Research Article Global Minimization of Nonsmooth Constrained Global …downloads.hindawi.com/journals/mpe/2014/563860.pdf · nonsmooth and smooth constrained global optimization

Research ArticleGlobal Minimization of Nonsmooth Constrained GlobalOptimization with Filled Function

Wei-xiang Wang1 You-lin Shang2 and Ying Zhang3

1 Department of Mathematics Shanghai Second Polytechnic University Shanghai 201209 China2Department of Mathematics Henan University of Science and Technology Luoyang 471003 China3Department of Mathematics Zhejiang Normal University Jinhua 321004 China

Correspondence should be addressed to Wei-xiang Wang zhesx126com

Received 17 May 2014 Accepted 29 August 2014 Published 25 September 2014

Academic Editor Sergio Preidikman

Copyright copy 2014 Wei-xiang Wang et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

A novel filled function is constructed to locate a global optimizer or an approximate global optimizer of smooth or nonsmoothconstrained global minimization problems The constructed filled function contains only one parameter which can be easilyadjusted during the minimization The theoretical properties of the filled function are discussed and a corresponding solutionalgorithm is proposed The solution algorithm comprises two phases local minimization and filling The first phase minimizes theoriginal problem and obtains one of its local optimizers while the second phase minimizes the constructed filled function andidentifies a better initial point for the first phase Some preliminary numerical results are also reported

1 Introduction

Science and economics rely on the increasing demand forlocating the global optimization optimizer and thereforeglobal optimization has become one of the most attractiveresearch areas in optimization However the existence ofmultiple local minimizers that differ from the global solutionconfronts us with two difficult issues that is how to escapefrom a localminimizer to a smaller one and how to verify thatthe current minimizer is a global one These two issues makemost of global optimization problems unsolvable directly byclassical local optimization algorithms Up to now variouskinds of new theories and algorithms on global optimizationhave been presented [1ndash3] In general global optimizationmethods can be divided into two categories stochasticmethods and deterministic methodsThe stochastic methodsare usually probability based approaches such as geneticalgorithm and simulated annealing methodThese stochasticmethods have their advantages but their shortages are alsoobvious such as being easily trapped in a local optimizerDeterministic methods such as filled function method [4ndash7]

tunneling method [8] and stretching function method [9]can however often skip from the current local minimizer toa better one

The filled function approach initially proposed forsmooth optimization by Ge and Qin [4] and improved in[5ndash7] is one of the effective global optimization approachesIt furnishes us with an efficient way to use any local opti-mization procedure to solve global optimization problemThe essence of filled function method is to construct afilled function and then to minimize it to obtain a betterinitial point for the minimization of the original prob-lem The existing filled function methods are usually onlysuitable for unconstrained optimization problem More-over it requires the objective function to be continuouslydifferentiable and the number of local minimizers to befinite But in practice optimization problems may be non-smooth and often have many complicated constraints andthe number of local minimizers may also be infinite Todeal with such situation in this paper we extend filledfunction methods to the case of nonsmooth constrainedglobal optimization and propose a new filled function

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 563860 5 pageshttpdxdoiorg1011552014563860

2 Mathematical Problems in Engineering

method The proposed filled function method combinesfilled functionmethod for unconstrained global optimizationwith the exterior penalty function method for constrainedoptimization

The paper is organized as follows In Section 2 a newfilled function is proposed and its properties are discussedIn Section 3 a corresponding filled function algorithm isdesigned and numerical experiments are performed Finallyin Section 4 some conclusive remarks are given

In the rest of this paper the generalized gradient of anonsmooth function 119891(119909) at the point 119909 isin 119883 is denotedby 120597119891(119909) and the generalized directional derivative of 119891(119909)

in the direction 119889 at the point 119909 is denoted by 1198910(119909 119889) Theinterior the boundary and the closure of the set 119878 are denotedby int 119878 120597119878 and 119888119897119878 respectively

2 A New One-Parameter FilledFunction and Its Properties

21 Problem Formulation Consider the following nons-mooth constrained global optimization problem (119875)

min119909isin119878

119891 (119909) (1)

where 119878 = 119909 isin 119883 | 119892119894(119909) le 0 119894 isin 119868 119883 sub 119877119899 is a

box set 119891(119909) and 119892119894(119909) 119894 isin 119868 are Lipschitz continuous with

constants 119871119891and 119871

119892119894 119894 isin 119868 respectively and 119868 = 1 119898

is an index set For simplicity the set of local minimizersfor problem (119875) is denoted by 119871(119875) and the set of the globalminimizers is denoted by 119866(119875)

To proceed we assume that the number of minimizers ofproblem (119875) is infinite but the number of different functionvalues at the minimizers is finite

Definition 1 A function 119875(119909 119909lowast) is called a filled function of119891(119909) at a local minimizer 119909lowast if it satisfies the following

(1) 119909lowast is a strictly local maximizer of 119875(119909 119909lowast) on 119883

(2) for any 119909 isin 1198781119909lowast or 119909 isin 119883 119878 one has 0 notin 120597119875(119909 119909lowast)

where 1198781= 119909 isin 119878 | 119891(119909) ge 119891(119909lowast)

(3) if 1198782= 119909 isin 119878 | 119891(119909) lt 119891(119909lowast) is not empty then there

exists a point 1199092isin 1198782such that 119909

2is a local minimizer

of 119875(119909 119909lowast)

Definition 1 guarantees that when any local search proce-dure for unconstrained optimization is used to minimize theconstructed filled function the sequences of iterative pointwill not stop at any point at which the objective functionvalue is larger than 119891(119909lowast) If 119909lowast is not a global minimizerthen a point 119909 with 119891(119909) lt 119891(119909lowast) could be identified in theprocess of the minimization of 119875(119909 119909lowast) Then we can get abetter local minimizer of 119891(119909) by using 119909 as an initial pointBy repeating these two steps we could finally obtain a globalminimizer or an approximate globalminimizer of the originalproblem

22 A New Filled Function and Its Properties We propose inthis section a one-parameter filled function as follows

119875 (119909 119909lowast 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max (0 119892119894(119909))]

2

minus arg (1 +1003817100381710038171003817119909 minus 119909lowast

10038171003817100381710038172)

+ 119902[min (0max (119891 (119909) minus 119891 (119909lowast) 119892119894(119909) 119894 isin 119868))]

3

(2)

where 119902 gt 0 is a parameter and 119909lowast is the current localminimizer of 119891(119909)

Theorems 2ndash4 show that when parameter 119902 gt 0 is suitablylarge the function 119875(119909 119909lowast 119902) is a filled function

Theorem 2 Assume that 119909lowast isin 119871(119875) then 119909lowast is a strictly localmaximizer of 119875(119909 119909lowast 119902)

Proof Since 119909lowast isin 119871(119875) there is a neighborhood 119874(119909lowast 120575) of119909lowast with 120575 gt 0 such that 119891(119909) ge 119891(119909lowast) and 119892

119894(119909) le 0 119894 isin 119868

for all 119909 isin 119874(119909lowast 120575) cap 119878 We consider the following two cases

Case 1 For all 119909 isin 119874(119909lowast 120575) cap 119878 and 119909 = 119909lowast we havemin(0max(119891(119909) minus 119891(119909lowast) 119892

119894(119909) 119894 isin 119868)) = 0 and so

119875 (119909 119909lowast 119902) = minus1

119902(119891 (119909) minus 119891 (119909lowast))


10038171003817100381710038172)

lt minus arg 1 = 119875 (119909lowast 119909lowast 119902)

(3)

Case 2 For all 119909 isin 119874(119909lowast 120575) cap (119883 119878) there exists at least oneindex 119894

0isin 119868 such that 119892

1198940(119909) ge 0 it follows that

min (0max (119891 (119909) minus 119891 (119909lowast) 119892119894(119909) 119894 isin 119868)) = 0 (4)

Thus

119875 (119909 119909lowast 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max(0 119892119894(119909))]

2


10038171003817100381710038172)


(5)

The above discussion indicates that 119909lowast is a strictly localmaximizer of 119875(119909 119909lowast 119902)

Theorem 3 Assuming that 119909lowast isin 119871(119875) then for any 119909 isin 1198781

119909 = 119909lowast or 119909 isin 119883119878 it holds that 0 notin 120597119875(119909 119909lowast1 119902) when 119902 gt 0

is big enough

Mathematical Problems in Engineering 3

Proof For any 119909 isin 1198781 119909 = 119909lowast or 119909 isin 119883 119878 we have

min(0max(119891(119909) minus 119891(119909lowast) 119892119894(119909) 119894 isin 119868)) = 0 and so

119875 (119909 119909lowast 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max(0 119892119894(119909))]

2


10038171003817100381710038172)

(6)

Thus

120597119875 (119909 119909lowast 119902)

sub minus1

119902[119891 (119909) minus 119891 (119909lowast) +

119898

sum119894=1

max (0 119892119894(119909))]

times (120597119891 (119909) +119898

sum119894=1

120582119894120597119892119894(119909))

minus2 (119909 minus 119909lowast)

1 + (1 + 119909 minus 119909lowast2)2

(7)

where 0 le 120582119894le 1 119894 isin 119868Therefore when 119902 gt 0 is big enough

it holds that

⟨120597119875 (119909 119909lowast 119902) 119909 minus 119909lowast

119909 minus 119909lowast⟩

le1

119902[119871119863 +

119898

sum119894=1

max119909isin119883

1003817100381710038171003817119892119894 (119909)1003817100381710038171003817][119871

119891+119898

sum119894=1

119871119892119894]

minus21003817100381710038171003817119909 minus 119909lowast

1003817100381710038171003817

1 + (1 + 119909 minus 119909lowast2)2

lt 0

(8)

which implies that 0 notin 120597119875(119909 119909lowast 119902)

Theorem 4 Assume that 119909lowast isin 119871(119875) but 119909lowast notin 119866(119875) and119888119897 int 119878 = 119888119897119878 If 119902 gt 0 is suitably large then there exists a point1199090isin 119878 such that 119909

0is a local minimizer of 119875(119909 119909lowast 119902)

Proof By the conditions there exists an 1199092isin int 119878 such that

119891(1199092) lt 119891(119909lowast) 119892

119894(1199092) lt 0 119894 isin 119868

Then for any 119909 isin 120597119878

119875 (119909 119909lowast 119902) = minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max (0 119892119894(119909))]

2


10038171003817100381710038172)

119875 (1199092 119909lowast 119902) = minus

1

119902[119891(1199092) minus 119891(119909lowast)]

2


10038171003817100381710038172)

+ 119902[max(119891(1199092) minus 119891(119909lowast) 119892

119894(1199092) 119894 isin 119868)]

3

(9)

Since as 119902 rarr +infin

119875 (119909 119909lowast 119902) ge minus120587

2 119875 (1199092 119909lowast 119902) 997888rarr minusinfin (10)

then when 119902 gt 0 is suitably big we have

119875 (1199092 119909lowast 119902) lt 119875 (119909 119909lowast 119902) forall119909 isin 120597119878 (11)

Assume that the function 119875(119909 119909lowast 119902) reaches its global mini-mizer over 119878 at 119909

0 Since 119878120597119878 is an open set then 119909

0isin 119878120597119878

and it holds that

min119909isin119878

119875 (119909 119909lowast 119902) = min119909isin119878120597119878

119875 (119909 119909lowast 119902)

= 119875 (1199090 119909lowast 119902) le 119875 (119909

2 119909lowast 119902)

(12)

In the following we will prove that

119891 (1199090) lt 119891 (119909lowast) 119892

119894(1199090) lt 0 119894 isin 119868 (13)

which leads to the resultThe proof is by contradiction Suppose that

119891 (1199090) ge 119891 (119909lowast) 119892

119894(1199090) lt 0 119894 isin 119868 (14)

Then as 119902 rarr +infin

119875 (1199090 119909lowast 119902) 997888rarr minus arg (1 +

10038171003817100381710038171199090 minus 119909lowast10038171003817100381710038172) gt minusinfin (15)

which implies that when 119902 gt 0 is suitably big we have

119875 (1199090 119909lowast 119902) gt 119875 (119909

2 119909lowast 119902) (16)

This is a contradiction

3 Algorithm and Numerical Examples

Based on the properties of the proposed filled function wenowgive a corresponding filled function algorithmas follows

31 Filled Function Algorithm FFAM

Initialization Step

(1) Set a disturbance constant 120572 = 01(2) Select an upper bound of 119902denoted by 119902

119906and set 119902

119906=

108(3) Select directions 119890

119896 119896 = 1 2 119897 with integer 119897 =

2119899 where 119899 is the number of variables(4) Set 119896 = 1

Main Step

(1) Start from an initial point 119909 minimize the problem(119875) by implementing a nonsmooth local search pro-cedure and obtain the first localminimizer119909lowast

1of119891(119909)

(2) Let 119902 = 1


(3) Construct a filled function at 119909lowast1

119875 (119909 119909lowast1 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast

1) +119898

sum119894=1

max(0 119892119894(119909))]

2


1

10038171003817100381710038172)

+ 119902[min (0max (119891 (119909) minus 119891 (119909lowast1) 119892119894(119909) 119894 isin 119868))]

3

(17)

(4) If 119896 gt 119897 then go to (7) Else set 119909 = 119909lowast1

+ 120572119890119896as an

initial point minimize 119875(119909 119909lowast 119902) by implementing anonsmooth local search algorithm and obtain a localminimizer 119909

119896

(5) If 119909119896isin 119878 then set 119896 = 119896 + 1 and go to (4) Else go to

(6)(6) If 119909

119896meets 119891(119909

119896) lt 119891(119909lowast

1) then set 119909 = 119909

119896and

119896 = 1 Start from 119909 as a new initial point minimizethe problem (119875) by using a local search algorithmand obtain another local minimizer 119909lowast

2of 119891(119909) with

119891(119909lowast2) lt 119891(119909lowast

1) Set 119909lowast

1= 119909lowast2and go to (2) Else go to

(7)(7) Increase 119902 by setting 119902 = 10119902(8) If 119902 le 119902

119906 then set 119896 = 1 and go to (3) Else the

algorithm stops and 119909lowast1is taken as a global minimizer

of the problem (119875)

At the end of this section we make a few remarks below(1) Algorithm FFAM is comprised of two stages local

minimization and filling In stage 1 a local minimizer 119909lowast1

of 119891(119909) is identified In stage 2 filled function 119875(119909 119909lowast1 119902) is

constructed and then minimized Stage 2 terminates whenone point 119909

119896isin 1198782is located Then algorithm FFAM

reenters into stage 1 with 119909119896as an initial point to find a new

minimizer 119909lowast2of 119891(119909) (if such one minimizer exists) and so

onThe above process is repeated until some certain specifiedstopping criteria are met and then the last local minimizer isregarded as a global minimizer

(2)Themotivation andmechanism behind the algorithmFFAM are given below

In Step (3) of the Initialization Step we can choosedirections 119890

119896 119896 = 1 2 119897 as positive and negative unit

coordinate vectors For example when 119899 = 2 the directionscan be chosen as (1 0) (0 1) (minus1 0) and (0 minus1)

In Step (1) and Step (6) of the Main Step we can obtain alocal optimizer of the problem (119875) by using any nonsmoothconstrained local optimization procedure such as Bundlemethods and Powellrsquos method In Step (4) of the MainStep we can minimize the proposed filled function by usingHybrid Hooke and Jeeves-Direct Method for NonsmoothOptimization [10] Mesh Adaptive Direct Search Algorithmsfor Constrained Optimization [11] and so forth

(3) The proposed filled function algorithm can also beapplied to smooth constrained optimization Any smoothlocal minimization procedure in the minimization phase

Table 1 Computational results with initial point (minus1 minus1 minus1)

119896 119909119896

119909lowast119896

119891(119909lowast119896)

1 (minus1 minus1 minus1) (minus19802 minus00132 minus00006) minus194102 (11931 06332 minus11932) (19889 minus00001 minus00111) minus59477

can be used such as conjugate gradient method and quasi-Newton method

32 Numerical Examples Weperform the numerical tests forthree examples All the numerical tests are programmed inFortran 95 In nonsmooth case we search for the local mini-mizers by using Hybrid Hooke and Jeeves-Direct Method forNonsmooth Optimization [10] and theMesh Adaptive DirectSearch Algorithms for Constrained Optimization [11] Insmooth case we use penalty function method and conjugategradient method to get the local minimizers

The following are the three examples and their numericalresults And the symbols used in the tables are explainedbelow

119896 the iteration number in finding the 119896th localminimizer119909119896 the 119896th new initial point in finding the 119896th local

minimizer119909lowast119896 the 119896th local minimizer

119891(119909lowast119896) the function value of the 119896th local minimizer

Problem 1 Consider

min 119891 (119909) = minus11990921+ 11990922+ 11990923minus 1199091

st 11990921+ 11990922+ 11990923minus 4 le 0

min 1199092minus 1199093 1199093 le 0

(18)

Algorithm FFAM succeeds in finding an approximate globalminimizer 119909lowast = (19889 minus00001 minus00111)119879 with 119891(119909lowast) =minus59477 The computational results are given in Table 1

Problem 2 Consider

min 119891 (119909) = minus20 exp(minus02radic10038161003816100381610038161199091

1003816100381610038161003816 +10038161003816100381610038161199092

10038161003816100381610038162

)

minus exp(cos (2120587119909

1) + cos (2120587119909

2)

2) + 20

st 11990921+ 11990922le 300

21199091+ 1199092le 4

minus 30 le 119909119894le 30 119894 = 1 2

(19)

Algorithm FFAM succeeds in finding a global minimizer119909lowast = (0 0)119879 with 119891(119909lowast) = minus27183 The computationalresults are listed in Table 2


Table 2 Computational results with initial point (minus1 minus1)

119896 119909119896

119909lowast119896

119891(119909lowast119896)

1 (minus1 minus1) (minus150000 00000) 571642 (minus10585 05165) (00001 minus02094) minus036903 (00007 minus00435) (00000 00000) minus27183

Table 3 Computational results with initial point (0 0)

119896 119909119896

119909lowast119896

119891(119909lowast119896)

1 (0 0) (06116 34423) minus405412 (21653 22546) (23295 31780) minus55081

Problem 3 Consider

min 119891 (119909) = minus1199091minus 1199092

st 1199092le 211990941minus 811990931+ 811990921+ 2

1199092le 411990941minus 321199093

1+ 881199092

1minus 96119909

1+ 36

0 le 1199091le 3

0 le 1199092le 4

(20)

Algorithm FFAM succeeds in finding a global minimizer119909lowast = (23295 31780)119879 with 119891(119909lowast) = minus55081 This problemis taken from [12] We give this numerical example here toillustrate that algorithm FFAM is also suitable for smoothconstrained global optimization The computational resultsare given in Table 3

4 Conclusion

In this paper we present a new filled function for bothnonsmooth and smooth constrained global optimizationand investigate its properties The filled function containsonly one parameter which can be readily adjusted in theprocess of minimization We also design a correspondingfilled function algorithm Moreover in order to demonstratethe performance of the proposed filled function method wemake three numerical tests The preliminary computationalresults show that the proposed filled function approach ispromising

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

Thisworkwas supported by theNNSF of China (nos 11471102and 11001248) the SEDF under Grant no 12YZ178 theKey Discipline ldquoApplied Mathematicsrdquo of SSPU under noA30XK1322100 and NNSF of Zhejiang (no LY13A010006)

References

[1] P M Pardalos and H E Romeijn Eds Handbook of GlobalOptimization Volume 22 Heuristic Approaches Kluwer Aca-demic Publishers Dordrecht The Netherlands 2002

[2] P M Pardalos and H E Romeijn Eds Handbook of GlobalOptimization vol 62 of Nonconvex Optimization and Its Appli-cations Kluwer Academic Publishers Dordrecht The Nether-lands 2002

[3] R Horst and P M Pardalos Eds Handbook of Global Opti-mization vol 2 ofNonconvex Optimization and Its ApplicationsKluwer Academic Publishers Dordrecht The Netherlands1995

[4] R P Ge and Y F Qin ldquoA class of filled functions for findingglobal minimizers of a function of several variablesrdquo Journal ofOptimization Theory and Applications vol 54 no 2 pp 241ndash252 1987

[5] W Wang Y Yang and L Zhang ldquoUnification of filled functionand tunnelling function in global optimizationrdquo Acta Mathe-maticaeApplicatae Sinica English Series vol 23 no 1 pp 59ndash662007

[6] Z Xu H Huang P M Pardalos and C Xu ldquoFilled functionsfor unconstrained global optimizationrdquo Journal of Global Opti-mization vol 20 no 1 pp 49ndash65 2001

[7] Y-J Yang and Y-L Shang ldquoA new filled function method forunconstrained global optimizationrdquo Applied Mathematics andComputation vol 173 no 1 pp 501ndash512 2006

[8] A V Levy and A Montalvo ldquoThe tunneling algorithm for theglobal minimization of functionsrdquo SIAM Journal on Scientificand Statistical Computing vol 6 no 1 pp 15ndash29 1985

[9] Y Wang and J Zhang ldquoA new constructing auxiliary functionmethod for global optimizationrdquo Mathematical and ComputerModelling vol 47 no 11-12 pp 1396ndash1410 2008

[10] C J Price B L Robertson and M Reale ldquoA hybrid Hookeand Jeevesmdashdirect method for non-smooth optimizationrdquoAdvanced Modeling and Optimization vol 11 no 1 pp 43ndash612009

[11] C Audet and J Dennis ldquoMesh adaptive direct search algorithmsfor constrained optimizationrdquo SIAM Journal on Optimizationvol 17 no 1 pp 188ndash217 2006

[12] C A Floudas and P M Pardalos A Collection of Test Prob-lems for Constrained Global Optimization Algorithms SpringerBerlin Germany 1990

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Stochastic AnalysisInternational Journal of


method The proposed filled function method combinesfilled functionmethod for unconstrained global optimizationwith the exterior penalty function method for constrainedoptimization

The paper is organized as follows In Section 2 a newfilled function is proposed and its properties are discussedIn Section 3 a corresponding filled function algorithm isdesigned and numerical experiments are performed Finallyin Section 4 some conclusive remarks are given

In the rest of this paper the generalized gradient of anonsmooth function 119891(119909) at the point 119909 isin 119883 is denotedby 120597119891(119909) and the generalized directional derivative of 119891(119909)

in the direction 119889 at the point 119909 is denoted by 1198910(119909 119889) Theinterior the boundary and the closure of the set 119878 are denotedby int 119878 120597119878 and 119888119897119878 respectively

2 A New One-Parameter FilledFunction and Its Properties

21 Problem Formulation Consider the following nons-mooth constrained global optimization problem (119875)

min119909isin119878

119891 (119909) (1)

where 119878 = 119909 isin 119883 | 119892119894(119909) le 0 119894 isin 119868 119883 sub 119877119899 is a

box set 119891(119909) and 119892119894(119909) 119894 isin 119868 are Lipschitz continuous with

constants 119871119891and 119871

119892119894 119894 isin 119868 respectively and 119868 = 1 119898

is an index set For simplicity the set of local minimizersfor problem (119875) is denoted by 119871(119875) and the set of the globalminimizers is denoted by 119866(119875)

To proceed we assume that the number of minimizers ofproblem (119875) is infinite but the number of different functionvalues at the minimizers is finite

Definition 1 A function 119875(119909 119909lowast) is called a filled function of119891(119909) at a local minimizer 119909lowast if it satisfies the following

(1) 119909lowast is a strictly local maximizer of 119875(119909 119909lowast) on 119883

(2) for any 119909 isin 1198781119909lowast or 119909 isin 119883 119878 one has 0 notin 120597119875(119909 119909lowast)

where 1198781= 119909 isin 119878 | 119891(119909) ge 119891(119909lowast)

(3) if 1198782= 119909 isin 119878 | 119891(119909) lt 119891(119909lowast) is not empty then there

exists a point 1199092isin 1198782such that 119909

2is a local minimizer

of 119875(119909 119909lowast)

Definition 1 guarantees that when any local search proce-dure for unconstrained optimization is used to minimize theconstructed filled function the sequences of iterative pointwill not stop at any point at which the objective functionvalue is larger than 119891(119909lowast) If 119909lowast is not a global minimizerthen a point 119909 with 119891(119909) lt 119891(119909lowast) could be identified in theprocess of the minimization of 119875(119909 119909lowast) Then we can get abetter local minimizer of 119891(119909) by using 119909 as an initial pointBy repeating these two steps we could finally obtain a globalminimizer or an approximate globalminimizer of the originalproblem

22 A New Filled Function and Its Properties We propose inthis section a one-parameter filled function as follows

119875 (119909 119909lowast 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max (0 119892119894(119909))]

2


10038171003817100381710038172)

+ 119902[min (0max (119891 (119909) minus 119891 (119909lowast) 119892119894(119909) 119894 isin 119868))]

3

(2)

where 119902 gt 0 is a parameter and 119909lowast is the current localminimizer of 119891(119909)

Theorems 2ndash4 show that when parameter 119902 gt 0 is suitablylarge the function 119875(119909 119909lowast 119902) is a filled function

Theorem 2 Assume that 119909lowast isin 119871(119875) then 119909lowast is a strictly localmaximizer of 119875(119909 119909lowast 119902)

Proof Since 119909lowast isin 119871(119875) there is a neighborhood 119874(119909lowast 120575) of119909lowast with 120575 gt 0 such that 119891(119909) ge 119891(119909lowast) and 119892

119894(119909) le 0 119894 isin 119868

for all 119909 isin 119874(119909lowast 120575) cap 119878 We consider the following two cases

Case 1 For all 119909 isin 119874(119909lowast 120575) cap 119878 and 119909 = 119909lowast we havemin(0max(119891(119909) minus 119891(119909lowast) 119892

119894(119909) 119894 isin 119868)) = 0 and so

119875 (119909 119909lowast 119902) = minus1

119902(119891 (119909) minus 119891 (119909lowast))


10038171003817100381710038172)


(3)

Case 2 For all 119909 isin 119874(119909lowast 120575) cap (119883 119878) there exists at least oneindex 119894

0isin 119868 such that 119892

1198940(119909) ge 0 it follows that

min (0max (119891 (119909) minus 119891 (119909lowast) 119892119894(119909) 119894 isin 119868)) = 0 (4)

Thus

119875 (119909 119909lowast 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max(0 119892119894(119909))]

2


10038171003817100381710038172)


(5)

The above discussion indicates that 119909lowast is a strictly localmaximizer of 119875(119909 119909lowast 119902)

Theorem 3 Assuming that 119909lowast isin 119871(119875) then for any 119909 isin 1198781

119909 = 119909lowast or 119909 isin 119883119878 it holds that 0 notin 120597119875(119909 119909lowast1 119902) when 119902 gt 0

is big enough




119875 (119909 119909lowast 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max(0 119892119894(119909))]

2


10038171003817100381710038172)

(6)

Thus

120597119875 (119909 119909lowast 119902)

sub minus1

119902[119891 (119909) minus 119891 (119909lowast) +

119898

sum119894=1

max (0 119892119894(119909))]

times (120597119891 (119909) +119898

sum119894=1

120582119894120597119892119894(119909))


1 + (1 + 119909 minus 119909lowast2)2

(7)


it holds that



le1

119902[119871119863 +

119898

sum119894=1

max119909isin119883

1003817100381710038171003817119892119894 (119909)1003817100381710038171003817][119871

119891+119898

sum119894=1

119871119892119894]


1003817100381710038171003817

1 + (1 + 119909 minus 119909lowast2)2

lt 0

(8)





119891(1199092) lt 119891(119909lowast) 119892

119894(1199092) lt 0 119894 isin 119868

Then for any 119909 isin 120597119878

119875 (119909 119909lowast 119902) = minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max (0 119892119894(119909))]

2


10038171003817100381710038172)

119875 (1199092 119909lowast 119902) = minus

1

119902[119891(1199092) minus 119891(119909lowast)]

2


10038171003817100381710038172)

+ 119902[max(119891(1199092) minus 119891(119909lowast) 119892

119894(1199092) 119894 isin 119868)]

3

(9)


119875 (119909 119909lowast 119902) ge minus120587






0isin 119878120597119878

and it holds that

min119909isin119878

119875 (119909 119909lowast 119902) = min119909isin119878120597119878

119875 (119909 119909lowast 119902)

= 119875 (1199090 119909lowast 119902) le 119875 (119909

2 119909lowast 119902)

(12)


119891 (1199090) lt 119891 (119909lowast) 119892

119894(1199090) lt 0 119894 isin 119868 (13)


119891 (1199090) ge 119891 (119909lowast) 119892

119894(1199090) lt 0 119894 isin 119868 (14)





119875 (1199090 119909lowast 119902) gt 119875 (119909

2 119909lowast 119902) (16)





Initialization Step


119906and set 119902

119906=


119896 119896 = 1 2 119897 with integer 119897 =


Main Step


1of119891(119909)

(2) Let 119902 = 1



119875 (119909 119909lowast1 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast

1) +119898

sum119894=1

max(0 119892119894(119909))]

2


1

10038171003817100381710038172)


3

(17)


+ 120572119890119896as an


119896


(6)(6) If 119909

119896meets 119891(119909

119896) lt 119891(119909lowast

1) then set 119909 = 119909

119896and


2of 119891(119909) with

119891(119909lowast2) lt 119891(119909lowast

1) Set 119909lowast





















119896 119909119896

119909lowast119896

119891(119909lowast119896)








Problem 1 Consider

min 119891 (119909) = minus11990921+ 11990922+ 11990923minus 1199091

st 11990921+ 11990922+ 11990923minus 4 le 0

min 1199092minus 1199093 1199093 le 0

(18)


Problem 2 Consider


1003816100381610038161003816 +10038161003816100381610038161199092

10038161003816100381610038162

)

minus exp(cos (2120587119909

1) + cos (2120587119909

2)

2) + 20

st 11990921+ 11990922le 300

21199091+ 1199092le 4

minus 30 le 119909119894le 30 119894 = 1 2

(19)




119896 119909119896

119909lowast119896

119891(119909lowast119896)



119896 119909119896

119909lowast119896

119891(119909lowast119896)

1 (0 0) (06116 34423) minus405412 (21653 22546) (23295 31780) minus55081

Problem 3 Consider

min 119891 (119909) = minus1199091minus 1199092

st 1199092le 211990941minus 811990931+ 811990921+ 2

1199092le 411990941minus 321199093

1+ 881199092

1minus 96119909

1+ 36

0 le 1199091le 3

0 le 1199092le 4

(20)


4 Conclusion




Acknowledgments


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119875 (119909 119909lowast 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max(0 119892119894(119909))]

2


10038171003817100381710038172)

(6)

Thus

120597119875 (119909 119909lowast 119902)

sub minus1

119902[119891 (119909) minus 119891 (119909lowast) +

119898

sum119894=1

max (0 119892119894(119909))]

times (120597119891 (119909) +119898

sum119894=1

120582119894120597119892119894(119909))


1 + (1 + 119909 minus 119909lowast2)2

(7)


it holds that



le1

119902[119871119863 +

119898

sum119894=1

max119909isin119883

1003817100381710038171003817119892119894 (119909)1003817100381710038171003817][119871

119891+119898

sum119894=1

119871119892119894]


1003817100381710038171003817

1 + (1 + 119909 minus 119909lowast2)2

lt 0

(8)





119891(1199092) lt 119891(119909lowast) 119892

119894(1199092) lt 0 119894 isin 119868

Then for any 119909 isin 120597119878

119875 (119909 119909lowast 119902) = minus1

119902[119891(119909) minus 119891(119909lowast) +

119898

sum119894=1

max (0 119892119894(119909))]

2


10038171003817100381710038172)

119875 (1199092 119909lowast 119902) = minus

1

119902[119891(1199092) minus 119891(119909lowast)]

2


10038171003817100381710038172)

+ 119902[max(119891(1199092) minus 119891(119909lowast) 119892

119894(1199092) 119894 isin 119868)]

3

(9)


119875 (119909 119909lowast 119902) ge minus120587






0isin 119878120597119878

and it holds that

min119909isin119878

119875 (119909 119909lowast 119902) = min119909isin119878120597119878

119875 (119909 119909lowast 119902)

= 119875 (1199090 119909lowast 119902) le 119875 (119909

2 119909lowast 119902)

(12)


119891 (1199090) lt 119891 (119909lowast) 119892

119894(1199090) lt 0 119894 isin 119868 (13)


119891 (1199090) ge 119891 (119909lowast) 119892

119894(1199090) lt 0 119894 isin 119868 (14)





119875 (1199090 119909lowast 119902) gt 119875 (119909

2 119909lowast 119902) (16)





Initialization Step


119906and set 119902

119906=


119896 119896 = 1 2 119897 with integer 119897 =


Main Step


1of119891(119909)

(2) Let 119902 = 1



119875 (119909 119909lowast1 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast

1) +119898

sum119894=1

max(0 119892119894(119909))]

2


1

10038171003817100381710038172)


3

(17)


+ 120572119890119896as an


119896


(6)(6) If 119909

119896meets 119891(119909

119896) lt 119891(119909lowast

1) then set 119909 = 119909

119896and


2of 119891(119909) with

119891(119909lowast2) lt 119891(119909lowast

1) Set 119909lowast





















119896 119909119896

119909lowast119896

119891(119909lowast119896)








Problem 1 Consider

min 119891 (119909) = minus11990921+ 11990922+ 11990923minus 1199091

st 11990921+ 11990922+ 11990923minus 4 le 0

min 1199092minus 1199093 1199093 le 0

(18)


Problem 2 Consider


1003816100381610038161003816 +10038161003816100381610038161199092

10038161003816100381610038162

)

minus exp(cos (2120587119909

1) + cos (2120587119909

2)

2) + 20

st 11990921+ 11990922le 300

21199091+ 1199092le 4

minus 30 le 119909119894le 30 119894 = 1 2

(19)




119896 119909119896

119909lowast119896

119891(119909lowast119896)



119896 119909119896

119909lowast119896

119891(119909lowast119896)

1 (0 0) (06116 34423) minus405412 (21653 22546) (23295 31780) minus55081

Problem 3 Consider

min 119891 (119909) = minus1199091minus 1199092

st 1199092le 211990941minus 811990931+ 811990921+ 2

1199092le 411990941minus 321199093

1+ 881199092

1minus 96119909

1+ 36

0 le 1199091le 3

0 le 1199092le 4

(20)


4 Conclusion




Acknowledgments


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119875 (119909 119909lowast1 119902)

= minus1

119902[119891(119909) minus 119891(119909lowast

1) +119898

sum119894=1

max(0 119892119894(119909))]

2


1

10038171003817100381710038172)


3

(17)


+ 120572119890119896as an


119896


(6)(6) If 119909

119896meets 119891(119909

119896) lt 119891(119909lowast

1) then set 119909 = 119909

119896and


2of 119891(119909) with

119891(119909lowast2) lt 119891(119909lowast

1) Set 119909lowast





















119896 119909119896

119909lowast119896

119891(119909lowast119896)








Problem 1 Consider

min 119891 (119909) = minus11990921+ 11990922+ 11990923minus 1199091

st 11990921+ 11990922+ 11990923minus 4 le 0

min 1199092minus 1199093 1199093 le 0

(18)


Problem 2 Consider


1003816100381610038161003816 +10038161003816100381610038161199092

10038161003816100381610038162

)

minus exp(cos (2120587119909

1) + cos (2120587119909

2)

2) + 20

st 11990921+ 11990922le 300

21199091+ 1199092le 4

minus 30 le 119909119894le 30 119894 = 1 2

(19)




119896 119909119896

119909lowast119896

119891(119909lowast119896)



119896 119909119896

119909lowast119896

119891(119909lowast119896)

1 (0 0) (06116 34423) minus405412 (21653 22546) (23295 31780) minus55081

Problem 3 Consider

min 119891 (119909) = minus1199091minus 1199092

st 1199092le 211990941minus 811990931+ 811990921+ 2

1199092le 411990941minus 321199093

1+ 881199092

1minus 96119909

1+ 36

0 le 1199091le 3

0 le 1199092le 4

(20)


4 Conclusion




Acknowledgments


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119896 119909119896

119909lowast119896

119891(119909lowast119896)



119896 119909119896

119909lowast119896

119891(119909lowast119896)

1 (0 0) (06116 34423) minus405412 (21653 22546) (23295 31780) minus55081

Problem 3 Consider

min 119891 (119909) = minus1199091minus 1199092

st 1199092le 211990941minus 811990931+ 811990921+ 2

1199092le 411990941minus 321199093

1+ 881199092

1minus 96119909

1+ 36

0 le 1199091le 3

0 le 1199092le 4

(20)


4 Conclusion




Acknowledgments


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Documents

Research Article Global Minimization of Nonsmooth Constrained Global …downloads.hindawi.com/journals/mpe/2014/563860.pdf · nonsmooth and smooth constrained global optimization