De««y
ALFRED
The
P.
WORKING PAPER
SLOAN SCHOOL OF MANAGEMENT
probabilistic
Part
I:
minimum
spajining tree problem
complexity and combinatorial properties
Dimitris Bertsimcis
Sloan W.P. No. 2057-88
August 1988
MASSACHUSETTS
INSTITUTE OF TECHNOLOGY
50 MEMORIAL DRIVE
CAMBRIDGE, MASSACHUSETTS 02139
The
probabilistic
Part
I:
minimum spanning
tree
problem
complexity and combinatorial properties
Dimitris Bertsimas
Sloan W.P. No. 2057-88
August 1988
Abstract
In this paper
classical
we consider a natural
miaimum spanning
probabilistic
tree
minimum spanning
we consider the
case where not
probabilistic variation of the
(MST) problem, which we
PMST
and
problem (PMST). In particular,
all
the points are deterministically
find a closed
pected length of a given spanning
we prove
that the problem
is
tree.
NP —
PMST
problem with the
sign problem and
in
discuss the apfor the ex-
Based on these expressions
complete.
MST
We
form expression
some combinatorial properties of the problem,
of the
the
tree
present, but are present with certain probability.
plications of the
call
We
further examine
establish the relation
problem and the network de-
examine some cases where the problem
is
solvable
polynomial time.
Key words:
Probabilistic combinatorial optimization problems,
ning tree, network design,
NP — completeness.
minimum span-
Introduction
1
The
classicad
minimum spanning
in combinatorial optimization.
tree
(MST) problem
plays an important role
possesses the matroidal property which
It
lows the greedy algorithm to solve the problem optimally, and thus
summary
prototype for problems solvable in polynomial time. For a
properties and algorithms for
From a
tion,
its
practical point of view,
problem.
we
the
of its
Steiglitz
[7].
has important applications in transporta-
communications, distribution systems,
In this paper
and
solution see Papadimitriou
it
it is
al-
etc.
consider a natural probabilistic variation of this classical
In particular,
we consider the case where not
all
the points are
deterministically present, but are present with certain probability. Formally,
given a weighted graph
each subset
S
of V,
minimum expected
G =
{V,E) and a probability of presence p{S)
we want
to construct
an a priori spanning tree of
length in the following sense: on any given instance of the
problem delete the vertices and
their adjacent edges
vertices provided that the tree remains connected.
an a priori spanning tree of
minimum spanning
of the
is
T
PMST
tree
minimum expected
(PMST)
among
length
is
for all
i
G
V
if
in
Figure
every node
is
is
part of a
the probabilistic
1.
If
the a priori tree
tree
becomes
present with probability Pi
then the problem reduces to the classical
This paper
of finding
problem. In order to clarify the definition
problem, consider the example
easily observe that
the set of absent
The problem
and nodes 2,7,9 are the only ones not present, the
One can
for
MST
more general investigation
Ti.
=
1
problem.
of the properties of
combinatorial optimization problems when instances are modified probabilis-
Figure
The
PMST
[4]
asymptotic theorems
[1]
Ph.D
thesis of
on the probabilistic traveling salesman problem (PTSP), where he
posed the problem, examined some of
PTSP
methodology
Interest in this class of problems started with the
tically.
Jaillet
1:
are derived
in the plane. In
and the
its
combinatorial properties and proved
Bertsimas
results of Jaillet
[4]
[1]
further properties of the
are sharpened.
Bertsima^
includes also results on the probabilistic vehicle routing problem, defined
in Jaillet
and Odoni
and probabilistic
[5]
facility location
problems. To our
knowledge, the
PMST
ture despite
intrinsic interest as well as its applicability. In
which
the
its
problem
hcis
never been defined before in the
a sequel of the present paper,
is
PMST
we perform
and prove that surprisingly the
PMST
litera-
Bertsimas
[2],
probabilistic analysis of
problem
equivalent to the strategy of re-optimization, in which
is
asymptotically
we re-optimize every
instance of the problem.
In the next section
while in section 3
for the
we
discuss
some applications
we address the question
PMST
problem,
of finding an explicit expression
expected length of an a priori tree T.
4
of the
In section 4
we
investigate
we prove
the complexity of the problem and
the problem with
all
weights equal
MST
simplicity of the
problem
is
is
NP — complete,
which
in
view of the
a quite surprising result. In section 5 we
examine some combinatorial properties
PMST
that even a restricted version of
of the
problem (bounds, relation of
and MST, relation with re-optimization strategies and the network
design problem).
In section 6
the problem and find
some
we
exploit these combinatorial properties of
which are solvable
in
polynomial
time.
The
2
Discussion and Applications of the
PMST
special cases
final section includes
some concluding remarks.
Problem
It is
natural to ask
why
the
PMST
problem
is
interesting.
that the problem defines an efficient strategy to update
tree solutions
when problem's
is
first
minimum
observe
spanning
instances are modified probabilistically because
of the absence of certain nodes from the graph.
Ejp, where Tp
We
the optimal a priori tree.
We
Then
denote this strategy with
in the instance 5,
i.e.
when
only nodes in the set
S
are present, the strategy produces a tree Tp{S) with
length LjpiS), which
is
the length of the tree that connects nodes from the
set
S
of present nodes using parts of Tp. In the context of this discussion the
letter
Two
1.
S
denotes the strategy used.
possible other strategies are:
A
re-optimization strategy Sa/sTi in which
ning tree
(MST)
we
find the
minimum
of the set of present nodes in every instance.
span-
We
denote with Lmst{S) the length of the
2.
A
MST
of the nodes in the set 5.
re-optimization strategy T^steiner-, in which
we
find the
Steiner tree of the set of present nodes in every instance.
minimum
We
denote
with Lsteiner{S) the length of the Steiner tree of the nodes in the set
S, using possibly nodes from the set
Remark: The above
V—
S.
definition of the re-optimization strategy
^steiner
ap-
network, as opposed to the case where the
plies only for the case of a fixed
points are located in the Euclidean plane. In this case
Lsteiner{S)
is
the
length of the Steiner tree in the plane of the points from the set S.
Why
don't
we use
these re-optimization strategies Ea/st, ^steineR',
rather than the strategy Sj-p
we
are proposing?
Concerning the 'E,steiner strategy,
because the tree connecting the
set
S
it is
clear that
Lsteiner{S)
<
using only parts of the tree Tp
Ltj,{S),
is
also
a solution to the Steiner problem. The disadvantage of the Steiner strategy
is
that
which
we have
is
to solve an
NP — hard problem in every instance, something
problem instances.
feasible only for small
With the
strategy Sa/st
it
is
clear that
0(|5|^), using the greedy algorithm, but
it
we can compute Lmst{S)
is
not clear that
in
LmstIS) <
LxpiS). In fact in section 5 we construct examples where the probabilistic
strategy Sj^
we prove
we
are proposing
is
better than the Ea/st- Furthermore, in
[2]
that asymptotically under reasonable probabilistic assumptions the
probabilistic strategy
Ej
is
at least as
good as the Ea/st-
What
is
more important
is
real-time strategy to modify the solution
strategy satisfies this criterion, since the tree
T{S) can
be found
0(n) time
as follows:
1.
Start with the a priori tree T.
2.
Until there are no
if
i
an unmarked
G 5 mark
The
3.
Since
we
it;
instances
we need a
modified.
PMST
find
when the
applications
sire
Clearly, the
in
many
the fact that in
unmarked
leaf in T;
else delete
resulting tree
leaves in T:
is
i
from T.
the tree T{S).
are only looking at every node at most once, this
is
an 0(n) algo-
rithm. Note that the two re-optimization strategies are superlinear. In addition,
we may not have the computer
resources to re-optimize.
important motivation in favor of the Ejp strategy
is
An
even more
that this strategy does
not change the underlying network structure, while both the re-optimization
strategies can result in a completely different
problem instances, by adding new edges
nication network for example,
to create
new communication
it
may
network structure
cind deleting old ones. In a
the
PMST
let
links for each
us consider
commu-
be very expensive or even impractical
problem instance.
After this discussion of the various strategies available
stances are modified,
for different
some potential areas
when problem
in-
of application of
problem. Consider for example a VLSI environment. Suppose on
a circuit, there are n processors subject to failure and processor
i
becomes
inactive with probability
pi.
Then we would
connect the active pro-
like to
cessors using a spanning tree structure, which minimizes the manufacturing
Communication
cost.
means that the
cessors
this
two active processors through some inactive pro-
of
inactive processors allow communication.
example changing the underlying network structure
PMST
strategy
is
is
Since in
impractical, the
a good solution to the problem.
communication network, nodes may represent communication cen-
In a
ters, arcs represent
tion costs
among
communication
centers.
The
links
and
link costs are the
probability of failure
blocked communication in center
i.
If
p, is
communica-
the probability of
the centers axe blocked, they can only
be used to establish communication between unblocked centers. Then the
problem
is
the
of finding an a priori
PMST
A more
unusual application of the
problem
following:
minimum expected
PMST
problem
is
in the
area of or-
For instance, a rather intriguing paradigm for the
in the area of organizational structure design
might be the
Suppose the n points that we wish to interconnect represent our
agents or spies in a foreign country.
They
will
undertake in the future a
of missions, each mission involving a different subset of agents.
in
cost
problem.
ganizational structures.
PMST
network structure of
our context,
is
We
an instance of the problem.
A
series
mission,
are looking for an a priori
organizational structure in which, for obvious rea.sons, each agent will
know
only the people immediately above or below him/her in the structure; this
implies a spanning-tree-like structure.
point
i
is
The
probability
the a priori probability that agent
any random mission undertaken
bj'
i
will
p,
associated with
have to participate
in
the network. For any given mission, only
that part of the organization which
is
necessary to interconnect
participating in that particuleir mission
T of
Figure
1
activated. (For example,
is
represents the network and agents
and
j
j
is
will
1
the tree
represents the network
be activated.) The distance between points
interpreted as the cost or risk of exposure incurred
must communicate or work with each other. Given
and the distance matrix
if
6 and 8 must be in-
1, 3, 5,
volved in a particular mission, the tree Ti of Figure
and subset of agents that
the agents
all
when agents
p, for
for all possible pairs (t,j), the
=
i
PMST
i
i
and
1,2, ... ,n
gives the or-
ganizational structure which, in the expected value sense, minimizes the risk
random
of exposure of the network on a
In a routing context, a companj'
tions, using a tree-like structure.
mission.
may want
The
cost
an a
having a demand at location
priori
spanning tree
expected transportation
for the
p,
represents the proba-
The company would
i.
demand
loca-
between demand locations can
represent transportation cost and the probability
bility of
demand
to connect all
locations, such as to
like to find
minimize the
cost.
Other areas of potential application can be
in the areas of transportation
and of strategic planning.
One might
object that
all
the examples
idealization of reality. Nevertheless, the
in
many
we have
PMST
is
discussed represent some
a generic problem, which
applications where a particular type of randomness
be a more appropriate model than the
classical
question of finding a spanning tree which
than a solution which
is
is
MST.
It
PMST
present can
also addresses the
optimal on the average, rather
optimal on a particular instance.
characteristic therefore of the
is
problem
is
that
it
is
The
essential
a more global
problem than the
is
MST
PMST
problem; as well the optimal solution to the
a robust solution.
Unfortunately, as
we prove
in section 4,
one pays
for these nice properties
(robustness, globality) by changing the complexity of the problem radically.
While the
MST
problem
is
easily solvable, the
The Expected Length
3
PMST
problem
is
NP — hard.
of a Given Spanning
Tree
As we noted
in
the previous section the
PMST
strategy for updating spanning tree solutions
modified probabilistically
in
T we
which connects nodes from the
example
of T. For
in
Figure
when problem
efficient
instances axe
response to the absence of certain nodes from
the graph. Given an a priori tree
tree
problem defines an
1,
S =
define
set
S
Lt{S) to be the length
of the
of present nodes using only parts
{1,3,5,6,8} and Lt{S)
is
the length of
the tree T^.
Then
if
the set
S
of points present has probability p{S), the
problem can be
defined formally as follows:
Problem
definition:
Given a graph
E
—* R,
a.
G =
(V, E), not necessarily complete, a cost function c
probability function p
:
2^ —*
[0, 1]
we want
to find a tree
T
:
that
minimizes the expected length E[Lt]'-
E[Lt]
=
E PiS)LT{S),
scv
10
(1)
where the summation
is
taken over
subsets of V, the instances of the
all
problem.
at this level of generality
Note that
we can model dependencies cimong the
An
probabilities of presence of sets of nodes.
additional observation
with this formulation one would need 0(n2"), {\V\
We would
expected length of a given tree T.
The question we address
efficiently.
we can compute
functions p{S)
we
define
h{S)
= Pr{
If
Theorem
none of the nodes
is
E[Lt]
present
}
for
=
that
n) effort to compute the
be able to compute E[Lt]
in this section
efficiently
in S
like to
=
is
is
for
which probability
a given tree T.
X^rcv-sPC-^)? then
1
Given an a
T
priori tree
E[Lt]
where Ke^V
—
obtained from
=
E
A'e are
T
its
expected length
c(e){l
-
h{K,)
is
given by the expression
- h{V - /C) +
/i(V)},
(2)
the subsets of nodes contained in the two subtrees
by removing the edge
e (see
Figure
2).
Proof:
Given a tree
E[Lt]-
By
T
let
how much each edge
e
^
T
the definition of the problem only the edges in
this expectation. If
A{Ke) =
us consider
at least
we
contributes to
T
contribute in
define the events:
one node
in A'e is present,
then the contribution of every edge
e is
c{e)Pr{A{K,)r]A{V-K\)},
because the edge
e is
used
if
and only
if
11
there exists at least one node present
V-K,
Figure
in A'e
and
at least
2:
The
one node present
E\Lt]
sets
in
K^,
V—
V-
K,
Ke- As a result,
= X: c{e)Pr{A{K,) n A{V -
K,)}.
But
Pr{.4(A'e)n/l(V-A'e)}
=
But
we
1
-
Pr{.4^(A',)}
since
= Pr{[A'{K,)UA%V-K,)Y} = l-Pr{A%K,)UA%V-K,)}
- Pr{A'{V -
Pr{A%K\)] = Prjnone
A',)}
+ Pr{A'{h\) D A'{V -
of the nodes in
K^
is
present}
A%)}.
=
h{Ke),
easily obtain (2). •
Thus
if
instead of the probability function p{S)
h{S) we can compute E[Lt]
for
compute h{S)
we can
in 0(1), since
any given tree
T
we
in
are given the function
0(n), assuming we can
find all the sets A'e for all e in
12
0(n) by
An
starting the computation at the leaves.
in practice,
when
is
the nodes are present independently.
an explicit expression
Theorem
If
node
tree
T
interesting case,
for
and important
Then we can
find
E[Lt]-
2
is
i
present with probability
p,,
then the expectation E[Lt] of a given
given by the expression
is
£[ir]
= E^(^){i- n(i-p.)}{i-
n
(i-p.)}-
(3)
Proof:
In this case, because nodes are present independently
h{S)
=
l[{l-p,).
Substituting the above expression in
From
0(|5|).
0(n)
we can compute E[Lt]
(3)
By
in
we
easily obtain (3). •
0{n^), since
we can compute h{S)
as follows:
=
Let a
2.
Until node set
-
n.ev'(l
is
if i is
a
add
to the set
I
E[Lt]
in
organizing the computation carefully, we can compute E[Lt] in
1.
3.
(2),
leaf, let
=
P.)
:
let a,
=
MARKED=set
Let
1;
empty:
a,-
=
(1
-
p.)
MARKED
Ee=(.,;)eT c{e){l
-
UjeMARKED,
;
delete
a,){\
-
13
i
(.,»eT Oj;
from T.
a/a,).
of leaves.
An
important special case
E[Lt]
If
we
is
when
= E^(^){1 -
=
pi
p
for all
-Pr''}{l -
(1
Then E[Lt] becomes
i.
(1
-pr''^''}.
(4)
define
^{k)^{i-{i-p)''}{i-{i-pr-'},
(5)
then
E[LT]
=
j:c{e)4>i\IQ).
(6)
eer
Based on these closed form expressions we
PMST
the decision version of the
c(e)
is
=
1 is
NP — complete.
An
4
we prove
p,
=
=
PMST
1
for all
i
and
some key combinatorial properties
and
of
problem.
of the
PMST
Problem
that the simplest possible case of the
with equal weights c(e)
p
additional importance of the expressions (6)
The Complexity
In this section
prove in the next section that
problem, even with
that they will assist us in deriving
the optimal solution to the
will
p,
=
p
is
NP
—
complete.
PMST
We
problem
first
define
formally the decision version of this restricted problem.
The
Restricted
Instance
number
Problem (RPMST)
Given a graph
:
<
p,
Question
PMST
:
Is
p
<
1
G=
=
1
for all e
£ E,
a.
and a bound B.
there a spanning tree
E[Lt]
=
{V,E), costs c(e)
Z
^(^){1
-
(1
-
T
for
G
with
P)"^'''}{1
e6T
14
-
(1
-
P)""
"^'•'}
< ^?
rational
In order to prove that the
some properties
RPMST
of the function
problem
</>(/:)
=
(1
—
NP — complete
is
x*)(l
—
a:""*),
x
we
will
= 1 —p
need
defined
in (5).
Proposition 3
The
function
<
1.
If
2.
(i>{k
3.
34>i^)
it
has the following properties:
<f>{k)
m
<
+ m) <
-
f
then
,
+
(f){k)
<l>{k)
<
4>{m).
<f){Tn).
>0.
2(f>{i)
Proof:
These properties follow
from elementary algebraic manipulations as
easily
follows:
1.
-
(f>{k)
2
+
-
-r
2
<i){m)
=
+
2.
(j){k)
3.
3<j^(3)
x"-^)(l
(x'"
-
x*)(l
-
x"-'"-'^)
<
m
m
^
<
(1
-
+ x(l - x^) + x2(l - x)] >
0.
RPMST
is
<
if
A:
and
4-
n.
4>{m)
-
=
-
4>{k
=
2(^(4)
+ 2x^ -
(1
-
x*)(l
x^)[l
-
x"-3)
+ m) =
3(1
3j3)
-
=
(1
-
x"-'*)(l
-
-
x"')(l
-
x)[l
2(1
+
-
x""''-'")
x'')(l
-
>
0.
x""")
>
•
We now
have
all
the required tools to prove that the
NP — complete.
Theorem
The
4
RPMST
problem
is
NP - complete.
15
problem
.
Proof:
RPMST
Cleaxly,
belongs to the class
NP,
since given a tree
pute E[Lt] in polynomial time (0(n)) and compare
In order to prove the completeness of the
B.
NP - complete
to
[3])
problem
EXACT COVER BY
it
T we
can com-
with the given bound
problem we
will
reduce the
3-SETS (Garey and Johnson
it.
EXACT COVER BY
Instance
A
:
3-SETS (E-3C)
S =
family
{di,.
.
.
,a,] of 3-element subsets of a set
C =
.,C3c}.
{Ci,.
Question
Is
:
there a subfamily 5i
C 5
of pairwise disjoint sets such that
Given an instance of the E-3C problem, we define the following instance
of the
RPMST
problem:
G={V,E),
V = RliSuC,
R=
r
{oo,.
=
+
3
E=
.
.
,ar},
3c,
{(a„ ao),
2
=
1,
.
.
.
,
r)
{r
+
(f>{k)
=
(1
Zc)<j>{\)
-
x'=)(l
As an example
2,5
=
4,
Let
-
if
T"-''),
G 5} U
{(<7, c),
c
G a},
1,
5 =
x=l-p,n = r + l+s + 3c.
{{ci, 02,03}, {02,03,05}, {02,04,05}, {04,05, ce}}, c
be a feasible {E{Lt]
show that
<t), ct
c4,[A),
the corresponding graph
T
We now
+
{(gq,
< p<
p arbitrary rational with
B=
U
if
E[Lt\
is
< B)
<
presented in Figure
3.
spanning tree of G. Clearly
B, then {ao,a) 6
16
T
=
for all
(a,, oq)
G T.
a £ S. Suppose
Figure
E[Lt]
> B.
=
example
We
si
in Figure
there exists
(see Figure
of nodes in
=
4a, g,
4a).
C—
l,g^
some a £
for
We
i
S.
We
G 5 and
j
show that
C
such that
E.
define
{j} that are adjacent to
=
will
/
in T.
In the
2.
also define
=
number
the
vertices
From
5i
T
G
number
the
^ T,
Since {ao.a)
(ao,t),(i, j),(i.cr)
gi
^ T
there exists only one (ao,cr)
first
RPMST
Equivalent instances of E-3C and
3:
from
C
in
nodes
T
[l
these definitions
+ 252 + 353 =
We now
E[Lt]
of
=
3c
-
=
5 —
{i,cr] in
T
that are adjacent to exactly
/
0,1,2,3).
we
5,
in
get
-
<7^
-
1 =J>
53
=
-(3c
- 2^2 -
s^
-
g,
- g„ -
I).
(7)
write an expression for E[Lt]-
r<^{l)
+ {3c-g,-g„-l)4>{l)-\-s,<i>{2) + S2<f>{3)-\-S34>{^) + 4>{9^+9c + 3)+
17
where the
first
term (r^(l))
the second term
in
C
is
is
from the contributions of the r edges
from the contributions of the edges connecting the nodes
except the ones that are connected with
3), (f){2-\-ga),
i,(T,
and the terms
<f){g,
4>{\+g„) are from the contributions of the edges (ao,t),
respectively.
(a,, oq),
(t,
+ g,, +
j), {j,cr)
Then
E[Lt] >
B=
+
(r
+
3c)4>{l)
c4>{i)
^
si4>i2)+s2(f>iZ)+s3(l>{4)+4>{9,+ga+^)+H^+ga)-\-Hi-\-g.)-H^)-c<f>{i)
>
0.
Substituting (7) we get
E[Lt] >
B^
-
\s,[Z<t>{2)
<?i(4)]
+
^52[3<;5(3)
-
2(^(4)]
+
^[3<^(5.+5.+3)-(5.+5.)<^(4) + <^(2+5.)] + ^[2<^(2+5.)-<^(4)] + [<^(l+5.)-<i(l)]>0.
Using proposition 3 we can
positive
a tree, there exist
is
Then
if
Tk-\
in
If
,
i
we add the edge
(oo,
k nodes in
T
tree
T
cri
E S and
(gq, <^k)
which there are only k
we denote the
li,,
the terms in
j
I
in
not connected to ao,
Uj.u^^ be the
number
^
C
[
]
are strictly
nodes
(see Figure
<7i,
.
.
.
,
(j,
cr^-i
ak)
we
get a
new
tree
not connected with oq.
order to represent the fact that there are
we claim that
>
E[Lt,_,].
of nodes in the subtrees
respectively (see also Figure 4b).
4b). Since
such that {ao,i),{i,j),{j,(Tk) € T.
and delete the edge
—
with Tk
^T
),..., (accrj.)
E[Lt,]
Let
all
and thus E[Lt\ > B.
Suppose now that there are
T
check that
eaisily
The
from nodes
i,j,crk
contribution of edges in Tk,Tk-i that
18
Figure
4:
The
cases (oq,
<t)
^ T and
(oq, ctj),
.
.
,
.
^T
(oq, (Tk)
are not involved in the cycle created by adding the edge (cq,
the same.
(Tk) is
Then
E[Lt,]-E[Lt,_,]
=
(^(u.
(f>{u,
where
<f){u,
+
+
1
Uj
+ l + Uj + H-u„, + l) + (/>(uj + H-u^, + l) + <A(u^, + l)-
+
+
1
+
t/j
+
1
+
t;<,^
+
-
-
<j>{u,
+
1)
1), 4>{uj
+
1 4- u,,;^
1)
+
<?i(ti^,
+
1),
1), 4>{u„^
+
1) are
the
contributions in Tk of (oo,i), (i,i) and {j,(Tk) respectively. Similarly in Tk-i
the terms
<f){u,
+ l + Uj + \),
{ao,<^k) respectively.
<f){u,
+1+
E[Lt^_^].
Uj
+
1)
By
and
4>{uj
+ l) and
proposition 3
4>{uj
+1+
u„^
we need
u,
+ l)
are from {ao,i),
we have that
+
l)
Note that we have used the
proposition 3 to hold
<f>{u„^
>
<t>{uj
fact r
=
<f>{ui
+
5
1).
+
+ l + Uj + H-ti<,* + 1 <
19
(i,
j)
and
+ l-\-Uj-\-l-\-u^^ + \) >
As a
result,
E[Lt^]
>
3c, since in order for
5
+ 3c <
^
=
i±l±2±3£.
As a
Making
{O'Oi'^k)-
=
We
will
it
now show
one missing arc
we have already proved
(oq, cti),
T
to be feasible all edges (ao,(7,)
e T.
that
using the quantities
5i
find:
>.. > E[LtA.
E[Lt,_,]
follows that for the tree
The expected
we
> B.
E[Lt]
By
>
E[Lt,]
tree Ti has only
that in this case E[Lj^]
Therefore,
decreases by adding one missing arc
this trajisformation inductively
E[Lt]
But since the
T
expected cost of
result, the
+
T
=
is
(r
•<=>E>3C has a solution.
=
1,2,3) defined above
(/
+
2.':2
cost of
E[Lt]
si
< B
353
0,
=
3c,
5o
+
+
5i
52
+
we have
53
=
S.
then given by
+
+
3c)<^(l)
+
s,4>{2)
S2<f>{3)
+
53<^(4).
Thus
E[Lt]
<B^
^-s,[3<p{2)
From
proposition
3, 3(^(2)
inequality (8) holds
is
if
+
5i<;5(2)
-
,p{A)]
<^(4)
and only
52<^(3)
and hence the
RPMST
(53
+
^52[3^(3)
>
and
if
Sj
=
equivalent to E-3C having a solution.
solution,
+
problem
20
52
-
-
-
0.
2<^(4)
>
and hence
Thus E[Lt] <
is
<
<
2(^(4)]
3(?i(3)
=
c)<^(4)
B
NP — complete.
<»
(8)
0.
53
As a
=
c,
result,
which
•O E-3C has a
•
We
can add some insight to why the problem
As p
following remarkable fact.
is
easily solvable.
<i>{k)
—
What
is
the limit as
-
(1
-
=
(1
hard by noticing the
PMST approciches
p — 0? In this case
the
1
^
is
-
the
MST
which
>
- p'k{n - k).
(1
-
p)"-*)
\Ke\)
is
the objective function of another
p)'){l
.\s a result,
—
The expression Hegr
c(e)|/Ve|(n
famous problem, the
NETWORK DESIGN PROBLEM
on a
tree,
which
is
defined as follows:
NETWORK DESIGN PROBLEM
Instance
Question
sum
the
A
:
:
Is
graph
G=
there a spanning tree
T
for
G
such that,
^
W{{u,v)]) denotes
—
|A'e|).
NP — complete
in
Johnson, Lenstra and Rinnooy
problem approaches
why
as
p
The network
^
an
the problem
hard. In fact,
originally led us to suspect that the
PMST
Kan
is
it is
problem
is
[6].
f{T)
=
Thus the
PMST
this observation that
hard.
PMST
NP — complete. We now
21
e that
problem which gives some
proved that the restricted version of the
on a non-complete graph
T, then
design problem on a tree was proved
NP — complete
is
in
W{{u,v})<B?
XIegjc(e)|A'el(7i
We have
and a bound B.
by considering the contribution of every edge
easily seen
intuition as to
if
E
on the path joining u and v
of the weights of the edges
f{T)=
It is
£
{V, E), a weight c(e) for each e
with equal costs
prove that even
if
the
graph
complete, but the costs are either small or large, the problem
is
is still
hard.
The
PMST
Instance
A
:
probability
Question
Theorem
The
complete graph
<p<
p,
Is
:
problem on a complete graph
A'„,
a cost c(e) € {1,M}, a bound
B
and a
1.
there a spanning tree
T
with E[Lt]
< Bl
5
PMST
problem on a complete graph
is
NP — complete.
Proof:
Clearly the problem
in
is
NP because of the closed form expressions we have
found. To prove that the problem
in the proof of
theorem
4.
complete we use the same reduction as
is
remaining edges but with very high
if
make the graph complete we add
In order to
=
cost, i.e c(e)
we include any edge
of this type, its contribution
= 5+
can not be included in the
c(e)<^(l)
1, i.e.
it
the
rp^^^
{r+3c)iii)+c<t>{4)+i
_
would be
c{e)4>{\Kt\)
>
tree. Therefore, the proof
remains unchanged since edges with large costs never appear in a tree with
E[Lt] < B.
•
In section 6
by exploiting some combinatorial properties of the problem,
we examine some
special Ccises of the
PMST
solved in polynomial time. For example
with
all
costs equal the
problem
indicate that the problem
costs are
1
or
is
M or the graph
is
hard
is
in
which the problem can be
we prove that
solvable in 0(n).
if
in
The previous theorems
either the graph
is
complete and the
non-complete but the costs are equal.
combine these two requirements (complete graph, equal
becomes
a complete graph
easy.
22
costs) the
If
we
problem
Combinatorial and Functional Properties
5
PMST
of the
we examine the case with equal
In this section
case
we
p.
In this
are trying to find a spanning tree that minimizes the expression
f{p)
^ nun Mp) =
min{ Yl c{em\K,\)].
Functional Properties of the
5.1
=
probabilities p,
Expression
(9)
is
(9)
PMST
clearly a function of the coverage probability p. For different
values of p the corresponding optimal probabilistic trees which minimize (9)
We first
are different.
address the question of specifying the properties of the
we have seen
function f{p).
From
difficult to find
/(p) for a particular value of p, but can
the results of section 4
properties of this function which will give
call
some
we
would be
that
it
find
some global
insight into the
problem?
We
these properties functional because they are related to the function /(p).
Some
initial
observations are stated in the following proposition.
Proposition 6
The function
np >
2
it is
f{p)
is
continuous, increasing, piecewise differentiable.
also concave
if
the costs are positive.
Proof:
We
examine the properties of the function
<^,(p)^(i-(i-p)'=)(i-(i-pr*^).
23
For
We
can easily check that
=
^mp)
(1
- p)'-'m -
- rr'') +
(1
n(i
- pr-"(i -
(i
- p)')] >
o,
[n(n-l)(l-p)"-*-Jt(Jt-l)-
^Mp) =
_(„ _ k){n
^
-k-l){\- pr-^'jil - p)'-\
k
can be easily checked that j^4>k{p)
if
np >
it
is
Thus the function /r(p)
2.
a polynomial and furthermore
since
it
is
a weighted
function f{p)
of a finite
is
is
sum
concave
number
<
for
np >
/jzCpa)
it
>
<
=
2
and ^4>\{p) <
2
is
continuous and differentiable since
is
increasing and concave for
>
and continuous, since
P
^
Pi+i for
p2
if
fip,)
=
fTXPi)-,^
/(Pj)- Finally there
We
Between successive breakpoints
some
T,.
2,
Therefore, the
0).
it is
is
=
the
minimum
li2, then f{p\)
a finite
number
which can possibly minimize /(p). Thus the function /(p) has a
of breakpoints.
np >
and continuous functions. Furthermore, /(p)
of concave
<
fT:i{pi)
for all
with positive weights (c(e)
increasing because for pi
/ti(Pi)
it
<
2;
k=l.
(n-l)(l-p)"-3(2-np),
It
>
Hence /(p)
is
p,, p,+i,
finite
=
/(p)
=
of trees,
number
/r,(p),Pi
<
piecewise differentiable. •
can now combine the above proposition 6 and our previous observa-
tions that as p
p close to
1
is
—
>
the
1
the
PMST
MST, and
as
tends to the
p
^
MST,
the optimal
i.e.
the optimal tree for
PMST
is
the solution to
the network design problem, to sketch a possible graph of the function /(p)
in
Figure
5.
24
MST
fip)
\
Lmst
Network Design
solution
Figure
5:
The
PMST
problem
as a function of the coverage probability
25
p
Bounds
5.2
for the
PMST
Based on the above functional properties of f{p) and some properties of
<i>{k)
from proposition 3 we can prove the following proposition.
Proposition 7
If
Tp
is
PMST
the optimal
-
max[plA,sT,p(l
and Lj
is
the length of the tree T, then
-pr-')LT,] < E[Lt,] <
(1
(1
-
(l
- py^^fL^fST.
(10)
Proof:
From
the concavity of the function f{p)
we
get that
/(p)>p/(1) + (1-p)/(0)=pImst,
where
clearly L},fST
solution of
PMST
=the length
for
p
we
=
of the
minimum spanning
proposition 3
From
the closed form expression (6) for E[Lt]
Since E[LTp]
=
is
the
get
<l>{l)j:c{t)
< E[Lmst]^ we
<
E[Lt]
ecisily
=
we
find
^c(e)^(|A'el)
PMST
<
<;5(L^J)Ir.
derive (10). •
Exploiting these bounds, we address the question of
as a solution to the
which
1.
From
^(1)It
tree,
problem. The following
the bounds (10).
Proposition 8
26
is
how good
is
the
MST
an obvious corollary of
T,
Figure
6:
The
E[Lmst] - E[Lt^] ^
(1
trees Ti,T2
-
-
p)(i
(1
-
p)lfJ-i)
(11)
Proof:
< E[Lmst] <
Since E[Lt,]
> pLmst we
E[Lt^]
becomes 0(n).
<?(LiJ)^mst
which
is
(1
-
—+
not informative.
intuition that the
for
p large enough (say p
>
MST
However, as p
In fact, the following
1/2
l,...,n and c(e)
example
1,2,
.
.
.
,n
+
1
=
—
satisfies
and T2
then clearly Tj
is
the
is
^
(z,
z
+
1).
27
=
oo
»
example confirms our
If
PMST
+
the tree Ji
the star tree rooted at node n
E[Lt,]
—
problem.
1)
=
1,
i
=
Note that the cost function
the triangle inequality.
MST. Then
MST
problem,
and n
>
can be a very poor solution to the
2 for all e
the
)
PMST
Consider a complete graph A'„+i with cost function: c{i,i
in this
the bound
•
consistent with our intuition.
is
and
p)^?J)i^A/sr,
a good approximation for the solution of the
is
bound
this
-
(1
can easily derive (11). Note that as p
These bounds indicate that
solution
<
(2n
-
+
1
1)<?S(1)
is
the path
(see Figure
and E[Lt,]
6),
=
2E.li Hi) = "(1 +
number). Then
if
(1
Tp
- pT) -
is
the minimal
"(1
E\Lmst\ ^ E[Lr,]
>
E[Lt,]
If
p
=
-
2tlz£l±Eiiz£lLitlz£i: (assuming n
+
PMST, we
(1
E[Lt,]
-
p)")
(2n
^ for some constant a
>
2
we
-
-
an even
obtain
oUrEllEilzfll^iirEn
l)p(l
-
(1
easily obtain as
E[L\jst]
is
n
- p)n)
—
oo that
>f)(n).
E[Lt,]
Thus from
(11)
we always have
E[L\{st]
=
0(n),
E[Lt,,
and we have found an example
for
which
-m^ As a
result,
we conclude
that the
bound
''<''
(11)
is
the best possible.
Furthermore, we can address the opposite question.
PMST
solution to the
Similarly
MST
How
good
is
the
problem?
we can show
Proposition 9
Ljp
— Lmst
1 — P
^
- p(l-(l-p)"-i)
Lmst
Proof:
Inequality (12) follows from the inequality (10) as follows:
p(l
-
(1
-p)"-^)Ztp < E[Lt,] < Lmst.*
28
(12)
Relation of the
5.3
PMST problem and Re-optimization
Strategies
We
an
have suggested in section 2 the idea that the
efficient strategy to
lems,
when problem
PMST
update the solution to minimum spanning tree prob-
instances are modified probabilistically because of the
absence of certain nodes from the graph.
In section 2
we introduced two
other alternative strategies Sa/sj and ^steineRi in which
mum spanning
nodes
the
in
MST
tree
problem defines
(MST)
every instance.
If
or the
minimum
we
find the mini-
Steiner tree of the set of present
Lmst{S) and LsTElNERiS) denote the length
or the Steiner tree respectively of the nodes in the set S,
of
we then
define the expectation of these re-optimization strategies as follows:
EI^mst] =
E PiS)LMSTiS),
(13)
^ p{S)LsTEINERiS),
(14)
scv
E[T,stEINEr]
=
scv
be the probability that only nodes in
where p{S) was defined
earlier to
present. In this section,
we address the question
of
we have
to
it is
difficult to find
compute a sum
PMST
of 0(2") terms. Instead,
we
will find
Proposition 10
every node
is
independently present with probability p then
rrv
,
^
strategy.
a closed form expression for E[T,mst], since
the E[Lmst]-
If
are
comparing the expectation
of the re-optimization strategies with the expectation of the
In general
S
np+(l-p)"-l
n —
29
1
,
a bound on
Proof:
E[Uist] =
E
P'(l
-
We
Dk =
define
Escv,|S|=fc
claim that
We
will
S,
Lmst{S) and thus
= Ep'(l-pr'Z?fc.
(16)
prove the above claim by backward induction. Consider the n sets
= V-{i}. Then
Lmst{S,) + ciKj) >
LmsHV) = Lmst
because the tree created by adding the edge
feasible solution to the instance V.
it
LmsHS).
SCV,|S|=it
£:pA/5r]
We
E
P)""*
fc=2
holds for any edge in the
edge
(?',
j)
We
MST, we
the one with the
is
Summing
over
all
i
we
minimum
cost
1
=
i
choose for every
—
6
{i,j) to the
apply (18) for
from the MST, such that the n
one edge
V(e, j)
i
MST,
MST
1
.
.
.
on
(IS)
S,
is
a
n and since
the corresponding
edges {i,j) are distinct, and
among
all
edges in the
MST.
get
Yl^MST{Sr) + Y^c{iJ) > uLmst«=i
i=\
In order to choose n
leeLst
1.
in cost
—
1
edges
we perform the
Find the edge
e'
(f,
j) to be distinct
following algorithm:
with smallest cost c(e*).
30
and the one remaining the
r
Until the node set
2.
if
I
is
a leaf in the
non-empty,
is
MST
l({i,j,) 7^ e* delete
then
Since there are n
—
1
let e"
be their corresponding edge.
edges {i,j) that are distinct, then
"
^c(z, j) = La/st +
c(e*)
<
1
+ ——r)LMST'
(1
"
•=i
As a
'
result,
"
I>n-i
= y^^MsriS,) >
r—
(n
-
1
1
Consider now the
A,
let A,,j
=
A,
t
=
-
Adding with respect
(l)
{j}.
to
i
subsets of
-)Lmst =
n —
1=1
all
MST.
I.
For the two remaining nodes
3.
be the unique edge in
let (i, j,)
\'
— 2)
—Lmstn — 1
n{n
1
of cardinality k, Ai, A-2,.
.
.,
Af
For
Arguing as before
we
get
E^mst(A.,)>^^^Z)..
(19)
But,
E^A^5T(A,.,)
since in the
subsets of
V
summation
in (20)
of cardinality k
-
= (n-fc +
we count each
1,
n
-
[k
—
l)(n
k
+
I
l)Z),_i,
distinct subset of the [k-ij
times.
find
31
—
K
+
(20)
1)
Combining
(19), (20)
we
Applying (21) inductively we easily
obtciin (17)
Then from
(16), (17)
we
find
> tp'i^-Pr-'^f^fjLMST.
^[Ea/st]
Therefore,
a-pT-l LmST-
np +
^[^MSTl
Note that
as n
Ei^Msr] >
+
As
Cj, ci
will
As a
<
E[Ltp]- Let
C2
<
be shown
n —
\
1
oc the bound becomes
^
<
not clear that E[E\fST]
It is
c,
—
d.
.
.
.
<
G=
c„.
£'[i^Tp]-
In fact,
we
give an example where
A'„ with c{i,j)
=
the star tree rooted at node
1.
{V,E) be a complete graph
MST
Then, the
in proposition 12
is
the optimal
PMST
is
the same star tree.
result,
E[Lt^ =
In this
example we
P(l
-
(1
-
p)"-^)[(n
-
l)ci
+ f: c,].
be able to find a closed form expression for E[Emst]
will
by exploiting the special structure of the cost function.
present and the
1st,...,
i
—
node
is
1th nodes are not present then the optimal tree
is
the star tree rooted at node
form expression
for
From
i.
E[Emst]
If
the
ith.
we can write a
this observation
closed
'•
n-l
E[^mst] =
^p{^ —
py~^E[LT,\node
i
is
present],
1=1
where E[Lx^] means the expected length
rooted at node
i
with leaves
z
+
1,
.
.
.
,
in the
n. Since
32
PMST
E[Lt,
\i
sense of the star tree
is
present]
=
pLr,
=
p[{n
— 1 — i)c, 4- IZfc=,+i
some algebraic manipulations, we
after
C/t],
easily find
that
El^xfsr]
=
pf2c,\pin
- 0(1 - py-' +
1
-
-
(1
p)'-'].
t=i
Choosing
Ci
= we
I
find
£[Ir,l
= "'"^''-V + n-? + 0((l-pr).
=
E[^mst]
Letting np
=
and n
c
^[^Tp]
Thus
c<
6
n
as
—
>
oc
—
oo
^
(c/2
>
-Ei-Z^jp]
we
p)").
see
+
-
1
£[EA/sr] -^ cn/2.
3/c)n,
> E[Emst]
Some
for c
>
3,
but E[LTf,]
<
£[Ea/5x] for
Special Cases
we
exploit
some
proved
in section 5, to find
PMST
problem
graph and c(e) G
{1,
The Role
first
of the combinatorial properties,
some
special cases in
which were
which we can solve the
polynomial time. In section 4 we have seen that the more
in
restricted versions of the
The
"% + 0((1 -
3.
In this section
6.1
"'"^f
M}
PMST
in a
problem with
c(e)
complete graph are
=
1
in a
non-complete
NP — complete.
of the Star Tree
natural question concerns the complexity of the problem
combine the above
restrictions,
i.e.
when we
when we have a complete graph with
33
all
costs c(e)
=
1.
We
prove a more general theorem, which includes this case
and chaxacterizes the optimal
Theorem
11
In the case
the
MST
solution.
where
problem
=
p,
p
for all
t
€ V, whenever the optimum solution
a star tree T., then T.
is
also the solution to the
is
of
PMST
problem.
Proof:
For
all
T
trees
E[LT]
from proposition
Since T.
is
3.
=
Y,<^)^i\^<'\)^<f>i^)LT
But
by assumption the
MST Lj >
Lj. for
all trees.
Combining the
above inequalities
E[Lt.]
Therefore, the star tree T. solves the
Theorem
star tree T..
<
E[Lt].
PMST
problem. •
11 characterizes the optimal solution
But are there interesting examples
in
whenever the
which T,
is
the
MST
is
a
MST?
Proposition 12
In the following
11, r. is the
examples the
MST
is
a star tree T. and thus, by theorem
PMST.
1.
A
complete graph, with c{i,j)
=
c,
2.
A
complete graph, with c{i,j)
=
c.Cj, c,
34
+
Cj.
>
0.
3.
4.
A
complete graph, with c{i,j)
di
=
A
complete graph, with c{i,j)
=
+
c,
Cj
d,dj,w\t\i ci
-|-
=
mine, and
min<fi.
—
min(ci,Cj).
Proof:
Consider an arbitrary tree
Ci
=
min,
one
at least
Lt.,
i.e.
Since
c,.
T,
T is connected
MST,
the
Without
.
Then Lj =
ij is 1.
is
T
cost
its
C2+c,j
+
.
with T, rooted
Lt =
is
c(2, 12)
.-\-Cn+c,„
.
we assume
loss of generality
in
node
>
+
-
•
•+c(n,
i^),
(n — l)ci-|-C2 +
.
.
that
where
.+c„
=
1.
With exactly the same argument we can prove
that T,
is
the
MST
in the
other ca^es. •
A
MST
corollary of theorem 11
is
that in a complete graph with c(e)
a star tree T. and thus T.
optimal solution can be found
The Case
6.2
is
p,
^
in
T.,
whenever T.
case
p, 7^
Theorem
If
Pj?
The
PMST.
Hence,
the
in this case the
pj
have shown that the optimum
the
also the
1
0(n) time.
We
is
is
=
MST. Does
PMST,
in the case p,
=
p,
is
this result continue to hold
a star tree
even in the
following theorem answers this question.
13
the probability of presence of node
star tree T, rooted at
node
1,
then T.
i
is p,,
is
Proof:
35
the
pi
=
minjp, and the
PMST.
MST
is
a
From
=
\
— Pi and
without
n
(i-n^.)(i-
because n.6A-<-{i}
we assume that
loss of generality
x.)-(i-xi)(i-nx.)=(x,-
<
^i
1
and
Xi
>
>
i,
n.€V-A'«
Ml
the assumption that Lj.
<
i^r
we
E[Lt]
star tree T.
As a
is
corollary, in a
rooted at node
^.)>o,
a;,.
eeT
/,
the
PMST.
find that
>
E[Lt.].
•
complete graph with c(e)
where
pi
=
min,
=
the
1
PMST
the star tree
is
p,.
investigate next the conditions under which the star tree T., which was
optimal
for certain special cases,
the cost function
We
is
n
^o(i-
But
Sensitivity Analysis
6.3
We
/^e-
-f[{l - p.))Y:<e).
i=2
The
n
G
1
result,
E[Lt] >
By
(i-p.)}.
xGY-Ke
i€K^
e€T
Let Xi
n
= i:<^){i- n(i-p.)}{i-
£:[lt]
As a
T
the closed form expression (3) for ajiy tree
is
remains optimal in the case
p,
=
p when
arbitrary.
define a node in a tree to be an outer
one, an inner node
if
the degree of the node
node
is
if
the degree of the node
two or more.
outer nodes from a tree of n nodes then the remaining graph
formed by inner nodes. This tree
will
If
is
we erase
again a tree
be referred to as the inner tree
36
all
Tj. In
some nodes
the tree Tj, there again must be
of degree
one and these nodes
axe called extreme inner nodes.
Theorem
14
Let a,b,c be the costs of three sides of any triangle,
nodes, in the n
positive
t
<
t
— node
>
network (n
4),
<
with a
b
any
i.e.
<
If
c.
set of three
there exists a
with
(1
-p)(i _
(1
-p)L?J-^)Vp(L^J
-
+ tb>
c
i)(i
-
(1
-p)"-')
=
0(^),
such that
a
for all triangles in the network,
then there exists a
PMST
which
is
a star
tree.
Proof:
Note that the smaller the value
t
=
it
Q,
then
of
restricts all sides of
it
/,
the more restrictive the inequality.
any triangle to be of equal length.
reduces to the regular triangle inequality.
less
than
1,
this condition
is
Since the value of
<
=
1
must be
t
stronger than the triangle inequality,
If
If
i.e.
more
restrictive.
It
is
sufficient to
any spanning
cost. So, let
Let
A'',
tree,
T
which
is
of inner nodes in
not a star tree, without increasing the expected
be any spanning tree which contains at least two inner nodes.
be an extreme inner node in
node. Since
A^^ is
be of degree one
7,
show that we can reduce the number
T
with a neighbor Np which
an extreme inner node,
in T. Call these
all its
is
an inner
neighbors except
Np must
nodes Ni,... Nk-i- This
where the distance between Np and Ni
37
is
denoted by
is
c,.
shown
in
Figure
no<
> k-i
Figure
Let us construct a
that the nodes Ni
tree T\,
Ng
is
=
{i
The
7:
new spanning
I
,
.
.
.
,
k
—
I)
tree
tree Ti
T
which
the sajne as
is
are connected to
Np
T
except
directly. In the
new
no longer an inner node. Thus the number of inner nodes has
decreased by one.
We
show that
will
E[LtA < E[Lt].
If
we apply
tree T.
this idea recursively to the trees TijTj,.
which
exactly the
is
same
both
T
and
Ti,
-
E[Lt,]
= Ml -
we need only
—
a;
=
1
a:')(l
-
i"-*)
for the part to the right oi Np. If
E[Lt]
we
.,
will finally get a
Since the part of the tree to the
a star tree.
for
.
p,
left of
Np
is
calculate the expected cost
then
+ X: a.(l -
ar)(l
- x"-^-
1=1
fc-i
-
- x"-M -
E
x){l
-
x"-^).
net decrease of expected cost in changing from
T
to Ti
6o(l
x)(l
c.(l
-
1=1
The
^"^
xCl
-
E[ir]-^[lTj = (l-x)(l-x"-^)$:[a.-c + 6or
(it«=i
3S
is
x*-Mfl
'^
1)(1
-
1"-*=-^
-x)(l -x"-i)-
The decrease
be positive
will
,
,
""
For a fixed n, x(l
when k
-
each term
is
positive,
x(l-x^-M(l-x"-^-^)
namely
^,'^
^ ^(fc-l)(l-x)(l-x"-i) -
x''-''){\
largest, that
is
if
-
x"-''-i)/(^
when k =
is
[n/2j.
-
-
1)(1
Thus
x)(l
it is
(1-p)(1-(1-p)ItJ-1)^
-
x""^)
sufficient to
is
smallest
have
^_
^
••
"•^S(LtJ-i)(i-(i-p)'^-')Assuming
a,
<
<
60
c,
gives the strongest inequality,
ment
of the theorem. •
7
Some Concluding Remarks
We have seen that
and we have the
a natural probabilistic variation of a classical combinatorial
problem has the potential to model various practical situations,
alternative
way
probabilistically
its
to
offers
an
update solutions to problem instances which are modified
and leads to very
The
deterministic counterpart.
problem was proved
that the
state-
to
be
NP
different properties in
comparison with
simplest possible version of the
— complete^m
PMST
sharp contrast with the fact
MST problem is solved by a greedy, most
straightforward algorithm.
Surprisingly, our analysis of the combinatorial properties of the problems
established
some
interesting connections with the network design
and naturally with the MST.
tends to
0,
the
PMST
problem
In particular^ as the probability of presence
p
approaches the solution to the network design prob-
lem. This limiting behavior suggests the idea of solving the network design
problem
as a sequence of
PMST
problems, which
39
is
a topic of future research.
At a
final step,
we examined the
the solution of the
PMST
problem under some conditions.
As a general conclusion,
rial
special role of the star tree, which can be
probabilistic variations of classical combinato-
optimization problems raise interesting and entirely
pared with their deterministic counterparts and
ing of the properties of the probabilistic
terministic problems,
BlS
it
new
questions com-
in addition,
understand-
problem can add insight to
de-
was the case with the network design problem.
Acknowledgments
I
would
useful
like to
thank
comments and
my
thesis advisor Professor
interesting discussions.
Patrick Jaillet for his constructive remarks.
I
Amedeo Odoni
want
also to
for several
thank Professor
This work was partially sup-
ported from the National Science Foundation under grant ECS-8717970.
References
[1]
Bertsimas D., (1988), "Probabilistic Combinatorial Optimization Problems",
Ph.D
thesis,
Massachusetts Institute of Technology, Cambridge,
Massachusetts.
[2]
Bertsimas D., (1988), "The Probabilistic
lem, Part
II:
Minimum Spanning
Tree Prob-
Probabilistic Analysis and Asymptotic Results", submitted
for publication.
40
Garey M. and Johnson D., (1979), Computers and
[3]
Guide to the Theory
of
NP-Completeness, Freeman, Sam
Jaillet P., (1985), "Probabilistic
[4]
Intractability:
A
Fraaicisco.
Traveling Salesman Problems", (Ph.D.
Thesis) Technical Report No. 185, Operations Research Center, Massachusetts Institute of Technology, Cambridge, Massachusetts.
Jaillet P.
[5]
and Odoni A., (1988), The Probabilistic Vehicle Routing Prob-
lem, in Vehicle Routing: Methods and Studies, edited by B.L. Golden
and A. A. Assad, North Holland, Amsterdam.
[6]
Johnson D., Lenstra J.K., A. Rinnooy Kan, (1978), "The Complexity of
the Network Design Problem", Networks,
[7]
8,
279-285.
Papadimitriou C.H. and Steiglitz K., (1982), Combinatorial Optimization:
Algorithms and Complexity, Prentice-Hall,
41
kSkl
Ui|5
New
Jersey.
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