Applied Mathematics, 2011, 2, 13721377 doi:10.4236/am.2011.211193 Published Online November 2011 (http://www.SciRP.org/journal/am) Copyright © 2011 SciRes. AM A Nonmonotone Filter Method for Minimax Problems* Qi Zhao, Nan Guo Department of Basi c , Jiangsu University of Science and Technology, Zhangjiagang, China Department of Basi c , Nanjing Institute of Tech nology, Nanjing, China Email: johnzqzq@163.com Received September 18, 2011; revised October 21, 2011; accepted October 28, 2011 Abstract In this paper, we propose a modified trustregion filter method algorithm for Minimax problems, which based on the framework of SQPfilter method and associated with the technique of nonmonotone method. We use the SQP subproblem to acquire an attempt step, and use the filter to weigh the effect of the attempt step so as to avoid using penalty function. The algorithm uses the Lagrange function as a merit function and the nonmonotone filter to improve the effect of the algorithm. Under some mild conditions, we prove the global convergence. Keywords: Minimax Problem, Nonmonotone, Global Convergence, Filter Methods 1. Introduction Consider the following Minimax problem: 1 min max() i xR im x (1) where (): n i xR R is a twice continuously differen tiable function. The problem (1) can be transformed into the following problem below: min .() i t stf xt 0 (2) The Minimax problem is one of the most important nondifferentiable optimization problems. It does not only have broader applications in engineering design, electronic microcircuits programming, game theory and so on, but also has very close relationship with nonlinear equations, mutiobject programming, nonlinear progra mmming, etc. There are some methods e.g., line search method SQP method, trust region method and the active set method, for solving Minimax problems. C. Chara lambous and A.R. Conn [1] proposed the line search method. A. Vardi [2] presented the trust region method with the activeset methods. There are many other effec tive algorithms, see Z. B. Zhu [3], L. Gao [4], J. L. Zhou [5] ,Y. Xue [6]. Recently, the filter method for nonlinear programming has broader applications and good numerical effects, see [712]. The major filter methods are of two kinds: line search and trustregion methods. R. Fletcher proposed the global convergent SQPfilter trustregion method [9], based on this idea, Huang [13] proposed a filter method for Minimax problems. In [14], Ulbrich S. used the Lagrange function to replace the fun ction and gave the local superlinear convergence proof of the SQPfilter trustregion method. The nonmonotone technique can improve the effect of the algorithm, relax the accept criteria of the attempt step. Recently, Su [15] and Shen [16] presented the idea of using nonmonotone filter methods for nonlinear pro gramming. Motivated by their idea, we present a modi fied filtermethod for Minimax problems. The algorithm uses the Lagrangian function instead of the function it self as a merit function, and combines it with a non monotone filter technique to improve the effect of the algorithm. Consider the SQP subproblem of problem (2): 1 min( )2 () .() 1 tT kk i i qs ssHs fx tfxs t s (3) We use the following notations: 1 (,), {1,2,,}, xt n ssR Im (1,1, ,1)T e 12 (,)(), ()( (),(),()) T m hxtf xtefxfxfxfx (1)(1) knn HR is a symmetric matrix, and it is the approximate Hessian matrix of the subproblem (3). *This work was supported by Chinese NSF Grant 10201026.
Q. ZHAO ET AL.1373 Remark: k is updated by the Powell’s safeguard BFGS update form ula. This paper is organized as follows. The new algorithm is described in Section 2. Basic assumptions and some important lemmas are given in Section 3. The analysis of the global convergence is given in Sections 4 and 5. 2. Algorithm Now we introduce some definitions about the filter used in this paper. Definition 1 : [14] Lagrange f unction: (,, )(,), T lxtt hxt Constrain violation functio n: 2 2 2 (, ,)(,)(,) T thxt hxt For simplicity, we just use the following notations: (,, ),(,, ), tlxtl (,,),(,,) jjjjjjj tlxtl Definition 2 : A pair (,) kk l obtained on iteration is said to do minate another pair k (,) ll l if and only if and kl k ll l A filter set is a list of pairs (,) j l such that no pair dominates any other. We denote the set by for each iteration k. k Similar to the definition in Fletcher. and Leyffer, S. [9], a point ((, ),) ts can be accept to (,) kkk l if , or , ll (,) (,) jkkk ll Here we use the nonmonotone filter idea in [14] a point ((, ),) ts can be accept to (,) kkk l if 0()1 max kr rmk or ()1 , 0 max, mk kkr r ll kr l ) (4) where (,)(, kr krkk k ll , 01 1 (0) 0,0() min(1)1, mmkmkM ()1 , 01, mk kr r ,, (0,1),0, 1 krkrM 1 Update of the fi lt er set If (,) kk l is added to the filter, then the new filter set is updated as follows: 1:( ,)(,),min,0 kkk jjkjkjk ll ll Definition 3 : [14 ] () (), kk predsq s ()1 , 0 () max, ,,() mk l kkkr r xt kk kr k rared sl lxstss l (),, (), lxt kkkk k aredsll xstss where () k is the Lagrange multiplier of the subprob lem (3). In order to improve both the feasibility and optimality if 1/2 ():(), 0 T kkkkk pred spred sh (5) then we require () () l k rared spred s k (6) and call it is a ftype iteration. If 1/2 () kk k pred s (7) then we add (,) kk l to the filter set and update the filter set, calling it is a htype iteration . ()k If the subproblem (3) is not compatible or 1/21 ,0,0, k 1 we also call it is a htype iteration . ()k Now we describe the detailed algorithm below: Step 0 (Initialization) Give ,,,,(0,1), ,0, min ,: , 0k0 set the initial filter set . o Step 1: Solve subproblem (3) to get an attempt step s. Step 2: If the solution of (3) is not compatible or 1/21 ,0, k 1 Then add (,) kk l to the filter set and update the filter set, let: 11 (),(, )(,) kkkk xx txt 1 ki im tmaxf k go to Step 1; If s , stop; else go to step 3. Step 3: If (4) fails, then , go to Step 1; else go to Step 4. :0.5 Step 4: If (5) holds but (6) fails, th en , go to step 1, else go to Step 6. :0.5 Step 5: If (5) fails, add (,) kk l to the filter set and update the filter to k1k , else go to Step 6. Step 6: : k s , 11 :,: ) x kkkk 1 , t kkkk ( xst ts s k to 1k ,min update . , go to Step 1. rk 2 totep 5 arcalled inn tio sic Assumptions and Lemmas A1: All iterations :1kk Rema . Step Se er circle itera n steps, while Step 1, Step 6 are called outer circle steps. . Ba3 irst we make the following assumptions: F(,) kk t remain in a close and Copyright © 2011 SciRes. AM
Q. ZHAO ET AL. 1374 bounded convex set . A2: The functions () i x are twice continuously dif ferentible. A3: The matrix k is bounded, kH M. Lemma 1: If thtion of the su then (,) e solubproblem (3) is s = 0, kk t is KKT point of p. troblem (2)he Proof: It is not difficult to get the result by using the first order necessary optimality conditions. Lemma 2: All the elements (,) j l in the filter set k satisfy 0 j . Proof. Suppose, by contradictit the result is not e, then thuson, tha truere mit a t ex(,) jk l and 0 j , which means (,) j t is a feasible solution of the pro blem (2), and 21 1,0 T jj h ; but 0 j s j so 1/2 0 j j () jj pred spredsed on tate () j s of the filter set, , ba mechanism he upd (,) j l radicti can’t be added to on.Thus, the Lemma the filter set, which is a cont olds is sufficiently large enough: is proven. Lemma 3: If {} k l is bounded below and one of the following hwhen k 10()1 max kkr rmk ()1 11 , 0 , , mk kkkkr rkr ll max l then lim 0 kk . Proof. We consider te following two cases: e 1: h ma Cas 10()1 r rmk x , kk ently larfo ough. r k suffici Let ((lk ge en ax 0()1rm k ))m,(( )1( )) kr k mklkk th en 11 1)1 0() 1 (( 1))axmax (()),)(()) 0( m max( kr kr k rmk k lk lklk rm )) which implies ((lk converges. By (()))1)),(0,1)lk k(((ll we get ((lk)) 0,k S o 1(())0 klk 1k ge en th . Case 2: )1 1k l ( , 0 max ,, mk k krkr r l l fo ently larough. will showat, for all 1 k r r r k suffici First we 1k 1k 0 1 krk r ll 0 l (8) We prove (8) by inction. When 1 du ll1k, 10 10 l , assume ho ove it also holds for 1 that (8) lds for 1, 2,, k, we want to pr 1. k We consider the following two cases: (a), 0 max ,kkkrk rr lll ()1mk 110 1 k kkkrk r ll l (b) r ()1 ()1 ,, 00 max, mk mk kkkrkrr rk r lll Let () 1pmk , then k k 1 0 1 ,0 1 01 11 ,0 01 1 12 ,1 01 1 1 ,0 1 ,0 , 0 p ktk t pkt ktrktk tr kp k krrk rrkp kp k krr rrkp kp kprkp r p kt kt tt l l l l l l 1,ktk l 1 10 ,1 0 kp p r r p ktk tk t By the fact that 1 ,, 01,, 0 p kr krr t 1 1 01 1 11 01 1 10 kp k rrkp k rk r kk rr r r l l krk ll r So (8) is true. Since is bounded below, so Thus the Lemma is proven. Some assumptions are needed for Lagrange multiplier es {} k l 1,0, rk rk timates () k . A4: There exits constants ,0 yL MM, all of the La grange mu() k satisfies: ltiplier estimates 0() kiy ,1,2,, Mi m 2 ()(, ) kkk L ii shxtsM Lemma 4: Under the assumptions A1A4e have , w 2224 2 1 (,)( 1) 4 kk f hxt smnM (9) 22 224 4 1 (,),()(1) 4 : kk kf L xt s smnM mM M 4 (10) Proof. Use the Taylor Expansion: 22 () () () xt ik kk ik1 11 () ( 1) 22 T ik xT x ik f fx xstst fx s sf ysnM Copyright © 2011 SciRes. AM
Q. ZHAO ET AL.1375 which implies (9) holds. 2 2 24 (,) ,()(,) (()(((,))) kk kkk T kkk xt s shxt s shxtsM where 2222 1(1) 4 f mMmn M which implies (10) holds. Thus, the Lemma is proven. 4. The Well Definedness of the Algorithm Lemma 5: Let (,) min jj k kj l , when 4k , (,) ,() kk k tss can be accept to . k Proof. From the definition of k , Use the result of (10), when 4k have , we so (,) ,()max(,), kkkr rkrk xt s sl , 0()1 k k rm k (,) ,() kk k tss can be accepted to . k Thus, the oven. Lemma 6:umptions Lemma is pr Under the assA1A4, if (,)xt oblem ) (,) is a feasible point ,but not a KKT point of the pr ere must exit a (2 then thneighborhood N of** t, and constants ,ˆ ,ˆ >0 e satisfying (11), the sub for all (,) kk xt N , and all ave a sothproblem must h lution, and 11 2 k , 1/2 21/2ˆ ˆ() kk (11) 1/2 1 () 3 kk pred s (12) ()() l k rared spred s k Proof. From the results of Lemma7 in [13], there must exit a neighborhood of (13) ()(,), l kkk raredsx ts ()s (14) N ** (,) t, and some constants ,, 0 , for all and (, kk xt)N 2 kk (,)hx t , s aion, athe subproblem (3) ha solutnd 2 () ((, )) kk k ared sh x ts 1, ()(), 2 kkk predared ss (15) next we will prove, there must exit , when (11) holds, then ()spred ˆˆ , 11 2 k and (12)(14) holds. If 11 2 k 1/2 (,) T kkk k hx t and the results of (15) we can deduce 1/2 , )() kkkk prex tpreds 1/2 k () ( 11 T kk d sh 36 If , then 11 122 k , from 1/2 ) k 1( 6 then 1/2 1 ()(,)3 T kkkk predsh xt k So if we take 1 1 ˆ,,6 1 max then 1/2 1 () k pred s3k , (12) holds. From the definition of () l k rareds 2 ()()()(, )) ()() ( , )) ( 1) 2 1 () lTT kkkkkkk TT kk kk T kk f rared sared shs hxts ared shts hMmnM and by Taylor Ex p ansion we kn ow kk s hx ky ared s 2 1 ()()( 1) 2 kk ared spred snMH . So we can deduce 2 () () lT kkkk rared sared sh 2 11 (1) (1) 22 (1 )() 11 (1)(1) 22 1 () (1) 3 11 (1)(1) 22 Hy f k Hy f k Hy f nMMmnM pred s nMMmnM pred s nMMmnM () k pred s By simple calculation we know that if 1 1 (1) 3: 11 (1)(1) 22 Hy f nMMmnM ()() l kk rared spred s then 4 ()(, ),() 1 3 l kkk raredsxtss M k Copyright © 2011 SciRes. AM
Q. ZHAO ET AL. 1376 if we take So 1 3 2 1 3: M then take then (13)(14) holdThus, the Lemma Lemma 7: Under the assumptions A1A4, the inner on terminates finitely. a KKT point, other ()(, ),() l kkkk raredsx tss . From the dis cu s s ab o ve we kn o w if we 12 ˆmax, is proven. s. iterati Proof. If 0s, the algorithm terminates at if the inner iteration does not terminate finitely, then the rule for d ecreasing ensures wise, 0. We consider the following two cases: Case 1: 0 k When 0 , 1/2 1 k cannot hold, so the al gorithm will enter a restoration phrase at step 2 and ter minates finitely. Case 2: 0 k This means (, kk xt) is a feasible po f satisfie int. When , is 0 1 1/2 21/24 ˆ ˆ 0min, kM k k the result of Lemma 6, the subproblem (3) must beible a2). From Lemma 5 and (14) we know that From compatnd (1(14) holds (,) kk ts can be accept to (,) kkk l . So all thitions for ftype step are satisfied and the inner iteration terminates successfully. Lemma 8: e cond 5. Global Convergence Under the assumptions A1A4, if the algo rithm does not terminate finitely, and there are infinite oints added to the filter set, then lim 0 pkk Proof. Let ()1 max,max, mk ll , 0 0( )1 kr kkkrkrk t rmk We consider the following two cases: Case 1: 1, k l k om the resultk sufficiently la then frwe know for krge, s of Lemma 3, lim 0 kk a subset Case 2:ue, define as follows: the first element of is the fiex w if Case 1 is not tr1 k1 rst ind hich satisfies k , and the next element k is thhie index w, jk ch satisfies jk m Ca . From the definition of the filter set, we can deduce that: 1 , kkk ll k From the proof of the Lema 3, se 2, we know that 1 0, kk lim . For kkj , it is obviously jk 0 . Thus, the Lemma is proven. Theorem 1: Under the assumptions A1A4, if the al an accumulation point which is a KKToint. Proof. We discuss it in two cases: se 1: theree iterations. gorithm doesn’t terminate finitely,then there must exit p 1) Ca are infinite htyp From Lemma 8 we know lim 0 kk , from the up date mechanism of the filter set, there must exit a subset S, 1, kkk kS . Without loss of generality,we can assume that **kSk k lim(,)(, ) txt from which we know ** (,) tpoint. If, suppose, by contradict is a feasible ion , ** (,) t point , then from Lemmas 6 and 7, we know that if: is not a KKT kk k 1 1/2 21/24 ˆ ˆmin ,:M . (15) ns foried and sfully Then all the cond itio ftype step are satisf the inner iteration terminates succes. Because 0 kk , the upper bound will be 1/2 21/2 greater than twice the lower bound . Initially a value ˆ() kk min is chosen, succes halving sively will evcate a value in the in te entually (a) lo rval (15), or (b) locate a value to the right of this inter val. It is obviously that ()0pred s u k k om the of s, if (b) is true, note that nder case (a). Frptimality o creases, sowe ()preds is nondecreasing if k know in () 0 k preds , which means a ftyp e iteration will occur, and it is a contradiction. 2) Case 2: there are only iterations. That means for kK sufficiently large, no filter en tries are made and finite htype () () lT kkk rared spred () l k rared s ks h 1/2 k . Similar to the proof of Lmma 3 we can deduce: e 1() ma k Km K l1 min(, )1/2 xkrM k rrK rK l k r yh Because is bo unded below, we know that min(, ) 1() max kkrMT kKmKrKr rr rK ll pred {} k l () ()0, 0, . kk k k k predspred sh k T Without loss of generality,we can assume that Copyright © 2011 SciRes. AM
Q. ZHAO ET AL. Copyright © 2011 SciRes. AM 1377 476. doi:10.1007/BF00939377 ** lim(,)(, ) kk txt from which we know ** (,) t ** (,) is a se, on the contrary, feasible point.If, suppo t is not a KKT point .Note that for ,kK kK from Lemma know that 6 and Lemma 7 we when 1 1/2 21/24 ˆ ˆmin ,: kk K M then te l [6] Y. Xue, “A SQP Method for Minimax Problems,” in Chinese, Journal of System Science and Math Science, Vol. 22, No. 3, 2002, pp. 355364. [7] R. Fletcher and S. Leyffer, “Nonlinear Programming without a Penalty Function,” Mathematical Programming, Vol. 91, No. 2, 2002, pp. 239269. doi:10.1007/s101070100244 he iteration is ftype and terminates successfully. Because the right side is a constant, and theft side tends to zero, the upper bound will be greater than twice the lower bound 1/2 2 ˆk en in [8] R. Fletcher, N. I. M. Gould, S. Leyffer, P. L. Toint and A. Wächter, “Global Convergence of a TrustRegion SQP Filter Algorithm for General Nonli SIAM Journal on Optimizationear Programming,” n, Vol. 13, No. 3, 2002, pp. 1/2 k , and wh ner iteration terminates we must have 635659. doi:10.1137/S1052623499357258 [9] R. Fletcher, S. Leyffer and P. L. Toint, “On the Global Convergence of a FilterSQP Algorithm,” SIAM Journal on Optimization, Vol. 13, No. 1, 2002, pp. 4459. doi:10.1137/S105262340038081X 1ˆ min, 2 min so 11 ˆ () min, 32 min k pred s ,this is a contradiction [10] A. Wächter and L. T. Biegler, “Line Search Filter Meth ods for Nonlinear Programming: Motivation and Global Convergence,” SIAM Journal on Optimization, Vo No. 1, 2005, pp. 131. Th 6. References C. Charalambous and A. R. C to Solve the Minimax Problem on Numerical Analysis, Vol. 15, No. 1, 1978, pp. 162 187. doi:10.1137/0715011 l. 16, 1052623403426556doi:10.1137/S to () ()predspred sus, the theorem 0. T kkk kh is proven. [11] A. Wächter and L. T. Biegler, “Line Search Filter Meth ods for Nonlinear Programming: Local Convergence,” SIAM Journal on Optimization, Vol. 16, No. 1, 2005, pp. 3248. doi:10.1137/S1052623403426544 [1] onn, “An Efficient Method Directly,” SIAM Journal [12] R. Fletcher, S. Leyffer and P. L. Toint, “A Brief History of Filter Methods,” SIAG/OPT Views and News, Vol. 18, No. 1, 2006, pp. 212. [13] L. P. Huang, “A Filter Method for Minim di, “New Minmax Algorithm,” Journal of Optimi zation Theory and Applications, Vol. 75, No. 3, 1992, pp. 0.1007/BF00940496 ax Problems,” erlinear Convergence of Trust Re . Pu, “A Nonmonotone Filter Trust Re [2] A. Varin Chinese, Post Graduate Thesis, Suzhou University, Suzhou, 2009. [14] S. Ulbrich, “On the Sup 613634. doi:1 ] Z. B. Zhu, “An Improved SQP Algorithm for Solving gion SQPFilter Algorithm,” Mathmatical Programming, Vol. 100, Series B, 2004, pp. 217245. [15] K. Su and D. G [3 Minimax problems,” Applied Mathematics Letters, Vol. 22, No. 1, 2009, pp. 464469. doi:10.1016/j.aml.2008.06.017 [4] Y. H. Yu and L. Gao, “Nongion Method for Nonlinear Constrained Optimization,” Journal of Computational and Applied Mathematics, Vol. 22, No. 1, 2009, pp. 230239. monotone Line Search Algo rithm for Constrained Minimax Problems,” Journal of Optimization Theory and Applications, Vol. 115, No. 2, 2002, pp. 419446. doi:10.1023/A:1020896407415 doi:10.1016/j.cam.2008.01.013 [16] R. Fletcher, S. Leyffer and C. G. Shen, “Nonmonotone Filter Method for Nonlinear Optimization,” Argonne Na tional Laboratory, Lemont, 2009 [5] J. L. Zhou and A. L. Tits, “Nonmonotone Line Search Method for Minimax Problems,” Journal of Optimization Theory and Applications, Vol. 76, No. 3, 1993, pp. 455 .
