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7
LARGE SCALE GEOMETRIC LOCATION PROBLEMS II:
STOCHASTIC LIMIT PROPERTIES
by
Mordecai Haimovich*
OR 131-85
*
Revised October 1985
University of Chicago, Graduate School of Business
Abstract
Given a set of N points (customers) in the plane, the K-median problem
is concerned with finding a set of K points (centers) so as to minimize the
sum of distances between customers and their closest center. Suppose that the
customers are independent random samples from a distribution with bounded support. This paper shows that as N and K tend to infinity while K/N tends
to a constant a
(possibly zero), the value functional is almost surely
proportional to NK-1 /2 with a proportionality constant a1/2(ac). It also
(i) shows that,as a function of a, (ca) is convex; and (ii) provides an
approximation of 6(a) in the neighborhood of a = 0 and a = 1. As a
approaches 0, the problem coincides asympotically with an underlying
deterministic continuous problem.
outline
1.
Introduction
2.
Asymptotic Continuity (K/N + 0)
3.
Asymptotic Separability (K/N - a > 0)
4.
Conclusions
5.
References
Acknowledgements:
This research was supported by Grants ECS-7926625 and ECS-8316224 from the
Systems Theory and Operations Research Division of the National Science Foundation.
The author is indebted to Professor Thomas L. Magnanti for his encouragement and for many helpful suggestions.
__
-_
_
1.
INTRODUCTION
Let
X = {Xl,X2 ,...xN}
be a set of points in the plane, frequently re-
ferred to as customers or demand points.
The K-median problem is the
problem of finding another set of points
C = {Cl,C 2 ,...,CK},
referred to
as centers (or medians), that minimizes the total sum of distances between
customers and their closest center.
Let
DK(X) = Min ( Z
C
x£X
min
cC
x - c
)
(.)
be the optimal value (or cost) functional of the problem.
two variants of the problem:
We will consider
the restricted version in which centers can
be located only at customer locations (i.e.,
which proposes no such constraint.
CcX) and the free version
Since many of the results and derivations
in this work are not version specific, we will use the same notation for both
versions.
Moreover, our assertions and derivations apply to both, unless
explicitly indicated otherwise.
While simple to state, the K-median problem seems rather hard to solve,
especially for large values of
N
and
K.
In fact, Papadimitriou [Pa]
showed that the problem is NP-complete (in the terminology of contemporary
computational complexity theory [GJI), which while short of proof, provides
strong evidence of the problem's (asymptotic) computational intractibility.
This fact gives rise to research into the design and analysis of computationally
efficient approximation algorithms (heuristics).
While to the best of our
knowledge no heuristics with a guaranteed percentage error (i.e.,
-optimal) are
known for this problem, it is still likely that in practice certain heuristics
This deviates from our notations in [Ha] where D
is defined on measures and
according to which we would have DK( E 6 )rather than D (X), where
x
is
the unit mass (measure) at
x.
X
K
x
_^___^W___nllllllla
-2-
will perform well in most situations, or on the average.
To check such
assertions, one can use empirical tests, or instead use the standard
mathematical tool concerning this type of assertions, i.e., probability
theory.
The prototype of the probabilistic approach is Karp's analysis of
partitioning algorithms for the Travelling Salesman problem [Ka], which
is based on the
asymptotic order of the optimal value for problems
defined on N cities that are chosen independently aid randomly from a fixed
planar region.
This asymptotic growth rate which implies (and is in turn
implied by) the asymptotic optimality of partitioning heuristics, was first
established by Beardwood Halton and Hammersley
BHH].
NK-1
development, Fisher and Hochbaum [FH] established an
order for the K-median problem for
N
In a subsequent
/2
asymptotic
customers that again are independent
random points of a fixed planar region.
Their analysis supports the asymptotic
optimality of a certain aggregation heuristic when K grows no faster than log N.
The present study was motivated by the desire to sharpen the results
in [FH] as well as to extend the range of the analysis to encompass arbitrary (relative) growth rates of K.
We show that the asymptotic be-
havior of the problem manifests two different convergence modes.
The
first one, associated with what we call asymptotic continuity, occurs whenever K/N tends to 0 as N tends to infinity, i.e.,K = o(N).
In this case,
we show that the problem almost surely converges (in terms of value and the
optimality of solutions) to an underlying continuous deterministic problem,
and if K itself tends to infinity, then the problem almost surely follows
the associated asymptotic behavior of the underlying deterministic continuous problem, as analyzed in [Hati
and [HM1].
Accordingly, the optimal
I_-I_
YC
CII-·-
-I-
_
_
I
-3-
2
value almost surely tends2 to
('2/3
3/2
-1/2
(6)(f
d>)
NK/
where
(6)0O.377 is
the expected distance between a random point of a unit area regular
hexagon and its center, and
f
is the density of the absolutely continuous
part of the common probability distribution of the customers.
Moreover,
one can use asymptotic approximations for the solution of the underlying
deterministic problem based on the asymptotic formula for aggregate center
allocation as given in [Ha]
as indicated in [HM1].
and the asymptotic hexagonal partitioning property
Somewhat weaker results of this nature have been inde-
pendently established by Papadimitriou [Pa] whose analysis is for the uniform
distribution and requires the stronger assumption K = o(N/logN).
Zemel [Z]
has extended Papadimitriou's analysis to the K-center problem which is obtained by replacing the operator
Ez X
in (1.1) by
Max x.
The second convergence mode, associated with what we call asymptotic
separability, occurs whenever K/N tends to a constant a > 0 as N tends to
infinity.
In this case we show that for the uniform distribution on the unit
square the optimal value almost surely tends 2 to
function of a alone.
We also establish bounds for
close approximations of it when
(a)V where
B(a),
(a) is a
as well as very
is in the neighborhood of 0 or of 1.
This
convergence result implies (and is implied by) the asymptotic optimality
of partitioning procedures of the type employed by Karp [Ka].
While these
results are similar in nature to those of Beardwood et al. [BHH] for the
TSP and other problems as discussed in Subsection 3.5, the techniques in
[BHH]
are not adequate for the K-median problem.
Independently, Steele [SI] and
Hochbaum and Steele [HS] have obtained results on the free and restricted
2More precisely, the ratio between the optimal value and the given expression tends to 1.
-4versions of the K-median problem.
Our analysis provides a sharper rate of
convergence for the deviation probabilities (that is, for any given
the probability that the ratio between the optimal value ad
from
(ac) by more than
VI
> 0
differs
), which, incidently, answers a question posed
by Hochbaum 3 about a stronger mode of convergence.
We also evaluate the
limiting behavior of the asymptotic constant
as
1.
3(a)
a
approaches 0 or
In addition, our approach has the potential of being extended to non-
uniform distributions.
We discuss and derive asymptotic continuity in Section 2 and asymptotic
separability in Section 3.
In contrast to the expository style in [Ha],
which considered the K-median problem through a set of formal properties
it satisfies,
the treatment here, though generalizable to other situations,
is far more concrete.
The probabilistic apparatus we use is rather elementary:
Tchebychev's
inequality as applied to second or fourth order moments coupled with the
Borel-Cantelli's Lemma, in a manner not dissimilar to the standard textbook
derivation of the strong law of large number for independent identically
distributed random variables with finite fourth moments.
these rudimentary tools can be quite powerful.
3
Private communication.
As we will see,
-52.
Asymptotic Continuity (K/N
o)
Large scale K-median problems where the number of customers Der facility is very high, i.e.,
K = o(N)
can be approximated by aggregation that
aggregates/unifies customers that are close together, or by "continuation,"
i.e., the use of a smooth continuous distribution to represent the essentially discrete demand.
While Fisher and Hochbaum
approach, we will use the second.
FH] take the first
This section is devoted to establishing
the asymptotic accuracy of the continuous deterministic approximation (and
thus also the asymptotic optimality of heuristics based on it)
large-scale discrete problems with
2.1
for random
K = o(N).
Results
Let
square
p
2
S = [0, 11
and let
f
of independent p-distributed random points.
set of the first
TcR
N
points.
X (T) = 1 if
,
with support in the unit
be the density function of its absolutely
continuous part which we assume is not null.
for all
R2
be a Borel probability measure on
Let
Xi £
6X
2
T
Let
Let
X 1, X2,... be a sequence
XN = {X1,XX2 ...
N
be the point unit mass at
6X (T) = 0
and
} be the
X
1'
otherwise).
(i.e.,
Follow-
ing [Ha], we will define the value function for a finite-positive regular
Borel measure
w
with a bounded support as
DK()
where
H mi
ICI is the cardinality of
min
x-c
dw
C; the restricted version restricts
C
to
N
N
Consequently, DK(X )=DK( E 6X ) (which is an abuse of
i
i=l
N
1
notation). Now, PN
N
E
X
is the empirical probability measure
i=l
i
determined by the first N points of the sequence. By linearity of
the support of
w.
integration we have
-6-
DK(PN) =
DK(XN)
PN
As is well-known (see [Ch] for example),
almost surely (a.s.) as
g: Rd
R2
-
R, f gdp n
+
N
o (i.e., for every bounded continous function
-
Weak convergence of Borel measures on
f gdp a.s.).
is in turn known (see [Du]) to imply uniform weak convergence relative
to any uniformly bounded and equicontinuous family of functions
Sup If gdp N - f gdp - 0 a.s.).
g£F
Observing that the family F
DK(XN
1
show that
may change with
)
(i.e.,
N,
and
DK(PN)
o,
N
definitely as
K =
i.e.,
by
K
N).
min IxcEC
bounded on
as
c,
S
ICI = K - 1
Since
DK(p)
+ m
(where
increases in-
is
O(K
) (see [FH] or
KIDK(pN) - DK(p)! + 0 a.s. as
K = o(N) (i.e., when
K/N - 0
F = {g: g(x) =
is neither equicontinuous nor uniformly
mere weak convergence of
and
N
K
Note, however, that the family of functions
ca).
K
+ 0
In the case where
N + oo, which we shall prove for the case
-
I
our notation will suppress,
,
[Hal) a meaningful result here will be
N
DK(P)
however, this result becomes meaningless as both
converge to G.
DK(p)
defined by
= ID(PN) -
DK(P )
though, the indexing of
as
F
F = {g: g(x) = Mlin flx - cll,
cCC
can easily
S,
one
on
bounded
uniformly
and
, CcS} is equicontinuous
C #
K
p
weakly converges to
PN
to
p
is not sufficient
What is needed is convergence of the empirical
for proving this result.
measure in a stronger sense as established by Sublemma 2' in the proof of
Lemma 2 below.
The main results of this section are summarized by
THEOREM 1:
(i)
If
K(N)
N
O
as
N
+
(i.e.,
K = o(N),
then
-1~
I
Iu··
I
I
-7-
DK(X )
N
-
K
=
DK(P)
0
almost surely
N-to
(ii)
e
Let
<
Any algorithm which for some
0.
>
solutions for demand p
b-optimal
can be modified by reallocating an arbitrarily
(though fixed) fraction of the centers4 to yield solutions that are
small
almost surely asymptotically c-optimal for the demand
If in addition
(iii)
K
where
finds
6(6) =
K(N) + O
lim K
D(X
N-N
K
1
1
(- + 4
n 3)-
=
as
(6)(
N -+ o,
then
2/3
D3/2
f2/3 d)3/2
Z
i=l
X
almost surely
23
Remarks:
(1)
The reallocation of a small fraction of the centers, referred to
in the statement of part (ii) above, consists of placing an arbitrarily
small fraction, say 0.01, of the centers according to a uniform grid (e.g.
a square grid) before proceeding to solve the 0.99K-median problem for demand
p.
This device which is essential to our proof can be avoided, however, if
is absolutely continuous with a density
Sup
Sf(x) < m; e.g., if
just the square
(2)
S,
p
f
is uniform on
satisfying,
S.
can serve as the support of
Inf
p
Sf(x) > 0,
(Any polygon, rather than
p).
6(6) as given above in (ii) is the expected distance between a
(uniformly distributed) random point of a regular hexagon of unit area and
its center.
4
For the restricted version there is an additional modification involving the
replacement of solution centers for demand p, by the X.'s that are closest
to them.
J
--
11
11~-
--
I~-
-8-
(3)
(X1 C X2
While the theorem is stated for a sequence of nested demands
C ...) (it is an "incremental model" in Weide's [We] terminology),
the proof below does not make any use of the associated dependency between
N
DK(X
)
s.
Let
= K' DK(XN
DEV
)
- K~ DK(p).
Almost sure convergence in the
subsequent proofs follows always from the Borel-Cantelli Lemma (i.e., uses
N Prob( DEVN
> e) <
for any
X
individual distributions of the
> 0) and thus depends exclusively on the
DK(XN)s.
Consequently the theorem actually
holds for arbitrary (non-nested) sequences of problem instances, each of
which is generated by independent random points.
In particular, it holds for
the case where the instances in the sequence are generated independently (the
"independent model" in Weide's [We] terminology).
This mode of convergence
which by Borel-Cantelli's Lemma (both halves) is equivalent to summability
(over the whole sequence) of probabilities of
deviation from the limit,
is known also as "complete convergence" [HR].
The rest of this section is devoted to proving Theorem 1.
The
technicalities, while not entirely trivial, may be too specifically tailored
to be of general interest, and the reader may wish to proceed directly to
Section 3.
2.2
Derivation
.otational Conventions:(1)
We need a notation for
the value of suboptimal solutions.
,et, ten,
D(X,C)
=
Z
xEX
i
ccC
-
;;lip
D(p,Cq
min
c!
x-cll
1_11__11
___1__*
-9-
(2)
In order to enhance clarity, we will append the argument "(c)"
, can be thought
to random entities when appearing in "pointwise" statements.
of as an element
£ Q where (,
F, P) is some implicit underlying proba-
bility space that we will not otherwise invoke.
(3)
For a random variable Y,
2(y)
(Y) is
Note a (Y) ~ (a2 ()) 2 '
its fourth central moment.
LEMMA 1 (Free version only):
For any sequence {K(N)}N>l'
K2
K
D (XN)
lim sup [N
N-
is its variance while
pK(
)
] < 0 almost surely.
Proof:
Fix
0 < 6 < 1.
If
K < 1/6, let
K = K
and let
centers solving the K-median problem with demand
let
K = K -
r6/-K1 2
in the partition of
and let
S
into
K-median problem with demand
1K D(XN(w))
-
be the set of
p; otherwise, if
K < 1/6,
consists of the centers of the subsquares
CK
[1 2
p.
CK
subsquares and of the centers solving the
Now,
KK(P) = UN(w) + VN(W ) + WK(N)
(L.1)
+ ZKN)
where
UN(w) = K
VN(w)
DK(XN(w))
KN
D(XN(w),C)
K= D(XN(w),CK) - KD(pCK
)
(L1.2)
WK(N) = K(p,CK) - KD_(p)
ZK(N)
K'D(p) - K
K(P)
Z~~K(N
__1111_
_
II__·_^
1_
-10-
UN(w) < 0 for all w, since (by definition) DK(XN(w))
WK(N)
D(XN(w),CK).
0 since the set of centers that (for demand p) yield
~~~~K(N)
Also,
cost D(p)
K
is a subset of CK.
Now VN(w) may be rewritten as
VN(w) =
where
Yj,K(N)(w) = K(min
Y
(
w)
=l YjK(N)
JIXj(w) - c
(L1.3)
- D(p,CK))
cCK
The Y
K(N)' j
1,2,...,N are independent identically distributed zero
=
mean random variables.
Also all the moments of
point in the support of
in
CK,
and thus
p
is at distance
1/2
Y
1
< K
j,K '
-
are finite since every
at most from some center
1
<
2' [,/l
[
Yj k
V
Now emulating the standard proof of the strong law of large numbers
for variables with finite fourth order moments, we first note that
a (VN )
and thus for any
Prob (IVN)
£
N4 [No(
Z
)
+ 6( )c
2
(Y
1)]
= (N2 )
1,K
> 0 we obtain (using Tchebychev's inequality) that
) = O(N
>
(YK
) and thus that
Prob(IVNI > E)
<
N=l
which, using the Borel Cantelli Lemma, implies that lim
N-o
surely.
VN = 0 almost
-YI111111--
P-ll
1I^I
·._._
-11-
It follows, then, from (L1.1) that,
lim sup(W- D(XN)
KDK(p))
Finally, noting that for
K
1/6,
li
sup ZK
K
=
K
K
a.s.
K
K
-iK - 2K
1
1 - 26 '
<
we have
(1- 26 ) K2 D (p) - K2 DK(p)
(N)
|
and recalling (Theorem 1 in [Ha])
if
K >- 1/6
if
K < 1/6
the convergence of KDK(P) when K
A,
we deduce that ZK(N) can be made arbitrarily small by choosing 6 small
enough, which concludes the proof.
LEMMA 2:
If K(N)
- N
0 as N
lim inf[K
Proof:
(i.e.,K = o(N)) then,
DK(XN) - KK(p)]
O0almost surely.
Similarly to the proof of Lemma 1, fix 0 < 6 < 1.
CK,N()
If
K < 1/6,
let
be a set of centers solving the K-median problem for the first
customers,
let
D
CK,N(W)
X1(X), X2 (w),
... XN()
and
let
K = K; otherwise, , if
include also the centers of the partition of
congruent subsquares, and let
K = K + r
S
into
N
K - 1/6,
r/i2
1 2.
Now,
Kn D
K(
(w)
)
FN(w) + GN()
+
(N)
L2.1)
I--·_I_---__III__I___
Il^ll-___C ^___
-12-
where
N
= K- DK(X (w)) - K
EN()
FN()
=
FN(W)
D(N(w),C
)
NK D((w),CKN(w)) - KD(p,CK (w))
K,N
D(
(w),CK,
N
(L2.2)
K D(p,CK N(w)) - KD(p)
GN(w)
- K K(p)
HK(N)= KIDK(P)
By definition,
EN(X)
possible to show that
K(N)
ZK(N))
O
HK(N)
and
GN( )
O0 for all
.
Also it is
(as it was in the proof of Lemma 1 for
can be made arbitrarily close to
0
by choosing a sufficiently
small 6.
The proof will be complete, then, if we show that
lim inf FN
N-*co
>
0
(L2.3)
a.s.
To prove (L2.3) consider first, for M > 0, the partition of R into
MK
MK
MS
I[J4/H12 subsquares, S1 , S , ..., S
th
i
MK
subsquare S
M
let pK
,
tomer will lie in S
let yMK be the center of the
be the probability that any particular cus-
(i.e.,
i
MK
2
i
= f
dp), and let NM N(w) denote the
iS
L
(N )
actual number of customers, out of the first N, that are in S
K½'
N
D((),C
M(w))=
KN N
min
j=l CCK N(W)
KI 1 2r1r1
N
i=l
Now,
IXj(w) - c
I·~icl
U£M
MK
Ml'
mintlX.(w)
- yi
M
+ Yi
-cli
j:Xj ()S M
_-111
-
_
_I
__I
II_
I__Y
I__IX
-
-13-
-il1 2 N'N(w)
> K
2
i=1
I
i=l
-
cl
-2
i
IlyMK - clI
yi
N N min
,N
CCK
,N()
T1
41
2
CCKN(W)
N
N(w)
> K
MK
Iy
(min
1
1
2M
-
and similarly,
KD(p,CNK( )
I
i=l
K
jMK
Si
lix - cl dp
min
CaCN,K(W)
1
k[,,
2
K
P.
I
i=l
MK
i
min
-cfl
M
CCN,K(w)
Thus,
NM.'
I
FN(w) > K
(w)
(
N
N
i=l
Now for 0 <
P.
' )
C£CK,
_4
IlyMX - clI
min
(L2.4)
M
N
< 1,define
IM'&(w) = {1 < i < [f/-F1
2
(1 -
PMK
) - 1) PMK min
Ilyi
1
CKN
cCKN
cli
N
:
·N
N_
(L2.5)
and,thus,
FN(w) > K %
((1 iWIN
- K
iM
(W)
NF
isN
()
pMK min
I yM K - cI
- 49
(L2.6)
M
CK,N
_
_1__
__1_
11
· 1
--
---
I
I
_
·
I
-14-
Noting that
MK
min
M1y
C
i
N
cCK~N
C '
CK
FNM) >-c-&
1
~-
1
1
26
=1
and recalling that
Z P
i
1,
we have
Choosing
Mi
( IY
- &Ni
(
1
-
(L2.7)
)
sufficiently small and M sufficiently large can make the
first and third terms on the right hand side of (L2.7) arbitrarily small.
To complete the proof, then, it is enough to prove:
Sublemma 2':
If
lim K(N) = o
N
lim (
i
N-xo i I
then, for every 0 < £ < 1 and M > 0,
MK )=
0
(L2.8)
a.s.
N
Proof (of sublemma):
Let N be a Poisson distributed random variable with mean N.
Consider
IN
(w) =
1 <i [
N N~~~1
12 : N M(w) < NPMK(1 1
N M is the number of
where N'
customers out of the first
)}
(w) that lie in
i
We first prove that (L2.8) holds if the summation is over IM'(w)
N
rather
than over
over IM'(w)
rather
than
I N (w)
Let
Let
+
Claim:
for some
> 1 + 4- > 1
NMW). be the probability that
i
N,M,
Qi
1
i
TN
N
(w).
NpMK(N)
zN
1
_
for all N and i
.4.1111-1
IIIP----L-IIIII
-II_-------
I
-
---
-15-
t(1-&)X
N,M,
Qi
Proof:
1
n
-
e
MK
= NP.
where
n-
1
n=n=
O
< e-xe/P
pl (l-s)xA
(e e
(1i
pi
=
for any p > 1
)
(Note that
2
eAX/P pl(l-)AJ = [1 + - + (A/p)
2!
p
(1-)A
t
I
x t(1-E)AI
A2
2!
Now let p = 1 +
* I(1-a)A!+
/2 then,
1
1-1/p
-
1+/2
(1 +
p(1-)
e
e
1
< (1 +
4 )
i
M
)
1+a
2
1+/2e > es /4 >
> I +-
-
pMK (N)
N,M,a
Now let Yi
1I(l+a
e
/2(1-a)
-NP.
QNiM,
1
1+£/2
e
(1 + S/Z)
a2
Thus
1lL(1-a)A]
+ ... ]p
+
(1
+
E
4
MK
(QED Claim)
)
if i
-M,if
i
I
(W)
otherwise
___LIIO__IUIUULIYLIILI^
-_
11111
1^._11_1--
-16-
P
Obviously,
i£I
MK
)
YN M
I
'
(w)
and thus for any
i=l
(w)
E(yN,M,
E(
M,& pi
icl
N
MK
I,']
I
i=l
<
i=1
2
-NP.
Q
P.
1
MK
p
-NP M
+
MK
P.1
1
i:i
MK
+
MK> z
i N('-1)
N(O-1)
< N(e-
1)
+
(
I
-NPi
1
PMK
P.))e-
z
i =N(6-1)
fVW)- Z
+ e-z
e -z
as N -
- 1)
N(
NP.
1) implies
(Note that Pi >-
1
e-Z
< e
.
.)
Taking z large enough we get, then
limE(
PMK) = 0 .
N->o:
iI N'
Letting a
denote the
of yM,
1
(i = 1, 2, ..., [~]Z2)
th
(L2.9)
central moment we have, using the independence
z,
-17-
I2
[VR 2 N,M)<[
a4(
MP
K) =
N
4(
YiM
)
3(
i=l
i=l
1N
<3(
3(i=l
E
i=1
2
E[(yN,M,)2])
E
o2(yN,M,e))2 +
i
'
+
E [ (yN
I
u 4 (yNM,')
=1
i=l
) ]4
i=l
NpMK
M
-NP MNp
.
(p M K ) 20
<3(
1
)2
+
i=l
i=l
(pMK) 4o
1
i=1
MK
_
-NP.
=3(
M -5/6 (] pMK
i:P. <N / 6 5
1I
2
+
1
6
3 pMN
i:P
>N-5/
NpMK
+ ( pMK< -5/6 (pMK)4e
1
i:P. <N
1
-NP.
+
1
<3([
Recalling that
fV1
2
]12N
(PMK)
1
5/6
-5/6
PM.
i:1
-N1/6 )2 + ([
3(
-10/61
- 10/
+ 6
= 1 +
MK
1 2 0 /6 +
2N-20/6
+
4
0
1 )
N1 /
is negligible) and that
2/4 > 1 (thus,
= 0(K) = o(N), we conclude that
20
10
4
4(
-ME
M
PMK )
) +
0(N
(N
6)
0(N-4/3)
iNi
By Tchebychev's inequality we have, then, for every
Y4(
E8 p. MK
Prob(
i,1
N
P.I)I >
E(
<
IM e
> 0,
MK
P1 )
= 0(N - 4/3)
04
ieN
Consequently, this probability is summable over
N = 1, 2, ...
thus the Borel-Cantelli lemma and (L2.9) imply that
and
(L2.10)
-18MK
PK converges almost surely to O, which is the limit of its expeciINI
tation.
It remains to extend the result to the case where the number of
customers is N rather than a Poisson random variable with mean N.
PMK
8IN
N
M,
(w)
For every
MX
=
EM/2
i&
N
P
(W)
1
I
E(W)-
isI 'N)
(L2. 11)
P i
M/2
-IN
(w
M, &
-=M)&/2
IN (w) - IN'
(w), we have
i
N (W)
<
(1
-
)PMKN
<
(1
-
/2)PMKN
N.'
(w)
and therefore,
pMKN <
2i
MN(w) - NM,N()
i
1
Substituting, then, in (L2.11) we get
iN
,£N
N
'(w)
1
<
(/2
M
i
MK
M
Pi
(w)
z
- )1/2
isIN
EN
ii
M/2
()-I
,/2(w)
N
-
1
2N(w) -N
N
(L2.12)
We have just proven that the first term on the right hand side of (L2.12)
converges almost surely to
0.
That the second term
converges almost surely to 0 is well-known.
> 6) <
for all 6 > 0.
2
(
N
-)
Prob(NN N
N=1
This, rather than just a.s. convergence, is
(In fact,
needed here to justify our claim in Remark 2 regarding the applicability of the results to problem sequences of arbitrary dependence
-19-
structure.
The proof of this is straight forward.)
is complete, and so is the proof of Lemma 2.
The proof of the sublemma
a
We are ready to proceed now with the proof of the Theorem:
Proof (of Theorem 1):
(i)
(ii)
Consider first the free version.
This result is an immediate corollary of Lemmas 1 and 2.
First assume an algorithm yielding optimal solutions for the problem
with demand
p,
=
i.e.
.
Note that the proof of Lemma 1 is
essentially based on a heuristic in which
[-]2
centers are
= 6K
placed on a square grid while the remaining centers solve the
(K -
[/-i]2) -median problem for demand
p.
It is shown there that
the difference between the cost attained by this heuristic (when
divided by
NK-Z)
and
KiDK(p)
is almost surely asymptotically
upper bounded by a number that tends to
0
as
6 + O.
Since Lemma 2
shows that almost surely we cannot achieve a cost asymptotically
better than
KDK(p),
it follows that for any given
e > 0
we can
find
6 > 0
so that the heuristic is almost surely asymptotically
E-optimal.
Using
for demand
p,
for any given
(iii)
-optimal (
> 0),
rather than optimal solutions,
in the proof of Lemma 1, yields a similar consequence
e
>
.
This result follows from (i) and Corollary 1.2 in [HM1].
We turn now to the restricted version of the K-median problem, i.e., the
centers must be a subset of the
Xj's.
(For demand
p,
version requires that the centers lie in the support of
the restricted
p).
The arguments
above do not apply since Lemma 1 was restricted to the free version.
We show
below (Lemma 1R) that Lemma 1 is valid also in the restricted version under
the additional condition that
K = o(N).
_11 1_1_
-20-
(i) follows from Lemmas 1R and 2;
Analogously to the previous case then:
(ii) follows from a heuristic implicit in the proof of Lemma 1R and (iii)
follows from (i),
LEMMA 1R:
from Corollary C2 of [Ha] and Corollary 1.2 of [HM1].
K(N)/N + 0
If
N +
as
,
then the assertion of Lemma 1 remains
valid for the restricted version of the K-median problem.
Proof:
0 < 6 < 1.
Fix
There are sets
U
and
V
both finite unions of rectangles (with sides parallel
to the coordinate axes) so that
p({x
({x e u:
where
f(x) > 6}) < 6
S - U:
(L1R.1)
(L1R.2)
f(x) < 6}) < 66
ps(S - V) < 6
(L1R.3)
i(V) < 2
(L1R.4)
denotes the singular part of the measure
ps
This follows from the
p.
regularity of finite Borel measures in Euclidean space (for an explicit argument
of [Ha, appendix A]).
see, for example, the proof of the "claim" in Lemma A
Consider now the partitioning of
S
into
[/i12
congruent subsquares (which we
will refer to as "primary" subsquares) and define the following subsets of
U
(i)
U,
the union of subsquares contained
(ii)
U,
the union of subsquares intersecting
U
V,
the union of subsquares intersecting
V
W,
the union of subsquares with more than
(iii)
(iv)
X 1 (w), X2 (w), ...,
S:
in
N/(63 K)
customers (out of
XN(w))
_^___·_1_11_
1_1__1_11__1·1___11_II
-21-
Let
25
be the total number of rectangles in the sets
2 6-13, [then for all
K > L
(U
U
and let
L
- U) + (V
- V) < 66
(L1R.5)
U
and
V
are less
and thus the number of subsquares intersecting their boundaries is less
4%/(1/r[/rK1) + 4
than
V
we have,]
(This follows from the fact that the combined perimeters of
than 4
and
4W(/[K1 + 1)
/i-K1
2
= 4
+ 1)
([/i-K1
and their total area is less than
which is smaller than
66
for
K > L = 25926-13)
and
thus, in particular
p(U - U) < 66
It follows from (L1R.A) and (L1R.5) that for
(62 + 266) [/-K12
63 K
are no more than
V
or
U - U
subsquares in
V
subsquares in
and
W.
(L1R.6)
K > L,
(U - U).
there are less than
It is also clear that there
For each of the subsquares in
we consider an additional partitioning into
r1/612
congruent
(secondary) subsquares.
The total number of subsquares (both primary and
secondary) is less than
[/ -l2
close to
36K,
Now, for
are possible for
K
L
let
= K
6
(Tighter bounds, arbitrarily
small enough).
and let
CK
be a set of
solving the restricted K-median problem for demand
K - 17 [,/i12
and let
CK
primary subsquares.
center in CK
where for
Let
CKN()
by a customer location
K > L
CK,N()
Xi(w)
p
K
points (centers)
p; otherwise, if K > L,
be the union of a set of
R-median problem (either version) for demand
[ 12
+ 63 K) [1/612
+ ((62 + 26 6)f/-1
< (1 + 2(1 + 62) 62 [1/612) r/-K12 < 17 [/-K 12
W,
R
let
points solving the
and the set of centers of the
be the set obtained by replacing each
(1
i
N) that is closest to it,
is augmented by additional customer locations so that
each secondary subsquare that contains a customer location contains also a point
-22-
of
CK,N(w).
By our construction
CKN()
consists of no more than
K
customer
locations and is thus a feasible (suboptimal) solution to the restricted K-median
problem for
N
customers located at
Xl(w), X2 (w),
..., XN(w).
Now, the
expression in the assertion of Lemma 1 can be decomposed as follows:
K[N DK(XN(w)) - DK(p)
]
= EN(W) + FN( ) + GN(W) + HK(
(L1R.7)
where:
EN(w) = - [DK(XN(w)) - D(XN(w), CK N(M))
KN(W)
= - [D(XN(X),
FN()
N
CKN(w) - D(XN(w), CK)]
(L1R.8)
GN(W)
= K½[k D(XN(w), CK) - D(p,
HK(N) = K[D(p, CK) - DK(P)
Note the
DK
and
D
]
CK)
< K[DR(p) - DK(P) ]
in the above expressions correspond to values of the
restricted version of the K-median and
K-median
problems, respectively.
To show that the upper limit of the left hand side of (L1R.7) is almost
surely
0, it is enough to demonstrate it for each of its components in (L1R.8) or
at least to show that their upper limits can be made arbitrarily small by
sufficiently small
By definition
6.
EN(w)
0
proof of the same fact for
the convergence of
VN
KDK(p)
can show (as we did for
for all
lim GN = 0
(a.s.)
K +
(Theorem 1 and Appendix B of [Ha]) we
in the proof of Lemma 1) that
arbitrarily small by choosing sufficiently small
is almost surely
0(6).
HK(N)
can be made
6.
To conclude the proof we will show that the upper limit of
N + a,
following the
(see (L1.2)) in the proof of Lemma 1. Recalling
when
ZK(N)
w.
FN,
as
-23-
K
For
L
we have by the triangle inequality,
K()
FN( ) N < N N.
min
- max
IIx -
cl I
x XN(w)
cECK
< Li oN (w)
(L1R.9)
where
PN(W)
min
max
xEXN(w)
CECUKMC K
For
IIx - c II
K > M, we consider separately the partial customer populations:
xN,
1
= xN n (W u V u (U - U))
XN'2 = X
n (S - U - V)
XN,3 = XN n (U - W)
Clearly,
XN = XN, 1
where for
u XN, 2 U XN,3
and
FN(w)
+ FN,2()
FNi()
(L1R. 10)
+ FN,3(W)
i = 1, 2, 3
K)
=
N [D(XN'i(W), CKN(W)) - D(Xi (),
FN,i(W)
CK)J
Clearly, for all
FN l(W) K0
FNi l() <N
.
1
1
r A-K I
rl1/SI]
(L1R. 11)
and
F,
(W)
'
.
XN'
2 ()(I
1
F/i I
N,2()I
1
N
L2
(L1R.12)
Before we bound
FN 3
note that the inclusion of the primary subsquare
centers in CK
implies that no customer is further than
nearest center
in CK.
1/(iri/1-K)
We can thus assert that every center
customers outside a circle of radius
1/(/r/6K1])
customers in no more than 9 subsquares.
from the
ccCK
serves no
around it, and thus serves
Replacing
ceCK
by the customer closest
to it might (by the triangle inequality) cost the customers served by
additional distance of up to
Let
'..,
Q 1, Q2 '
rN i(W)
lIx - clI,
but no more than
1/(/[
1).
xsX (w)
be the list of the primary subsquares and define
Q[/12
= min
minN
c an
max
lx - II ,
min
1/(/2 [/6K]
} .
(L1R.13)
ceCK Qi xEX (w)
rN,i(w)
is a bound to the increase in distance, due to replacing
CK,N(w),
for any customer served by a center in Qi.
customers in U - W (i.e.,
by
All the centers serving
XN,3 customers) lie in U. The number of
XN,3
customers served by all the centers in any given subsquare
most
9
3-
CK
Qi
is at
From all that we may conclude that:
6K
K
FN,3(W)
<N
F
½
N
9 3- rN
N,i()
-
i:QicU
< 9F
-5/2
1
r
N,i)
i:QicU
< 11
-5/2 i
~r
N
(W2
M
½
(L1R.14)
,
iQiCU
where
t is the number of subsquares
(and since
/K
5 for
Qi
in V, and since
K > L = 25i 2 S- 13 ) we have
9I-
t
9
[/i-K12
< 11.
-25-
From (L1R.9) - (L1R.12) and (L1R.14) we have
FN()
N( W) +6
0
N
N,2w) + 6 5/2
X
rN, i(w)2
(L1R.15)
i:Q.cU
Note now that
any
e > 0
center
UK<LC K
there is a
c
uK LCK
for all such
c
is a finite subset of the support of
8 > 0
such that a circle of radius
has probability (p-measure) greater than
Prob(minN
lix - cll > ) < (1 - )N
p.
For
around any
0
, and thus
from which we conclude
xeX
that
Prob(pN > c) <
and thus summing over
N
L(L - 1)(1 -
)N
and employing the Borel-Cantelli lemma we have
PN + O
a.s.
(L1R.16)
By the strong law of large numbers we know that
1IxN,2
But, letting
1
Pc
singular parts of
+ Pc ({x
=
IXN n (S - U - V)j + p (S - U - V) a.s.
and
p
p,
denote, respectively, the absolutely continuous and
we have
p(S - U - V) = p({x
S - U - V:
f(x) < 6}) + ps({x e S - U - V:
S - U - V:
(L1R.17a)
f(x) < 6})
f(x)
6})
which by
(L1R.1) and (L1R.3) yields
p(S - U - V) < 6 + 6 + 6 = 36 .
Finally, note that each subsquare
around which we have a circle of radius
customers.
Qi c U
rN i(w)
The intersection of this circle with
side no shorter than
r
(L1R.17b)
contains a center from
CK
that is empty of
Qi
contains a square with
(w) (consider the square, with sides parallel to
,i
the coordinate axes, that is inscribed by this circle, then at least one of
-
,/
-26-
its corners and thus one of its quadrant subsquares must be inside
otherwise
rN, i
(L1R.13).
Consequently
must be larger than
U
1/i(2 [/r/1)
Qi'
since
in contradiction to
contains a disjoint union of no more than
empty (of customers) squares with cumulative area
rN
[/i12
()2.
Now,
i:QicU
if, as in Lemma 2, we use a refined partition into
arbitrarily large (but fixed)
M,
[A-il2
subsquares for
then we can by employing sublemma 2',
show
that the probability (i.e. p-measure) of the empty squares converges almost
surely to
0
v({x
f(x) < 6}) < 266 ,
U:
and thus, noting that by (L1R.2) and (L1R.6)
we may conclude that
lim sup
N
r
2 < 266
rN
26
a.s.
a.s.
i:QicU
and thus that
lim sup
65/2
(
rrNi 2
2
a.s.
(L1R.18)
i:QicU
Plugging (L1R.16), (L1R.17a, b) and (L1R.18) in (L1R.15) yields
lim sup FN
and since
6
0(6)
a.s.
(L1R.19)
can be made arbitrarily small the proof is complete.
Note, again that as in the rest of this paper, complete convergence is
O
essentially implicit in all the a.s. convergence statements in the proof
above.
-27-
3.
Asymptotic Separability (K/N + a > 0)
The analysis of the previous section and the resulting asymptotic
continuity apply only when
K = o(N).
In this section we introduce another
regime of asymptotic behavior, one that is associated with
N/K
does not necessarily approach
+o
K -+
(where
as in the previous section).
This
type of behavior which we will call separability, corresponds to the fact
that when
K
is large, then the optimal solution (and the value) may be
approximated by the solutions to (and the sum of the values of) the components in a partition of the problem, according to some geographic lines.
More specifically, suppose we have a collection of customers uniformly
scattered in a square, or for that matter suppose we have a uniform demand
distribution in a square, and consider a partition of the square into two
congruent rectanges.
Then as
K
grows, the optimal solution can be
approximated by the sum of the separate solutions for K/2-median problems
in the rectangles.
Similarly, if we partition the region into
gruent components, then as
K/L
grows (note
L
may change with
L
conK), the
overall solution and value can be approximated by the sum of the solutions of the components.
This of course suggests the applicability of a
partitioning ("divide and conquer") heuristic as the one introduced by
Karp [Ka] for the Travelling Salesman Problem (TSP).
a problem with
K
medians, we solve
L
problems with
Instead of solving
K/L
medians.
While the separability increases (i.e., the error decreases) as
so does the computational effort.
computational effort.
Once a
K/L
K/L
grows,
One must trade off the accuracy and the
was chosen, the relative error of the
heuristic (i.e., the ratio between the error and the optimal value) does
not depend on
K.
All these features of separability/partitioning are
-28analogous to those explored by Karp for the TSP.
Separability can be
established for the deterministic continuous version of the problem, and
is implicit in the proof of Theorem 1 of [Ha]
(specifically consult Lemma 4
and the preliminary discussion preceding the theorem).
Using the deter-
ministic seDarability and the asymptotic continuity results of Section 2,
one can easily establish separability for the case where
K + c
N/K
+ o
and
simultaneously.
In this section we are interested in establishing asymptotic sepa-
rability for the case where N/K
thus N
oXas well).
1/a for some a > 0 and K -,
X
(and
'For this case, we cannot use the deterministic
separability and we prove stochastic limit results that establish almost
sure asymptotic separability (and thus support the viability of the
partitioning approach).
3.1
Results
As in Section 2,
XN
will denote the first
N
points in an infinite
sequence of independent uniformly distributed random points of the unit
square.
The following theorem summarizes the results of this section.
THEOREM 2:
(i)
If
(N)
a >
as N - -,
there is a non-random function
lim N-1/2 DK(XN)
N-+O
where
then
= B(c)
: (0,1] -+ [0,)
so that,
almost surely
is convex, monotonely decreasing and has the following bounds and
asymptotics
lim ~c)= 1
-*+ 0
where
(a)
a-
(1 +
o~~~~~~~
n3)
2
(T2.1
(T2.1)
-29-
lim
a-*l
1*
)_
)
where
1(a)
1
-erf(ii"n(l/(2
-)))
(a
2a-1
2
In the free version, (a) <
In the restricted version
(1/)
n(/(2a-l))
(T3.1)
0 (a) for all
B(c)
>
2 erf (/n(T/7))
-acx(l/n)Zn(l/1) for all a .
(Here erf(x)
(ii)
into
2
For any
e-
dt.)
> 0,there is an
subsquares where
M > 0
such that partitioning the square
=Lv7T] and solving a separate
problem in each is almost surely asymptotically
K/ 2-median
-optimal.
Remarks:
(1)
Note that here as in the previous section, the a.s. convergence
always follows from the application of Borel-Cantelli Lemma to the whole
sequence; that is, from summability of the deviation probabilities (Prob
(IN-1/ 2 DK(XN)-0(a)I>e)) over N=1,2,3,..., and thus is essentially complete convergence.
(2)
It is quite conceivable that via a Lagrangian technique as used
in Subsection 3.6 and 3.7, one can obtain a non-uniform counterpart of
the theorem.
The rest of this section (Subsection 3.2-3.7) is devoted to the proof
of these results.
The techniques are more interesting than those of
Section 2 and the discussion (mostly Subsection 3.5) contains some
methodological and comparative remarks.
-30-
3.2
A Poisson Window Model
Consider a unit density Poisson point process in the plane.
Let
N (i,j) denote the number of points occurring in the square S (i,j) =
[(i-1)7im, im)x[(j-l)ii,
jm) and for m
1/a let Z (i,j) denote the value
>
of the Lamj-median problem on the points in
S (i,j).
sionally suppress the (1,1) in this notation, i.e., N
2
For m,n > l/a, let Z - L[nmJ; thus [O,Zv~m)
C[O,i)2
For i=j=l, we will occa- N (l,l), Z
- Z (1,1).
C[O,(+l)m) 2.
2
Zn ,the value of the Lan]-median problem on [O, n) ,is clearly not higher
than the cost attained by the crude partitioning heuristic that assigns
2
the points in [O,vn)
2
separate
to the combined centers of the (exact) solutions of
jm]-median problems, one on each of the squares S (ij)
i,j=l, ...,Z. That is
Z
<
Z
Z (i,j) + 3m(N-Nn
(3.2.1)
where the right most term corresponds to the fact that each point in
[O, (+l)))
[0,,)
2
.
< 3; of some center in
is within a distance of 2/2/
For the same reason,Z
< 3
n
N
n
and thus (substituting 1/a and
m for m and n respectively):
(3.2.2)
Zm < (3/vra)Nm
The basic convergence result here is
LEMMA 3:
(i)
lim
n-o
There is a constant
E(I
Z
nn
> 0 such that
- BI) =
This can be easily extended to L
(L1 - convergence)5
convergence for all q>l.
-31-
(ii)
Z
Prob( Z >
integer n>l/a
n n-
+ £)< co
for all
£
> 0
Remark:
The constant
To make this dependence
is, of course, dependent on a.
explicit when necessary, we will denote it
3(a).
Proof:
For notational simplicity, let V (i,j)
m
W
-
-
(i),
m m
V Z
n
n n
and
(N -N 2 ), (3.2.1) can be restated as
n n
2
2m
1
V <
Z V(i,j) + W
nn
2
m
n
2
ij=1
By the independence and stationarity of the Poisson process, and by
the translation invariance of the value functional, Dk, it follows that.
the V (i,j)'s are independent identically distributed random variables,
m
which according to (3.2.2) are dominated by Poisson random variables and
thus have finite fourth moments.
Following the standard derivation of
the strong law of large numbers for the case of finite fourth moments 6, we
have for all
£
> 0,Prob
I2
is summable over n=1,2,....
respectively,
-
n
4
Vm(ij)-E V i>)=O( _)=O(n
2
Now the mean and variance of W
(n-k2m) = O(n
1/
) and (_
n
)
3/2
for all
) which again is summable over n=1,2,.,.
are,
>0,Prob (W >)=
n
. We conclude, then, that
>0
Z
Prob(V
n>l/a
> E Vm + s)< o
or, using the Borel-Cantelli Lemma, that
6
) which
(n- 2m)=O(n - 3 / 2 ) which
implies, using Tchebychev's inequality, that for all
O(n
2
There is an example in the proof of Lemma 1.
(L3.1)
-32-
lim sup V
By (3.2.2),
m
< E V
< (
Z )
=E(
E V
n-
im
for all m > 1/c
a.s.
m
(m2 + m) < 9 (1-ka).
m
such that EVn -
by Lemma 4,to follow, there is a finite constant
i.e., part (i) of Lemma 3 is established.
E V +,
n
Consequently,
+O
Since this result implies that
we have, recalling (L3.1), part (ii) as well.
Li
< E V
almost surely for
LEMMA 4: Let suPn E(Vn ) < o and let lim sup V
1
all m > 1.- Then V n- converges in L
I<(EIVn
Since IE Vn<EIVn
Proof:
n-+
to some constant S.
1/2
) , the first part of the hypothesis implies
that supnlE VnI< - and thus that lim inf E Vn is a finite constant, say B.
n
inf E V = 0 and by the second
Then by definition lim
n
n
Now let Vn = Vn-S.
n--_ +
part of the hypothesis lim sup V n ' 0 a.s. Let Vn
Then, almost
= max(0, V ).
n
surely
0 < lim sup V+ = lim sup(max(O,Vn))=max(O, lim sup Vn) < 0
n
n
i.e.,
'
+ 0 almost surely and noting that E[(V+)
n
<E(V ) and that the
n
Finally, note that
latter are uniformly bounded in n,we get E Vn + 07.
IV I = 2V
n
- V , and thus
n
lim nsup E(IVnl)
= 2 lir
n
7
For simplicity, let U
2
Prob (T )E(U ITn)<supEU
2
-+
sup E(V+) - lim inf E V = 20-0 = 0
n
n
n
n
- Vand let T
nM T
M.
n
clearly implies li
sup E U <£.
n
S.
p
T
1/2
/
Thus, E(UnlT )<[E((unlTn)]/2 < [M/Prob(Tn)]
and therefore EU < £ + Prob(T )E(U
--
denote the event Un >
<
2
2
n
IT ) < £ + Prob(T )1/2 M1/2, and Prob (T )--O
n
n
-33-
3.3
A Conversion Based on Asymptotic Continuity in Center Density
The following "continuity" property converts the results from the
convenient "fixed ratio" (i.e.,
K = [an]) and Poisson window model
to the case where the ratio just converges to a constant and where the
demand points are the first n points of an infinite sequence of independent, uniformly distributed points in [0,1) 2 .
Recall our notation, in
which X n is this set of points.
LEMMA 5:
Let K(n) be a sequence of integers satisfying
lim K(n)/n =
>0.
Then,
(i)
lim
En-1/2DK(Xn)_-
= 0
n-
(ii)
where
Proof:
Z
Prob(n-1/2DK(Xn) >
n>l/2c
B+ ) <
for all
>0
B(=3(a)) is the same constant as in Lemma 3.
Note that in general,
K -K >n -n >0 implies D
(X
s1
) < D
nn
(X
<2
)
(L5.1)
n
(This follows from the fact that solving the K2 median problem on X 22
1
and then assigning all the n-n 2 points of Xn 2-X
as centers, yields a
n
cost equal to DK (X ), but which by definition cannot be lower than the
n
2
optimal value of the K-median (K>K2+nl-n2) problem on X .)
Now fix
>0; since K/n-a as n-~,
we may for asymptotic considerations
assume that K>(a-e)n and thus that K-[(a-3)nl>2en.
(1-2)n < N(l-)n
(i.e., if 2n > n-N(
Therefore if,
- n
> 0) then we also have K-[(a-3)n>n-N(
which by (L5.1) implies that
N
(L5.2)
)>0
-34-
As in Subsection 3.2, we assume that {Nt}t>l is a family of Poisson random
variables with
ENt=t.
The probability that (L5.3) will be violated is, then, lower than
the probability of (L5.2) being violated, which is
stant
>1.
0(
-n
) for some con-
(The proof of this last fact is rather straightforward, and
is essentially equivalent to the proof of the "claim" in Sublemma 2' in
the proof of Lemma 2).
Now,
n
/DK(Xn) -(c)
= n
[DK(Xn) - D[(a_3£)nX
N(1-)n
(
-3)
1-n
r(a-3K nI
(+ £) 1/2
+ [(l-)1/2 B(i__
(
)
-
)n-3
3( )]
(
)
D
(L5.4)
The probability that the first term on the right hand side of (L5.4) is
positive is
0(0
-n
), which is summable over
n=1,2, ....
part of the expected value of this term is at most
The positive
O(n /2-
n
) (since
when positive, the expression within the brackets is clearly bounded by
n).
For the second term in the right hand side of (L5.4), note that
(l-E)n/
D( _3)n
(X(1-)n)
1
(-3n
n Zn , where n' = (l-C)n
has a distribution identical to that of
and where the center density is
a'
=
1-
Hence, by Lemma 3(i) the expected value of the expression within the brackets
of this term converges to 0, while the probability that it is at least
for any
1
> 0
is summable over n'=1,2,... .
1
Noting, however, that this
summability followed from an O(n'-3/2) (which is equivalent to 0(n-3/2))
-35-
bound on the probability, we deduce summability over n as well.
tends to 0 so does the third
is continuous by Lemma 6 below, as
since
Finally,
term in the right hand side of (L5.4).
We may conclude, then, that
E(n /2 DK(Xn)-3(c))
lim
O0
(L5.5)
n-o
max (O,a)) and
(where a
00
-1/2
n
DK(X ) >
Z Prob(n
n=l
+ £1) <
()
(L5.6)
n
To show that the expected value of the negative part of 1/DK(Xn
) -(()
converges to 0, we repeat the same analysis, this time comparing DK(X)
N
K
As above, we get that the probability that
to D (a+3 ) 1 (X (l+)n).
N
nI (X (l+£)n)
D [(a+3E)
(+3)n]
will be violated is
(e-n) for some
follows step by step.
K
< DK(Xn
)
> 1, and the rest of the analysis
(The only difference being that we do not get an
inverted inequality counterpart to (L5.6) since there is no such counterpart in Lemma 3 itself.
We will get it however later in Lemma 10 below).
8 is obviously, non-increasing in a.
LEMMA 6:
Proof:
We furthermore have
B is convex (and thus also continuous) in a
(a>0)
First, we refine our notation to include a as an index
(a
superscript), e.g.,Za (rather than Z ) is the value of the [n]-median
problem
in
on
the
points
[
For any positive a and a
problem on the points in [0,n)
.
For any positive a1 and 0.29 we have
D
-36-
1/2(a
4n
2
<
n
(1,1) + Z (1,2) + Z2 (2,1) + Z 2 (2,2)
n
n
n
Dividing by 4n and taking expectations we get in the limit using
Lemma 3(i)
1
2
2
)
1
) <
(P(a)+(ac)+ (a2)+(C
which implies the convexity of
3.4
=
) )2
1
((a
+
)1
(a 2) )
DO
3.
More About the Asymptotic Constants
(a)
We next derive further information regarding bounds and asymptotically
tight approximations of the function
: (0,1] + [0,o).
LEMMA 7:
(i)
(ii)
lim
a-0
()
(a )
o(c
=
where
1
lim
-
(iii)
For the free median version,
(iv)
For the restricted median version,
1
J3J3
1
where 81 (a) = 1 erf (Zn(/(2
>
n3
3 4
(a)
(~
a-*l (X 3I1
()
/2 [(1 +
( a)
erf(/Zn(l/))
(ca)
))
2a-1
o (a)
for all
a.
for all a.
Remarks:
- Another lower bound to
and Hochbaum [FH].
(a) may be obtained from the analysis of Fisher
-37-
-
Our notation does not distinguish between the free and restricted versions of the problem, the latter having a slightly higher value of
Parts (i)-(ii) are valid, though, for both versions.
.
Parts (iii)-(iv)
may be true in a non (version) specific form, but proving this may be
rather involved.
-
Note that the cost of restricted version is less than twice the cost of
the free version ([HM2],
Lemma 6) and thus (iii) and (iv) imply looser
bounds for the version that is not specified in the assertion.
-
Part (ii)
(and the monotonicity of
) clearly implies that
is
a < 1 (both versions)
strictly positive for
Proof:
(i)
Note first that following the derivation of (3.2.2) we have
8
1/2
N
-1/2
K
N
DK(X ) < 3NK
.
Therefore, N
DK(X ) is uniformly bounded and the
a.s. convergence in Theorem 1 (iii) implies
L1
convergence and for uniform
distribution on [0,1) 2, we get
lim
N-
N
oN
E(DK(XN)
K
=
(6)
+ 3x4))3
for all sequences {K(N)}N, 1 such that Ko, but K/N-+
Consider now any positive sequence
let
19'
Oa2
...
(i.e., N/K-o)
converging to 0 and
2' ... be the associated asymptotic constants (i.e.,
By virtue of Lemma 5(i), for each
for all
1,
(L7.1)
j
there is a constant
M
j=3(aoj)).
such that
N >_M
1/2 IN-
3
8partitioning [0,1) 2 into
2 E(D
(XN))-
[L.N1j
Sj
<
/j
(L7.2)
K1/2] 2 subsquares and assigning a center in each
subsquare (containing at least one customer) guarantees, since all customers
K1/2 -1
1/2 -1
are within v/7 [/
j
of a center, a cost of no more than V'2 N LK /
2
T NK-1/2 < 3 NK-1/2
-38Now let
N
= 1, N
1
j
= max (Mj, [Nj
1
/ajl),
and
This construction ensures that both
j=2,3,....
infinity while
K./N.
tends to 0.
= [ajNjN
JJ
K
Nj
and
for
increase to
K
Consequently (T,7.1) implies that
1/2
K.
N.1/
NJ
E(DK
(X
) tends to
(6)
N.
K.
'
3
3
Q(6) as well.
(iii)
and,
p
is uniform
K
1/2
(see Corollary (1.2) of [HM1]).
(iv)
DK(P)
approaches
tends to
(6)
as
K
To avoid needless complications, in parts (iv) and (ii),
to center distance.
N -[an]
n
[HM1].
we sacrifice
It is convenient to consider
is the (L1 ) limit of
B
the Poisson window model.
approaches
It also follows from Theorem 2 of
some rigor by omitting unnecessary details.
the
Bj
This result is an immediate consequence of Lemma 1, Lemma 5, and the
fact that if
+oo
therefore, by (L7.2), a.
-
n
Z , the average customer
n
This average is clearly improved if we assume that all
customers that are not centers themselves have their closest
neighbor as a center, and if we further assume that these customers are those
with the smaller "closest neighbor distances"
(CNDs from now on).
With this
assumption, the average distance is a truncated average of the CNDs where
the highest [anJ or (asymptotically speaking) the upper
a
fraction is
eliminated from the summation.
r
Now for an arbitrary point, the probability of its CND being more than
2
is e
(which is the probability of having no occurrences within a
circle of radius
r).
The probability distribution function of the CND,
2
then, is
Now let
F(r) = l-e7
R
2
and the corresponding density is
v(l/)n(l/).
Then
e-
f(r) = 2rer
= a; loosely speaking, an a
of the Poisson points have a CND larger than
R.
.
fraction
The truncated average is,
R
2
then, (asymptotically) equal to the truncated expectation
f r(2rre- r
)dr
0
which after little work gives the asserted bound.
-39(ii)
Classify all possible service sets (i.e., sets of customers sharing
a common center) according to their size, as singletons, doublets, triplets
etc.
Let
fl,f
,f2
3,...
denote the fraction of customers belonging to each
category, then, clearly
fl +f
+f
+f... = 1
(L7.3)
1
fl + 2 f+
It is easy to verify that
1
3f
=
+ ...
c
fl = 2a-1, f2 = 2(1-a), f3 = f4 = 0
is a possible
distribution.
The value of the best solution with this choice is just the sum of the
shortest
(1-a)N
change argument).
first
intercustomer distances (as can be seen by a simple interThis value can be approximated by half the sum of the
2(1-a)N CNDs, the reason being that if
R
= /(l/T,).n(l/(2o-l)T, then
while the probability that a random customer will have a closest neighbor
-7TR 2
within distance R1 is l-e
= 1-(2a-1) = 2(1-c), and the probability that
-7TR 2
-TR2
it will have two or more neighbors within this distance is l-e
- e
rR
= 1-(2a-l)(
rule).
+
n(1/(2a-l)))
which is
o(l-ca) for a-1 (using L'hospital's
In other words, when we list the 2(1-)N customers with smallest CND,
and draw an edge for each closest neighbor relation, we will have close to
(1-ca)N disjoint edges, the total length of which is no greater than the best
value for
f
2a-1, f2 = 2(1-x).
The non disjoint edges can be separated at
negligible cost using remaining singletons.
As seen by the argument used in (iv), the (asymptotic) average customer
to center distance in such a solution is equal to the truncated expectation
12
for
1
r(2
Tre )dr
which after little work yield the expression asserted
1. (The "1/2" reflects the fact that when summing the CNDs each doublet
is counted twice.)
-40-
To complete the proof,it is enough to show that any solutions with
fj # 0 j > 3 are asymptotically inferior to the one just described and
evaluated.
The cost of a j-"multiplet" (whether in the free or the restricted
version of the problem) is at least the diameter of the multiplet and
thus at least as large as the (j-l) th closest neighbor distance (denoted
CND(j-1)) from two of its members.
2
gj l(r) = 2r(rr
)j-le
.
The probability density of CND(j-1) is
As in (iv), we can obtain a lower bound (on
the asymptotic average customer to center distance) by taking truncated
expectations.
is smaller than
in
a,
Let
Rj_ 1
Rj_1
be such that the probability that
is
CND(j-1)
fj/j
(which is, recall, the part of j-multiplets
Rj(r)dr = f/j. A lower
the overall center fraction) i.e.,f
0
j_
l(r)dr
1
bound on the contribution of these multiplets to the (asymptotic)
average is
Rj-lrg
(r)dr.
(for all j > 2), and thus R
Note that as
1
tends to 1, all f
tends to 0 as well.
tend to 0
We may use a first order
Taylor approximation of the integrands gjl(r) and rgjl(r) for small r.
Evaluating the first integral to solve for Rj_ 1 in terms of f
stituting in the second integral yields
Cjf
1+1/(2j-2) (where
a constant) as the associated truncated expectation.
and subCj
is
An overall lower
bound will be obtained by minimizing jE 2 Cjfj 1/(2j2) subject to (L7.3)
Z (l-l/j)f.=l-a. It is not hard to see that
j>2
as (l-a) tends to 0 the minimizing fraction distribution satisfies f2 Z 1-a
or, equivalently, subject to
Z f. = o(l-a). Therefore, asymptotically we cannot improve over
j'_3 J
a solution based on singletons and doublets alone, since as shown
and
above this type of solution asymptotically achieves the lower bound.
a
-41-
3.5
Methodological Remarks
While we have established L
- convergence, and thus also conver-
gence in probability, in Lemmas 3 and 5, we have not yet obtained a.s. convergence or specific rates of convergence of the deviation probabilities.
More precisely, we have obtained bounds on the upper deviation probabilities 9
which are summable over n,but have not obtained any bounds for the lower
.
deviation probabilities
This asymmetry is due to the fact that the re-
sults were based on the subadditive relationship (3.2.1).
If we had a
reversed "superadditive" inequality of that type, then we would have similar
bounds on the lower deviation probabilities.
Both sub and super additive
relationships arise in many geometric graph optimization problems, including
the Travelling Salesman problem, Steiner's Tree
Spanning Tree problem.
roblem, and the Minimum
And indeed they were used by Beardwood Halton and
Hammersley [BHE] to obtain a.s. convergence and by Karp [Ka]
relative error in partitioning algorithms.
to bound the
Steele [S2] refined the BHH
convergence result by establishing (via completely different methods) the
summability of the (absolute) deviation probabilities (complete convergence)l
9Prob(n- /2DK(Xn) >
+£)
1(Prob(n-1/2D(Xn) < 8-)
11
The convergence rate of the deviation probabilities established in [S2]
is better than what is needed for summability. Such convergence rate can
be also derived given two sided inequalities by the standard "even order moment/Tchebyclhev inquality" technique using fourth order moments of sums of
independent-variables, as used in Lemma 3 above to bound the upper deviation
probabilities.
-42-
Unfortunately, it seems hard to establish a superadditivity inequality
here, noting that unlike these geometric graph optimization problems, here
the overall solution in disjoint regions of the plane interacts not only
through the arcs crossing their boundaries, but also through the allocation
of the centers, and this interaction cannot be bounded by simple local
(along the boundaries) perturbation arguments.
We will overcome this diffi-
culty by the standard trick of optimization theory in such situations.
Namely, we will add Lagrange multipliers to remove the constraint on the
overall number of centers.
We will use, then, the X-cost median problem
which is concerned with the evaluation of the functional
FL(X)
min (DK(X) +
K
XK)
(3.5.1)
For this functional, we will establish both super and sub additive
relationships and then derive summable bounds for the absolute deviation
probabilities.
We will then use the convexity of
correspondence between the limits of
DK(Xn)
and
(ao)to establish the
FX(X n ) (this is-
essentially an asymptotic "strong duality" relationship between them.)
To get the sought after summability of the lower deviation probabilities
in our problem.
Finally, before proceeding with the proof, we want to point out that
to prove a.s. convergence for either the "Poisson window" or the "partial
uniform sequence" models one can still manage using subadditivity alone
using the beautiful subadditive ergodic theory of Kingman [Ki].
And in-
deed, Steele [S1], Hochbaum and Steele [HS] and the present author (in
independent unpublished work), have done so.
These proofs did not
-43-
establish summability of the deviation probabilities for the full sequence,
but rather for a subsequence.
The resulting a.s. convergence is then ex-
tended, using monotonicity or continuity arguments, to the rest of the
sequence.
3.6
The
-Cost Median Problem
Consider the X-cost median problem as defined in (3.5.1).
it first in the Poisson window model.
We analyze
Using a definition similar to
that we introduced in Section 3.2, define
YX
n
Z
as the value of the X-cost
median problem on the occurrences of a unit density Poisson process in
the square [0, v')
Observe that this definition is equivalent to
.
Y
n
= min{Z
n
+
(cn)} .
(3.6.1)
As in the proof of Lemma 6, we have refined the notation of
the index a since the latter is a variable.
(3.6.1) for any
X > 0, an
a>l may be possible when
to include
Note that in the solution to
is always an integer.
N ,
Z
Note also that values of
the number of points in [O,v)
, exceeds n,
its expectation (and X is sufficiently small).
Adapting the "Z (i,j)"
m
denote the value of the
translates of [0,
notation of Section 3.2, we use
Y (ij)
m
to
X-cost median problem in the corresponding
/2) . Recall also that
Ln7- ij.
Like the development
of (3.2.1), we get the following subadditive relationship.
Yn
n
Z
Ym(i,j)
+ X(Nn-Ng2
n 'mm )
m
(3.6.1)
Here, the right most term is obtained by assigning a center to each point
in [O,V)
- [0,Qm)
.
For a similar reason, Y
n
'N
n
.XN
-44-
Unlike the previous case, we now have a suDeradditive relationship
as well
Y (i,j) - 4nm
YA > E
m
n -. .
1,3=1
1 /4
-
(3.6.2)
N m
n
Verification of (3.6.2):
It is enough to verify that the terms subtracted from the right hand
required if we
side represent a bound on the additional cost (beyond Yn)
n
want to prohibit customers from crossing the boundaries of the S (i,j)
squares.
(This is a "partitioning cost").
To establish this result, add
centers, tentatively on the partition boundaries, spaced evenly at intervals
of length m
/4.
We will consider them as double centers, one for each side
of the boundary, and pay according X for each.
There are 2
partitioning
lines each of length /n, thus the total number of such centers is
2-2Qrn/m-
/
1
4nm -
/
.
The additional cost due to centers, then, is
the first term subtracted on the right hand side of (3.6.2).
Now in the
free version of the problem customers that crossed boundaries can switch
1 -1/4
to the closest of these centers with an added cost of less than 1 m
2
(which can be clearly obtained if at the point of crossing the boundary
the customer rather than continuing to the center on the other side turns
and goes to the closest of the centers that we installed on the boundary).
Now for the restricted version keep in mind the same reassignment of
crossover customers, but move each of the boundary centers from the
boundary to that of the customers (re) assigned to it which is closest
to the boundary.
This customer (as are the rest of the customers as-
asigned to this boundary center) is within
2
-m
from the boundary
center in a direction parallel to the boundary and thus within m/4 in that
direction from any of its other "companion" customers.
In the direction
-45-
orthogonal to the boundary, however, it is even closer to the other
customers than was the boundary center.
any crossover customer is at most m
more than N
-1/.4
.
Therefore, the cost of switching-for
Noting that there are certainly no
such customers, we get the second subtraction on the right
hand side of (3.6.2).
With (3.6.1) and (3.6.2) in place,we are ready to state.
LEMMA 8:
(i)
There is a constant y > 0 (dependent on X) such that,
lim
El 1 YX
Y- = 0
n-woo
co1
Z Prob(
n=l
(ii)
n
X
Y
n
y- > £)
<
X
for all
>
0
Proof:
(i)
(ii)
The summability of the upper deviation probabilities,
Prob(
Y
-
n
follows exactly as (i) of Lemma 3.
-
n
>
), follow also exactly (ii) of Lemma 3.
summability of the lower deviation probabilities, Prob (
-
n
For the
Y
n
- y < -
),
we also proceed similarly, the only difference being that the remainder
term (the subtractions in (3.6.2) divided by n) converges
O(m
/4
this time to
) rather than to 0 as the analogous term (W ) did in the proof of
Lemma 3.
Letting m grow indefinitely (as we do anyhow) nullifies this
remainder.
FI
Remark:
Note that by continuity arguments, one can, by analogy to the development of Lemma 5, convert this result to the "uniform independent sequence"
model.
12
Note also that the superadditivity enables us to extend even to
In the appropriate sense of summability of deviation probabilities.
-46-
the nonuniform case, where if
f
is the density of the absolutely
continuous part, we obtain, similarly to results in [BHH] and [S1], con-
vergence to
3.7
f
f1/2 dl rather than to y.
The Last Nail
To transform the convergence result in Lemma 8 (ii) from the
-cost
median problem back into a convergence result for the a-density median
problem, we need to show that for every given a we have a X such that
the asymptotic center density associated with the
will converge to a while
n
Y
n
-cost median problem
- Xa will converge to
. Using the current
vocabulary of constrained optimization theory, this result is classified
as strong durality.
side of an
(That is, for every
iven value of the right hand
nequality/equality constraint in the problem, there exists a
value for a Lagrange multiplier so that the solution of the Lagran-ean
problem coincides with that of the original problem for the secified
value of the right hand side).
The standard proof of such property when it exists is by showing
that the optimal value is a convex function of the right hand side of the
constraint.
Here, indeed, our proof is based on the convexity of
(the asymptotic minimal value) in a (the normalized number of centers
allowed) which was established in Lemma 6.
To make use of the relations
between the asymptotic constants,we first have to insure the commutativity
between the limit operator and the minimization operator.
LEMMA 9:
Proof:
y(X) = min(B(a) +
a)
By definition (3.6.1) we have, dividing by n,
1 YX <
Za +
n n - n
n
a
for all
a
-47Taking expectations and using Lemmas 3(i) and 5(i), we get in the limit
(as n tends to infinity) y(X) < B(a) + Xa for all a and thus
y(X)
min(3(a) + Xa)
.
To get the converse is a bit more involved.
for i = 0,1,2,...,Z.
n
(L9.1)
For
Then for all n,
Y
min ( Z
nn
n
>
min ((ai)
i
+
)
1
--
i
+ Xai)-z
1
Zn
-B
The first inequality follows from the fact that if a
minimizing
nn
Z
+ h,
then
n
j1
1 acj
z
+ XI =
n n
j-1
1
n
cxj.
Z
n
< a
a
(i)
*
-
(L9.2)
is the value of a
for some j 13 and thus by the
*
non-increasing monotonicity of Z
>
1, let a. = i/k
>
n
X
+ Xj
in
Z
n
>
n
, implying-
n
Za
n
+
a
Taking the expectation of both sides in (L9.2) and letting n tend to
infinity, we get in the limit (by Lemmas 8(i) and 3(i) and by virtue of
the finite number of c.i's)
y()
and letting
>
+ Xa )-X
i
i
Q
min ((a.)
i
>
min ((a)
-
+ Xa) -
-*owe get the converse of (L9.1), completing the proof.
At last, we get to
00
LEMMA 10:
Proof:
Z Prob (
n=l
Z
n(a)
- £) <
<
for all
By (3.6.1) we have for all X
1
Z
n
>
1 Y
n n
Xa
13We exclude the very marginal possibility of
We exclude the very marginal possibility of
everyw.There
cx> 1.
> 0
o
-48-
and since by Lemma 8,Z Prob (1 Y < y(X) - ) <
n n
n
Z Prob ( Z < y(X) - X - )
n n
n
<
for all
>O, we have
for all
>O .
(L10.1)
Now by Lemma 9,
y(X) = min(3(9) +
e)
14
of 6(a) at a yields (since
Choosing X as any subgradient
min(B(e) + Xt)
e
is convex)
= 6(a) + Xa
i.e.,
y(X) - Xa
= S(a)
D
which when substituted in (LlO.1) completes the proof.
To complete the proof of part (i) of Theorem 2, note that the proof of
Lemma 5 will convert this last result into the uniform independent sequence
model, used in the assertion of the Theorem.
For the associated Poisson window model, part (ii) of the Theorem is
a rather straightforward consequence of Lemmas 3 and 10.
The result can
then be extended to the "uniform independent sequence" case by again
using the arguments of (the "continuity") Lemma 5.
(a) is most probably differentiable but we have not proved it.
-494.
Conclusion
We have found that in randomly generated large scale K-median problems
we have two types of ergodic (or "large numbers") convergence processes.
These processes referred to as asymptotic continuity and asymptotic
separability enhance the analysis of the problems in two ways.
First they
permit us to predict aggregate features of the solution,- like the optimal
value and the (aggregate) distribution of centers.
Second, they enhance the
computational tractibility of such problems by suppressing the significance
of detailed information about customer locations (in the case of asymptotic
continuity) and by suppressing the significance of interactions between elements of the solution in spatially separated regions (in the ease of asympotic
separability).
So one is tempted to conclude here (as in other instances
of probabilistic analysis) that the benefits of an exact combinatorial
optimization model for very large problems are limited, as the large
numbers act to suppress the problem's combinatorial elements.
___IIIIIII__IIII__
-LI-·------
I·
-50-
5.
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[Ch]
Chung, K.L. (1974).
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[Du]
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of Mathematics, M.I.T., Cambridge, MA.
[FH]
Fisher, M.L. and D.S. Hochbaum. (1980). "Probabilistic Analysis
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[GJ]
Garey, M.R. and D.S. Johnson. (1979).
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Haimovich, M. (1985). "Large Scale Geometric Location Problems I:
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[HS]
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[HR]
Hsu, P.L. and H. Robbins. (1947). "Complete Convergence and the
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[Ka]
Karp, R.M. (1977). "Probabilistic Analysis of Partitioning
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[Ki]
Kingman, J.F.C. (1976). "Subadditive Processes", in Lecture Notes
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[Pa]
Papadimitriou, C.H. (1980). "Worst Case and Probabilistic
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