et al. [31] (see also [8]) where they have been solved after having written
them under the form of quasi-variational inequalities. The algorithm used by these authors is
nothing else than our Algorithm 2 when f (x, y) = F (x), y − x. In this paper we consider
these three quasi-variational inequality problems and we solve them by using each of our
six algorithms with the equilibrium function f (x, y) = F (x), y − x corresponding to a
quasi-variational inequality problem. Doing so we can compare the numerical behavior of
our six algorithms with the Zhang and al. Algorithm 2 given in [31, p. 101].
Example 2 Consider a two-person game whose QVI formulation involves the function F =
(F1 , F2 ) and the multivalued mapping K defined for each x = (x1 , x2 ) ∈ IR2 by
F1 (x1 , x2 ) = 2x1 + 83 x2 − 34,
F2 (x1 , x2 ) = 2x2 + 54 x1 − 24.25,
K1 (x2 ) = {y1 ∈ IR | 0 ≤ y1 ≤ 10, y1 ≤ 15 − x2 },
K2 (x1 ) = {y2 ∈ IR | 0 ≤ y2 ≤ 10, y2 ≤ 15 − x1 }.
The set of solutions is composed of the point (5, 9)T and the line segment [(9, 6)T , (10, 5)T ].
The numerical results obtained for this example are listed in Tables 3 and 4 where different
starting points are chosen. The computed solution up to 10−6 is (5, 9)T for all the chosen
starting points except for the point (10, 0)T where the solution (10, 5)T is obtained.
Example 3 Consider an oligopolistic market in which five firms supply a homogeneous product in a noncooperative fashion. The function F = (F1 , . . . , F5 ) is defined for each x ∈ IR5 ,
by
(1/βi ) xi
5000 (1/η) xi
− 1 , i = 1, . . . , 5
+
Fi (x) = ci +
τi
Q
ηQ
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J Glob Optim
Table 4 The results of Example 2 for Algorithms 2, 2a, and 2b
Starting point
Number of iterations
CPU (s)
Algorithm 2
Algorithm 2a
Algorithm 2b
Algorithm 2
Algorithm 2a
Algorithm 2b
(0, 0)T
236
187
188
7.75
6.34
4.32
(10, 0)T
(10, 10)T
297
245
246
9.37
8.12
5.18
(0, 10)T
188
146
145
6.01
4.85
2.78
(5, 5)T
297
244
245
9.35
7.95
5.10
Table 5 The results of Example 3 for Algorithms 1, 1a, and 1b
Starting point
Number of iterations
CPU (s)
Algorithm 1 Algorithm 1a Algorithm 1b Algorithm 1 Algorithm 1a Algorithm 1b
(10, 10, 10, 10, 10)T 36
24
24
1.31
0.82
0.60
(50, 50, 50, 50, 50)T 44
27
27
1.68
0.96
0.53
Table 6 The results of Example 3 for Algorithms 2, 2a, and 2b
Starting point
Number of iterations
CPU (s)
Algorithm 2 Algorithm 2a Algorithm 2b Algorithm 2 Algorithm 2a Algorithm 2b
(10, 10, 10, 10, 10)T
96
65
66
3.07
2.23
1.39
(50, 50, 50, 50, 50)T 110
70
71
3.54
2.35
1.28
where Q = 1≤i≤5 x i . The reader is referred to the paper by Zhang et al. [31] for more
details about this problem and the different values of the parameters ci , βi , τi and η. On the
other hand, the sets Ki (x−i ), i = 1, . . . , 5, are defined as follows
xj ≤ N }.
Ki (x−i ) = {xi ∈ IR | 1 ≤ xi ≤ 150, xi +
j =i
The numerical results obtained for this example are listed in Tables 5 and 6 for two different
starting points. The computed solution up to 10−6 is the same for the two starting points:
(36.9325, 41.8181, 43.7066, 42.6592, 39.1790).
Example 4 In this example, the set K2 (x1 ) in Example 2 is replaced by K2 (x1 ) = {y2 ∈
IR | 0 ≤ y2 ≤ 10}. The corresponding numerical results are given in Tables 7 and 8. As for
Example 2, the computed solution up to 10−6 is (5, 9)T for all the chosen starting points
except for the point (10, 0)T where the solution (10, 5)T is obtained.
For each example and for each test the algorithm is terminated when x k − y k < ε for an
iterate x k . For each of our computational experiments the following values of the parameters
have been used: ε = 10−6 , γ = 1.99, α = 0.5 and c = 0.5. Furthermore, for each test the
number of iterations and the corresponding CPU times in seconds are reported. The symbol
“” means that the number of iterations exceeds 1000.
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J Glob Optim
Table 7 The results of Example 4 for Algorithms 1, 1a, and 1b
Starting point
Number of iterations
CPU (s)
Algorithm 1
Algorithm 1a
Algorithm 1b
Algorithm 1
Algorithm 1a
Algorithm 1b
(0, 0)T
282
201
201
9.14
6.26
4.25
(10, 0)T
155
207
130
5.06
6.59
2.62
(10, 10)T
218
177
177
6.65
5.71
3.87
(0, 10)T
258
195
195
8.21
6.10
4.03
(5, 5)T
281
123
123
8.70
3.93
2.45
Table 8 The results of Example 4 for Algorithms 2, 2a, and 2b
Starting point
Number of iterations
CPU (s)
Algorithm 2
Algorithm 2a
Algorithm 2b
Algorithm 2
Algorithm 2a
Algorithm 2b
(0, 0)T
298
246
246
9.45
7.81
5.26
(10, 0)T
305
248
247
9.68
7.79
5.21
(10, 10)T
198
168
168
6.23
5.28
3.42
(0, 10)T
190
146
146
5.98
4.64
2.84
(5, 5)T
289
237
237
9.18
7.64
4.76
From these preliminary results, it seems that the numerical behavior of Algorithms 1a, 1b,
and 2a, 2b is better than the one of Algorithms 1 and 2, respectively. Furthermore the number
of iterations used by Algorithms 1a and 1b to get a solution is very similar. The difference
between them lies in the CPU time which is systematically smaller for the second one. A
similar behavior can be observed for Algorithms 2a and 2b.
8 Conclusion
In this paper we have presented a very general algorithm for solving the class of quasi-equilibrium problems. Then we have studied and proven the convergence of six new implementable
versions of this algorithm based on the extragradient method and on two different linesearches. Only very preliminary numerical results have been reported. Many other test problems
should be considered and other choices of linesearches and parameter values should be studied to improve the performance of these algorithms. This could be the subject of future
researches.
Acknowledgments
The authors thank the two referees for their helpful suggestions for improving this paper.
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