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Linear Complementarity Problems and their Sources

The Linear Complementarity Problem (LCP) (q, M) is defined as

follows:

Given a real nn matrix M and an n-vector q, find z Rn suchthat

z0, q+ M z0, zT(q+ M z) = 0.

Define the mapping F(z) :=q+ M z.

ThenFis an affine transformation from Rn into itself.

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Some notation

Given the LCP (q, M)we write

FEA(q, M) ={z :q+ M z0, z 0}

and

SOL(q, M) ={z :q+ M z0, z0, zT(q+ M z) = 0.}

These are thefeasible set (region)andsolution setof the LCP(q, M),

respectively.

Note that ifFEA(q, M)=, it is a closed polyhedral set.

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How do such problems arise?

Optimality criterion for Linear Programming(LP)

Consider the LP

(P)minimize cTxsubject to Ax b

x 0.

According to the theory of linear programming, a vector x is optimal

for (P) if and only if it is feasible and there exists a vector ysuch that

yT

A cT

, y0, yT

(Ax b) = 0, (yT

A cT

)x= 0.

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Now arrange these conditions as follows:

u = c + ATy 0

v = b + Ax 0

x 0, y0

xTu= 0, yTv= 0.

Next define

w=

u

v

, q=

c

b

, M=

0 AT

A 0

, z =

x

y

.

The optimality conditions of the LP then become the LCP(q, M).

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Optimality conditions for Quadratic Programming(QP)

Consider the QP

(P)

minimize cTx + 12

xTQx

subject to Ax bx 0.

According to the Karush-Kuhn-Tucker (KKT) Theorem, if the vector

x is a local minimizer for (P), there exists a vector y such that

c+QxATy0, y0, yT(Axb) = 0, xT(c+QxATy) = 0.

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If we assemble these conditions along with the feasibility of the vector

x, we obtain the LCP (q, M)where

w= u

v , q=

c

b , M=

Q AT

A 0 , z =

x

y .

Remark. The above necessary conditions of optimality for QP are also

sufficient for (global) optimality whenQis positive semi-definite.

Note that ifQis positive semi-definite, then so is

M= Q AT

A 0 .

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Bimatrix Games as LCPs

The initial set up

LetA andB denote twom nmatrices.

These are payoff matrices for Players I and II, respectively.

Letm={x Rm+ :eTx= 1}andn={yRn+:e

Ty= 1}.

If x m and y n, the expected lossesof Players I and II are,respectively:

xTAy and xTBy.

Let(A, B)denote the corresponding two person game.

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Nash Equilibrium Point of (A,B)

The pair (x, y) m n is a Nash Equilibrium Point(NEP) for(A, B)if

(x)TAy xTAy for all x m

(x)TBy (x)TBy for allyn

It is crucial to note that given(x, y) mn, each of the vectorsx, y is optimal in a simple linear program defined in terms of the

other. The LPs are:

minimize(Ay)Tx subject to eTx= 1, x 0

and

minimize(BTx)Ty subject to eTy= 1, y0

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Let E be the m n matrix whose entries are all 1. For a suitablescalar >0, all the entries of the matrices A + EandB+ Eare

positive.

It is easy to see that (A, B)and(A + E,B + E)have the same

equilibrium points (if any).

Thus, it is not restrictive to assume that A andB are (elementwise)

positive matrices.

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Now consider the LCP

u =em+ Ay0, x 0, xTu= 0

v =en+ BTx 0, y0, yTv = 0

In this case, we have

w=

u

v

, q=

em

en

, M=

0 A

BT 0

, z=

x

y

,

where

em= (1, . . . , 1) Rm en= (1, . . . , 1) R

n.

We wish to show that

to every solution of this LCP, there corresponds a Nash equi-

librium point of(A, B)and vice versa.

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The correspondences are as follows:

If(x, y)is a Nash equilibrium of(A, B), then

(x, y) = (x/(x)TBy, y/(x)TAy)

solves the LCP (q, M)given above.

If(x, y)solves the LCP (above), then

(x, y) = (x/eTmx, y/eTny

)

is a Nash equilibrium point for (A, B).

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A Market Equilibrium Problem

Here we seek to determine prices at which there is a balance between

supplies and demands.

The supply sideminimize cTx

subject to Ax bBx r

x 0

The demand side

r=Q(p) =Dp + d

Equilibration

p =

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Formulation as an Equilibrium Problem

y =c ATv BT 0, xT0, (x)Ty = 0

u =b + Ax 0, v 0, (v)Tu = 0

=r + Bx 0, 0, ()T = 0

SubstituteDp

+ dforr

and

forp

.

Then we get the LCP (q, M)with

q=

cbd

M=

0 AT BTA 0 0

B 0 D

and z=

xv

What sort of matrix is D? Having it be negative (semi)definite andsymmetric would be nice.

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Convex Hulls in the Plane

Given {(xi, yi)}n+1i=0 R

2 find the extreme pointsand the facetsof

their convex hull and the order in which they appear.

First find the lower envelopeof the convex hull.

Ifxi = xj and yi yj, we can ignore (xj, yj) without changing thelower envelope.

Thus, assume x0 < x1 < < xn < xn+1. In practice, this wouldrequiresorting.

The lower envelope is a piecewise linear convex function f(x), thepointwise maximum of all convex functionsg(x)such thatg(xi) yi

fori= 0, 1, . . . , n + 1.

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Defineti=f(xi) and letzi=yi ti, fori= 0, 1, . . . , n + 1.

Note thatz0=zn+1= 0.

If(xi, yi)is a breakpoint, then ti=yi andzi= 0.

The segment of the lower envelope between (xi1, ti1) and (xi, ti)

has a different slope than the segment between (xi, ti) and (xi+1, ti+1).

Since f(x) is convex, the former (left-hand) segment must have a

smaller slope than the latter (right-hand) segment.

Hence strict inequality holds in

ti ti1xi xi1

ti+1 tixi+1 xi

.

Ifzi>0, then(xi, yi)cannot be a breakpoint off(x).

In that case, equality holds in the inequality above.

8/12/2019 040504

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The vector z = {zi}ni=1 must solve the LCP (q, M) where q Rn

andMRnn are defined by

qi = i i1 and mij =

i1+ i ifj =i,

i ifj =i + 1,

j ifj =i 1,

0 otherwise,

and where

i= 1/(xi+1 xi)andi=i(yi+1 yi)fori= 0, . . . , n .

This LCP has a unique solution.

The matrix M associated with this LCP has several nice propertieswhich can be exploited to produce very efficient solution procedures.