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Projection iterative method for solving general variational inequalities. Abdellah Bnouhachem, Muhammad. Aslam Noor, Mohamed Khalfaoui &. Sheng Zhaohan ...
Projection iterative method for solving general variational inequalities

Abdellah Bnouhachem, Muhammad Aslam Noor, Mohamed Khalfaoui & Sheng Zhaohan Journal of Applied Mathematics and Computing ISSN 1598-5865 J. Appl. Math. Comput. DOI 10.1007/s12190-012-0581-9

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Author's personal copy J Appl Math Comput DOI 10.1007/s12190-012-0581-9

JAMC

C O M P U TAT I O N A L M AT H E M AT I C S

Projection iterative method for solving general variational inequalities Abdellah Bnouhachem · Muhammad Aslam Noor · Mohamed Khalfaoui · Sheng Zhaohan

Received: 28 January 2012 © Korean Society for Computational and Applied Mathematics 2012

Abstract In this paper, we suggest and analyze a new projection iterative method for solving general variational inequalities by using a new step size. We also prove the global convergence of the proposed method under some suitable conditions. Some preliminary numerical experiments are included to illustrate the advantage and efficiency of the proposed method. Keywords General variational inequalities · Self-adaptive rules · Projection method · Co-coercive operators Mathematics Subject Classification (2000) 49J40 · 65N30

1 Introduction General variational inequalities involving two operators were introduced and studied by Noor [22] in 1988. It have been shown that a wide class of problems with application in various branches of pure and applied sciences can be studied via the general variational inequalities in a unified and general framework, see [1–6, 11, 22–28] and A. Bnouhachem () · S. Zhaohan School of Management Science and Engineering, Nanjing University, Nanjing, 210093, P.R. China e-mail: [email protected] A. Bnouhachem Ibn Zohr University, ENSA, BP 1136, Agadir, Morocco M.A. Noor Mathematics Department, COMSATS Institute of Information Technology, Islamabad, Pakistan M. Khalfaoui Ecole Supérieure de Technologie de Salé, Mohamed 5 University, Agdal-Rabat, Morocco

Author's personal copy A. Bnouhachem et al.

the references therein. This field is dynamic and is experiencing an explosive growth in both theory and applications: as consequence, several numerical techniques including projection and its variant form Wiener-Hopf equations are being developed for solving various classes of variational inequalities and related optimization problems. Projection methods represent important tools for finding the approximate solutions of the general variational inequalities. The main idea of this technique is to establish the equivalence between the variational inequalities and the fixed point problem by using the concept of projection. The idea of this technique can be traced back to Lions and Stampacchia [31]. This alternative equivalent formulation has played a significant and fundamental part in developing various projection iterative methods for solving the general variational inequalities. It is well known that the convergence of projection iterative methods requires that the operator must be strongly monotone and Lipschitz continuous. Due to these strict conditions, many applications of the projection iterative methods are rule out. To overcome these drawbacks and deficiencies, several modifications of the projection methods are suggested; see e.g. [8] for more details. The extragradient method [17, 18] overcome this difficulty by performing an additional forward step and a projection at each iteration according to double projection. This method can be viewed as predictor-corrector method. It convergence requires only that a solution exists and the monotone operator is Lipschitz continuous. When the operator is not Lipschitz continuous or when the Lipschitz continuity constant is not known, the extragradient method and its variant forms require an Armijo-like line search procedure to compute the step size with a new projection need for each trial, which leads to expansive computation. To overcome these difficulties, several modified projection methods have been developed for solving the variational inequalities, see [1–32] and the references therein. Inspired and motivated by the research going on in this area, we suggest and analyze a new projection iterative method for solving the general variational inequalities by using a new direction with a new step size. We also analyze the convergence criteria of the proposed method under some suitable conditions. Some numerical examples are given to illustrate the efficiency of the proposed method along wit its comparison with methods of Bnouhachem and Noor [3, 6]. As a special case, we obtain a new projection method for solving the variational inequalities. Results obtain in this paper may be viewed as an improvement and refinement of the previously known methods.

2 Preliminaries and basic results Let K be a nonempty closed convex subset of R n . For given operators finding u∗ ∈ R n such that g(u∗ ) ∈ K and   ∗   T u , g(v) − g u∗ ≥ 0, ∀v ∈ R n , g(v) ∈ K. (2.1) Problem (2.1) is called the general variational inequality, which first introduced and studied by Noor [22] in 1988. For the applications, formulation and numerical methods of general variational inequalities (2.1), we refer the reader to [1, 3, 11, 23–27].

Author's personal copy Projection iterative method for solving general variational inequalities

If g ≡ I , then the problem (2.1) is equivalent to finding u∗ ∈ K such that   ∗  T u , v − u∗ ≥ 0, ∀v ∈ K,

(2.2)

which is known as the classical variational inequality introduced and studied by Stampacchia [31] in 1964. For the applications, formulation, sensitivity analysis and numerical methods for variational inequalities, see [1–32] and the references therein. We also need the following basic concepts and results. Lemma 2.1 Let K be a closed convex set in R n . For a given z ∈ R n , u ∈ K satisfies the inequality u − z, v − u ≥ 0,

∀v ∈ K,

if and only if u = PK z, where PK is the projection of R n onto a closed convex set K. It is well known that the projection operator satisfies the following properties.   z − PK (z), PK (z) − v ≥ 0, ∀z ∈ R n , v ∈ K, (2.3)   PK (u) − PK (v) ≤ u − v , ∀u, v ∈ R n , (2.4)  2   PK (z) − v  ≤ z − v 2 − z − PK (z)2 , ∀z ∈ R n , v ∈ K. (2.5) Using Lemma 2.1, one can easily show that the general variational inequality (2.1) is equivalent to the fixed point problem. This result is due to Noor [22]. Lemma 2.2 u∗ ∈ R n is solution of problem (2.1) if and only if u∗ ∈ R n satisfies the relation:        (2.6) g u∗ = PK g u∗ − ρT u∗ , ρ > 0. From Lemma 2.2, it is clear that u is solution of (2.1) if and only if u is a zero point of the function   r(u, ρ) := g(u) − PK g(u) − ρT (u) . The next lemma shows that r(u, ρ) is a non-decreasing function with respect to ρ. Lemma 2.3 ([1]) For all u ∈ R n and ρ > ρ > 0, it holds that      r u, ρ  ≥ r(u, ρ).

(2.7)

Bnouhachem and Noor [3] suggested and analyzed the following iterative scheme for solving the general variational inequalities (2.1).

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Algorithm 2.1 For given uk ∈ R n : g(uk ) ∈ K and ρk > 0, compute        g w k = PK g uk − ρk T uk ,

(2.8)

such that       k     ρk T u − T w k  ≤ δ g uk − g w k ,

0 < δ < 1.

(2.9)

And the new iterate is defined by        g uk+1 (αk ) = PK g uk − αk ρk T w k , where      ε k := ρk T w k − T uk ,       D uk , ρk := g uk − g w k + ε k ,          φ uk , ρk := g uk − g w k , D uk , ρk

(2.10) (2.11) (2.12)

and αk :=

φ(uk , ρk ) .

D(uk , ρk ) 2

(2.13)

Inspired and motivated by the research going in this direction, we proposed a new projection iterative method for solving general variational inequalities by using a new direction with a new step size αk , which is the main motivation of this paper. Throughout this paper, we make following assumptions. Assumptions • g is homeomorphism on H i.e., g is bijective, continuous and g −1 is continuous. • T is g-co-coercive on R n i.e., there exists a constant c > 0 such that 2   T (u) − T (v), g(u) − g(v) ≥ cT (u) − T (v) ,



∀u, v ∈ R n .

• The solution set of problem (2.1) denoted by S ∗ , is nonempty.

3 Projection iterative methods In this section, we suggest and analyze a new method for solving general variational inequality (2.1) by using a new direction with a new step size αk , which is the main motivation of this paper. Step 1. For given uk ∈ R n : g(uk ) ∈ K and ρk > 0, compute         g w k = PK g uk − ρk T uk

(3.1)

Author's personal copy Projection iterative method for solving general variational inequalities

and      ε k = ρk T w k − T uk , such that

    k   ε  ≤ δ g uk − g w k ,

(3.2)

0 < δ < 1.

(3.3)

        d uk , ρk := g uk − g w k + ρk T w k ,

(3.4)

Step 2. Set

and for αk > 0 the new iterate is defined by        g uk+1 (αk ) = PK g uk − αk d uk , ρk .

(3.5)

Latter, we show how to choose a suitable step length αk > 0 to force convergence. Remark 3.1 (3.3) implies that      k       g u − g w k , ε k ≤ δ g uk − g w k 2 ,

0 < δ < 1.

(3.6)

Remark 3.2 Since g(w k ) ∈ K, it follows from (2.1) that     ∗  k  T u , g w − g u∗ ≥ 0.

(3.7)

Under the assumption that T is g-co-coercive, we have     k  k T w , g w − g u∗ ≥ 0.

(3.8)

           k  k T w , g u − g u∗ ≥ T w k , g uk − g w k .

(3.9)

It follows from (3.8) that

Since g(uk+1 (αk )) ∈ K, from (2.3), we have               k g w − g uk + g uk − g uk+1 (αk ) , g uk − ρk T uk − g w k ≥ 0. (3.10) Then           k g u − g uk+1 (αk ) , g uk − g w k − ρk T uk            ≥ g uk − g w k , g uk − g w k − ρk T uk .

(3.11)

How to choose values of αk to ensure that g(uk+1 (αk )) is closer to the solution set than g(uk ). For this purpose, we define     2     2 Θ(αk ) = g uk − g u∗  − g uk+1 (αk ) − g u∗  .

(3.12)

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Theorem 3.1 Let u∗ be a solution of problem (2.1) and uk ∈ R n : g(uk ) ∈ K. Then we get

where

         2 Θ(αk ) ≥ 2αk g uk − g w k , D uk , ρk − αk2 D uk , ρk      2   + g uk − g uk+1 (αk ) − αk D uk , ρk           + 2αk g uk+1 (αk ) − g u∗ , g uk − g w k

(3.13)

       D uk , ρk := g uk − g w k + ε k .

(3.14)

Proof From (2.5) and (3.5) we obtain     2     2 Θ(αk ) = g uk − g u∗  − g uk+1 (αk ) − g u∗      2      2  ≥ g uk − g u∗  − g uk − g u∗ − αk d uk , ρk      2   + g uk − g uk+1 (αk ) − αk d uk , ρk             2 = 2αk g uk − g u∗ , d uk , ρk + g uk − g uk+1 (αk )         − 2αk g uk − g uk+1 (αk ) , d uk , ρk . From the definition of d(uk , ρk ) (see (3.4)), we have    2  Θ(αk ) ≥ g uk − g uk+1 (αk )             − 2αk g uk − g uk+1 (αk ) , g uk − g w k + ρk T w k            + 2αk g uk − g u∗ , g uk − g w k + ρk T w k 2            = 2αk ρk g uk − g u∗ , T w k + g uk − g uk+1 (αk )         − 2αk ρk g uk − g uk+1 (αk ) , T w k          + 2αk g uk+1 (αk ) − g u∗ , g uk − g w k        = 2αk ρk g uk − g u∗ , T w k     2   + g uk − g uk+1 (αk ) − αk D uk , ρk         + 2αk g uk − g uk+1 (αk ) , D uk , ρk   2        − αk2 D uk , ρk  − 2αk ρk g uk − g uk+1 (αk ) , T w k          + 2αk g uk+1 (αk ) − g u∗ , g uk − g w k             = 2αk ρk g uk − g u∗ , T w k + g uk − g uk+1 (αk )   2 2  − αk D uk , ρk  − αk2 D uk , ρk 

Author's personal copy Projection iterative method for solving general variational inequalities

           + 2αk g uk − g uk+1 (αk ) , g uk − g w k − ρk T uk          + 2αk g uk+1 (αk ) − g u∗ , g uk − g w k .

(3.15)

Applying (3.9) to the first term in the right side of (3.15), we obtain        Θ(αk ) ≥ 2αk ρk g uk − g w k , T w k       2 2   + g uk − g uk+1 (αk ) − αk D uk , ρk  − αk2 D uk , ρk             + 2αk g uk − g uk+1 (αk ) , g uk − g w k − ρk T uk          (3.16) + 2αk g uk+1 (αk ) − g u∗ , g uk − g w k . Adding (3.11) (multiplied by 2αk ) to (3.16) and using the definition of D(uk , ρk ), we obtain          2 Θ(αk ) ≥ 2αk g uk − g w k , D uk , ρk − αk2 D uk , ρk      2   + g uk − g uk+1 (αk ) − αk D uk , ρk           + 2αk g uk+1 (αk ) − g u∗ , g uk − g w k 

and the lemma is proved. Lemma 3.1 Let u∗ ∈ S ∗ be a solution point of (2.1), we have

   ∗  k   k      k ρk  g uk − g w k 2 . g u − g u ,g u − g w ≥ 1 − 4c

(3.17)

Proof Since u∗ is a solution, it follows from (2.1) that   k     g w − g u∗ , ρk T u∗ ≥ 0

(3.18)

and from (2.3), we have   k             g w − g uk + g uk − g u∗ , g uk − ρk T uk − g w k ≥ 0.

(3.19)

Adding (3.18) and (3.19), we get   k                 g u − g u∗ − g uk − g w k , g uk − g w k − ρk T uk − T u∗ ≥ 0 and consequently   k            g u − g u∗ + ρk T uk − T u∗ , g uk − g w k     2          ≥ g uk − g w k  + ρk g uk − g u∗ , T uk − T u∗ .

(3.20)

Using the g-co-coercivity of T and by a simple manipulation, it follows from (3.20) that

Author's personal copy A. Bnouhachem et al.

  k       g u − g u∗ , g uk − g w k        2  2 ≥ g uk − g w k  + ρk cT uk − T u∗           − ρk T uk − T u∗ , g uk − g w k    k  k 2 √   k   ∗    = g u −g w +  ρk c T u − T u   k 2 ρk   k   2 1 ρk   k  g u −g w  − − g u − g w k   2 c 4c

    ρk  g uk − g w k 2 . ≥ 1− 4c



Now, we consider the last two terms on the right-hand side of (3.13), we have   k     g u − g uk+1 (αk ) − αk D uk , ρk 2          + 2αk g uk+1 (αk ) − g u∗ , g uk − g w k       2 = g uk − g uk+1 (αk ) − αk D uk , ρk             − 2αk g uk − g uk+1 (αk ) − αk D uk , ρk , g uk − g w k        − 2αk2 D uk , ρk , g uk − g w k          + 2αk g uk − g u∗ , g uk − g w k          ≥ 2αk g uk − g u∗ , g uk − g w k            2 − 2αk2 D uk , ρk , g uk − g w k − αk2 g uk − g w k 



    2 ρk 2  − αk g uk − g w k  ≥ 2αk 1 − 4c        − 2αk2 D uk , ρk , g uk − g w k , where the last inequality follows from (3.17). Then from (3.13), we obtain        Θ(αk ) ≥ 2αk g uk − g w k , D uk , ρk   2        − αk2 D uk , ρk  + 2 D uk , ρk , g uk − g w k



    2 ρk 2  + 2αk 1 − − αk g uk − g w k  . 4c

(3.21)

4 Convergence analysis In this section, we prove some useful results which will be used in the consequent analysis and then investigate the strategy of how chose the new step size αk . For this purpose, we define

Author's personal copy Projection iterative method for solving general variational inequalities

       Φ(αk ) := 2αk g uk − g w k , D uk , ρk   2        − αk2 D uk , ρk  + 2 g uk − g w k , D uk , ρk . Note that Φ(αk ) is a quadratic function of αk and it reaches its maximum at αk∗ =

g(uk ) − g(w k ), D(uk , ρk )

D(uk , ρk ) 2 + 2g(uk ) − g(w k ), D(uk , ρk )

(4.1)

and   Φ αk∗ =

(g(uk ) − g(w k ), D(uk , ρk ))2

D(uk , ρk ) 2 + 2g(uk ) − g(w k ), D(uk , ρk )        = αk∗ g uk − g w k , D uk , ρk .

(4.2)

In the next theorem we show that αk∗ and Φ(αk∗ ) are lower bounded away from zero, whenever g(uk ) = g(w k ) and it is one of the keys to prove the global convergence results. Theorem 4.1 For given uk ∈ R n : g(uk ) ∈ K and ρk > 0, then we have the following αk∗ ≥

1 4

(4.3)

and      1  2 Φ αk∗ ≥ (1 − δ)g uk − g w k  . 4

(4.4)

Proof It follows from (3.6) that   k         2       g u − g w k , D uk , ρk = g uk − g w k  + g uk − g w k , ε k     2 ≥ (1 − δ)g uk − g w k  . (4.5) From 0 < δ < 1 and (3.3), we have   k         2       g u − g w k , D uk , ρk = g uk − g w k  + g uk − g w k , ε k  2       1   = g uk − g w k  + g uk − g w k , ε k 2  2 1   + g uk − g w k  2   2       1   ≥ g uk − g w k  + g uk − g w k , ε k 2 1  2 + ε k  2  2 1 = D uk , ρk  2

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and thus αk∗ =

1 g(uk ) − g(w k ), D(uk , ρk ) ≥ . k 2 k k k

D(u , ρk ) + 2g(u ) − g(w ), D(u , ρk ) 4 

Using (4.2), (4.3) and (4.5) directly we obtained (4.4). Let ρk satisfy ∞



k=0

k=0

0 < ρl ≤ inf ρk ≤ sup ρk ≤ ρu < 4c, αk∗∗

= min{(1 − ρ4ck ), αk∗ }. For fast convergence, we take we take the new step size as a relaxation factor γ ∈ [1, 2) and set the step-size αk in (3.5) by αk = γ αk∗∗ . If αk = γ (1 − ρ4ck ), it follows from (3.21) that

    ρk 2  g uk − g w k 2 Θ(αk ) ≥ γ (2 − γ ) 1 − 4c

    ρu 2  g uk − g w k 2 . ≥ γ (2 − γ ) 1 − 4c

(4.6)

If αk = γ αk∗ , it follows from (3.21) and Theorem 4.1 that

       Θ(αk ) ≥ γ αk∗ 2 g uk − g w k , D uk , ρk   2        − γ αk∗ D uk , ρk  + 2 D uk , ρk , g uk − g w k        = γ (2 − γ )αk∗ g uk − g w k , D uk , ρk ≥ γ (2 − γ )

    (1 − δ)  g uk − g w k 2 . 4

(4.7)

Then from Theorem 3.1, (4.6) and (4.7) there is a constant 

 ρu 2 (1 − δ) c := γ (2 − γ ) min 1 − >0 , 4c 4 such that   k+1              g u (αk ) − g u∗ 2 ≤ g uk − g u∗ 2 − cg uk − g w k 2

∀u∗ ∈ S ∗ . (4.8)

Then, we have      k+1           g u (αk ) − g u∗  ≤ g uk − g u∗  ≤ · · · ≤ g u0 − g u∗ . Since g is homeomorphism and from the above inequality, it is easy to verify that the sequence uk is bounded. Theorem 4.2 The sequence {uk } generated by the proposed method converges to a solution point of problem (2.1).

Author's personal copy Projection iterative method for solving general variational inequalities

Proof It follows from (4.8) that ∞    k   g u − g w k 2 < ∞, k=0

which means that

     lim g uk − g w k  = 0.

k→∞

(4.9)

Since g is homeomorphism, we have   lim uk − w k  = 0, k→∞

consequently {w k } is also bounded. Since r(uk , ρ) is a non-decreasing function of ρ, it follows from ρk ≥ ρl that   k    k  r w , ρ l  ≤  r w , ρ k          = g w k − PK g w k − ρk T w k              (using (3.1) and (3.2)) = PK g uk − ρk T w k + ε k − PK g w k − ρk T w k        (using (2.4)) ≤ g uk − g w k + ε k       (using (3.3)) ≤ (1 + δ)g uk − g w k  and from (4.9), we get

  lim r w k , ρl = 0.

k→∞

(4.10)

Let u¯ be a cluster point of {w k } and the subsequence {w kj } converges to u. ¯ Since r(u, ρ) is a continuous function of u, it follows from (4.10) that   r(u, ¯ ρl ) = lim r w kj , ρl = 0. j →∞

According to Lemma 2.2, u¯ is a solution point of problem (2.1). Note that inequality (4.8) is true for all solution point of problem (2.1), hence we have   k+1       g u (αk ) − g(u) (4.11) ¯ , ∀k ≥ 0. ¯  ≤ g uk − g(u) Since {g(w kj )} → g(u) ¯ and g(uk ) − g(w k ) → 0, for any given ε > 0, there is an l > 0, such that      k    g w l − g(u) (4.12) ¯  < ε/2 and g ukl − g w kl  < ε/2. Therefore, for any k ≥ kl , it follows from (4.11) and (4.12) that            k      g u − g(u) ¯  ≤ g ukl − g(u) ¯  ≤ g ukl − g w kl  + g w kl − g(u) ¯  a, AAT u∗ + Ac = a (otherwise u∗ = 0 is the trivial solution). Therefore, we test the problem with a = ρk Ac and ρk ∈ (0, 1). In all tests we take δ = 0.95 and γ = 1.98. All iterations start with u0 = (1, . . . , 1)T and stopped whenever r(uk , 1) ∞ ≤ 10−3 . The iteration numbers and the computational time for the both methods with different dimensions are given in the following tables. Tables 1 and 2 show that the proposed method is more efficient. Numerical results indicate that the proposed method can be save about 19–30 percent of the number of iterations.

Author's personal copy A. Bnouhachem et al. Table 1 Numerical results for problem (5.3) with ρ0 = 10−2

Table 2 Numerical results for problem (5.3) with ρ0 = 10−3

Dimension of the problem

The proposed method

The method in [3]

No. It.

No. It.

CPU (sec.)

CPU (sec.)

n = 100

97

0.05

121

0.09

n = 150

214

0.65

297

0.78

n = 200

148

0.82

194

1.06

n = 250

316

3.87

437

4.72

Dimension of the problem

The proposed method

The method in [3]

No. It.

No. It.

CPU (sec.)

CPU (sec.)

n = 100

7

0.01

10

0.02

n = 150

18

0.05

27

0.07

n = 200

11

0.07

16

0.091

n = 250

25

0.31

35

0.43

5.2 Numerical experiments II We consider the nonlinear complementarity problems: u ≥ 0,

T (u) ≥ 0,

uT T (u) = 0,

(5.4)

where T (u) = D(u) + Mu + q, D(u) and Mu + q are the nonlinear part and linear part of T (u) respectively. We form the linear part in the test problems similarly as in Harker and Pang [10]. The matrix M = AT A + B, where A is an n × n matrix whose entries are randomly generated in the interval (−5, +5) and a skew-symmetric matrix B is generated in the same way. The vector q is generated from a uniform distribution in the interval (−500, 500) in Table 3, and q ∈ (−500, 0) in Table 4. In D(x), the nonlinear part of T (u), the components are chosen to be Dj (u) = dj ∗ arctan(uj ), where dj is a random variable in (0, 1). The test results for problems (5.4) are reported in Tables 3–4. k is the number of iterations and l denotes the number of evaluations of mapping T . Tables 3–4 show that the proposed method is very efficient algorithm even for large-scale classical NCP. 5.3 Numerical experiments III In this subsection, we apply the proposed method to the traffic equilibrium problems and present corresponding numerical results. Consider a network [N, L] of nodes N and directed links L, which consists of a finite sequence of connecting links with a certain orientation. Let a, b, etc., denote the links, and let p, q, etc., denote the paths. We let ω denote an origin/destination

Author's personal copy Projection iterative method for solving general variational inequalities Table 3 Numerical results for problem (5.4) with q ∈ (−500, 500) Dimension of the problem

The method in [3] k

l

CPU (sec.)

The method in [6] k

l

CPU (sec.)

The proposed method k

l

CPU (sec.)

n = 100

254

528

0.063

201

420

0.21

134

298

0.031

n = 200

384

798

0.11

274

586

0.31

186

411

0.032

n = 500

476

988

0.23

285

615

0.57

234

518

0.12

n = 600

393

817

0.26

256

546

0.61

190

424

0.15

n = 800

392

815

0.46

238

511

0.81

187

420

0.25

n = 1000

447

929

1.85

248

535

1.73

208

448

0.91

Table 4 Numerical results for problem (5.4) with q ∈ (−500, 0) Dimension of the problem

The method in [3] k

The method in [6]

l

CPU (sec.)

k

l

The proposed method

CPU (sec.)

k

CPU (sec.)

l

n = 100

519

1077

0.12

365

781

0.31

284

618

0.031

n = 200

756

1570

0.15

482

1034

0.53

356

789

0.047

n = 500

1049

2173

0.54

646

1383

1.31

474

1016

0.26

n = 600

951

1973

0.68

684

1387

1.67

417

884

0.31

n = 800

854

1770

1.06

547

1181

2.21

385

827

0.52

n = 1000

1029

2133

4.31

613

1325

4.22

450

924

1.89

Fig. 1 An illustrative example of given directed network and the O/D pairs

(O/D) pair of nodes of the network and Pω denotes the set of all paths connecting O/D pair ω. Note that the path-arc incidence matrix and the path-O/D pair incidence matrix, denoted by A and B, respectively, are determined by the given network and O/D pairs. To see how to convert a traffic equilibrium problem into a variational inequality, we take into account a simple example depicted in Fig. 1.

Author's personal copy A. Bnouhachem et al.

For the given example in Fig. 1, the path-arc incidence matrix A and the path-O/D pair incidence matrix B have the following forms: No. link A=

1

2

3

4

5

0 ⎜1 ⎜ ⎝0 0

0 0 0 1

1 0 0 0

0 0 1 0

0 1⎟ ⎟, 0⎠ 1



No. O/D pair



B=

ω1 ⎛ 1 ⎜1 ⎜ ⎝0 0

ω2 ⎞ 0 0⎟ ⎟. 1⎠ 1

Let up represent the traffic flow on path p and fa denote the link load on link a, then the arc-flow vector f is given by f = AT u. Let dω denote the traffic amount between O/D pair ω, which must satisfy  dω = up . p∈Pω

Thus, the O/D pair-traffic amount vector d is given by d = B T u. Let t (f ) = {ta , a ∈ L} be the vector of link travel costs, which is a function of the link flow. A user traveling on path p incurs a (path) travel cost θp . For given link travel cost vector t, the path travel cost vector θ is given by   θ = At (f ) and thus θ (x) = At AT u . Associated with every O/D pair ω, there is a travel disutility λω (d). Since both the path costs and the travel disutilities are functions of the flow pattern u, the traffic network equilibrium problem is to seek the path flow pattern u∗ such that u∗ ≥ 0, where

and thus

T    u − u∗ T u∗ ≥ 0,

  Fp (u) = θp (u) − λω d(u) ,

∀u ≥ 0

(5.5)

∀ω, p ∈ Pω

    T (u) = At AT u − Bλ B T u .

We apply the proposed method to the example taken from [21] (Example 7.5 in [21]), which consisted of 25 nodes, 37 links and 6 O/D pairs. The network is depicted in Fig. 2. For this example, there are together 55 paths for the 6 given O/D pairs and hence the dimension of the variable u is 55. Therefore, the path-arc incidence matrix A is a 55 × 37 matrix and the path-O/D pair incidence matrix B is a 55 × 6 matrix. The user cost of traversing link a is given in Table 5.

Author's personal copy Projection iterative method for solving general variational inequalities

Fig. 2 A directed network with 25 nodes and 37 links Table 5 The link traversing cost functions ta (f ) in the example t1 (f ) = 5 · 10−5 f14 + 5f1 + 2f2 + 500

4 + 6f + f + 300 t20 (f ) = 3 · 10−5 f20 20 21

t2 (f ) = 3 · 10−5 f24 t3 (f ) = 5 · 10−5 f34 t4 (f ) = 3 · 10−5 f44 t5 (f ) = 6 · 10−5 f54

+ 4f2 + 4f1 + 200

4 + 4f + f + 400 t21 (f ) = 4 · 10−5 f21 21 22

+ 3f3 + f4 + 350

4 + 6f + f + 500 t22 (f ) = 2 · 10−5 f22 22 23

+ 6f4 + 3f5 + 400

4 + 9f + 2f + 350 t23 (f ) = 3 · 10−5 f23 23 24

+ 6f5 + 4f6 + 600

4 + 8f + f + 400 t24 (f ) = 2 · 10−5 f24 24 25

t6 (f ) = 7f6 + 3f7 + 500

4 + 9f + 3f + 450 t25 (f ) = 3 · 10−5 f25 25 26

t7 (f ) = 8 · 10−5 f74 + 8f7 + 2f8 + 400

4 + 7f + 8f + 300 t26 (f ) = 6 · 10−5 f26 26 27

t8 (f ) = 4 · 10−5 f84 + 5f8 + 2f9 + 650 t9 (f ) = 10−5 f94 + 6f9 + 2f1 0 + 700

4 + 8f + 3f + 500 t27 (f ) = 3 · 10−5 f27 27 28

t10 (f ) = 4f10 + f12 + 800

4 + 3f + f + 450 t29 (f ) = 3 · 10−5 f29 29 30

4 + 7f + 4f + 650 t11 (f ) = 7 · 10−5 f11 11 12

4 + 7f + 2f + 600 t30 (f ) = 4 · 10−5 f30 30 31

t12 (f ) = 8f12 + 2f13 + 700

4 + 8f + f + 750 t31 (f ) = 3 · 10−5 f31 31 32

4 + 7f + 3f + 600 t13 (f ) = 10−5 f13 13 18

4 + 8f + 3f + 650 t32 (f ) = 6 · 10−5 f32 32 33

t14 (f ) = 8f14 + 3f15 + 500

4 + 9f + 2f + 750 t33 (f ) = 4 · 10−5 f33 33 31

4 + 9f + 2f + 200 t15 (f ) = 3 · 10−5 f15 15 14

4 + 7f + 3f + 550 t34 (f ) = 6 · 10−5 f34 34 30

t16 (f ) = 8f16 + 5f12 + 300

4 + 8f + 3f + 600 t35 (f ) = 3 · 10−5 f35 35 32

4 + 7f + 2f + 450 t17 (f ) = 3 · 10−5 f17 17 15

4 + 8f + 4f + 750 t36 (f ) = 2 · 10−5 f36 36 31

t18 (f ) = 5f18 + f16 + 300

4 + 5f + f + 350 t37 (f ) = 6 · 10−5 f37 37 36

4 + 7f + 650 t28 (f ) = 3 · 10−5 f28 28

t19 (f ) = 8f19 + 3f17 + 600

The disutility function is given by λω (d) = −mω dω + qω

(5.6)

and the coefficients mω and qω in the disutility function of different O/D pairs for this example are given in Table 6. The test results for problems (5.5) for different ε are reported in Table 7, k is the number of iterations and l denotes the number of evaluations of mapping T . Table 7 shows that the new method is more flexible and efficient to solve traffic equilibrium problem. Moreover, it demonstrates computationally that the new method

Author's personal copy A. Bnouhachem et al. Table 6 The O/D pairs and the parameters in (5.6) of the example

Table 7 Numerical results for different ε

(O,D) pair ω (1, 20) (1, 25) (2, 20) (3, 25) (1, 24) (11, 25) mω

1

6

10

5

7

9



1000

800

2000

6000

8000

7000

|Pω |

10

15

9

6

10

5

The proposed method

The method in [6]

Different ε

k

l

CPU (sec.)

k

CPU (sec.)

l

10−5

202

439

0.078

222

485

0.18

10−6

261

562

0.031

291

634

0.21

10−7

319

688

0.016

362

788

0.26

10−8

377

818

0.016

427

929

0.31

10−9

439

946

0.018

499

1084

0.36

is more effective than the method presented in [6] in the sense that the new method needs fewer iteration and less evaluation numbers of T , which clearly illustrate its efficiency. Acknowledgements Abdellah Bnouhachem would like to thank Prof. Omar Halli, Rector, Ibn Zohr University, for providing excellent research facilities.

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