Journal of Global Optimization, 29: 455–477, 2004. 455
© 2004 Kluwer Academic Publishers. Printed in the Netherlands.
Scalarization and Nonlinear Scalar Duality for Vector
Optimization with Preferences that are not necessarily
a Pre-order Relation
A.M. RUBINOV
1
and R.N. GASIMOV
2
1
SITMS, University of Ballarat, Victoria 3353, Australia (e-mail: amr@ballarat.edu.au)
2
Department of Industrial Engineering, Osmangazi University, Bademlik 26030, Eski¸ sehir, Turkey
(e-mail: kasimov@ogu.edu.tr)
(Received: 1 August 2003; accepted: 13 August 2003)
Abstract. We consider problems of vector optimization with preferences that are not necessarily
a pre-order relation. We introduce the class of functions which can serve for a scalarization of
these problems and consider a scalar duality based on recently developed methods for non-linear
penalization scalar problems with a single constraint.
Key words. duality, preferences, scalarization, vector optimization.
1. Introduction
Problems of vector (multi-criteria) optimization arise when there are some differ-
ent criteria for the choice of a preferable object. As a rule it is assumed that the
totality of these criteria forms a pre-order relation. The theory of vector optimiza-
tion with respect to (w.r.t.) pre-order relation is well developed (see, for example,
[5, 11]). However, often we get preferences that form a relation, which is not a
pre-order. Let us give some simple examples. Assume that we have m> 1 crite-
ria (objective functions) f
1
f
m
defined on a set X. Each element x ∈ X can be
estimated by a vector of numbers f
1
xf
m
x. Usually it is assumed that x
is more preferable that yx y if f
i
x f
i
y for all i ∈ I = 1m. Clearly
is a pre-order relation. However, sometimes we need different kind of prefer-
ences, which are either weaker or stronger than . For example, let m> 2 and
I
1
= 2m, I
m
= 1m − 1. Consider preferences
1
defined in the follow-
ing way: x
1
y if either f
i
x f
i
y for i ∈ I
1
or f
i
x f
i
y for i ∈ I
m
. The
preferences
1
are weaker than (i.e. x y implies x
1
y) and these preferences
are not transitive, so
1
is not a pre-order relation. Consider now another prefer-
ences
2
. We say that x
2
y if f
i
x f
i
y for all i ∈ I and either f
1
x − f
1
y
f
2
x − f
2
y or f
2
x − f
2
y f
3
x − f
3
y. Clearly
2
is stronger than
and
2
is not a pre-order relation. Both relations
1
i = 1 2 have the following
structure: x
i
y means that the vector f
1
x − f
1
yf
m
x − f
m
y belongs