An improved approximation algorithm for the
metri unapaitated faility loation problem
Maxim Sviridenko
IBM T. J. Watson Researh Center, Yorktown Heights, P.O. Box 218, NY 10598,
USA;
svirius.ibm.om,
WWW home page:
http://www.researh.ibm.om/people/s/sviri/sviridenko.html
We design new approximation algorithm for the metri unapaitated faility loation problem. This algorithm is of LP rounding
type and is based on a rounding tehnique developed in [5{7℄.
Abstrat.
1
Introdution
In the unapaitated faility loation problem we are given a set of potential
faility loations F and a set of demand points D. Eah point i 2 F has an
assoiated opening ost fi . One faility point i has been opened it an provide
an unlimited amount of ertain ommodity. A lient j 2 D has a demand of
ommodity that must be shipped from one of the opened failities. We assume
that transportation osts of demand transferred from faility point i to demand
point j are proportional to the distane ij between them. We assume that is
a metri, i.e. ii = 0 for all i 2 F , ij = ji for all i; j 2 D [F and ij + jk ik
for all i; j; k 2 D[F . The goal is to determine a subset of potential faility points
that minimizes the sum of total opening osts of failities at these loations and
total servie osts of all demand points from opened failities.
This problem an be written as the following mixed integer program
XX
X
min
ij xij + fi yi ;
(1)
i2F j 2D
X
i2F
i2F
xij = 1; j 2 D;
xij yi ; i 2 F ; j 2 D;
xij 0; i 2 F ; j 2 D;
yi 2 f0; 1g; i 2 F :
(2)
(3)
(4)
(5)
The unapaitated faility loation problem (UFLP) (another popular name
is the simple plant loation problem) is one of the most studied and known
optimization problems. There are hundreds of papers on dierent aspets of this
problem (see the survey hapter of Cornuejols, Nemhauser & Wolsey in [31℄ and
2
short survey paper on polynomially solvable subases of the problem by Ageev &
Beresnev[3℄). In this paper we are mainly interested in approximation algorithms
with proven performane guarantees. We say that an algorithm has performane
guarantee 1 if it always delivers a solution within a fator of of optimum.
We also all suh an algorithm as -approximation algorithm.
The rst approximation algorithm for UFLP was built by Cornuejols, Fisher
& Nemhauser [16℄. They obtain (1 e 1 )-approximation algorithm for the maximization variant of UFLP. Reently, Ageev & Sviridenko [4℄ obtained an 0:828approximation algorithm for this problem. The rst approximation algorithm
for the minimization version of the general UFLP ( might not satisfy the triangle inequality in this ase) was built by Hohbaum [21℄. Her algorithm has
performane guarantee at most ln n + 1. Slightly better algorithm was reently
disovered by Young [33℄. And it is easy to prove by a redution from the set
over problem that the general UFLP is at least as hard to approximate as
the set over problem. On the other side Feige [17℄ proved that for any " > 0
there is no (1 ") ln n-approximation algorithm for the set over problem unless
NP 6 DT IME (nlog log n ).
The most interesting speial ase of the general UFLP is the metri UFLP.
In this problem distane funtion satises the triangle inequality. The rst
approximation algorithm for this problem was obtained by Shmoys, Tardos &
Aardal [32℄ using lustering and ltering tehniques due to Lin & Vitter [26℄.
The performane guarantee of their algorithm is 3.16. Guha & Khuller [18℄ obtain 2.408-approximation algorithm by ombining Shmoys, Tardos & Aardal's
algorithm with the greedy algorithm. They also proved that there is no an
approximation algorithm with performane guarantee better than 1:463 unless
NP 6 DT IME (nlog log n ) by the redution from the set over problem. Chudak
& Shmoys [12, 14℄ improved the Shmoys, Tardos&Aardal's algorithm by using
dependent randomized rounding and information about the dual linear program
to the linear programming relaxation of the problem (1)-(5). Their original performane guarantee [12℄ was 1 + 2e 1 1:736 but using some additional triks
they slightly improved it to 1:732 [14℄. Charikar & Guha [10℄ obtained 1:728approximation algorithm by ombining the primal-dual algorithm, the greedy
algorithm and the Chudak & Shmoys's algorithm. Jain, Mahdian and Saberi
[22℄ obtained 1.61-approximation algorithm using tehnique developed by Mahdian et al. [27℄. Reently, ombining few dierent approahes Mahdian, Ye &
Zhang [28℄ obtained 1.52-approximation algorithm.
There are also few very interesting approximation algorithm for the metri UFLP based on dierent ideas with worse performane guarantees: loal
searh algorithms was analyzed by Korupolu et al. [25℄, Arya et al. [9℄ and
Charikar & Guha[10℄, primal-dual algorithms was studied by Jain&Vazirani[23℄
and Charikar&Guha[10℄. Mettu & Plaxton [29℄ analyzed an algorithm very similar with Jain & Vazirani's one but their analysis doesn't use the primal-dual
tehnique. The advantage of suh algorithms is that all of them are ombinatorial and very fast. All these tehniques, developed originally for UFLP, was
applied to dierent generalizations of this problem like apaitated faility loa-
3
tion problem, multilevel faility loation problem and others [32, 23, 15, 13, 1, 25,
30℄.
Several authors onsidered the metri UFLP in speial metri spaes. Arora et
al. [8℄ obtain a polynomial time approximation sheme (PTAS) for the UFLP on
the plane, later Kolliopoulos&Rao [24℄ improved running time of this PTAS and
they also generalized the result to the Eulidean spaes of onstant dimension.
Ageev [2℄ obtain a PTAS for the UFLP on planar graphs for instanes of UFLP
satisfying some tehnial ondition.
The main result of this paper is an 1:582-approximation algorithm for the
metri UFLP. The result is based on few main ideas. We use lustering tehnique
developed in works [26, 32, 12, 14℄. We also use rounding tehnique developed
in [5{7℄ for the problems with ardinality onstraint, these tehnique allows
us to derandomize randomized algorithms without atually onstruting them.
For example, assume that you want to onstrut a randomized algorithm whih
delivers an approximate solution with some expeted value whih is an analytial
expression but you don't know how to do that. In some ases it is enough to know
the analytial expression for the expetation and have some onavity property
for this expression to onstrut a deterministi algorithm giving a solution of
value no more than expeted one.
2
Struture Lemma
Consider a linear programming relaxation of the problem (1)-(5) obtained by
relaxing the integrality onstraints (5) with the onstraints
yi 0; i 2 F :
(6)
Let (x ; y ) be an optimal solution of the linear programming relaxation and
let LP be the value of this solution. The following lemma is an extension of
Lemma 4 in [14℄.
Lemma 1. Suppose (x; y) is a feasible solution to the linear program (1)-(4),(6)
for a given instane of the unapaitated faility loation problem I and let 1
be some xed onstant. Then we an nd, in polynomial time, an equivalent
instane I~ and a feasible solution (~xij ); (~yi ) suh that both solutions have the
same frational servie and faility osts. The new instane I~ diers only by
replaing eah faility loation by at most dejDj + 1 opies of the same loation
and the new feasible solution (~xij ); (~yi ) has three additional properties:
1. x~ij = y~i for all i 2 F~ , j 2 D suh that x~ij > 0,
2. y~i 1=; i 2 F~,
3. For any demand point j there is a permutation j of faility points F~ and
faility point sj(j) suh that 1 j 2 j : : : nj (we omit supersript j
P (j )
from j ) and si=1
y~i = 1=.
Proof. We start with F~ = F and (~xij ); (~yi ) = (x ); (y ) then on eah iteraij
i
tion we will hange an instane of the problem by adding new opies for some
failities, deleting the old ones and hanging the urrent feasible solution.
4
1. Pik any faility i whih violate the rst property, i.e. there is a demand
point j suh that 0 < x~ij < y~i . Let j0 be a demand point with smallest value of
x~ij among all demand points with x~ij > 0. Instead of faility i reate two new
failities i1 and i2 with the same loation and opening ost and set y~i1 = x~ij0
and y~i2 = y~i x~ij0 . Then after that set x~i1 j = x~ij0 and x~i2 j = x~ij x~ij0 for
all demand points j suh that x~ij > 0. After that repeat the proess until the
solution satises property 1 of the lemma. During the proess, eah faility in
our original instane an be opied at most jDj 1 times.
2. The seond property is proved in a similar way. If y~i > 1= we dene a set
Si of P
at most de opies of faility i and orresponding variables y~k ; k 2 Si suh
that k2Si y~k = y~i and 0 y~k 1=, we also split the demand x~ij between
new failities for any demand point j with x~ij > 0.
3. For eah demand point j 2 D x any permutation j of the new set of
failities suh that 1 j 2 j : : : n j . Let k = s(j ) be the faility point
P (j )
P (j )
y~i > 1= then we
with smallest index s(j ) suh that si=1
y~i 1=. If si=1
dene two new failities k 0 and k 00 instead of faility k . The faility assignment
P (j )
y~i 1= and y~k0 = y~k y~k00 .
variables are dened as follows y~k00 = si=1
We also modify the permutation j : tj = tj for t = 1; : : : ; s(j ) 1, sj(j ) =
k0 ; sj(j)+1 = k00 and tj = tj 1 for t = s(j ) + 2; : : : ; n + 1. All demands assigned
to the faility k are splited now between two new failities k 0 and k 00 . All other
permutations w ; w 6= j are also modied aordingly. If k = uw then tw = tw
for t = 1; : : : ; u 1, uw = k 0 ; uw+1 = k 00 and tw = tw 1 for t = u + 2; : : : ; n + 1.
We repeat this proedure for eah demand point j .
So, after appliation of this lemma we have a new solution for the instane of the
problem where we allow to loate few idential failities at the same point. New
solution has the same ost and any integral solution produed by the rounding
proedure for the new instane an be easily transformed into a feasible solution
of original problem with the same or smaller value of objetive funtion (we just
delete redundant opies of the same faility from approximate solution).
3
Approximation algorithm: General idea
On the rst step of the algorithm we solve linear programming relaxation (1)(4),(6).
On the seond step we hoose a parameter 2 [1; +1) from some probability
distribution g () dened later (atually, g () will have nonzero value only when
2 [1; 1:71566℄). After that we onstrut a new instane (~x; y~) of the problem by
applying the algorithm from the Lemma 1 and dene a new solution yi = y~i , i 2
F~ whih is feasible by the seond property from the Lemma 1. For onveniene,
we redene F = F~ . New demand assignment variables xij are dened in an
obvious way by using third property from the Lemma 1: xij = yi for losest s(j )
failities and xij = 0 for all other failities, where sj(j ) is a faility dened in
the third property in the Lemma 1.
5
On the third step we dene a set of d nonoverlapping lusters K1 ; : : : ; Kd
(i.e. Kp \ Kq = ; for any p; q = 1; : : : ; d) where eah luster isPa set of failities.
Eah luster K = K1 ; : : : ; Kd satises the following equality i2K yi = 1 . The
exat desription of the lustering proedure will appear in the next setion.
On the fourth step for eah demand point j we dene a funtion Fj (z1 ; : : : ; zn ; )
of n = jFj variables zi 2 [0; 1℄ and variable 1. Note that the denition of
the funtions depends on a frational solution (x; y) and lustering dened on
the third step. These funtions
Psatisfy the following properties for any frational
solution z1 ; : : : ; zn suh that i2K zi = 1 for any luster K = K1 ; : : : ; Kd :
P
R
P
g()d 1:582LP f
y
1. 1+1
F
(
y
;
:
:
:
;
y
;
)
+
i
i
j
1
n
i2F
j 2D
2. for any luster K = K1 ; : : : ; Kd with at least one frational faility in it there
exist two frational failities p; q 2 K suh that the funtions 'j ("; p; q ) =
Fj (z1 ; : : : ; zp + "; : : : ; zq "; : : : ; zn; ) are onave funtions of variable " 2
[ minfzp ; 1 zq g; minf1 zp ; zq g℄
3. the funtions j ("; p) = Fj (z1 ; : : : ; zp + "; : : : ; zn ; ) are onave funtions of
variable " 2 [ zp ; 1 zp ℄ for any faility point p 2 F n [di=1 Ki
4. for any integral solution z1 ; : : : ; zn suh that there is at least one integrally
open faility in eah luster K1 ; : : : ; Kd the value of Fj (z1 ; : : : ; zn ; ) is an
upper bound on the distane from the demand point j to the losest open
faility in the solution z1 ; : : : ; zn
The rth step of approximation algorithm is the so-alled "pipage rounding".
Let z1 ; : : : ; zn be a urrent frational solution. In the beginning of the rounding
proedure zi = yi ; i 2 F and at the end z1 ; : : : ; zn is an approximate frational
solution with exatly one integrally open faility in eah luster
P K = K1 ; : : : ; Kd.
During the rounding we always maintain the property that i2K zi = 1 for any
luster K .
If there is a luster K with a frational faility inside then we hoose two
frational failities p and q from luster K suh that all funtions 'j ("; p; q )
are onave (Property 2). Let zi (") = zi if i 2 F n fp; q g, zp (") = zp + " and
zq (") = zq ". Let "1 = minfzp; 1 zq g and "2 = minf1 zp ; zq g. The funtion
X
X
G(") =
'j ("; p; q) + fi zi ()
j 2D
i2F
is a onave funtion of variable " on the interval [ "1 ; "2 ℄ and therefore it is
minimized when " = "1 or " = "2 , in partiular minfG( "1 ); G("2 )g G(0).
Let " 2 f "1 ; "2 g be the value of " minimizing the funtion G("). We dene
zi = zi (" ) and repeat the "pipage" step for the new solution z until all lusters
ontain a faility integrally assigned to it. Note that on eah pipage step the
number of frational variables is dereasing.
The sixth and the last step is the nal rounding. In the end of the previous
pipage step we had a frational
solution suh that eah luster K ontains a
P
faility i with zi = 1 and i2K zi = 1. We apply a simpler rounding proedure
sine we do not have frational failities inside of lusters.
6
We hoose any frational faility p. Let zi (") = zi if i 2 F n fpg and zp (") =
zp + ". Let "1 = zp and "2 = 1 zp . The funtion
G(") =
X
j 2D
j ("; p) +
X
i2F
fi zi ()
is onave (Property 3) and therefore it is minimized on the endpoints of the
interval [ "1 ; "2 ℄. Let " 2 f "1 ; "2 g be the value of " minimizing the funtion
G("). We dene zi = zi (" ) for i 2 F and repeat rounding step again.
In the end of the rounding proedure we obtain an integral solution z1 ; : : : ; zn
suh that
X
X
X
X
Fj (z1 ; : : : ; zn ; ) + fi zi Fj (y1 ; : : : ; yn ; ) + fi yi :
j 2D
i2F
j 2D
i2F
Moreover the Property 4 guarantees that Fj (z1 ; : : : ; zn ; ) is an upper bound on
the distane from the demand point j to the losest open faility in the solution
z1 ; : : : ; zn and therefore by the Property 1 the solution z1 ; : : : ; zn is an 1.582approximation solution for the metri unapaitated faility loation problem.
4
Approximation algorithm: Details
In this setion we desribe the lustering proedure and give the denition for
funtions Fj then in the next few setions we show that Fj satisfy properties
(1)-(4).
Let N (j ) = fi 2 Fjxij > 0g, i.e. N (j ) is a set of failities frationally
serving j in the modied frational
solution, let Rj () = maxi2N (j ) ij be a
P
radius of N (j ). Let Cj () = i2N (j ) ij xij be a servie ost of demand point
j in the modied frational solution (x; y). Notie that Cj () is P
a noninreasing
funtion of and in general Cj () an be smaller then Cj (1) = i2N 1 (j ) ij xij .
Let D0 = ; and let j1 2 D be a demand point suh that Rj1 () + Cj1 () =
minj 2D (Rj () + Cj ()). Add j1 to D0 . Delete j1 and all demand points j suh
that N (j ) \ N (j1 ) 6= ; from D (we will all suh demand points as demand
points assigned to luster enter j1 ). Delete the set of faility points N (j1 )
from F and repeat the lustering proedure until D is empty. This lustering
proedure denes a set of lusters K1 = N (j1 ); : : : ; Kd = N (jd ) and a set of
luster enters D0 = fj1 ; : : : ; jd g. Notie that by onstrution N (j ) \ N (j 0 ) = ;
for any two dierent
luster enters j; j 0 2 D0 and by the third property from
P
the Lemma 1 i2K yi = 1 for eah luster K = K1 ; : : : ; Kd .
We now dene funtions Fj (z1 ; : : : ; zn ; ) for all demand points j . Variable
zi always represents the fration of faility assigned to faility loation i 2 F .
Originally, i.e. in the moment when we dene funtions Fj zi = yi = y~i .
Assume that demand point j was assigned to some luster enter jk . For eah
demand point j let j be the permutation on the set of all faility loation
points F suh that 1 j 2 j : : : n j (we omit supersript j from j )
7
1
and Nj1 = [m
k=1 fk g where m = jNj j. The funtion Fj (z1 ; : : : ; zn ; ) is dened
1
as follows: if Njk Nj then Fj (z1 ; : : : ; zn ; ) =
!
m
mY1
Y
(1 zk ) zm + Rj (1) (1 zk );
1 j z1 + 2 j (1 z1 )z2 + : : : + m j
k=1
k=1
if Njk n Nj1 6= ; then Fj (z1 ; : : : ; zn ; ) =
1 j z1 + 2 j (1 z1 )z2 + : : : + m j
m
Y
k=1
(1
mY1
k=1
!
(1
zk ) zm +
P
!
i2Njk nNj1 ijk zi
;
zk ) jjk + P
i2Njk nNj1 zi
(7)
where jjk is a distane between demand point j and luster enter jk .
The intuition behind this denitions is that if Njk n Nj1 6= ; then Fj is an
upper bound for the expetation of the distane from demand point j to the
losest open faility in the following random proess. Open a faility at the
faility point i 2 Nj1 at random independently with probability zi , if we didn't
open a faility in Nj1 during this proess then open one failityPfrom Njk n Nj1
at random suh that faility i is hosen with probability zi = i2Nj nNj1 zi . If
k
Njk Nj1 then notiing that m j = Rj (1) we an dene the similar random
proess. Open a faility at the faility point k at random independently with
probability zk for k = 1; : : : ; m 1, if this random proess didn't open any
faility we open a faility at the point m with probability 1.
The reason we use a deterministi rounding instead of randomized one is
that we don't know how to onstrut a random proess whih guarantee the
above expetations for all demand points. The randomized rounding due to Chudak&Shmoys[12, 14℄ gives similar expetations but they are a little bit dierent.
Instead of distane from demand point j to the faility loation point k they
have some average distane from j to the set of failities whih belong to the
same luster (see [12, 14℄ for details).
5
Proof of properties
To prove that our approximation algorithm has performane guarantee 1.582 we
need to prove four properties for funtions Fj .
Theorem 1. Funtion Fj (z ; : : : ; zn; ) satises the Property 2.
1
Proof. If there is a luster k with some frational failities in it then we hose
two failities p; q to be frational failities in the luster K = Njk with longest
distanes to the enter jk , i.e. pjk and qjk are the biggest among all distanes
between jk and other failities in Njk . We prove that for this hoie of p; q
8
the seond derivative of 'j ("; p; q ) is always negative for " 2 [ minfzp ; 1
zq g; minf1 zp; zq g℄ and therefore 'j ("; p; q) is a onave funtion on this interval
for every j 2 F . See next setion for details.
Property 3 follows from the fat that funtions j ("; p) are linear in " for
any index p 2 Nj1 . We now prove the Property 4. Let q be a losest to j opened
faility. We know that q 2 Nj1 [ Njk sine there is always an opened faility in
Njk . If q 2 Nj1 then zp = 0 for all failities p suh that 1 (p) < 1 (q), therefore
there is only one non-zero term in the expression for Fj (z1 ; : : : ; zn ; ) and this
term is equal to qj . If q 2 Njk n Nj1 then Fj (z1 ; : : : ; zn ; ) = jjk + qjk qj by
the triangle inequality.
To prove Property 1 we dene the density funtion g (x) as follows
Kex
1 x B;
g(x) = 0x; 3 ; ifotherwise,
where e is the base of the natural logarithm, B = 1:71566 and
ZB x ! 1
e
K=
dx
0:839268
3
1
x
is a onstant normalizing g (x).
Theorem 2.
0
1
Z +1 X
X
Fj (y1 ; : : : ; yn ; ) +
fi yi A g()d 1
j 2D
i2F
(
ZB
B (e =B e )
g()d;
max K +
B 1
1
ZB
ZB
B (1 e =B )g()d +
e g()d;
1
1
) ZB
!
ZB
g()d +
e g()d LP = LP 1
1
Proof. The proof of this theorem is rather ompliated from tehnial viewpoint,
so we put it into the Setions 7 and 8. It onsists of many auxiliary lemmas on
properties of Fj and the proof that g (x) is a solution of ertain integral equation
of Volterra type.
We ompute eah quantity in the above expression and the performane
guarantee of our algorithm numerially by Mathematia
maxf1:30404; 1:25777; 1:30404g + 0:27707 = 1:58111:
9
Note that our algorithm an be simply derandomized using the disretizing of
the probability spae and hoosing the best parameter from the disrete probability spae. The simplest way to do it is to split interval [1; B ℄ into suÆiently
many equally spaed intervals and then run the algorithm on eah of the endpoints of subintervals. Using this approah we are loosing small onstant " in
our performane guarantee whih an be made arbitrary small by dereasing the
length of subintervals.
6
Property 2
Proof of theorem 1 Assume that we have the luster K = K1; : : : ; Kd with at
least one frational faility
P(then we must have at least two frational failities
due to the property that i2K zi = 1) and enter jk . Choose two frational failities p; q 2 K with biggest distanes pjk and qjk among all frational failities
in K .
Fix j 2 F . We onsider now few ases. Assume that j was assigned to the
luster enter jk . If Njk Nj1 then it is easy to see that the funtion 'j ("; p; q ) is
a quadrati polynomial in ". Moreover we will prove now that it has nonpositive
main oeÆient and therefore the funtion 'j ("; p; q ) is onave in this ase.
W.l.o.g. we assume that faility p is loser to j then faility q , i.e. 1 (p) <
1
(q), therefore we an expliitly write the main oeÆient in quadrati polynomial 'j ("; p; q ) as
0
1
Y
qj (1 zk )A "2
k2f1;:::;
m
X
t=
1 (q )+1
0
t j zt
0
Rj (1) 1 (q ) 1
k2f1;:::;t
1 (q )+1
gnf
g
1 (p); 1 (q )
0
zt
g
1 (p); 1 (q )
(1
Y
k2f1;:::;
t=
1
Y
0
qj "2 m
X
g
1 (p)
Y
k2f1;:::;mgnf
0
1
gnf
1 (q ) 1
tY1
k=
(1
1 (q )+1
gnf
g
1 (p)
1
zk )A
(1
1
zk )A "2
1
zk )A "2 (1
1
zk )A m
Y
k=
1 (q )+1
(1
1
zk )A = 0:
If Njk n Nj1 6= ; then if both p; q 2 Njk n Nj1 then 'j ("; p; q ) is a linear funtion
sine only the last term in the expression (7) is nononstant and +" and " are
anelled in denominator of
P
i2N nN 1 ijk zi (")
P jk j
:
i2Nj nNj1 zi (")
k
10
If both p; q 2 Nj1 then we laim again that 'j ("; p; q ) is a quadrati polynomial in " with nonpositive main oeÆient sine zi (") = zi for i 2 Njk n Nj1 . The
proof is very similar to the ase when Njk Nj1 . We just should use the fat
that
P
P
i2Njk nNj1 (ijk + jjk )zi
i2Njk nNj1 ijk zi (")
P
=
jjk + P
i2Njk nNj1 zi (")
i2Njk nNj1 zi
P
i2N nN 1 ij zi
P jk j
i2Njk nNj1 zi
m j :
The last inequality follows from the fat that ij m j for any i 2 Njk n Nj1 .
Note that this ase overs the ase when j was not assigned to the luster enter
jk but p; q 2 Nj1 if Nj1 ontains only p or only q and j was not assigned to the
luster enter jk then 'j ("; p; q ) is a linear funtion in ".
We now onsider the most diÆult ase when q 2 Nj1 \ Njk and p 2 Njk n Nj1 .
In this ase the nonlinear term of 'j ("; p; q ) is
P
!
m
Y
i2Njk nNj1 ijk zi (")
P
:
(8)
(1 zk ("))
i2Nj nNj1 zi (")
k=1
k
If number of remaining frational failities in the set Njk
then we assume that
P
i2N nN 1 ijk zi (")
P jk j
= pjk :
i2Nj nNj1 zi (")
n Nj
1
is exatly one
k
The reason we stress it out is that we an have a division by zero during rounding
if we would treat the above expression as funtion of ". Therefore 'j ("; p; q ) is a
linear funtion in this ase.
So without loss of generality we onsider the ase when there are at least two
1
frational
Q failities in the set Njk nNj . Dividing (8)
Q by the positive onstant fator
pjk k2f1;:::;mgnf 1 (q)g (1 zk (")) = pjk k2f1;:::;mgnf 1 (q)g (1 zk ) we
obtain that nonlinear term (8) is proportional to
P
" + i2Nj nNj1 ijk zi =pjk !
Pk
:
(1 zq + ")
" + i2Nj nNj1 zi
k
P
Let A = 1 zq , B = i2Nj nNj1 ijk zi =pjk and C = i2Nj nNj1 zi . We want
k
k
to prove onavity of the funtion (A + ")(B + ")=(C + "). Note that C + " > 0
sine " 2 [ minfzp ; 1 zq g; minf1 zp ; zq g℄ and there are at least two frational
failities in the set Njk n Nj1 . The rst derivative of this funtion is
P
A+" B+"
+
C +" C +"
(A + ")(B + ")
(C + ")2
11
and the seond derivative is equal to
2(A + ")(B + ")
2
+
C +"
(C + ")3
2(2" + A + B )
(C + ")2
We laim that
(C + ")2 + (A + ")(B + ") (2" + A + B )(C + "):
And therefore the seond derivative is nonpositive when " 2 [ minfzp ; 1
zq g; minf1 zp; zq g℄. Indeed, the above inequality is equivalent to the inequality
C 2 + AB AC + BC , whih is equivalent to B (A C ) C (A C ). We prove
it by proving two inequalities A C and B C .
P
The inequality A C is equivalent to 1 zq i2Nj nNj1 zi whih holds
k
P
sine q 2 Njk \ Nj1 and 1 = i2Nj zi .
k
P
P
The inequality B C is equivalent to i2Nj nNj1 ijk zi pjk ( i2Nj nNj1 zi )
k
k
whih holds sine we hose p; q to be furthest faility points from the luster
enter jk . Therefore the seond derivative of 'j ("; p; q ) is negative when " 2
[ minfzp ; 1 zq g; minf1 zp ; zq g℄, this implies the statement of the lemma.
7
Property 1: tehnial lemmas
We now prove few tehnial lemmas on the auxiliary properties of Fj . Let rj (t) =
Rj (1=t) and reall that Rj (1=t) is a smallest number suh that a ball of this radius
around demand point j ontains failities with total sum of faility assignment
variables of least t. The rst tehnial lemma establishes the onnetion between
rj (t); Rj () and Cj ().
Lemma 2.
Cj () = Z 1=
Z +1
Rj ( )
rj (t)dt = d
2
0
Proof. Shmoys, Tardos and Aardal (Lemma 10, [32℄) proved this statement for
= 1, our proof is very similar. Reall that 1 j : : : Pmj . TheP
funtion rj (t)
is a step funtion and it is equal to k j for every t 2 ( sk=11 ys ; ks=1 ys ℄ (we
P
P
use the fat that yi = xij ). Let q be an index suh that 1= 2 ( sq=11 ys ; qs=1 ys ℄.
Hene,
! q 1
q 1
X
X
xs j + s j xs j =
Cj () = q j 1
q j
1
q 1
X
s=1
x
s j
!
+
s=1
q 1
X
s=1
s=1
!
s j xs j = Z 1=
0
rj (t)dt:
The seond equality is trivial sine we just should hange variable t into variable
= 1=t.
12
The next lemma is basially a reformulation of some property of expetation
used in [12, 14℄.
Lemma 3. In the ase when Njk n Nj 6= ; for some luster enter jk and some
1
demand point j assigned to it, the following inequality holds
P
i2N nN 1 ijk zi
Rj (1) + Rj () + Cj ():
jjk + P jk j
i2Njk nNj1 zi
Proof. If there is a faility q 2 Nj1 \ Njk suh that qjk Cjk () then by triangle
inequality jjk jq + qjk and sine jq Rj (1) and ijk Rjk () we obtain
P
i2N nN 1 ijk zi
jjk + P jk j
Rj (1) + Cjk () + Rjk () Rj (1) + Cj () + Rj ()
i2Nj nNj1 zi
k
where the last inequality follows from the denition of luster enters.
If qjk > Cjk () for all failities q 2 Nj1 \ Njk then
P
P
i2N nN 1 ijk zi
P jk j
i2Njk nNj1 zi
i2N ijk zi
P jk
= Cjk ():
i2Njk zi
Using the fat that by triangle inequality jjk
statement of the lemma.
Rj (1) + Rjk () we obtain the
Lemma 3 implies that
Fj (y1 ; : : : ; yn ; ) 1 j y1 + 2 j (1 y1 )y2 + : : : +
m j
Lemma 4.
mY1
k=1
(1
!
m
Y
yk ) ym + (1
k=1
yk ) (Rj (1) + Rj () + Cj ()) :
1 j y1 + 2 j (1 y1 )y2 + : : : + m j
Rj (1)
m
Y
k=1
(1
yk ) Z
1
0
mY1
k=1
(1
!
yk ) ym +
e t rj (t)dt + Rj (1)e Proof. Reall that the way we modied the solution (x ; y ) in the Setion 2
that yi = yi = xij . Let Y1 = y1 ; Y2 = (1 y1 )y2 ; : : : ; Ym =
guarantees
Qm
Qm 1
y1 ; Z2 =
k=1 (1 yk ) ym and Ym+1 = k=1 (1 yk ). Let Z1 = 1 e
P
P
m 1 m e y1 e (y1 +y2 ) ; : : : ; Zm = e k=1 yk e k=1 yk and Zm+1 = e :
We laim that these two sequenes of numbers satisfy the following properties:
13
P +1
Pm+1
P1. m
Z =1,
i
=1 Yi =
P i=1 i
P
P2. ki=1 Yi ki=1 Zi ; k = 1; : : : ; m.
The rst
follows diretly from the denition of sequenes and the
P property
= 1. We now derive the seond property:
fat that m
y
k=1 k
k
X
i=1
Yi = 1
k
Y
(1
i=1
yi ) 1 e Pk
i=1 yi
=
k
X
i=1
Zi
Pk
Q
i=1 pi for 0 where we applied the well-known inequality ki=1 (1 pi ) e
pi 1.
R
Sine rj (t) is a step funtion and ab e t dt = e t jba we obtain
Z1
e t rj (t)dt = 1 j Z1 + 2 j Z2 + : : : + mj Zm :
0
Therefore, what we really need to prove is the inequality
1 j Y1 + 2 j Y2 + : : : + m j Ym + Rj (1)Ym+1 1 j Z1 + 2 j Z2 + : : : + m j Zm + Rj (1)Zm+1 :
(9)
We laim that the inequality (9) holds for any sequenes of positive numbers Y
and Z satisfying properties P 1; P 2 and any sequene 0 1 j 2 j : : : m j Rj (1), the similar inequality was proved by Hardy, Littlewood and Polya
[20℄.
Lemma 5. For any real number B 1 the following inequality holds
Z
0
1
e t rj (t)dt B (1 e =B )
Z 1=B
0
rj (t)dt+
B
B
1
(e =B e )
Z
1
1
=B
rj (t)dt
Proof. The proof is a diret appliation of the following integral inequality. We
are given two monotone funtions g (x) and r(x) on the interval (a; b). If funtion
g(x) is noninreasing and funtion r(x) is nondereasing then
R b
R b
Zb
g
(
x
)
dx
r
(
x
)
dx
a
a
:
(10)
g(x)r(x)dx a
b a
The above inequality is a speial ase of the Chebyhev Inequality (see inequality
236 for an integral version or Theorem 43 for the nite version of this inequality
in [20℄). Note also that Chebyhev Inequality we mentioned above is dierent
from the well-known probabilisti inequality with the same name.
Lemma 5 using the inequality (10), we just split the integral
R 1 To prove
t
e rj (t)dt into two integrals, one on the interval [0; 1=B ℄ and another on
0
the interval [1=B; 1℄ and apply the inequality (10) to eah of these integrals.
14
Lemma 6 (Chudak&Shmoys[14℄).
X
j 2D
Rj (1) LP We reformulated this lemma using our notations. The proof uses the onnetion
between Rj (1) and dual variable orresponding to the demand point j in the dual
LP to the linear relaxation (1)-(4),(6). We omit the proof from this version of the
paper. The following lemma an be heked by a straightforward omputation.
Lemma 7. The funtion f (x) = xex is a solution of the following integral equa3
tion
x f (x) +
xe
2
Zx
1
te t f (t)dt = 1:
From above lemmas we derive the following upper bound on Fj (y1 ; : : : ; yn ; )
for any j 2 D:
Theorem 3.
Fj (y1 ; : : : ; yn; ) B (1 e =B )
B
B 1
8
(e =B
e )
ZB
Rj ( )
2
1
Z +1
Rj ( )
2
B
d +
d + e Rj (1) + Rj () + Z +1
Rj ( )
2
d
Property 1: proof of the Theorem 2
We want to estimate above the following quantity
0
1
Z +1 X
X
Fj (y1 ; : : : ; yn ; ) +
fi yi A g()d
j 2D
1
i2F
using optimal value of linear relaxation LP .
By Theorem 3
Z
1
+
Z +1
1
1
=B )
Fj (y1 ; : : : ; yn ; )g()d Z +1
Rj ( )
B
=B
)
ZB
Rj ( )
d +
d
(e
e
2
B 1
2
1
Z +1
Rj ( )
+e Rj () + d
+
R
(1)
g()d = :
j
2
B (1 e
B
We will simplify this expression using Fubini's Theorem (swithing the order of
integration)
Z +1
Z +1
Rj ( )
=B )g ()d d +
=
B
(1
e
2
B
1
15
Z +1
ZB
B
Rj ( )
1
(e =B
e )g()d d ++
Z +1
e Rj ()g()d+
B
1
1
1
Z Z +1
Z +1
Rj ( )
e g()d =
e g()d d + Rj (1)
2
1
1
1
Renaming variable to in the third integral we ontinue
Z +1
Z +1
Rj ( )
e g()d
F ( )d + Rj (1)
=
2
1
1
2
where
F ( ) =
Z +1
B
B 1
1
(e =B
if 1 B and
Z +1
B (1
F ( ) =
e )g()d + 2 e g( ) +
e =B )g()d + 2 e g( ) +
1
Z
1
Z
1
e g()d;
e g()d;
if > B: Using Lemma 7 and the fat that g ( ) = 0 if > B we simplify the
expressions for F ( ):
(R B
B (e =B e )g ()d + K;
if 1 B;
F ( ) = R1B B 1
RB
=B
B (1 e
)g ()d + 1 e g ()d; if > B:
1
Therefore, F ( ) is a step funtion and
(Z B
ZB
F ( ) max
B (1 e =B )g()d +
e g()d;
1
ZB
1
B
=B
)g ()d + K
)
(e
e
:
1
We are ready now to prove Theorem 2
0
1
Z +1 X
X
Fj (y1 ; : : : ; yn ; ) +
fi yi A g()d 1
B
1
j 2D
Z
+1
X
Rj ( )
j 2D
1
2
F ( )d + Rj (1)
i2F
Z +1
1
Z +1
1
e
g ()d +
g()d
X
i2F
fi yi we ontinue applying Lemmata 2 and 6
Z +1
Z +1
X
X
max F ( )
Cj (1) +
e g()d LP +
g()d fi yi 1
1
1
j 2D
i2F
Z +1
Z +1
max max F ( );
g()d +
e g()d LP :
1
1
1
16
9
Conlusion
In this paper we obtained an approximation algorithm for the metri unapaitated faility loation problem using new rounding tehnique and many new
tehnial lemmas. The most interesting new tehnial lemma is the Lemma 4
whih is a renement of Lemma 17 from [14℄. The rounding tehnique we used
in this paper is rather general and an be implemented in the similar way for
other more general faility loation problems like fault tolerant faility loation
[19℄ or faility loation with outliers [11℄.
The performane guarantee in this paper an be easily improved by using
more numerial methods. Instead of using Lemma 5 we ould hoose the density
funtion g (x) as saled solution of the following integral equation
ZB
Zx
t=x
2
x
te f (t)dt + x e f (x) + te t f (t)dt = 1:
1
1
Unfortunately, this equation does not seem to have an analyti solution, it an
be solved by disretizing an interval [1; B ℄ and solving a orresponding system
of linear algebrai equations. Notie that numerial omputations show that for
big values of onstant B the funtion g (x) beomes negative and therefore we
should be areful in hoosing a value for B . Sine all our numerial experiments
gave a onstant worse then reently announed 1.52 [28℄ we deided to omit
the desription of more exat estimation of the performane guarantee of our
algorithm.
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