Chapter 1 LINEAR COMPLEMENTARITY PROBLEM, ITS ...

2 Chapter 1. Linear Complementarity Problem, Its Geometry, and Applications
The only data in the problem is the column vector q and the square matrix M. So we
will denote the LCP of nding w 2 R n , z 2 R n satisfying (1.1) by the symbol (q M).
It is said to be an LCP of order n. 8In
an9LCP of8order n there are 2n variables. As
a speci c example, let n =2,M = >: 2 1>,
q = >:
1 2
;5
9
>. This leads to the LCP
;6
w1 ; 2z1; z2 = ;5
w2 ; z1;2z2 = ;6:
w1w2z1z2 > = 0 and w1z1 = w2z2 =0:
The problem (1.2) can be expressed in the form of a vector equation as
w1
8
9
8
9
8
9
(1:2)
>: 1>
+ >: w2
0
0>
+ >: z1
1
;2>
+ >: z2
;1
;1>
= >:
;2
;5>
;6
(1:3)
w1w2z1z2 > = 0 and w1z1 = w2z2 =0 (1:4)
In any solution satisfying (1.4), at least one of the variables in each pair (wjzj),
has to equal zero. One approach for solving this problem is to pick one variable from
each of the pairs (w1z1), (w2z2) and to x them at zero value in (1.3). The remaining
variables in the system may be called usable variables. After eliminating the zero
variables from (1.3), if the remaining system has a solution in which the usable variables
are nonnegative, that would provide a solution to (1.3) and (1.4).
Pick w1, w2 as the zero-valued variables. After setting w1, w2 equal to 0 in (1.3),
the remaining system is
-2 () -1
z1
8
9
8
9
8
8
>: ;2>
+ >: z2
;1
;1>
= >:
;2
;5>
=
;6
-1 ( -2)
q
2
z1 > = 0 z2 > = 0
q1
9
9
8
8
>: q1
9
> = q
q2
9
(1:5)
Figure 1.1 A Complementary Cone
Equation (1.5) has a solution i the vector q can be expressed as a nonnegative
linear combination of the vectors (;2 ;1) T and (;1 ;2) T . The set of all nonnegative