Noname manuscript No.
(will be inserted by the editor)
A fixed point optimization algorithm for the equilibrium
problem over the fixed point set and its applications
Trinh Ngoc Hai
Received: date / Accepted: date
Abstract We discuss the equilibrium problem for a continuous bifunction over the fixed
point set of a firmly nonexpansive mapping. We then present an iterative algorithm, which
uses the firmly nonexpansive mapping at each iteration, for solving the problem. The algorithm is quite simple and it does not require monotonicity and Lipschitz-type condition on
the equilibrium function. At the end of the paper, we present a numerical example and an
application to power control in CDMA data networks.
Keywords Equilibrium problem · Nash equilibrium · variational inequality problem · fixed
point problem · firmly nonexpansive mapping · CDMA network
Mathematics Subject Classification (2000) 65 K10 · 65 K15 · 90 C25 · 90 C33
1 Introduction
In recent years, equilibrium problems (EP) is an important subject that recently has been
considered in many research papers. It is well known that various classes of optimization,
variational inequality, Kakutani fixed point, Nash equilibrium in noncooperative game theory and minimax problems can be formulated as an equilibrium problem of the form (EP)
[5].
The typical form of equilibrium problems is formulated by means of Ky Fan’s inequality
due to Ky Fan’s contribution to this field and is given as [5]:
find x∗ ∈ C such that f (x∗ , y) ≥ 0 for all y ∈ C,
EP( f ,C)
where C is a nonempty closed convex subset in Rn and f : C ×C → R is a bifunction such
that f (x, x) = 0 for all x ∈ C. The set of solutions of EP( f ,C) is denoted by Sol( f ,C).
If f (x, y) = ⟨F(x), y − x⟩ where F is a mapping from C to C, then problem EP( f ,C)
becomes the following variational inequality:
find x∗ ∈ C such that ⟨F(x∗ ), y − x∗ ⟩ ≥ 0 for all y ∈ C,
V I(F,C)
Trinh Ngoc Hai
School of Applied Mathematics and Informatics, Hanoi University of Science and Technology, 1st Dai Co
Viet street, Hanoi, Viet Nam
E-mail: [email protected]
2
Trinh Ngoc Hai
The set of solutions of V I(F,C) is denoted by Sol(F,C).
It is well-known that x∗ is solution of V I(F,C) if and only if it is the fixed point of the
mapping PrC (I − λ F), that is, x∗ = PrC (x − λ F(x∗ )), where λ > 0 and PrC is Euclidean
projector on C. Under the assumptions that F is strongly monotone and Lipschitz continuous, the mapping PrC (I − λ F) is strictly contractive over C, hence the sequence (xk )k∈N
generated by the projected gradient algorithm
{
x0 ∈ C
xk+1 = PrC (xk − λ F(xk ))
converges to the unique solution x∗ of V I(F,C) [51].
If F is monotone and Lipschitz, the projected gradient algorithm may not be convergent.
For example, suppose C = R2 and F is a rotation with π2 angle. It is obvious that F is monotone and Lipschitz. However, since ∥xk+1 ∥ ≥ ∥xk ∥ for all k, the sequence (xk )k∈N generated
by the projected gradient algorithm does not converge to the origin - the unique solution of
V I(F,C).
In order to deal with this situation, Korpelevich introduced in [21] an extragradient algorithm:

 x0 ∈ C

yk = PrC (xk − λ F(xk ))

 k+1
x
= PrC (xk − λ F(yk )).
Under the assumptions that F is L− Lipschitz and monotone, λ ∈ (0, L1 ), the sequences
(xk )k∈N and (yk )k∈N convergence to the same point x∗ ∈ Sol(F,C).
This extragradient algorithm has been extended to equilibrium problem in [29]:

0

 x ∈C
k
y = argmin{λ f (xk , y) + 12 ∥y − xk ∥ : y ∈ C}

 k+1
x
= argmin{λ f (yk , y) + 12 ∥y − xk ∥ : y ∈ C}
Under assumptions that f is pseudomonotone and Lipschitz-type continuous, the authors
showed that the sequence (xk )k∈N converges to a solution of EP( f ,C).
To avoid the Lipschitzian condition, the Armijo-backtracking linesearch has been introduced in [45] to solve V I(F,C). The authors used a hyperplane separating xk from the
solution set. Then the new iterate xk+1 is the projection of xk onto this hyperplane. This
method is also extended for pseudomonotone equilibrium problems in [1].
Since all the above methods require monotonicity or pseudomonotonicity of function f ,
a natural question arises: Is it possible to solve equilibrium problems without the monotone
and Lipschitz conditions on f .
To answer this question, we introduce an algorithm to the following equilibrium problem
over the fixed point set: given a continuous function f : Rn × Rn → R satisfying f (x, x) = 0
for all x ∈ Rn and a firmly nonexpansive mapping T : Rn → Rn ,
find a point x∗ ∈ Fix(T ) such that f (x∗ , y) ≥ 0 for all y ∈ Fix(T ),
EP( f , Fix(T ))
where Fix(T ) = {x ∈ Rn : T x = x}. The solution set of this problem is denoted by Sol( f , Fix(T )).
We note that, with T = PrC , the problem EP( f , Fix(T )) becomes problem EP( f ,C). Moreover, in many cases, we deal with the equilibrium problems, of which constraint set C is
implicitly given. Then, the basis method can not be applied effectively.
Equilibrium problem over the fixed point set and its applications
3
H. Iiduka and I. Yamada proposed in [14] a subgradient-type method for solving the
problem (EP( f , Fix(T ))):
Step 0. Choose ε1 ≥ 0, λ1 > 0 and x1 ∈ Rn arbitrarily, and let ρ1 := ∥x1 ∥ and k = 1.
Step 1. Given xk ∈ Rn and ρk ≥ 0, choose εk ≥ 0 and λk > 0
– Find a point yk ∈ Kk := {x ∈ Rn : ∥x∥ ≤ ρk + 1} which satisfies
f (xk , yk ) ≥ 0 and max f (y, xk ) ≤ f (yk , xk ) + εk .
y∈Kk
– Choose ξk ∈ ∂
f (yk , .)(xk )
x
k+1
arbitrarily and compute
= T (x − λk f (yk , xk )ξk ) and ρk+1 = max{ρk , ∥xk+1 ∥}.
k
Step 2. Update k := k + 1, and go to Step 1.
The convergence of this algorithm was proved under suitable assumptions. One of them
is the boundedness of the sequence (ξk )k∈N .
In [18] H. Iiduka considered the variational inequality problem over the fixed point set:
Given C is a nonempty closed convex subset in Rn and F : C → C is a continuous operator,
T : C → C is a firmly nonexpansive mapping. The variational inequality problem over the
fixed point set can be formulated as
find a point x∗ ∈ Fix(T ) such that ⟨F(x), y − x⟩ ≥ 0 for all y ∈ Fix(T )
The solution set of this problem is denoted by V I(F, Fix(T )). For solving this problem,
H. Iiduka proposed a fixed point optimization algorithm:
Step 0. Choose x1 ∈ C, λ1 ∈ (0, ∞) and α1 ∈ [0, 1) arbitrarily, and set n := 1.
Step 1. Given xk ∈ C, choose λk ∈ (0, ∞), αk ∈ [0, 1), and compute xk+1 as follows:
{
yk := T (xk − λk F(xk )),
(1)
xk+1 := PrC (αk xk + (1 − αk )yk ).
Step 2. Update n := n + 1, and go to Step 1.
To prove the convergence of this algorithm, the condition: V I(F, Fix(T )) ⊂ Ω := {x ∈
Fix(T ) : f (xk , x) ≤ 0, ∀k ≥ k0 } is needed.
The main goal of this paper is to extend the fixed points optimization algorithm for
solving the problem (EP( f , Fix(T ))). The convergence of algorithm will be proved without
the condition V I(F, Fix(T )) ⊂ Ω := {x ∈ Fix(T ) : f (xk , x) ≤ 0, ∀k ≥ k0 }.
The rest of this paper is organized as follows. Section 2 briefly explains the necessary
mathematical background. Section 3 presents the fixed point optimization algorithm and
proves that it converges to a solution of Problem (EP( f , Fix(T ))) under certain assumptions.
Numerical results are provided in Section 4.
2 Mathematical preliminaries
A function g : Rm → R is said to be τ −Hölder continuous if there exist Q > 0 and τ ∈ (0, 1]
such that |g(x) − g(y)| ≤ Q∥x − y∥τ for all x, y ∈ Rm . If τ = 1 then g is said to be Lipschitz
continuous. It is obvious that any Hölder continuous function is continuous.
A fixed point of mapping T : Rn → Rn , is a point, x ∈ Rn , satisfying T (x) = x. The set
Fix(T ) = {x ∈ Rn : T (x) = x} called the fixed point set of T . A mapping T : Rn → Rn is said
to be nonexpansive if ∥T (x) − T (y)∥ ≤ ∥x − y∥ for all x, y ∈ Rn . Any nonexpansive mapping
is also continuous. We summarize some properties of the fixed point set of a nonexpansive
mapping in the the following proposition:
4
Trinh Ngoc Hai
Proposition 1 (see [11]) Let C be nonempty, closed convex subset of Rn and T : C → C be
nonexpansive mapping. Then
(a) Fix(T ) is closed and convex;
(b) If C is bounded then Fix(T ) is nonempty.
A mapping T : Rn → Rn is said to be firmly nonexpansive if ∥T (x) − T (y)∥2 ≤ ⟨x −
y, T (x) − T (y)⟩ for all x, y ∈ Rn . Mapping T is firmly nonexpansive if and only if it can be
formulated as T = 21 I + 12 N where I is identity and N is some nonexpansive mapping. It is
well-known that any firmly nonexpansive mapping is also nonexpansive.
Given a nonempty closed convex set C in Rn . The metric projection onto C is defined as
PrC : Rn → C, PrC (x) = argmin{∥x − y∥ : y ∈ C}. The metric projection also can be defined
by relation:
x∗ ∈ C satisfied x∗ = PrC x ⇔ x∗ ∈ C satisfied ⟨x − x∗ , y − x∗ ⟩ ≤ 0 for all y ∈ C,
and therefore PrC is firmly nonexpansive with Fix(PrC ) = C. We summarize some properties of the nonexpansive mapping in the following proposition:
Proposition 2 (see [51])
(a) Let Ti : C → C be nonexpansive mappings (i = 1, 2, . . . , m). Then both Tm Tm−1 · · · T1 and
m
∑m
i=1 wi Ti are also nonexpansive where wi ∈ [0, 1] and ∑i=1 wi = 1.
∩
/
(b) Let Ti : C → C (i = 1, 2, . . . , m) be nonexpansive mappings satisfying m
i=1 Fix(Ti ) ̸= 0.
∩
m
m
Then Fix(∑m
i=1 wi Ti ) = i=1 Fix(Ti ) where wi ∈ (0, 1] and ∑i=1 wi = 1.
(c) T : C → C is firmly nonexpansive if and only if 2T − I is nonexpansive. Moreover, for
given firmly nonexpansive mappings Ti : C → C (i = 1, 2, . . . , m) and wi ≥ 0 satisfying
m
∑m
i=1 wi = 1, ∑i=1 wi Ti is firmly nonexpansive.
We need the following technical lemma.
Lemma 1 (see [49]) Suppose that (αk ) and (βk ) are sequences of nonnegative real numbers
such that
αk+1 ≤ αk + βk , k ≥ 1
where ∑∞
k=1 βk < ∞. Then the sequence (αk ) is convergent.
3 Fixed point optimization algorithm
Assumption 1 We assume
(A1) C ⊂ Rn is a nonempty, closed convex set.
(A2) T : C → C is firmly nonexpansive mapping with Fix(T ) ̸= 0.
/
(A3) f : C ×C → R is τ −Hölder continuous bifunction satisfying f (x, x) = 0 for all x ∈ Rn .
Function f (x, .) convex ∀x ∈ C.
This paper discusses the following equilibrium problem over fixed point set:
Problem 1 Under Assumption 1, we are interested in
finding a point x∗ ∈ Fix(T ) such that f (x∗ , y) ≥ 0 for all y ∈ Fix(T ),
Equilibrium problem over the fixed point set and its applications
5
For solving Problem 1, we investigate the asymptotic behavior of the sequence (xk ) generated by the following algorithm:
Algorithm 1 (Fixed point optimization algorithm)
Step 0. Choose x1 ∈ C , (λk ) ∈ (0, ∞) and (αk ) ∈ [0, 1) arbitrarily. Set k := 1.
Step 1. Given xk , compute xk+1 as follows:

1
k
k
k

 y = argmin{λk f (x , y) + 2 ∥y − x ∥ : y ∈ C}
zk = T (yk )

 k+1
x
= αk xk + (1 − αk )zk .
Step 2. Update k := k + 1, and go to Step 1.
Theorem 1 Assume that
(i) Sol( f ,C) is nonempty.
(ii) There exists k0 ∈ N such that the set Ω := {x ∈ Fix(T ) : f (xk , x) ≤ 0, ∀k ≥ k0 } is
nonempty .
1
2−τ
(iii) Sequences (αk ) and (λk ) satisfy lim supk→∞ αk < 1, ∑∞
< ∞.
k=1 λk
Then the sequences (xk ) and (zk ) generated by Algorithm 1 have following properties:
(a) For every x ∈ Ω , limk→∞ ∥xk − x∥ exists. Two sequences (xk ) and (zk ) are bounded.
(b) limk→∞ ∥xk − zk ∥ = 0 and limk→∞ ∥xk − T (xk )∥ = 0.
(c) If ∥xk − zk ∥ = o(λk ) then (xk ) converges to x̂ ∈ Sol( f , Fix(T )).
Proof (a) Since
1
yk = argmin{λk f (xk , y) + ∥y − xk ∥2 : y ∈ C}
2
(
)
1
0 ∈ ∂ λk f (xk , .) + ∥. − xk ∥2 + δC (.) (yk )
2
we have
where δC is the index function onto C. There exist w ∈ ∂ f (xk , .)(yk ) and v ∈ NC (yk ) such
that
0 = λk w + yk − xk + v.
We obtain
⟨v, x − yk ⟩ ≤ 0 ∀x ∈ C
and hence
⟨xk − yk − λk w, x − yk ⟩ ≤ 0 ∀x ∈ C.
From w ∈ ∂ f (xk , .)(yk ), it follows that
λk ( f (xk , x) − f (xk , yk )) ≥ λk ⟨w, x − yk ⟩
≥ ⟨xk − yk , x − yk ⟩ ∀x ∈ C.
This implies that
⟨yk − xk , xk − x⟩ ≤ λk ( f (xk , x) − f (xk , yk )) − ∥yk − xk ∥2 ∀x ∈ C.
(2)
Applying (2) with x := xk we have
∥xk − yk ∥2 ≤ −λk f (xk , yk ) ≤ λk | f (xk , yk )|
(3)
6
Trinh Ngoc Hai
On other hand, from τ −Hölder continuity of function f , it follows that there exist Q > 0
and τ ∈ (0, 1] such that
| f (xk , yk )| = | f (xk , yk ) − f (xk , xk )| ≤ Q∥(xk , yk ) − (xk , xk )∥τ = Q∥xk − yk ∥τ
(4)
Combining (3) and (4) we have
1
∥xk − yk ∥ ≤ (Qλk ) 2−τ
and
2
λk | f (xk , yk )| ≤ (Qλk ) 2−τ .
1
2−τ
Since ∑∞
< ∞, it implies that limk→∞ ∥xk − yk ∥ = 0 and ∑∞
k=1 λk
k=1
For all x ∈ Ω ⊂ Fix(T ) we have
√
λk | f (xk , yk )| < ∞.
∥xk+1 − x∥2 = ∥αk xk + (1 − αk )zk − x∥2
≤ αk ∥xk − x∥2 + (1 − αk )∥T (yk ) − x∥2
≤ αk ∥xk − x∥2 + (1 − αk )∥yk − x∥2
≤ αk ∥xk − x∥2 + (1 − αk )(∥yk − xk ∥2 + ∥xk − x∥2 + 2⟨yk − xk , xk − x⟩)
≤ ∥xk − x∥2 + (1 − αk )(∥yk − xk ∥2 + 2(λk ( f (xk , x) − f (xk , yk )) − ∥yk − xk ∥2 ))
= ∥xk − x∥2 − (1 − αk )∥yk − xk ∥2 + 2λk (1 − αk )( f (xk , x) − f (xk , yk ))
≤ ∥xk − x∥2 + 2λk | f (xk , yk )|.
k k
k
Since ∑∞
k=1 λk | f (x , y )| < ∞, by lemma 1, we have limk→∞ ∥x − x∥ exists for all x ∈ Ω .
This implies sequence (xk ) bounded. As T is firmly nonexpansive, it follows that (zk ) is also
bounded.
(b) By definition of firmly nonexpansive function, for all x ∈ Fix(T ), we have
∥zk − x∥2 = ∥T (yk ) − T (x)∥2 ≤ ⟨yk − x, zk − x⟩.
Applying ⟨a, b⟩ = 12 (a2 + b2 − (a − b)2 ) ∀a, b ∈ C we have
1
∥zk − x∥2 ≤ (∥yk − x∥2 + ∥zk − x∥2 − ∥yk − zk ∥2 ).
2
Hence
∥zk − x∥2 ≤ ∥yk − x∥2 − ∥yk − zk ∥2
= (∥yk − xk ∥2 + ∥xk − x∥2 + 2⟨yk − xk , xk − x⟩)
− (∥yk − xk ∥2 + ∥xk − zk ∥2 + 2⟨yk − xk , xk − zk ⟩)
= ∥xk − x∥2 − ∥xk − zk ∥2 + 2⟨yk − xk , zk − x⟩ ∀x ∈ Fix(T ).
(5)
Then, we have
∥xk+1 − x∥2 ≤ αk ∥xk − x∥2 + (1 − αk )∥zk − x∥2
≤ αk ∥xk − x∥2 + (1 − αk )(∥xk − x∥2 − ∥xk − zk ∥2 + 2⟨yk − xk , zk − x⟩)
= ∥xk − x∥2 + (1 − αk )(−∥xk − zk ∥2 + 2⟨yk − xk , zk − x⟩).
(6)
Equilibrium problem over the fixed point set and its applications
7
This implies that
(1 − αk )∥xk − zk ∥2 ≤ ∥xk − x∥2 − ∥xk+1 − x∥2 + 2(1 − αk )⟨yk − xk , zk − x⟩.
Let x ∈ Ω . From ∥xk − yk ∥ → 0, αk → 0 and existence of limk→∞ ∥xk − x∥ it implies that
∥xk − zk ∥ → 0. Next, we have
∥zk − T (xk )∥ = ∥T (yk ) − T (xk )∥ ≤ ∥yk − xk ∥.
Since ∥yk − xk ∥ → 0, we obtain that ∥zk − T (xk )∥ → 0. From
∥xk − T (xk )∥ ≤ ∥xk − zk ∥ + ∥zk − T (xk )∥,
it implies that ∥xk − T (xk )∥ → 0.
(c) The boundedness of (xk ) guarantees the existence of a subsequence (xik ) of (xk ) such
that limi→∞ xki = x̂. From 0 = limi→∞ ∥xki − T (xki )∥ = ∥x̂ − T (x̂)∥, it implies that x̂ ∈ Fix(T ).
Next we shall prove that x̂ ∈ Sol( f , Fix(T )). It follows from (5) and (2) that for all
x ∈ Fix(T )
∥zk − x∥2 ≤ ∥xk − x∥2 − ∥xk − zk ∥2 + 2⟨yk − xk , zk − xk ⟩ + 2⟨yk − xk , xk − x⟩
≤ ∥xk − x∥2 + 2⟨yk − xk , zk − xk ⟩ + 2(λk ( f (xk , x) − f (xk , yk )) − ∥yk − xk ∥2 )
≤ ∥xk − x∥2 + 2⟨yk − xk , zk − xk ⟩ + 2λk ( f (xk , x) − f (xk , yk )).
Hence,
0 ≤ ∥xk − x∥2 − ∥zk − x∥2 + 2⟨yk − xk , zk − xk ⟩ + 2λk ( f (xk , x) − f (xk , yk ))
≤ (∥xk − x∥ + ∥zk − x∥)(∥xk − zk ∥) + 2⟨yk − xk , zk − xk ⟩ + 2λk ( f (xk , x) − f (xk , yk )).
Since λk > 0 ∀k ≥ 1, we have
0 ≤ (∥xk − x∥ + ∥zk − x∥)
∥xk − zk ∥
⟨yk − xk , zk − xk ⟩
+2
+ 2( f (xk , x) − f (xk , yk )).
λk
λk
Let k := ki → ∞. From assumption ∥xk − zk ∥ = o(λk ), the boundedness of (xk ) and (zk ) and
f (xk , yk ) → 0 we have
0 ≤ f (x̂, x) ∀x ∈ Fix(T ).
Hence, x̂ ∈ Sol( f , Fix(T )) ⊂ Fix(T ).
From (6), we have
∥xk+1 − x∥2 ≤ ∥xk − x∥2 + (1 − αk )(−∥xk − zk ∥2 + 2⟨yk − xk , zk − x⟩)
≤ ∥xk − x∥2 + K∥yk − xk ∥
√
≤ ∥xk − x∥2 + K λk | f (xk , yk )|,
√
where K := sup{2∥zk − x∥ : k ≥ 1} < ∞. Since ∑∞
λk | f (xk , yk )| < ∞, applying lemma 1,
k=1
we obtain that the limit limk→∞ ∥xk − x∥ exists ∀x ∈ Fix(T ). It implies that
lim ∥xk − x̂∥ = lim ∥xki − x̂∥ = 0
k→∞
That is, xk → x̂ ∈ Sol( f , Fix(T )).
i→∞
⊓
⊔
8
Trinh Ngoc Hai
Remark 1 When we choose f (x, y) = ⟨F(x), y − x⟩, where F : C → C is a continuous operator, we have the fixed point optimization algorithm for the variational inequality problem
over the fixed point set (1), which is proposed in [18]. However, not as in [18], the convergence of the algorithm is obtained without the condition: V I(F, Fix(T )) ⊂ Ω := {x ∈
Fix(T ) : ⟨xk − x, F(xk )⟩ ≥ 0, ∀k ≥ k0 }
Remark 2 The condition ∥xk − zk ∥ = o(λk ) is satisfied when we choose suitable parameters
λk (see Example 1). Analogously to [18], the numerical results in Example 1 show that the
condition ∥xk − zk ∥ = o(λk ) is not satisfied with a fast diminishing constant sequence such
as λk = k1β , (β > 2). Hence, we will use a slowly diminishing constant sequence such as
λk = k1β , β ∈ (1, 2).
4 Numerical examples
In this section, we present some numerical examples for Algorithm 1.
Example 1 Let C = R7 , f (x, y) = ⟨Ax + By + c, y − x⟩ where


3 1 −2 3 4 2 0
−1 −4 3 0 2 4 2 


 3 1 −3 2 −2 2 −3



A=
 1 1 2 −4 3 1 0  ,
0 2 0 1 3 2 3


1 3 2 0 1 3 1
2 1 3 0 1 2 4


 
−3 0 1 −2 1 1 2
5
 2 3 −1 2 1 1 0 
7


 
 2 0 4 −1 0 1 3 
9


 
 , c = −8
−1
0
1
−5
2
2
−1
B=


 
1 3 1 0 3 1 2
6


 
 0 2 1 2 −1 4 0 
 10 
3 2 1 0 0 1 4
9
and mapping T : R7 → R7 defined by
T (x) = PrD (x)
where PrD is the metric projection onto D = {x ∈ R7 : ∥x − (1, 0, 0, 0, 0, 0, 0)∥ ≤ 1}. We note
that f is not pseudomonotone and T is firmly nonexpansive mapping. Matrix B is positive
defined, hence function f (x, .) is convex for all x ∈ R7 . Choose x1 = (1, 2, 0, 1, 2, 0, 1), αk =
α , α ∈ (0, 1) and λk = k1β , β > 1. To check if the condition ∥xk − yk ∥ = o(λk ) is satisfied, we
shall investigate the asymptotic behavior of the sequence uk =
∥xk −zk ∥
λk . It is seen from Figure
1 and Figure 2 that when α := 1/2 and β = 1.1, 1.2, 1.5, (uk )k∈N converges to 0 and when
β = 2.0, 2.3 the sequence (uk )k∈N does not converges to 0. Moreover, since ∥xk ∥ ≤ 1 and
∥yk ∥ ≤ 1, we have | f (xk , yk )| = |⟨Axk + Byk + c, yk − xk ⟩| ≤ (∥A∥ + ∥B∥ + ∥c∥)∥yk − xk ∥1 .
1
Choose λk = k1.1
, αk = 12 . From above argument, it implies that all conditions of Theorem 1
are satisfied. Applying Algorithm 1 for problem EP( f , Fix(T )), we have the result in Table
1. We use stopping criteria: ∥xk+1 − xk ∥ ≤ ε with ε = 10−4 .
Equilibrium problem over the fixed point set and its applications
9
3.5
α =1/2, β=1.1
α =1/2, β=1.2
α =1/2, β=1.5
3
2.5
uk
2
1.5
1
0.5
0
0
10
1
2
10
3
10
10
Number of iterations
Fig. 1 The condition ∥xk − zk ∥ = o(λk ) is satisfied when α := 1/2 and β = 1.1, 1.2, 1.5.
14
α=1/2, β= 1.1
α=1/2, β= 2.0
α=1/2, β= 2.3
12
10
u
k
8
6
4
2
0
0
10
1
2
10
3
10
10
Number of iterations
Fig. 2 The condition ∥xk − zk ∥ = o(λk ) is not satisfied when α := 1/2 and β = 2.0, 2.3.
Example 2 In this example, we will apply Algorithm 1 to the power-control problem for
code-division multiple-access (CDMA) systems. We use the model, which was introduced
in [15, 18]. Consider a network with n users. Let I := 1, 2, ..., n be the set of users and p :=
(p1 , p2 , . . . , pn )T is the transmit power of users. Let Ck := [Pkmin , Pkmax ] where Pmax > Pmin ≥
0 and put
C := ∏ Ck .
k∈I
The signal-to-interference-plus-noise ratio (SINR) of kth user can be expressed by a function
of p as following γk : C → R for all p := (p1 , p2 , . . . , kn )T ∈ C, γk (p) =
pk h2k
, where
σ 2 + N1 ∑ j̸=k p j h2j
hk ∈ R is the channel gain for the kth user, σ 2 > 0 is the noise power, and N > 0 is processing
gain.
Suppose that the utility of kth user is a function of p:
Uk (p) =
L
Rk g(γk (p)),
M
10
Trinh Ngoc Hai
Iter
k= 1
k= 2
k= 3
k= 4
k= 5
k= 6
k= 7
k= 8
k= 9
···
k= 294
x1k
1
0.8860
0.9951
1.1698
1.2806
1.1363
1.2673
1.5777
1.4517
x2k
2
0.9624
0.5601
0.1519
0.0116
0.0755
-0.0334
-0.0054
0.1127
x3k
0
0.2404
0.1580
-0.1328
-0.1271
-0.0874
-0.1687
-0.1185
-0.0914
x4k
1
0.8279
0.8728
0.7895
0.8228
0.8984
0.8811
0.6517
0.7416
x5k
2
0.9488
0.6141
0.1756
0.0109
-0.0311
-0.0436
-0.0601
-0.1805
x6k
0
-0.0731
-0.0855
-0.1399
-0.1813
-0.1673
-0.1228
-0.1143
-0.1616
x7k
1
0.2504
0.2100
0.0175
-0.0397
-0.0317
0.0127
0.0581
-0.0233
∥xk+1 − xk ∥
1.6880
0.5444
0.7217
0.2554
0.1850
0.2001
0.3933
0.2490
0.1762
0.8530
-0.2444
-0.4909
0.7094
-0.1133
-0.2882
-0.2801
9.9444.10−5
Table 1 Sequence (xk )k∈N converges to x∗ = (0.8530, −0.2444, −0.4909, 0.7094, −0.1133, −0.2882, −0.2801) ∈
Sol( f , Fix(T )).
where L and M are the number of information bits and the total number of bits in a packet,
respectively, Rk stands for the transmission rate for the kth user, and g(γ ) := (1 − e−γ )M is
the approximate packet success rate (PSR). Let
D :=
∩
Dk , where Dk := {p ∈ Rn : γk (p) ≥ δk } (k ∈ I)
k∈I
where δk > 0 (k ∈ I) is the required SINR for the kth user in the network. Let
f (p, q) = ∑ (Uk (p) −Uk (pk̂ , qk ))
(7)
k∈I
for all p, q ∈ C, where (pk̂ , qk ) := (p1 , p2 , . . . , pk−1 , qk , pk+1 , . . . , pn )T ∈ C. We have to choose
the transmit power p∗ ∈ C in order to maximize the utility of users. Moreover, each user must
achieve the required SINR. That is, find p∗ ∈ Sol( f ,C ∩ D).
However, the set C ∩ D can be empty, for example, when the noise σ 2 is large or one
of the users is too far from base station. In order to avoid this drawback, consider the the
generalized convex feasible set [51], CΦ , defined by
CΦ := {p̂ ∈ C : Φ (p̂) = min Φ (p)}
p∈C
where Φ (p) := 12 ∑k∈I wk d(p, Dk )2 (p ∈ Rn ), wk ∈ (0, 1) (k ∈ I) with ∑k∈I wk = 1 and
d(p, Dk ) := min{∥p − q∥ : q ∈ Dk } (k ∈ I, p ∈ Rn ). When C ∩ D ̸= 0/ we have CΦ = C ∩ D.
So CΦ is a generalization of C ∩ D. Since CΦ is the set of all minimizers of Φ over C, it
cannot be expressed explicitly. Hence, we can not solve EP( f ,CΦ ) directly. We define the
mapping N : Rn → Rn :
[
]
N(p) := PrC
∑ wk PrDk (p)
(p ∈ Rn ),
(8)
k∈I
where PrC is the metric projection onto C. Then, N is nonexpansive and Fix(N) = CΦ . Let
T (p) = 21 p + 12 N(p). It can be seen that mapping T is firmly nonexpansive and Fix(T ) =
Fix(N) = CΦ . We will apply the Algorithm 1 for problem EP( f , Fix(T )).
As in [18] we assume that L = 100, M = 100, Rk = 104 bits/second, (k ∈ I), N = 100
and σ 2 = 10 × 10−14 watts. Suppose that, for all k ∈ I, Pkmin = 0.1 watts and Pkmax = 1
Equilibrium problem over the fixed point set and its applications
11
watts. The initial transmit power of all user is 0.1 watts and hk :=
0.3
,
dk2
(k ∈ I), where dk is
the distance from the kth user to the base station. Suppose d1 := 310m, d2 := 460m, d3 :=
570m, d4 := 660m, d5 := 740m, d6 := 810m, d7 := 880m, d8 := 940m, d9 := 1, 000m and also
wk = 19 (k ∈ I). The required SINR for the kth user is δk = 1, (k ∈ I). Note that in this case,
C ∩ D = 0/ because C ∩ D9 = 0.
/ It is obvious that function f is Lipschitz continuous on
C ×C and we can choose λk =
1
.
(k+50)1.1
1
2−τ
The condition ∑∞
< ∞ is satisfied. We use
k=1 λk
1 1
αk = 10
, 2 . To check if covergence condition ∥xn − zn ∥ = o(λn ) is satisfied, we consider the
behavior of the sequence, (uk )k≥1 , defined by
uk :=
∥xk − zk ∥
λk
It is seen from Figure 3 that limk→∞ uk = 0. That means the condition ∥xn − zn ∥ = o(λn ) is
5
αk=1/10
4.5
αk=1/2
4
3.5
uk
3
2.5
2
1.5
1
0.5
0
0
10
1
2
10
10
3
10
Number of iterations
Fig. 3 The condition ∥xk − zk ∥ = o(λk ) is satisfied .
satisfied. Choose λk =
1
, αk
(k+50)1.1
1
10 . Applying Algorithm 1 for problem EP( f , Fix(T )),
use stopping criteria: ∥xk+1 − xk ∥ ≤ ε with ε = 10−4 .
=
we have the result in Figure 4. We
From Figure 4 we can see that the transmit power of the 1st user is low and the transmit power of the 9th user is high; in other words, transmitted powers are high when users
are far from the base station. The algorithm stop after 175 iterations. The sequence (xk )
convergences to the solution x∗ of EP( f , Fix(T )),
x∗ = (0.1309, 0.1000, 0.1243, 0.2472, 0.4008, 0.5801, 0.8101, 1.0000, 1.0000).
5 Conclusion
In this paper, we have proposed the fixed point optimization algorithm for the equilibrium
problem over fixed point set of firmly nonexpansive. The proposed algorithm does not require the monotonicity of bifunction. However, some convergence conditions are needed.
The proposed problem can be applied for the equilibrium problem over set C, where C does
12
Trinh Ngoc Hai
1
1st user
5th user
9th user
0.9
Transmit power [W]
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
20
40
60
80
100
120
140
160
180
Number of iterations
Fig. 4 The transmit power of 1st user, 5th user and 9th user.
not necessarily have explicit form. Finally, we have applied the algorithm to the power control problem for CDMA network and have presented the numerical examples for the transmit power. Numerical results have shown that with suitable choosing of parameters, the
convergence conditions are satisfied and the proposed algorithm succeeds in approximating
a solution of the proposed equilibrium problem.
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