Operations Research Letters 36 (2008) 271–275

OperationsResearchLetters

www.elsevier.com/locate/orl

A note on the minmax regret centdian location on trees

Eduardo CondeDepartamento de Estadística e Investigación Operativa, Facultad de Matemáticas, Universidad de Sevilla,

Campus Universitario de Reina Mercedes, 41012 Sevilla, Spain

Received 13 November 2006; accepted 30 May 2007Available online 1 July 2007

Abstract

The minmax regret optimization model of the doubly weighted centdian location on trees is considered. Assuming that both types of weights,demands and relative importance of the customers, are partially known through interval estimates, an exact algorithm of complexity O(n3 log n)

is derived. This bound is improved in some special cases.© 2007 Elsevier B.V. All rights reserved.

Keywords: Robust optimization; Location; Centdian criterion

1. Introduction

The problem of locating a new facility on a transportationtree in which each node represents a different customer hasbeen considered in the literature according to different costmodels. Most of the cost models define a monotone nonde-creasing function of travel distances between the customersand the new server as the objective function that must beminimized. The median and center models are particular ver-sions of such cost functions. The optimal locations of thesetwo models exhibit good properties of efficiency and equity,respectively. Unfortunately, both models drive to, in general,different optimal locations. In order to reconcile both tenden-cies, the convex combination of the two criteria was introducedby Halpern in [4] who called the new cost function, centdiancriterion.

In this paper, we will study the centdian location model ona tree when the demands of the customers and its weights ofrelative importance are only partially known. Specifically, weassume that a set of interval estimates of these parameters isgiven and each possible set of values for the parameters is con-sidered as a possible scenario under which the proposed loca-tion must have the best possible behavior. One of the criteria

E-mail address: educon@us.es.

0167-6377/$ - see front matter © 2007 Elsevier B.V. All rights reserved.doi:10.1016/j.orl.2007.05.009

used in the literature in these context is the minmax regretcriterion (see e.g. [1–3,6] and references therein).

The hypotheses considered in our minmax regret locationmodel are listed below.

Hypothesis 1.1. (a) The location of the new server will bemade in any point on a link of an undirected transportationtree T = (N, E), where N is the set of nodes a1, . . . , an and Eis the set of n − 1 links between given pairs of nodes. We willdenote a feasible location x by the constraint x ∈ T .

(b) The length of each link (ai, aj ) is a positive value lij > 0which is assumed to be known.

(c) Every node ai represents the location of a given customerwhose demand can be any value ws

i ∈ [w−i , w+

i ](w−i �0). The

index s is used here to denote a possible demand scenario.(d) The relative importance associated to each node ai will

be any value usi ∈ [u−

i , u+i ](u−

i �0) under the scenario s.

According to these hypotheses, the median and center costfunctions are given respectively, by

Zm(x, s) =n∑

i=1

wsi d(x, ai)

and

Zc(x, s) = maxi=1...n

{usi d(x, ai)},′

272 E. Conde / Operations Research Letters 36 (2008) 271–275

under the scenario s, where by d(x, ai) we have denoted thetravel distance between the node ai and the location x of thenew server.

The doubly weighted centdian objective function under thescenario s is defined (see e.g. [8]) by

F(x, s) = Zm(x, s) + Zc(x, s).

Given a location x ∈ T we will measure its behavior underthe scenario s by comparing its centdian cost F(x, s) with theoptimal value under this scenario, that is,

R(x, s) = F(x, s) − F ∗(s),

where

F ∗(s) = miny∈T

F (y, s). (1)

We call this measure, R(x, s), the regret of x under the scenarios and it represents the opportunity loss incurred because thelocation x has been selected without knowing the true scenarios that takes place.

For the sake of simplicity in notation, we will denote the setof scenarios by S, a parametrization of the cartesian productof the intervals [w−

i , w+i ], [u−

i , u+i ] for each i = 1, . . . , n. The

maximum regret over the set of possible scenarios, that is,

Z(x) = maxs∈S

R(x, s) (2)

is the objective function that will be minimized in our opti-mization model given by

minx∈T

Z(x) ROBCENT(T , w, u).

Here, the label used, ROBCENT, makes reference to ROBustCENTdian model in virtue of the robustness properties thatexhibit the optimal solutions of the minmax regret optimizationmodels (see [6]).

The deterministic version of the problem ROBCENT(T , w, u),that is, the problem generated when the intervals defined by thevectors w and u are degenerated, can be exactly solved in lin-ear time, O(n), by the algorithm of Tamir et al. [8]. However,in order to solve some instances of ROBCENT(T , w, u) thathave been recently considered in the literature, superlinear-time algorithms are required. For instance, the robust medianproblem, that corresponds to ROBCENT(T , w, 0) can be solvedin O(n log2 n) by the algorithm of Averbakh and Bermanproposed in [3] and the best known algorithm for the robustcenter problem (ROBCENT(T , 0, u)) needs O(n2) elementaryoperations to have an optimal solution, [1].

In the following section, some techniques developed by Aver-bakh and Berman in [1,3] are used in order to give an algo-rithm of O(n3 log n) complexity for ROBCENT(T , w, u). Wealso determine a subtree containing at least one of the optimalsolution of this problem that can be used to prune some of thebranches of the original tree without eliminating the whole setof the optimal solutions. In the last section we analyze some

interesting special cases of the general problem that allow usto decrease the previous upper bound on the complexity.

2. General weights

The evaluation of the maximum regret of a feasible loca-tion x, Z(x), defined in (2), requires solving a maximizationproblem over the feasible set of scenarios. The efficiency withwhich this evaluation can be done, influences in a crucial way,the numerical behavior of the algorithms developedd to solvethe robust median and the robust center problems [1,3]. Inthese papers, a drastic reduction of the set of feasible scenar-ios that must be considered in the calculation of Z(x) wasmade. In particular, in these two problems one only needs toconsider a subset of cardinality O(n) of scenarios in orderto compute the maximum regret of a feasible location. Fromnow on, we will refer to this reduction as a sufficient subset ofscenarios.

As it is easy to see, combining the sufficient subsets of sce-narios proposed in [1,3] for the robust minmax versions of themedian and center locations on trees we can have a sufficientset of scenarios for the robust centdian model. First of all, wegive, for the sake of completeness, the sufficient subsets of sce-narios for these two problems.

Let e be a link of T, and let T1 and T2 be the two subtrees inwhich T is split by deleting e. We construct the feasible scenariogiven by selecting the demand upper bound, w+

i , of each nodein T1 and its demand lower bound, w−

i , for each node in T2.In this way, one can construct two demand scenarios for eachlink e, by interchanging the role of the subtrees T1 and T2. Lets1,2(e) and s2,1(e) be both scenarios and Sm, the whole set ofthese scenarios when all the different links, e, have been takeninto account. The set Sm, that has (at most) a cardinality of2(n − 1) scenarios, has been shown [3] to be a sufficient set ofscenarios for the robust median location on trees.

On the other hand, let Sc be the set of the n scenarios ob-tained by selecting one of the upper bound on the relative im-portance u+

i , associated to the demand point ai and the lowerbounds u−

j , for the rest of the customers. It was shown in [1],that Sc is a sufficient set of scenarios for the minmax cen-ter location on a tree. Combining both results we have thefollowing

Proposition 2.1. The set Sm ×Sc is a sufficient set of scenariosto evaluate the maximum regret of the centdian cost Z(x), thatis

Z(x) = maxs∈Sm×Sc

{F(x, s) − F ∗(s)}.

Proof. Using [2] and [3] one has that

Zm(x, s) + Zc(x, s) − Zm(y, s) − Zc(y, s)

� maxs∈Sm×Sc

{Zm(x, s) + Zc(x, s) − Zm(y, s) − Zc(y, s)},

∀s ∈ S, x, y ∈ T .

E. Conde / Operations Research Letters 36 (2008) 271–275 273

Hence,

maxs∈S

maxy∈T

{Zm(x, s) + Zc(x, s) − Zm(y, s) − Zc(y, s)}

� maxy∈T

maxs∈Sm×Sc

{Zm(x, s)+Zc(x, s)−Zm(y, s)−Zc(y, s)}

= maxs∈Sm×Sc

maxy∈T

{Zm(x, s)+Zc(x, s)−Zm(y, s)−Zc(y, s)}

� maxs∈S

maxy∈T

{Zm(x, s) + Zc(x, s) − Zm(y, s) − Zc(y, s)},

x ∈ T .

The last expression is equal to the first one, then every inequalityin this chain is an identity and we have the result. �

Following Proposition 2.1, in order to evaluate Z(x) we needall the values F ∗(s) for s ∈ Sm × Sc. Taking into account thatthe deterministic centdian problem on a tree can be solved inO(n) time by the algorithm of Tamir et al. [8], and havingO(n2) different scenarios in the sufficient set Sm ×Sc, the finalcomplexity grows to O(n3) (distances d(x, ai) for all ai ∈ N

can be obtained in O(n) time) that is, the following is verified.

Property 2.1. The maximum regret function Z(x) can be eval-uated in O(n3).

The computational effort spent to obtain the values F ∗(s)for the set Sm × Sc can also be used to determine a subtreeof T that contains, at least, one minimum of Z(x). Let C bea set containing at least one optimal solution y(s) ∈ T foreach sufficient scenario s ∈ Sm × Sc, that is, one has thatF(y(s), s) = F ∗(s) and let us take TC, the envelope tree of Cin T given by

TC =⋃

x,y∈CP(x, y),

where P(x, y) represents the path on T joining x and y. Thesubtree TC can be obtained as a by-product of the process ofdetermining the set C. In fact, if the tree T is rooted at somedistinguished node, say a1, we only need to collect the pointsy(s) furthest from a1 in every branch of the rooted tree. Oncethe set C has been constructed the ending piece of each branch,from the furthest point y(s) to the leaf of each branch, can betrimmed since it does not contain any point of C.

Property 2.2. The subtree TC contains, at least, one optimalsolution of the problem ROBCENT(T , w, u).

Proof. Given x ∈ T , let s(x) be a worst scenario for x, thatis, Z(x)=F(x, s(x))−F ∗(s(x)). By construction, there existsy(s(x)) ∈ C such that F(y(s(x)), s(x)) = F ∗(s(x)). In partic-ular, given y ∈ T such that x ∈ P(y(s(x)), y), by convexityof the function y −→ F(y, s(x)) over the path P(y(s(x)), y)

one has that

F(y, s(x))�F(x, s(x)),

R(y, s(x)) = F(y, s(x)) − F ∗(s(x))�Z(x),

Z(y)�Z(x).

Now, let y /∈ TC, and let x be the leaf of the subtree TC nearestto the location y, that is, x ∈ C such that

x ∈ P(y, z) ∩ TC, ∀z ∈ C,

then, we have that Z(x)�Z(y) and hence the result has beenshown. �

The path convexity (see e.g. [5]) of the function F(·, s)allows us to determine a link of the tree T that contains, at least,an optimal solution of ROBCENT(T , w, u) by means of thealgorithm 1 of Averbakh and Berman [3]. Hence, using Prop-erty 2.1, this optimal link can be determined in O(n3 log n)

(see [3]). At this point, it is interesting to mention that we caninitialize the algorithm 1 of [3] by the subtree TC, known tocontain at least an optimal solution (Property 2.2), instead of Twith no increase in the computational effort. This can mean, inpractice, a substantial reduction in the computational time ofexecution of the algorithm.

Finally, in order to obtain an optimal solution of the robustcentdian problem, one must solve the minimization of Z(x)

over the link that contains the optimal solution. The functionZ(x), can be expressed on any link of T as the maximum of n3

linear function using Proposition 2.1 and the expression

F(x, s) = maxj=1...n

{n∑

i=1

wsi d(x, ai) + us

j d(x, aj )

}.

Hence, a minimum of Z(x) can be obtained in O(n3) by thealgorithm of Megiddo [7].

For the sake of completeness, we include here the algorithmproposed in [3] for the determination of the optimal edge wherethe initialization has been modified by using Property 2.2.

Algorithm 2.1.

0: Input: T = (N, E), w±i , u±

i for each ai ∈ N .1: Initialization: Calculate an optimal

solution of problem (1) for each s ∈ Sm × Sc.Let C be the set of such optimal solutions.Take T 1 = TC.

2: Iteration k = 1, 2, . . .: Find a centroid xk of

T k . Evaluate Z(xk) and let s(xk) and y(xk) suchthat

Z(xk) = F(xk, s(xk)) − F(y(xk), s(xk)).

If y(xk) = xk , STOP, xk is an optimal solution(see [3]). In other case set T k+1 as {xk} unionwith the connected component of T k\{xk}that contains y(xk).

274 E. Conde / Operations Research Letters 36 (2008) 271–275

3: Output : If T k+1 is contained in a singleedge of T, STOP, this edge contains anoptimal solution of our problem. In othercase, go to Iteration k + 1.

Theorem 2.1. Problem ROBCENT(T , w, u) can be solved inO(n3 log n) by Algorithm 2.1.

3. Special cases

In this section we will consider particular instances ofROBCENT(T , w, u) that can be solved with a smaller compu-tational bound than the general case.

3.1. Case: u−i = 0, u+

i = u, ∀i

In this case, we assume that the relative importance weightfor each customer can fluctuate between zero and a commonupper bound u. This represents the situation in which the deci-sion makers cannot distinguish between the importance of thedifferent customers and admit that the exact value of impor-tance for each customer can vary in the same range of weights.In particular, the optimal location we are looking for must havea good behavior in the potential scenario s in which some ofthe customers have no influence (us

i =0) in the equity criterion.Let us consider the model ROBCENT(T , w, u0), where u0 is

defined by u−i =0 and u+

i =u > 0 for each i =1, . . . , n. In thissituation, using Proposition 2.1 one can express the maximumregret Z(x) as follows

Z(x) = maxj=1...n

maxs∈Sm

{n∑

i=1

wsi d(x, ai) + ud(x, aj )

− miny∈T

[n∑

i=1

wsi d(y, ai) + ud(y, aj )

]}. (3)

Taking into account that

miny∈T

[n∑

i=1

wsi d(y, ai) + ud(y, aj )

]

is the median problem on a tree, there will exist, at least, anoptimal solution in one of the nodes of T. Hence, we can rewriteZ(x) as the following maximum:

Z(x) = maxj=1...n

maxy∈N

�(x, y),

where

�(x, y) = maxs∈Sm

{n∑

i=1

wsi d(x, ai) + ud(x, aj )

−n∑

i=1

wsi d(y, ai) − ud(y, aj )

}.

Finally, using Lemma 2 of [3] it follows that

maxy∈N

�(x, y)

can be evaluated in O(n log n) and then, we have the following:

Property 3.1. If u−i = 0 and u+

i = u > 0, ∀i, the maximumregret of a location x, Z(x), can be evaluated in O(n2 log n).

Following the same procedure described in the last section,we can use Algorithm 1 of [3] in order to find a link contain-ing, at least, one optimal location. Now, taking into accountthe complexity bound given in Property 3.1, one has that thisoptimal link can be found in O(n2 log2 n).

Finally, in order to obtain a minimum of Z(x) over the op-timal edge, we can use the result of Averbakh and Berman[3, Section 4], where a set of, at most, n linear function, whosecoefficients can be obtained in linear time, is proposed in orderto evaluate the maximum

maxs∈Sm

{n∑

i=1

wsi d(x, ai) + ud(x, aj )

− miny∈T

[n∑

i=1

wsi d(y, ai) + ud(y, aj )

]}.

Hence, having in mind expression (3) one has that Z(x) can bewritten in the optimal edge as a maximum of, at most, O(n2)

linear function whose coefficients can be obtained in O(n2)

time. A minimum of this type of functions can be determinedin O(n2) time, using the algorithm of Megiddo [7]. Then wehave the following

Theorem 3.1. If u−i = 0, u+

i = u > 0, ∀i, the problemROBCENT(T , w, u0) can be solved in O(n2 log2 n).

3.2. Case: u−i = u+

i = u, ∀i

The instance of the problem ROBCENT(T , w, u) consideredhere assumes that there is no uncertainty about the weightsmeasuring the relative importance of the customer, in fact, allthe customers have the same importance weight u. When u in-creases, the optimal locations for our model will be closer tothe center of the tree. Hence, this model gives a robust gener-alization of the centdian solution where the uncertainty is onlyconsidered in demands. As we will see in what follows, forthis model it is possible to improve the general upper boundon the computational effort needed in order to find an optimalsolution. To do it we will use some known properties (see [4]),verified by the optimal solutions of the deterministic centdianmodel.

Let c be the absolute center of the tree, that is, the optimalsolution of

miny∈T

max{d(x, ai)}, (4)

E. Conde / Operations Research Letters 36 (2008) 271–275 275

we have the following:

Proposition 3.1. If u−i = u+

i = u for every i = 1 . . . n then, itis verified that

F(x, s) =n∑

i=1

wsi d(x, ai) + ud(x, c) + uZc(c, s).

Proof. It is known [4] that the only solution of (4) is the middlepoint of the path joining the two farthest nodes of T, let us saya1 and a2. If we add c to the set of nodes, we can split thetree T into subtrees defined by each connected component ofT \{c}, including c in every subtree.

It is verified that none of these subtrees contain the twonodes a1 and a2 that determine the absolute center, because cis properly contained in the path joining both nodes.

On the other hand, given x ∈ T it is true that

d(x, ai)�d(x, c) + d(c, ai)

�d(x, c) + maxj=1...n

{d(c, aj )} ∀i.

However, since x belongs to one of the subtrees defined by cand having in mind that this subtree cannot contain the twonodes a1 and a2, one has that

d(x, c) + maxj=1...n

{d(c, aj )} = max{d(x, a1), d(x, a2)} ∀i.

Hence,

maxi=1...n

{d(x, ai)} = d(x, c) + maxj=1...n

{d(c, aj )},

that is,

Zc(x, s) = ud(x, c) + uZc(c, s). �

Proposition 3.1 implies that the robust centdian problem canbe solved by means of a robust median problem on a modifiedtree T ′, such as it happens in the deterministic model (see [4]).

In the cases in which the absolute center c lies in the inte-rior of a given edge of T, let us say (ar , as), we take T ′ byeliminating such an edge, and adding c to the set of nodes and(ar , c), (c, as) to the set of edges. In other case, c will matchwith a given node of T, in this situation we do not modify thestructure of the original tree, that is, we take T ′ = T .

Finally, we will take as the demand interval associated to thenew node [w′−

c , w′+c ] = [u, u], remaining invariant the rest of

the demand intervals, that is, [w′−i , w

′+i ] = [w−

i , w+i ]. In the

case in which c matches with one of the nodes of T, let us sayar , we will modify its original demand interval by [w′−

r , w′+r ]=

[w−r + u, w+

r + u].

Proposition 3.2. If u−i = u+

i = u, ∀i, the problem ROBCENT(T , w, u) is equivalent to the robust median problem ROBCENT(T ′, w′, 0).

Proof. Following Proposition 3.1 we have that

F ∗(s) = uZc(c, s) + miny∈T

(Zm(y, s) + ud(y, c)).

Hence,

R(x, s) = Zm(x, s) + ud(x, c) − miny∈T

(Zm(y, s) + ud(y, c)),

that is, the maximum regret Z(x) that must be mini-mized reduces to the objective function of the problemROBCENT(T ′, w′, 0). �

Finally, we conclude with:

Theorem 3.2. If u−i = u+

i = u > 0, ∀i = 1 . . . n, the problemROBCENT(T , w, u) can be solved in O(n log2 n).

Proof. Determining the absolute center c is a task that requiresO(n) elementary operations [7]. With c the modified tree T ′and the new vector of bounds for the demand intervals, w′, canbe built. Finally, it is known [3] that the robust median problemROBCENT(T ′, w′, 0) can be solved in O(n log2 n). �

Acknowledgments

This research has been supported by the Spanish Ministryof Education and Science and FEDER Grant No. MTM2006-04393.

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