A new algorithm for solving planar multiobjective location problems involving the Manhattan norm

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1 A new algorithm for solving planar multiobjective location problems involving the Manhattan norm Shaghaf Alzorba a, Christian Günther a, Nicolae Popovici b and Christiane Tammer a a Martin Luther University Halle-Wittenberg, Faculty of Natural Sciences II, Institute for Mathematics, Halle (Saale), Germany b Babeş-Bolyai University, Faculty of Mathematics and Computer Science, Cluj-Napoca, Romania January 26, 2016 Abstract This paper is devoted to the study of unconstrained planar multiobjective location problems, where distances between points are defined by means of the Manhattan norm. By identifying all nonessential objectives, we develop an effective algorithm for generating the whole set of efficient solutions. We prove the correctness of this algorithm and present some computational results, obtained by implementing the algorithm in MAT- LAB. Keywords Multiobjective optimization; Location problem; Manhattan norm; Scalarization; Nonessential objective 1 Introduction Multiobjective location problems, as well as their scalarizations, may be found in the literature in many variants. Indeed, the objective functions and the constraints depend on the specific practical applications, as for instance: urban development planning, engineering, logistics, economics or radiation therapy. For a detailed study of special location problems and corresponding algorithms we refer the reader to Love, Morris, & Wesolowsky [15], Hamacher [13], Göpfert et al. [11], Nickel, Puerto, & Rodriguez-Chia [18] [19] and references therein. Corresponding author. Tel.: +49 (345) ; fax: +49 (345) address: christiane.tammer@mathematik.uni-halle.de (Chr. Tammer) 1

2 Usually, the planar multiobjective location problem is formulated as d(x, a 1 ) f(x) := v-min, d(x, a p x S ) where a 1,..., a p R 2 are given points in the plane, d : R 2 R 2 R is a distance function, while S R 2 is the feasible domain, defined by certain constraints. Such location problems have been intensively studied in the last decades (see, e.g., Wendell, Hurter Jr., & Lowe [24], Chalmet, Francis, & Kolen [2], Lowe et al. [16], Gerth & Pöhler [10], Pelegrin & Fernandez [20], and Ward [23]). The decision maker may define the distance function d by means of an appropriate norm : R 2 R, i.e., d(x, x ) := x x for all x, x R 2. In particular, many interesting results have been obtained for the following two classical polyhedral norms, defined for any x = (x 1, x 2 ) R 2 by x 1 := x 1 + x 2 (Manhattan norm); x := max{ x 1, x 2 } (Maximum norm). In this paper we focus only on unconstrained (i.e., S := R 2 ) planar multiobjective location problems, where the distance function d is defined by means of the Manhattan norm 1. However, our results can be easily adapted to the dual norm or even to more general block-norms, by an appropriate linear transformation. The aim of our paper is twofold: first, to characterize the nonessential objectives (i.e., the objective associated to those points among a 1,..., a p which can be removed without changing the set of efficient solutions) and second, to develop an effective algorithm, which generates the whole set of efficient solutions and also decomposes this set into a finite number of (possibly degenerated) rectangles. The paper is organized as follows. In Section 2 we recall some basic notions of vector optimization and we formulate the planar multiobjective location problem as a multiobjective optimization problem. Also, we recall an important result concerning the efficient solutions generated by the weighted sum scalarization method (Lemma 3). In Section 3 we characterize the set of efficient solutions by combining the above-mentioned result with the approach proposed by Gerth & Pöhler [10]. Also, similarly to Alzorba, Günther, & Popovici [1], we characterize the set of weakly efficient solutions in terms of the so-called rectangular hull of a set with respect to the Maximum norm (Corollary 8). The possibility of reducing the number of objectives is investigated in Section 4. By considering four ordering cones of R 2, namely the quadrants of the usual coordinate system, we derive an interesting characterization of nonessential objectives in terms of minimal elements of the set {a 1,..., a p } with respect to these cones (Theorem 11 and its Corollary 12). In order to compute these 2

3 minimal elements we adapt the Jahn-Graef-Younes method for solving discrete vector optimization problems (Algorithm 1). Sections 5, 6 and 7 represent the practical part of the paper. We start by formulating our Rectangular Decomposition Algorithm (Algorithm 2), which generates the whole set of efficient solutions, by providing a decomposition of this set into a finite family of (possibly degenerated) rectangles. Then we prove its correctness and present some illustrative computational results, obtained by implementing the algorithm in MATLAB. We conclude with some remarks in Section 8. 2 Preliminaries In this section we recall some basic notions and results concerning vector optimization and multiobjective location theory. As usual, we denote by R and N the sets of real and natural numbers, respectively. For notational convenience, we also denote R + := [0, + [, R := ], 0] and, for any n N, we introduce the index set I n := N [1, n] = {1,..., n}. The interior, the closure, the boundary and the convex hull of any set S R n will be denoted by int S, cl S, bd S and conv S, respectively. Consider a finite family A := {a 1,..., a p } of points in the plane R 2, a i := (a i 1, a i 2), i I p, representing some a priori given facilities. Throughout the paper we assume that card A = p 2, i.e., a 1,..., a p are pairwise distinct. The multiobjective location problem associated to A consists in finding new facilities, i.e., points of R 2, which minimize (simultaneously) the distances to all given points of A. In our paper the distances will be induced by the Manhattan norm 1. Hence, the multiobjective location problem associated to A is formulated as x a 1 1 f A (x) :=... v-min, (P A) x a p x R 2 1 where the minimization is understood in the sense of vector optimization, according to the definitions below. Definition 1. Let K R n be a pointed closed convex cone, i.e., K ( K) = {0}, cl K = K and K = K + K = R + K. Then, given any set Y R n, we define the set of minimal elements of Y with respect to K by MIN(Y, K) := {y 0 Y (y 0 K) Y = {y 0 } } (1) = {y 0 Y (y 0 K) Y \{y 0 } = }. Moreover, assuming that K is solid, i.e., int K is nonempty, we also define the set of weakly minimal elements of Y with respect to K as WMIN(Y, K) := {y 0 Y (y 0 int K) Y = }. 3

4 By Definition 1, when K is solid, we obviously have MIN(Y, K) WMIN(Y, K) Y bd Y. (2) Also, it is a simple exercise to check that MIN(Y, K) = MIN(Y \{y}, K) for all y Y \MIN(Y, K). (3) Definition 2. Let f A [R 2 ] := {f A (x) x R 2 } R p be the image of the feasible set R 2 by the objective function f A : R 2 R p and let K = R p + be the standard ordering cone of the real linear space R p. The set of (Pareto) efficient solutions of the multiobjective location problem (P A ) is defined by Eff(P A ) := {x 0 R 2 f A (x 0 ) MIN(f A [R 2 ], R p + )} while the set of weakly efficient solutions of (P A ) is given by WEff(P A ) := {x 0 R 2 f A (x 0 ) WMIN(f A [R 2 ], R p + )}. As a direct consequence of the first inclusion in (2) we obtain Eff(P A ) WEff(P A ). Moreover, for any x 0 R 2, we have: { i x 0 Eff(P A ) x R 2 Ip : x a i 1 x 0 a i 1, s.t. j I p : x a j 1 < x 0 a j 1 ; x 0 WEff(P A ) x R 2 s.t. i I p : x a i 1 < x 0 a i 1. In what follows we will apply the classical weighted sum scalarization method for generating efficient solutions of (P A ). For every weight vector λ := (λ 1,..., λ p ) int R p + let us consider the scalar optimization problem λ, f A (x) := p i=1 λ i x a i 1 min x R 2 (P λ,a ) and denote the set of its optimal solutions by λ-eff(p A ). motivated by the fact that This notation is λ-eff(p A ) := argmin λ, f A (x) Eff(P A ). (4) x R 2 Indeed, it is well-known that the points of λ-eff(p A ) are properly efficient solutions of (P A ) in the sense of Geoffrion [9]. For more details we refer the reader to the books by Luc [17], Ehrgott [4], Göpfert et al. [11] and Jahn [14]. In preparation for the next lemma, we focus on the coordinates of the given points a 1,..., a p R 2 and define A k := {a 1 k,..., ap k } for each k I 2. 4

5 Denoting p k := card A k card A = p for each k I 2, there exist uniquely determined numbers u 1 < < u p1 and v 1 < < v p2 such that A 1 = {u 1,..., u p1 } and A 2 = {v 1,..., v p2 }. (5) Then, the Cartesian product A 1 A 2 := {(u i, v j ) i I p1 be seen as a set of grid points in R 2, such that and j I p2 } can A A 1 A 2. (6) The following result (see Hamacher [13, Satz 2.3]) plays an important role for deriving properties of the set of properly/weakly efficient solutions of the problem (P A ) (see Corollary 8) and for the proof of the correctness of the proposed algorithm formulated in Section 5. Lemma 3. For every weight vector λ int R p +, the set λ-eff(p A) is an axisparallel rectangle, whose vertices are adjacent grid points, which may be degenerated into a line segment or a singleton. More precisely, λ-eff(p A ) has one of the following four forms: 1 {(u i, v j )} for some (i, j) I p1 I p2 ; 2 [u i, u i+1 ] {v j } for some (i, j) I p1 1 I p2 (p 1 2); 3 {u i } [v j, v j+1 ] for some (i, j) I p1 I p2 1 (p 2 2); 4 [u i, u i+1 ] [v j, v j+1 ] for some (i, j) I p1 1 I p2 1 (p 1, p 2 2). 3 Structure of the sets of weakly and (properly) efficient solutions The geometrical and topological structure of the sets of weakly and (properly) efficient solutions of certain multiobjective convex optimization problems including multiobjective location problems, has been intensively studied (see, e.g., Lowe et al. [16], Durier & Michelot [3], Luc [17] or Popovici [21], and references therein). Characterizations of weakly and (properly) efficient solutions of multiobjective location problems can be found for instance in the papers by Wendell, Hurter Jr., & Lowe [24], Chalmet, Francis, & Kolen [2], Lowe et al. [16], or Gerth & Pöhler [10]. In order to characterize the sets WEff(P A ) and Eff(P A ), we will follow the approach proposed in [10], which is based on the dual of the Manhattan norm, namely the Maximum norm. Let us denote by B (x, r) := {x R 2 x x r} the closed ball (w.r.t. Maximum norm) centered at any point x R 2 of radius r > 0. Definition 4. Let D R 2 be a nonempty bounded set. The set N (D) := { B (x, r) x R 2, r > 0, D B (x, r) } is called the rectangular hull of D (w.r.t. Maximum norm). 5

6 The following two results represent the counterparts of similar results previously obtained for the Manhattan norm (instead of the Maximum norm) by Alzorba, Günther, & Popovici [1, Lemma 4.2 and Theorem 4.3]. Lemma 5. If D R 2 is nonempty and compact, then we have N (D) = N ({x, x }). x,x D Theorem 6. The set of all weakly efficient solutions of the multiobjective location problem (P A ) is given by WEff(P A ) = N (A). In order to adapt the approach proposed by Gerth & Pöhler [10] for characterizing the set Eff(P A ), we define certain sets related to the structure of the subdifferential of the Manhattan norm. For each i I p, let s 1 (a i ) := {x R 2 x = (x 1, x 2 ), x 1 < a i 1, x 2 < a i 2}, s 2 (a i ) := {x R 2 x = (x 1, x 2 ), x 1 > a i 1, x 2 > a i 2}, s 3 (a i ) := {x R 2 x = (x 1, x 2 ), x 1 > a i 1, x 2 < a i 2}, s 4 (a i ) := {x R 2 x = (x 1, x 2 ), x 1 < a i 1, x 2 > a i 2}, and then, for every r I 4, consider the (possibly empty) set S r := {x N (A) k I p : x s r (a k )} = N (A) k I p s r (a k ). (7) The following result is a counterpart of the characterization of efficient solutions obtained by Gerth & Pöhler [10] (where the authors considered multiobjective location problems defined by the Maximum norm instead of the Manhattan norm). Theorem 7. The set of efficient solutions of the multiobjective location problem (P A ) is given by Eff(P A ) = [ (cl S 1 cl S 2 ) ( (N (A)\S 1 ) (N (A)\S 2 ) )] [ (cl S 3 cl S 4 ) ( (N (A)\S 3 ) (N (A)\S 4 ) )]. We conclude this section by highlighting some useful properties of the solution sets Eff(P A ) and WEff(P A ). Corollary 8. The following assertions hold: 1 A Eff(P A ), i.e., the given facilities are efficient solutions of (P A ). 2 Eff(P A ) = λ int R p + λ-eff(p A), hence every efficient solution of (P A ) is properly efficient. 3 Eff(P A ) can be represented as a finite union of (possibly degenerated) axis-parallel rectangles, hence Eff(P A ) is compact. 6

7 4 WEff(P A ) = N (A) = [u 1, u p1 ] [v 1, v p2 ] with u 1, u p1, v 1, v p2 given by (5), where the intervals may degenerate into singletons whenever p 1 = 1 or p 2 = 1. Proof. Assertion 1 follows from Definition 2, taking into account that {a i } = argmin x R 2 x a i 1 for all i I p. The equality stated at 2 is a classical result. Actually, the inclusion follows from (4), while the inclusion holds since the Manhattan norm is polyhedral, as pointed out by Wendell, Hurter Jr., & Lowe [24]. Assertion 3 is a straightforward consequence of Lemma 3, which shows actually that the set Eff(P A ) can be represented as a union of (at most) (p 1 1)(p 2 1) + (p 1 1)p 2 + p 1 (p 2 1) + p 1 p 2 axis-parallel rectangles. Therefore, Eff(P A ) is closed and bounded, i.e., a compact subset of R 2. By Definition 4 and (6) we can easily deduce that N (A) = [u 1, u p1 ] [v 1, v p2 ], hence assertion 4 follows from Theorem 6. 4 Reducing the number of objectives In certain real-life applications the number p of objectives is very large. In order to develop an effective algorithm for solving such location problems, we will first identify the nonessential objectives. Recall that an objective is said to be nonessential (or redundant) if the set of efficient solutions does not change when it is removed from the multiobjective optimization problem (see, e.g., the pioneering paper by Gal & Leberling [8] or the more recent one by Gal & Hanne [7]). In this section we will characterize the largest set of points which can be eliminated from A without altering the set Eff(P A ) of all efficient solutions of the multiobjective location problem (P A ). By removing all nonessential objectives we obtain a reduced location problem, which can be seen as a subproblem of (P A ). Therefore, the results presented by us in this section may open the way for further investigations, concerning the decomposition of multiobjective problems into subproblems (see, e.g., Ward [23], Wendell et al. [24], Ehrgott & Nickel [5], Popovici [21], or Engau & Wiecek [6]). Lemma 9. If a i A (i I p ) satisfies the relation a i t I 4 cl s t (a jt ) for some a j 1,..., a j 4 A\{a i } (j 1,..., j 4 I p \{i}), then the following hold: 1 WEff(P A ) = WEff(P A\{a i }). 2 Eff(P A ) = Eff(P A\{a i }). 7

8 Proof. In order to prove 1 observe that, according to Theorem 6, we have WEff(P A ) = N (A) and N (A\{a i }) = WEff(P A\{a i }). Thus, it suffices to prove that N (A) = N (A\{a i }). The inclusion N (A) N (A\{a i }) holds by Lemma 5, since A\{a i } A. In order to prove the inclusion N (A) N (A\{a i }) consider an arbitrary x N (A). Thanks to Lemma 5, there exist two points a t 1, a t 2 A, i.e., t 1, t 2 I p, such that x N ({a t 1, a t 2 }). We distinguish four possible cases: Case 1 : If a t1 a i a t 2, then we have {a t 1, a t 2 } A\{a i }, hence x N ({a t 1, a t 2 }) N (A\{a i }) by Lemma 5. Case 2 : If a t 1 = a i = a t 2, then we have x N ({a t 1, a t 2 }) = {a i }, i.e., x = a i. Since x = a i cl s 1 (a j 1 ) cl s 2 (a j 2 ) = N ({a j 1, a j 2 }) and {a j 1, a j 2 } A\{a i }, we get x N ({a j 1, a j 2 }) N (A\{a i }) by Lemma 5. Case 3 : If a t1 a i = a t 2, then we have x N ({a t 1, a i }) and there exists r I 4 such that a i cl s r (a t 1 ). Without loss of generality we can suppose that r = 1. By hypothesis, there exists a point a j 2 A\{a i } such that a i cl s 2 (a j 2 ). We observe that a i cl s 1 (a t 1 ) cl s 2 (a j 2 ), and therefore the inclusion N ({a t 1, a i }) N ({a t 1, a j 2 }) holds. Recalling that a t 1, a j 2 A\{a i }, it follows that N ({a t 1, a j 2 }) N (A\{a i }) by Lemma 5. Consequently, we have x N (A\{a i }). Case 4 : If a t 1 = a i a t 2, then we can deduce that x N (A\{a i }) similarly to the previous case. Thus, in all cases we have x N (A\{a i }), hence N (A) N (A\{a i }). In order to prove 2 we will apply Theorem 7 twice, for A and for A\{a i }. Actually, since N (A) = N (A\{a i }) (as already seen in the proof of 1 ), it suffices to show that S r = Ŝr for all r I 4, where S r is given by (7) while Ŝ r := N (A\{a i }) s r (a k ). (8) k I p\{i} Let r I 4 be arbitrarily chosen. The inclusion Ŝr S r is obvious. For proving the inclusion S r Ŝr, consider any x S r. By (7) there exists j I p such that x s r (a j ). If j i, then we obviously have x Ŝr by (8). In what follows assume that j = i. By hypothesis there exists some j r I p \{i} such that a i cl s r (a jr ). Recalling the construction of the sets s r ( ), we deduce that x s r (a i ) cl s r (a jr ). Since the set s r (a i ) is open, it follows that x s r (a i ) int cl s r (a jr ) = s r (a jr ). Thus we have S r Ŝr. Consider now four pointed convex cones of R 2, namely the quadrants of the usual coordinate system, labelled as follows: K 1 := R + R + ; K 2 := R R ; K 3 := R R + ; K 4 := R + R. Note that K 2 = K 1 and K 4 = K 3. Thus, it will be convenient to introduce the permutation ψ : I 4 I 4, defined by ( ) ( ) =. ψ(1) ψ(2) ψ(3) ψ(4)

9 It is easy to see that, for all i I p and t I 4, the following hold: s t (a i ) = a i int K t = a i + int K ψ(t) ; cl s t (a i ) = a i K t = a i + K ψ(t). (9) In what follows we will consider the minimal elements of A = {a 1,..., a p } with respect to the ordering cones K 1,..., K 4. Note that MIN(A, K t ) is nonempty for every t I 4, since A is nonempty and finite. Lemma 10. Let i I p. The following assertions are equivalent: 1 There exist j 1,..., j 4 I p \{i} such that a i t I 4 cl s t (a jt ). 2 There exist k 1,..., k 4 I p \{i} such that a kt cl s t (a i ) for all t I 4. 3 There is no t I 4 such that a i MIN(A, K t ), i.e., a i A \ t I 4 MIN(A, K t ). Proof. We will prove the chain of implications 1 = 2 = 3 = 1. First, assume that 1 holds and consider any j 1,..., j 4 I p \{i} such that a i t I 4 cl s t (a jt ). Then, by (9) we have a i cl s t (a jt ) = a jt K t, hence a jt a i +K t = a i K ψ(t) = cl s ψ(t) (a i ), for any t I 4. Since ψ is a permutation on I 4, we infer that a j ψ(t) cl s ψ(ψ(t)) (a i ) for any t I 4. Taking into account that ψ 1 = ψ, i.e., ψ(ψ(t)) = t for all t I 4, it follows that a kt cl s t (a i ), where k t := j ψ(t) I p \{i}, for any t I 4. Thus 2 holds. Now, assume that 2 holds and suppose to the contrary that 3 does not hold. Then, there exists some t I 4 such that a i MIN(A, K t ). By 2 we can choose an index k t I p \{i} (i.e., a kt A\{a i }), such that a kt cl s t (a i ). By (9), we infer that a kt (a i K t ) A\{a i }. In view of Definition 1, it follows that a i / MIN(A, K t ), a contradiction. Finally, assume that 3 holds. Then, for every index t I 4 we have a i / MIN(A, K t ), hence (a i K t ) A\{a i } =. In other words, for each t I 4 there is j t I p \{i} such that a jt (a i K t ) A\{a i }, which shows that a i a jt + K t = a jt K ψ(t) = cl s ψ(t) (a jt ), according to (9). Therefore we have a i t I 4 cl s ψ(t) (a jt ). Taking into account that ψ is a permutation on I 4, we infer 1. Example 1. Consider the set A = {a 1,..., a 8 } R 2, where a 1 = (14, 12); a 2 = (10, 9); a 3 = (18, 10); a 4 = (12, 4); a 5 = (6, 2); a 6 = (4, 8); a 7 = (2, 6); a 8 = (8, 7). 9

10 By using (1) for Y := A and each cone K {K 1,..., K 4 }, it is a simple exercise to check that MIN(A, K 1 ) = {a 5, a 7 }; MIN(A, K 2 ) = {a 1, a 3 }; MIN(A, K 3 ) = {a 3, a 4, a 5 }; MIN(A, K 4 ) = {a 1, a 2, a 6, a 7 }. Therefore we have A \ t I 4 MIN(A, K t ) = {a 8 }. (10) By (10) and Lemma 10 (3 = 1 ) there exist j 1,..., j 4 I 7 such that a 8 t I 4 cl s t (a jt ). (11) Indeed, as we can see in Figure 1 (left picture), we can choose the indices j 1 = 1, j 2 = 7, j 3 = 6 and j 4 = 4, which satisfy (11). Therefore, by the proof of Lemma 10 (1 = 2 ) it follows that a kt cl s t (a i ) for all t I 4, where k 1 := j 2 = 7, k 2 := j 1 = 1, k 3 := j 4 = 4 and k 4 := j 3 = 6. This property is illustrated in Figure 1 (right picture). On the other hand, by Corollary 8 (4 ) and Lemma 9 (1 ), we have WEff(P A ) = WEff(P A\{a 8 }) = [2, 18] [2, 12]. Moreover, by Lemma 9 (2 ) we have Eff(P A ) = Eff(P A\{a 8 }). This set will be computed later on (see Figure 3). Figure 1: Illustration of Lemma 10 (1 and 2 ) for the set A defined in Example 1. Theorem 11. Let i I p. Then the following equivalence is true: a i A \ t I 4 MIN(A, K t ) Eff(P A ) = Eff(P A\{a i }). Proof. The implication = follows from Lemma 9 and Lemma 10. In order to prove the implication =, define the sets S r and Ŝr for all r I 4 by (7) and (8), respectively. Assume that Eff(P A ) = Eff(P A\{a i }) and 10

11 suppose to the contrary that there exists r I 4 such that a i MIN(A, K r ). Without loss of generality we can consider r = 1, hence a i MIN(A, K 1 ). In view of Definition 1 and (9), it follows that (a i K 1 ) A\{a i } = cl s 1 (a i ) A\{a i } =. (12) Consider the following two index sets (associated to a i A): J 1 := {j I p a j 1 > ai 1} and J 2 := {j I p a j 2 > ai 2}. Note that these sets are nonempty. Indeed, suppose to the contrary that J 1 =. Then, we should have A ], a i 1 ] R, which together with (12) yields A\{a i } ], a i 1 ] ]ai 2, + [. Since A\{ai } is a finite set, we infer that N (A\{a i }) ], a i 1 ] ]ai 2, + [, hence ai / N (A\{a i }). However, by Corollary 8 (1 ) and Theorem 7 (applied for A\{a i } in the role of A) we deduce that a i Eff(P A ) = Eff(P A\{a i }) N (A\{a i }), a contradiction. Similarly we can prove that J 2 is nonempty. Since both J 1 and J 2 are nonempty and finite, we can define the point ( δ1 + a i 1 x = ( x 1, x 2 ) :=, δ 2 + a i ) 2 R by means of the real numbers δ 1 := min{a j 1 j J 1} = min{a j 1 aj A, a j 1 > ai 1}; δ 2 := min{a j 2 j J 2} = min{a j 2 aj A, a j 2 > ai 2}. By the construction of x and recalling (12), we can easy deduce that: x s 2 (a i ); (13) ( x K 1 ) A\{a i } =. (14) In particular, (14) shows that there is no point a k A\{a i } (i.e., k I p \{i}) such that x a k + K 1 = a k K 2 = cl s 2 (a k ). Consequently, we have x / cl Ŝ2. (15) The following cases may occur: Case 1 : If s 2 (a i ) A\{a i }, then there is j I p \{i} (i.e., a j A\{a i }) such that a j s 2 (a i ). In this case, by the definition of x, we have x s 1 (a j ). (16) Now we distinguish two subcases: Subcase 1.1 : If there exist a h, a k A\{a i } (i.e., h, k I p \{i}) such that a h cl s 3 (a i ) and a k cl s 4 (a i ), then (12) yields a h, a k / cl s 1 (a i ). Moreover, by definition of x it follows that x 1 < a h 1 and x 2 > a h 2 as well as x 1 > a k 1 and x 2 < a k 2. By Lemma 5 and relations (13) and (16), we deduce that x N ({a h, a k }) N (A) and x s 4 (a h ) s 3 (a k ) s 2 (a i ) s 1 (a j ). Therefore we have x S 1 S 2 S 3 S 4. By Theorem 7 we conclude that x Eff(P A ). On the 11

12 other hand, since a h, a k A\{a i }, we also have x N ({a h, a k }) N (A\{a i }) according to Lemma 5 (applied for A\{a i }). Since a j A\{a i }, we infer from (16) that x Ŝ1. The latter relation together with (15) allows us to deduce by Theorem 7 (applied for A\{a i }) that x / Eff(P A\{a i }). We conclude that Eff(P A ) Eff(P A\{a i }), a contradiction. Subcase 1.2 : If cl s 3 (a i ) A\{a i } = or cl s 4 (a i ) A\{a i } =, then we can suppose without any loss of generality that cl s 4 (a i ) A\{a i } =. Then, by relation (12) we deduce that (cl s 1 (a i ) cl s 4 (a i )) A\{a i } =. It follows that A\{a i } ]a i 1, + [ R, hence ai 1 < ak 1 for all k I p\{i}. In particular, we have a i 1 < aj 1. Consequently, we have ai / N (A\{a i }). By Theorem 7 (applied for A\{a i }), we conclude that a i / Eff(P A\{a i }). However, by assertion 1 of Corollary 8, we infer a i Eff(P A ), contradicting the hypothesis that Eff(P A ) = Eff(P A\{a i }). Case 2 : If s 2 (a i ) A\{a i } =, then we consider two subcases: Subcase 2.1 : If there exist a h, a k A\{a i } (i.e., h, k I p \{i}) such that a h cl s 3 (a i ) and a k cl s 4 (a i ), then a h, a k / cl s 1 (a i ) by (12). Also, we have x N ({a h, a k }) N (A). In view of (13), it follows that x N (A) s 2 (a i ), hence x S 2. Also, since s 2 (a i ) A\{a i } =, we have x / S 1. Therefore, we can conclude by Theorem 7 that x / Eff(P A ). On the other hand, since a h, a k A\{a i }, we have x N ({a h, a k }) N (A\{a i }). Thus, x / S 1 entails x / Ŝ1. Furthermore, similarly to Subcase 1.1, it is easy to check that x s 3 (a h ) s 4 (a k ). Hence, the relation x Ŝ3 Ŝ4 holds. Recalling (15), we deduce that x (N (A\{a i })\Ŝ1) (N (A\{a i })\Ŝ2) Ŝ3 Ŝ4. By Theorem 7 we infer x Eff(P A\{a i }), which contradicts again the hypothesis that Eff(P A ) = Eff(P A\{a i }). Subcase 2.2 : If cl s 3 (a i ) A\{a i } = or cl s 4 (a i ) A\{a i } =, then observe that one of the sets cl s 3 (a i ) A\{a i } and cl s 4 (a i ) A\{a i } should be nonempty, because of the assumption of Case 2 and the fact that card A 2. Without loss of generality we can suppose that cl s 4 (a i ) A\{a i } =, hence there exists a point a h cl s 3 (a i ) A\{a i }. It is easy to check that a i 1 < ah 1 and a i / N (A\{a i }). Therefore, following the same lines as in Subcase 1.2, we arrive to a contradiction. Remark 1. In contrast to Theorem 11, the equivalence a i A \ t I 4 MIN(A, K t ) WEff(P A ) = WEff(P A\{a i }) may be false for some i I p, although the implication = is always true by Lemma 9 and Lemma 10. As a counterexample for the implication =, consider the set A := {a 1, a 2, a 3 } R 2, where a 1 := (0, 0), a 2 := (0, 1) and a 3 := (1, 1). We have WEff(P A ) = [0, 1] [0, 1] = WEff(P A\{a 2 }), but a 2 MIN(A, K 4 ). In what follows we consider the reduced multiobjective location problem (PÃ), 12

13 associated to the following nonempty subset of the initial set A: Ã := t I 4 MIN(A, K t ). (17) Corollary 12. The multiobjective location problems (P A ) and (PÃ) have the same efficient solutions, i.e., Eff(P A ) = Eff(PÃ). (18) Proof. Since (18) trivially holds when A = Ã, we just have to study the case when A\Ã. Assume that A\Ã =: {ai 1,..., a in } with i 1,..., i n I p and card(a\ã) = n 1 (i.e., ai 1,..., a in are pairwise distinct). We will apply Theorem 11 recursively. First note that a i 1 A\Ã means a i 1 A \ MIN(A, K t ), (19) t I 4 according to (17). By applying Theorem 11 for i := i 1, we deduce that Eff(P A ) = Eff(P A\{a i 1 } ). (20) If n = 1, then (20) becomes the desired relation (18). Otherwise, if n 2, then consider the point a i 2 (A\{a i 1 })\Ã = (A\{ai 1 }) \ t I 4 MIN(A, K t ). Relation (19) shows that, for all t I 4, we have a i 1 A\MIN(A, K t ), hence MIN(A, K t ) = MIN(A\{a i 1 }, K t ), in view of (3). Therefore we have a i 2 (A\{a i 1 }) \ MIN(A\{a i1 }, K t ). t I 4 By applying Theorem 11 again, this time for A\{a i 1 } in the role of A and i := i 2 I p \{i 1 }, we deduce that Eff(P A\{a i 1 } ) = Eff(P A\{a i 1,a i 2 } ). (21) If n = 2, then (18) follows from (20) and (21). argument as above we deduce that Otherwise, by a similar Eff(P A\{a i 1,...,a i m 1 } ) = Eff(P A\{a i 1,...,a im} ) (22) for every m {3,..., n}. Finally, we conclude (18) from (20) (22). Remark 2. In the proof of Corollary 12 we have used the implication = of Theorem 11. Similarly, by using Lemma 9 and Lemma 10, we can prove that problems (P A ) and (PÃ) have the same weakly efficient solutions, i.e., WEff(P A ) = WEff(PÃ), On the other hand, the implication = of Theorem 11 shows that a further reduction of (PÃ) is not possible, since the points of A that can be eliminated without altering the set of efficient solutions should belong to Ã. Moreover, Theorem 11 will play a key role in Section 6, for proving that Algorithm 2 generates all efficient solutions of (P A ) (see Theorem 20). 13

14 In Section 5 we will develop a new algorithm for computing the set of all efficient solutions of the multiobjective location problem (P A ) via the reduced problem (PÃ). In order to apply (18) in practice, we need an effective procedure to generate the set Ã, i.e., the minimal elements of the finite set A = {a 1,..., a p } with respect to each cone K {K 1, K 2, K 3, K 4 }. Among various methods existing in the literature for solving discrete vector optimization problems, we will adapt the following Graef-Younes method with backward iteration, proposed by Jahn [14, Algorithm 12.20]: Algorithm 1 (Jahn-Graef-Younes method). Input: The set A = {a 1,..., a p } (representing the a priori given facilities) and any ordering cone K {K 1, K 2, K 3, K 4 }. Step 1. Let i := 1 and define u i := a i and U := {u i }. Step 2. For j = 2,..., p: If a j / U + K, then let i := i + 1, u i := a j and U := U {u i }. Endfor Step 3. Define T := {u i }. If i > 1, then go to Step 4, else go to Output. Step 4. For j = 1,..., i 1: If u i j / T + K, then let T := T {u i j }. Endfor Output: The set MIN(A, K) := T of all minimal elements of A w.r.t. K. 5 Formulation of the Rectangular Decomposition Algorithm In this section we present a new algorithm for computing the set Eff(P A ) of all efficient solutions of the multiobjective location problem (P A ). In contrast to other algorithms known in the literature (see, e.g., Wendell et al. [24], Chalmet et al. [2], or Gerth & Pöhler [10]), our algorithm aims to eliminate the nonessential objectives in a first phase using Algorithm 1 and thereafter to generate a well structured representation of the set Eff(P A ). This set Eff(P A ) can be used as input for other algorithms, as for instance to minimize/maximize an additional cost function over the efficient solution set of the multiobjective location problem (P A ). In the following we formulate the Rectangular Decomposition Algorithm for generating the set Eff(P A ). The correctness of this algorithm is shown in Section 6. Algorithm 2 (Rectangular Decomposition Algorithm). Input: The set A := {a 1,..., a p } (representing the a priori given facilities). 14

15 Step 1. For every r I 4 compute the set T r := MIN(A, K r ) by means of Algorithm 1. Step 2. Construct the reduced set à = {ã 1,..., ã q } := T r, r I 4 where ã 1 = (ã 1 1, ã1 2 ),..., ãq = (ã q 1, ãq 2 ) and q := card( r I 4 T r ). Step 3. For every k I 2 define à k := {ã 1 k,..., ãq k } and denote q k := card Ãk (note that q k q). Sort the elements of à 1 and Ã2 in ascending order and delete duplicated values, i.e., determine the numbers u 1 < < u q1 and v 1 < < v q2 such that à 1 = {u 1,..., u q1 } and Ã2 = {v 1,..., v q2 }. Construct the set of grid points in R 2 : à 1 Ã2 := {(u i, v j ) i I q1 and j I q2 }. Among these grid points, define e 1 := (u 1, v 1 ); e 2 := (u q1, v q2 ); e 3 := (u q1, v 1 ); e 4 := (u 1, v q2 ). Step 4. For every r I 4 define the (possibly empty) set T r := T ψ(r) (e r + int K r ). Step 5. Construct the sets C 1,..., C q1 by the iterative procedure described below, initially letting them empty, i.e., C i := for all i I q1. For i = 1,..., q 1 : For j = 1,..., q 2 : (1 ) Define bool 1 := 0 and bool 2 := 0. If (u i, v j ) T 1 K 1, then let bool 1 := 1. If bool 1 = 1, then check: If (u i, v j ) T 2 K 2, then let bool 2 := 1. If bool 2 = 1, then go to (3 ), else go to (2 ). 15

16 (2 ) If bool 1 = 0, then let bool 1 := 1, else check: If (u i, v j ) T 1 int K 1, then let bool 1 := 0. Let bool 2 := 0. If bool 1 = 1, then let bool 2 := 1 and check: If (u i, v j ) T 2 int K 2, then let bool 2 := 0. If bool 2 = 1, then go to (3 ), else continue with the next iteration. (3 ) Let bool 1 := 0 and bool 2 := 0. If (u i, v j ) T 3 K 3 holds, then let bool 1 := 1. If bool 1 = 1, then check: If (u i, v j ) T 4 K 4, then let bool 2 := 1. If bool 2 = 1, then go to (5 ), else go to (4 ). (4 ) If bool 1 = 0, then bool 1 := 1, else check: If (u i, v j ) T 3 int K 3, then let bool 1 := 0. Let bool 2 := 0. If bool 1 = 1, then bool 2 := 1 and check: If (u i, v j ) T 4 int K 4, then let bool 2 := 0. If bool 2 = 1, then go to (5 ), else continue with the next iteration. (5 ) Let C i := C i {v j } and continue with the next iteration. Endfor Endfor Step 6. For all i I q1 define the numbers c i := min C i and c i := max C i. 16

17 If q 1 = 1, then define the sets and go to Output. R 1 := {u 1 } [c 1, c 1 ] R 2 := For all i I q1 1 define the numbers Then, define the sets R 1 := R 2 := c i := max{c i, c i+1 } and c i := min{c i, c i+1 }. q 1 1 i=1 conv{(u i, c i), (u i, c i ), (u i+1, c i ), (u i+1, c i)}. If c 1 < c 1, then let R 1 := R 1 ({u 1} [c 1, c 1 ]). If c 1 > c 1, then let R 1 := R 1 ({u 1} [c 1, c 1]). If c q1 < c q 1 1, then let R 1 := R 1 ({u q 1 } [c q1, c q 1 1 ]). If c q1 > c q 1 1, then let R 1 := R 1 ({u q 1 } [c q 1 1, c q 1 ]). If q 1 > 2, then check: For i = 2,..., q 1 1: If c i < min{c i 1, c i }, then let R 1 := R 1 ({u i} [c i, min{c i 1, c i }]). If c i > max{c i 1, c i }, then let R 1 := R 1 ({u i} [max{c i 1, c i }, c i]). If c i < c i 1, then let R 1 := R 1 ({u i} [c i, c i 1 ]). If c i 1 < c i, then let R 1 := R 1 ({u i} [c i 1, c i ]). Endfor Output: The set Eff(P A ) = Eff(PÃ) := R 1 R 2 the initial multiobjective location problem (P A ). of all efficient solutions of Remark 3. According to the procedure described at Step 6, the set R 1 is either empty or it is a union of vertical line segments, while the set R 2 is a union of axis parallel rectangles, which may degenerate into horizontal line segments, 17

18 unless it is empty (when q 1 = 1). Consequently, we may also provide an explicit decomposition of the set Eff(P A ) into a family of (possibly degenerated) axis parallel rectangles. 6 Analysis of the Rectangular Decomposition Algorithm In order to prove that Algorithm 2 generates the set Eff(PÃ), we start by describing this set by using Theorem 7 (for à in the role of A) as Eff(PÃ) = [(cl S 1 cl S 2 ) ( N (Ã)\( S 1 S 2 ) )] [ (cl S 3 cl S 4 ) ( N (Ã)\( S 3 S 4 ) )] (23), where, for every r I 4, S r := N (Ã) i I q s r (ã i ) = i I q (N (Ã) s r(ã i )). (24) The next lemma allows us to interpret (23) in terms of the sets T 1,..., T 4 introduced at Step 4. In preparation of this result, observe that T r := MIN(A, K r ) = MIN(Ã, K r), r I 4, (25) as a consequence of (3). Also, note that all points of à are grid points, i.e., à Ã1 Ã2 (26) while the particular grid points e 1,..., e 4 defined at Step 3 are the vertices of the rectangular hull N (Ã) and we have N (Ã) = (e r + K r ) = conv{e 1,..., e 4 }. (27) r I 4 Lemma 13. For every r I 4 the following relations hold: S r = N (Ã) ( T r int K r ); (28) cl S r = N (Ã) ( T r K r ). (29) Proof. For any r I 4 define the (possibly empty) set Ĩ r := {i I q ã i T r }. (30) By definition of s r ( ), we have T r int K r = i Ĩr (ã i int K r ) = i Ĩr s r (ã i ) and T r K r = i Ĩr (ã i K r ) = i Ĩr cl s r (ã i ). Therefore, in order to prove (28) and (29), we just have to show that the following relations hold: S r = i Ĩr (N (Ã) s r(ã i )); (31) cl S r = i Ĩr (N (Ã) cl s r(ã i )). (32) 18

19 In view of (24), relation (31) can be rewritten as i I q (N (Ã) s r(ã i )) = i Ĩr (N (Ã) s r(ã i )). (33) Since I q Ĩr, the inclusion in (33) holds trivially. In order to prove the reverse inclusion, consider any point x i I q (N (Ã) s r(ã i )). Then x N (Ã) s r(ã j ) for some index j I q. Since the set à = {ãi i I q } is nonempty and compact, it satisfies the domination property with respect to the ordering cone K ψ(r) (see, e.g., Luc [17] or Göpfert et al. [11]): à MIN(Ã, K ψ(r)) + K ψ(r). (34) From (25) and (34) it follows that there is k I q such that ã j ã k + K ψ(r) and ã k MIN(Ã, K ψ(r)) = T ψ(r). (35) In view of (9), it follows that s r (ã j ) = ã j int K r ã k + K ψ(r) int K r = ã k K r int K r = ã k int K r = s r (ã k ), hence x N (Ã) s r(ã j ) N (Ã) s r(ã k ). (36) By (27) and (36) we get x N (Ã) er + K r and x s r (ã k ) = ã k int K r, hence ã k x + int K r e r + K r + int K r = e r + int K r. By the second part of (35) we infer that ã k T ψ(r) (e r + int K r ) = T r, which means that k Ĩr according to (30). Therefore (36) ensures that x j Ĩr (N (Ã) s r(ã j )). We conclude that inclusion in (33) holds, which ends the proof of (31). Now let us prove (32). The set Ĩr being finite, relation (31) entails cl S r = i Ĩr cl(n (Ã) s r(ã i )). (37) Without loss of generality we can assume that Ĩr. Then, for any i Ĩr we have ã i T r = T ψ(r) (e r + int K r ) = MIN(Ã, K ψ(r)) (e r + int K r ) à (e r + int K r ) in view of (25). Since {e r, ã i } N (Ã) and ãi e r + int K r, it follows that 1 2 (er + ã i ) (e r + int K r ) (ã i int K r ) = int N ({e r, ã i }) int N (Ã) int s r(ã i ), which shows that int N (Ã) int s r(ã i ). Taking into account that N (Ã) and s r(ã i ) are convex, we infer by a classical argument in convex analysis (see, e.g., Rockafellar [22, Theorem 6.5]) that cl(n (Ã) s r(ã i )) = cl N (Ã) cl s r(ã i ). (38) The desired relation (32) follows by (37) and (38), since N (Ã) is closed. 19

20 Remark 4. Given any point x N (Ã), it is easy to deduce by (23) and Lemma 13 that x Eff(PÃ) if and only if the following two conditions hold: and x ( T 1 K 1 ) ( T 2 K 2 ) or x / ( T 1 int K 1 ) ( T 2 int K 2 ) (39) x ( T 3 K 3 ) ( T 4 K 4 ) or x / ( T 3 int K 3 ) ( T 4 int K 4 ). (40) Our next result shows that the sets C 1,..., C q1, generated at Step 5, can be used to determine those efficient solutions of the reduced location problem (PÃ), which are also grid points (i.e., belong to Ã1 Ã2). To this aim, we introduce the so-called construction set C := {(u i, v j ) v j C i } = ({u i } C i ). (41) (i,j) I q1 I q2 i I q1 Theorem 14. For every i I q1 we have C i = j I q2 {v j (u i, v j ) Eff(PÃ)}. (42) Consequently, the construction set contains those efficient solutions of the reduced location problem, which are also grid points, i.e., we have C = (Ã1 Ã2) Eff(PÃ). (43) Proof. For every i I q1, the set C i has been defined within Step 5 as C i := j I q2 {v j x := (u i, v j ) satisfies (39) and (40)}. (44) Since {(u i, v j ) (i, j) I q1 I q2 } = Ã1 Ã2 N (Ã), relation (42) follows from (44) in view of Remark 4. Consequently, (43) holds. Remark 5. By applying Corollary 8 (1 ) for à in the role of A we infer that à Eff(PÃ). Therefore, (26) and (43) show that à C. Consequently, for every i I q1 the set C i is nonempty, hence c i := min C i and c i := max C i are well defined at Step 6. Example 2. Consider the set A = {a 1,..., a 8 } defined in Example 1. We have already computed the sets MIN(A, K r ) for all r I 4 (see Figure 1), so by applying Algorithm 2 at Step 1 we recover T 1 = {a 5, a 7 }, T 2 = {a 1, a 3 }, T 3 = {a 3, a 4, a 5 } and T 4 = {a 1, a 2, a 6, a 7 }. At Steps 2 and 3 we obtain à = A\{a8 } = {ã i i I 7 } with ã i := a i for all i I 7, then Ã1 = {u i i I 7 } and Ã2 = {v j j I 7 }, where u 1 = 2, u 2 = 4, u 3 = 6, u 4 = 10, u 5 = 12, u 6 = 14, u 7 = 18, v 1 = 2, v 2 = 4, v 3 = 6, v 4 = 8, v 5 = 9, v 6 = 10, v 7 =

21 Here we have e 1 = (2, 2), e 2 = (18, 12), e 3 = (18, 2) and e 4 = (2, 12). Thus, at Step 4 we find T 1 = T 2, T2 = T 1, T3 = T 4 and T 4 = T 3. By means of these sets, at Step 5 we obtain C 1 = {v 3 }, C 2 = {v 3, v 4 }, C 3 = {v 1, v 2, v 3, v 4 }, C 4 = {v 2, v 3, v 4, v 5 }, C 5 = {v 2, v 3, v 4, v 5, v 6 }, C 6 = {v 5, v 6, v 7 } and C 7 = {v 6 }. By (43) we deduce that the set of those efficient solutions of the reduced location problem (PÃ), which are also grid points, is C = i I 7 ({u i } C i ) (see Figure 2, where à C in view of Remark 5). Figure 2: The construction set for the set A in Example 1. In order to explain the procedure used in Step 6 for generating the whole set Eff(PÃ) we need four preliminary results. Lemma 15. Let R be an axis-parallel (possibly degenerated) rectangle. If all vertices (extreme points) of R belong to Eff(PÃ), then R Eff(PÃ). Proof. Without loss of generality we can assume that R is not a singleton. When R is degenerated into a (horizontal or vertical) line segment, the conclusion follows from a known result by Wendell et al. [24, Corollary 1]. Consequently, when R is a non-degenerated rectangle, then its horizontal edges are subsets of Eff(PÃ). Moreover, since any point x of R belongs to a vertical line segment whose extreme points are located on the horizontal edges of R, it follows that x Eff(PÃ). Lemma 16. Assume that q 1, q 2 > 1. If (u i, v j ), (u i+1, v k ) C for some i I q1 1, j I q2 and k I q2 \{j} (i.e., v k v j ), then we have (u i+1, v j ) C or (u i, v l ) C for some l I q2 \{j} such that min{j, k} l max{j, k} (i.e., v j < v l v k or v k v l < v j, depending on the values of the given indices, j < k or k < j, respectively). Proof. Follows as a particular instance of a known result by Wendell et al. [24, Corollary 3]. According to Step 6 of Algorithm 2 we consider in the following results the elements c i := max{c i, c i+1 } and c i := min{c i, c i+1 } for q 1 > 1 and for all i I q

22 Lemma 17. Assume that q 1 > 1. For every i I q1 1, we have (u i, c i), (u i, c i ), (u i+1, c i ), (u i+1, c i) C. (45) Proof. We first note that (41) and (43) yield (u i, c i ), (u i, c i ) {u i } C i C Eff(PÃ); (46) (u i+1, c i+1 ), (u i+1, c i+1 ) {u i+1 } C i+1 C Eff(PÃ). (47) Without any loss of generality we suppose in what follows that max{c i, c i+1 } = c i, i.e., c i = c i c i+1 ; min {c i, c i+1 } = c i, i.e., c i = c i c i+1. Under these assumptions, we can easily deduce by (46) that (u i, c i) = (u i, c i ) C and (u i, c i ) = (u i, c i ) C. (48) In order to show that (u i+1, c i ) Eff(P ) we distinguish two situations. If à c i = c i+1, then we have (u i+1, c i ) = (u i+1, c i+1 ) Eff(PÃ) by (47). Otherwise, if c i < c i+1, then q 2 > 1 and there exist j, k I q2, j < k, such that c i = v j and c i+1 = v k. Thus we have (u i, v j ) = (u i, c i ) C by (46) and (u i+1, v k ) = (u i+1, c i+1 ) C by (47). Moreover, since c i = max C i, there is no l I q2 such that v l > c i = v j and (u i, v l ) C. Therefore, by Lemma 16 we can conclude that (u i+1, c i ) = (u i+1, c i ) = (u i+1, v j ) C. (49) Similarly, for proving that (u i+1, c i ) Eff(P ), we should analyze two possible situations. If c i = c i+1, then (u i+1, c i ) = (u i+1, c i+1 ) Eff(PÃ) by à (47). If c i > c i+1, then c i = v j and c i+1 = v k for some j, k I q2 such that j > k. By (46) and (47) it follows that (u i, v j ) = (u i, c i ) C and (u i+1, v k ) = (u i+1, c i+1 ) C. On the other hand, since c i = min C i, there is no l I q2 such that v l < c i = v j and (u i, v l ) C. Lemma 16 yields By (48), (49) and (50) we infer (45). (u i+1, c i) = (u i+1, c i ) = (u i+1, v j ) C. (50) Lemma 18. Assume that q 1 > 1. For every i I q1 1, we have max{c i, c i+1 } = c i c i = min{c i, c i+1 }. Proof. For any i I q1 1 it holds (u i, c i ), (u i+1, c i ) C by Lemma 17, hence c i c i and c i c i+1. Thus we have c i min{c i, c i+1 } = c i. Remark 6. According to Lemma 18, if q 1 > 1, then at Step 6 we have R 2 = ([u i, u i+1 ] [c i, c i ]). i I q1 1 22

23 In preparation for the next results, we introduce the set R 1 := i I q1 ({u i } [c i, c i ]). (51) Remark 7. By construction of R 1 within Step 6 and by (51), we have C R 1 R 1. Lemma 19. The following relation holds: R 1 R 2 = R 1 R 2. (52) Proof. In view of Remark 7, in order to prove (52) it suffices to show that R 1 R 1 R 2. (53) First note that for q 1 = 1 we have R 1 = R 1, hence (53) holds. Now assume that q 1 > 1. According to Lemma 18, we have [c 1, c 1 ] = [c 1, c 1] [c 1, c 1] [c 1, c 1 ]; [c q1, c q1 ] = [c q1, c q 1 1] [c q 1 1, c q 1 1] [c q 1 1, c q1 ]. On the other hand, by construction of R 1 and R 2 at Step 6 we have ({u 1 } [c 1, c 1]) ({u 1 } [c 1, c 1 ]) R 1; ({u q1 } [c q1, c q 1 1]) ({u q1 } [c q 1 1, c q1 ]) R 1; ({u 1 } [c 1, c 1]) ({u q1 } [c q 1 1, c q 1 1]) R 2. Therefore the following inclusion holds: ({u 1 } [c 1, c 1 ]) ({u q1 } [c q1, c q1 ]) R 1 R 2. (54) If q 1 = 2, then the desired inclusion (53) follows by (51) and (54). If q 1 > 2, then consider any i I q1 \ {1, q 1 }. We just have to prove that {u i } [c i, c i ] R 1 R 2. (55) Consider the following (possibly degenerated) rectangles: R 1 := [u i 1, u i ] [c i 1, c i 1] and R 2 := [u i, u i+1 ] [c i, c i ]. (56) In view of Remark 6, we have Relations (56) and (57) show in particular that R 1 R 2 R 2. (57) ({u i } [c i 1, c i 1]) ({u i } [c i, c i ]) = R 1 R 2 ; (58) ({u i } [c i 1, c i 1]) ({u i } [c i, c i ]) R 2. (59) 23

24 By Lemma 18 we have c i c i 1 c i 1 c i and c i c i c i c i, hence c i min{c i 1, c i } max{c i 1, c i } c i. If c i < min{c i 1, c i }, then by construction of R 1 at Step 6 we have {u i } [c i, min{c i 1, c i}] R 1. Otherwise, if c i = min{c i 1, c i }, then the interval [c i, min{c i 1, c i }] reduces to a singleton, namely {c i 1 } or {c i }, and we infer by (59) that {u i } [c i, min{c i 1, c i}] {u i } {c i 1, c i} R 2. Similarly, when max{c i 1, c i } < c i we have {u i } [max{c i 1, c i }, c i ] R 1 according to the construction of R 1 at Step 6. max{c i 1, c i } = c i it follows by (59) that On the other hand, when {u i } [max{c i 1, c i }, c i ] {u i } {c i 1, c i } R 2 Thus, in all possible situations we have: ({u i } [c i, min{c i 1, c i}]) ({u i } [max{c i 1, c i }, c i ]) R 1 R 2. (60) Note that [c i, c i ] = [c i, min{c i 1, c i }] [min{c i 1, c i }, max{c i 1, c i }] [max{c i 1, c i }, c i]. Therefore, (60) allows us to deduce (55) by proving that {u i } [min{c i 1, c i}, max{c i 1, c i }] R 1 R 2. (61) To this aim we distinguish two cases: Case 1 : If R 1 R 2, then by (58) we have [c i 1, c i 1 ] [c i, c i ], hence [min{c i 1, c i }, max{c i 1, c i }] = [c i 1, c i 1 ] [c i, c i ]. By (59) we infer {u i } [min{c i 1, c i}, max{c i 1, c i }] = {u i } ([c i 1, c i 1] [c i, c i ]) R 2, which yields the desired relation (61). Case 2 : If R 1 R 2 =, then by (58) we have c i < c i 1 or c i 1 < c i. If c i < c i 1, then c i < c i 1 and c i < c i 1, hence min{c i 1, c i } = c i and max{c i 1, c i } = c i 1. Therefore, it is easy to deduce that [min{c i 1, c i}, max{c i 1, c i }] = [c i, c i ] [c i, c i 1] [c i 1, c i 1]. (62) Following the construction of R 1 at Step 6, we infer {u i } [c i, c i 1] R 1. (63) Hence by (59), (62) and (63) we conclude (61). Similarly, when c i 1 < c i we have min{c i 1, c i } = c i 1 and max{c i 1, c i } = c i. Thus, it follows [min{c i 1, c i}, max{c i 1, c i }] = [c i 1, c i 1] [c i 1, c i] [c i, c i ]. (64) According to Step 6, we have Consequently by (59), (64) and (65) it follows (61). {u i } [c i 1, c i] R 1. (65) 24

25 We are now ready to show that the whole set of efficient solutions of problem (P A ) can be recovered by means of the sets R 1 and R 2 generated at Step 6 of Algorithm 2. Theorem 20. The following representation holds: Eff(P A ) = Eff(PÃ) = R 1 R 2. Proof. In view of Corollary 12 and Lemma 19, it suffices to prove that Eff(PÃ) = R 1 R 2. (66) For proving inclusion in (66), let x R 1 R 2 be arbitrarily chosen. Case 1 : If x R 1, then there exists i I q 1 such that x {u i } [c i, c i ]. (67) Observe that (u i, c i ), (u i, c i ) {u i } C i C Eff(PÃ), according to (41) and (43). Therefore, by Lemma 15 we have {u i } [c i, c i ] Eff(PÃ), hence x Eff(PÃ) by (67). Case 2 : If x R 2, then q 1 > 1 and according to Remark 6 we can find an index i I q1 1 such that Due to Lemma 17 and (43), we know that x [u i, u i+1 ] [c i, c i ]. (68) (u i, c i), (u i, c i ), (u i+1, c i ), (u i+1, c i) Eff(PÃ). (69) Hence, by (68), (69) and Lemma 15 we deduce that x Eff(PÃ). Consequently inclusion in (66) holds. In order to prove the reverse inclusion in (66), we observe first that, according to Corollary 8 (2 ) applied for à in the role of A, we have Eff(P à ) = λ int R q + λ-eff(p à ). Therefore, it suffices to prove that for any λ int Rq + the following inclusion holds: λ-eff(pã) R 1 R 2. (70) To this aim we distinguish the following four cases, according to Lemma 3 (1 4 ) applied for à in the role of A: Case 1 : If λ-eff(pã) is a singleton, then there is (i, j) I q1 I q2 with λ-eff(pã) = {(u i, v j )} Ã1 Ã2. Since we have also λ-eff(pã) Eff(PÃ), we infer by (43) and Remark 7 that λ-eff(pã) C R 1. Case 2 : If λ-eff(pã) is a horizontal line segment, then q 1 > 1 and there is (i, j) I q1 1 I q2 such that λ-eff(pã) = [u i, u i+1 ] {v j }. Since c i v j c i, we conclude that λ-eff(pã) [u i, u i+1 ] [c i, c i ] R 2. Case 3 : If λ-eff(pã) is a vertical line segment, then q 2 > 1 and there is (i, j) I q1 I q2 1 with λ-eff(pã) = {u i } [v j, v j+1 ] {u i } [c i, c i ] R 1. Case 4 : If λ-eff(pã) is a non-degenerated rectangle, then q 1, q 2 > 1 and there is (i, j) I q1 1 I q2 1 such that λ-eff(pã) = [u i, u i+1 ] [v j, v j+1 ]. Since c i v j < v j+1 c i, we infer that λ-eff(p à ) [u i, u i+1 ] [c i, c i ] R 2. Thus (70) holds in all cases. 25

26 Example 3. Consider once again the set A = {a 1,..., a 8 } introduced in Example 1. By means of the sets C 1,..., C 7 constructed within Example 2, we apply the procedure described at Step 6 to generate the sets R 1 = ( {u 3 } [v 1, v 2 ] ) ( {u 6 } [v 6, v 7 ] ) ; R 2 = ( [u 1, u 2 ] {v 3 }) ( [u 2, u 3 ] [v 3, v 4 ] ) ( [u 3, u 4 ] [v 2, v 4 ] ) ( [u 4, u 5 ] [v 2, v 5 ] ) ( [u 5, u 6 ] [v 5, v 6 ] ) ( [u 6, u 7 ] {v 6 } ). Finally, we obtain the representation of the set of all efficient solutions Eff(P A ) = Eff(PÃ) = R 1 R 2 as the union of eight rectangles, among which four are degenerated into line segments (see Figure 3). Figure 3: Output of Algorithm 2 for the set A considered in Examples 1, 2, 3. 7 Computational results The Rectangular Decomposition Algorithm (Algorithm 2) is conceived to solve general multiobjective location problems of type (P A ) in absence of any (a priori given) information about the existence/number of nonessential objectives. Therefore we have implemented two variants of this algorithm, namely: Variant I (the implicit form of the algorithm) already formulated in Section 5, and Variant II, obtained from the implicit form by bypassing the procedure described at Steps 1 and 2 (identification of all essential objectives and elimination of the nonessential ones) and by letting à := A in Step 3. Both variants have been implemented in MATLAB and tested on various location problems of type (P A ). Our computational experiments show that Variant I is highly effective when the number of nonessential objectives, i.e., p q, is large. In what follows we present some numerical results obtained by solving several location problems with p {2 4, 2 6, 2 8, 2 10, 2 12, 2 14 } on a Core i x 3.20GHz CPU computer. For every p we have considered several test problems of type (P A ) with randomly generated points of A and we solved them by both variants. The averaged running times (in seconds) are listed in Table 1. Moreover, by Variant I we have also determined the number of essential objectives for each of these test problems, i.e., the cardinality of the reduced set à generated at Step 2. The averaged values of q are also listed in Table 1. 26

27 p q I II Table 1: Running times in seconds. Figure 4 contains two screen captures obtained in MATLAB for a test problem (P A ) with p = 16, which represent the rectangular decompositions of Eff(P A ) generated by Variant I (left picture) and Variant II (right picture). Figure 4: Rectangular decomposition of Eff(P A ) by Variant I and Variant II. 8 Concluding remarks The Rectangular Decomposition Algorithm developed in this paper is a new effective numerical method for solving multiobjective location problems of type (P A ), where A is a finite set of points in R 2 representing a priori given facilities, while the distance-type objective functions are defined by means of the Manhattan norm. Its mathematical background mainly relies on a dual characterization of efficient solutions of (P A ) given by Gerth & Pöhler [10], several structural properties of the set Eff(P A ) established by Wendell et al. [24], and a new characterization of this set in terms of minimal elements of A with respect to four ordering cones of R 2. The latter characterization leads to an implementable procedure, embedded in the first phase of the Rectangular Decomposition Algorithm, based on the Jahn-Graef-Younes method for solving discrete vector optimization problems. In comparison with other known algorithms for solving problems of type (P A ), our algorithm has several special features. First of all it identifies all nonessential objectives, hence the running time needed to generate Eff(P A ) can be reduced drastically, by eliminating these objectives. In contrast to certain algorithms which produce one efficient solution or a part of the set Eff(P A ) corresponding to a particular choice of weights or other scalarization parameters, our algorithm provides a well structured representation of the whole set Eff(P A ) as a finite union of line segments and rectangles. This 27