# Generalized Complementarity Problem with Three Classes of Generalized Variational Inequalities Involving⊕ Operation

Authors: Rais Ahmad; Iqbal Ahmad; Zahoor Ahmad Rather; Yuanheng Wang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this study, we introduce and study a generalized complementarity problem involving XOR operation and three classes of generalized variational inequalities involving XOR operation. Under certain appropriate conditions, we establish equivalence between them. An iterative algorithm is defined for solving one of the three generalized variational inequalities involving XOR operation. Finally, an existence and convergence result is proved, supported by an example.


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## Body

## 1. Introduction

It is well known that the many unrelated free boundary value problems related to mathematical and engineering sciences can be solved by using the techniques of variational inequalities. In a variational inequality formulation, the location of the free boundary becomes an intrinsic part of the solution, and no special devices are needed to locate it. Complementarity theory is an equally important area of operations research and application oriented. The linear as well as nonlinear programs can be distinguished by a family of complementarity problems. The complementarity theory have been elongated for the purpose of studying several classes of problems occurring in fluid flow through porous media, economics, financial mathematics, machine learning, optimization, and transportation equilibrium, for example, [1–5].The correlations between the variational inequality problem and complementarity problem were recognized by Lions [6] and Mancino and Stampacchia [7]. However, Karamardian [8, 9] showed that both the problems are equivalent if the convex set involved is a convex cone. For more details on variational inequalities and complementarity problems, refer to [6, 10–12].The exclusive “XOR,” sometimes also exclusive disjunction (short: XOR) or antivalence, is a Boolean operation which only outputs true if only exactly one of its both inputs is true (so, if both inputs differ). There are many applications of XOR terminology, that is, it is used in cryptography, gray codes, parity, and CRC checks. Commonly, the⊕ symbol is used to denote the XOR operation. Some problems related to variational inclusions involving XOR operation were studied by [13–16].Influenced by the applications of all the above discussed concepts in this study, we introduce and study a generalized complementarity problem involving XOR operation with three classes of generalized variational inequalities involving XOR operation. Some equivalence relations are established between them. An existence and convergence result is proved for one of the three types of generalized variational inequalities involving XOR operation. For illustration, an example is provided.


## 2. Some Basic Concepts and Formulation of the Problem

Throughout this study, we assumeE to be real ordered Banach space with norm ⋅ and E∗ be its dual space. Suppose that d is the metric induced by the norm, 2E (respectively, CBE) is the family of nonempty (respectively, closed and bounded) subsets of E. The Hausdorff metric D.,. on CBE is defined as(1)DA,ℬ=maxsupx∈Adx,ℬ,supy∈ℬdA,y,∀A,ℬ∈CBE,where dx,ℬ=infy∈ℬdx,y, and dA,y=infx∈Adx,y.LetC be a pointed closed convex positive cone in E, and t,x denotes the value of the linear continuous function t∈E∗ at x.The following definitions and concepts are required to achieve the goal of this study, and most of them can be found in [17, 18].Definition 1.
The relation “≤” is called the partial order relation induced by the cone C, that is, x≤y if and only if y−x∈C.Definition 2.
For arbitrary elementsx,y∈E, if x≤y (or y≤x) holds, then x and y are said to be comparable to each other (denoted by x∝y).Definition 3.
For arbitrary elementsx,y∈E, lubx,y and glbx,y mean the least upper bound and the greatest upper bound of the set x,y. Suppose lubx,y and glbx,y exist, then some binary operations are defined as(i)
x∨y=lubx,y(ii)
x∧y=glbx,y(iii)
x⊕y=x−y∨y−x(iv)
x⊙y=x−y∧y−x
The operations∨,∧, ⊕, and ⊙ are called OR, AND, XOR, and XNOR operations, respectively.Proposition 1.
Let⊕ be an XOR operation and ⊙ be an XNOR operation. Then, the following relations hold:(i)
x⊙x=0, x⊙y=y⊙x(ii)
ifx∝0, then −x⊕0≤x≤x⊕0(iii)
0≤x⊕y, if x∝y(iv)
Ifx∝y, then x⊕y=0 if and only if x=y(v)
x⊕y=y⊕x(vi)
x⊕x=0(vii)
0≤x⊕0(viii)
Ifx≤y and u≤v, then x+u≤y+v(ix)
Ifx∝y, then x⊕0⊕y⊕0≤x⊕y⊕0=x⊕y, for all x,y,u,v∈E and λ∈ℝProposition 2.
LetC be a cone in E; then, for each x,y∈E, the following relations hold:(i)
0⊕0=0=0(ii)
x∨y≤x∨y≤x+y(iii)
x⊕y≤x−y(iv)
Ifx∝y, then x⊕y=x−yDefinition 4.
LetA:E⟶E be a single-valued mapping, then(i)
A is said to be a comparison mapping, if x∝y, then Ax∝Ay, x∝Ax, and y∝Ay, for all x,y∈E(ii)
A is said to be a strongly comparison mapping, if A is a comparison mapping and Ax∝Ay, if and only if x∝y, for any x,y∈EDefinition 5.
Letf:E⟶ℝ∪+∞ be a proper functional. A vector ω∗∈E∗ is called subgradient of f at x∈domf, if(2)ω∗,y−x≤fy−fx,for ally∈E.
The set of all subgradients off at x is denoted by ∂fx. The mapping ∂f:E⟶2E∗ defined by(3)∂fx=ω∗∈E∗:ω∗,y−x≤fy−fx,for ally∈Eis called subdifferential of f.Definition 6.
The resolvent operatorJρ∂f associated with ∂f is given by(4)Jρ∂fx=I+ρ∂f−1x,for allx∈E,where ρ>0 is a constant, and I is the identity operator.
It is well known that the resolvent operatorJρ∂f is single-valued as well as nonexpansive.Definition 7.
A mappingf:C⟶ℝ is said to be(i)
Positive homogeneous if, for allα>0 and x∈C, fαx=αfx(ii)
Convex, ifx,y∈C and all λ∈0,1(5)fλx+1−λy≤λfx+1−λfy.Definition 8.
A multivalued mappingF:C⟶2E∗∖∅ is said to be(i)
Upper semicontinuous atx∈C if, for every open set V containing Fx, there exists an open set U containing x such that FU⊆V, where E∗ is equipped with ω∗ topology(ii)
Upper semicontinuous onC if it is upper semicontinuous at every point of C(iii)
Upper hemicontinuous onC if its restriction to line segments of C is upper semicontinuous(iv)
Monotone if, for everyx,y∈C(6)t1−t2,y−x≥0,for allt1∈Fy,t2∈Fx.Definition 9.
A multivalued mappingF:E⟶2E is said to be D-Lipschitz continuous, if there exists a constant λDF>0 such that(7)DFx,Fy≤λDFx−y,for allx,y∈E.Definition 10.
A multivalued mappingF:E⟶2E is said to be relaxed Lipschitz continuous, if there exists a constant k>0 such that(8)w1−w2,x−y≤−kx−y2,for allw1∈Fx,w2∈Fy.
LetF:C⟶2E∗∖∅ be a multivalued mapping with nonempty values and f:C⟶ℝ∪+∞ be a proper functional. We consider the following generalized complementarity problem involving XOR operation.
Findx¯∈C, t¯∈Fx¯ such that(9)t¯,x¯⊕fx¯=0,t¯,y⊕fy≥0,∀y∈C.
We denote bySC⊕ the solution set of generalized complementarity problem involving XOR operation (9).
We mention some special cases of problem (9) as follows.(i)
If we replace⊕ by + and f by f:C⟶ℝ, then problem (9) reduces to the problem of finding x¯∈C and t¯∈Fx¯ such that(10)t¯,x¯+fx¯=0,t¯,y+fy≥0,∀y∈C.Problem (10) is called generalized f complementarity problem, introduced and studied by Huang et al. [19].(ii)
Iff≡0, then problems (9) as well as (10) reduce to the problem of finding x¯∈C and t¯∈Fx¯ such that(11)t¯,x¯=0,t¯,y≥0,∀y∈C.Problem (11) can be found in [20, 21].
We remark that for suitable choices of operators involved in the formulation of (9), a number of known complementarity problems can be obtained easily, for example, [17, 22–24].
Simultaneously, we also study the following three types of generalized variational inequalities involving XOR operation.(1)
Findx¯∈C such that(12)∃t¯∈Fx¯,∀y∈C:t¯,y−x¯⊕fy−fx¯≥0;(2)
Findx¯∈C such that(13)∀y∈C,∃t¯∈Fx¯:t¯,y−x¯⊕fy−fx¯≥0;(3)
Findx¯∈C such that(14)∀y∈C,∀t∈Fy:t,y−x¯⊕fy−fx¯≥0.
We denote the solution set of (12) by S1⊕, (13) by S2⊕, and (14) by S3⊕.
Many known variational inequality problems can be obtained from problems (12)–(14), for example, [25–29] and the references therein.


## 3. Equivalence Results

We establish the equivalence among problems (9), (12)–(14). First, we establish the equivalence between generalized complementarity problem involving XOR operation (9) and generalized variational inequality problem involving XOR operation (12).Theorem 1.
LetF:C⟶2E∗∖∅ be a multivalued mapping with nonempty values and f:C⟶ℝ∪+∞ be a proper functional. Then, the following statements are true:(i)
Ift¯,x¯∝fx¯, then SC⊕⊆S1⊕(ii)
Iff is positive homogeneous, then S1⊕⊆SC⊕Proof.
(i)
Letx¯∈SC⊕, then x¯∈C, and there exists t¯∈Fx¯ such that(15)t¯,x¯⊕fx¯=0,t¯,y⊕fy≥0.Sincet¯,x¯∝fx¯, by (iv) of Proposition 1, we have(16)t¯,x¯=fx¯,Also ast¯,y⊕fy≥0,t¯,y⊕fy⊕fy≥0⊕fy,which implies that(17)t¯,y≥fy.By using (16) and (17), we have(18)t¯,y−x¯=t¯,y−t¯,x¯≥fy−fx¯,t¯,y−x¯⊕fy−fx¯≥fy−fx¯⊕fy−fx¯,that is,(19)t¯,y−x¯⊕fy−fx¯≥0,which implies thatx¯∈S1⊕. So, we have SC⊕⊆S1⊕.(ii)
Letx¯∈S1⊕, then x¯∈C, and there exists t¯∈Fx¯ such that(20)t¯,y−x¯⊕fy−fx¯≥0,∀y∈C.
SinceC is a pointed closed convex positive cone, clearly y=2x¯∈C and y=1/2x¯∈C. Putting y=2x¯ in generalized variational inequality involving XOR operation (12) and using positive homogenity of f, we get(21)t¯,y−x¯⊕fy−fx¯≥0,t¯,2x¯−x¯⊕f2x¯−fx¯≥0,t¯,x¯⊕fx¯≥0.
Now, puttingy=1/2x¯ in generalized variational inequality involving XOR operation ((12)) and using positive homogenity of f, we get(22)t¯,y−x¯⊕fy−fx¯≥0,t¯,y−x¯⊕fy−fx¯⊕fy−fx¯≥0⊕fy−fx¯,which implies that(23)t¯,y−x¯≥fy−fx¯,t¯,12x¯−x¯≥f12x¯−fx¯,t¯,−12x¯≥−12fx¯,thus,(24)t¯,x¯≤fx¯,t¯,x¯⊕fx¯≤fx¯⊕fx¯=0,that is,(25)t¯,x¯⊕fx¯≤0.
Combining (21) and (25), we have(26)t¯,x¯⊕fx¯=0.
From generalized variational inequality involving XOR operations (12) and (16), we have(27)t¯,y−x¯⊕fy−fx¯≥0,t¯,y−x¯⊕fy−fx¯⊕fy−fx¯≥0⊕fy−fx¯,which implies that(28)t¯,y−x¯⊕0≥0⊕fy−fx¯,t¯,y−x¯≥fy−fx¯,t¯,y−t¯,x¯≥fy−fx¯,t¯,y−fx¯≥fy−fx¯,t¯,y≥fy,t¯,y⊕fy≥fy⊕fy=0,thus, we have t¯,y⊕fy≥0. So, we have x¯∈SC⊕. That is, S1⊕⊆SC⊕.Theorem 2.
The following statements are true.(i)
S1⊕⊆S2⊕(ii)
IfF is monotone, then S2⊕⊆S3⊕(iii)
IfF is upper hemicontinuous and f is convex, then S3⊕⊆S2⊕Proof.
(i)
Is trivial(ii)
Letx¯∈S2⊕. Then, for all y∈C, there exists t¯∈Fx¯ such that(29)t¯,y−x¯⊕fy−fx¯≥0.SinceF is monotone, for every y∈C,t∈Fy, and using the above inequality, we have(30)t−t¯,y−x¯≥0,t,y−x¯≥t¯,y−x¯,t,y−x¯⊕fy−fx¯≥t¯,y−x¯⊕fy−fx¯≥0,which implies thatt,y−x¯⊕fy−fx¯≥0. Thus, x¯∈S3⊕.(iii)
Suppose that the conclusion is not true. Then, there existsx¯∈C such that x¯∈S3⊕ and x¯∉S2⊕. Then, for some y∈C and t∈Fx¯, we have(31)t,y−x¯⊕fy−fx¯<0.
SinceF is upper hemicontinuous and f is convex, setting xλ=λy+1−λx¯ and taking λ⟶0, we have(32)tλ,y−x¯⊕fy−fx¯<0,∀tλ∈Fxλ,tλ,y−x¯⊕fy−fx¯⊕fy−fx¯<0⊕fy−fx¯,which implies that(33)tλ,y−x¯<fy−fx¯,tλ,xλ−x¯<fxλ−fx¯,tλ,xλ−x¯⊕fxλ−fx¯<fxλ−fx¯⊕fxλ−fx¯,thus,(34)tλ,xλ−x¯⊕fxλ−fx¯<0,which contradicts that x¯∈S3⊕. Thus, x¯∈S2⊕, and (iii) is true.Remark 1.
If we replace⊕ by + and dropping the concepts related to ⊕ operation, then with slight modification in Theorems 1 and 2, one can obtain some results of Huang et al. [19]. Additionally, for suitable choices of operators in Theorems 1 and 2, one can obtain some results of Farajzadeh and Harandi [30].


## 4. Existence and Convergence Result

In this section, we first establish the equivalence between the generalized variational inequality problem involving XOR operation (12) and a nonlinear equation. Based on this equivalence, we construct an iterative algorithm for solving generalized variational inequality problem involving XOR operation (12).Lemma 1.
The generalized variational inequality problem involving XOR operation (12) admits a solution x¯,t¯, x¯∈C and t¯∈Fx¯, if and only if the following relation is satisfied:(35)x¯=Jρ∂fx¯+ρt¯,where ρ>0 is a constant, Jρ∂f=I+ρ∂f−1 is the resolvent operator associated with f, and I is the identity operator.Proof.
From the definition of resolvent operatorJρ∂f associated with f and relation (35), we have(36)x¯=Jρ∂fx¯+ρt¯=I+ρ∂f−1x¯+ρt¯,which implies that x¯+ρt¯∈x¯+ρ∂fx¯, that is,(37)t¯∈∂fx¯.
By the definition of subdifferential operator∂fx¯ and (37), we have(38)fy−fx¯≥t¯,y−x¯.
Using (vi) of Proposition1, we have(39)t¯,y−x¯⊕fy−fx¯≥t¯,y−x¯⊕t¯,y−x¯,t¯,y−x¯⊕fy−fx¯≥0.
Thus, the generalized variational inequality problem involving XOR operation (12) is satisfied.
Conversely, suppose that generalized variational inequality problem involving XOR operation (12) is satisfied. That is,(40)t¯,y−x¯⊕fy−fx¯≥0,t¯,y−x¯⊕t¯,y−x¯⊕fy−fx¯≥0⊕t¯,y−x¯,that is, fy−fx¯≥t¯,y−x¯, which implies that(41)t¯∈∂fx¯,ρt¯∈ρ∂fx¯,x¯+ρt¯∈x¯+ρ∂fx¯,x¯+ρt¯∈I+ρ∂fx¯,x¯=I+ρ∂f−1x¯+ρt¯,x¯=Jρ∂fx¯+ρt¯,that is, the relation (35) is satisfied.
Based on Lemma1, we develop the following iterative algorithm for solving the generalized variational inequality problem involving XOR operation (12).Iterative Algorithm 1.
LetC⊂E be a pointed closed convex positive cone. Suppose that t¯n∝t¯n−1, for n=1,2,…. Let for x¯0∈C, there exists t0∈Fx¯0, such that(42)x¯1=1−αx¯0+αJρ∂fx¯0+ρt0¯.
Sincet¯0∈Fx¯0∈CBE, by Nadler [31], there exists t¯1∈Fx¯1, using (iv) of Proposition 2, and as t¯0∝t¯1, we have(43)t¯0⊕t¯1=t¯0−t¯1≤DFx¯0,Fx¯1.
Continuing this way, compute the sequencesx¯n and t¯n by the following scheme:(44)x¯n+1=1−αx¯n+αJρ∂fx¯n+ρt¯n,(45)t¯n⊕t¯n−1=t¯n−t¯n−1≤DFx¯n,Fx¯n−1,for n=1,2,…, where x¯n∈C, t¯n∈Fx¯n can be chosen arbitrarily, α∈0,1, D.,. is the Hausdorff metric on CBE, and ρ>0 is a constant.
Now, we prove our main result.Theorem 3.
LetE be a real ordered Banach space and C be a pointed closed convex positive cone in E with partial ordering “≤.” Let f:C⟶ℝ∪+∞ be a functional such that the resolvent operator Jρ∂f associated with f is strongly comparison and continuous. Suppose that F:C⟶2E∗∖∅ is a multivalued mapping such that F is the relaxed Lipschitz continuous with constant k>0 and D-Lipschitz continuous with constant λDF>0. Let x¯n∝x¯n−1 and t¯n∝t¯n−1, where t¯n∈Fx¯n and t¯n−1∈Fx¯n−1, n=1,2,…, such that for ρ>0, the following condition is satisfied:(46)ρ−kλDF2<kλDF2.Then, the sequencesx¯n and t¯n strongly converge to x∗ and t∗, respectively, the solutions of generalized variational inequality problem involving XOR operation (12).Proof.
Sincex¯n+1∝x¯n, for n=1,2,…, using (iii) of Proposition 1, we evaluate(47)0≤x¯n+1⊕x¯n=1−αx¯n+αJρ∂fx¯n+ρt¯n⊕1−αx¯n−1+αJρ∂fx¯n−1+ρt¯n−1≤1−αx¯n⊕x¯n−1+αJρ∂fx¯n+ρt¯n⊕Jρ∂fx¯n−1+ρt¯n−1.
From (47), it follows that(48)x¯n+1⊕x¯n=1−αx¯n⊕x¯n−1+αJρ∂fx¯n+ρt¯n⊕Jρ∂fx¯n−1+ρt¯n−1≤1−αx¯n⊕x¯n−1+αJρ∂fx¯n+ρt¯n⊕Jρ∂fx¯n−1+ρt¯n−1.
Asx¯n∝x¯n−1, t¯n∝t¯n−1, obviously, x¯n+ρt¯n∝x¯n−1+ρt¯n−1, for n=1,2,…. Since the resolvent operator Jρ∂f is strongly comparison, we have(49)Jρ∂fx¯n+ρt¯n∝Jρ∂fx¯n−1+ρt¯n−1.
Using above facts, (iv) of Proposition2 and nonexpansiveness of Jρ∂f, (48) becomes(50)x¯n+1−x¯n≤1−αx¯n−x¯n−1+αJρ∂fx¯n+ρt¯n−Jρ∂fx¯n−1+ρt¯n−1≤1−αx¯n−x¯n−1+αx¯n+ρt¯n−x¯n−1+ρt¯n−1=1−αx¯n−x¯n−1+αx¯n−x¯n−1+ρt¯n−t¯n−1.
Since the multivalued mappingF is the relaxed Lipschitz continuous with constant k>0, D-Lipschitz continuous with constant λDF>0, and using (45) of Iterative Algorithm 1, we have(51)x¯n−x¯n−1+ρt¯n−t¯n−12=x¯n−x¯n−12+2ρt¯n−t¯n−1,x¯n−x¯n−1+ρ2t¯n−t¯n−12≤x¯n−x¯n−12−2ρkx¯n−x¯n−12+ρ2λDF2x¯n−x¯n−12=1−2ρk+ρ2λDF2x¯n−x¯n−12,thus,(52)x¯n−x¯n−1+ρt¯n−t¯n−1≤1−2ρk+ρ2λDF2x¯n−x¯n−1=θx¯n−x¯n−1,where θ=1−2ρk+ρ2λDF2.
Combining (50) and (52), we have(53)x¯n+1−x¯n≤1−αx¯n−x¯n−1+αθx¯n−x¯n−1≤1−α+αθx¯n−x¯n−1,thus, we have(54)x¯n+1−x¯n≤γnx¯1−x¯0,where γ=1−α+αθ. Hence, for m>n>0, we have(55)x¯n−x¯m≤∑i=nm−1x¯i+1−x¯i≤x¯1−x¯0∑i=nm−1γi.
It is clear from condition (46) that 0<γ<1, and consequently, we have x¯n−x¯m⟶0, as n⟶∞. Thus, x¯n is a Cauchy sequence in E, and as E is complete, x¯n⟶x∗∈E, as n⟶∞. From (45) of Iterative Algorithm 1, we have(56)t¯n⊕t¯n−1=t¯n−t¯n−1≤DFx¯n,Fx¯n−1≤λDFx¯n−x¯n−1,thus, t¯n is also a Cauchy sequence in E such that t¯n⟶t∗∈E, as n⟶∞. Now, we will show that x∗,t∗ is a solution of generalized variational inequality problem involving XOR operation (12). As x¯n⟶x∗,t¯n⟶t∗, and resolvent operator Jρ∂f is continuous, we can write(57)x∗=limn⟶∞x¯n+1=limn⟶∞1−αx¯n+αJρ∂fx¯n+ρt¯n=1−αlimn⟶∞x¯n+αJρ∂flimn⟶∞x¯n+ρlimn⟶∞t¯n=1−αx∗+αJρ∂fx∗+ρt∗.
Thus, the relation (35) is satisfied. It remains to show that t∗∈Fx∗. Since t¯n∈Fx¯n, we have(58)dt∗,Fx∗≤t∗−t¯n+dt¯n,Fx∗≤t∗−t¯n+DFx¯n,Fx∗≤t∗−t¯n+λDFx¯n−x∗⟶0,asn⟶∞.
Hencedt∗,Fx∗⟶0, t∗∈Fx∗ as Fx∗∈CBE. By Lemma 1, x∗∈C, t∗∈Fx∗ is a solution of generalized variational inequality problem involving XOR operation (12). This completes the proof.Remark 2.
Combining Theorems1 and 3, we assert that the solution x¯∈C, t¯∈Fx¯ of generalized variational inequality involving XOR operation (12) is also a solution of generalized complementarity problem involving XOR operation (9).


## 5. Numerical Example

In this section, we construct a numerical example in support of Theorem3. Finally, the convergence graphs and the computation tables are provided for the sequences generated by Iterative Algorithm 1.Example 1.
LetE=E∗=ℝ with the usual inner product and norm. Let C=x¯∈ℝ:0≤x¯≤1 be a pointed closed convex positive cone in ℝ. Let f:C⟶ℝ∪+∞ be a functional, ∂f:ℝ⟶2ℝ be the subdifferential of f, F:C⟶2ℝ∖∅ be a multivalued mapping, and Jρ∂f be the resolvent operator associated with f such that(59)fx¯=2x¯2+1,Fx¯=−x¯7,∀x¯∈C.
Then,(60)∂fx¯=4x¯,Jρ∂fx¯=x¯1+4ρ,∀x¯∈C.
One can easily verify that the resolvent operatorJρ∂f¯ is a strongly comparison mapping and continuous.
Forx¯,y∈C, w1∈Fx¯, and w2∈Fy, we have(61)w1−w2,x¯−y=−x¯7+y7,x¯−y=−17x¯−y2≤−110x¯−y2,that is,(62)w1−w2,x¯−y≤−110x¯−y2.
Thus,F is the relaxed Lipschitz continuous with constant k=1/10.
Also,(63)DFx¯,Fy=maxsupx¯∈Fx¯dx¯,Fy,supy∈FydFx¯,y≤max−x¯7+y7,−y7+x¯7=17maxx¯−y,x¯−y≤17x¯−y≤15x¯−y,that is,(64)DFx¯,Fy≤15x¯−y.
Thus,F is the D-Lipschitz continuous with constant λDF=1/5.
Let us takeρ=1, then for k=1/10 and λDF=1/5, the condition (46)(65)ρ−kλDF2<kλDF2,is satisfied.
Furthermore, forρ=1 and α=1/3, we obtain the sequences x¯n and t¯n generated by the Iterative Algorithm 1 as(66)x¯n+1=1−αx¯n+αJρ∂fx¯n+ρt¯n=23x¯n+115x¯n+t¯n,where t¯n∈Fx¯n, and thus, t¯n=−x¯n/7. It is clear that the sequence x¯n converges to x∗=0, and consequently, the sequence t¯n also converges to t∗=0.
For initial valuesx¯0=5,10, and 15, we have the following convergence graphs, which ensure that the sequences x¯n and t¯n converge to 0. Two computation tables are provided for the iterations (Tables 1 and 2) of the sequences x¯n and t¯n (Figures 1, and 2).Table 1
The values ofxn with initial values x0=5, x0=10, and x0=15.
No. ofForx0=5Forx0=10Forx0=15Iterationxnxnxnn = 151015n = 23.619047619047627.2380952380952410.8571428571429n = 32.619501133786855.239002267573707.85850340136055n = 41.896019868264773.792039736529535.68805960479430n = 51.372357237982122.744714475964234.11707171394635n = 60.9933252389203891.986650477840782.97997571676117n = 70.7189782681709481.437956536341902.15693480451284n = 100.2726394162605420.5452788325210840.817918248781626n = 140.07483173525287480.1496634705057500.224495205758624n = 180.02053917470100880.04107834940201770.0616175241030265n = 210.007788536662174760.01557707332434950.0233656099865243n = 250.002137730932324920.004275461864649840.00641319279697477n = 260.001547310008158990.003094620016317980.00464193002447697n = 27000n = 28000Table 2
The values oftn with initial values x0=5, x0=10, and x0=15.
No. ofForx0=5Forx0=10Forx0=15Iterationtntntnn = 1−0.714285714285714−1.42857142857143−2.14285714285714n = 20.1020408163265310.2040816326530610.306122448979592n = 3−0.0145772594752187−0.0291545189504373−0.0437317784256560n = 40.002082465639316950.004164931278633900.00624739691795085n = 5−0.000297495091330993−0.000594990182661986−0.000892485273992979n = 64.24992987615704e − 058.49985975231408e − 050.000127497896284711n = 7−6.07132839451006e − 06−1.21426567890201e-05−1.82139851835302e − 05n = 10−1.23904661112450e − 073.54013317464143e − 085.31019976196215e − 08n = 14−5.16054398635777e − 111.47444113895936e − 112.21166170843905e − 11n = 18−2.14933110635476e − 146.14094601815645e − 159.21141902723467e − 15n = 216.26627144709842e − 17−1.79036327059955e − 17−2.68554490589932e − 17n = 252.60985899504307e − 205.21971799008614e − 207.82957698512922e − 20n = 26−3.72836999291867e − 21−7.45673998583735e − 21−1.11851099787560e − 20n = 27000n = 28000Figure 1
The convergence graph of the sequencex¯n with initial values x0=5, x0=10, and x0=15.Figure 2
The convergence graph of the sequencet¯n with initial values x0=5, x0=10, and x0=15.


## 6. Conclusion

In this study, we introduce and study a generalized complementarity problem involving XOR operation with three classes of generalized variational inequalities involving XOR operation. Some equivalence relations are established between them. Finally, a generalized variational inequality problem involving XOR operation (12) is solved in real ordered Banach spaces. A numerical example is constructed with convergence graphs and computation tables for illustration of our main result.We remark that our results may be further extended using other tools of functional analysis.



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