Heuristic and Special Case Algorithms for Dispersion Problems
Author(s): S. S. Ravi, D. J. Rosenkrantz and G. K. Tayi
Reviewed work(s):
Source: Operations Research, Vol. 42, No. 2 (Mar. - Apr., 1994), pp. 299-310
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HEURISTICAND SPECIALCASE ALGORITHMS
FOR DISPERSIONPROBLEMS
and G. K. TAYI
S. S. RAVI, D. J. ROSENKRANTZ
Universityat Albany-SUNY,Albany,New York
(ReceivedAugust 1991;revisionreceivedJune 1992;acceptedJune 1992)
The dispersion problem arises in selecting facilities to maximize some function of the distances between the
facilities.The problemalso arisesin selectingnondominatedsolutionsfor multiobjectivedecisionmaking.It is known
to be NP-hardundertwo objectives:maximizingthe minimumdistance(MAX-MIN)betweenany pairof facilitiesand
maximizingthe averagedistance (MAX-AVG).We consider the question of obtaining near-optimalsolutions. For
MAX-MIN, we show that if the distancesdo not satisfythe triangleinequality,there is no polynomial-timerelative
approximationalgorithmunless P = NP. When the distancessatisfythe triangleinequality,we analyze an efficient
heuristicand show that it provides a performanceguaranteeof two. We also prove that obtaininga performance
guaranteeof less than two is NP-hard.For MAX-AVG,we analyzean efficientheuristicand show that it providesa
performanceguaranteeof four when the distancessatisfythe triangleinequality.We also presenta polynomial-time
algorithmfor the 1-dimensionalMAX-AVGdispersionproblem.Using that algorithm,we obtain a heuristicwhich
providesan asymptoticperformanceguaranteeof ir/2 for the 2-dimensionalMAX-AVGdispersionproblem.
any problemsin locationtheorydeal with the
placementof facilitieson a networkto minimize somefunctionof the distancesbetweenfacilities
or betweenfacilitiesand the nodes of the network
(Handlerand Mirchandani1979). Such problems
model the placementof "desirable"
facilitiessuch as
warehouses,hospitals,and fire stations. However,
therearesituationsin whichfacilitiesareto be located
to maximizesome functionof the distancesbetween
pairsof nodes. Such location problemsare referred
to as dispersionproblems (Chandrasekharan
and
Daughety1981,Kuby1987,ErkutandNeuman1989,
1990,and Erkut1990)becausethey modelsituations
in which proximityof facilitiesis undesirable.One
example of such a situationis the distributionof
businessfranchisesin a city (Erkut).Otherexamples
of dispersionproblemsarisein the contextof placing
"undesirable"
(also called obnoxious)facilities,such
as nuclearpowerplants,oil storagetanks,andammunitiondumps(Kuby 1987,Erkutand Neuman 1989,
1990, and Erkut 1990). Such facilitiesneed to be
spreadout to the greatestpossibleextent so that an
accidentat one of the facilitieswill not damageany
of the others.The conceptof dispersionis also useful
in the context of multiobjectivedecision making
(Steuer 1986). When the numberof nondominated
solutionsis large,a decisionmakermay be interested
M
in selectinga manageable
collectionof solutionswhich
are dispersedas far as possiblewith respectto the
objectivefunctionvalues.Otherapplicationsof facility dispersionarediscussedin Erkut(1990)andErkut
and Neuman(1989, 1990).
Analytical models for the dispersion problem
assumethatthe givennetworkis represented
by a set
V = {vI, v2, ...,
Vn} of n nodes with nonnegative
distance(alsocallededge weight)betweeneverypair
of nodes.The distancesareassumedto be symmetric
andso the networkcanbe thoughtof as an undirected
completegraphon n nodeswitha nonnegativeweight
on eachedge.The weightof the edge {vi,vjI(i $ j) is
denoted by w(vi, vj). We assume that w(vi, vi) = 0 for
1 s i s n. The objectiveof the dispersionproblemis
to locatep facilities(p s n) amongthe n nodesof the
network,withat most one facilitypernode,suchthat
some functionof the distancesbetweenfacilitiesis
maximized.Two of the optimalitycriteriaconsidered
in the literature(Kuby 1987, Erkut1990,and Erkut
and Neuman 1989) are MAX-MIN (i.e., maximize
the minimum distancebetweena pair of facilities)
and MAX-AVG(i.e., maximizethe averagedistance
betweena pair of facilities).Under either criterion,
the problemis knownto be NP-hard,even whenthe
distancessatisfythe triangleinequality(Wang and
Kuo 1988,HansenandMoon 1988,andErkut1990).
Subjectclassifications:Analysisof algorithms:computationalcomplexity.Facilities/equipment
planning:discretelocation.Programming:
heuristic.
Areaof review:OPTIMIZATION.
OperationsResearch
Vol. 42, No. 2, March-April 1994
299
0030-364X/94/4202-0299$01.25
? 1994 Operations Research Society of America
300 / RAVI, ROSENKRANTZAND TAYI
Althoughmany researchershave studiedthe dispersion problem(see Erkut and Neuman 1989 for a
survey and an extensive bibliography),except for
White(1991), the questionof whetherthereare efficient heuristicswith provablygood performancehas
not been addressed.This question forms the main
focus of this paper.We showthat if the distancesdo
not satisfythe triangleinequality,then for any constantK 3 1, no polynomialtime algorithmcan providea performance
guaranteeof K forthe MAX-MIN
dispersionproblemunless P = NP. When the distances satisfy the triangleinequality,we analyze a
knownheuristicand provethat it providesa performanceguaranteeof 2 for the MAX-MINdispersion
problem.This is an improvementover the performanceguaranteeof 3 provenin White(1991)usinga
differentheuristic.(This improvementwas obtained
independentlyby White 1992.)We also showthatno
polynomial-timealgorithm can provide a performanceguaranteeof lessthan2 unlessP = NP. Wealso
analyzean efficientheuristicfor the MAX-AVGdispersionproblemwith triangleinequality,and prove
that it providesa performanceguaranteeof 4. An
efficientalgorithmfor the 1-dimensionalMAX-MIN
dispersion problem is presented in Wang and
Kuo. We provide an efficient algorithm for the
1-dimensionalMAX-AVGdispersionproblem.We
also showhow this algorithmcan be usedto obtaina
heuristicwithan asymptoticperformance
guaranteeof
1.571for the 2-dimensionalMAX-AVGdispersion
problem.
The remainderof this paperis organizedas follows.
Section 1 containsthe formaldefinitionsand a discussionof the previousworkon the dispersionproblem. Sections2 and 3 addressthe dispersionproblem
under the MAX-MIN and MAX-AVG criteria,
respectively.Section 4 presentsour results for the
1- and 2-dimensionaldispersionproblems.Section5
containstables which summarizeprior results,our
contributions,and open problems.
1. DEFINITIONSAND PREVIOUSWORK
We begin with the specificationsof the MAX-MIN
and MAX-AVGdispersionproblemsin the formatof
Gareyand Johnson(1979).
MAX-MIN
FacilityDispersion(MMFD)
Instance:A set V = Iv1, v2, ..., VnI of n nodes, a
nonnegativedistancew(vi, v;) for each pair vi, v; of
nodes,and an integerp suchthat 2 < p < n.
Requirement: Find a subset P = Ivil, vi2,...,
of V with IPI = p, such that the objectivefunction
f(P) = minxpA w(x, y)} is maximized.
MAX-AVGFacility Dispersion (MAFD)
Instance:As in MMFD.
Requirement:Find a subset P=
{vil, vi2,...,
vij of
Vwith IPI = p, suchthatthe objectivefunction
g(P)
2
=
P(P
-
1)XyeP
W(X,A)
is maximized.
The objectivefunctionfor MAFD has the above
formbecausethe numberof edgesamongthe nodes
in P is p(p - 1)/2. Note thatmaximizingthe average
distance is equivalentto maximizingthe sum of
the distances.We point out that maximizingthe
averagewould sometimesproducesolutionswhich
are far from the optimum with respect to the
MAX-MINcriterionand vice versa.
The distancesspecifiedin an instanceof MMFDor
MAFD satisfy the triangle inequality if for any three
distinctnodes vi, vj, and Vk, w(vi, vJ) + w(vj, VOk)
w(vi, Vk). The set P of nodes at which an algorithm
places the p facilities is called a placement. Given a
placementP for an MMFD instance,the quantity
f(P) defined by
f(P)
=
minIw(x, y)j
(1)
xYEP
is called the solution value correspondingto P. Simi-
larly,given an MAFD instanceand a placementP,
to P is defined
the solutionvalueg(P) corresponding
by
2
(2)
g(P) =
i w(x, y).
p
-
1) XJYEP
Both MMFD and MAFD are known to be NPhard,even when the edgeweightssatisfythe triangle
inequality(Wangand Kuo 1988, Hansenand Moon
1988, and Erkut 1990). Much of the work on the
dispersionproblemreportedin the literature(see the
bibliographyin Erkutand Neuman 1989) falls into
two categories.Papersin the firstcategorydeal with
branch-and-boundalgorithms and heuristics (see
Kuby 1987, Erkut, Baptie and von Hohenbalken
1990, Erkut 1990, Erkutand Neuman 1989, 1990,
andthe referencescitedtherein).However,exceptfor
White(1991), only experimentalstudiesof the performanceof the heuristicshavebeen reported.White
(1991)presentsa heuristicforMMFDwhenthe nodes
are points in d-dimensionalEuclideanspace and
the distancebetweena pairof pointsis theirEuclidean
AND TAYI
RAVI, ROSENKRANTZ
distance. He shows that the heuristic always produces
a placement whose solution value is within a factor of
3 of the optimum solution value. In the next section,
we improve that result by considering a different
heuristic which guarantees a placement whose solution value is within a factorof 2 of the optimal solution
value for any instance of MMFD in which the distances satisfy the triangle inequality. We also show
that unless P = NP, no polynomial-time algorithm
can provide a better performanceguarantee.
Papers in the second category deal with restricted
versions and variants of MMFD and MAFD. For
example, 1- and 2-dimensional versions of MMFD
were studied by Wang and Kuo. They present a polynomial algorithm for the 1-dimensional MMFD and
prove that the 2-dimensional MMFD is NP-hard. We
note that problems MMFD and MAFD defined above
are discrete in nature because each facility must
be placed at one of the given nodes. Researchers
have also considered continuous versions of the
dispersion problems (Church and Garfinkel 1978,
Chandrasekharan and Daughety 1981, and Tamir
1991) where facilities may be placed at any point on
the edges of a given network. In Chandrasekharanand
Daughety a polynomial algorithm is presented for the
continuous version of MMFD on tree networks.
Churchand Garfinkelpresenta polynomial algorithm
for locating one facility on an edge of a connected
(but not necessarilycomplete) network to maximize a
weighted sum of the distances from the facility to the
nodes of the network. Tamir presents an improved
algorithmfor the same problem. In addition, he establishes the NP-hardness of the continuous versions of
MMFD and MAFD and presents results concerning
performanceguaranteesfor the continuous version of
MMFD. Dasarathy and White (1980) assume the
nodes of the network to be points in k-dimensional
space and consider the problem of finding a point
within a given convex polyhedron to maximize the
minimum Euclidean distance between the point and
the nodes. They present polynomial algorithms for
k = 2 and k = 3. For k = 2, this problem is referred
to as the largestemptycircle(LEC)in the computational geometry literature (see subsection 6.4 of
Preparataand Shamos 1985). The LEC problem can
be solved in O(n log n) time, which is known to be
optimal (Preparataand Shamos). A problem similar
to LECbut with a differentdistance function is studied
in Melachrinoudis and Cullinane (1986). A weighted
version of the problem for k = 2 is studied in Erkut
and Oncu (1991). For a discussion of other variants,
we refer the reader to Erkut and Neuman (1989,
1990).
/
301
In this paper, we consider only the discrete versions
of the dispersion problems. Our focus is on the analysis of heuristicsfor MMFD and MAFD. By a heuristic
we meana polynomial-time
approximation
algorithm
which produces feasible, but not necessarily optimal,
solutions. Heuristics are commonly classified as absolute or relativedepending on the types of performance
guaranteesthat can be establishedfor them (Horowitz
and Sahni 1984). An absolute approximation algorithm guaranteesa solution that is within an additive
constant of the optimal value for every instance of the
problem. A relative approximation algorithm guarantees a solution that is within a multiplicativeconstant of the optimal value for every instance of the
problem. It is easy to show, using the technique presented in Garey and Johnson (pp. 138-139), that there
are no absolute approximation algorithms for MAFD
or for MMFD, unless P = NP. So we restrict our
attention to the study of relative approximation
algorithms.
2. NEAR-OPTIMALSOLUTIONSTO MAX-MIN
FACILITYDISPERSION
We first consider MMFD without requiring the distances to satisfy the triangle inequality and prove a
negative result concerning relative approximation
algorithms.
Theorem 1. If thedistancesare not requiredto satisfy
the triangleinequality,then thereis no polynomialtime relativeapproximationalgorithmfor MMFD
unlessP = NP.
Proof. Suppose that A is a polynomial-time approximation algorithmwhich provides a performanceguarantee of K , 1 for MMFD. We show that A can be
used to devise a polynomial-time algorithm for a
problem which is known to be NP-complete. This
contradicts the assumption that P ? NP and, hence,
will establish the theorem.
The known NP-complete problem used here is
CLIQUE, whose definition is as follows (Garey and
Johnson).
Problem CLIQUE
Instance:An undirectedgraph G(N, E) and a positive
integer J < INI.
Question: Does G contain a clique of size : J (i.e., is
there a subset N' C N such that IN' 1 > J and every
pair of vertices in N' is joined by an edge in E)?
Consider an arbitraryinstance of CLIQUE defined
by a graph G(N, E) and integer J. Let n = INI and
302 /
RAvI, ROSENKRANTZ
AND TAYI
{XI, x2, ...,
Xn We constructan instanceof
MMFD (without triangle inequality) as follows. The
node set V= {v1, v2, . . . , Vn of the MMFD instance
is in one-to-one correspondence with N. The number
of facilities p is set equal to J and the distances are
defined as follows. Let w(v,, vj) = K + 1 if {xi, xj; is
in E; otherwise,let w(vi, v;)= 1. Clearly,this construction can be carried out in polynomial time. We will
show that for the resulting MMFD instance, the solution value of a placement produced by A is greater
than 1 iff G has a clique of size J.
Suppose that G has a clique of size J. Let
$xi, xi2, ..., xiJ denote the vertices of the clique.
Consider the placement P = {Ivi, vi2,. . . , viJ. By our
definition of the distances, the weight of every edge in
P is equal to K + 1. Therefore, the solution value of
the placement P is also K + 1. Since A provides a
performanceguarantee of K, the solution value of the
placement returned by A is at least (K + 1)/K, which
is greater than 1.
Now suppose that G does not have a clique of size
J. In this case, notice that no matter which subset of
p = J nodes is chosen as the placement, there will
always be at least one pair of nodes vi and v; with
w(vi, v>)= 1. Therefore,the solution value corresponding to any placement is at most 1. In particular,the
solution value of a placement produced by A is also
at most 1. Thus, by merely comparing the solution
value of the placement produced by A with 1, we can
solve an arbitraryinstance of the CLIQUE problem.
This completes the proof.
N=
Even though Theorem 1 provides a strong negative
result, it is not applicable in many practicalsituations
because distances often satisfy the triangle inequality.
Therefore,it is of interest whether there is an efficient
relative approximation algorithm for MMFD when
the distances satisfy the triangle inequality
(MMFD-TI). The remaining theorems in this section
precisely characterize the performance guarantees
obtainable for MMFD-TI.
A greedy heuristic (which we call GMM) for
MMFD-TI is shown in Figure 1. This heuristic is
essentially the same as the "furthest point outside
the neighborhood" heuristic, described in Steuer
(Chapter 11), using a different format. An experimental study of the performanceof this heuristic is carried
out in Erkut and Neuman (1990). In describing this
heuristic, we use P to denote the set of nodes at which
GMM places the p facilities. The heuristic begins by
initializing P to contain a pair of nodes in V which
are joined by an edge of maximum weight. Subsequently, each iteration of GMM chooses a node v
Step 1. Let vi and uj be the endpoints of an edge of maximum weight.
Step 2. P - {xtu,vj}.
Step 3. while ( P1 < p ) do
begin
a. Find a node v E V - P such that min {w(
, v')} is maximum
among the nodes in V - P.
b. Pe- P tt}
end
Step 4. Output P.
Figure 1. Details of heuristic GMM.
from V - P such that the minimum distance from v
to a node in P is the largest among all the nodes in
V - P. In each step, ties are broken arbitrarily.
Heuristic GMM terminates when IP I = p. The solution value of the placement P produced by GMM is
equal to minx and w(x, y)I.
We now present an example to illustratethe GMM
heuristic. Consider an MMFD-TI instance with five
nodes (denoted by Vt, V2, V3, V4, and V5)and let p = 3.
The edge weights which satisfy the triangle inequality
are: w(v1,
v2)=
3,
W(V2, V3) =
W(V2, V4) = W(V2, V5) =
1, and all other edges are of weight 2. To begin, GMM
will place two of the facilities at vt and v2 because
w(v1, v2) = 3 is the maximum edge weight. Now, no
matter where the third facility is placed, the solution
value of the placement is 1 because each of the remaining nodes (V3, V4, and v5) has an edge of weight 1 to
v2. However, an optimal placement consists of the
three nodes V3,V4, and vs and has a solution value of
2. Thus, for this example, the solution value produced
by GMM differs from the optimal value by a factor
of 2. Our next theorem shows that the performance
of GMM is never worse. (This result was obtained
independentlyby White 1992.) Moreover,we will also
show (Theorem 3) that unless P = NP, no polynomialtime heuristic can provide a better performance
guarantee.
Theorem2. Let I be an instanceof MMFD-TI. Let
OPT(I)andGMM(I)denote,respectively,
thesolution
valuesof an optimalplacementand thatproducedby
GMMforthe instanceL ThenOPT(I)/GMM(I)< 2.
Proof. Consider the set-valued variable P in the
description of GMM. Let f(P) = minxyp w(x, y)}.
We will show by induction that the condition
f(P) 3_:OPT(I)12
(3)
holds after each addition to P. Since GMM(I) = f(P)
after the last addition to P, the theorem then follows.
Since the first addition inserts two nodes joined by
AND TAYI /
RAVI. ROSENKRANTZ
303
Our next theoremshows that if P ? NP, GMM
provides the best possible performanceguarantee
obtainablein polynomialtime for MMFD-TI.
an edge of the largestweightinto P, (3) clearlyholds
afterthe firstaddition.So, assumethat the condition
holdsafterk additionsto P for some k - 1. We will
prove that the condition holds after the (k + l)st
additionto P as well.
Theorem 3. If P $ NP, no polynomial-timerelative
vI} denote an
optimalplacement.For convenience,we use /* for
guarantee of (2 - e)for any e > Ofor MMFDT-TI.
To that end, let P*
=
{v*, v!,
...,
OPT(I). The followingobservationis an immediate
consequenceof the definitionof the solution value
to a placementforan MMFDinstance.
corresponding
Observation1. For everypair vi, v7 of distinctnodes
in P*, w(v', v*) > 1*.
Let Pk = {x1, x2, ..., Xk+II denote the set P after
k additions.(Note that IPk I = k + 1, becausethe first
additioninsertstwo nodesinto P.) SinceGMM adds
at leastone more node to P, the followingis a trivial
observation.
Observation 2.
PkI
=k+
1 <p.
For each vi E P* (I < i s p), define S*=
{ E VIw(v0', u) < 1*/2}.That is, SP'is the set of all
nodeswhosedistancesfromvi*areless than1*/2.The
followingclaimprovidestwo usefulpropertiesof these
sets.
ClaimI
a. For 1 < i < p, SP'is nonempty.
b. For i #1, SP'and Sj*aredisjoint.
Proof of Claim 1
Part a: This is obvious, because vi*E S#' for 1 <
i < p.
f 0 for some i # j. Let
Partb: SupposethatS# n Sj*
u E SP'U Si*. Thus, w(v#, u) < 1*/2 and w(vj*,u) <
1*/2. By Observation1, w(v', vj*), 1*. These three
inequalitiestogetherimplythatthe triangleinequality
does not hold for the three nodes u, vi* and vj*.
Claimlb follows.
We now continuewiththe mainproof.SincePk has
less than p nodes (Observation2) and there are p
Sp*,there must be at least
one set, say Sr*(for some r, 1 < r < p), such that
Pk n Sr*- 0. Therefore,by the definitionof S*, we
must have for each u E Pk, w(vr*,u) : 1*/2. Since
is availablefor selectionby GMM, and GMM
rV*
selectsa node v E V - Pk for whichminced
pkw(v, v')
is a maximumamongthe nodesin V - Pk, it follows
disjoint sets So*,St, ...,
that (3) holds even afterthe (k + 1)stadditionto P.
Thiscompletesthe proofof Theorem2.
approximationalgorithmcan providea performance
Proof. Weusea constructionsimilarto thatpresented
in the proofof Theorem1,exceptthattheedgeweights
are chosen as follows. Let w(vi, v;) = 2 - e/2 if {xi, xj;
is in E; otherwise,let w(vi,vj)= 1. Usingthe factthat
0 < e < 1,it is easyto verifythatthe resultingdistances
satisfythe triangleinequality.The proofthatthe solution value of a placement produced by A for
MMFD-TI is greaterthan 1 if G has a clique of
sizeJ is virtuallythe sameas thatof Theorem1.
3. NEAR-OPTIMALSOLUTIONSTO MAX-AVG
FACILITYDISPERSION
In this section,we discussa relativeapproximation
algorithmfor MAFD under the triangleinequality
assumption(MAFD-TI).Thisheuristic,whichwe call
GMA, is shown in Figure2. It is identicalto the
GMM heuristicof Figure1, exceptthat in Step 3a,
we choose a node v E V - P for which Zv'eP w(v, v')
is maximumamongall the nodesin V - P. Note that
the solutionvalue of the placementP producedby
GMA is equal to 2/p(p - 1)
X,3EP
w(x, y).
An experimentalstudy of this heuristicis carried
out in ErkutandNeuman(1990).We focuson determining the performanceguarantee provided by
this heuristic.Our next theorem shows that GMA
is indeed a relative approximationalgorithmfor
MAFD-TI. Before presenting that theorem, we
introducesome notationwhich will also be used in
Section 4. Let A and B be disjoint subsets of V.
Define W(A)= X,,y w(x, y) and W(A,B) =
y). (Note that W(A, B) = W(B, A).)
xeA,yEB w(x,
Also, for x E A, let W(x, B) = Z,,B w(x, y). We now
Step 1. Let v; and oj be the endpointsof an edge of maximumweight.
Step 2. P <- {vi, o;}.
Step 3. while ( JP1< p ) do
begin
a. Finda nodev E V - P suchthatX:w ( vo,v') is maximum
EP
amongthe nodes in V - P.
b. PNP{P}.
end
Step 4. OutputP.
Figure 2. Details of heuristic GMA.
304 / RAVI, ROSENKRANTZAND TAYI
give two lemmaswhichare used severaltimes in the
proofof the performance
boundforthe GMAheuristic. The first is an immediateconsequenceof the
pigeonhole principle(Roberts1984).
Lemma 1. Let A and B be nonemptyand disjoint
subsetsof V. Thenthereis a node x E A such that
W(x,B) > W(A,B)/IA 1
Lemma 2. Given an instance of MAFD-TI, let A
and B be nonemptyand disjointsubsets of V with
IB I : 2. ThenW(A,B) : IAIW(B)/(IBI - 1).
Proof. Let v be an arbitrarynode in A. Let IB I = t
and B = lb,, b2, ..., b1[.Since t = IBI : 2, there is
at least one edge in B. For each edge bin,bjl in
B(i #1), we haveby the triangleinequality,
w(v, bi)+ w(v, bj), w(b, bj) 1 < i< js t. (4)
If we sum the inequalitiesshownin (4), the left-hand
side of the sum is (t - 1)W(v,B) becauseeach edge
weight w(v, bi)(1s i - t) appearsexactly (t - 1)times
in the sum. The right-handside of the sum is simply
W(B). Therefore,we get (t - 1) W(v, B) > W(B), or
W(v,B) - W(B)/(t - 1) = W(B)/(IBI - 1). (5)
Since inequality(5) holds for each v E A, we get
W(A,B) -Al W(B)/(IBI -1).
Theorem 4. Let I be an instanceof MAFD-TI. Let
thesolution
OPT(I)and GMA(I) denote,respectively,
valuesof an optimalplacementand thatproducedby
GMAfor the instanceL ThenOPT(I)/GMA(I)< 4.
Proof. Weshowby inductionthataftereachaddition,
the average weight of an edge in P is at least
OPT(I)/4.
The statementis clearlytrueafterthe firstaddition
(whichbringstwo nodes into P) becausean edge of
maximumweightis addedto P. So, assumethatp >
3 and the statementholds afterk additionsfor some
k > 1. We will provethat the statementholds after
the (k + 1)stadditionas well.
For convenience,we use /* for OPT(I). Let Pk
denotethe setP afterk additions.We havethe following two-partobservation.The firstpartis due to the
factthatGMAaddsat leastone morenodeto P. The
secondis an immediateconsequenceof the inductive
hypothesis.
Observation3
a. W(Pk)= k + /P12 */.
b. W(Pk) :,; k(k + 1)/2 1*/4.
To provetheinductivehypothesisforPk,1, it suffices
to show that thereis a node x E V - Pk such that
W(x, Pk) > (k + 1)1*/4,becausethis conditionin
conjunctionwith Observation3b implies that the
averageedgeweightis at least1*/4afterthe (k + 1)st
additionto P as well. We now state this condition
formallyas a claimand presentits proof.
Claim 2. There is a node x E V - Pk such that
W(x, Pk) > (k + 1)1*/4.
Proofof Claim2. LetP* denotethe set of p nodesin
an optimalplacement.Bythe definitionof 1*,we have
-
W(P*) = p(P 2 1)1*
(6)
We have two cases to consider, dependingupon
whetheror not Pk andP* aredisjoint.
Case1. (Pk andP* aredisjoint.)We applyLemma2
with Pk as the set A and P* as the set B. (We can do
so becausePk and P* aredisjointand IP*I = p > 3.)
We get
W(Pk, P*) > IPk I W(P*)/(P
=
-
1)
(7)
(k + 1)pl*12(using 6).
Since W(Pk,P*) = W(P*, Pk), we have W(P*,Pk) >
(k + 1)pl*12. Now, by Lemma 1, there must be a
nodex E P* suchthat
W(x, Pk) > W(P*,
Pk)/P
> (k + 1)1*/2 (using7).
Thus, the node x satisfiesa conditionwhichis even
strongerthanthatrequiredby Claim2.
Case 2. (Pk and P* are not disjoint.) Let Y = P* n Pk
and let X = P* - Y. Since Y is nonempty, we must
have:
IXI <p - 1.
(8)
Since X and Y are disjoint and P* = X U Y, we have
W(P*)
= AP
-
2
1) I*
= W(X) + W(Y) + W(X, Y).
(9)
We havetwo subcasesdependingon IlXI.
Case2a. (IXI < 1.) Note thatX ? 0, otherwise,Y =
P* = Pk and so IPkl = p, contradicting
Observation
la. Therefore,in this subcase,we need to consider
only the possibilitythat IXI = 1. Then, Pk = Y and
so Pk consistsof p - 1 nodes from P*. Let x be the
AND TAYI /
RAVI, ROSENKRANTZ
node in X. SinceGMA selectsa node v E V - Pk for
whichXV'epk w(v, v') is a maximumamongthe nodes
in V- Pk, and x is availablefor selection,it follows
that GMA will producean optimalsolutionin this
subcase.
Case 2b. (IXI > 1 (i.e., IXI , 2).) Here, from (9),
we can conclude that either W(X) : W(P*)/2 or
W(Y) + W(X,Y) - W(P*)/2.We considerthesetwo
possibilitiesseparately.
305
Let Q = Pk - Y. Note that Q and Y are disjointand
so IQI = IPkI - Iyl = (k + 1)- IYI. We apply
Lemma2 with Q as the set A and Y as set B (recall
that IYI : 2). We get
(
IYI)w(Y)
W(Q, Y) (k + 1 -- |y)(2
(12)
(IYI 1)
SincePk = Q U Y and the sets Q and Y are disjoint,
we also have,
W(Pk)? W(Y) + W(Q, Y)
Case2b.i. (W(X) > W(P*)/2)We use Lemma2 with
Pk as set A and X as set B (we can do so because
W(X,Pk) = W(Pk, X)
-
> (kl+ 1) W(X)
-
1) (
] (from 12)
l)
W((Y) I
(13)
> XI W(Pk)/(IPkI- 1)
(10)
From Lemma 1, there is a node x E X such that
W(x, Pk)> W(X,Pk)/ lXI. Combiningthis observation with (10), we get
APp- 1)
l
W(X, Pk)
(k + 1) pp 1)1*/4 (using 6).
IXI(IXI
IY-l
We now applyLemma2 with X as the set A and Pk
as the set B. We get,
(k + 1
IXI -) W(P*)/2
W(X, Pk)
+
W(Y)
IXI > 2) toget
=
IXl W(Pk)/k(fromObservationla)
>
IXIW(Y)/(IYI - 1) (from 13)
(IY1
[ W(P*)/2 - W(X,Pk)]
(from(ll)
)*4
)1*14.
(14)
Rearranging
(14) we get,
Sincep - 1 > IXI (from 8), the quantityp(p - 1)/
(IXI(IXI - 1)) is greaterthan 1. Therefore,we get
W(x, Pk) > (k + 1)1*/4 as required.
Case b.ii. (W(Y) + W(X, Y) > W(P*)/2)Firstnote
that if IYi = 1, then W(Y) = 0 and so W(X, Y) >
W(P*)/2. Since Y C Pk and the edge weightsare
nonnegative,we must have W(X, Pk) > W(X, Y).
1X +
WKXPk)
JYJ
I
1
>1 |XYl
W(P*)12.
Noting that IXI + IY = P*I = p, substitutingfor
W(P*)from(6),andeliminatingthecommondenominatorIYI - 1, we get
W(X, Pk)(p - 1)
IXI p(p - 1)1*/4.
Therefore, W(X, Pk) 2 W(P*)/2 = p(p - 1)1*/4, and
That is, W(X, Pk)
by Lemma1, theremustbe a nodex E X suchthat
and Lemma 1, we concludethat theremust be node
x E X suchthat W(x,Pk) , pl*/4 > (k + 1)1*/4(from
Observationla).
This completesthe proofof Claim2 and also that
of Theorem4.
Theorem4 showsthat the performanceguarantee
providedby GMA is no worsethan 4. However,the
guaranteemaywellbe less.Thefollowingresultshows
thatthe guaranteecannotbe lessthan2.
A
W(X, Pk)
|P -
1*/4
>
pl*/4 (from8)
>
(k + 1)1*/4(fromobservationla)
and so Claim2 holds when IYl = 1. Therefore,for
the remainderof this proof,we assumethat IYi > 2.
Sincein this subcasewe areassumingthat W(Y) +
W(X, Y) > W(P*)/2, and as observed above,
W(X, Y) < W(X,Pk), we have
W(
> W(P)
2
- W(X,Pk)
(11)
IXI pl*/4. From this inequality
Theorem 5. For any instance I of MAFD-TI, let
thesolution
OPT(I)and GMA(I)denote,respectively,
valuesof an optimalplacementand thatproducedby
GMA for the instanceL For any e > 0, thereis an
instance I, for which OPT(IE)/GMA(IJ)> 2 -
e.
306 / RAVI, ROSENKRANTZAND TAYI
Proof. Consider the MAFD instance I described
below. This instancehas a total of 2p nodes and p
facilitiesareto be located.Forconveniencein description, we partitionthe set of 2p nodes into threesets
calledX, Y,and Z. The setX hasp nodes(denotedby
xi, x2, . . ., xp), the set Y has p - 2 nodes (denoted
by Yi, Y2, . . ., YP-2), and Z has two nodes (denoted by
z1 and z2).The edge weights are chosen as follows. For
any distinctnodesxi and xj E X, w(xi,xj) = 2. Also
w(z1, z2) = 2. All other edge weights are 1. It is
to verifythat the distancessatisfythe
straightforward
triangleinequality.The setX is an optimalplacement
and its solutionvalue (OPT(I)) is 2 (becauseevery
edge in X has a weightof 2). We can forceGMA to
place the first two facilities at z1 and z2 because
w(zI, Z2)= 2 is a maximumedgeweight.It is easy to
verifythatin the subsequent(p - 2) steps,GMA can
be forcedto chooseall the (p - 2) nodesfromthe set
Y. Thus, we force GMA to return Z U Y as the
placement.The solutionvalue(GMA(I))corresponding to this placementis givenby
4.1. A Polynomial-Time Algorithm for 1D-MAFD
Ourpolynomial-time
algorithmfor 1D-MAFDis also
basedon a dynamicprogramming
approach.It runs
in O(max{nlog n, pnj) time. In the developmentof
the dynamicprogramming
formulationfor this problem, we use the notationintroducedin Section3. In
studying the formulation,the reader should bear
in mindthatmaximizingtheaveragedistanceis equivalentto maximizingthe sum of the distances.
Webeginby sortingthepointsinto increasingorder.
Let V = {x,, x2, .. ., x"} denote the points in sorted
order. Consider a point x; and an integer k s
min(j,p). Foreachsuchcombinationof xj and k, the
dynamicprogramming
algorithmconsiderschoosing
k points from {xl, x2, . . ., xjI to maximize a certain
quantitydescribedbelow.Let C = A U B be a set of p
points consisting of a subset A C {xi, . . ., xj such
that IAl = k, and a subset B C Ixjl, . . ., x"}
such that IB I = p - k. For each pairof pointsxu in
A andx, in B, we have
w(xU,xv) = w(xj,xU)+ W(xj,xe).
GMA(I)=
P p- 1)2
2 +)
because in the placement,one edge (namely, that
between z, and z2) is of weight 2 and each remaining
edge is of weight 1. A bit of simplificationshows
that GMA(I) = 1 + 2/(p(p - 1)). The ratio
OPT(I)/GMA(I) is given by
OPT(I)
GMA(I)
2
1+ 2/(p(p -1))
closeto 2 by
Clearly,the ratiocan be madearbitrarily
choosingp to be sufficientlylarge.
4. DISPERSIONPROBLEMSIN ONE AND TWO
DIMENSIONS
The 1-dimensionaldispersionproblemsarerestricted
versions of MMFD and MAFD, where the node
set V consistsof a set of n points (denotedby xl,
x2, ...
, xn) on a line. Thus w(xi, xj) = Ixi - xjl. We
denote these problems by 1D-MMFD and
1D-MAFD,respectively.Similarly,in the case of the
2-dimensional dispersion problems (denoted by
2D-MMFDand 2D-MAFD,respectively),the node
set V is a set of n points in ,2 and the distance
betweena pairof pointsis the Euclideandistance.It
is knownthat1D-MMFDcanbe solvedin polynomial
time using a dynamic programmingapproachand
that2D-MMFDis NP-hard(WangandKuo).Accordingly,we consider1D-MAFDand 2D-MAFDin this
subsection.
(15)
If we sum (15) overthe pointsin B, we get
W(xu, B) = (p - k)w(xj, xu) + W(xj, B).
(16)
If we sum ( 16) overthe pointsin A, we get
W(A, B) = (p
-
k)W(xj,A) + kW(xj,B).
(17)
Hence,
W(C)
=
W(A U B)
= W(A) + W(B) + W(A, B)
= W(A) + (p-k)W(xj,
A)
+ W(B) + kW(xj, B).
(18)
The optimizationgoal of 1D-MAFDis to maximize
W(C).Supposethatfor a givenk, we wantto choose
a setC maximizingW(C),butsubjectto theconstraint
thatA containsk pointsand B containsp - k points.
Equation(18) shows that underthis constraint,the
choice of B has no effect on the choice of A. For a
given subsetA 5 xi, . . , xjI such that IAI= k, define
fkj(A) = W(A) + (p
-
k)W(xj, A).
(19)
Let OPTkj be a k-element subset of lx1, . . ., xj4 that
maximizesfkj.Note that OPTp,,is the solutionto the
1D-MAFDprobleminstance.The dynamicprogramming algorithm computes all the values of
and then uses these values to find an
fkj(OPTkj),
optimalsolutionto the 1D-MAFDinstance.Thealgorithm finds the values of fk,j(OPTk,j) by considering
AND TAYI
RAVI, ROSENKRANTZ
increasingvalues of j. For a given value of j, the
algorithmconsidersincreasingvaluesof k.
Now considerhow to computefkj(OPTkj). Note
that OPTkj eitherincludesthe pointxj or excludesit.
if xi 4 OPTk,j, then OPTkj must be OPTk,j]1, for
whichwe have,using(19),
fkj(OPTk,j} I)
= W(OPTkj1l) + (p
k) W(x,,
-
OPTk,j11)
= W(OPTk,j-1) + (p - k) W(x,-1, OPTk,j
X)
+ k(p - k)(xj - xj- )
=
fk,j-,(OPTk,j-l) +
k(p - k)(xj - x-1).
(20)
If xj E OPTk,j,then OPTk,jmustbe OPTk1,,j1l Ufjxf,
and againusing(19) we have,
fk,j(OPTk-l,j-l
U fxj)
= W(OPTk-l,j1,)+ (p - k)(W(xj-1,OPTklj_1)
/ 307
Theorem 6. An optimalsolutionto any instanceof
1D-MAFD givenby a set V of n pointsandan integer
p < n can be obtainedin O(maxfn log n, pnj) time.
4.2. A Heuristic for 2D-MAFD
It is open whether2D-MAFDis NP-hard.Note that
GMA (Section3) providesa performanceguarantee
of 4 for 2D-MAFD.Here,we firstpresenta heuristic
for 2D-MAFDwhich providesa performanceguaranteeof 4(,I2 - 1) - 1.657and then show how this
heuristiccan be modifiedto obtainanotherheuristic
guarantee
whichprovidesan asymptoticperformance
of r/2 - 1.571.Theseheuristicsuse our polynomial
algorithmfor 1D-MAFD.
We assumethatan instanceof 2D-MAFDis given
by a set V = lVI, v2, . . ., v") of n points(whereeach
point vi is specifiedby a pair of coordinates(xi, yi))
andan integerp < n. The stepsof thisheuristic(called
PROJECTL4) are shown in Figure 4. The performance guaranteeprovidedby PROJECTL4 is indicatedin the followingtheorem.
+ (k - 1)(x -xj-x)) + W(xj-1,OPTklj_1)
Theorem 7. Let I be an instanceof 2D-MAFD. Let
? (k - )(xj - xj,)
the
denote,respectively,
OPT(I)and PROJECTA4(I)
solutionvaluesof an optimalplacementand thatproduced by PROJECTL4 for the instance I. Then
= W(OPTk-l,1j_) + (p
-
k + 1)W(Xj1,
OPTkjlj1l)
OPT(I)/PROJECTI4(I) < 4(V/
+ (k - l)(p - k + 1)(xj- x,-,)
=fk-,1j-l(OPTk-1,j-l)
+ (k - l1)(p - k + l1)(x, - xj-,).
(21)
Equations (20) and (21) show that given the
and the values of
sets OPTkjI and OPTklj-1,
we can comfk,j-l(OPTk,j-1) and fk-,j-l(OPTk-lj-l),
puteOPTkjandfkj(OPTkj). To completethedynamic
formulation,we notethatthe boundary
programming
conditions are OPToj= 0, foj(OPToj) = 0(1 j <n),
OPT1,I = Ixi1, and fi,I(OPT,,1) = 0.
The detailsof the algorithmareshownin Figure3.
The algorithmfirst computes (Step 4) the entries
of the arrayF (which correspondsto the function
f in the above dynamicprogrammingformulation)
and then uses these entriesto constructan optimal
placementP (Step 5). This implementationobviates
the need for the sets OPTkj used in the formulation.
The runningtime of the algorithmis O(n log n) for
sortingplus O(pn) to carryout the dynamicprogramming (thereare O(pn) entriesto compute,and each
entrycan be computedin constanttime). Thus, the
overallrunningtime is O(max(nlog n, pn)).
The abovediscussionis summarizedin the followingtheoremwhichwillbe usedin the nextsubsection.
-
1).
Proof. Recall that maximizingaveragedistance is
equivalentto maximizingthe sum of the distances.So
we will presentthe proofin termsof the sum of the
distancesbetweenpairsof points.
Let P* be an optimalplacement.For convenience,
we use the term edge to referto the line segment
between a pair of (distinct) points in P*. For an edge
e E P*, let l(e) denote its length (i.e., the Euclidean
distancebetween the end points of e). Note that
OPT(I) =
Zeep*
l(e).
Given
an
edge
e,
let
lx(e),ly(e),l,(e), and lw(e)denote the magnitudesof
the projectionsof e on the X, Y, V and W axes,
respectively.We havethe followingclaim.
Claim 3. lx(e)+ ly(e)+ lQ(e)+ lw(e), (1 + V2)l(e).
Proofof Claim3. Sincewe areconsideringthe sum of
the projections,assumewithoutloss of generalitythat
the angle0 of e with respectto the X axis is between
00
and 450, as shown in Figure 5. From that figure,we
have
lx(e)= l(e)cos0, ly(e)= l(e)sin 0,
lQ(e)= l(e)cos(45? - 0), and lw(e)= l(e)sin(45? - 0).
308 / RAVI, ROSENKRANTZAND TAYI
(*- - In the following, arrayF represents the function f in the formulation. - -*)
Step 1. Sort the given points, and let {x, x2,
x,} denote the points in increasing order.
Step 2. for j:= I to n do
F [OJ]
-- 0;
Step 3. F [1,1] - 0.
Step 4. (*- - Compute the value of an optimal placement forj =2 to n do
for k:= I to min (p,j) do
begin
1] + k(p - k)(xj- xj-);
F[k- I,j-I] + (k - I)(p - k + I)(xj-xj
tI - F[k,j-
t2
if
tl >
I);
then (*- - do not include xj -
t2,
F[k,J] <t-
else (*- - Includexj F[k,J]
<-
t2;
end;
Step 5. (*- - Construct an optimal placement P - {xx}; k - p; j -- n;
while k > I do
begin
if F[k,J] = F[k- I, j-1] + (k- 1)(p-k+
I)(xj-xj-1),
then (*- - xj to be included in optimal placement begin
P <- P u {xj;
k
-
k - I;
end;
j*-j-
I;
end;
Step 6. Output P.
Figure 3. Details of the algorithm for 1D-MAFD.
Step 1. Obtain the projections of the given set V of points on each of the four axes defined by the
equations y = 0 (X axis), y = X(Vaxis),
{=
0 (Yaxis), and y = -{(W~axis).
Step 2. Find optimal solutions to each of the four resulting instances of ID-MAFD.
Step 3. Return the placement corresponding to the best of the four solutions found in Step 2.
Figure 4. Details of heuristic PROJECT_4.
v
Y
W
____
_______
0
-
Figure 5. Proof of claim 1.
Let s(e) = l,(e) + ly(e) + l,(e) + lw(e). Using well
known trigonometric identities and the fact that sin
450 = cos 45? = I/N/, it is not difficult to verify that
s(e)
=
l(e)[sin 0 + (1 + ./2)cos 0].
(22)
From (22), it is easy to verify that the minimum value
of s(e) occurs when 6 = 00 or 0 = 450 and that this
minimum value is (1 + 2I5)I(e).This completes the
proof of Claim 3.
We now continue with the main proof. Note that
Claim 3 holds for each edge in P*. So if we sum up
RAVI, ROSENKRANTZ
AND TAYI
the resultof Claim3 overall the edgesin P*, we get
E Ix(e) + ly(e) + I(e) + lw(e)
eEP*
(1+ A)
+ V)OPT(I).
(23)
If we let Sx = Caps lx(e), and use analogousdefinitions for S, So, and SW,we get from (23),
Sx + Sy + S, + S, , (1 + V;2)OPT(I).
Fromthe lastinequality,it followsthat
(24)
+ N/2)
OPT(I).
max(Sx, Sy, SunS)
(25)
SincePROJECU4 choosesthe best placementfrom
optimalsolutionsforthe four1D-MAFDinstances,it
followsthat PROJECT14(I), max (Sx, Sy, Sv, Sw).
We thushavefrominequality(25),
OPT(I)
PROJECT.A(I)
(O < 0 < ir/k) be the angleof e with respectto the
X1axis.Let li(e) denotethe magnitudeof the projection of e on the Xi axis (1 < i < k), and let s(e) =
0
Xk=, li(e). It is easy to verify that
-k12
k12- 1-
I(e)
eEP*
-(1
/ 309
s(e) =I(e) [
cos(rr/k -0) + E cos(rir/k + 0)]. (27)
calculationsshowthat
Tedious,but straightforward,
s(e) = 1(e) [sin 0 + sin(r/k- 0)]
(28)
From(28),it is easyto verifythatthe minimumvalue
of s(e) occurswhen 0 = 0 or 0 = ir/k and that this
minimumvalue is l(e)/tan(r/2k). Thus s(e) ; I(e)/
tan(r/2k). The remainderof the proof to show that
OPT(I)/P,(I) < k tan(r/2k) is virtuallythe same as
of Theorem7 (keepingin mind that the numberof
axesis k insteadof 4).
We note that Theorem8 generalizesthe boundof
Theorem 7 in that choosing e = 4(V2 - 1) - ir/2
4/(1 + VJ) = 4(12
-
1)
as indicatedin the statementof the theorem.
The performance guarantee provided by
PROJECT4 is approximately1.657. By projecting
the given set of pointson more axes,it is possibleto
achieve an asymptotic performanceguaranteeof
ir/2 ~ 1.571, as shownbelow.
Theorem8. Let I be an instanceof 2D-MAFD and
let OPT(I) denote the solutionvalue of an optimal
placementfor L Then,for anyfixed e > 0 thereis a
polynomial-timeapproximationalgorithmP, which
producesa placementwith solutionvaluePE(I)such
that OPT(I)/PE(I) < (Xr/2+ e).
Proof. Givene > 0, let k be the smallestevenpositive
integersatisfyingthe condition
k tan(r/2k) s r/2 + E.
Such a k exists because limka
(26)
k tan(r/2k) =
r/2, and this limit is approachedfrom above.
AlgorithmP, firstprojectsthe givenset of pointson k
axes (denoted by X1, X2, .. . , Xk) such that the angle
betweenany pair of successiveaxes is ink. It then
solvesthe 1D-MAFDproblemon each of the k axes
andoutputsthe bestof the placementsfound.In view
of (26), Theorem 8 would follow by provingthat
OPT(I)/Pe(I)is bounded by k tan(r/2k).
The proofis verysimilarto that of Theorem7. Let
P* denotean optimalplacementandconsideran edge
e of lengthl(e) in P*. Withoutloss of generality,let
(~0.0861) leadsto projectionon k = 4 axes.Also, to
achievea performance
guaranteeof 1.571(anapproximate value of gr/2), 80 projectionsare needed;in
general,the number of projectionsgoes to oo, as
e approaches
0.
5. CONCLUSIONS
The results of this paper, prior results,and open
problemsaresummarizedin twotables.TableI shows
the complexityresultsfor solvingtheseproblemsoptimally, while Table II shows performanceguarantee
results for heuristics.In these tables, prior results
are indicated through appropriatecitations; our
resultsare indicatedby specifyingthe corresponding
theorems.
ACKNOWLEDGMENT
An extendedabstractcontainingsome of the results
in this paperappearsin the Proceedingsof the 1991
Workshop on Algorithms and Data Structures
(WADS-91), Ottawa, Canada, August 1991. The
researchof the first authorwas supportedby NSF
grantCCR-89-05296,and the researchof the second
authorwas supportedby NSF grantsCCR-88-03278
and CCR-90-06396.The researchof the thirdauthor
was supported by the Faculty Research Award
Programof the Universityat Albany.
We thankProfessorBinayK. Bhattacharya
(Simon
FraserUniversity,Burnaby,Canada)for bringingthe
paperby Wangand Kuo (1988)to our attention.We
310
/
AND TAYI
RAVI, ROSENKRANTZ
Table I
ComplexityResults for DispersionProblems
Problem
General
TriangleInequality
MAX-AVG
MAX-MIN
NP-hard
(Hansenand Moon 1988)
NP-hard
(Hansenand Moon 1988)
NP-hard
(Erkut1990)
NP-hard
(Erkut1990)
ID Version
0 (max Ipn, n log nJ)
0 (max Ipn, n log n )
2D Version
(Wangand Kuo 1988)
NP-hard
(Wangand Kuo 1988)
(Theorem6)
Open
Table II
PerformanceGuaranteeResults for NP-hardDispersionProblems
MAX-AVG
MAX-MIN
UpperBound
(BestKnown
Guarantee)
LowerBound
(IntrinsicLimit,
AssumingP # NP)
General
-
TriangleInequality
2
(Theorem2)
2
(Theorem2)
No guaranteedratio
(Theorem1)
2
(Theorem3)
Open
Problem
2D Version
areindebtedto the refereesfortheircarefulreadingof
our manuscriptandtheirvaluablecomments.In particular,one of the refereessuggestedpresentingthe
conclusionsin Section5 in the formof tables.
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