Abstract. A complementarity problem with a continuous mapping f from

Transcription

1 Homotopy Continuation Methods for Nonlinear Complementarity Problems Masakazu Kojima y Nimrod Megiddo z Toshihito Noma x January 1989 Abstract. A complementarity problem with a continuous mapping f from the n-dimensional Euclidean space R n into itself can be written as the system of equations F (x y) =0 and (x y) 0: Here F is the mapping from R 2n into itself dened by F (x y) =(x 1 y 1 x 2 y 2 x n y n y ; f(x)) for every (x y) 0: Under the assumption that the mapping f is a P 0 -function, we study various aspects of homotopy continuation methods that trace a trajectory consisting of solutions of the family of systems of equations F (x y) =t(a b) and (x y) 0 until the parameter t 0 attains 0. Here (a b) denotes a 2n-dimensional constant positive vector. We establish the existence of a trajectory which leads to a solution of the problem, and then present a numerical method for tracing the trajectory. We also discuss the global and local convergence of the method. Parts of this research were done while the rst author was visiting at the IBM Almaden Research Center. Partial support from the Oce of Naval Research under Contract N C-0820 is acknowledged. y Department of Information Sciences, Tokyo Institute of Technology, Oh-Okayama, Meguro, Tokyo 152, Japan. Supported by Grant-in-Aids for Co-operative Research ( ) of The Ministry of Education, Science and Culture z IBM Research Division, Almaden Research Center, San Jose, CA , USA, and School of Mathematical Sciences, Tel Aviv University, Tel Aviv, Israel. x Department of Systems Science, Tokyo Institute of Technology, Oh-Okayama, Meguro, Tokyo 152, Japan. 1

2 1. Introduction Let R n denote the n-dimensional Euclidean space. We use the notation R n for the nonnegative orthant fx 2 R n : x 0g and R n for the positive orthant fx 2 R n : x > 0g of R n. The complementarity problem CP[f] with respect to a continuous mapping f : R n! R n (see, for example, Cottle [1], Karamardian [6], Kojima[7], Lemke and Howson [15], etc.) is dened to be the problem of nding a z 2 R 2n such that z = (x y) 0 y = f(x) and x i y i =0 (i =1 2 n): Under the nonnegativity condition z =(x y) 0, thecomplementarity condition x i y i =0(i =1 2 n) can be rewritten as the condition that the inner product x y = x T y is equal to zero. We say that the CP[f] islinear if the mapping f is a linear mapping of the form f(x) =Mx q for some n n matrix M and q 2 R n, and nonlinear otherwise. A feasible solution is a z =(x y) 2 R 2n satisfying the nonnegativity condition z =(x y) 0 and the equality y = f(x). To distinguish a solution of the CP[f] from a feasible solution, we often call a solution of the CP[f] acomplementary solution. We use the symbols S [f] for the set of all the feasible solutions, and S [f] for the set of all the strictly positive feasible solutions: S [f] = fz =(x y) 2 R 2n : y = f(x)g S [f] = fz =(x y) 2 S [f] : (x y) > 0g: This paper studies homotopy continuation methods for nonlinear complementarity problems, which were originally developed for linear programs (Gonzaga [3], Kojima, Mizuno and Yoshise [12], Monteiro and Adler [20], Renegar [23], Vaidya [26], etc.), and then extended to linear complementarity problems (Kojima, Mizuno and Yoshise [13], Megiddo [17]) and nonlinear complementarity problems (Kojima, Mizuno and Noma [10 11]). See also Jarre [5], Mehrotra and Sun [18], Monteiro and Adler [21], Ye [27] for extensions to quadratic programs. A common basic idea of the algorithms in this class is tracing the path of centers (or analytic centers) of polytopes which leads to solutions. This idea was proposed by Sonnevend [25], and the rst polynomial time algorithm in this class was given by Renegar [23]. We also refer to Megiddo [17] who generalized the idea to linear complementarity problems and, in particular, to linear programs in the primal-dual setting. We describe an outline of the homotopy continuation method for the CP[f]. Let X = diag x denote the n n diagonal matrix with the coordinates of a vector x 2 R n. Dene the mapping F from R 2n into R n R n by F (z) =(Xy y ; f(x)) for every z =(x y) 0 (1) to rewrite the CP[f] into the system of equations: F (z) =0 and z =(x y) 0: (2) 2

3 Let c =(a b) 2 R n R n. Consider the family of systems of equations with a nonnegative real parameter t: F (z) =tc and z =(x y) 0: (3) Obviously, the system (3) with the parameter t = 0 coincides with the system (2) or the CP[f]. Let C = ftc : t>0g: Under certain assumptions, the system (3) has a unique solution z(t) for each positive t such that z(t) iscontinuous in the parameter t hence the set F ;1 (C) =fz 2 R 2n : F (z) =tc for some t>0g = fz(t) :t>0g forms a trajectory, a one-dimensional curve. Furthermore, z(t) leads to a solution of the system (2) as t tends to 0. Suppose that we knowapoint z(t 1 ) 2 F ;1 (C) inadvance. Thus, if we start from the known point z(t 1 ) and trace the trajectory F ;1 (C) until the parameter t attains zero, we get a solution of the system (2) or a complementary solution of the CP[f]. There are several questions arising from the continuation method described above. Theoretically, wehave to establish the existence of a trajectory consisting of solutions of the system (3). We have to study the limiting behavior of the trajectory as the parameter t approaches zero. In addition, we need to show how to prepare an initial point z(t 1 ) 2 F ;1 (C) aswell as how to trace the trajectory F ;1 (C) numerically. Global and local convergence of the method should be discussed too. These questions have been answered partially for some special cases. We rst consider the case where f is a linear mapping, f(x) =Mx q, witha positive semi-denite matrix M, i.e., x (Mx) 0 for every x 2 R n. As a special case of (3), consider the family of systems of equations with a nonnegative real parameter t: F (z) =t(e 0) and z =(x y) 0: (4) Here e =(1 1) 2 R n. Suppose that the set S [f] of all the strictly positive feasible solutions of the CP[f] is nonempty. Then the set of the solutions z of the system (4) with the positive t's forms a trajectory fz(t) :t>0g which converges to a complementary solution of the CP[f] ast tends to zero (Megiddo [17]). In this case the trajectory can be regarded as a generalization of the path of centers of a system of linear inequalities. The algorithm given by Kojima, Mizuno and Yoshise [13] computes a complementary solution of the CP[f] by tracing the path of centers numerically. The system (4) can be rewritten as y = Mx q Xy = te and z =(x y) 0: Hence the solution z(t) of the system (4) is restricted to running in the relative interior S [f] of the xed feasible region S [f]. This lacks the exibility inchoosing an initial 3

4 point. Theoretically, we can construct an articial problem which has an initial point z 0 2 S [f] suciently close to the path of centers (See Section 6 of [13]). However the magnitude of such a theoretical initial point is too large for implementation on computers (Lustig [16], Mizuno, Yoshise and Kikuchi [19]). The family of systems of equations (3) gives us more freedom in choosing initial points than the family (4). We consider now more general nonlinear cases. If the system (3) has a solution for every t>0, we must have tb 2 B [f] fu 2 R n : u = y ; f(x) for some (x y) > 0g for every t>0. Hence it is necessary to take avector b 2 R n such that tb 2 B [f] for every t > 0. When we describe a numerical method in Section 5, we will assume b > 0 to meet this necessary condition. In fact, by the denition of the sets S [f] andb [f], we see that if the CP[f] has a strictly positive feasible solution, i.e., S [f] 6=, then there exists an (~x ~y) > 0 such that0 = ~y ; f(~x) hence R n fu 2 Rn : u = y ; f(~x) for some y 2 Rn gb [f] : This ensures that tb 2 B [f] for every t>0. It should be noted that even in this case, we can start from any x 1 2 R n in the x-space by taking appropriate y 1 2 R n and c =(a b) 2 R n such thata =(x 1 1y 1 1 x 1 2y 1 2 x 1 ny 1 n)andb = y 1 ; f(x 1 ) > 0. This exibility in choosing initial points is very important especially when we apply the continuation method with the use of the family (3) to nonlinear problems where nding a feasible solution is generally as dicult as solving them. Kojima, Mizuno and Noma [10 11] presented two conditions to ensure the existence of a trajectory consisting of solutions of the family of systems of equations (3): Condition 1.1. (Kojima, Mizuno and Noma [10]) The mapping f is a uniform P -function, i.e., there exists a positive number such that max i (x i ; y i )(f i (x) ; f i (y)) kx ; yk 2 for every x y 2 R n : Condition 1.2. (Kojima, Mizuno and Noma [11]) (i) The mapping f is a monotone function, i.e., (x ; y) (f(x) ; f(y)) 0 for every x y 2 R n : (ii) The set S [f] of all the strictly positive feasible solutions of the CP[f] is nonempty. 4

5 Remark 1.3. When f is a linear mapping from R n into itself with an nn matrix M,it is monotone if and only if M is a positive semi-denite matrix, and a uniform P -function if and only if M is a P -matrix, i.e., all the principal minors of M are positive (Fiedler and Ptak [2]). It is well-known that the Karush-Kuhn-Tucker optimality conditions for linear and convex quadratic programs can be formulated as a positive semi-denite linear complementarity problem. Conditions 1.1 and 1.2 above can be unied as follows: Lemma 1.4. If Condition 1.1 or 1.2 holds then Condition 1.5 does. Condition 1.5. (i) f is a P 0 -function, i.e., for every x y 2 R n with x 6= y, there is an index i such that x i ; y i 6=0 and (x i ; y i )(f i (x) ; f i (y)) 0: (ii) The set S [f] of all the strictly positive feasible solutions is nonempty. (iii) The set F ;1 (D) =fz =(x y) 2 R 2n : F (z) 2 Dg is bounded for every compact (i.e., bounded and closed as a subset of R 2n ) subset D of R n B [f]. Remark 1.6. As we have already seen, R n B [f] follows from (ii) of Condition 1.5. Hence, Condition 1.5 implies that the set F ;1 (D) =fz =(x y) 2 R 2n : F (z) 2 Dg is bounded for every bounded subset D of R 2n.We will often use this fact later. After listing in Section 2 some symbols and notation, which will be used throughout this paper, we give a proof of Lemma 1.4 in Section 3. In Section 4 we show some basic properties of the mapping F : R 2n! Rn Rn dened by (1). Specically, we establish under Condition 1.5 the existence of the unique trajectory F ;1 (C) leading to solutions of the CP[f]. This result is an extension of the results given by Kojima, Mizuno and Noma [10 11]. In Section 5 we present an algorithm for tracing the trajectory. This algorithm is a modication and extension of the primal-dual algorithm given by Kojima, Mizuno and Yoshise [12] for linear programs. See also the papers [ ]. Although we can apply the algorithm to linear complementarity problems, the main emphasis will be placed on nonlinear cases. In Section 6 we show the global convergence property of the algorithm. In Section 7 we discuss the local convergence property of the algorithm under a nondegeneracy condition, and present a modied algorithm which has a locally quadratic convergence property. 5

6 2. Symbols and Notation R n : the n-dimensional Euclidean space. R n = fx 2 R n : x 0g : the nonnegative orthant ofr n. R n = fx 2 Rn : x > 0g : the positive orthant ofr n. e =(1 1) 2 R n. f : a continuous mapping from R n into itself. X = diag x : the n n diagonal matrix with the coordinates of a vector x 2 R n. F (z) = (Xy y ; f(x)) for every z =(x y) 2 R 2n. CP[f] (the complementarity problem) : Find a z =(x y) 2 R 2n such that F (z) =0 z 0: S [f] = fz =(x y) 2 R 2n : y = f(x)g : the feasible region of the CP[f]. S [f] = fz =(x y) 2 R 2n : y = f(x)g. B [f] = fu 2 R n : u = y ; f(x) for some z =(x y) > 0g. c =(a b) 2 R n B [f]. It will be assumed that c 2 R 2n and kck =1. C = ftc : t>0g. U R 2n [f0g : a closed convex cone whose interior contains the half-line C. U 1 = fu 2 U : c u =1g. U(t) = fu 2 U : c u tg (t 0). 2 (0 1), 2 (0 1) : constant scalars such that(1) <1. 3. Proof of Lemma 1.4 First we deal with the case where the mapping f satises Condition 1.1. The statement (i) of Condition 1.5 directly follows from the denition of a uniform P -function. It has been shown in [10] that F maps R 2n onto Rn Rn homeomorphically. In particular there exists an (x y) 2 R 2n such that F (x y) =(e 0). Recall that e =(1 1) 2 R n. Since x i y i =1 (i =1 2 n), we have (x y) 2 R 2n hence (x y) 2 S [f]. Thus we have shown (ii) of Condition 1.5, i.e., S [f] 6=, or equivalently, 0 2 B [f]. By a similar argument, we can easily show thatb [f] =R n.ifdis a compact subset of R n Rn, then F ;1 (D) is also compact since F is a homeomorphism. Thus, (iii) of Condition 1.5 is satised. Now we consider the case where the mapping f satises Condition 1.2. The statements (i) and (ii) of Condition 1.5 follow immediately. To show (iii) of Condition 1.5, assume, on the contrary, that the set E = F ;1 (D) =f(x y) 2 R 2n : F (x y) 2 Dg 6

7 is unbounded for some compact subset D of R n B [f]. Then we can take a sequence of points f(x k y k ) 2 E : k =1 2 g such that and lim k!1 k(xk y k )k = 1 lim k!1 (yk ; f(x k )) = u for some u 2 B [f]: Since B [f] isanopensubsetofr n,we can nd a ~u 2 B [f] such that y k ; f(x k ) ~u for every suciently large k. By the denition of the set B [f], there is an (~x ~y) 2 R 2n satisfying ~y ; f(~x) =~u. Furthermore, (X k y k y k ; f(x k )) lies in the bounded set D for every k, where X k = diag x k.sowe can nd positive numbers and such that x k y k and ~x (y k ; f(x k ) ; ~u ~y) (k =1 2 ): Hence, for every suciently large k, we have 0 (x k ; ~x) (f(x k ) ; f(~x)) (by (i) of Condition 1.2) = (x k ; ~x) (y k ; (y k ; f(x k ) ; ~u ~y)) (by ~y ; ~u = f(~x) ) = x k y k ; ~x y k ; x k (y k ; f(x k ) ; ~u ~y)~x (y k ; f(x k ) ; ~u ~y) x k y k ; ~x y k ; x k ~y ~x (y k ; f(x k ) ; ~u ~y) ; ~x y k ; x k ~y : Thus we have obtained (since x k 0 and y k ; f(x k ) ; ~u 0 ) ~x y k x k ~y for every suciently large k. Since (x k y k ) 2 R 2n (k =1 2 ), the inequality above ensures that the bounded set f(x y) 2 R 2n : ~x y x ~y g contains (x k y k ) for every suciently large k. But this contradicts the fact that lim k!1 k(x k y k )k = 1. This completes the proof of Lemma The Existence of the Trajectory F ;1 (C) In the remainder of the paper we assume Condition 1.5. The main assertion of this section is Theorem 4.4, which establishes that the set F ;1 (C) consisting of the solutions of the system (3) for all positive t forms a trajectory leading to solutions of the CP[f]. This result will give a theoretical basis to the homotopy continuation method described 7

8 in Section 5. To prove the theorem, we need three lemmas. The rst two lemmas ensure the existence and uniqueness of a solution of the system of equations for every (a b) 2 R n B [f], where F (z) =(a b) and z =(x y) 2 R 2n (5) B [f] =fu 2 R n : u = y ; f(x) for some (x y) 2 R 2n g: Lemma 4.1. The mapping F is one-to-one on R 2n. Proof: Assume on the contrary that F (x 1 y 1 )=F (x 2 y 2 ) for some distinct (x 1 y 1 ) (x 2 y 2 ) 2 R 2n. Then f(x 1 ) ; f(x 2 )=y 1 ; y 2 and x 1 i y1 i = x2 i y2 i > 0 (i =1 2 n): Since the mapping f is a P 0 -function, we can nd an index k such that x 1 k 6= x 2 k and 0 (x 1 k ; x 2 k)(f k (x 1 ) ; f k (x 2 )) = (x 1 k ; x 2 k)(y 1 k ; y 2 k): We may assume without loss of generality thatx 1 k >x2 k. Then the inequality above implies that yk 1 yk. 2 This contradicts the equality x 1 kyk 1 = x 2 kyk 2 > 0. Lemma 4.2. The system (5) has a solution for every (a b) 2 R n B [f]. Proof: Let (a b) 2 R n B [f]. It follows from b 2 B [f] that^y ; f(^x) =b for some (^x ^y) 2 R 2n. Let ^a =(^x 1^y 1 ^x 2^y 2 ^x n^y n ) 2 R n. Nowwe consider the family of systems of equations with the parameter t 2 [0 1]: F (x y) = ((1 ; t)^a ta b) and (x y) 2 R 2n : (6) Let t 1 be the supremum of ^t's such that the system (6) has a solution for every t 2 [0 ^t ]. Then there exists a sequence f(x k y k t k )g of solutions of the system (6) such that lim k!1 t k = t. Since the right-hand side ((1 ; t)^a ta b) of the system (6) lies in the compact convex subset D = f((1 ; t)^a ta b) :t2 [0 1]g of R n B [f] for all t 2 [0 1], (iii) of Condition 1.5 ensures that the sequence f(x k y k )g is bounded. Hence we may assume that it converges to some (x y). By the continuity of the mapping F, the point (x y t) satises the system (6). Hence if t = 1 then the desired result follows. Assume on the contrary that t <1. Then we have x i y i =(1; t)^a i ta i > 0 for every i =1 2 n. Hence (x y) 2 R 2n. It follows from Lemma 4.1 that the mapping F is a local homeomorphism at (x y). (See the domain invariance theorem in Schwartz [24].) Hence the system (6) has a solution for every t suciently close to t. This contradicts the denition of t. 8

9 Lemma 4.3. (i) R n B [f]. (ii) F maps R 2n onto R n B [f] homeomorphically. Proof: As we have already seen in Section 1, the assertion (i) follows from (ii) of Condition 1.5. We will show the assertion (ii). By the denition, we immediately see F (R 2n ) R n B [f], and R n B [f] F (R 2n ) by Lemma 4.2. Hence F maps R 2n onto Rn B [f]. By Lemma 4.1, the continuous mapping F is one-to-one on the open subset R 2n of R2n.Thus (ii) follows from the domain invariance theorem (see Schwartz [24]). We remark here that if a continuous mapping f satises Condition 1.1 then F maps R 2n onto R n R n homeomorphically (Kojima, Mizuno and Noma [10]). Now we are ready to establish the existence of the trajectory F ;1 (C) consisting of solutions of the system (3) for all positive t. Theorem 4.4. Let c =(a b) 2 R n Rn, and C = ftc : t>0g. (i) For every t>0, the system (3) has a unique solution z(t), which is continuous in t hence the set F ;1 (C) =fz(t) :t>0g forms a trajectory. (ii) For every t 0 > 0, the subtrajectory fz(t) :0<t<t 0 g is bounded hence there isat least one limiting point of z(t) as t! 0. (iii) Every limiting point of z(t) as t! 0 isacomplementary solution of the CP[f]. (iv) If f is a linear mapping of the form f(x) =Mx q, then z(t) converges to a solution of the CP[f] as t! 0. Proof: By (i) of Lemma 4.3, we rst observe that tc 2 C R n R n R n B [f] for every t>0. Hence the assertion (i) follows from (ii) of Lemma 4.3. If we take D = ftc :0<t<t 0 g,we see by Remark 1.6 that the set F ;1 (D) =fz 2 R 2n : F (z) 2 Dg = fz(t) :0<t<t 0 g is bounded. Thus we obtain (ii). By the continuity of the mapping F,ifz is a limiting point ofz(t) ast! 0, we have F (z) =0 and z 0 hence z is a complementary solution of the CP[f]. Thus we have shown (iii). Finally, to see the assertion (iv), we will utilize some result on real algebraic varieties. We call a subset V of R m areal algebraic variety if there exist a nite number of polynomials g i (i =1 2 k)such that V = fx 2 R m : g i (x) =0(i =1 2 k)g: 9

10 We know that a real algebraic variety has a triangulation (see, for example, Hironaka [4]). That is, it is homeomorphic to a locally nite simplicial complex. Let V = f(x y t) 2 R 2n1 : y = Mx q tb x i y i = ta i (i =1 2 n)g: Obviously, the set V is a real algebraic variety, so it has a triangulation. Let z =(x y) be a limiting point ofz(t) ast! 0. Then the point v =(x y 0) lies in V. Since the triangulation of V is locally nite, we can nd a sequence ft p > 0g andasubset of V which is homeomorphic to a one-dimensional simplex such that lim p!1 tp =0 p!1 lim z(t p )=z z(t p ) 2 (p =1 2 ): But we know that V \R 2n1 coincides with the one-dimensional curve f(z(t) t):t>0g. Thus, the subset f(z(t) t):t p1 t t p g of the curve must be contained in the set for every p, since otherwise is not arcwise connected. This ensures that z(t)converges to z as t! 0. Remark 4.5. One of the referees suggested another proof of the assertion (iv) of the theorem using the well-known result that every real algebraic variety contains only nitely many connected components. Indeed, for every >0, the algebraic variety V \f(x y t) 2 R 2n1 : k(x y t) ; vk 2 2 g has nitely many connected components. It follows that z(t) must be within distance of v for all suciently small t>0. 5. A Numerical Method for Tracing the Trajectory F ;1 (C) In the previous section, we have shown the existence of the trajectory F ;1 (C) leading to solutions of the CP[f] forevery c 2 R R and C = ftc : t>0g. In general, the trajectory F ;1 (C) is nonlinear, so that exact tracing is dicult even if we knowan initial point on the trajectory. Of course, exact tracing is not necessary since our aim is only to get an approximate solution of the CP[f]. We willcontrol the distance from the trajectory in such away thatkf (z) ; tck tends to zero as the right-hand side of the system (3) tends to zero along the half-line C = ftc 2 R n : t>0g. For this purpose, we will introduce a \cone-neighborhood" U of the half-line C. Todevelop a numerical method that traces the trajectory F ;1 (C), we further assume in the remainder of the paper: Condition 5.1. (i) The mapping f associated with the CP[f] is continuously dierentiable. (ii) c =(a b) 2 R 2n and kck = 1 hence the half-line C = ftc : t>0g lies in R2n. (iii) U R 2n [f0g is a closed convex cone whose interior int U contains the half-line C. (iv) We know a point z 1 =(x 1 y 1 ) such thatf (z 1 ) 2 U in advance. 10

11 It is always possible to choose c =(a b), U and (x 1 y 1 ) satisfying (ii), (iii) and (iv) of Condition 5.1. For example, choose x 1 > 0, y 1 > 0 and b 0 > 0 such thatb 0 = y 1 ;f(x 1 ). Let a 0 =(x 1 1y 1 1 x 1 2y 1 2 x 1 ny 1 n). Let c =(a 0 b 0 )=k(a 0 b 0 )k, and be a positive number such that c i >(i =1 2 2n). Dene U = fu : ku ; tck t for some t 0g. Then the set of c =(a b), U and (x 1 y 1 ) satises Condition 5.1. The lemma below shows some properties of the neighborhood U of the half-line C, which will be utilized in the succeeding discussions. Lemma 5.2. (i) The set U 1 = fu 2 U : c u =1g is bounded. (ii) There exists a positive number such that ku ; (cu)ck (cu) for each u 2 U. (iii) There isapositive number such that if ku ; tck t for some t>0 then u 2 int U. Proof: (i) One can easily see that the set fu 2 R 2n : c u =1g, which contains the set U 1, is bounded because c 2 R 2n. (ii) Since the set U 1 = fu 2 U : c u =1g is bounded, there is a positive number such that the ball B = fu 2 R 2n : ku ; ck g contains the set U 1. Let u 2 U. Obviously c u 0 because c u 2 R 2n.Ifcu > 0 then the point u=(c u) belongs to the set U 1 B hence the inequality ku ; (c u)ck (c u) follows. Now suppose that u 2 U and c u =0. Thenu = 0 because c 2 R 2n and u 2 R 2n. Hence the inequality above holds trivially. (iii) By Condition 5.1, the point c lies in the interior int U of the cone U. Hence, we can nd a positive number such thatintu contains the ball B 0 = fu 2 R 2n : ku ; ck g. Suppose that ku ; tck t for some t>0. Then u=t belongs to the ball B 0. Since U isacone,wehave u 2 int U. The set F ;1 (U) will serve as an admissible region in whichwe will generate a sequence fz k 2 R 2n g, toapproximate the trajectory F ;1 (C), such that lim k!1 c F (z k )=0. Such a sequence fz k g leads to complementary solutions of the CP[f] aswe will see in the theorem below. Theorem 5.3. Suppose fz k 2 F ;1 (U)g is a sequence such that lim k!1 c F (z k )=0. Then the sequence fz k g is bounded and any limiting point of the sequence isacomplementary solution of the CP[f]. 11

12 Proof: Since F (z k ) 2 U (k =1 2 ) holds from the assumption, we seeby Lemma 5.2 that kf (z k ) ; (c F (z k ))ck (c F (z k )) (k =1 2 ): Hence lim k!1 F (z k )=0. Furthermore, the sequence ff (z k )g is bounded. By Remark 1.6, so is the sequence fz k g. Therefore we see from the continuity of the mapping F that F (z) =0forany limiting point z of the sequence fz k g. Assuming that we are at some point z 2 F ;1 (U) \ R 2n,i.e.,z 2 F ;1 (U ;f0g), we show how to generate a new point z 2 F ;1 (U) such that c F (z) < c F (z). This process corresponds to one iteration of the algorithm described below. Let 2 (0 1) and >0 be xed such that (1 ) <1: (7) Let 2 (0 ], and t = c F (z) > 0. We apply a Newton iteration with a step length 2 (0 1] to the system of equations: F (z) =tc at the point z. That is, we solvethe Newton equation, the system of linear equations in the variable vector z DF (z)z = F (z) ; tc: (8) Here DF(z) denotes the Jacobian matrix of the mapping F at z. We callz the Newton direction. The step length will be determined later by an inexact line search such that c F (z ; z) ((1 ; )(1 ))t: Thus we dene a new point z 2 R 2n by z = z ;z: The lemma below ensures that the 2n 2n coecient matrix on the left hand side of the system (8) is nonsingular whenever (x y) 2 R 2n. Hence the system (8) consistently and uniquely determines the Newton direction z = DF(z) ;1 (F (z) ; tc). Lemma 5.4. (i) The Jacobian matrix Df(x) is a P 0 -matrix at every x 2 R n, i.e., for every nonzero u 2 R n, there is an index i such that u i 6=0 and u i [Df(x)u] i 0: (ii) The Jacobian matrix DF(z) is nonsingular at every z =(x y) 2 R 2n. Proof: (i) Let x 2 R n and 0 6= u 2 R n.we consider a sequence fx (1=k)u : k = 1 2 g.for every k =1 2 there is an index i such that 1 k u i 6= 0 and 1 k u i(f i (x 1 k u) ; f i(x)) 0: 12

13 Since the index set f1 2 ng is nite, we can nd an index i such that the relation above holds for this i and innitely many k's. For such i and k, wehave u i 6=0 and u i [Df(x)u] i o( 1 k )=1 k 0: Here o(h)=h! 0ash! 0. Taking the limit as k!1,we obtain u i 6=0 and u i [Df(x)u] i 0: (ii) Let z =(x y) 2 R 2n. The Jacobian matrix DF(z) is written as DF (z) = Y ;Df(x)! X I where X = diag x, Y = diag y, andi stands for the n n identity matrix. To see that the matrix DF (z) is nonsingular, assume on the contrary that! u DF (z) = 0 v or Yu Xv = 0 and ; Df(x)u v = 0 for some nonzero (u v) 2 R 2n. It follows that u 6= 0 and Df(x)u = ;X ;1 Yu: : Hence This contradicts (i). u 6= 0 and u i [Df(x)u] i = ; y iu 2 i x i (i =1 2 n): Remark 5.5. From (ii) of Lemma 4.3 and (ii) of Lemma 5.4, we see that F maps R 2n onto R n B [f] dieomorphically. Recall that 2 (0 1) and >0 are constants satisfying (7), and that z is a unique solution of the Newton equation (8) at z 2 F ;1 (U) with the parameter 2 (0 ]. Assume for the time being that z = z ; z 2 F ;1 (U) for every suciently small nonnegative. Ideally, wewant tochoose the step length ^ such that F (^z) = F (z ; ^z) 2 U c F (^z) = min fc F (z ; z) : 2 [0 1]g: 13

14 However, the computation of the exact value of the ideal step length ^ is generally impossible in a nite number of steps. In the algorithm presented below, we will use an inexact line search: Find the smallest nonnegative integer ` such that F (z ; `z) 2 U (9) c F (z ; `z) (1 ; `) `(1 ) t: (10) Here 2 (0 1) denotes a constant and ` stands for the `'th power of. Now we are ready to describe the algorithm. ALG1[U ]. Step 0. Let t 1 = c F (z 1 )andk =1. Step 1. Let z = z k and t = t k. Step 2. Compute the direction z by solving the Newton equation (8). Step 3. Let ` be the smallest nonnegative integer satisfying (9) and (10). Dene Step 4. Replace k by k 1. Go to Step 1. z k1 = z ; `z and t k1 = c F (z k1 ): If it happens that t k = 0 for some k in the algorithm above, then c F (z k ) = 0 hence F (z k ) = 0. In this case we may stop the algorithm because we have obtained z k as a complementary solution of the CP[f]. So it is implicitly assumed in the algorithm above that t k > 0 for every k. 6. Global and Monotone Convergence Let f(z k t k )g be a sequence generated by the ALG1[U ]. We will show in Theorem 6.2 that lim k!1 t k = 0 hence, by Theorem 5.3, the sequence fz k g is bounded and any limiting point of the sequence is a complementary solution of the CP[f]. For this purpose we prove the lemma below: Lemma 6.1. Suppose that z 2 F ;1 (U) and t = c F (z) > 0. Let 2 (0 1) and >0 be constants satisfying (7), and let z be the solution of the Newton equation (8) at z with the parameter 2 (0 ]. (i) Dene = maxf 2 [0 1] : z ; z 2 R 2n g (11) e() = F (z ; z) ; F (z)df (z)z for every 2 [0 ]: (12) 14

15 Then (ii) Dene Then and lim ke()k= = 0 (13)!0 F (z ; z) = (1 ; )F (z)(tc e()=) for every 2 (0 ] (14) c F (z ; z) (1 ; )t (t ke()k=) for every 2 (0 ]: (15) =supf 0 2 (0 ] : ke()k= < [minf g]t for every 2 (0 0 ]g: (16) for every 2 (0 ). 0 < 1 (17) ke()k= < [minf g]t (18) F (z ; z) 2 int U (19) c F (z ; z) < ((1 ; )(1 )) t (20) Proof: (i) It should be noticed that 2 (0 1]. The relation (13) follows directly from the continuous dierentiability of the mapping F. By the denition, F (z ; z) =F (z) ; DF(z)z e() for every 2 [0 ]: Since z is the solution of the Newton equation (8), we alsohave F (z) ; DF (z)z =(1; )F (z)tc for every 2 [0 ]: Hence the equality (14) follows from these two equalities. Taking the inner product of each side of (14) and the vector c, wehave that c F (z ; z) = (1 ; )c F (z)(tkck 2 c e()=) (1 ; )t (t ke()k=) for every 2 (0 ]. Thus we have shown the inequality (15). (ii) The inequality (17) follows from the denitions (11), (16) of, and the relation (13). The inequality (18) is obvious by the denition (16) of,too. Let 2 (0 ). By Lemma 5.2, the point (tc e()=) lies in int U. Since the point F (z ; z) isa convex combination of the point F (z) in the convex set U and the point (tc e()=) in int U, it lies in int U. Thus we have shown (19). The inequality (20) follows from (15) and (18). Lemma 6.1 guarantees that we can consistently nd the smallest nonnegative integer ` satisfying (9) and (10) at Step 3 of ALG1[U ]. We arenow ready to prove the global and monotone convergence property ofalg1[u ]. 15

16 Theorem 6.2. Let 2 (0 1) and > 0 be constants satisfying (7). Suppose that 2 (0 ] and 2 (0 1). Let f(z k t k )g be asequence generated bythealg1[u ]. (i) The sequence ft k g is monotone decreasing and converges to zero ask!1. (ii) The sequence fz k g is bounded and its limiting points are complementary solutions of the CP[f]. Proof: In view of Theorem 5.3, it suces to prove the assertion (i). By applying Lemma 6.1 at each z = z k,we see that t k >t k1 (k =1 2 ). Hence the sequence ft k g is monotone decreasing. Since each t k is nonnegative, there exists a nonnegative number ~t to which the sequence converges. If ~t = 0,we obtain the desired result. Assume on the contrary that ~t >0. Dene the compact subset V = fu 2 U : ~t c u t 1 g of R 2n (see Lemma 5.2). Then we see by (ii) of Lemma 4.3 that the set F ;1 (V ) which contains the sequence fz k g is a compact subset of R 2n since V R 2n R n B [f]. Taking a subsequence if necessary, wemay assume that the sequence fz k g converges to some ~z 2 F ;1 (V ). Then it is easily seen that ~t = c F (~z). Now, applying Lemma 6.1 to the point ~z, we can nd a positive number ~ such that for every 2 (0 ), ~ F (~z ; z) g 2 int U c F (~z ; z) g < ((1 ; )(1 )) ~t : Here g z denotes the Newton direction determined by the equation (8) with z = ~z and t = ~t. On the other hand, the Jacobian matrix DF (z) is nonsingular and continuous at z = ~z (see (ii) of Lemma 5.4). This implies that the Newton direction generated at the k'th iteration, z k, converges to g z. Therefore, for a nonnegative integer ` such that ` 2 (0 ~ ), we have F (z k ; `z k ) 2 int U and c F (z k ; `z k ) < (1 ; `)`(1 ) t k for every suciently large k. Let `k be the nonnegative integer determined at Step 3 of the k'th iteration in ALG1[U ]. Then, for every suciently large k, we see that `k ` hence t k1 (1 ; `k ) `k (1 ) t k (1 ; `)`(1 ) t k : This contradicts the fact that the sequence ft k g converges to ~t. 16

17 7. Local Convergence We will assume the condition below in addition to Conditions 1.5 and 5.1 throughout this section, which is divided into two subsections, 7.1 and 7.2. Subsection 7.1 is devoted to a locally linear convergence property ofalg1[u ]. In Subsection 7.2 we will modify ALG1[U ] to get a locally quadratic convergence. Condition 7.1. (i) At each complementary solution z =(x y) ofthecp[f], the set of the columns I i (i 2 I (y)) and [Df(x)] j (j 2 I (x)) forms a basis of R n. Here I i denotes the i'th column of the n n identity matrix I, [Df(x)] j the j'th column of the n n Jacobian matrix Df(x) of the mapping f, I (y) = fi : y i > 0g, and I (x) =fj : x j > 0g. (ii) The Jacobian matrix Df(x) ofthemappingf is Lipschitz continuous on each bounded subset E R n, i.e., there is a positive constant such that kdf(x 2 ) ; Df(x 1 )kkx 2 ; x 1 k for every x 1 x 2 2 E where kak denotes the matrix norm max fkawk : w 2 R n kwk =1g for every n n matrix A. We note that (i) of Condition 7.1 implies the strict complementarity, i.e., x i = 0 if and only if y i > 0(i =1 2 n). By using the well-known implicit function theorem, we can also derive the local uniqueness of each solution of the CP[f] from (i) of Condition 7.1. Furthermore we will see in Lemma 7.3 below that the CP[f] has a unique solution. 7.1 Locally Linear Convergence Now we state the locally linear convergence of ALG1[U ]. Theorem 7.2. Let 2 (0 1) and > 0 be constants satisfying (7). Suppose that 2 (0 ] and 2 (0 1). Let f(z k t k )g be asequence generated bythealg1[u ]. Then there isapositive number K such that t k1 (1 )t k for every k K: We will prove a series of lemmas which leads us to Theorem 7.2. Lemma

18 (i) The Jacobian matrix DF(z) of the mapping F is nonsingular at every z =(x y) 2 R 2n [ F ;1 (0). (ii) The CP[f] has a unique solution. (iii) There arepositive constants and such that for every z 2 F ;1 (U(t 1 )) and for every z 1 z 2 2 F ;1 (U(t 1 )). (iv) There is a positive constant such that kdf(z)k (21) kdf (z) ;1 k (22) kf (z 2 ) ; F (z 1 )k kz 2 ; z 1 k (23) kz 2 ; z 1 k kf (z 2 ) ; F (z 1 )k (24) kf (z 2 ) ; F (z 1 ) ; DF(z 1 )(z 2 ; z 1 )kkz 2 ; z 1 k 2 for every z 1 z 2 2 F ;1 (U(t 1 )). Here U(t 1 )=fu 2 U : c u t 1 g. Proof: (i) Recall the denition (1) of the mapping F : R 2n! R n R n. If z 2 R 2n [ F ;1 (0) then we have either or z =(x y) 2 R 2n F (z) =0 z =(x y) 2 R 2n : Note that z is a complementary solution of the CP[f] in the latter case. Wehaveshown in Lemma 5.4 that the Jacobian matrix DF (z) is nonsingular at every z 2 R 2n. Now suppose that F (z) =0 and z 2 R 2n. By (i) of Condition 7.1, we can easily verify that if DF(z)w = 0 for some w 2 R 2n then w = 0 hencedf(z) is nonsingular. (ii) By Theorem 4.4, we know that the CP[f] has a complementary solution. On the other hand, by applying the implicit function theorem (see, for example, Ortega and Rheinboldt [22]) to the system (3) at each complementary solution z of the CP[f] and t =0,we see that the unique trajectory F ;1 (C), whose existence is ensured by Theorem 4.4, converges to z as t! 0. Hence the solution of the CP[f] must be unique. (iii) Let z be the unique solution of the CP[f]. Since U(t 1 ) U R 2n [f0g (R n B [f]) [f0g 18

19 we know F ;1 (U(t 1 )) R 2n [fz g. Noting (ii) of Lemma 4.3 and extending slightly the argument in (ii) above, we can show thatf is a homeomorphism between F ;1 (U(t 1 )) and U(t 1 ). On the other hand, the set U(t 1 ) is compact by Lemma 5.2. Therefore the set F ;1 (U(t 1 )) is compact, too. Let W be the convex hull of the compact set F ;1 (U(t 1 )). Then W is also compact and W R 2n [fz g.wehave seenby (i) above that the Jacobian matrix DF(z) is nonsingular at every z in the set R 2n [fz g.thus, by the continuity of the Jacobian matrix DF (z), there exist positive numbers and such that (21) Holds for every z 2 W and that (22) holds for every z 2 F ;1 (U(t 1 )). Since the sets W and U(t 1 ) are convex, the inequalities (23) and (24) follow from (21), (22) and Theorem of [22]. (iv) It follows from (ii) of Condition 7.1 that the Jacobian matrix DF(z) islipschitz continuous on the bounded convex set W, i.e., there is a positive number such that kdf (z 2 ) ; DF(z 1 )k2kz 2 ; z 1 k for every z 1 z 2 2 W: The assertion (iv) follows from Theorem of [22]. Lemma 7.4. Let z 2 F ;1 (U(t 1 )), t = c F (z) > 0, and let z be the solution of the Newton equation (8) at z with the parameter 2 (0 ]. Dene e() and by (12) and (16), respectively. Then ke()k (( 1; )t) 2 for every 2 [0 ]: (25) Proof: Let 2 [0 ). Since z is the solution of the Newton equation (8), we have kzk = kdf(z) ;1 (F (z) ; tc)k Thus we have seen that kdf(z) ;1 k (kf (z) ; tck (1; )tkck) (kf (z) ; tck (1; )t) (by Lemma 7.3 and kck =1) (t (1; )t) (by Lemma 5.2): kzk ( 1; )t: (26) On the other hand, by the relation (19), (20) in Lemma 6.1 and the continuity, wesee for every 2 [0 ]that F (z ; z) 2 U c F (z ; z) ((1 ; )(1 )) t t t 1 hence z ; z 2 F ;1 (U(t 1 )): 19

20 Therefore, by the denition (12) of e() and (iv) of Lemma 7.3, we have ke()k kzk 2 : The desired inequality follows from (26) and the inequality above. For each t>0 and 2 (0 ], dene For every 2 (0 ], let ^(t ) = min ( ) minf g (( 1; )) 2 t 1 : s() = maxft 0 0:^(t )=1 for every t 2 (0 t 0 ]g = minf g (( 1; )) : (27) 2 Here and are the positive constants that were introduced in Lemma 5.2. Lemma 7.5. Let z 2 F ;1 (U(t 1 )), t = c F (z) > 0, and let z be the solution of the Newton equation (8) at z with the parameter 2 (0 ]. Then for every 2 [0 ^(t )]. F (z ; z) 2 U (28) c F (z ; z) ((1 ; )(1 )) t (29) Proof: Dene e() and by (12) and (16), respectively. If ^(t ) then the relations (28) and (29) follow from (19), (20) in Lemma 6.1 and the continuity. Thus it suces to show that^(t ).Wemay assume <1. Suppose that ke( )k= <[minf g]t: (30) Then from the denition (16) of and the continuity ofe() it follows that =. Just as we derived (19) from (18) in the proof of Lemma 6.1, we see from (30) that F (z ; z) 2 int U: This contradicts the denition (11) of since = <1. Thus we have shown that ke( )k= [minf g]t: By the inequality (25) in Lemma 7.4 and the inequality above, we have hence ^(t ). [minf g]t (( 1; )t) 2 20

21 Lemma 7.6. Let z 2 F ;1 (U(t 1 )), t = c F (z) > 0, and let z be the solution of the Newton equation (8) at z with the parameter 2 (0 ]. Ift s() then F (z ; z) 2 U c F (z ; z) (1 )t: Proof: By the denition (27) of s() and Lemma 7.5, the relations (28) and (29) hold for every 2 [0 1]. We arenow ready to prove Theorem 7.2. We have already shown in Theorem 6.2 that the sequence ft k g converges to zero as k!1. Let K be a positive integer such that t k s() for every k K. Then, by Lemma 7.6, the inequality in the theorem holds for every k K. This completes the proof of Theorem A Modication of ALG1[U ] and Its Locally Quadratic Convergence The integer K in Theorem 7.2 depends on the positive number, but we can take any positive value of 2 (0 ] although K may diverge as tends to zero. This suggests that the sequence ft k g converges to zero at least super-linearly if we suitably decrease the value of the parameter as the iteration proceeds. In fact, we can modify the algorithm so that the sequence converges to zero quadratically. Let z 2 F ;1 (U(t 1 )) and t = c F (z) > 0. In the remainder of the section we denote the solution of the Newton equation (8) at z with the parameter 2 (0 ]by z(). We will be concerned with the following two sets of relations: and F (z ; `z( )) 2 U (31) c F (z ; `z( )) (1 ; `) `(1 ) t (32) F (z ; z( m )) 2 U (33) c F (z ; z( m )) (1 ) m t: (34) Here ` and m represent the `'th power of 2 (0 1) and the m'th power of 2 (0 1), respectively. Nowwe are in position to state a modication of ALG1[U ] whose local convergence will be investigated later. ALG2[U ]. Step 0. Let t 1 = c F (z 1 )andk =1. Step 1. Let z = z k and t = t k. 21

22 Step 2. Let z( ) be the direction determined by the Newton equation (8) with the parameter =.Let ` be the smallest nonnegative integer satisfying the relations (31) and (32). If ` = 0 then go to Step 4. Step 3. Dene z k1 = z ; `z( ): Go to Step 5. Step 4. Let m be the largest positive integer such that the relations (33) and (34) hold for all m =1 2 m. Dene z k1 = z ; z( m ): Step 5. Let t k1 = c F (z k1 ). Replace k by k 1. Go to Step 1. Let f(z k t k )g be a sequence generated by ALG2[U ]. Then, in a way similar to the proof of Theorem 6.2, we can show that the sequence f(z k t k )g converges to (z 0). Here z is the unique solution of the CP[f] (see (ii) of Lemma 7.3). Furthermore, ALG2[U ] has a locally quadratic convergence property aswe will see in Theorem 7.8 below. Lemma 7.7. Let z 2 F ;1 (U(t 1 )) and t = c F (z) > 0. Suppose that t s( ), where s :(0 ]! R is dened by (27). Then the relations (31) and (32) hold for ` =0, i.e., the relations (33) and (34) hold for m =1.Let m be the largest positive integer such that the relations (33) and (34) hold for m =1 2 m. Then c F (z ; z( m )) (( 1))2 (1 ) t 2 : minf g Proof: Let ~m be the largest integer such that t s( m ) for every m =1 2 ~m. Then~m 1 since t s( ). From Lemma 7.6, we seethat the relations (33) and (34) hold for m =1 2 ~m. Hence ~m m. On the other hand, it follows from the denition of ~m that t > s( ~m1 ) = ~m1 minf g ( 1; ~m1 ) 2 or equivalently ~m < ( 1; ~m1 ) 2 minf g t : 22

23 Since ~m m, wehave Therefore we obtain m < (( 1))2 t minf g : c F (z ; z( m )) (1 ) m t < (( 1))2 (1 ) t 2 : minf g Theorem 7.8. Let f(z k t k )g be asequence generated bythealg2[u ]. Then the convergence of the sequence f(z k t k )g to (z 0) is locally quadratic. More precisely, t k1 (( 1))2 (1 ) (t k ) 2 (35) minf g kz k1 ; z k 2 (( 1)) 3 (1 ) kz k ; z k 2 (36) minf g for every suciently large k. Here z is the unique solution of the CP[f]. Proof: As wehave noted above, the sequence f(z k t k )g converges to (z 0) as k!1. Let K 0 be a positive integer such thatt k s( )forevery k K 0. Then, the inequality (35) (for every k K 0 ) follows directly from Lemma 7.7. Furthermore, we have kz k1 ; z k kf (z k1 )k (by (24) and F (z )=0 ) (kf (z k1 ) ; t k1 ck kt k1 ck) (t k1 t k1 ) (by Lemma 5.2 and kck =1) = ( 1)t k1 (( 1))3 (1 ) (t k ) 2 (by (35)) minf g (( 1))3 (1 ) kf (z k )k 2 (since t k = c F (z k ) and kck =1) minf g (( 1))3 (1 ) 2 kz k ; z k 2 (by F (z )=0 and (23)) minf g for every k K 0.Thus we have shown (36). 23

24 8. Concluding Remarks We have formulated the complementarity problem CP[f] as a system of equations with a nonnegativity condition on a variable vector z 2 R m : F (z) =0 and z 0 (37) and proposed a homotopy continuation method, ALG1[U ], founded on a oneparameter family of systems of equations: F (z) =tc and z 0: Here m = 2n. Supposing that the CP[f] satises Condition 1.5 and (i) of Condition 5.1, we have seen that the system above enjoys the following properties: (a) We can choose a c > 0, a closed convex cone U R m [f0g and a point z1 2 F ;1 (U) such that (ii), (iii) and (iv) of Condition 5.1 hold. (b) R m F (R m ). (c) F maps R m onto F (Rm ) dieomorphically. (d) the set F ;1 (D) is bounded for every bounded subset D of R m. Generally, ALG1[U ] computes an approximate solution of the system (37) if all the conditions above are satised. This may remind the readers of applications of ALG1[U ] to some other problems which are converted into systems of the form (37). The algorithms ALG1[U ] andalg2[u ] aswell as their global and local convergence results in this paper can apply to linear complementarity problems satisfying Condition 1.5. We could modify the ALG1[U ] to derive the pathfollowing algorithm ([13 21]) which solves linear complementarity problems with positive semi-denite matrices in O( p nl) iterations. Furthermore we could prove the globally linear convergence of the modied algorithm when it is applied to linear complementarity problems with P -matrices. These results on the positive semi-denite and P -matrix cases, which were presented in the original version of the paper [8] but cut in the revised version, will be further extended to a wider subclass of linear complementarity problems with P 0 -matrices in the paper [9] where we explore a unied approach ([14]) toboththe path-following algorithm ([13 21]) and the potential reduction algorithm ([14]) for linear complementarity problems. Acknowledgment The authors are grateful to Prof. Takuo Fukuda who suggested using the powerful result on real algebraic varieties given in the paper Hironaka [4] for the proof of (iv) of Theorem

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