Journal of Mathematical Economics 4 (1977) U-86.0 North-Holland Publishing Company

PIECEWISE LINEAR EXCHANGE EQUILIBRIUM

David GALE

University of California, Berkeley, and Center for Advanced Study in the Behavioral Sciences, Stanford, CA, USA

Received January 1976

It is known that if a pure exchange model in which all consumers have linear utility functions has an equilibrium it has one which is rational in the initial data. Examples show on the other hand that this is no longer true in general if utility functions are merely piecewise linear. In this note we observe that for the case of two traders the rationality result carries over and we sketch a finite procedure for calculating the equilibrium.

1. Introduction

In Eaves (1976) a finite procedure has been given for calculating an equilibrium for a pure exchange economy when all agents have linear utility functions. As a corollary to this result it follows that if the original data of the problem are given by rational numbers then the equilibrium prices and allocations can also be chosen to be rational. On the other hand, an example due to Mas-Cole11 shows that if the utility functions of agents are only piecewise linear then this rationality result no longer holds. The Mas-Cole11 example involves three agents, each with a utility function of the constant proportions type.

It is the purpose of the present note to show that there is one situation in which the rationality result does carry over to the piecewise linear case, namely when there are only two agents. The case may be of somewhat more than academic interest. Dantzig in connection with energy studies has proposed an international trade model in which each country is trying to maximize a linear social welfare function subject to linear constraints. More precisely, in an n-good world each country wants to solve a linear program of

maximizing cx ,

subject to Ax = y, x 2 0,

where x is the usual vector of activity levels and y is the n-vector of goods available to the country. Now if one defines u(y) to be the value of this program then it is well known that u is a concave piecewise linear function defined on the

82 D. Gale, Piecewise linear exchange equilibrium

cone generated by the columns of A. Thus, the Dantzig model is a special case of a piecewise linear model and our present analysis would apply to the two-country case of such a model.

As in the case of the linear model, the proof of rationality strongly suggests that there is a finite procedure for finding a competitive equilibrium. Such a procedure will be sketched in our final section.

2. The m-trader model

A concave piecewise linear function is defined to be the minimum of a finite number of linear functions defined on a convex polyhedron. It is easy to see that any such function ui can be written as

where

u’(x) = max r,

Se’-A’x s b’.

(1)

(2)

Here e’ and b’ are given vectors and A’ is a given matrix. The superscript i indicates that ui is the utility function of the ith trader, Ti. In an n-good world each matrix A’ will have n columns and x represents the trade proposed by Ti,

where we may think of good j as being demanded or supplied according as the jth entry xj of x is positive or negative. We denote by ej and bi, the jth entry of e’ and b’, and by A;, the jth column of A’. In order for the maximum (1) to exist it suffices fore’ to have at least one positive entry. This will be assumed from here on.

Definition. An equilibrium is a set of m+ 1 n-vectors (p, xi), i = 1, . . ., m, such that

xi maximizes U’(X), (utility maximization), (3)

subject to

PX 6 0, (budget inequality),

and

c xi = 0, (balance of supply and demand).

[Note that (4) and (5) implypx’ = 0 for all i.]

(4)

(5)

We say that T, is unsatiated at the above equilibrium if there exists x such that U(X) > u(X’). We first prove:

D. Gale, Piecewise linear exchange equilibrium 83

Lemma 1. If T, is unsatiated ut the equilibrium (p, xi) then there exists a non- negative m-vector vi and a positive number Ei, such that

V’A’ =p, (6)

viei = ei, (7)

ci(viei) - viAXi = &i, gi = u(2). (8)

Proof. From the definition of equilibrium it follows that (c$, xi) is a solution of the linear program of maximizing 5 subject to (2) and (4). From the Duality Theorem it therefore follows that there exists a non-negative vector v and a non- negative scalar A, such that

VA’ = Ap, (6’)

ve’ = 1, (7’)

[‘(ve’)-VA’*’ = vb”. (8’)

To establish (6)-(8) we need only show that 1 is positive for then we can define vi = v/A and ai = I/A. By contradiction if 1 = 0 then from (6’) VA’ = 0 so from (7’) and (8’) t’ = vb’. On the other hand, for all (<, x) satisfying (2) we have, multiplying by v, that t 5 vbi, hence U(X) 5 vb’ = 5’ = u(xi) so Ti would be satiated contrary to assumption. n

Suppose vi is a vector satisfying Lemma 1 and let S, = {j 1 vi > O}. It follows in a familiar way that (8) is equivalent to

tie:- A$x’ = bj, for all j E S’. (2’)

Finally define R$ = (v 1 v E R,f and vj = 0 forj # S’} . Now suppose (p, xi) is an equilibrium. If Ti is unsatiated let vi satisfy Lemma 1,

and let

and let Xi = (X 1 3g such that (5, X) satisfy (2) and (2’)},

Pi= {p/p = VA’ where ai= ve’ and VER.$}.

If Ti is satiated define

P = Ix I u(x) = maxI.

84 D. Gale, Piecewise linear exchunge equilibrium

and

Finally let

P = fi P’. l=l

Clearly the sets Xi and P are convex polyhedra.

Lemma 2. If p E P and xi E x’ and (4) and (5) are satisjied, then (p, xi) is an equilibrium.

Proof It is only necessary to establish (3), so suppose (& x) satisfies (2). It must be shown that U(X) 5 u(x’). If trader i is satiated then this follows at once. If not, since p E Pi we have for some v E R$ p = VA’ where vei = Ed. Multiplying (2) by v gives vb’ 2 #-px 2 @, but if x satisfies (4) then 5 5 v’b’/e’. On the other hand, since xi E Xi andp E Pi we know that there exists {‘, so that (c’, xi) satisfies (2) and (2’) and hence (8), but VA’ = p and px’ = 0 from (4) and (5), so from (7) and (8) we have t’ = v’b’/s’ and hence 5’ 2 <, and therefore u(x’) 2 u(x) for all x satisfying (4). n

3. The case of two traders

For the case m = 2 equilibrium condition (5) reduces to

x1 = -x2 . (9)

We therefore define W = X1 n (- X2), where x’ is as defined in the previous section. Lemma 2 then asserts that there is an equilibrium if and only if there is X in X and p in P such that pX = 0. The problem is therefore reduced to:

Lemma 3. If P and X are convex polyhedra defined by rational inequalities and there exists a pair (p, x), p E P, x E X such that px = 0, then there exists a rational such pair.

Proof By the basic facts on structure of polyhedra X has a set of generators

x1,..., Z,andX,,. . .,R,, such that every x in Xcan be written in the form

(10)

and C”= 11, = 1. Geometrically this says that X is the vector sum of a convex polytope and a convex cone.

D. Gale, Piecewise linear exchange equilibrium 85

Similarly P has generators pi, . . .,p, and pl, . . ., &. Now consider all scalar products X,~j, X#i, X#j, R#j. If all (m +n)(r+s) of these are positive (negative) then from (10) so are all scalar products px for p in P and x in X, so we could not have px = 0. If not, there are essentially three cases which by appropriate reindexing can be expressed as :

(0 %lsl < 0, E,P, > 0.

But then if Z1pz 2 0 we have Z,(@, +(l -,Q,) = 0 for some rational 1, while if X,g, 5 0 then (Lx1 + (1 - ~)xz)pz = 0 for some rational 1.

All Zdj < 0, but say ZiJj > 0.

Then Z,& +Lyi) = 0 for some non-negative rational L.

(iii) All Zsj < 0, all Fuji < 0, but say R,j$ > 0.

Then (x1 +X,)(7, -k Izi;J = 0 for some non-negative rational 1. So in all cases there is a rational solution. n

4. Sketch of algorithm

Associated with the 2-trader case we consider the following problem: For each 0 E [0, l]

maximize et1 + (1 - t3)e2,

subject to

t’e’-A’2 S b’,

lj2e2-A2x2 5 b2,

x,+x2 = 0.

Letx, = xandx, = - x. These constraints simplify to

rf,e’-A’x 5 b’,

t2e2+A2x 5 b2.

The dual problem is to

minimize v’b’ +v2b2,

D. Gale, Piecewise linear exchange equilibrium

subject to

de1 = 0 ,

v2e2 = (l-8),

u’A’-v2A2 = 0.

If we now define p = v’A’ = -v2A2, one easily sees that the variables p, v, x, 5 which solve this problem form a system satisfying the condition of Lemma 1 as well as the balance of supply and demand (5). This will provide an equilibrium provided the conditionp*x = 0 is also satisfied. Now the algorithm for parametric programming enables one to obtain solutions to the above problem for all 19 on [0, 11. For each 8 one obtains in fact an optimal basis and if this basis is unique one also obtains unique values of x andp. For a certain finite set of critical values of ($0 5 8, < 8, . . . < f& S 1, however, there is more than one optimal basis and associated with each such basis one gets different values of the vectors p and x. The algorithm involves looking at successive optimal bases and computing the scalar product px associated with them at successive critical values of 8. If all such scalar products are of the same sign for 0 < 8 < 1 then there is no unsatiated equilibrium. However, as soon as two successive scalar products have opposite signs one can by choosing the suitable convex combination as in Lemma 3 find a pair such that px = 0, giving the desired equilibrium.

Reference

Eaves, B. Curtis, 1976, A finite algorithm for the linear exchange model, Journal of Mathe- matical Economics 3, 198-204.