Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 172060 6 pageshttpdxdoiorg1011552013172060

Research ArticleSymmetric SOR Method for AbsoluteComplementarity Problems

Javed Iqbal and Muhammad Arif

Department of Mathematics Abdul Wali Khan University Mardan Mardan 23200 Pakistan

Correspondence should be addressed to Muhammad Arif marifmathsyahoocom

Received 21 April 2013 Revised 3 August 2013 Accepted 19 August 2013

Academic Editor Filomena Cianciaruso

Copyright copy 2013 J Iqbal and M Arif This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study symmetric successive overrelaxation (SSOR) method for absolute complementarity problems Solving this problem isequivalent to solving the absolute value equations Some examples are given to show the implementation and efficiency of themethod

1 Introduction

Absolute complementarity problem seeks real vectors 119909 ge 0

and 119860119909 minus |119909| minus 119887 ge 0 such that

⟨119909 119860119909 minus |119909| minus 119887⟩ = 0 (1)

where 119860 isin 119877119899times119899 and 119887 isin 119877

119899 The complementarity the-ory was introduced and studied by Lemke [1] and Cottleand Dantzig [2] The complementarity problems have beengeneralized and extended to study a wide class of problemswhich arise in pure and applied sciences see [1ndash9] andthe references therein Equally important is the variationalinequality problem which was introduced and studied in theearly sixties

In this paper we suggest and analyze SSOR [5] methodfor absolute complementarity problemwhich was introducedby Noor et al [10] The convergence analysis of the proposedmethod is considered under some suitable conditions Weshow that the absolute complementarity problems are equiv-alent to variational inequalities Results are very encouragingThe ideas and the technique of this paper may stimulatefurther research in these areas

Let 119877119899 be the finite dimension Euclidean space whosethe inner product and norm are denoted by ⟨sdot sdot⟩ and sdot

respectively For a given matrix 119860 isin 119877119899times119899 a vector 119887 isin 119877119899 we

consider the problem of finding 119909 isin 119870lowast such that

119909 isin 119870lowast

119860119909 minus |119909| minus 119887 isin 119870lowast

⟨119860119909 minus |119909| minus 119887 119909⟩ = 0

(2)

where 119870lowast = 119909 isin 119877119899

⟨119909 119910⟩ ge 0 119910 isin 119870 is the polar coneof a closed convex cone 119870 in 119877

119899 and |119909| will denote thevector in 119877119899 with absolute values of components of 119909 isin 119877

119899We remark that the absolute value complementarity problem(2) can be viewed as an extension of the complementarityproblem considered by Karamardian [6]

Let 119870 be a closed and convex set in the inner productspace 119877119899 We consider the problem of finding 119909 isin 119870 suchthat

⟨119860119909 minus |119909| minus 119887 119910 minus 119909⟩ ge 0 forall119910 isin 119870 (3)

The problem (3) is called the absolute value variationalinequality which is a special form of the mildly nonlinearvariational inequalities [11] If 119870 = 119877

119899 then the problem (3)is equivalent to find 119909 isin 119877119899 such that

119860119909 minus |119909| minus 119887 = 0 (4)

To propose and analyze algorithms for absolute complemen-tarity problems we need the following definitions

Definition 1 119861 isin 119877119899times119899 is called an 119871-matrix if 119887

119894119894gt 0 for 119894 =

1 2 119899 and 119887119894119895le 0 for 119894 = 119895 119894 119895 = 1 2 119899

2 Journal of Applied Mathematics

Definition 2 If 119860 isin 119877119899times119899 is positive definite then

(i) there exists a constant 120574 gt 0 such that

⟨119860119909 119909⟩ ge 1205741199092

forall119909 isin 119877119899

(5)

(ii) there exists a constant 120573 gt 0 such that

119860119909 le 120573 119909 forall119909 isin 119877119899

(6)

2 Absolute Complementarity Problems

To propose and analyze algorithm for absolute complemen-tarity problems we need the following results

Lemma 3 (see [12]) Let119870 be a nonempty closed convex set in119877119899 For a given 119911 isin 119877119899 119906 isin 119870 satisfies the inequality

⟨119906 minus 119911 119906 minus V⟩ ge 0 V isin 119870 (7)

if and only if

119906 = 119875119870119911 (8)

where 119875119870is the projection of 119877119899 onto the closed convex set 119870

Lemma 4 (see [10]) If119870 is the positive cone in 119877119899 then 119909 isin 119870is a solution of absolute variational inequality (3) if and only if119909 isin 119870 is the solution of the complementarity problem (2)

The next result proves the equivalence between varia-tional inequality (3) and the fixed point

Lemma 5 (see [10]) If119870 is closed convex set in119877119899 then 120588 gt 0119909 isin 119870 satisfies (3) if and only if 119909 isin 119870 satisfies the relation

119909 = 119875119870(119909 minus 120588 [119860119909 minus |119909| minus 119887]) (9)

where 119875119870is the projection of 119877119899 onto the closed convex set 119870

Now using Lemmas 4 and 5 the absolute complementar-ity problem (2) can be transformed to fixed-point problem as

119909 = 119875119870(119909 minus 120588 [119860119909 minus |119909| minus 119887]) (10)

Theorem 6 (see [10]) Let 119860 isin 119877119899times119899 be a positive definite

matrix with constant 120572 gt 0 and continuous with constant120573 gt 0 If 0 lt 120588 lt 2(120574 minus 1)(120573

2

minus 1) 120573 gt 1 120574 gt 1 thenthere exists a unique solution 119909 isin 119870 such that

⟨119860119909 minus |119909| minus 119887 119910 minus 119909⟩ ge 0 forall119910 isin 119870 (11)

where119870 is a closed convex set in 119877119899

To define the projection operator 119875119870 we consider the

special case when 119870 = [0 119888] is a closed convex set in 119877119899 asfollows

Definition 7 (see [3]) Let 119870 = [0 119888] is a closed convex set in119877119899 Then the projection operator 119875

119870119909 is defined as

(119875119870119909)119894= min max (0 119909

119894) 119888119894 119894 = 1 2 119899 (12)

Lemma 8 (see [3]) For any 119909 and 119910 in 119877119899 the following factshold

(i) 119875119870(119909 + 119910) le 119875

119870119909 + 119875119870119910

(ii) 119875119870119909 minus 119875119870119910 le 119875119870(119909 minus 119910)

(iii) 119909 le 119910 rArr 119875119870119909 le 119875119870119910

(iv) 119875119870119909 + 119875

119870(minus119909) le |119909| with equality if and only if

minus119888 le 119909 le 119888

Now one splits the matrix 119860 as

119860 = 119863 minus 119871 minus 119880 (13)

where 119863 is the diagonal matrix and 119871 and 119880 are strictly lowerand strictly upper triangular matrices respectively Let 0 lt 120596 lt

2 using (13) one suggests the SSOR method for solving (3) asfollows

Algorithm 9 Consider the following

Step 1 Choose an initial vector 1199090isin 119877119899 and a parameter 120596 isin

119877+ Set 119896 = 0

Step 2 Calculate

119909119896+1

= 119875119870(119909119896minus 119863minus1

times [minus120596119871119909119896+1

+ (120596 (2 minus 120596)119860 + 120596119871) 119909119896

minus120596 (2 minus 120596) (1003816100381610038161003816119909119896

1003816100381610038161003816+ 119887)] )

(14)

Step 3 If 119909119896+1

= 119909119896 then stop else set 119896 = 119896 + 1 and go to

step 2

Algorithm 10 Consider the following

Step 1 Choose an initial vector 1199090isin 119877119899 and a parameter 120596 isin

119877+ Set 119896 = 0

Step 2 Calculate

119909119896+1

= 119875119870(119909119896minus 119863minus1

times [minus120596119880119909119896+1

+ (120596 (2 minus 120596)119860 + 120596119880) 119909119896

minus 120596 (2 minus 120596) (1003816100381610038161003816119909119896

1003816100381610038161003816+ 119887)] )

(15)

Step 3 If 119909119896+1

= 119909119896 then stop else set 119896 = 119896 + 1 and go to

step 2Nowwe define an operator 119892 119877119899 rarr 119877

119899 such that 119892(119909) =120585 where 120585 is the fixed point of the system

120585 = 119875119870(119909 minus 119863

minus1

times [minus120596119871120585 + (120596 (2 minus 120596)119860 + 120596119871) 119909

minus 120596 (2 minus 120596) (|119909| + 119887)] )

(16)

We also assume that the set

120593 = 119909 isin 119877119899

119909 ge 0 119860119909 minus |119909| minus 119887 ge 0 (17)

Journal of Applied Mathematics 3

of the absolute complementarity problem is nonempty Toprove the convergence of Algorithm 9 we need the followingresult

Theorem 11 Consider the operator 119892 119877119899

rarr 119877119899 as defined

in (16) Assume that119860 isin 119877119899times119899 is an 119871-matrix Also assume that

0 lt 120596 le 1 Then for any 119909 isin 120593 it holds that

(i) 119892(119909) le 119909

(ii) 119909 le 119910 rArr 119892(119909) le 119892(119910)

(iii) 120585 = 119892(119909) isin 120593

Proof To prove (i) we need to prove that

120585119894le 119909119894 119894 = 1 2 119899 (18)

with 120585119894satisfying

120585119894= 119875119870(119909119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895(120585119895minus 119909119895) + 120596 (2 minus 120596)

times(119860119909 minus |119909| minus 119887)119894

]

]

)

(19)

To prove the required result we use mathematical inductionFor this let 119894 = 1

1205851= 119875119870(1199091minus 119886minus1

11120596 (2 minus 120596) (119860119909 minus |119909| minus 119887)

1) (20)

Since 119860119909 minus |119909| minus 119887 ge 0 0 lt 120596 le 1 therefore 1205851le 1199091

For 119894 = 2 we have

1205852= 119875119870(1199092minus 119886minus1

22

times [minus12059611987121(1205851minus 1199091) + 120596 (2 minus 120596)

times(119860119909 minus |119909| minus 119887)2] )

(21)

Here 119860119909 minus |119909| minus 119887 ge 0 0 lt 120596 le 2 11987121ge 0 and 120585

1minus 1199091le 0

This implies that 1205852le 1199092

Suppose that

120585119894le 119909119894

for 119894 = 1 2 119896 minus 1 (22)

we have to prove that the statement is true for 119894 = 119896 that is

120585119896le 119909119896 (23)

Consider

120585119896= 119875119870(119909119896minus 119886minus1

119896119896

times[

[

minus120596

119896minus1

sum

119895=1

119871119896119895(120585119895minus 119909119895) + 120596 (2 minus 120596)

times(119860119909 minus |119909| minus 119887)119896

]

]

)

= 119875119870(119909119896minus 119886minus1

119896119896

times [minus120596 (1198711198961(1205851minus 1199091) + 1198711198962(1205852minus 1199092)

+ sdot sdot sdot + 119871119896119896minus1

(120585119896minus1

minus 119909119896minus1

))

+ 120596 (2 minus 120596) (119860119909 minus |119909| minus 119887)119896] )

(24)

Since 119860119909 minus |119909| minus 119887 ge 0 0 lt 120596119896le 2 119871

1198961 1198711198962 119871

119896119896minus1ge 0

and 120585119894le 119909119894for 119894 = 1 2 119896 minus 1 from (24) we can write

120585119896le 119909119896 (25)

Hence (i) is provedNow we prove (ii) for this let us suppose that 120585 = 119892(119909)

and 120601 = 119892(119910) We will prove that

119909 le 119910 997904rArr 120585 le 120601 (26)

As

120585 = 119875119870(119909 minus 119863

minus1

times [minus120596119871120585 + (120596 (2 minus 120596)119860 + 120596119871) 119909

minus 120596 (2 minus 120596) (|119909| + 119887)] )

(27)

so 120585119894can be written as

120585119894= 119875119870( minus 119886

minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895120585119895+ 120596119886119894119894119909119894+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119894119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

4 Journal of Applied Mathematics

= 119875119870((1 minus 120596) 119909

119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895120585119895+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119894119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

(28)

Similarly for 120601119894we have

120601119894= 119875119870((1 minus 120596) 119910

119894minus 119886minus1

119894119894

times[

[

minus 120596

119894minus1

sum

119895=1

119871119894119895120601119895+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119910119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1119895 = 119894

119880119894119895119910119895minus 120596 (2 minus 120596)

1003816100381610038161003816119910119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

(29)

For 119894 = 1 we have

1206011= 119875119870((1 minus 120596) 119910

1minus 119886minus1

11120596 (2 minus 120596)

times[

[

minus

119899

sum

119895=1 119895 = 119894

1198801119895119910119895minus10038161003816100381610038161199101

1003816100381610038161003816minus 1198871

]

]

)

ge 119875119870((1 minus 120596) 119909

1minus 119886minus1

11120596 (2 minus 120596)

times[

[

minus

119899

sum

119895=1 119895 = 119894

1198801119895119909119895minus10038161003816100381610038161199091

1003816100381610038161003816minus 1198871

]

]

)

= 1205851

(30)

Since 1199101ge 1199091 therefore minus|119910

1| le minus|119909

1| Hence it is true for

119894 = 1 Suppose it is true for 119894 = 1 2 119896 minus 1 we will prove itfor 119894 = 119896 for this consider

120601119896= 119875119870((1 minus 120596) 119910

119896minus 119886minus1

119896119896

times[

[

119896minus1

sum

119895=1

119871119896119895120601119895+ (120596 minus 120596 (2 minus 120596))

times

119896minus1

sum

119895=1

119871119896119895119910119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119896119895119910119895minus 120596 (2 minus 120596)

1003816100381610038161003816119910119896

1003816100381610038161003816

minus 120596 (2 minus 120596) 119887119896

]

]

)

ge 119875119870((1 minus 120596) 119909

119896minus 119886minus1

119896119896

times[

[

119896minus1

sum

119895=1

119871119896119895120585119895+ (120596 minus 120596 (2 minus 120596))

times

119896minus1

sum

119895=1

119871119896119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119896119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119896

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119896

]

]

)

= 120585119896

(31)

Since 119909 le 119910 and 120585119894le 120601119894for 119894 = 1 2 119896 minus 1 hence it is true

for 119896 and (ii) is verifiedNext we prove (iii) that is

120585 = 119892 (119909) isin 120593 (32)

Let 120582 = 119892(120585) = 119875119870(120585minus119863

minus1

[120596119871(120582minus120585)+120596(2minus120596)(119860120585minus |120585|minus

119887)]) from (i) 119892(120585) = 120582 le 120585 Also by definition of 119892 120585 = 119892(119909) ge

0 and 120582 = 119892(120585) ge 0Now

120582119894= 119875119870(120585119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895(120582119895minus 120585119895) + 120596 (2 minus 120596)

times (119860120585 minus10038161003816100381610038161205851003816100381610038161003816minus 119887)119894

]

]

)

(33)

Journal of Applied Mathematics 5

For 119894 = 1 1205851ge 0 by definition of 119892 Suppose that (119860120585 minus |120585| minus

119887)119894lt 0 so

1205821= 119875119870(1205851minus 119886minus1

11120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)1)

gt 119875119870(1205851) = 1205851

(34)

which contradicts the fact that 120582 le 120585 Therefore (119860120585 minus |120585| minus119887)119894ge 0Now we prove it for any 119896 in 119894 = 1 2 119899 Suppose the

contrary (119860120585 minus |120585| minus 119887)119894lt 0 then

120582119896= 119875119870(120585119896minus 119886minus1

119896119896

times[

[

minus120596

119896minus1

sum

119895=1

119871119896119895(120582119895minus 120585119895) + 120596 (2 minus 120596)

times (119860120585 minus10038161003816100381610038161205851003816100381610038161003816minus 119887)119896

]

]

)

(35)

As it is true for all 120572 isin [0 1] it should be true for 120572 = 0 Thatis

120582119896= 119875119870(120585119896minus 119886minus1

119896119896120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)119896)

gt 119875119870(120585119896) = 120585119896

(36)

which contradicts the fact that 120582 le 120585 So (119860120585 minus |120585| minus 119887)119896ge 0

for any 119896 in 119894 = 1 2 119899Hence 120585 = 119891(119909) isin 120593

Now we prove the convergence criteria of Algorithm 9when the matrix 119860 is an 119871-matrix as stated in the next result

Theorem 12 Assume that 119860 isin 119877119899times119899 is an 119871-matrix and 0 lt

120596 le 1Then for any initial vector 1199090isin 120593 the sequence 119909

119896 119896 =

0 1 2 defined by Algorithm 9 has the following properties

(i) 0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2

(ii) lim119896rarrinfin

119909119896= 119909lowast is the unique solution of the absolute

complementarity problem

Proof Since 1199090isin 120593 by (i) of Theorem 11 we have 119909

1le 1199090

and 1199091isin 120593 Recursively usingTheorem 11 we obtain

0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2 (37)

From (i) we observe that the sequence 119909119896 is monotone

bounded therefore it converges to some 119909lowast isin 119877119899+satisfying

119909lowast

= 119875119870(119909lowast

minus 119863minus1

times [minus120596119871119909lowast

+ 120596 (2 minus 120596) (119860 + 120596119871) 119909lowast

minus120596 (2 minus 120596) (1003816100381610038161003816119909lowast1003816100381610038161003816+ 119887)] )

= 119875119870(119909lowast

minus 119863minus1

120596 (2 minus 120596)

times [119860119909lowast

minus1003816100381610038161003816119909lowast1003816100381610038161003816minus 119887] )

(38)

Hence 119909lowast is the solution of the absolute complementarityproblem (2)

Note The convergence of Algorithm 10 has the same steps asgiven inTheorems 11 and 12

3 Numerical Results

In this section we consider several examples to show theefficiency of the proposedmethodsThe convergence of SSORmethod is guaranteed for 119871-matrices only but it is alsopossible to solve different type of systems All the experimentsare performedwith Intel(R) Core 2times 21 GHz 1 GBRAM andthe codes are written in MATLAB 7

Example 13 (see [10]) Consider the ordinary differentialequation

1198892

119909

1198891199052minus |119909| = (1 minus 119905

2

) 0 le 119909 le 1

119909 (0) = 0 119909 (1) = 1

(39)

The exact solution is

119909 (119905) =

07378827425 sin (119905)minus3 cos (119905)+3minus1199052 119909 lt 0

minus07310585786119890minus119905

minus02689414214119890119905

+1+1199052

119909 gt 0

(40)

We take 119899 = 10 the matrix 119860 is given by

119886119894119895=

minus242 for 119895 = 119894

121 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(41)

The constant vector 119887 is given by

119887 = (

120

121

117

121

112

121

105

121

96

121

85

121

72

121

57

121

40

121

minus14620

121

)

119879

(42)

Here119860 is not an119871-matrixThe comparison between the exactsolution and the approximate solutions is given in Figure 1

In Figure 1 we see that the SSOR method convergesrapidly to the approximate solution of absolute complemen-tarity problem (2) as compared to GAOR method

In the next example we compare SSOR method withiterative method by Noor et al [13]

Example 14 (see [13]) Let the matrix 119860 be given by

119886119894119895=

8 for 119895 = 119894

minus1 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(43)

Let 119887 = (6 5 5 5 6)119879 the problem size 119899 ranging from

4 to 1024 The stopping criteria are 119860119909 minus |119909| minus 119887 lt 10minus6

6 Journal of Applied Mathematics

0 50 100 150 200 250 300Number of iterations

2-no

rm o

f res

idua

l

GAORSSOR

10minus8

10minus6

10minus4

10minus2

100

102

104

Figure 1 Comparison between GAOR method and SSOR method

Table 1

119899

Iterative method SSOR methodNumber of iterations TOC Number of iterations TOC

4 10 00168 8 00018 11 0018 8 000116 11 0143 9 000132 12 3319 9 000164 12 7145 9 0015128 12 11342 9 0020256 12 25014 10 1051512 12 98317 10 61301024 13 534903 10 126242

We choose initial guess 1199090as 1199090= (0 0 0)

119879 The com-putational results are shown in Table 1

In Table 1 TOC denotes the total time taken by CPUTherate of convergence of SSOR method is better than that ofiterative method [13]

4 Conclusion

In this paper we have discussed symmetric SOR method forsolving absolute complementarity problem The comparisonwith other methods showed the efficiency of the methodThe results and ideas of this paper may be used to solve thevariational inequalities and related optimization problems

References

[1] C E Lemke ldquoBimatrix equilibrium points and mathematicalprogrammingrdquoManagement Science vol 11 pp 681ndash689 1965

[2] R W Cottle and G B Dantzig ldquoComplementary pivot theoryof mathematical programmingrdquo Linear Algebra and its Applica-tions vol 1 no 1 pp 103ndash125 1968

[3] B H Ahn ldquoIterative methods for linear complementarity prob-lems with upperbounds on primary variablesrdquo MathematicalProgramming vol 26 no 3 pp 295ndash315 1983

[4] RW Cottle J-S Pang and R E StoneTheLinear Complemen-tarity Problem Computer Science and Scientific ComputingAcademic Press Boston Mass USA 1992

[5] M Dehghan and M Hajarian ldquoConvergence of SSORmethodsfor linear complementarity problemsrdquo Operations ResearchLetters vol 37 no 3 pp 219ndash223 2009

[6] S Karamardian ldquoGeneralized complementarity problemrdquo Jour-nal of OptimizationTheory and Applications vol 8 pp 161ndash1681971

[7] Y Li and P Dai ldquoGeneralized AOR methods for linear com-plementarity problemrdquo Applied Mathematics and Computationvol 188 no 1 pp 7ndash18 2007

[8] O L Mangasarian ldquoAbsolute value equation solution via con-cave minimizationrdquo Optimization Letters vol 1 no 1 pp 3ndash82007

[9] O L Mangasarian ldquoSolution of symmetric linear complemen-tarity problems by iterative methodsrdquo Journal of OptimizationTheory and Applications vol 22 no 4 pp 465ndash485 1977

[10] M A Noor J Iqbal K I Noor and E Al-Said ldquoGeneralizedAOR method for solving absolute complementarity problemsrdquoJournal of Applied Mathematics vol 2012 Article ID 743861 14pages 2012

[11] M A Noor On variational inequalities [PhD thesis] BrunelUniversity London UK 1975

[12] M A Noor ldquoOn merit functions for quasivariational inequali-tiesrdquo Journal of Mathematical Inequalities vol 1 no 2 pp 259ndash268 2007

[13] M A Noor J Iqbal K I Noor and E Al-Said ldquoOn an iterativemethod for solving absolute value equationsrdquo OptimizationLetters vol 6 no 5 pp 1027ndash1033 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Journal of Applied Mathematics

Definition 2 If 119860 isin 119877119899times119899 is positive definite then

(i) there exists a constant 120574 gt 0 such that

⟨119860119909 119909⟩ ge 1205741199092

forall119909 isin 119877119899

(5)

(ii) there exists a constant 120573 gt 0 such that

119860119909 le 120573 119909 forall119909 isin 119877119899

(6)

2 Absolute Complementarity Problems

To propose and analyze algorithm for absolute complemen-tarity problems we need the following results

Lemma 3 (see [12]) Let119870 be a nonempty closed convex set in119877119899 For a given 119911 isin 119877119899 119906 isin 119870 satisfies the inequality

⟨119906 minus 119911 119906 minus V⟩ ge 0 V isin 119870 (7)

if and only if

119906 = 119875119870119911 (8)

where 119875119870is the projection of 119877119899 onto the closed convex set 119870

Lemma 4 (see [10]) If119870 is the positive cone in 119877119899 then 119909 isin 119870is a solution of absolute variational inequality (3) if and only if119909 isin 119870 is the solution of the complementarity problem (2)

The next result proves the equivalence between varia-tional inequality (3) and the fixed point

Lemma 5 (see [10]) If119870 is closed convex set in119877119899 then 120588 gt 0119909 isin 119870 satisfies (3) if and only if 119909 isin 119870 satisfies the relation

119909 = 119875119870(119909 minus 120588 [119860119909 minus |119909| minus 119887]) (9)

where 119875119870is the projection of 119877119899 onto the closed convex set 119870

Now using Lemmas 4 and 5 the absolute complementar-ity problem (2) can be transformed to fixed-point problem as

119909 = 119875119870(119909 minus 120588 [119860119909 minus |119909| minus 119887]) (10)

Theorem 6 (see [10]) Let 119860 isin 119877119899times119899 be a positive definite

matrix with constant 120572 gt 0 and continuous with constant120573 gt 0 If 0 lt 120588 lt 2(120574 minus 1)(120573

2

minus 1) 120573 gt 1 120574 gt 1 thenthere exists a unique solution 119909 isin 119870 such that

⟨119860119909 minus |119909| minus 119887 119910 minus 119909⟩ ge 0 forall119910 isin 119870 (11)

where119870 is a closed convex set in 119877119899

To define the projection operator 119875119870 we consider the

special case when 119870 = [0 119888] is a closed convex set in 119877119899 asfollows

Definition 7 (see [3]) Let 119870 = [0 119888] is a closed convex set in119877119899 Then the projection operator 119875

119870119909 is defined as

(119875119870119909)119894= min max (0 119909

119894) 119888119894 119894 = 1 2 119899 (12)

Lemma 8 (see [3]) For any 119909 and 119910 in 119877119899 the following factshold

(i) 119875119870(119909 + 119910) le 119875

119870119909 + 119875119870119910

(ii) 119875119870119909 minus 119875119870119910 le 119875119870(119909 minus 119910)

(iii) 119909 le 119910 rArr 119875119870119909 le 119875119870119910

(iv) 119875119870119909 + 119875

119870(minus119909) le |119909| with equality if and only if

minus119888 le 119909 le 119888

Now one splits the matrix 119860 as

119860 = 119863 minus 119871 minus 119880 (13)

where 119863 is the diagonal matrix and 119871 and 119880 are strictly lowerand strictly upper triangular matrices respectively Let 0 lt 120596 lt

2 using (13) one suggests the SSOR method for solving (3) asfollows

Algorithm 9 Consider the following

Step 1 Choose an initial vector 1199090isin 119877119899 and a parameter 120596 isin

119877+ Set 119896 = 0

Step 2 Calculate

119909119896+1

= 119875119870(119909119896minus 119863minus1

times [minus120596119871119909119896+1

+ (120596 (2 minus 120596)119860 + 120596119871) 119909119896

minus120596 (2 minus 120596) (1003816100381610038161003816119909119896

1003816100381610038161003816+ 119887)] )

(14)

Step 3 If 119909119896+1

= 119909119896 then stop else set 119896 = 119896 + 1 and go to

step 2

Algorithm 10 Consider the following

Step 1 Choose an initial vector 1199090isin 119877119899 and a parameter 120596 isin

119877+ Set 119896 = 0

Step 2 Calculate

119909119896+1

= 119875119870(119909119896minus 119863minus1

times [minus120596119880119909119896+1

+ (120596 (2 minus 120596)119860 + 120596119880) 119909119896

minus 120596 (2 minus 120596) (1003816100381610038161003816119909119896

1003816100381610038161003816+ 119887)] )

(15)

Step 3 If 119909119896+1

= 119909119896 then stop else set 119896 = 119896 + 1 and go to

step 2Nowwe define an operator 119892 119877119899 rarr 119877

119899 such that 119892(119909) =120585 where 120585 is the fixed point of the system

120585 = 119875119870(119909 minus 119863

minus1

times [minus120596119871120585 + (120596 (2 minus 120596)119860 + 120596119871) 119909

minus 120596 (2 minus 120596) (|119909| + 119887)] )

(16)

We also assume that the set

120593 = 119909 isin 119877119899

119909 ge 0 119860119909 minus |119909| minus 119887 ge 0 (17)

Journal of Applied Mathematics 3

of the absolute complementarity problem is nonempty Toprove the convergence of Algorithm 9 we need the followingresult

Theorem 11 Consider the operator 119892 119877119899

rarr 119877119899 as defined

in (16) Assume that119860 isin 119877119899times119899 is an 119871-matrix Also assume that

0 lt 120596 le 1 Then for any 119909 isin 120593 it holds that

(i) 119892(119909) le 119909

(ii) 119909 le 119910 rArr 119892(119909) le 119892(119910)

(iii) 120585 = 119892(119909) isin 120593

Proof To prove (i) we need to prove that

120585119894le 119909119894 119894 = 1 2 119899 (18)

with 120585119894satisfying

120585119894= 119875119870(119909119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895(120585119895minus 119909119895) + 120596 (2 minus 120596)

times(119860119909 minus |119909| minus 119887)119894

]

]

)

(19)

To prove the required result we use mathematical inductionFor this let 119894 = 1

1205851= 119875119870(1199091minus 119886minus1

11120596 (2 minus 120596) (119860119909 minus |119909| minus 119887)

1) (20)

Since 119860119909 minus |119909| minus 119887 ge 0 0 lt 120596 le 1 therefore 1205851le 1199091

For 119894 = 2 we have

1205852= 119875119870(1199092minus 119886minus1

22

times [minus12059611987121(1205851minus 1199091) + 120596 (2 minus 120596)

times(119860119909 minus |119909| minus 119887)2] )

(21)

Here 119860119909 minus |119909| minus 119887 ge 0 0 lt 120596 le 2 11987121ge 0 and 120585

1minus 1199091le 0

This implies that 1205852le 1199092

Suppose that

120585119894le 119909119894

for 119894 = 1 2 119896 minus 1 (22)

we have to prove that the statement is true for 119894 = 119896 that is

120585119896le 119909119896 (23)

Consider

120585119896= 119875119870(119909119896minus 119886minus1

119896119896

times[

[

minus120596

119896minus1

sum

119895=1

119871119896119895(120585119895minus 119909119895) + 120596 (2 minus 120596)

times(119860119909 minus |119909| minus 119887)119896

]

]

)

= 119875119870(119909119896minus 119886minus1

119896119896

times [minus120596 (1198711198961(1205851minus 1199091) + 1198711198962(1205852minus 1199092)

+ sdot sdot sdot + 119871119896119896minus1

(120585119896minus1

minus 119909119896minus1

))

+ 120596 (2 minus 120596) (119860119909 minus |119909| minus 119887)119896] )

(24)

Since 119860119909 minus |119909| minus 119887 ge 0 0 lt 120596119896le 2 119871

1198961 1198711198962 119871

119896119896minus1ge 0

and 120585119894le 119909119894for 119894 = 1 2 119896 minus 1 from (24) we can write

120585119896le 119909119896 (25)

Hence (i) is provedNow we prove (ii) for this let us suppose that 120585 = 119892(119909)

and 120601 = 119892(119910) We will prove that

119909 le 119910 997904rArr 120585 le 120601 (26)

As

120585 = 119875119870(119909 minus 119863

minus1

times [minus120596119871120585 + (120596 (2 minus 120596)119860 + 120596119871) 119909

minus 120596 (2 minus 120596) (|119909| + 119887)] )

(27)

so 120585119894can be written as

120585119894= 119875119870( minus 119886

minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895120585119895+ 120596119886119894119894119909119894+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119894119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

4 Journal of Applied Mathematics

= 119875119870((1 minus 120596) 119909

119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895120585119895+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119894119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

(28)

Similarly for 120601119894we have

120601119894= 119875119870((1 minus 120596) 119910

119894minus 119886minus1

119894119894

times[

[

minus 120596

119894minus1

sum

119895=1

119871119894119895120601119895+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119910119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1119895 = 119894

119880119894119895119910119895minus 120596 (2 minus 120596)

1003816100381610038161003816119910119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

(29)

For 119894 = 1 we have

1206011= 119875119870((1 minus 120596) 119910

1minus 119886minus1

11120596 (2 minus 120596)

times[

[

minus

119899

sum

119895=1 119895 = 119894

1198801119895119910119895minus10038161003816100381610038161199101

1003816100381610038161003816minus 1198871

]

]

)

ge 119875119870((1 minus 120596) 119909

1minus 119886minus1

11120596 (2 minus 120596)

times[

[

minus

119899

sum

119895=1 119895 = 119894

1198801119895119909119895minus10038161003816100381610038161199091

1003816100381610038161003816minus 1198871

]

]

)

= 1205851

(30)

Since 1199101ge 1199091 therefore minus|119910

1| le minus|119909

1| Hence it is true for

119894 = 1 Suppose it is true for 119894 = 1 2 119896 minus 1 we will prove itfor 119894 = 119896 for this consider

120601119896= 119875119870((1 minus 120596) 119910

119896minus 119886minus1

119896119896

times[

[

119896minus1

sum

119895=1

119871119896119895120601119895+ (120596 minus 120596 (2 minus 120596))

times

119896minus1

sum

119895=1

119871119896119895119910119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119896119895119910119895minus 120596 (2 minus 120596)

1003816100381610038161003816119910119896

1003816100381610038161003816

minus 120596 (2 minus 120596) 119887119896

]

]

)

ge 119875119870((1 minus 120596) 119909

119896minus 119886minus1

119896119896

times[

[

119896minus1

sum

119895=1

119871119896119895120585119895+ (120596 minus 120596 (2 minus 120596))

times

119896minus1

sum

119895=1

119871119896119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119896119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119896

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119896

]

]

)

= 120585119896

(31)

Since 119909 le 119910 and 120585119894le 120601119894for 119894 = 1 2 119896 minus 1 hence it is true

for 119896 and (ii) is verifiedNext we prove (iii) that is

120585 = 119892 (119909) isin 120593 (32)

Let 120582 = 119892(120585) = 119875119870(120585minus119863

minus1

[120596119871(120582minus120585)+120596(2minus120596)(119860120585minus |120585|minus

119887)]) from (i) 119892(120585) = 120582 le 120585 Also by definition of 119892 120585 = 119892(119909) ge

0 and 120582 = 119892(120585) ge 0Now

120582119894= 119875119870(120585119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895(120582119895minus 120585119895) + 120596 (2 minus 120596)

times (119860120585 minus10038161003816100381610038161205851003816100381610038161003816minus 119887)119894

]

]

)

(33)

Journal of Applied Mathematics 5

For 119894 = 1 1205851ge 0 by definition of 119892 Suppose that (119860120585 minus |120585| minus

119887)119894lt 0 so

1205821= 119875119870(1205851minus 119886minus1

11120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)1)

gt 119875119870(1205851) = 1205851

(34)

which contradicts the fact that 120582 le 120585 Therefore (119860120585 minus |120585| minus119887)119894ge 0Now we prove it for any 119896 in 119894 = 1 2 119899 Suppose the

contrary (119860120585 minus |120585| minus 119887)119894lt 0 then

120582119896= 119875119870(120585119896minus 119886minus1

119896119896

times[

[

minus120596

119896minus1

sum

119895=1

119871119896119895(120582119895minus 120585119895) + 120596 (2 minus 120596)

times (119860120585 minus10038161003816100381610038161205851003816100381610038161003816minus 119887)119896

]

]

)

(35)

As it is true for all 120572 isin [0 1] it should be true for 120572 = 0 Thatis

120582119896= 119875119870(120585119896minus 119886minus1

119896119896120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)119896)

gt 119875119870(120585119896) = 120585119896

(36)

which contradicts the fact that 120582 le 120585 So (119860120585 minus |120585| minus 119887)119896ge 0

for any 119896 in 119894 = 1 2 119899Hence 120585 = 119891(119909) isin 120593

Now we prove the convergence criteria of Algorithm 9when the matrix 119860 is an 119871-matrix as stated in the next result

Theorem 12 Assume that 119860 isin 119877119899times119899 is an 119871-matrix and 0 lt

120596 le 1Then for any initial vector 1199090isin 120593 the sequence 119909

119896 119896 =

0 1 2 defined by Algorithm 9 has the following properties

(i) 0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2

(ii) lim119896rarrinfin

119909119896= 119909lowast is the unique solution of the absolute

complementarity problem

Proof Since 1199090isin 120593 by (i) of Theorem 11 we have 119909

1le 1199090

and 1199091isin 120593 Recursively usingTheorem 11 we obtain

0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2 (37)

From (i) we observe that the sequence 119909119896 is monotone

bounded therefore it converges to some 119909lowast isin 119877119899+satisfying

119909lowast

= 119875119870(119909lowast

minus 119863minus1

times [minus120596119871119909lowast

+ 120596 (2 minus 120596) (119860 + 120596119871) 119909lowast

minus120596 (2 minus 120596) (1003816100381610038161003816119909lowast1003816100381610038161003816+ 119887)] )

= 119875119870(119909lowast

minus 119863minus1

120596 (2 minus 120596)

times [119860119909lowast

minus1003816100381610038161003816119909lowast1003816100381610038161003816minus 119887] )

(38)

Hence 119909lowast is the solution of the absolute complementarityproblem (2)

Note The convergence of Algorithm 10 has the same steps asgiven inTheorems 11 and 12

3 Numerical Results

In this section we consider several examples to show theefficiency of the proposedmethodsThe convergence of SSORmethod is guaranteed for 119871-matrices only but it is alsopossible to solve different type of systems All the experimentsare performedwith Intel(R) Core 2times 21 GHz 1 GBRAM andthe codes are written in MATLAB 7

Example 13 (see [10]) Consider the ordinary differentialequation

1198892

119909

1198891199052minus |119909| = (1 minus 119905

2

) 0 le 119909 le 1

119909 (0) = 0 119909 (1) = 1

(39)

The exact solution is

119909 (119905) =

07378827425 sin (119905)minus3 cos (119905)+3minus1199052 119909 lt 0

minus07310585786119890minus119905

minus02689414214119890119905

+1+1199052

119909 gt 0

(40)

We take 119899 = 10 the matrix 119860 is given by

119886119894119895=

minus242 for 119895 = 119894

121 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(41)

The constant vector 119887 is given by

119887 = (

120

121

117

121

112

121

105

121

96

121

85

121

72

121

57

121

40

121

minus14620

121

)

119879

(42)

Here119860 is not an119871-matrixThe comparison between the exactsolution and the approximate solutions is given in Figure 1

In Figure 1 we see that the SSOR method convergesrapidly to the approximate solution of absolute complemen-tarity problem (2) as compared to GAOR method

In the next example we compare SSOR method withiterative method by Noor et al [13]

Example 14 (see [13]) Let the matrix 119860 be given by

119886119894119895=

8 for 119895 = 119894

minus1 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(43)

Let 119887 = (6 5 5 5 6)119879 the problem size 119899 ranging from

4 to 1024 The stopping criteria are 119860119909 minus |119909| minus 119887 lt 10minus6

6 Journal of Applied Mathematics

0 50 100 150 200 250 300Number of iterations

2-no

rm o

f res

idua

l

GAORSSOR

10minus8

10minus6

10minus4

10minus2

100

102

104

Figure 1 Comparison between GAOR method and SSOR method

Table 1

119899

Iterative method SSOR methodNumber of iterations TOC Number of iterations TOC

4 10 00168 8 00018 11 0018 8 000116 11 0143 9 000132 12 3319 9 000164 12 7145 9 0015128 12 11342 9 0020256 12 25014 10 1051512 12 98317 10 61301024 13 534903 10 126242

We choose initial guess 1199090as 1199090= (0 0 0)

119879 The com-putational results are shown in Table 1

In Table 1 TOC denotes the total time taken by CPUTherate of convergence of SSOR method is better than that ofiterative method [13]

4 Conclusion

In this paper we have discussed symmetric SOR method forsolving absolute complementarity problem The comparisonwith other methods showed the efficiency of the methodThe results and ideas of this paper may be used to solve thevariational inequalities and related optimization problems

References

[1] C E Lemke ldquoBimatrix equilibrium points and mathematicalprogrammingrdquoManagement Science vol 11 pp 681ndash689 1965

[2] R W Cottle and G B Dantzig ldquoComplementary pivot theoryof mathematical programmingrdquo Linear Algebra and its Applica-tions vol 1 no 1 pp 103ndash125 1968

[3] B H Ahn ldquoIterative methods for linear complementarity prob-lems with upperbounds on primary variablesrdquo MathematicalProgramming vol 26 no 3 pp 295ndash315 1983

[4] RW Cottle J-S Pang and R E StoneTheLinear Complemen-tarity Problem Computer Science and Scientific ComputingAcademic Press Boston Mass USA 1992

[5] M Dehghan and M Hajarian ldquoConvergence of SSORmethodsfor linear complementarity problemsrdquo Operations ResearchLetters vol 37 no 3 pp 219ndash223 2009

[6] S Karamardian ldquoGeneralized complementarity problemrdquo Jour-nal of OptimizationTheory and Applications vol 8 pp 161ndash1681971

[7] Y Li and P Dai ldquoGeneralized AOR methods for linear com-plementarity problemrdquo Applied Mathematics and Computationvol 188 no 1 pp 7ndash18 2007

[8] O L Mangasarian ldquoAbsolute value equation solution via con-cave minimizationrdquo Optimization Letters vol 1 no 1 pp 3ndash82007

[9] O L Mangasarian ldquoSolution of symmetric linear complemen-tarity problems by iterative methodsrdquo Journal of OptimizationTheory and Applications vol 22 no 4 pp 465ndash485 1977

[10] M A Noor J Iqbal K I Noor and E Al-Said ldquoGeneralizedAOR method for solving absolute complementarity problemsrdquoJournal of Applied Mathematics vol 2012 Article ID 743861 14pages 2012

[11] M A Noor On variational inequalities [PhD thesis] BrunelUniversity London UK 1975

[12] M A Noor ldquoOn merit functions for quasivariational inequali-tiesrdquo Journal of Mathematical Inequalities vol 1 no 2 pp 259ndash268 2007

[13] M A Noor J Iqbal K I Noor and E Al-Said ldquoOn an iterativemethod for solving absolute value equationsrdquo OptimizationLetters vol 6 no 5 pp 1027ndash1033 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 3

of the absolute complementarity problem is nonempty Toprove the convergence of Algorithm 9 we need the followingresult

Theorem 11 Consider the operator 119892 119877119899

rarr 119877119899 as defined

in (16) Assume that119860 isin 119877119899times119899 is an 119871-matrix Also assume that

0 lt 120596 le 1 Then for any 119909 isin 120593 it holds that

(i) 119892(119909) le 119909

(ii) 119909 le 119910 rArr 119892(119909) le 119892(119910)

(iii) 120585 = 119892(119909) isin 120593

Proof To prove (i) we need to prove that

120585119894le 119909119894 119894 = 1 2 119899 (18)

with 120585119894satisfying

120585119894= 119875119870(119909119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895(120585119895minus 119909119895) + 120596 (2 minus 120596)

times(119860119909 minus |119909| minus 119887)119894

]

]

)

(19)

To prove the required result we use mathematical inductionFor this let 119894 = 1

1205851= 119875119870(1199091minus 119886minus1

11120596 (2 minus 120596) (119860119909 minus |119909| minus 119887)

1) (20)

Since 119860119909 minus |119909| minus 119887 ge 0 0 lt 120596 le 1 therefore 1205851le 1199091

For 119894 = 2 we have

1205852= 119875119870(1199092minus 119886minus1

22

times [minus12059611987121(1205851minus 1199091) + 120596 (2 minus 120596)

times(119860119909 minus |119909| minus 119887)2] )

(21)

Here 119860119909 minus |119909| minus 119887 ge 0 0 lt 120596 le 2 11987121ge 0 and 120585

1minus 1199091le 0

This implies that 1205852le 1199092

Suppose that

120585119894le 119909119894

for 119894 = 1 2 119896 minus 1 (22)

we have to prove that the statement is true for 119894 = 119896 that is

120585119896le 119909119896 (23)

Consider

120585119896= 119875119870(119909119896minus 119886minus1

119896119896

times[

[

minus120596

119896minus1

sum

119895=1

119871119896119895(120585119895minus 119909119895) + 120596 (2 minus 120596)

times(119860119909 minus |119909| minus 119887)119896

]

]

)

= 119875119870(119909119896minus 119886minus1

119896119896

times [minus120596 (1198711198961(1205851minus 1199091) + 1198711198962(1205852minus 1199092)

+ sdot sdot sdot + 119871119896119896minus1

(120585119896minus1

minus 119909119896minus1

))

+ 120596 (2 minus 120596) (119860119909 minus |119909| minus 119887)119896] )

(24)

Since 119860119909 minus |119909| minus 119887 ge 0 0 lt 120596119896le 2 119871

1198961 1198711198962 119871

119896119896minus1ge 0

and 120585119894le 119909119894for 119894 = 1 2 119896 minus 1 from (24) we can write

120585119896le 119909119896 (25)

Hence (i) is provedNow we prove (ii) for this let us suppose that 120585 = 119892(119909)

and 120601 = 119892(119910) We will prove that

119909 le 119910 997904rArr 120585 le 120601 (26)

As

120585 = 119875119870(119909 minus 119863

minus1

times [minus120596119871120585 + (120596 (2 minus 120596)119860 + 120596119871) 119909

minus 120596 (2 minus 120596) (|119909| + 119887)] )

(27)

so 120585119894can be written as

120585119894= 119875119870( minus 119886

minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895120585119895+ 120596119886119894119894119909119894+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119894119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

4 Journal of Applied Mathematics

= 119875119870((1 minus 120596) 119909

119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895120585119895+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119894119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

(28)

Similarly for 120601119894we have

120601119894= 119875119870((1 minus 120596) 119910

119894minus 119886minus1

119894119894

times[

[

minus 120596

119894minus1

sum

119895=1

119871119894119895120601119895+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119910119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1119895 = 119894

119880119894119895119910119895minus 120596 (2 minus 120596)

1003816100381610038161003816119910119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

(29)

For 119894 = 1 we have

1206011= 119875119870((1 minus 120596) 119910

1minus 119886minus1

11120596 (2 minus 120596)

times[

[

minus

119899

sum

119895=1 119895 = 119894

1198801119895119910119895minus10038161003816100381610038161199101

1003816100381610038161003816minus 1198871

]

]

)

ge 119875119870((1 minus 120596) 119909

1minus 119886minus1

11120596 (2 minus 120596)

times[

[

minus

119899

sum

119895=1 119895 = 119894

1198801119895119909119895minus10038161003816100381610038161199091

1003816100381610038161003816minus 1198871

]

]

)

= 1205851

(30)

Since 1199101ge 1199091 therefore minus|119910

1| le minus|119909

1| Hence it is true for

119894 = 1 Suppose it is true for 119894 = 1 2 119896 minus 1 we will prove itfor 119894 = 119896 for this consider

120601119896= 119875119870((1 minus 120596) 119910

119896minus 119886minus1

119896119896

times[

[

119896minus1

sum

119895=1

119871119896119895120601119895+ (120596 minus 120596 (2 minus 120596))

times

119896minus1

sum

119895=1

119871119896119895119910119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119896119895119910119895minus 120596 (2 minus 120596)

1003816100381610038161003816119910119896

1003816100381610038161003816

minus 120596 (2 minus 120596) 119887119896

]

]

)

ge 119875119870((1 minus 120596) 119909

119896minus 119886minus1

119896119896

times[

[

119896minus1

sum

119895=1

119871119896119895120585119895+ (120596 minus 120596 (2 minus 120596))

times

119896minus1

sum

119895=1

119871119896119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119896119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119896

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119896

]

]

)

= 120585119896

(31)

Since 119909 le 119910 and 120585119894le 120601119894for 119894 = 1 2 119896 minus 1 hence it is true

for 119896 and (ii) is verifiedNext we prove (iii) that is

120585 = 119892 (119909) isin 120593 (32)

Let 120582 = 119892(120585) = 119875119870(120585minus119863

minus1

[120596119871(120582minus120585)+120596(2minus120596)(119860120585minus |120585|minus

119887)]) from (i) 119892(120585) = 120582 le 120585 Also by definition of 119892 120585 = 119892(119909) ge

0 and 120582 = 119892(120585) ge 0Now

120582119894= 119875119870(120585119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895(120582119895minus 120585119895) + 120596 (2 minus 120596)

times (119860120585 minus10038161003816100381610038161205851003816100381610038161003816minus 119887)119894

]

]

)

(33)

Journal of Applied Mathematics 5

For 119894 = 1 1205851ge 0 by definition of 119892 Suppose that (119860120585 minus |120585| minus

119887)119894lt 0 so

1205821= 119875119870(1205851minus 119886minus1

11120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)1)

gt 119875119870(1205851) = 1205851

(34)

which contradicts the fact that 120582 le 120585 Therefore (119860120585 minus |120585| minus119887)119894ge 0Now we prove it for any 119896 in 119894 = 1 2 119899 Suppose the

contrary (119860120585 minus |120585| minus 119887)119894lt 0 then

120582119896= 119875119870(120585119896minus 119886minus1

119896119896

times[

[

minus120596

119896minus1

sum

119895=1

119871119896119895(120582119895minus 120585119895) + 120596 (2 minus 120596)

times (119860120585 minus10038161003816100381610038161205851003816100381610038161003816minus 119887)119896

]

]

)

(35)

As it is true for all 120572 isin [0 1] it should be true for 120572 = 0 Thatis

120582119896= 119875119870(120585119896minus 119886minus1

119896119896120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)119896)

gt 119875119870(120585119896) = 120585119896

(36)

which contradicts the fact that 120582 le 120585 So (119860120585 minus |120585| minus 119887)119896ge 0

for any 119896 in 119894 = 1 2 119899Hence 120585 = 119891(119909) isin 120593

Now we prove the convergence criteria of Algorithm 9when the matrix 119860 is an 119871-matrix as stated in the next result

Theorem 12 Assume that 119860 isin 119877119899times119899 is an 119871-matrix and 0 lt

120596 le 1Then for any initial vector 1199090isin 120593 the sequence 119909

119896 119896 =

0 1 2 defined by Algorithm 9 has the following properties

(i) 0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2

(ii) lim119896rarrinfin

119909119896= 119909lowast is the unique solution of the absolute

complementarity problem

Proof Since 1199090isin 120593 by (i) of Theorem 11 we have 119909

1le 1199090

and 1199091isin 120593 Recursively usingTheorem 11 we obtain

0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2 (37)

From (i) we observe that the sequence 119909119896 is monotone

bounded therefore it converges to some 119909lowast isin 119877119899+satisfying

119909lowast

= 119875119870(119909lowast

minus 119863minus1

times [minus120596119871119909lowast

+ 120596 (2 minus 120596) (119860 + 120596119871) 119909lowast

minus120596 (2 minus 120596) (1003816100381610038161003816119909lowast1003816100381610038161003816+ 119887)] )

= 119875119870(119909lowast

minus 119863minus1

120596 (2 minus 120596)

times [119860119909lowast

minus1003816100381610038161003816119909lowast1003816100381610038161003816minus 119887] )

(38)

Hence 119909lowast is the solution of the absolute complementarityproblem (2)

Note The convergence of Algorithm 10 has the same steps asgiven inTheorems 11 and 12

3 Numerical Results

In this section we consider several examples to show theefficiency of the proposedmethodsThe convergence of SSORmethod is guaranteed for 119871-matrices only but it is alsopossible to solve different type of systems All the experimentsare performedwith Intel(R) Core 2times 21 GHz 1 GBRAM andthe codes are written in MATLAB 7

Example 13 (see [10]) Consider the ordinary differentialequation

1198892

119909

1198891199052minus |119909| = (1 minus 119905

2

) 0 le 119909 le 1

119909 (0) = 0 119909 (1) = 1

(39)

The exact solution is

119909 (119905) =

07378827425 sin (119905)minus3 cos (119905)+3minus1199052 119909 lt 0

minus07310585786119890minus119905

minus02689414214119890119905

+1+1199052

119909 gt 0

(40)

We take 119899 = 10 the matrix 119860 is given by

119886119894119895=

minus242 for 119895 = 119894

121 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(41)

The constant vector 119887 is given by

119887 = (

120

121

117

121

112

121

105

121

96

121

85

121

72

121

57

121

40

121

minus14620

121

)

119879

(42)

Here119860 is not an119871-matrixThe comparison between the exactsolution and the approximate solutions is given in Figure 1

In Figure 1 we see that the SSOR method convergesrapidly to the approximate solution of absolute complemen-tarity problem (2) as compared to GAOR method

In the next example we compare SSOR method withiterative method by Noor et al [13]

Example 14 (see [13]) Let the matrix 119860 be given by

119886119894119895=

8 for 119895 = 119894

minus1 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(43)

Let 119887 = (6 5 5 5 6)119879 the problem size 119899 ranging from

4 to 1024 The stopping criteria are 119860119909 minus |119909| minus 119887 lt 10minus6

6 Journal of Applied Mathematics

0 50 100 150 200 250 300Number of iterations

2-no

rm o

f res

idua

l

GAORSSOR

10minus8

10minus6

10minus4

10minus2

100

102

104

Figure 1 Comparison between GAOR method and SSOR method

Table 1

119899

Iterative method SSOR methodNumber of iterations TOC Number of iterations TOC

4 10 00168 8 00018 11 0018 8 000116 11 0143 9 000132 12 3319 9 000164 12 7145 9 0015128 12 11342 9 0020256 12 25014 10 1051512 12 98317 10 61301024 13 534903 10 126242

We choose initial guess 1199090as 1199090= (0 0 0)

119879 The com-putational results are shown in Table 1

In Table 1 TOC denotes the total time taken by CPUTherate of convergence of SSOR method is better than that ofiterative method [13]

4 Conclusion

In this paper we have discussed symmetric SOR method forsolving absolute complementarity problem The comparisonwith other methods showed the efficiency of the methodThe results and ideas of this paper may be used to solve thevariational inequalities and related optimization problems

References

[1] C E Lemke ldquoBimatrix equilibrium points and mathematicalprogrammingrdquoManagement Science vol 11 pp 681ndash689 1965

[2] R W Cottle and G B Dantzig ldquoComplementary pivot theoryof mathematical programmingrdquo Linear Algebra and its Applica-tions vol 1 no 1 pp 103ndash125 1968

[3] B H Ahn ldquoIterative methods for linear complementarity prob-lems with upperbounds on primary variablesrdquo MathematicalProgramming vol 26 no 3 pp 295ndash315 1983

[4] RW Cottle J-S Pang and R E StoneTheLinear Complemen-tarity Problem Computer Science and Scientific ComputingAcademic Press Boston Mass USA 1992

[5] M Dehghan and M Hajarian ldquoConvergence of SSORmethodsfor linear complementarity problemsrdquo Operations ResearchLetters vol 37 no 3 pp 219ndash223 2009

[6] S Karamardian ldquoGeneralized complementarity problemrdquo Jour-nal of OptimizationTheory and Applications vol 8 pp 161ndash1681971

[7] Y Li and P Dai ldquoGeneralized AOR methods for linear com-plementarity problemrdquo Applied Mathematics and Computationvol 188 no 1 pp 7ndash18 2007

[8] O L Mangasarian ldquoAbsolute value equation solution via con-cave minimizationrdquo Optimization Letters vol 1 no 1 pp 3ndash82007

[9] O L Mangasarian ldquoSolution of symmetric linear complemen-tarity problems by iterative methodsrdquo Journal of OptimizationTheory and Applications vol 22 no 4 pp 465ndash485 1977

[10] M A Noor J Iqbal K I Noor and E Al-Said ldquoGeneralizedAOR method for solving absolute complementarity problemsrdquoJournal of Applied Mathematics vol 2012 Article ID 743861 14pages 2012

[11] M A Noor On variational inequalities [PhD thesis] BrunelUniversity London UK 1975

[12] M A Noor ldquoOn merit functions for quasivariational inequali-tiesrdquo Journal of Mathematical Inequalities vol 1 no 2 pp 259ndash268 2007

[13] M A Noor J Iqbal K I Noor and E Al-Said ldquoOn an iterativemethod for solving absolute value equationsrdquo OptimizationLetters vol 6 no 5 pp 1027ndash1033 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Journal of Applied Mathematics

= 119875119870((1 minus 120596) 119909

119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895120585119895+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119894119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

(28)

Similarly for 120601119894we have

120601119894= 119875119870((1 minus 120596) 119910

119894minus 119886minus1

119894119894

times[

[

minus 120596

119894minus1

sum

119895=1

119871119894119895120601119895+ (120596 minus 120596 (2 minus 120596))

times

119894minus1

sum

119895=1

119871119894119895119910119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1119895 = 119894

119880119894119895119910119895minus 120596 (2 minus 120596)

1003816100381610038161003816119910119894

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119894

]

]

)

(29)

For 119894 = 1 we have

1206011= 119875119870((1 minus 120596) 119910

1minus 119886minus1

11120596 (2 minus 120596)

times[

[

minus

119899

sum

119895=1 119895 = 119894

1198801119895119910119895minus10038161003816100381610038161199101

1003816100381610038161003816minus 1198871

]

]

)

ge 119875119870((1 minus 120596) 119909

1minus 119886minus1

11120596 (2 minus 120596)

times[

[

minus

119899

sum

119895=1 119895 = 119894

1198801119895119909119895minus10038161003816100381610038161199091

1003816100381610038161003816minus 1198871

]

]

)

= 1205851

(30)

Since 1199101ge 1199091 therefore minus|119910

1| le minus|119909

1| Hence it is true for

119894 = 1 Suppose it is true for 119894 = 1 2 119896 minus 1 we will prove itfor 119894 = 119896 for this consider

120601119896= 119875119870((1 minus 120596) 119910

119896minus 119886minus1

119896119896

times[

[

119896minus1

sum

119895=1

119871119896119895120601119895+ (120596 minus 120596 (2 minus 120596))

times

119896minus1

sum

119895=1

119871119896119895119910119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119896119895119910119895minus 120596 (2 minus 120596)

1003816100381610038161003816119910119896

1003816100381610038161003816

minus 120596 (2 minus 120596) 119887119896

]

]

)

ge 119875119870((1 minus 120596) 119909

119896minus 119886minus1

119896119896

times[

[

119896minus1

sum

119895=1

119871119896119895120585119895+ (120596 minus 120596 (2 minus 120596))

times

119896minus1

sum

119895=1

119871119896119895119909119895minus 120596 (2 minus 120596)

times

119899

sum

119895=1 119895 = 119894

119880119896119895119909119895minus 120596 (2 minus 120596)

1003816100381610038161003816119909119896

1003816100381610038161003816

minus120596 (2 minus 120596) 119887119896

]

]

)

= 120585119896

(31)

Since 119909 le 119910 and 120585119894le 120601119894for 119894 = 1 2 119896 minus 1 hence it is true

for 119896 and (ii) is verifiedNext we prove (iii) that is

120585 = 119892 (119909) isin 120593 (32)

Let 120582 = 119892(120585) = 119875119870(120585minus119863

minus1

[120596119871(120582minus120585)+120596(2minus120596)(119860120585minus |120585|minus

119887)]) from (i) 119892(120585) = 120582 le 120585 Also by definition of 119892 120585 = 119892(119909) ge

0 and 120582 = 119892(120585) ge 0Now

120582119894= 119875119870(120585119894minus 119886minus1

119894119894

times[

[

minus120596

119894minus1

sum

119895=1

119871119894119895(120582119895minus 120585119895) + 120596 (2 minus 120596)

times (119860120585 minus10038161003816100381610038161205851003816100381610038161003816minus 119887)119894

]

]

)

(33)

Journal of Applied Mathematics 5

For 119894 = 1 1205851ge 0 by definition of 119892 Suppose that (119860120585 minus |120585| minus

119887)119894lt 0 so

1205821= 119875119870(1205851minus 119886minus1

11120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)1)

gt 119875119870(1205851) = 1205851

(34)

which contradicts the fact that 120582 le 120585 Therefore (119860120585 minus |120585| minus119887)119894ge 0Now we prove it for any 119896 in 119894 = 1 2 119899 Suppose the

contrary (119860120585 minus |120585| minus 119887)119894lt 0 then

120582119896= 119875119870(120585119896minus 119886minus1

119896119896

times[

[

minus120596

119896minus1

sum

119895=1

119871119896119895(120582119895minus 120585119895) + 120596 (2 minus 120596)

times (119860120585 minus10038161003816100381610038161205851003816100381610038161003816minus 119887)119896

]

]

)

(35)

As it is true for all 120572 isin [0 1] it should be true for 120572 = 0 Thatis

120582119896= 119875119870(120585119896minus 119886minus1

119896119896120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)119896)

gt 119875119870(120585119896) = 120585119896

(36)

which contradicts the fact that 120582 le 120585 So (119860120585 minus |120585| minus 119887)119896ge 0

for any 119896 in 119894 = 1 2 119899Hence 120585 = 119891(119909) isin 120593

Now we prove the convergence criteria of Algorithm 9when the matrix 119860 is an 119871-matrix as stated in the next result

Theorem 12 Assume that 119860 isin 119877119899times119899 is an 119871-matrix and 0 lt

120596 le 1Then for any initial vector 1199090isin 120593 the sequence 119909

119896 119896 =

0 1 2 defined by Algorithm 9 has the following properties

(i) 0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2

(ii) lim119896rarrinfin

119909119896= 119909lowast is the unique solution of the absolute

complementarity problem

Proof Since 1199090isin 120593 by (i) of Theorem 11 we have 119909

1le 1199090

and 1199091isin 120593 Recursively usingTheorem 11 we obtain

0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2 (37)

From (i) we observe that the sequence 119909119896 is monotone

bounded therefore it converges to some 119909lowast isin 119877119899+satisfying

119909lowast

= 119875119870(119909lowast

minus 119863minus1

times [minus120596119871119909lowast

+ 120596 (2 minus 120596) (119860 + 120596119871) 119909lowast

minus120596 (2 minus 120596) (1003816100381610038161003816119909lowast1003816100381610038161003816+ 119887)] )

= 119875119870(119909lowast

minus 119863minus1

120596 (2 minus 120596)

times [119860119909lowast

minus1003816100381610038161003816119909lowast1003816100381610038161003816minus 119887] )

(38)

Hence 119909lowast is the solution of the absolute complementarityproblem (2)

Note The convergence of Algorithm 10 has the same steps asgiven inTheorems 11 and 12

3 Numerical Results

In this section we consider several examples to show theefficiency of the proposedmethodsThe convergence of SSORmethod is guaranteed for 119871-matrices only but it is alsopossible to solve different type of systems All the experimentsare performedwith Intel(R) Core 2times 21 GHz 1 GBRAM andthe codes are written in MATLAB 7

Example 13 (see [10]) Consider the ordinary differentialequation

1198892

119909

1198891199052minus |119909| = (1 minus 119905

2

) 0 le 119909 le 1

119909 (0) = 0 119909 (1) = 1

(39)

The exact solution is

119909 (119905) =

07378827425 sin (119905)minus3 cos (119905)+3minus1199052 119909 lt 0

minus07310585786119890minus119905

minus02689414214119890119905

+1+1199052

119909 gt 0

(40)

We take 119899 = 10 the matrix 119860 is given by

119886119894119895=

minus242 for 119895 = 119894

121 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(41)

The constant vector 119887 is given by

119887 = (

120

121

117

121

112

121

105

121

96

121

85

121

72

121

57

121

40

121

minus14620

121

)

119879

(42)

Here119860 is not an119871-matrixThe comparison between the exactsolution and the approximate solutions is given in Figure 1

In Figure 1 we see that the SSOR method convergesrapidly to the approximate solution of absolute complemen-tarity problem (2) as compared to GAOR method

In the next example we compare SSOR method withiterative method by Noor et al [13]

Example 14 (see [13]) Let the matrix 119860 be given by

119886119894119895=

8 for 119895 = 119894

minus1 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(43)

Let 119887 = (6 5 5 5 6)119879 the problem size 119899 ranging from

4 to 1024 The stopping criteria are 119860119909 minus |119909| minus 119887 lt 10minus6

6 Journal of Applied Mathematics

0 50 100 150 200 250 300Number of iterations

2-no

rm o

f res

idua

l

GAORSSOR

10minus8

10minus6

10minus4

10minus2

100

102

104

Figure 1 Comparison between GAOR method and SSOR method

Table 1

119899

Iterative method SSOR methodNumber of iterations TOC Number of iterations TOC

4 10 00168 8 00018 11 0018 8 000116 11 0143 9 000132 12 3319 9 000164 12 7145 9 0015128 12 11342 9 0020256 12 25014 10 1051512 12 98317 10 61301024 13 534903 10 126242

We choose initial guess 1199090as 1199090= (0 0 0)

119879 The com-putational results are shown in Table 1

In Table 1 TOC denotes the total time taken by CPUTherate of convergence of SSOR method is better than that ofiterative method [13]

4 Conclusion

In this paper we have discussed symmetric SOR method forsolving absolute complementarity problem The comparisonwith other methods showed the efficiency of the methodThe results and ideas of this paper may be used to solve thevariational inequalities and related optimization problems

References

[1] C E Lemke ldquoBimatrix equilibrium points and mathematicalprogrammingrdquoManagement Science vol 11 pp 681ndash689 1965

[2] R W Cottle and G B Dantzig ldquoComplementary pivot theoryof mathematical programmingrdquo Linear Algebra and its Applica-tions vol 1 no 1 pp 103ndash125 1968

[3] B H Ahn ldquoIterative methods for linear complementarity prob-lems with upperbounds on primary variablesrdquo MathematicalProgramming vol 26 no 3 pp 295ndash315 1983

[4] RW Cottle J-S Pang and R E StoneTheLinear Complemen-tarity Problem Computer Science and Scientific ComputingAcademic Press Boston Mass USA 1992

[5] M Dehghan and M Hajarian ldquoConvergence of SSORmethodsfor linear complementarity problemsrdquo Operations ResearchLetters vol 37 no 3 pp 219ndash223 2009

[6] S Karamardian ldquoGeneralized complementarity problemrdquo Jour-nal of OptimizationTheory and Applications vol 8 pp 161ndash1681971

[7] Y Li and P Dai ldquoGeneralized AOR methods for linear com-plementarity problemrdquo Applied Mathematics and Computationvol 188 no 1 pp 7ndash18 2007

[8] O L Mangasarian ldquoAbsolute value equation solution via con-cave minimizationrdquo Optimization Letters vol 1 no 1 pp 3ndash82007

[9] O L Mangasarian ldquoSolution of symmetric linear complemen-tarity problems by iterative methodsrdquo Journal of OptimizationTheory and Applications vol 22 no 4 pp 465ndash485 1977

[10] M A Noor J Iqbal K I Noor and E Al-Said ldquoGeneralizedAOR method for solving absolute complementarity problemsrdquoJournal of Applied Mathematics vol 2012 Article ID 743861 14pages 2012

[11] M A Noor On variational inequalities [PhD thesis] BrunelUniversity London UK 1975

[12] M A Noor ldquoOn merit functions for quasivariational inequali-tiesrdquo Journal of Mathematical Inequalities vol 1 no 2 pp 259ndash268 2007

[13] M A Noor J Iqbal K I Noor and E Al-Said ldquoOn an iterativemethod for solving absolute value equationsrdquo OptimizationLetters vol 6 no 5 pp 1027ndash1033 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 5

For 119894 = 1 1205851ge 0 by definition of 119892 Suppose that (119860120585 minus |120585| minus

119887)119894lt 0 so

1205821= 119875119870(1205851minus 119886minus1

11120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)1)

gt 119875119870(1205851) = 1205851

(34)

which contradicts the fact that 120582 le 120585 Therefore (119860120585 minus |120585| minus119887)119894ge 0Now we prove it for any 119896 in 119894 = 1 2 119899 Suppose the

contrary (119860120585 minus |120585| minus 119887)119894lt 0 then

120582119896= 119875119870(120585119896minus 119886minus1

119896119896

times[

[

minus120596

119896minus1

sum

119895=1

119871119896119895(120582119895minus 120585119895) + 120596 (2 minus 120596)

times (119860120585 minus10038161003816100381610038161205851003816100381610038161003816minus 119887)119896

]

]

)

(35)

As it is true for all 120572 isin [0 1] it should be true for 120572 = 0 Thatis

120582119896= 119875119870(120585119896minus 119886minus1

119896119896120596 (2 minus 120596) (119860120585 minus

10038161003816100381610038161205851003816100381610038161003816minus 119887)119896)

gt 119875119870(120585119896) = 120585119896

(36)

which contradicts the fact that 120582 le 120585 So (119860120585 minus |120585| minus 119887)119896ge 0

for any 119896 in 119894 = 1 2 119899Hence 120585 = 119891(119909) isin 120593

Now we prove the convergence criteria of Algorithm 9when the matrix 119860 is an 119871-matrix as stated in the next result

Theorem 12 Assume that 119860 isin 119877119899times119899 is an 119871-matrix and 0 lt

120596 le 1Then for any initial vector 1199090isin 120593 the sequence 119909

119896 119896 =

0 1 2 defined by Algorithm 9 has the following properties

(i) 0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2

(ii) lim119896rarrinfin

119909119896= 119909lowast is the unique solution of the absolute

complementarity problem

Proof Since 1199090isin 120593 by (i) of Theorem 11 we have 119909

1le 1199090

and 1199091isin 120593 Recursively usingTheorem 11 we obtain

0 le 119909119896+1

le 119909119896le 1199090 119896 = 0 1 2 (37)

From (i) we observe that the sequence 119909119896 is monotone

bounded therefore it converges to some 119909lowast isin 119877119899+satisfying

119909lowast

= 119875119870(119909lowast

minus 119863minus1

times [minus120596119871119909lowast

+ 120596 (2 minus 120596) (119860 + 120596119871) 119909lowast

minus120596 (2 minus 120596) (1003816100381610038161003816119909lowast1003816100381610038161003816+ 119887)] )

= 119875119870(119909lowast

minus 119863minus1

120596 (2 minus 120596)

times [119860119909lowast

minus1003816100381610038161003816119909lowast1003816100381610038161003816minus 119887] )

(38)

Hence 119909lowast is the solution of the absolute complementarityproblem (2)

Note The convergence of Algorithm 10 has the same steps asgiven inTheorems 11 and 12

3 Numerical Results

In this section we consider several examples to show theefficiency of the proposedmethodsThe convergence of SSORmethod is guaranteed for 119871-matrices only but it is alsopossible to solve different type of systems All the experimentsare performedwith Intel(R) Core 2times 21 GHz 1 GBRAM andthe codes are written in MATLAB 7

Example 13 (see [10]) Consider the ordinary differentialequation

1198892

119909

1198891199052minus |119909| = (1 minus 119905

2

) 0 le 119909 le 1

119909 (0) = 0 119909 (1) = 1

(39)

The exact solution is

119909 (119905) =

07378827425 sin (119905)minus3 cos (119905)+3minus1199052 119909 lt 0

minus07310585786119890minus119905

minus02689414214119890119905

+1+1199052

119909 gt 0

(40)

We take 119899 = 10 the matrix 119860 is given by

119886119894119895=

minus242 for 119895 = 119894

121 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(41)

The constant vector 119887 is given by

119887 = (

120

121

117

121

112

121

105

121

96

121

85

121

72

121

57

121

40

121

minus14620

121

)

119879

(42)

Here119860 is not an119871-matrixThe comparison between the exactsolution and the approximate solutions is given in Figure 1

In Figure 1 we see that the SSOR method convergesrapidly to the approximate solution of absolute complemen-tarity problem (2) as compared to GAOR method

In the next example we compare SSOR method withiterative method by Noor et al [13]

Example 14 (see [13]) Let the matrix 119860 be given by

119886119894119895=

8 for 119895 = 119894

minus1 for

119895 = 119894 + 1 119894 = 1 2 119899 minus 1

119895 = 119894 minus 1 119894 = 2 3 119899

0 otherwise

(43)

Let 119887 = (6 5 5 5 6)119879 the problem size 119899 ranging from

4 to 1024 The stopping criteria are 119860119909 minus |119909| minus 119887 lt 10minus6

6 Journal of Applied Mathematics

0 50 100 150 200 250 300Number of iterations

2-no

rm o

f res

idua

l

GAORSSOR

10minus8

10minus6

10minus4

10minus2

100

102

104

Figure 1 Comparison between GAOR method and SSOR method

Table 1

119899

Iterative method SSOR methodNumber of iterations TOC Number of iterations TOC

4 10 00168 8 00018 11 0018 8 000116 11 0143 9 000132 12 3319 9 000164 12 7145 9 0015128 12 11342 9 0020256 12 25014 10 1051512 12 98317 10 61301024 13 534903 10 126242

We choose initial guess 1199090as 1199090= (0 0 0)

119879 The com-putational results are shown in Table 1

In Table 1 TOC denotes the total time taken by CPUTherate of convergence of SSOR method is better than that ofiterative method [13]

4 Conclusion

In this paper we have discussed symmetric SOR method forsolving absolute complementarity problem The comparisonwith other methods showed the efficiency of the methodThe results and ideas of this paper may be used to solve thevariational inequalities and related optimization problems

References

[1] C E Lemke ldquoBimatrix equilibrium points and mathematicalprogrammingrdquoManagement Science vol 11 pp 681ndash689 1965

[2] R W Cottle and G B Dantzig ldquoComplementary pivot theoryof mathematical programmingrdquo Linear Algebra and its Applica-tions vol 1 no 1 pp 103ndash125 1968

[3] B H Ahn ldquoIterative methods for linear complementarity prob-lems with upperbounds on primary variablesrdquo MathematicalProgramming vol 26 no 3 pp 295ndash315 1983

[4] RW Cottle J-S Pang and R E StoneTheLinear Complemen-tarity Problem Computer Science and Scientific ComputingAcademic Press Boston Mass USA 1992

[5] M Dehghan and M Hajarian ldquoConvergence of SSORmethodsfor linear complementarity problemsrdquo Operations ResearchLetters vol 37 no 3 pp 219ndash223 2009

[6] S Karamardian ldquoGeneralized complementarity problemrdquo Jour-nal of OptimizationTheory and Applications vol 8 pp 161ndash1681971

[7] Y Li and P Dai ldquoGeneralized AOR methods for linear com-plementarity problemrdquo Applied Mathematics and Computationvol 188 no 1 pp 7ndash18 2007

[8] O L Mangasarian ldquoAbsolute value equation solution via con-cave minimizationrdquo Optimization Letters vol 1 no 1 pp 3ndash82007

[9] O L Mangasarian ldquoSolution of symmetric linear complemen-tarity problems by iterative methodsrdquo Journal of OptimizationTheory and Applications vol 22 no 4 pp 465ndash485 1977

[10] M A Noor J Iqbal K I Noor and E Al-Said ldquoGeneralizedAOR method for solving absolute complementarity problemsrdquoJournal of Applied Mathematics vol 2012 Article ID 743861 14pages 2012

[11] M A Noor On variational inequalities [PhD thesis] BrunelUniversity London UK 1975

[12] M A Noor ldquoOn merit functions for quasivariational inequali-tiesrdquo Journal of Mathematical Inequalities vol 1 no 2 pp 259ndash268 2007

[13] M A Noor J Iqbal K I Noor and E Al-Said ldquoOn an iterativemethod for solving absolute value equationsrdquo OptimizationLetters vol 6 no 5 pp 1027ndash1033 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Applied Mathematics

0 50 100 150 200 250 300Number of iterations

2-no

rm o

f res

idua

l

GAORSSOR

10minus8

10minus6

10minus4

10minus2

100

102

104

Figure 1 Comparison between GAOR method and SSOR method

Table 1

119899

Iterative method SSOR methodNumber of iterations TOC Number of iterations TOC

4 10 00168 8 00018 11 0018 8 000116 11 0143 9 000132 12 3319 9 000164 12 7145 9 0015128 12 11342 9 0020256 12 25014 10 1051512 12 98317 10 61301024 13 534903 10 126242

We choose initial guess 1199090as 1199090= (0 0 0)

119879 The com-putational results are shown in Table 1

In Table 1 TOC denotes the total time taken by CPUTherate of convergence of SSOR method is better than that ofiterative method [13]

4 Conclusion

In this paper we have discussed symmetric SOR method forsolving absolute complementarity problem The comparisonwith other methods showed the efficiency of the methodThe results and ideas of this paper may be used to solve thevariational inequalities and related optimization problems

References

[1] C E Lemke ldquoBimatrix equilibrium points and mathematicalprogrammingrdquoManagement Science vol 11 pp 681ndash689 1965

[2] R W Cottle and G B Dantzig ldquoComplementary pivot theoryof mathematical programmingrdquo Linear Algebra and its Applica-tions vol 1 no 1 pp 103ndash125 1968

[3] B H Ahn ldquoIterative methods for linear complementarity prob-lems with upperbounds on primary variablesrdquo MathematicalProgramming vol 26 no 3 pp 295ndash315 1983

[4] RW Cottle J-S Pang and R E StoneTheLinear Complemen-tarity Problem Computer Science and Scientific ComputingAcademic Press Boston Mass USA 1992

[5] M Dehghan and M Hajarian ldquoConvergence of SSORmethodsfor linear complementarity problemsrdquo Operations ResearchLetters vol 37 no 3 pp 219ndash223 2009

[6] S Karamardian ldquoGeneralized complementarity problemrdquo Jour-nal of OptimizationTheory and Applications vol 8 pp 161ndash1681971

[7] Y Li and P Dai ldquoGeneralized AOR methods for linear com-plementarity problemrdquo Applied Mathematics and Computationvol 188 no 1 pp 7ndash18 2007

[8] O L Mangasarian ldquoAbsolute value equation solution via con-cave minimizationrdquo Optimization Letters vol 1 no 1 pp 3ndash82007

[9] O L Mangasarian ldquoSolution of symmetric linear complemen-tarity problems by iterative methodsrdquo Journal of OptimizationTheory and Applications vol 22 no 4 pp 465ndash485 1977

[10] M A Noor J Iqbal K I Noor and E Al-Said ldquoGeneralizedAOR method for solving absolute complementarity problemsrdquoJournal of Applied Mathematics vol 2012 Article ID 743861 14pages 2012

[11] M A Noor On variational inequalities [PhD thesis] BrunelUniversity London UK 1975

[12] M A Noor ldquoOn merit functions for quasivariational inequali-tiesrdquo Journal of Mathematical Inequalities vol 1 no 2 pp 259ndash268 2007

[13] M A Noor J Iqbal K I Noor and E Al-Said ldquoOn an iterativemethod for solving absolute value equationsrdquo OptimizationLetters vol 6 no 5 pp 1027ndash1033 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of