Lagrangian Duality for Multiobjective Programming Problems in Lexicographic Order
This paper deals with a constraint multiobjective programming problem and its dual problem in the lexicographic order. We establish some duality theorems and present several existence results of a Lagrange multiplier and a lexicographic saddle point theorem. Meanwhile, we study the relations between the lexicographic saddle point and the lexicographic solution to a multiobjective programming problem.
Duality assertions are very important in optimization researches from the theoretical as well as from the numerical point of view. Duality theorems in mathematical programming establish typical connections between a constrained minimization problem and a constrained maximization problem. The relationship is such that the existence of a solution to one of these problems ensures the existence of a solution to other, both having the same optimal value. In the past centuries, many authors have studied the duality problems of vector optimization problems; see, for example, [1–11] and reference therein.
On the other hand, it is well known that partial order plays an important role in multiobjective optimization theory. But partial efficient points are usually large, so that one needs certain additional rules to reduce them. One of the possible approaches is to utilize the lexicographic order, which is introduced by the lexicographic cone. The main reason is that the lexicographic order is a total ordering and it can overcome the shortcoming that not all points can be compared in partial order. The lexicographic order has been investigated in connection with its applications in optimization and decision making theory; see, for example, [12–19] and references therein. However, the lexicographic cone is neither open nor closed. Note that a lot of results for vector optimization problems are gotten under the assumption that the ordering cone is open or closed. Therefore, it is valuable to investigate multiobjective optimization problems in the lexicographic order. Konnov  first discussed the vector equilibrium problems in lexicographic order. Recently, Li et al.  studied the minimax inequality problem and have obtained minmax theorems in the lexicographic order.
The rest of the paper is organized as follows. In Section 2, we first recall some definitions and their properties. In Section 3, with respect to the lexicographic order, we first establish weak duality theorem and the Lagrangian multiplier rules for a multiobjective programming. In Section 4, we investigate a lexicographic saddle point of a vector-valued Lagrangian function and discuss the connection between the lexicographic saddle point and the lexicographic solution to a multiobjective programming problem.
2. Preliminaries and Notations
Throughout this paper, unless otherwise specified, let be an arbitrarily nonempty subset of a topology space and -dimensional Euclidean space. Let . Let be the space of continuous linear operator from to and . By denote the zero vector of .
For convenience, we set . As usual, for any and , will denote the th coordinate of with respect to the canonical basis . The lexicographic cone of is defined as the set of all vectors whose first nonzero coordinate (if any) is positive:
Note that the lexicographic cone is neither closed nor open. However, it is convex, pointed and . Thus the binary relation defined for any by is a total order on (i.e., it is reflexive, transitive, antisymmetric and any two vectors are comparable). The binary relation induced by is called a lexicographic order.
Now we recall the definitions of efficient points of a nonempty subset in the lexicographic order.
Definition 1 (see [13, 14]). Let be a nonempty subset. (i) A point is called a lexicographic maximal point of if ; that is, for any ; by , we denote the set of all lexicographic maximal points of .(ii) A point is called a lexicographic minimal point of if ; that is, for any ; by , we denote the set of all lexicographic minimal points of .
Obviously, if , is a single point set and so is .
Lemma 2 (see ). If is a nonempty compact set, then and .
Proposition 3. If are two nonempty compact subsets, then (i),(ii).
Proof. It suffices to verify (i) since (ii) is evident by . Indeed, let and . Then and ; and . Hence, and , and . Namely, and the proof is complete.
Definition 4. A vector-valued mapping is called lower semicontinuous if, for any , the set is closed in X.
Definition 5. A vector-valued mapping is called convex on if and only if, for any , , and , .
Consider the following multiobjective programming problem: where and , and .
Let denote the set of all feasible points for the multiobjective optimization problem (MP); that is, And let denote the set of all minimal points for on ; that is, Then, we write . Throughout this paper, we always assume that and .
A vector-valued Lagrangian function for (MP) is defined from by where ; that is, ; .
Then the Lagrangian dual problem is defined as the following multiobjective programming problem: where .
Definition 6. (i) A point is said to be a lexicographic minimal solution to (MP) if and . By denote the set of all lexicographic minimal solutions and the lexicographic minimal value to (MP), respectively.
(ii) A point is said to be a strong minimal solution to (MP) if and ; that is, . The set of all strong minimal solutions to (MP) is denoted by .
Remark 7. From , it is easy to verify that (i) of Definition 6 is equivalent and refers to the following concept: a point is said to be a lexicographic minimal solution to (MP) if and .
If is a strong solution to (MP), then is a lexicographic solution to (MP). However, the converse generally is not valid.
3. Duality Results
The following result shows that, with respect to lexicographic order, the objective value of any feasible point to the dual problem (DMP) is less than or equal to the objective value of any feasible point to the primal problem (MP).
Theorem 8 (weak duality theorem). Consider the primal problem (MP) given by (3) and its Lagrangian dual problem (DMP) given by (7). Let be a feasible point of (MP); that is, . Also, let be a feasible point of ; that is, . Then
Proof. Since , and , we obtain that ; that is, . Then, it follows from the definition of that and the proof is complete.
As a corollary of the previous theorem, we have the following result.
Corollary 9. Assume that . Then,
Before stating the Lagrangian multiplier rule to (MP) in lexicographic sense, we need the following result.
Lemma 10. Assume that . A point is a lexicographic solution for (MP) if and only if is a strong solution for (MP).
Proof. If is a lexicographic solution for (MP), then . Assume that . Then each is a strong solution for (MP), which implies that . Since is singleton, we have . That is, is a strong solution for (MP).
Conversely, suppose that is a lexicographic solution for (MP). Since , there exists which is a strong solution for (MP). By induction, we can show that which means that is a strong solution for (MP). Let for and . Obviously, . Noting that is a lexicographic solution for (MP), we have . Then, ; that is, Taking in the above inequality, we get On the other hand, since is a strong solution and . Therefore, and . Suppose that for . Obviously, . Similarly, we can verify that and . The proof is complete.
Theorem 11 (Lagrangian multiplier rule). Suppose that the following conditions are satisfied: (i) is a nonempty convex subset of ;(ii) is a -convex vector function; that is, , , are convex functions;(iii) is a -convex vector function; that is, , are convex functions;(iv) the Slater constraint qualification is satisfied; that is, there exists such that , where is the topology interior of ;(v). If is a lexicographic solution to (MP), then there exists such that
Proof. Let be a lexicographic solution of the problem (MP). By Lemma 10, is a strong solution for (MP). Then, for any ,
Then, the inequality system for some has no solution. Define the following set: The set is convex subset since , and are convex. Since the above inequality system has no solution, we have . By the standard separation theorem, there exists a nonzero vector such that That is,
Now, fix an . From the definition of , we have that and can be arbitrarily large. In order to satisfy (17), we must have . We will next show that . On the contrary, suppose that . By the Slater constraint qualification, there exists such that . Then, letting in (17), we have . But, since and is only possible if . Thus, it has been shown that , , , which is a contradiction.
After dividing (17) by and denoting , we obtain From (18), letting , we get . Since and , we obtain .
Similarly, we can show that there exist such that, for any , Set . Then, it follows from , (18), and (19) that Namely,
Thus, (21) and weak duality theorem (Theorem 8) together yield that And the proof is complete.
Now, we give an example to illustrate that Theorem 11 is applicable.
Example 12. Let , ; is defined by and , is defined by
Consider the following convex multiobjective programming problem: . Direct computation shows that , , and . Obviously, is a lexicographic solution of the multiobjective programming problem. Therefore, Theorem 11 is applicable. Indeed, by directly computing, there exists such that
From Theorem 11, we get a sufficient condition for the existence of Lagrangian multiplier rule for the problem (MP). But this is not a necessary condition, as the following example shows.
Example 13. Let , and is given as in Example 12. Let be defined by Clearly, the following problem: is a convex multiobjective programming. It follows from direct computation that , and , and is a unique lexicographic solution for (MP). Thus, the assumption (v) of Theorem 11 is not satisfied. However, we can verify that there exists such that, for any ,
In order to obtain another sufficient condition for the existence of Lagrangian multiplier rule, we consider the following assumptions.
Theorem 14 (Lagrangian multiplier rule). Assume that all conditions of Theorem 11 are satisfied except for the hypothesis (v) of Theorem 11 which is replaced by the following hypothesis: (v') is a strict convex function; that is, for any with and , .
If is a lexicographic solution to (MP), then there exists such that
Proof. Since is convex and is -convex, the set is convex. Noting that is strict convex, we obtain is singleton. Obviously, is a lexicographic solution to (MP). Following the same arguments as in Theorem 11, there exists such that , . Since is -convex and , we obtain that is a convex function (-convex function). By hypothesis , is a strict convex function. Thus, the solution to is also singleton. Let in Theorem 11. Therefore, is also a lexicographic solution to and . It completes the proof.
From Theorems 8 and 11, we have the following strong duality theorem. The following result shows that, under suitable assumptions, there is no duality gap between the primal and dual lexicographic optimal objective function values.
The following example is to illustrate that there is no duality gap if all conditions of Theorem 16 are satisfied.
Let is given as in Example 12. Let be defined by
Direct computation shows that And there exists such that .
4. Lexicographic Saddle Point
Now, we introduce the notion of lexicographic saddle point for the vector-valued Lagrangian map in terms of lexicographic order and give some optimality conditions.
Definition 18. A pair is said to be a lexicographic saddle point for the vector-valued Lagrangian function if, for all and ,
Theorem 19. Suppose that all conditions of Theorem 11 or Theorem 14 are satisfied. If is a lexicographic solution to (MP), then there exists such that is a lexicographic saddle point of the vector Lagrangian function .
Thus, we have verified the existence of a lexicographic saddle point for the vector-valued Lagrangian function under the appropriate conditions. We conclude this result by showing that the saddle point condition is sufficient for optimality for problem (MP).
Theorem 20. If is a lexicographic saddle point for the vector-valued Lagrangian function , then is a lexicographic solution for (MP), and .
Proof. Assume that is a lexicographic saddle point of . Then, from the first inequality of (33), we get
This implies that
We claim that . Otherwise, we suppose that . Since can be taken arbitrarily to be large, can be large enough, which contradicts (35).
Therefore, we also get since and . This implies that
On the other hand, by taking in (35), we can get Noting the reflexivity of the lexicographic order , (37) and (38) together yields
Since , is a feasible point of (MP). Let be any feasible point of (MP); that is, . Then, it follows from the second inequality of (33) and (39) that which implies that is a lexicographic solution to (MP).
The authors would like to express their deep gratitude to the anonymous referee and the associate editor for their valuable comments and suggestions which helped improve the paper. This research was partially supported by the National Natural Science Foundation of China (Grant no. 11201509).
J. Jahn, Mathematical Vector Optimization in Partially Ordered Linear Spaces, vol. 31 of Methoden und Verfahren der Mathematischen Physik, Peter Lang, Frankfurt, Germany, 1986.View at: MathSciNet
Y. Sawaragi, H. Nakayama, and T. Tanino, Theory of Multiobjective Optimization, vol. 176 of Mathematics in Science and Engineering, Academic Press, Orlando, Fla, USA, 1985.View at: MathSciNet
J. E. Martínez-Legaz, “Lexicographical order, inequality systems and optimization,” in System Modelling and Optimization, P. Thoft-Christensen, Ed., vol. 59 of Lecture Notes in Control and Information Sciences, pp. 203–212, Springer, Berlin, Germany, 1984.View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
V. V. Podinovskii and V. M. Gavrilov, Optimization with Respect to Sequentially Applied Criteria, Sovetskoe Radio, Moscow, Russia, 1976, (Russian).