On Smoothing Methods for the P0 Matrix Linear ... - Semantic Scholar
On Smoothing Methods for the P0 Matrix Linear Complementarity Problem Xiaojun Chen
Yinyu Yey
School of Mathematics Department of Management Sciences The University of New South Wales University of Iowa Sydney 2052, Australia Iowa City, IA, USA E-mail: [email protected] E-mail: [email protected]
5 March 1998
Abstract
In this paper, we propose a Big-M smoothing method for solving the P0 matrix linear complementarity problem. We study the trajectory de ned by the augmented smoothing equations and global convergence of the method under an assumption that the original P0 matrix linear complementarity problem has a solution.
Key words. linear complementarity problem, P0 matrix, smoothing algorithm. AMS subject classi cations. 65H10, 90C30, 90C33.
y
This author's work was supported by the Australian Research Council. This author is supported in part by NSF Grants DMI-9522507 and DMS-9703490.
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2
1 Introduction In this paper we consider the linear complementarity problem (LCP) tT s = 0; s = Mt + q; and t; s 0;
where M is an n n P0 matrix and q is an n dimensional vector. A matrix M 2 Rnn is called a P0 matrix if max t (Mt)i 0; for all t 2 Rn ; t 6= 0: 6 i
i:ti =0
A linear complementarity problem is called a P0 matrix LCP if the matrix M is a P0 matrix. The class of the P0 matrix LCP includes the monotone LCP and the P matrix LCP. The P0 matrix LCP has been studied extensively under additional conditions [5, 11]. A di erentiable function on Rn is called a P0 function if its Jacobian is a P0 matrix at every point in Rn. A nonlinear complementarity problem (NCP) is called a P0 function NCP if the involved function is a P0 function. Kojima, Megiddo and Noma [10] proved the existence of a trajectory in the interior of the feasible set of the P0 function NCP under some additional conditions. Their results in uenced the development of interior point methods and non-interior point methods, and led several continuation methods for solving P0 function NCP. Recently, Facchinei and Kanzow [7] applied regularization methods for solving a continuously di erentiable P0 function NCP under the following assumption. Assumption 1.1 The solution set of the P0 function NCP is nonempty and bounded. This assumption is weaker than that Kojima, Megiddo, Noma and Yoshise used in [10, 11]. Moreover it includes the monotone NCP with an interior point, and the P0 and R0 NCP [5]. After Facchinei-Kanzow's encouraging work, several algorithms and theoretical results on regularization methods for the P0 function NCP have been developed [14, 15, 16] under Assumption 1.1. In particular, Ravindran and Gowda [15] generalized the results of Facchinei and Kanzow [7] to a continuous P0 function variational inequality problem with box constraints. Facchinei and Kanzow [7] gave a counterexample to show that it is not possible to remove the boundedness assumption of the solution set for regularization methods for solving the P0 matrix LCP, and the P0 function NCP. In this paper, we study a \Big-M " smoothing method for the P0 matrix LCP under the following assumption, which removes the boundedness assumption of the solution set from Assumption 1.1. Assumption 1.2 The P0 matrix LCP has a solution. Big-M interior point methods have been studied for solving the monotone LCP [12]. The methods add one inequality, with a positive number ? as the right-hand-side bound, to bound the variables of the problem. If this inequality contains an original
3 solution, then the augmented problem has a solution and it is also a solution to the original problem. One can always set ? suciently big such that the inequality does contain a solution, assuming that it exists. However, the techniques used in Big-M interior point methods heavily rely on the monotone property, which cannot be carried over from the monotone LCP to the P0 matrix LCP. One di erence, for example, is that the existence of an interior feasible point implies the bounded solution set for the monotone LCP, but it is not held for the P0 matrix LCP.
Example 1.1
0 1
1
0 0 1 B M =@ 0 0
0 0 C B 1 A and q = @ 0 C A: 1 0 ?1 1 It is not dicult to verify that M is a P0 -matrix and that this LCP has the strictly feasible point (1; 1; 1; 1; 1; 1)T . However, the solution set of the LCP contains the unbounded line (x1 ; 0; 0; 0; 0; 1)T for all x1 0. The generalization of Big-M methods to the P0 matrix LCP is nontrivial, see [11]. In order to make the Big-M smooth path and its neighborhood be bounded, we have to slightly destroy the P0 property. Furthermore, in contrast with the trajectory analysis given by Kojima, Megiddo and Noma [10], the existence of suciently short central path is not guaranteed under Assumption 1.2, see Example 2.1. We use k k to denote k k1: We use e for a vector with all entries equal to 1 and I for a diagonal matrix with all diagonal entries equal to 1.
2 A Big-M smoothing model Let
0 M B N =@ 0
1
0 1
q 0 C B 1 0 A; p=@ 0 C A; T ? ?e ?1 ?1 where r = e ? Me ? q and ? n + 5 is suciently big. Let x = (t; ; ) 2 Rn+2 and y = (s; ; ) 2 Rn+2 . We consider the LCP(N; p) xT y = 0; y = Nx + p; and x; y 0: By the construction of the model, we have the following lemma. r
Lemma 2.1 1. LCP(N; p) has a feasible interior point t = e; = 1; = 1
2. If
(t; s)
s = e; = 1; = ? ? (n + 2) 1: is a solution of LCP(M; q) with eT t ?, then
(t; 0; ? ? eT t; s; 0; 0) and (t; 0; 0; s; 0; ? ? eT t )
4 are solutions of LCP(N; p). 3. If LCP(N; p) has a solution, then in every complementarity solution (t; ; ; s ; ; ) of LCP(N; p), (t; s) is a solution of LCP(M; q), = = 0 and + = ? ? eT t , 4. The feasible set of LCP(N; p) is bounded.
Notice that N is not a P0 matrix, since Nn+2;n+2 = ?1. Although we can easily construct a P0 matrix which satis es Results 1{3 of Lemma 2.1, e.g. set Nn+2;n+2 = 1, the resulting LCP may have a unbounded solution set. It seems to the authors that it is hard to construct a Big-M model for the P0 matrix LCP which has both the P0 property and the boundedness of the solution set. This contrasts with the monotone LCP, for which we always can construct a Big-M model having a bounded solution set without loss of the monotone property [12, 17]. Nevertheless, the matrix N is a block lower triangular matrix and its rst block is an (n + 1) (n + 1) P0 matrix, i.e., ~ 0 ! N N = ?eT ?1 where
N~ :=
M r
0
!
1 :
We will often use this fact later. In what follows, for simplicity, we use z := (x; y). It is easy to verify that the LCP(N; p) is equivalent to the following system of nonsmooth equations H0 (z ) :=
Nx + p ? y x ? max(x ? y; 0)
!
= 0:
To de ne a smoothing approximation function of H0, we employ two density functions 2 1 () = 2 ( + 4) 23 and (1 if ?4 0 2 () = 04 otherwise. Let
i(xi ; yi; ") = xi ?
Z (xi ?yi)=" ?1
(xi ? yi ? ")1()d;
and n+2(xn+2; yn+2; ") = xn+2 ?
Z (xn+2?yn+2 )=" ?1
i = 1; 2; : : : ; n + 1
(xn+2 ? yn+2 ? ")2()d:
5 Calculating the integral, we obtain q i(xi; yi; ") = 1 (xi + yi ? (xi ? yi)2 + 4"2); i = 1; 2; : : : n + 1 2 and
8 > xn+2 ? yn+2 ?4" < xn+2 xn+2 ? yn+2 0 n+2(xn+2 ; yn+2; ") = > yn+2 ? 2" : yn+2 ? 1 (xn+2 ? yn+2)2 ? 2" otherwise : 8"
Let " > 0. We de ne the smoothing approximation function of H0 (z) by Nx + p ? y (x; y; ")
H (z; ") :=
!
:
It was shown in [2] (also see [8]) that for the P0 matrix LCP if the corresponding smooth function is constructed by 1 only, then the smoothing function is continuously di erentiable in R2(n+2) and its Jacobian is nonsingular for every z 2 R2(n+2) , and every " > 0. However, the matrix N de ned in this section is not a P0 matrix, and we cannot avoid the singularity if we use 1 alone in . The reason is that if we use 1 in n+2, when xn+2 ? yn+2 = 0, H 0(z; ") becomes singular. In order to overcome the singularity, we use a nonsymmetric uniform density function in n+2. The following lemma shows that we can avoid the singularity by adjusting the smoothing parameter ".
Lemma 2.2 For every xed " > 0, H is continuously di erentiable in R2(n+2) , and kH (z; "1) ? H (z; "2)k 2("1 ? "2); for z 2 R2(n+2) ; "1 > "2 0: (2.1) Furthermore, let M be a P0 matrix. Then H 0 (z; ") is nonsingular at z 2 R2(n+2) if and only if xn+2 ? yn+2 = 6 ?2": Proof: The continuous di erentiability of H follows from that each component of is continuously di erentiable [3]. By Lemma 2.4 in [8], for i = 1; 2; :::; n + 1
Z1
j i(xi ; yi; "1) ? i(xi; yi; "2)j ( ?1 jj1()d)("1 ? "2) = 2("1 ? "2); and for i = n + 2
Z1
j i(xi ; yi; "1) ? (xi; yi; "2)j ( ?1 jj2()d)("1 ? "2 ) = 2("1 ? "2): Now we show the nonsigularity of H 0(z; "). Let di (xi ? yi ; ") =
Z (xi ?yi)=" ?1
1 ()d;
i = 1; 2; :::; n + 1
6 and Let
dn+2 (xn+2 ? yn+2 ; ") =
Z (xn+2 ?yn+2)=" ?1
2 ()d:
Dn+1 (x ? y; ") = diag(d1 (x1 ? y1 ; "); : : : ; dn+1(xn+1 ? yn+1; "))
and
D(x ? y; ") = diag(Dn+1 (x ? y; ); dn+2(xn+2 ? yn+2 ; ")): By the de nition of H (z; "), H 0 (z; ") =
!
N ?I I ? D(x ? y; ") D(x ? y; ") :
It is well known that H 0(z; ") is nonsingular if and only if I ? D(x ? y; ")(I ? N ) is nonsingular. Notice that N~ is a P0 matrix, and I ? D(x ? y; ")(I ? N ) ! ~ I n+1 ? Dn+1 (x ? y; ")(In+1 ? N ) 0 = ?d (x ? y ; ")eT 1 ? 2dn+2(xn+2 ? yn+2; ") : n+2 n+2 n+2 Since suppf1 g = R and N~ is a P0 matrix, In+1 ? Dn+1(x ? y; ")(In+1 ? N~ ) is nonsingular [8]. Hence H 0(z; ") is nonsingular if and only if dn+2(xn+2 ? yn+2; ") 6= 21 . By the de nition of dn+2(xn+2 ? yn+2; "), we have dn+2(xn+2 ? yn+2; ") 6= 21 if and only if xn+2 ? yn+2 6= ?2". This completes this lemma. We de ne the Big-M smooth path as a set of solutions z(") for all 1 " > 0, i.e., S" = fz : H (z; ") = 0; 1 " > 0g: We choose the following neighborhood around the Big-M smooth path: N = fz : kH (z; ")k c"; 1 " > 0g; where c is a positive constant. Lemma 2.3 If (z; ") satis es kH (z; ")k c"; (2.2) then xi + c" 0; yi + c" 0 (2.3) and (xi + c")(yi + c") 2(c + 1)(jxij + jyij)" + (4 + c2)"2; (2.4) for i = 1; 2; : : : ; n + 2. Moreover, the neighborhood N is bounded.
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Proof: Suppose (z; ) satis es (2.2). Then we have ! Nx + p ? y kH (z; ")k = k k c": (x; y; ")
Set u = (x; y; "): Then u 2 Rn+2 satis es c" ui ?c"; i = 1; 2; : : : ; n + 2:
By construction of (cf. Lemma 1 in [2]), we have (x ? u; y ? u; ") = 0:
(2.5)
Since 1 is continuous, symmetric and has an in nite support, by Theorem 2.1 in [4], we have xi ? ui > 0; yi ? ui > 0; i = 1; 2; :::; n + 1: Using c" ?ui , we obtain (2.3) for i = 1; 2; : : : ; n + 1: Now we show (2.3) for i = n + 2: By the de nition of and (2.5), we have the following equalities 0 = n+2(xn+2 ? un+2; yn+2 ? un+2; ") Z (xn+2 ?yn+2)=" (xn+2 ? yn+2 ? ")2()d: = xn+2 ? un+2 ? ?1
This implies that xn+2 ? un+2 0; since the integral part is nonnegative. To show yn+2 ? un+2 0, we assume on the contrary that yn+2 < un+2. Then xn+2 ? yn+2 > xn+2 ? un+2 0 and xn+2 ? un+2 =
Z (xn+2 ?yn+2)=" ?1
(xn+2 ? yn+2 ? ")2 ()d
Z0 1 = (x ? y ? ")d 4 ?4 n+2 n+2 = xn+2 ? yn+2 + 2";
which implies
yn+2 ? un+2 = 2" > 0: This contradicts the assumption that yn+2 ? un+2 < 0: Hence we have xn+2 ? un+2 0; and yn+2 ? un+2 0:
Using c" ?un+2, we obtain (2.3) for i = n + 2. Now we show (2.4). Suppose that xi yi. Then we have xi ? ui ? max(xi ? yi; 0) = min(xi ? ui ; yi ? ui ) = xi ? ui :
From Lemma 2.2 and (2.5), xi ? ui = j i(xi ? ui; yi ? ui; ") ? (xi ? ui)j 2":
8 By a simple manipulation, we obtain (xi + c")(yi + c") = (xi ? ui)(yi ? ui) + (xi + yi)(ui + c") ? u2i + c2"2 = (xi ? ui)(yi ? xi ) + (xi ? ui)2 + jxi + yij(ui + c") ? u2i + c2"2 2(jxij + jyij)" + 4"2 + 2jxi + yijc" + c2"2 2(1 + c)(jxi j + jyij)" + (4 + c2)"2; where the rst inequality follows from 0 xi ? ui 2" and ? c" ui c": The proof is similar for the case xi yi: Now, we show that N is bounded. Using c" Hn+2 (z; ") ?c";
we have c"
(Nx + p ? y)n+2 = ?eT x + ? ? yn+2 ?c":
This implies
? + c" eT x + yn+2 ? ? c": Since xi and yi are bounded below by (2.3), x and y cannot go to 1, and so N is bounded.
Theorem 2.1 Suppose that LCP(N; p) has a solution z . Then for every " > 0, there is a z" 2 R2(n+2) such that H 0 (z" ; ")T H (z"; ") = 0
and
q
kH (z"; ")k 2 2(n + 2)": q
(2.6) (2.7)
Proof: Let c be a constant satisfying c > 2 2(n + 2): We consider the following set: D(") = fz j kH (z; ")k c"g: By Lemma 2.2, at the solution z , we have
kH (z; ")k kH (z ; ") ? H0(z )k + kH0(z )k = 2": Hence z 2 D(") and so D(") is nonempty.
9 By Lemma 2.3, D(") is bounded. Let 1 (z ) = kH (z; ")k22 : 2 Since H (; ") is continuously di erentiable and D(") is nonempty and bounded, has a global minimum point z" in D("). Moreover, we have
kH (z"; ")k
kH (z"; ")k2 kqH (z; ")k2 2(n + 2)kH (z; ")k q 2(n + 2)(kH0(z )k + 2") q
= 2 2(n + 2)": Hence, (2.7) holds. Moreover, this implies z" 2 int D("). By [13, 4.1.3], we have 0 (z" ) = H 0 (z" ; ")T H (z" ; ") = 0:
We complete the proof. Theorem 2.1, together with Lemma 2.1, shows that if the LCP(M; q) has a solution, then there is a ? such that the LCP(N; p) has a solution and for every " > 0, we can nd a stationary point z" of 1 kH (z ; ")k2 " 2 2 which satis es q kH (z"; ")k 2 2(n + 2)": Hence following a path de ned by such stationary point as " ! 0, we can nd a solution of the LCP(M; q). By Lemma 2.1, if M is a P0 matrix, the rank of H 0(z"; ") is always greater than 2(n + 1) + 1. Moreover, H 0(z"; ") is nonsingular if and only if xn+2 ? yn+2 6= ?2". Hence, we can claim that H 0(z"; ") is nonsingular almost everywhere if M is a P0 matrix. From (2.6), if H 0(z"; ") is nonsingular then z" is a solution of H (z; ") = 0:
(2.8)
It is notable that Assumption 1.1 does not guarantee the existence of a solution of (2.8) for every " > 0. The following example shows that even if the P0 matrix LCP(M; q) has a unique strictly complementarity solution, the central path can be very short. Precisely, for any "0 > 0, changing an element in q can make that the central path does not exist for all " "0 even if the resulting P0 matrix LCP has a unique strictly complementarity solution. Furthermore, if the solution set is unbounded, the central path may not exist for all " > 0. On the other hand, the Big-M smooth path may be \much longer" than the central path.
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Example 2.1 Let
!
M = 00 ?11
q=
!
? 12 ;
where > 21 :
It is easy to verify that M is a P0 matrix and the LCP(M; q) has a unique solution (t ; s) = (0; 21 ; ? 12 ; 0): However for any 0 > 0 the complementarity level set f(t; s) : tT s 0 + ( ? 21 ); s = Mt + q; (t; s) > 0g contains the unbounded line 0 0 0 (k; ? ; ; ? ? 1 ) k k k 2 for all k > 0 =( ? 12 ): The central path
f(t; s) : ti si = "; i = 1; 2p; s = Mt + q; (t; s) > 0g = f( ?" t ; t2); t2 = 1 + 14 + 16" < g 2
does not exist for " ( ? 21 ): If = 21 , this problem has a unbounded solution set f(x1 ; 21 ; 0; 0); x1 0g: In this case the central path does not exist for all " > 0. Now we show that if we choose ? n + 5 = 7; then for all " 1, (2.8) has a solution, i.e. the Big-M smooth path exists for " 1: By Lemma 2.1 in [9] and Lemma 2.3, at the Big-M smooth path, (x; y) 0 and x1 y1 = x1 ( ? x2 + (2 ? )x3 ) = "2
1 2
1 2 x3 y3 = x23 = "2 y4 = ? ? x1 ? x2 ? x3 ? x4 Z (x4 ?y4)=" x4 = (x4 ? y4 ? ")2()d:
x2 y2 = x2 (? + x2 + x3 ) = "2
?1
We can calculate the point at the Big-M smooth path x3 = y3 = " 1
11
q
1 ? " + (1 ? ")2 + 16"2 x2 = 1 4 x1 =
"2
? x2 + (2 ? )"
1 1 2
y1 = ? x2 + (2 ? )x3 ; y2 = x2 + x3 ?
or
1 2
x4 = ? ? x1 ? x2 ? x3 ? 2" 2"; y4 = 2"
x4 = 0; y4 = ? ? x1 ? x2 ? x3 4"; where we use () = 0; for 62 [?4; 0] to calculate x4 and y4 .
There are two Big-M smooth paths which are bounded and converge to two solutions 1 1 1 z ;1 = (0; ; 0; ? ? ; ? ; 0; 0; 0) 2 2 2 and 1 1 1 z ;2 = (0; ; 0; 0; ? ; 0; 0; ? ? ); 2 2 2 respectively, as " ! 0. Both of them contain the solution of the original LCP(M; q) : (x1;1 ; x2;1 ; y1;1; y2;1) = (x1;2 ; x2;2; y1;2; y2;2) = (0; 12 ; ? 12 ; 0):
We cannot say that the solution of (2.8) is unique for every " > 0, since N is not a P0-matrix, Nevertheless, we can say that the solution is locally unique. This result is given by the following theorem.
Theorem 2.2 Suppose that M is a P0 matrix. Let z" be a solution of (2.8). Then its 2(n + 1) components (z" )i ; i = 1; 2; : : : ; n + 1; n + 3; : : : ; 2(n + 1) + 1, that is, ((x")i ; (y")i); i = 1; 2; : : : ; n + 1, are unique. Moreover the following results hold. 1. If H 0(z" ; ") is singular, then z" is the unique solution of (2.8). 2. If H 0 (z" ; ") is nonsingular, then z" is the unique solution of (2.8) in a neighborhood of z".
Proof: 1. Let x~ = (x1 ; : : : ; xn+1); y~ = (y1; : : : ; yn+1); z~ = (~x; y~) and ~ z ; ") = ( 1(x1 ; y1; "); : : : ; n+1(xn+1 ; yn+1; "))T : (~ Then z~ is a solution of
!
N~ x~ + p ? y~ : ~ (~z ; ")
(2.9)
Since N~ is a P0 matrix and ~ is given by 1 , by Theorem 2.3 in [4], z~" is the unique solution of (2.9).
12 1.
Assume that (2.8) has two solutions z"1 and z"2 . Then (z"1 )i = (z"2 )i, i = 1; : : : ; n + 1; n + 3; : : : ; 2(n + 1) + 1: By the structure of the (n + 2)th component of H , we obtain x1n+2 ? x2n+2 = yn2 +2 ? yn1 +2 : (2.10) Hence if x1n+2 = x2n+2, then yn1+2 = yn2+2. Without loss of generality, we assume on the contrary that x1n+2 > x2n+2 : Let (w) =
Z
w "
?1
(w ? ")2()d:
Then from n+2(x1n+2 ; yn1+2; ") = n+2(x2n+2; yn2+2; ") = 0; we have 0 < x1n+2 ? x2n+2 = (x1n+2 ? yn1+2) ? (x2n+2 ? yn2+2) = 0 ()(x1n+2 ? yn1+2 ? x2n+2 + yn2+2) = 2 0()(x1n+2 ? x2n+2 ); where 2 (x1n+2 ? yn1+2; x2n+2 ? yn2+2), and the last equality follows from (2.10). This implies that Z 0 () = " ()d = 1 2 2 ?1 which gives = ?2" and x1n+2 ? yn1 +2 > ?2" > x2n+2 ? yn2 +2 :
By Lemma 2.2, H 0(z"; ") is nonsingular. This is a contradiction. 2. Let G(z ) = z ? H 0 (z" ; ")?1H (z; "): Since H is continuously di erentiable, G is continuously di erentiable and G0 (z ) = I ? H 0(z" ; ")?1 H 0 (z; "):
Moreover the spectral radius of G0 (z") is zero. By Ostrowski Theorem [13], we claim that z" is the unique solution of (2.8) in a neighborhood of z": Results 1 and 2 of Theorem 2.2 may be explained by square functions. f (x) = x2 = 0 has a unique zero x = 0, which is multiple roots, and its derivative is singular at x. f (x) = x2 ? 1 has two zeros x;1 = 1 and x;2 = ?1, which are unique in (?1; 1) and (?1; 1) respectively, and its derivative is nonsingular at x;1 and x;2 .
3 Algorithm and its convergence In this section we propose an algorithm and prove its global convergence.
Algorithm 3.1 Given c 2, 2 (0; 1), and i 2 (0; 1) for i = 1; 2:
13
Step 0 (Initial Step )
Choose x0 ; y 0; "0 such that kH (z 0 ; "0)k c"0 and H 0(z 0 ; "0) is nonsingular.
Step 1 (Newton Step)
If H (z k ; "k ) = 0, set z k+1 = z k and go to Step 3. Otherwise, Let z k solve the equation H (z k ; "k ) + H 0 (z k ; "k )z k = 0: (3.1)
Step 2 (Line Search)
Let k be the maximum of the values 1; 1; 12 ; : : : such that
kH (zk + k zk ; "k )k (1 ? k )kH (zk ; "k )k:
(3.2)
Set z k+1 = z k + k z k .
Step 3 (" Reduction)
Let k be the maximum of the values 22 ; 23 ; : : : such that
kH (zk+1; (1 ? k )"k )k (1 ? k )c"k : (3.3) k +1 k k +1 If xkn+1 = (1 ? k )"k . Otherwise, set "k+1 = +2 ? yn+2 6= ?2(1 ? k )" ; set " (1 ? 2 k )"k : Algorithm 3.1 is similar to the smoothing method introduced by Burke and Xu [1]. The main di erence from the Burke-Xu method is that the de nition of "k+1 in Step 3 ensures the nonsingularity of H 0(zk+1 ; "k+1) for the P0 matrix LCP(M; q). It is easy to verify that if y0 = Nx0 + p, then yk = Nxk + p for all k 0. The following lemma shows that we can easily nd a starting point (z0 ; "0) satisfying these conditions in the initial step of Algorithm 3.1.
Lemma 3.1 Suppose that M is a P0 matrix. Let x0 = (e; 1; 0); y0 = (e; 1; ? ? (n +1)) and
1 "0 1: c+1 0 0 0 0 Then y = Nx + p 0, kH (z ; " )k c"0 , and H 0 (z 0 ; "0 ) is nonsingular.
(3.4)
Proof: Obviously y0 = Nx0 + p. Since ? n + 5, y0 0. Thus Hi(z0 ; "0) = 0; i =
1; 2; : : : ; n + 2. By a simple calculation, we have i(x0i ; yi0; "0) = 1 + 0 and n+2
(x0
n+2
; y0
n+2
; "0 )
=
Z0
?1
1 ()d = 1 ? "0 ; i = 1; 2; : : : ; n + 1;
Z ((n+1)??)="0 ?1
((n + 1) ? ? ? "0)2()d = 0;
14 where the last equality uses 2 () = 0 for ?4 and (3.4). Hence (z0 ; "0) satis es kH (z0 ; "0)k c"0: Moreover, from ? n + 5 and "0 1, we have ?2"0 > ?4 (n + 1) ? ? = x0n+2 ? yn0+2: Therefore H 0(z0 ; "0) is nonsingular by Lemma 2.2, since x0n+2 ? yn0+2 6= ?2"0.
Theorem 3.1 If M is a sequence fz k g satis es
P0 matrix, then Algorithm 3.1 is well de ned, and the
kH (zk ; "k )k c"k :
(3.5)
Proof: We prove this theorem by induction.
For k = 0, by Lemma 3.1, z0 = (e; 1; 0; e; 1; ? ? (n + 1)) satis es (3.5) and H 0 (z 0 ; "0 ) is nonsingular. We suppose that zk satis es (3.5) and H 0(zk ; "k ) is nonsingular. Then Step 1 is well de ned. If H (zk ; "k ) 6= 0, then zk 6= 0: Hence zk is a strictly decent direction of kH (; "k)k at zk , and so the line search procedure is nite by construction in Step 2. Step 3 is well de ned since if H (zk ; "k ) = 0 then zk+1 = zk and kH (zk+1; "k )k = 0 < c"k : Otherwise, by the construction of Step 2, kH (zk+1; "k )k < kH (zk ; "k )k c"k ; which implies that there is a nite number k > 0 such that (3.3) holds. By the construction of Step 3, H 0(zk+1 ; "k+1) is nonsingular. Now we show that (3.5) holds at (zk+1 ; "k+1). k +1 k If xkn+1 +2 ? yn+2 6= ?2(1 ? k )" ; then by construction of Step 3, (3.5) holds. Hence we only need to consider the case k +1 k xkn+1 +2 ? yn+2 = ?2(1 ? k )" ;
i.e., Notice that and Step 3 provides that
"k+1 = (1 ? 2 k )"k : "k > (1 ? 2 k )"k > (1 ? k )"k c"k+1
c(1 ? 2 k )"k c(1 ? k )"k H (z k+1 ; (1 ? k )"k ) ?c(1 ? k )"k :
(3.6)
15 By Result 3 of Proposition 1 in [2], for i = 1; 2; : : : ; n + 1, i(xki +1; yik+1; ) is strictly decreasing with respect to ", which gives
i(xki +1; yik+1; (1 ? k )"k )
c"k+1
> i (xki +1 ; yik+1; "k+1) = i(xki +1; yik+1; (1 ? 2 k )"k ) i(xki +1; yik+1; (1 ? k )"k ) + ( 2 ? 1) k "k > ?c(1 ? k )"k ? (1 ? 2 ) k "k = ?c"k + c k "k ? k "k + 2 k "k ?c"k + 2(c ? 1) k "k + 2 k "k > ?c"k + c 2 k "k = ?c"k+1;
where the third inequality follows from Result 4 of Proposition 1 in [2] (also see [8]), and the fourth inequality follows from (3.6). Hence (3.5) holds for i = 1; 2; : : : ; n +1: k +1 Let w = xkn+1 +2 ? yn+2 ; and (") =
Z
w "
?1
Then ! = ?2(1 ? k )"k < 0 and 0 (") = ?
Z
(w ? ")2()d: w "
?1
2 ()d 0:
Hence is monotonically increasing function. This implies that n+2 is monotonically decreasing and Lipschitz continuous with respect to the smooth parameter ". Hence we obtain c"k+1
k +1 k n+2(xkn+1 +2 ; yn+2 ; (1 ? k )" ) k k +1 n+2(xkn+1 +2 ; yn+2 ; (1 ? 2 k )" ) k +1 k k n+2(xkn+1 +2 ; yn+2 ; (1 ? k )" ) ? 2(1 ? 2 ) k " ?c(1 ? k )"k ? 2(1 ? 2 ) k "k = ?c"k + (c ? 2(1 ? 2 )) k "k = ?c"k + c 2 k "k + (c ? 2)(1 ? 2) k "k ?c(1 ? 2 k )"k = ?c"k+1;
where the third inequality follows from that n+2 is Lipschitz continuous with the Lipschitz constant 2 and c 2. (See Lemma 2.2). Therefore we have c"k+1 Hi (z k+1 ; "k+1) ?c"k+1 ; for i = n + 3; : : : ; 2(n + 2):
16 By the de nition of H , the parameter " is not involved in the rst n+2 components of H , i.e., Hi (z k+1 ; "k ) = Hi (z k+1 ; "k+1 ); for i = 1; 2; : : : n + 2: Hence (3.5) holds.
Theorem 3.2 Suppose that M is a P0 matrix. Then Algorithm 3.1 is well de ned. Let f(z k ; "k )g be a sequence generated by Algorithm 3.1. Then fz k g is remained in the bounded neighborhood N , and f"k g decreases monotonically in R++ . If furthermore for an accumulation point (z; ") of f(z k ; "k )g, we have xn+2 ? yn+2 = 6 ?2" or " = 0 then
lim "k = 0 and !1
k
lim H (z k ; "k ) = 0; !1
k
(3.7)
and for all accumulation points fz^; "^g of fz k ; "k g.
"^ = 0 and H (^z ; "^) = H0 (^z ) = 0:
(3.8)
Proof: By Theorem 3.1 and Lemma 2.3, Algorithm 3.1 is well de ned and fzk g is remained in a bounded set. Furthermore, by construction of Algorithm 3.1, 0 < "k+1 = (1 ? k )"k < "k or
0 < "k+1 = (1 ? 2 k )"k < "k : Hence f"k g is a monotonically decreasing sequence, and there is " such that lim "k = ": !1
k
If " = 0, then from Theorem 3.1, we have (3.7). Moreover, since f"k g is a monotonically decreasing sequence, (3.5) and (3.7) imply (3.8). Suppose on the contrary that " > 0: This implies k ! 0. Since fzk g is remained in the bounded neighborhood N , taking a subsequence if necessary, we may assume that the sequence fzk g converges to some z. Based on the " reduction step, we have kH (zk ; (1 ? 1 )"k?1)k > (1 ? 1 )c"k?1: 22 k?1
22 k?1
Since k ! 0, by passing limits on both sides, we have
kH (z; ")k c" > 0:
(3.9)
Since xn+2 ? yn+2 6= ?2", H 0(z ; ") is nonsingular by Lemma 2.2. Hence we can nd a unique solution z of the linear equations in Step 1. Furthermore, from (3.9) it is a strictly decent direction for kH (; ")k at z. As a result, the corresponding linear search step length and " reduction step length are both bound below by
17 a positive constant. Notice that the function H and its Jacobian H 0 are continuous in a neighborhood of (z; "). It follows that zk converges to z and therefore k must be uniformly bounded below by some positive constant for all large k. This contradicts to the assumption that k ! 0. Hence we must have "k ! 0, and so (3.7) and (3.8) hold. Let us denote
" = fz : kH (z; ")k c"g: By Lemma 2.3, the set " is bounded for every " > 0. Corollary 3.1 Suppose that the solution set S0 (M; q) of the P0 matrix LCP(M; q) is nonempty and bounded. Then there exist a big ? and "0 such that the sequence fzk g generated by Algorithm 3.1 converges to a solution of LCP(N; p). Proof: First we show that
"k "k?1 ; for "k "k?1: Suppose that z 2 "k . Then kH (z; "k?1)k kH (z; "k?1) ? H (z; "k )k + kH (z; "k )k 2("k?1 ? "k ) + c"k = c"k?1 ? (c ? 2)("k?1 ? "k )
c"k?1;
where we use c 2 and Lemma 2.2. Hence z 2 "k?1 : This, together with Theorem 3.1, implies that the sequence generated by Algorithm 3.1 remains in "0 : Let ? satisfy ? > 4eT t; for all (t; s) 2 S0(M; q): Then from Lemma 2.1, the solution set of the LCP(N; p) is given by
0 = f(t; 0; ? ? eT t; s; 0; 0); (t; 0; 0; s; 0; ? ? eT t); (t; s) 2 S0(M; q)g: Hence for a (t; s) 2 S0 (M; q), z;1 = (t; 0; ? ? eT t ; s; 0; 0) and z;2 = (t; 0; 0, s; 0; ? ? eT t) are solutions of LCP(N; p) and eT t + max min(jx ? y ? x;1 ? y;1jn+2; jx ? y ? x;2 ? y;2jn+2) z 2 0 = eT t + max jeT (t ? t)j z 2 0 ?4 + ?4 = ?2 : By the continuity of H (z; ") on ", for such ? there exists "0 2 (0; min(1; ?4 )) such that kH (z0; "0)k c"0 and for all z 2 "0 , ? eT t + 2"0 + zmax min(jx ? y ? x;1 + y;1jn+2; jx ? y ? x;2 + y;2jn+2): 2
"0
18 Let z 2 "0 . Without loss of generality we may assume jx ? y ? x;1 + y;1jn+2 jx ? y ? x;2 + y;2jn+2. Then
jxn+2 ? yn+2j 1 1 1 jxn;+2 ? yn;+21 j ? jxn+2 ? yn+2 ? xn;+2 + yn;+2 j T ;1 ;1 ? ? e t ? zmax min(jx ? y ? x + y jn+2; jx ? y ? x;2 + y;2jn+2) 2 "0 2"0 > 2"k for all k 1. Hence at any accumulation point (z ; ") generated by Algorithm 3.1, xn+2 ? yn+2 = 6 ?2". By Theorem 3.2, we complete the proof.
The proof of Corollary 3.1 shows that if the solution set of the P0 matrix LCP(M; q) is nonempty and bounded, then there exist ? and 0 such that for any z 2 ", " "0, H 0 (z" ; ") is nonsingular. Hence from Theorem 2.2 and Corollary 3.1, two Big-M smooth paths exist but they never cross each other, which converge to two solutions. The proof of Corollary 3.1 also suggests us to increase ? and restart Algorithm 3.1 when xn+2 ? yn+2 = ?2". For instance, we choose new starting point as x^0i = xi ; i = 1; 2; : : : ; n + 1 y^i0 = yi; i = 1; 2; : : : ; n + 1 1 x^0n+2 = xn+2 ? " 2 3 0 y^n+2 = yn+2 + "; 2 ?^ = ? + " "^0 = ": Then at the new starting point, the new LCP(N; p^) with the new ?^ satis es H^ i (^z 0 ; "^0 ) = Hi (z ; "); i = 1; 2; : : : ; 2(n + 1) + 1 and " Z ?2 1 0 0 0 ^ H2(n+2) (^z ; "^ ) = x^n+2 = xn+2 ? " = xn+2 + 2 4 ?4 (2 + )d = H2(n+2) (z; "); where we use x^0n+2 ? y^n0+2 ?4^"0 and 2 () = 0 for ?4: Hence we have H^ (^z 0 ; "^0 ) = H (z ; ") and "^0 = ": Furthermore, H^ 0(^z0 ; "^0) is nonsingular since x^0n+2 ? y^n0 +2 = xn+2 ? yn+2 ? 2" = ?4" = ?4^"0 :
At the end of this paper, we consider the Big-M smooth path for Example 1.1, which is a P0 matrix LCP but its solution set is unbounded.
19 In the Big-M smoothing model for Example 1.1, r = e ? Me ? q = 0. We choose ? = n + 5 = 8: At the Big-M smooth path, xi > 0; yi > 0 yi = (Nx + p)i ; xi yi = "2 ; i = 1; 2; 3; 4
and x5 =
Z (x5 ?y5)=" ?1
(x5 ? y5 ? ")2()d; y5 = ? ? x1 ? x2 ? x3 ? x4 ? x5 :
We can calculate this point for " 2 (0; 1]:
p
4 "2 4 "2 1 + 8 "2 + 1 p p x1 = 1; x3 = 1 + 8"2 + 1 1; 1; x2 = 4 1 + 8"2 + 1 x4 = " 1; and x5 = 0; or x5 = ? ? x1 ? x2 ? x3 ? x4 ? 2"; where we use 2 () = 0; for 62 [?4; 0] to calculate x5 : Two Big-M smooth paths never cross each other and converge to two solutions 1 1 1 1 z ;1 = (0; ; 0; 0; ? ? ; ; 0; ; 0; 0); 2 2 2 2 and 1 1 1 1 z ;2 = (0; ; 0; 0; 0; ; 0; ; 0; ? ? ); 2 2 2 2 respectively, as " ! 0. Both of them contain a strictly complementarity solution of the original LCP(M; q) (t ; s) = (0; 12 ; 0; 21 ; 0; 21 ):
Acknowledgement
The authors are grateful to M.S.Gowda, C.Kanzow, D.Sun and P.Tseng for their helpful comments.
References [1] J. Burke and S. Xu, \The global linear convergence of a non-interior pathfollowing algorithm for linear complementarity problem", to appear in Math. Oper. Res. [2] B. Chen and X. Chen, \A global and local superlinear continuation-smoothing method for P0 + R0 and monotone NCP," to appear in SIAM J. Optimization. [3] C. Chen and O.L. Mangasarian, \Smoothing methods for convex inequalities and linear complementarity problems", Math. Programming, 71 (1995), 51-69.
20 [4] X. Chen and Y. Ye, \On homotopy-smoothing methods for variational inequalities", AMR 96/39, Applied Mathematics Report, School of Mathematics, The University of New South Wales, Sydney 2052, December, (1996). [5] R.W. Cottle, J.-S. Pang and R.E. Stone, The Linear Complementarity Problem, Computer Science and Scienti c Computing, Academic Press Inc, San Diego, CA92101, (1990). [6] F. Facchinei, \Structural and stability properties of P0 nonlinear complementarity problems," DIS Working paper 12-97, Universita di Roma \La Sapienza", Roma, Italy, 1997. [7] F. Facchinei and C. Kanzow, \Beyond monotonicity in regularization methods for nonlinear complementarity problems", Dipartmento di Informatica e Sistemistica, Universita di Roma, Roma, Italy, (1997). [8] S.A. Gabriel and J.J. More, \Smoothing of mixed complementarity problems", in: M.C. Ferris and J.S. Pang, eds., Complementarity and Variational Problems: State of the Art, SIAM, Philadelphia, Pennsylvania, (1996), 105-116. [9] C. Kanzow, \Some noninterior continuation methods for linear complementarity problems", SIAM J. Matrix Anal. Appl. 17, (1996), 851-868. [10] M. Kojima, N. Megiddo, and T. Noma, \Homotopy continuation methods for nonlinear complementarity problems", Math. Oper. Res., 16 (1991), 754-774. [11] M. Kojima, N. Megiddo, T. Noma, and A. Yoshise, A uni ed approach to interior point algorithms for linear complementarity problems, LNCS 538, SpringerVerlag, (1991). [12] M. Kojima, S. Mizuno, and A. Yoshise, \A little theorem of the big M in interior point algorithms", Math. Programming 59 (1993), 361{375. [13] J.M. Ortega and W.C. Rheinboldt, Iterative Solution of Nonlinear Equations in Several Variables, Academic Press, New York, 1970. [14] H.-D. Qi, \A regularized smoothing Newton method for box constrained variational inequality problems with P0 -function", Research Report, Institute of Computational Mathematics and Scienti c/Engineering Computing, Chinese Academy of Sciences, P.O. Box 2719, Beijing, China, August, 1997. [15] G. Ravindran and M.S. Gowda, \Regularization of P0-function in box variational inequality problems", Research Report, Department of Mathematics and Statistics, University of Maryland Baltimore County, Baltimore, MD 21250, August, 1997.
21 [16] D. Sun, \A regularized Newton method for solving nonlinear complementarity problems", Applied Mathematics Report, School of Mathematics, The University of New South Wales, Sydney, July 1997. [17] Y. Ye, \On homogeneous and self-dual algorithms for LCP" Math. Programming 76 (1997) 211-222.
Yinyu Yey
School of Mathematics Department of Management Sciences The University of New South Wales University of Iowa Sydney 2052, Australia Iowa City, IA, USA E-mail: [email protected] E-mail: [email protected]
5 March 1998
Abstract
In this paper, we propose a Big-M smoothing method for solving the P0 matrix linear complementarity problem. We study the trajectory de ned by the augmented smoothing equations and global convergence of the method under an assumption that the original P0 matrix linear complementarity problem has a solution.
Key words. linear complementarity problem, P0 matrix, smoothing algorithm. AMS subject classi cations. 65H10, 90C30, 90C33.
y
This author's work was supported by the Australian Research Council. This author is supported in part by NSF Grants DMI-9522507 and DMS-9703490.
1
2
1 Introduction In this paper we consider the linear complementarity problem (LCP) tT s = 0; s = Mt + q; and t; s 0;
where M is an n n P0 matrix and q is an n dimensional vector. A matrix M 2 Rnn is called a P0 matrix if max t (Mt)i 0; for all t 2 Rn ; t 6= 0: 6 i
i:ti =0
A linear complementarity problem is called a P0 matrix LCP if the matrix M is a P0 matrix. The class of the P0 matrix LCP includes the monotone LCP and the P matrix LCP. The P0 matrix LCP has been studied extensively under additional conditions [5, 11]. A di erentiable function on Rn is called a P0 function if its Jacobian is a P0 matrix at every point in Rn. A nonlinear complementarity problem (NCP) is called a P0 function NCP if the involved function is a P0 function. Kojima, Megiddo and Noma [10] proved the existence of a trajectory in the interior of the feasible set of the P0 function NCP under some additional conditions. Their results in uenced the development of interior point methods and non-interior point methods, and led several continuation methods for solving P0 function NCP. Recently, Facchinei and Kanzow [7] applied regularization methods for solving a continuously di erentiable P0 function NCP under the following assumption. Assumption 1.1 The solution set of the P0 function NCP is nonempty and bounded. This assumption is weaker than that Kojima, Megiddo, Noma and Yoshise used in [10, 11]. Moreover it includes the monotone NCP with an interior point, and the P0 and R0 NCP [5]. After Facchinei-Kanzow's encouraging work, several algorithms and theoretical results on regularization methods for the P0 function NCP have been developed [14, 15, 16] under Assumption 1.1. In particular, Ravindran and Gowda [15] generalized the results of Facchinei and Kanzow [7] to a continuous P0 function variational inequality problem with box constraints. Facchinei and Kanzow [7] gave a counterexample to show that it is not possible to remove the boundedness assumption of the solution set for regularization methods for solving the P0 matrix LCP, and the P0 function NCP. In this paper, we study a \Big-M " smoothing method for the P0 matrix LCP under the following assumption, which removes the boundedness assumption of the solution set from Assumption 1.1. Assumption 1.2 The P0 matrix LCP has a solution. Big-M interior point methods have been studied for solving the monotone LCP [12]. The methods add one inequality, with a positive number ? as the right-hand-side bound, to bound the variables of the problem. If this inequality contains an original
3 solution, then the augmented problem has a solution and it is also a solution to the original problem. One can always set ? suciently big such that the inequality does contain a solution, assuming that it exists. However, the techniques used in Big-M interior point methods heavily rely on the monotone property, which cannot be carried over from the monotone LCP to the P0 matrix LCP. One di erence, for example, is that the existence of an interior feasible point implies the bounded solution set for the monotone LCP, but it is not held for the P0 matrix LCP.
Example 1.1
0 1
1
0 0 1 B M =@ 0 0
0 0 C B 1 A and q = @ 0 C A: 1 0 ?1 1 It is not dicult to verify that M is a P0 -matrix and that this LCP has the strictly feasible point (1; 1; 1; 1; 1; 1)T . However, the solution set of the LCP contains the unbounded line (x1 ; 0; 0; 0; 0; 1)T for all x1 0. The generalization of Big-M methods to the P0 matrix LCP is nontrivial, see [11]. In order to make the Big-M smooth path and its neighborhood be bounded, we have to slightly destroy the P0 property. Furthermore, in contrast with the trajectory analysis given by Kojima, Megiddo and Noma [10], the existence of suciently short central path is not guaranteed under Assumption 1.2, see Example 2.1. We use k k to denote k k1: We use e for a vector with all entries equal to 1 and I for a diagonal matrix with all diagonal entries equal to 1.
2 A Big-M smoothing model Let
0 M B N =@ 0
1
0 1
q 0 C B 1 0 A; p=@ 0 C A; T ? ?e ?1 ?1 where r = e ? Me ? q and ? n + 5 is suciently big. Let x = (t; ; ) 2 Rn+2 and y = (s; ; ) 2 Rn+2 . We consider the LCP(N; p) xT y = 0; y = Nx + p; and x; y 0: By the construction of the model, we have the following lemma. r
Lemma 2.1 1. LCP(N; p) has a feasible interior point t = e; = 1; = 1
2. If
(t; s)
s = e; = 1; = ? ? (n + 2) 1: is a solution of LCP(M; q) with eT t ?, then
(t; 0; ? ? eT t; s; 0; 0) and (t; 0; 0; s; 0; ? ? eT t )
4 are solutions of LCP(N; p). 3. If LCP(N; p) has a solution, then in every complementarity solution (t; ; ; s ; ; ) of LCP(N; p), (t; s) is a solution of LCP(M; q), = = 0 and + = ? ? eT t , 4. The feasible set of LCP(N; p) is bounded.
Notice that N is not a P0 matrix, since Nn+2;n+2 = ?1. Although we can easily construct a P0 matrix which satis es Results 1{3 of Lemma 2.1, e.g. set Nn+2;n+2 = 1, the resulting LCP may have a unbounded solution set. It seems to the authors that it is hard to construct a Big-M model for the P0 matrix LCP which has both the P0 property and the boundedness of the solution set. This contrasts with the monotone LCP, for which we always can construct a Big-M model having a bounded solution set without loss of the monotone property [12, 17]. Nevertheless, the matrix N is a block lower triangular matrix and its rst block is an (n + 1) (n + 1) P0 matrix, i.e., ~ 0 ! N N = ?eT ?1 where
N~ :=
M r
0
!
1 :
We will often use this fact later. In what follows, for simplicity, we use z := (x; y). It is easy to verify that the LCP(N; p) is equivalent to the following system of nonsmooth equations H0 (z ) :=
Nx + p ? y x ? max(x ? y; 0)
!
= 0:
To de ne a smoothing approximation function of H0, we employ two density functions 2 1 () = 2 ( + 4) 23 and (1 if ?4 0 2 () = 04 otherwise. Let
i(xi ; yi; ") = xi ?
Z (xi ?yi)=" ?1
(xi ? yi ? ")1()d;
and n+2(xn+2; yn+2; ") = xn+2 ?
Z (xn+2?yn+2 )=" ?1
i = 1; 2; : : : ; n + 1
(xn+2 ? yn+2 ? ")2()d:
5 Calculating the integral, we obtain q i(xi; yi; ") = 1 (xi + yi ? (xi ? yi)2 + 4"2); i = 1; 2; : : : n + 1 2 and
8 > xn+2 ? yn+2 ?4" < xn+2 xn+2 ? yn+2 0 n+2(xn+2 ; yn+2; ") = > yn+2 ? 2" : yn+2 ? 1 (xn+2 ? yn+2)2 ? 2" otherwise : 8"
Let " > 0. We de ne the smoothing approximation function of H0 (z) by Nx + p ? y (x; y; ")
H (z; ") :=
!
:
It was shown in [2] (also see [8]) that for the P0 matrix LCP if the corresponding smooth function is constructed by 1 only, then the smoothing function is continuously di erentiable in R2(n+2) and its Jacobian is nonsingular for every z 2 R2(n+2) , and every " > 0. However, the matrix N de ned in this section is not a P0 matrix, and we cannot avoid the singularity if we use 1 alone in . The reason is that if we use 1 in n+2, when xn+2 ? yn+2 = 0, H 0(z; ") becomes singular. In order to overcome the singularity, we use a nonsymmetric uniform density function in n+2. The following lemma shows that we can avoid the singularity by adjusting the smoothing parameter ".
Lemma 2.2 For every xed " > 0, H is continuously di erentiable in R2(n+2) , and kH (z; "1) ? H (z; "2)k 2("1 ? "2); for z 2 R2(n+2) ; "1 > "2 0: (2.1) Furthermore, let M be a P0 matrix. Then H 0 (z; ") is nonsingular at z 2 R2(n+2) if and only if xn+2 ? yn+2 = 6 ?2": Proof: The continuous di erentiability of H follows from that each component of is continuously di erentiable [3]. By Lemma 2.4 in [8], for i = 1; 2; :::; n + 1
Z1
j i(xi ; yi; "1) ? i(xi; yi; "2)j ( ?1 jj1()d)("1 ? "2) = 2("1 ? "2); and for i = n + 2
Z1
j i(xi ; yi; "1) ? (xi; yi; "2)j ( ?1 jj2()d)("1 ? "2 ) = 2("1 ? "2): Now we show the nonsigularity of H 0(z; "). Let di (xi ? yi ; ") =
Z (xi ?yi)=" ?1
1 ()d;
i = 1; 2; :::; n + 1
6 and Let
dn+2 (xn+2 ? yn+2 ; ") =
Z (xn+2 ?yn+2)=" ?1
2 ()d:
Dn+1 (x ? y; ") = diag(d1 (x1 ? y1 ; "); : : : ; dn+1(xn+1 ? yn+1; "))
and
D(x ? y; ") = diag(Dn+1 (x ? y; ); dn+2(xn+2 ? yn+2 ; ")): By the de nition of H (z; "), H 0 (z; ") =
!
N ?I I ? D(x ? y; ") D(x ? y; ") :
It is well known that H 0(z; ") is nonsingular if and only if I ? D(x ? y; ")(I ? N ) is nonsingular. Notice that N~ is a P0 matrix, and I ? D(x ? y; ")(I ? N ) ! ~ I n+1 ? Dn+1 (x ? y; ")(In+1 ? N ) 0 = ?d (x ? y ; ")eT 1 ? 2dn+2(xn+2 ? yn+2; ") : n+2 n+2 n+2 Since suppf1 g = R and N~ is a P0 matrix, In+1 ? Dn+1(x ? y; ")(In+1 ? N~ ) is nonsingular [8]. Hence H 0(z; ") is nonsingular if and only if dn+2(xn+2 ? yn+2; ") 6= 21 . By the de nition of dn+2(xn+2 ? yn+2; "), we have dn+2(xn+2 ? yn+2; ") 6= 21 if and only if xn+2 ? yn+2 6= ?2". This completes this lemma. We de ne the Big-M smooth path as a set of solutions z(") for all 1 " > 0, i.e., S" = fz : H (z; ") = 0; 1 " > 0g: We choose the following neighborhood around the Big-M smooth path: N = fz : kH (z; ")k c"; 1 " > 0g; where c is a positive constant. Lemma 2.3 If (z; ") satis es kH (z; ")k c"; (2.2) then xi + c" 0; yi + c" 0 (2.3) and (xi + c")(yi + c") 2(c + 1)(jxij + jyij)" + (4 + c2)"2; (2.4) for i = 1; 2; : : : ; n + 2. Moreover, the neighborhood N is bounded.
7
Proof: Suppose (z; ) satis es (2.2). Then we have ! Nx + p ? y kH (z; ")k = k k c": (x; y; ")
Set u = (x; y; "): Then u 2 Rn+2 satis es c" ui ?c"; i = 1; 2; : : : ; n + 2:
By construction of (cf. Lemma 1 in [2]), we have (x ? u; y ? u; ") = 0:
(2.5)
Since 1 is continuous, symmetric and has an in nite support, by Theorem 2.1 in [4], we have xi ? ui > 0; yi ? ui > 0; i = 1; 2; :::; n + 1: Using c" ?ui , we obtain (2.3) for i = 1; 2; : : : ; n + 1: Now we show (2.3) for i = n + 2: By the de nition of and (2.5), we have the following equalities 0 = n+2(xn+2 ? un+2; yn+2 ? un+2; ") Z (xn+2 ?yn+2)=" (xn+2 ? yn+2 ? ")2()d: = xn+2 ? un+2 ? ?1
This implies that xn+2 ? un+2 0; since the integral part is nonnegative. To show yn+2 ? un+2 0, we assume on the contrary that yn+2 < un+2. Then xn+2 ? yn+2 > xn+2 ? un+2 0 and xn+2 ? un+2 =
Z (xn+2 ?yn+2)=" ?1
(xn+2 ? yn+2 ? ")2 ()d
Z0 1 = (x ? y ? ")d 4 ?4 n+2 n+2 = xn+2 ? yn+2 + 2";
which implies
yn+2 ? un+2 = 2" > 0: This contradicts the assumption that yn+2 ? un+2 < 0: Hence we have xn+2 ? un+2 0; and yn+2 ? un+2 0:
Using c" ?un+2, we obtain (2.3) for i = n + 2. Now we show (2.4). Suppose that xi yi. Then we have xi ? ui ? max(xi ? yi; 0) = min(xi ? ui ; yi ? ui ) = xi ? ui :
From Lemma 2.2 and (2.5), xi ? ui = j i(xi ? ui; yi ? ui; ") ? (xi ? ui)j 2":
8 By a simple manipulation, we obtain (xi + c")(yi + c") = (xi ? ui)(yi ? ui) + (xi + yi)(ui + c") ? u2i + c2"2 = (xi ? ui)(yi ? xi ) + (xi ? ui)2 + jxi + yij(ui + c") ? u2i + c2"2 2(jxij + jyij)" + 4"2 + 2jxi + yijc" + c2"2 2(1 + c)(jxi j + jyij)" + (4 + c2)"2; where the rst inequality follows from 0 xi ? ui 2" and ? c" ui c": The proof is similar for the case xi yi: Now, we show that N is bounded. Using c" Hn+2 (z; ") ?c";
we have c"
(Nx + p ? y)n+2 = ?eT x + ? ? yn+2 ?c":
This implies
? + c" eT x + yn+2 ? ? c": Since xi and yi are bounded below by (2.3), x and y cannot go to 1, and so N is bounded.
Theorem 2.1 Suppose that LCP(N; p) has a solution z . Then for every " > 0, there is a z" 2 R2(n+2) such that H 0 (z" ; ")T H (z"; ") = 0
and
q
kH (z"; ")k 2 2(n + 2)": q
(2.6) (2.7)
Proof: Let c be a constant satisfying c > 2 2(n + 2): We consider the following set: D(") = fz j kH (z; ")k c"g: By Lemma 2.2, at the solution z , we have
kH (z; ")k kH (z ; ") ? H0(z )k + kH0(z )k = 2": Hence z 2 D(") and so D(") is nonempty.
9 By Lemma 2.3, D(") is bounded. Let 1 (z ) = kH (z; ")k22 : 2 Since H (; ") is continuously di erentiable and D(") is nonempty and bounded, has a global minimum point z" in D("). Moreover, we have
kH (z"; ")k
kH (z"; ")k2 kqH (z; ")k2 2(n + 2)kH (z; ")k q 2(n + 2)(kH0(z )k + 2") q
= 2 2(n + 2)": Hence, (2.7) holds. Moreover, this implies z" 2 int D("). By [13, 4.1.3], we have 0 (z" ) = H 0 (z" ; ")T H (z" ; ") = 0:
We complete the proof. Theorem 2.1, together with Lemma 2.1, shows that if the LCP(M; q) has a solution, then there is a ? such that the LCP(N; p) has a solution and for every " > 0, we can nd a stationary point z" of 1 kH (z ; ")k2 " 2 2 which satis es q kH (z"; ")k 2 2(n + 2)": Hence following a path de ned by such stationary point as " ! 0, we can nd a solution of the LCP(M; q). By Lemma 2.1, if M is a P0 matrix, the rank of H 0(z"; ") is always greater than 2(n + 1) + 1. Moreover, H 0(z"; ") is nonsingular if and only if xn+2 ? yn+2 6= ?2". Hence, we can claim that H 0(z"; ") is nonsingular almost everywhere if M is a P0 matrix. From (2.6), if H 0(z"; ") is nonsingular then z" is a solution of H (z; ") = 0:
(2.8)
It is notable that Assumption 1.1 does not guarantee the existence of a solution of (2.8) for every " > 0. The following example shows that even if the P0 matrix LCP(M; q) has a unique strictly complementarity solution, the central path can be very short. Precisely, for any "0 > 0, changing an element in q can make that the central path does not exist for all " "0 even if the resulting P0 matrix LCP has a unique strictly complementarity solution. Furthermore, if the solution set is unbounded, the central path may not exist for all " > 0. On the other hand, the Big-M smooth path may be \much longer" than the central path.
10
Example 2.1 Let
!
M = 00 ?11
q=
!
? 12 ;
where > 21 :
It is easy to verify that M is a P0 matrix and the LCP(M; q) has a unique solution (t ; s) = (0; 21 ; ? 12 ; 0): However for any 0 > 0 the complementarity level set f(t; s) : tT s 0 + ( ? 21 ); s = Mt + q; (t; s) > 0g contains the unbounded line 0 0 0 (k; ? ; ; ? ? 1 ) k k k 2 for all k > 0 =( ? 12 ): The central path
f(t; s) : ti si = "; i = 1; 2p; s = Mt + q; (t; s) > 0g = f( ?" t ; t2); t2 = 1 + 14 + 16" < g 2
does not exist for " ( ? 21 ): If = 21 , this problem has a unbounded solution set f(x1 ; 21 ; 0; 0); x1 0g: In this case the central path does not exist for all " > 0. Now we show that if we choose ? n + 5 = 7; then for all " 1, (2.8) has a solution, i.e. the Big-M smooth path exists for " 1: By Lemma 2.1 in [9] and Lemma 2.3, at the Big-M smooth path, (x; y) 0 and x1 y1 = x1 ( ? x2 + (2 ? )x3 ) = "2
1 2
1 2 x3 y3 = x23 = "2 y4 = ? ? x1 ? x2 ? x3 ? x4 Z (x4 ?y4)=" x4 = (x4 ? y4 ? ")2()d:
x2 y2 = x2 (? + x2 + x3 ) = "2
?1
We can calculate the point at the Big-M smooth path x3 = y3 = " 1
11
q
1 ? " + (1 ? ")2 + 16"2 x2 = 1 4 x1 =
"2
? x2 + (2 ? )"
1 1 2
y1 = ? x2 + (2 ? )x3 ; y2 = x2 + x3 ?
or
1 2
x4 = ? ? x1 ? x2 ? x3 ? 2" 2"; y4 = 2"
x4 = 0; y4 = ? ? x1 ? x2 ? x3 4"; where we use () = 0; for 62 [?4; 0] to calculate x4 and y4 .
There are two Big-M smooth paths which are bounded and converge to two solutions 1 1 1 z ;1 = (0; ; 0; ? ? ; ? ; 0; 0; 0) 2 2 2 and 1 1 1 z ;2 = (0; ; 0; 0; ? ; 0; 0; ? ? ); 2 2 2 respectively, as " ! 0. Both of them contain the solution of the original LCP(M; q) : (x1;1 ; x2;1 ; y1;1; y2;1) = (x1;2 ; x2;2; y1;2; y2;2) = (0; 12 ; ? 12 ; 0):
We cannot say that the solution of (2.8) is unique for every " > 0, since N is not a P0-matrix, Nevertheless, we can say that the solution is locally unique. This result is given by the following theorem.
Theorem 2.2 Suppose that M is a P0 matrix. Let z" be a solution of (2.8). Then its 2(n + 1) components (z" )i ; i = 1; 2; : : : ; n + 1; n + 3; : : : ; 2(n + 1) + 1, that is, ((x")i ; (y")i); i = 1; 2; : : : ; n + 1, are unique. Moreover the following results hold. 1. If H 0(z" ; ") is singular, then z" is the unique solution of (2.8). 2. If H 0 (z" ; ") is nonsingular, then z" is the unique solution of (2.8) in a neighborhood of z".
Proof: 1. Let x~ = (x1 ; : : : ; xn+1); y~ = (y1; : : : ; yn+1); z~ = (~x; y~) and ~ z ; ") = ( 1(x1 ; y1; "); : : : ; n+1(xn+1 ; yn+1; "))T : (~ Then z~ is a solution of
!
N~ x~ + p ? y~ : ~ (~z ; ")
(2.9)
Since N~ is a P0 matrix and ~ is given by 1 , by Theorem 2.3 in [4], z~" is the unique solution of (2.9).
12 1.
Assume that (2.8) has two solutions z"1 and z"2 . Then (z"1 )i = (z"2 )i, i = 1; : : : ; n + 1; n + 3; : : : ; 2(n + 1) + 1: By the structure of the (n + 2)th component of H , we obtain x1n+2 ? x2n+2 = yn2 +2 ? yn1 +2 : (2.10) Hence if x1n+2 = x2n+2, then yn1+2 = yn2+2. Without loss of generality, we assume on the contrary that x1n+2 > x2n+2 : Let (w) =
Z
w "
?1
(w ? ")2()d:
Then from n+2(x1n+2 ; yn1+2; ") = n+2(x2n+2; yn2+2; ") = 0; we have 0 < x1n+2 ? x2n+2 = (x1n+2 ? yn1+2) ? (x2n+2 ? yn2+2) = 0 ()(x1n+2 ? yn1+2 ? x2n+2 + yn2+2) = 2 0()(x1n+2 ? x2n+2 ); where 2 (x1n+2 ? yn1+2; x2n+2 ? yn2+2), and the last equality follows from (2.10). This implies that Z 0 () = " ()d = 1 2 2 ?1 which gives = ?2" and x1n+2 ? yn1 +2 > ?2" > x2n+2 ? yn2 +2 :
By Lemma 2.2, H 0(z"; ") is nonsingular. This is a contradiction. 2. Let G(z ) = z ? H 0 (z" ; ")?1H (z; "): Since H is continuously di erentiable, G is continuously di erentiable and G0 (z ) = I ? H 0(z" ; ")?1 H 0 (z; "):
Moreover the spectral radius of G0 (z") is zero. By Ostrowski Theorem [13], we claim that z" is the unique solution of (2.8) in a neighborhood of z": Results 1 and 2 of Theorem 2.2 may be explained by square functions. f (x) = x2 = 0 has a unique zero x = 0, which is multiple roots, and its derivative is singular at x. f (x) = x2 ? 1 has two zeros x;1 = 1 and x;2 = ?1, which are unique in (?1; 1) and (?1; 1) respectively, and its derivative is nonsingular at x;1 and x;2 .
3 Algorithm and its convergence In this section we propose an algorithm and prove its global convergence.
Algorithm 3.1 Given c 2, 2 (0; 1), and i 2 (0; 1) for i = 1; 2:
13
Step 0 (Initial Step )
Choose x0 ; y 0; "0 such that kH (z 0 ; "0)k c"0 and H 0(z 0 ; "0) is nonsingular.
Step 1 (Newton Step)
If H (z k ; "k ) = 0, set z k+1 = z k and go to Step 3. Otherwise, Let z k solve the equation H (z k ; "k ) + H 0 (z k ; "k )z k = 0: (3.1)
Step 2 (Line Search)
Let k be the maximum of the values 1; 1; 12 ; : : : such that
kH (zk + k zk ; "k )k (1 ? k )kH (zk ; "k )k:
(3.2)
Set z k+1 = z k + k z k .
Step 3 (" Reduction)
Let k be the maximum of the values 22 ; 23 ; : : : such that
kH (zk+1; (1 ? k )"k )k (1 ? k )c"k : (3.3) k +1 k k +1 If xkn+1 = (1 ? k )"k . Otherwise, set "k+1 = +2 ? yn+2 6= ?2(1 ? k )" ; set " (1 ? 2 k )"k : Algorithm 3.1 is similar to the smoothing method introduced by Burke and Xu [1]. The main di erence from the Burke-Xu method is that the de nition of "k+1 in Step 3 ensures the nonsingularity of H 0(zk+1 ; "k+1) for the P0 matrix LCP(M; q). It is easy to verify that if y0 = Nx0 + p, then yk = Nxk + p for all k 0. The following lemma shows that we can easily nd a starting point (z0 ; "0) satisfying these conditions in the initial step of Algorithm 3.1.
Lemma 3.1 Suppose that M is a P0 matrix. Let x0 = (e; 1; 0); y0 = (e; 1; ? ? (n +1)) and
1 "0 1: c+1 0 0 0 0 Then y = Nx + p 0, kH (z ; " )k c"0 , and H 0 (z 0 ; "0 ) is nonsingular.
(3.4)
Proof: Obviously y0 = Nx0 + p. Since ? n + 5, y0 0. Thus Hi(z0 ; "0) = 0; i =
1; 2; : : : ; n + 2. By a simple calculation, we have i(x0i ; yi0; "0) = 1 + 0 and n+2
(x0
n+2
; y0
n+2
; "0 )
=
Z0
?1
1 ()d = 1 ? "0 ; i = 1; 2; : : : ; n + 1;
Z ((n+1)??)="0 ?1
((n + 1) ? ? ? "0)2()d = 0;
14 where the last equality uses 2 () = 0 for ?4 and (3.4). Hence (z0 ; "0) satis es kH (z0 ; "0)k c"0: Moreover, from ? n + 5 and "0 1, we have ?2"0 > ?4 (n + 1) ? ? = x0n+2 ? yn0+2: Therefore H 0(z0 ; "0) is nonsingular by Lemma 2.2, since x0n+2 ? yn0+2 6= ?2"0.
Theorem 3.1 If M is a sequence fz k g satis es
P0 matrix, then Algorithm 3.1 is well de ned, and the
kH (zk ; "k )k c"k :
(3.5)
Proof: We prove this theorem by induction.
For k = 0, by Lemma 3.1, z0 = (e; 1; 0; e; 1; ? ? (n + 1)) satis es (3.5) and H 0 (z 0 ; "0 ) is nonsingular. We suppose that zk satis es (3.5) and H 0(zk ; "k ) is nonsingular. Then Step 1 is well de ned. If H (zk ; "k ) 6= 0, then zk 6= 0: Hence zk is a strictly decent direction of kH (; "k)k at zk , and so the line search procedure is nite by construction in Step 2. Step 3 is well de ned since if H (zk ; "k ) = 0 then zk+1 = zk and kH (zk+1; "k )k = 0 < c"k : Otherwise, by the construction of Step 2, kH (zk+1; "k )k < kH (zk ; "k )k c"k ; which implies that there is a nite number k > 0 such that (3.3) holds. By the construction of Step 3, H 0(zk+1 ; "k+1) is nonsingular. Now we show that (3.5) holds at (zk+1 ; "k+1). k +1 k If xkn+1 +2 ? yn+2 6= ?2(1 ? k )" ; then by construction of Step 3, (3.5) holds. Hence we only need to consider the case k +1 k xkn+1 +2 ? yn+2 = ?2(1 ? k )" ;
i.e., Notice that and Step 3 provides that
"k+1 = (1 ? 2 k )"k : "k > (1 ? 2 k )"k > (1 ? k )"k c"k+1
c(1 ? 2 k )"k c(1 ? k )"k H (z k+1 ; (1 ? k )"k ) ?c(1 ? k )"k :
(3.6)
15 By Result 3 of Proposition 1 in [2], for i = 1; 2; : : : ; n + 1, i(xki +1; yik+1; ) is strictly decreasing with respect to ", which gives
i(xki +1; yik+1; (1 ? k )"k )
c"k+1
> i (xki +1 ; yik+1; "k+1) = i(xki +1; yik+1; (1 ? 2 k )"k ) i(xki +1; yik+1; (1 ? k )"k ) + ( 2 ? 1) k "k > ?c(1 ? k )"k ? (1 ? 2 ) k "k = ?c"k + c k "k ? k "k + 2 k "k ?c"k + 2(c ? 1) k "k + 2 k "k > ?c"k + c 2 k "k = ?c"k+1;
where the third inequality follows from Result 4 of Proposition 1 in [2] (also see [8]), and the fourth inequality follows from (3.6). Hence (3.5) holds for i = 1; 2; : : : ; n +1: k +1 Let w = xkn+1 +2 ? yn+2 ; and (") =
Z
w "
?1
Then ! = ?2(1 ? k )"k < 0 and 0 (") = ?
Z
(w ? ")2()d: w "
?1
2 ()d 0:
Hence is monotonically increasing function. This implies that n+2 is monotonically decreasing and Lipschitz continuous with respect to the smooth parameter ". Hence we obtain c"k+1
k +1 k n+2(xkn+1 +2 ; yn+2 ; (1 ? k )" ) k k +1 n+2(xkn+1 +2 ; yn+2 ; (1 ? 2 k )" ) k +1 k k n+2(xkn+1 +2 ; yn+2 ; (1 ? k )" ) ? 2(1 ? 2 ) k " ?c(1 ? k )"k ? 2(1 ? 2 ) k "k = ?c"k + (c ? 2(1 ? 2 )) k "k = ?c"k + c 2 k "k + (c ? 2)(1 ? 2) k "k ?c(1 ? 2 k )"k = ?c"k+1;
where the third inequality follows from that n+2 is Lipschitz continuous with the Lipschitz constant 2 and c 2. (See Lemma 2.2). Therefore we have c"k+1 Hi (z k+1 ; "k+1) ?c"k+1 ; for i = n + 3; : : : ; 2(n + 2):
16 By the de nition of H , the parameter " is not involved in the rst n+2 components of H , i.e., Hi (z k+1 ; "k ) = Hi (z k+1 ; "k+1 ); for i = 1; 2; : : : n + 2: Hence (3.5) holds.
Theorem 3.2 Suppose that M is a P0 matrix. Then Algorithm 3.1 is well de ned. Let f(z k ; "k )g be a sequence generated by Algorithm 3.1. Then fz k g is remained in the bounded neighborhood N , and f"k g decreases monotonically in R++ . If furthermore for an accumulation point (z; ") of f(z k ; "k )g, we have xn+2 ? yn+2 = 6 ?2" or " = 0 then
lim "k = 0 and !1
k
lim H (z k ; "k ) = 0; !1
k
(3.7)
and for all accumulation points fz^; "^g of fz k ; "k g.
"^ = 0 and H (^z ; "^) = H0 (^z ) = 0:
(3.8)
Proof: By Theorem 3.1 and Lemma 2.3, Algorithm 3.1 is well de ned and fzk g is remained in a bounded set. Furthermore, by construction of Algorithm 3.1, 0 < "k+1 = (1 ? k )"k < "k or
0 < "k+1 = (1 ? 2 k )"k < "k : Hence f"k g is a monotonically decreasing sequence, and there is " such that lim "k = ": !1
k
If " = 0, then from Theorem 3.1, we have (3.7). Moreover, since f"k g is a monotonically decreasing sequence, (3.5) and (3.7) imply (3.8). Suppose on the contrary that " > 0: This implies k ! 0. Since fzk g is remained in the bounded neighborhood N , taking a subsequence if necessary, we may assume that the sequence fzk g converges to some z. Based on the " reduction step, we have kH (zk ; (1 ? 1 )"k?1)k > (1 ? 1 )c"k?1: 22 k?1
22 k?1
Since k ! 0, by passing limits on both sides, we have
kH (z; ")k c" > 0:
(3.9)
Since xn+2 ? yn+2 6= ?2", H 0(z ; ") is nonsingular by Lemma 2.2. Hence we can nd a unique solution z of the linear equations in Step 1. Furthermore, from (3.9) it is a strictly decent direction for kH (; ")k at z. As a result, the corresponding linear search step length and " reduction step length are both bound below by
17 a positive constant. Notice that the function H and its Jacobian H 0 are continuous in a neighborhood of (z; "). It follows that zk converges to z and therefore k must be uniformly bounded below by some positive constant for all large k. This contradicts to the assumption that k ! 0. Hence we must have "k ! 0, and so (3.7) and (3.8) hold. Let us denote
" = fz : kH (z; ")k c"g: By Lemma 2.3, the set " is bounded for every " > 0. Corollary 3.1 Suppose that the solution set S0 (M; q) of the P0 matrix LCP(M; q) is nonempty and bounded. Then there exist a big ? and "0 such that the sequence fzk g generated by Algorithm 3.1 converges to a solution of LCP(N; p). Proof: First we show that
"k "k?1 ; for "k "k?1: Suppose that z 2 "k . Then kH (z; "k?1)k kH (z; "k?1) ? H (z; "k )k + kH (z; "k )k 2("k?1 ? "k ) + c"k = c"k?1 ? (c ? 2)("k?1 ? "k )
c"k?1;
where we use c 2 and Lemma 2.2. Hence z 2 "k?1 : This, together with Theorem 3.1, implies that the sequence generated by Algorithm 3.1 remains in "0 : Let ? satisfy ? > 4eT t; for all (t; s) 2 S0(M; q): Then from Lemma 2.1, the solution set of the LCP(N; p) is given by
0 = f(t; 0; ? ? eT t; s; 0; 0); (t; 0; 0; s; 0; ? ? eT t); (t; s) 2 S0(M; q)g: Hence for a (t; s) 2 S0 (M; q), z;1 = (t; 0; ? ? eT t ; s; 0; 0) and z;2 = (t; 0; 0, s; 0; ? ? eT t) are solutions of LCP(N; p) and eT t + max min(jx ? y ? x;1 ? y;1jn+2; jx ? y ? x;2 ? y;2jn+2) z 2 0 = eT t + max jeT (t ? t)j z 2 0 ?4 + ?4 = ?2 : By the continuity of H (z; ") on ", for such ? there exists "0 2 (0; min(1; ?4 )) such that kH (z0; "0)k c"0 and for all z 2 "0 , ? eT t + 2"0 + zmax min(jx ? y ? x;1 + y;1jn+2; jx ? y ? x;2 + y;2jn+2): 2
"0
18 Let z 2 "0 . Without loss of generality we may assume jx ? y ? x;1 + y;1jn+2 jx ? y ? x;2 + y;2jn+2. Then
jxn+2 ? yn+2j 1 1 1 jxn;+2 ? yn;+21 j ? jxn+2 ? yn+2 ? xn;+2 + yn;+2 j T ;1 ;1 ? ? e t ? zmax min(jx ? y ? x + y jn+2; jx ? y ? x;2 + y;2jn+2) 2 "0 2"0 > 2"k for all k 1. Hence at any accumulation point (z ; ") generated by Algorithm 3.1, xn+2 ? yn+2 = 6 ?2". By Theorem 3.2, we complete the proof.
The proof of Corollary 3.1 shows that if the solution set of the P0 matrix LCP(M; q) is nonempty and bounded, then there exist ? and 0 such that for any z 2 ", " "0, H 0 (z" ; ") is nonsingular. Hence from Theorem 2.2 and Corollary 3.1, two Big-M smooth paths exist but they never cross each other, which converge to two solutions. The proof of Corollary 3.1 also suggests us to increase ? and restart Algorithm 3.1 when xn+2 ? yn+2 = ?2". For instance, we choose new starting point as x^0i = xi ; i = 1; 2; : : : ; n + 1 y^i0 = yi; i = 1; 2; : : : ; n + 1 1 x^0n+2 = xn+2 ? " 2 3 0 y^n+2 = yn+2 + "; 2 ?^ = ? + " "^0 = ": Then at the new starting point, the new LCP(N; p^) with the new ?^ satis es H^ i (^z 0 ; "^0 ) = Hi (z ; "); i = 1; 2; : : : ; 2(n + 1) + 1 and " Z ?2 1 0 0 0 ^ H2(n+2) (^z ; "^ ) = x^n+2 = xn+2 ? " = xn+2 + 2 4 ?4 (2 + )d = H2(n+2) (z; "); where we use x^0n+2 ? y^n0+2 ?4^"0 and 2 () = 0 for ?4: Hence we have H^ (^z 0 ; "^0 ) = H (z ; ") and "^0 = ": Furthermore, H^ 0(^z0 ; "^0) is nonsingular since x^0n+2 ? y^n0 +2 = xn+2 ? yn+2 ? 2" = ?4" = ?4^"0 :
At the end of this paper, we consider the Big-M smooth path for Example 1.1, which is a P0 matrix LCP but its solution set is unbounded.
19 In the Big-M smoothing model for Example 1.1, r = e ? Me ? q = 0. We choose ? = n + 5 = 8: At the Big-M smooth path, xi > 0; yi > 0 yi = (Nx + p)i ; xi yi = "2 ; i = 1; 2; 3; 4
and x5 =
Z (x5 ?y5)=" ?1
(x5 ? y5 ? ")2()d; y5 = ? ? x1 ? x2 ? x3 ? x4 ? x5 :
We can calculate this point for " 2 (0; 1]:
p
4 "2 4 "2 1 + 8 "2 + 1 p p x1 = 1; x3 = 1 + 8"2 + 1 1; 1; x2 = 4 1 + 8"2 + 1 x4 = " 1; and x5 = 0; or x5 = ? ? x1 ? x2 ? x3 ? x4 ? 2"; where we use 2 () = 0; for 62 [?4; 0] to calculate x5 : Two Big-M smooth paths never cross each other and converge to two solutions 1 1 1 1 z ;1 = (0; ; 0; 0; ? ? ; ; 0; ; 0; 0); 2 2 2 2 and 1 1 1 1 z ;2 = (0; ; 0; 0; 0; ; 0; ; 0; ? ? ); 2 2 2 2 respectively, as " ! 0. Both of them contain a strictly complementarity solution of the original LCP(M; q) (t ; s) = (0; 12 ; 0; 21 ; 0; 21 ):
Acknowledgement
The authors are grateful to M.S.Gowda, C.Kanzow, D.Sun and P.Tseng for their helpful comments.
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