Mathematical Programming Algorithms for TwoPath Routing
Problems with Reliability Considerations
April K.Andreas
J.Cole Smith
Department of Systems and Industrial Engineering
University of Arizona,Tucson,AZ 85721
Abstract
Most traditional routing problems assume perfect operability of all arcs and nodes.However,when
independent arc failure probabilities exist,a secondary objective must be present to retain some mea
sure of expected functionality,introducing nonlinear,nonconvex constraints.We examine the Robust
TwoPath Problem,which seeks to establish two paths between a source and destination node wherein at
least one path must remain fully operable with some threshold probability.We consider the case where
both paths must be arcdisjoint and the case where arcs can be shared between the paths.We begin by
proving the NPhardness of these problems,and then examine various strategies for solving the resulting
nonlinear integer program,including pruning,coe±cient tightening,lifting,and branchandbound par
titioning schemes.We discuss the advantages and disadvantages of these methods,and conclude with
computational results.
Subject Classi¯cations:
Programming,Integer:Nonlinear.Networks,Theory.Programming,Integer:
Branchandbound
Area of Review:
Optimization
1 Problem Framework
In this paper,we examine a class of routing problems on a directed graph G(N;A),where N is the set of
nodes f1;:::;ng and A is the set of arcs.Each arc (i;j) 2 A has a usage cost c
ij
and a reliability 0 < p
ij
· 1,
which denotes the probability that arc (i;j) successfully operates.These probabilities are assumed to be
independent.The fundamental problemthat we consider in this paper seeks to ¯nd the minimumcost pair of
directed paths from an origin node to a destination node,such that the probability that all arcs successfully
operate in at least one path is larger than some threshold value ¿,where 0 < ¿ < 1.In one variation of this
problem,we enforce a restriction that the paths must be arcdisjoint,while in the other variation,arcs may
be shared by both paths.Two paths are said to be arcdisjoint if no arc belongs to both of these paths.
Note that in the latter case,the failure of a shared arc causes both paths to fail.
Consider the example in Figure 1.Each of the thirteen arcs is labeled with its associated cost and reliability.
For this example,suppose ¿ = 0:65.If we require that the two paths must be arcdisjoint,then the optimal
solution uses paths ABFH and ADGH at a cost of 31,and a joint reliability of 72.6%.If we consider the
case in which the two paths can share arcs,then the best solution uses paths ABEH and ABFH,at a total
cost of only 27 and a joint reliability of 67%.
As an introduction to our problem,consider the single shortestpath problem that minimizes the sum of
arc usage costs,subject to the side constraint that the successful traversal of the path is greater than some
threshold value ¿.For 0 < ¿ < 1,it is not hard to show that this problem is ordinarily NPhard.(By
contrast,we will demonstrate that the twopath problem considered in this paper is strongly NPhard for
0 < ¿ < 1.)
The reliabilityconstrained shortestpath problemis related to several other singlepath problems that have
appeared in the literature.This problemfalls into the category of a shortestpath problemwith an additional
linear constraint,since the reliability condition can be converted to a linear constraint by a simple use of
logarithms.Such problems are referred to as resourceconstrained shortestpath problems in the literature,
since the side constraints on these problems can represent limitations on the consumption of resources as
the path is traversed.Joksch (1966) presented both a linear programming and a dynamic programming
approach for solving the shortestpath problem subject to a side constraint for a graph with nonnegative
costs and constraints.One type of method used to solve this problem applies approximation schemes,such
as those introduced by Hassin (1992),to a reduced graph.Another method is to enumerate paths (the
kshortest path problem) in order of increasing reduced cost and terminate when the constraint is met.This
is suggested by Handler and Zang (1980),who developed a dual algorithm for the constrained shortestpath
problem,developed further by Beasley and Christo¯des (1989).Mehlhorn and Ziegelmann (2000) suggest a
method of enumerating paths in order of increasing reduced cost combined with online pruning,and also
show the resourceconstrained shortestpath problem to be NPcomplete.
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v
B
v
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Figure 1:Robust TwoPath Problem Example with ¿ = 0:65.
Much additional work has been done in the realm of the constrained single shortestpath problem.Aneja,
Aggarwal,and Nair (1983) explore a labeling process to reduce the network size,eliminating arcs that
violate a side constraint by implementing Dijkstra's algorithm (Dijkstra 1959).Desrochers and Soumis
(1988) develop a labeling algorithm for the shortestpath problem with time windows,which the authors
state is also applicable for the shortestpath problem with a single side constraint.The idea of equity as a
side constraint is investigated by Gopalan,Batta,and Karwan (1990),who examine both loopingpath and
looplesspath problems.More recently,Chen,Song,and Sahni (2004),compare and improve upon methods
for discretization of link delay or link cost,and Dumitrescu and Boland (2001,2003) present an algorithm
that implements weightscaling and graph reduction through preprocessing techniques.
In related studies,Bard and Miller (1989) address an application in research and development project
selection where spending additional money on projects could increase their probability of technical success
and improve their performance.Dummy nodes are added for each project to represent success at di®erent
funding levels,and heuristics are used to discover the optimal solution.Zabarankin,Uryasev,and Pardalos
(2001) consider the problemof the optimal risk singlepath problemin the context of optimizing aircraft °ight
trajectory through a threat environment.Elimam and Kohler (1997) describe some unique applications of
the resourceconstrained shortestpath problem,such as the determination of optimal wastewater treatment
processes and thermal resistance of building structures.
Certain variations on the problemexamined in this paper have also been studied.Suurballe (1974) presents
a polynomialtime labeling algorithm to ¯nd k nodedisjoint paths between two nodes in a network,labeling
arcs positive or negative depending upon the optimal path in the network and using a technique called
arcsimple interlacing.Fortune,Hopcroft,and Wyllie (1980) provide an important characterization of NP
complete vertexindependent routing problems on directed graphs,showing via the graph homeomorphism
problem that ¯nding a path that includes a certain arc or vertex is strongly NPcomplete.
2
The problem studied here has to some extent existed before in the context of a network survivability
problem.In these problems,active and backup paths are not usually chosen in consideration of each others'
reliabilities.Instead,the assumption is made that at most one link in the network can fail at a time,so as
long as the two paths are edgedisjoint,if the active path goes down,the backup path can be used with full
reliability.Therefore,the objective is to minimize the total bandwidth reserved to meet a given demand,
allowing the sharing of the backup links among disjoint connections.Li et al.(2002) approach the problem
in terms of networking for multiprotocol label switched networks,developing extensions to the signaling
protocol to distribute additional link usage information,and compare their algorithm in terms of bandwidth
e±ciency and message overhead to two other wellknown distributed restoration path selection algorithms.
The authors also explain how to extend the algorithm to account for single node failures (multiple link
failures) and ¯ber span failures.Several other groups explore this problem,including Liu and Tipper (2001),
Xu,Qiao,and Xiong (2002),and Yee and Lin (1992).
In this paper,we examine both the case in which paths must be arcdisjoint and in which paths may
share common arcs.We refer to the former as the Reliable TwoPath Problem with Disjoint Arcs (RTPD),
and to the latter as the Reliable TwoPath Problem with Arc Sharing (RTPS).The arcdisjoint provision
may be required due to policy restrictions on backup paths (if one path fails,another path continues to
exist between the two nodes with some con¯dence),or due to bandwidth or capacity restrictions arising
in telecommunication and transportation applications.However,if these restrictions are not present,an
improved solution may exist in which paths share one or more arcs (at the risk of both paths failing when
any shared arc fails,although this risk would necessarily lie within the speci¯ed threshold tolerance).
We begin our discussion by examining the complexity of RTPD and RTPS in Section 2.In Section 3,we
develop a mathematical programming formulation for RTPD,using an analogy to the singlepath problem
with a side constraint.In Section 4 we extend the logic for modeling and solving RTPD to RTPS,and
examine the complications that arise when sharing arcs between paths is allowed.We provide computational
results of our methods in Section 5,and summarize our work in Section 6 with a discussion of areas for
future research.
2 Problem Complexity
In this section,we prove that both RTPD and RTPS are strongly NPhard,as opposed to the ordinary
NPhardness result for the reliablesinglepath case.Consider the decision problem3SAT,de¯ned as follows
(Garey and Johnson 1979).The inputs to a 3SAT decision problem are a set of V binary variables,which
can take on values of true or false,and a set of M clauses C
1
;:::;C
M
.(Assume that V ¸ 3 and M ¸ 3.)
Each clause contains three variable values,of which at least one must be satis¯ed by the variable value
assignments (e.g.,a clause might require that variable 1 is true,or variable 3 is false,or variable 4 is true).
The 3SAT decision problem seeks to ¯nd a set of variable values such that each clause is satis¯ed by at least
one variable assignment.For short,we can de¯ne v
i
to be the event that variable i is true,and
v
i
to be the
3
½¼
¾»
u
i
½¼
¾»
x
1
i
½¼
¾»
y
1
i
½¼
¾»
x
2
i
½¼
¾»
y
M
i
½¼
¾»
x
1
i
½¼
¾»
y
1
i
½¼
¾»
x
2
i
½¼
¾»
y
M
i
½¼
¾»
u
i+1








:::
:::
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H
H
H
Hj
H
H
H
H
Hj
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Figure 2:Example Lobe for Node i.
event that variable i is false.The foregoing example clause can thus be written as fv
1
;
v
3
;v
4
g.De¯ne the
following decision problem variation of problem RTPD.
Problem DRTPD:given a network G(N;A) with designated source and destination nodes,arc costs c
ij
and reliabilities p
ij
,8(i;j) 2 A,along with a cost goal of · and reliability goal of ¿,does there exist a
set of arcdisjoint paths from the source to the destination such that the total cost of the two paths is not
more than ·,while the probability that at least one path survives (i.e.,all arcs on at least one path remain
operational) is at least ¿?
Proposition 1.
Problem RTPD is strongly NPhard.
Proof.
ProblemDRTPDclearly belongs to NP,as a solution can be encoded as two di®erent O(n)length
paths,and both the cost summation and probability that at least one path survives can be evaluated in O(n)
time.To show that DRTPD is NPcomplete,we execute the following transformation from any arbitrary
instance of 3SAT to DRTPD,such that the 3SAT instance is a yesinstance if and only if the transformed
DRTPD instance is a yesinstance.
This transformation creates a topology that is virtually identical to that employed by Xu et al.(2004) for
a di®erent type of arcdisjoint routing problem.We create a\lobe"for each variable i = 1;:::;V,consisting
of nodes u
i
and u
i+1
,x
j
i
and y
j
i
for j = 1;:::;M,and
x
j
i
and
y
j
i
for j = 1;:::;M.Directed arcs are included
in this lobe to form two paths from u
i
to u
i+1
.The ¯rst of these paths consists of nodes u
i
,x
1
i
,y
1
i
,x
2
i
,:::,
y
M
i
,u
i+1
.The second path traverses nodes u
i
,
x
1
i
,
y
1
i
,
x
2
i
,:::,
y
M
i
,u
i+1
.Note that lobes for variables i and
i + 1 share node u
i+1
,for i = 1;:::;V ¡ 1.De¯ne Type I arcs as those that connect x
j
i
to y
j
i
and
x
j
i
to
y
j
i
,for i = 1;:::;V and j = 1;:::;M.All other arcs introduced thus far are called Type II arcs.Figure 2
depicts a lobe for node i.
Additionally,we add nodes w
j
I
and w
j
O
for j = 1;:::;M connected with Type III arcs (u
1
;w
1
I
),(w
j
O
;w
j+1
I
)
for j = 1;:::;M ¡ 1,and (w
M
O
;u
v+1
).For each clause j = 1;:::;M and for each element v
i
in C
j
,we
add Type III arcs (w
j
I
;x
j
i
) and (y
j
i
;w
j
O
),and for each element
v
i
in C
j
,we add Type III arcs (w
j
I
;
x
j
i
) and
(
y
j
i
;w
j
O
).
Set the cost of each Type I arc to 0,and its reliability to 1.Set the cost of each Type II arc to 0,and
its reliability to 1=2.Finally,set the cost of each Type III arc to 1 and its reliability to 1.Our cost goal is
4
· = 3M +1,and our reliability goal is ¿ = 1.
First,suppose that the 3SAT instance is a yesinstance.One DRTPD path begins with arc (u
1
;w
1
I
),and
ends with arc (w
M
O
;u
V +1
).This path moves from node w
j
I
to w
j
O
,for j = 1;:::;M,by ¯nding the variable
that satis¯es clause j.If v
i
satis¯es clause j,then the path uses arcs (w
j
I
;x
j
i
),(x
j
i
;y
j
i
),and (y
j
i
;w
j
O
) (these
are Type III,Type I,and Type III arcs,respectively),while if
v
i
satis¯es the clause j,the path uses arcs
(w
j
I
;
x
j
i
),(
x
j
i
;
y
j
i
),and (
y
j
i
;w
j
O
).Finally,this path moves from w
j
O
to w
j+1
I
for j = 1;:::;M ¡1 along the
Type III arc that connects these two nodes.The total reliability of this ¯rst path equals to 1 since only Type
I and Type III arcs are used,and hence the ¿ = 1 restriction is met regardless of the second path.However,
the cost of this path is 3M +1.We construct the second path by proceeding from u
i
to u
i+1
via nodes
x
1
i
,
y
1
i
,
x
2
i
,:::,
y
M
i
if v
i
is assigned a true value,and otherwise via nodes x
1
i
,y
1
i
,x
2
i
,:::,y
M
i
,in order to avoid
any arcs possibly used by the ¯rst path.This path has a cost of zero,since it does not use any Type III
arcs,and hence the cost goal of · is also satis¯ed.Therefore,the transformed DRTPD instance is also a
yesinstance.
Next,suppose that the DRTPD instance is a yesinstance.We must have one path in the solution
to DRTPD with a total reliability of 1 in order to meet the restriction that ¿ = 1.This can only be
done by using Type I and Type III arcs.Hence,this path must have used arcs (u
1
;w
1
I
),(w
j
O
;w
j+1
I
) for
j = 1;:::;M ¡1,and (w
M
O
;u
V +1
).For each j = 1;:::;M,the path must have traversed from w
j
I
to w
j
O
by arcs (w
j
I
;x
j
i
),(x
j
i
;y
j
i
),and (y
j
i
;w
j
O
) for some v
i
in clause C
j
,or by arcs (w
j
I
;
x
j
i
),(
x
j
i
;
y
j
i
),and (
y
j
i
;w
j
O
)
for some
v
i
in clause C
j
.Once again,this path has a total reliability of 1 and a total cost of 3M +1.The
second path must have used Type I and Type II arcs only,traveling from u
i
to u
i+1
for each i = 1;:::;V
either by nodes x
1
i
;y
1
i
;x
2
i
;:::;x
M
i
;y
M
i
,or by nodes
x
1
i
;
y
1
i
;
x
2
i
;:::;
x
M
i
;
y
M
i
.This implies that if we move from
w
j
I
to w
j
O
in the ¯rst path by using arcs (w
j
I
;x
j
i
),(x
j
i
;y
j
i
),and (y
j
i
;w
j
O
),we did not also use arcs (w
j
0
I
;
x
j
0
i
),
(
x
j
0
i
;
y
j
0
i
),and (
y
j
0
i
;w
j
0
O
) in the ¯rst path for j 6= j
0
,and vice versa,or else a second disjoint path using only
Type I and Type II arcs could not move from u
i
to u
i+1
.To construct a solution to the 3SAT instance,we
set v
i
to be true if the second path used nodes
x
1
i
;
y
1
i
;
x
2
i
;:::;
x
M
i
;
y
M
i
,and set v
i
to be false otherwise.Note
that each clause j = 1;:::;M is satis¯ed by v
i
if the ¯rst path uses arcs (w
j
I
;x
j
i
),(x
j
i
;y
j
i
),and (y
j
i
;w
j
O
),
for some v
i
2 C
j
,and is satis¯ed by
v
i
if the ¯rst path uses arcs (w
j
I
;
x
j
i
),(
x
j
i
;
y
j
i
),and (
y
j
i
;w
j
O
),for some
v
i
2 C
j
.
Thus,the 3SAT instance is a yesinstance if and only if the transformed DRTPD instance is a yes
instance.Moreover,since all numerical data used in this transformation is of polynomial size,we have
proven that DRTPD is strongly NPcomplete,and that RTPD is strongly NPhard.This completes the
proof.2
Proposition 2.
Problem RTPS is strongly NPhard.
The proof of this proposition is similar to Proposition 1.In this case,¿ must be set just less than 1,and
the probabilities must be carefully arranged so that the only feasible solution has the same form of the path
5
pair in the proof of Proposition 1.A full proof of this proposition appears in the Appendix.
3 Reliable Two Disjoint Path Problem
In this section we discuss the model formulation of RTPD,beginning in Section 3.1 with the case of
the singlepath problem having a threshold probability side constraint.For the singlepath problem,this
probability threshold is a linear side constraint,but when the problem is complicated by searching for two
paths,the requirement that at least one path survives with a threshold probability becomes nonlinear.We
cope with the di±culties imposed by this constraint in Section 3.2.In Section 3.3,we o®er strategies for
improving the model via preprocessing techniques that tighten the formulations introduced in Section 3.1.
3.1 Mathematical Programming Formulation
For the following problems,de¯ne the forward star FS(i) as the set of arcs that exit node i,and the reverse
star RS(i) as the set of arcs that enter node i,8i 2 N.(That is,FS(i) = fj 2 N:(i;j) 2 Ag and
RS(i) = fj 2 N:(j;i) 2 Ag.) As a preliminary step to our formulation,suppose we wish to ¯nd a path
from node 1 to node n of minimum cost,subject to the restriction that the probability of path survival is
at least ¿.For the following problem,let decision variables x
ij
equal 1 if arc (i;j) 2 A is used in the path,
and 0 otherwise.The formulation of this problem as a mixedinteger nonlinear program uses standard °ow
balance constraints to enforce the integrity of the path,plus the following reliability constraint:
Y
(i;j)2A
p
x
ij
ij
¸ ¿:(1)
Although (1) is a nonlinear constraint,we can linearize it in one of two ways.First,taking the log of both
sides of (1),we obtain the linear constraint
X
(i;j)2A
log (p
ij
) x
ij
¸ log (¿):(2)
An alternative technique is to de¯ne variables s
i
for each i 2 N to be the probability of successfully reaching
node i,given that the path visits node i.If the path does not visit node i,these variable values can
arbitrarily be determined.The reliability restriction can thus be equivalently formulated by replacing (1)
with the following set of constraints:
s
j
· p
ij
s
i
+(1 ¡x
ij
) 8(i;j) 2 A (3a)
s
1
= 1;s
n
¸ ¿:(3b)
Constraints (3) require that if x
ij
= 1 for some (i;j) 2 A,the probability of reaching node j 2 N is no more
than the probability of reaching its preceding node i 2 N,multiplied by the reliability of arc (i;j) 2 A.If for
some j 2 N,we have that x
ij
= 0 8i 2 RS(j),the path does not visit j and the value of s
j
is irrelevant.(This
technique is similar to one used by Pan,Charlton,and Morton (2003) for interdicting nuclear material.) The
6
bound s
n
¸ ¿ enforces the desired reliability constraint.Whereas the logarithmbased constraint (2) seems
to be a preferable constraint for use in the singlepath problem,obtaining the value s
n
(instead of log (s
n
))
via (3) becomes necessary in the twopath formulation.
We modify the reliablesinglepath formulation for the case in which at least one path must survive with
probability of at least ¿.Let binary decision variables x
q
ij
equal 1 if arc (i;j) is used in path q;8(i;j) 2
A;8q = 1;2,and let continuous decision variables s
q
i
equal the probability that path q successfully reaches
node i from node 1,for q = 1;2,given that path q visits node i.RTPD can be formulated as follows:
minimize
X
(i;j)2A
c
ij
¡
x
1
ij
+x
2
ij
¢
(4a)
subject to
X
j2FS(1)
x
q
1j
= 1 8q = 1;2 (4b)
X
j2FS(i)
x
q
ij
=
X
h2RS(i)
x
q
hi
8i 2 f2;:::;n ¡1g;8q = 1;2 (4c)
x
1
ij
+x
2
ij
· 1 8(i;j) 2 A (4d)
s
q
j
· p
ij
s
q
i
+(1 ¡x
q
ij
) 8q = 1;2;8(i;j) 2 A (4e)
s
1
1
= s
2
1
= 1 (4f)
s
1
n
+s
2
n
¡s
1
n
s
2
n
¸ ¿ (4g)
s
1
n
¸ s
2
n
(4h)
x
q
ij
2 f0;1g 8q = 1;2;8(i;j) 2 A:(4i)
The objective (4a) minimizes the cost of the two paths.Constraints (4b) and (4c) are path °ow balance
constraints,while (4d) enforces arc disjointness.The de¯nition of the svariables is stated by (4e) and (4f).
Constraint (4g) states that the probability that at least one path survives must be at least ¿.Note that (4h)
removes some symmetry present in the model since the feasible region for which s
1
n
· s
2
n
is identical to the
region where s
1
n
¸ s
2
n
.However,constraint (4g) induces a nonconvex feasible region in the (s
1
n
;s
2
n
) space,
even when integrality of the xvariables is relaxed,and presents several di±culties within a typical integer
programming branchandbound algorithm.
Remark 1.
One situation that can arise in our problem is that multiple pairs of paths exist that satisfy
the threshold probability constraint,such that the costs of these pairs of paths are equal.We may choose
to view this problem as a multiobjective optimization problem with a secondary objective that maximizes
the highest probability of success from among all lowestcost path pairs that satisfy the threshold reliability
constraint.To partially address this multiobjective view,we can subtract ®
¡
s
1
n
+s
2
n
¢
from the objective
function for some scalar ®.
In order to determine an appropriate value for ®,we assume that c
ij
¸ 1 8(i;j) 2 Aand c
ij
2 Z 8(i;j) 2 A,
where Z is the set of all integers.Given this assumption,a suboptimal solution objective exceeds the optimal
7
objective by at least one.Since we must ¯rst ¯nd the least expensive path pair that meets the threshold
probability,and then maximize the probability of those routes with equal cost,the contribution from the s
1
n
and s
2
n
terms must be strictly less than one.Since s
1
n
+s
2
n
· 2,an appropriate value for ® would be any
value in the range 0 < ® < 1=2.
Of course,even this additional term does not necessarily maximize the nonconvex term s
1
n
+s
2
n
¡s
1
n
s
2
n
,
which cannot be achieved using a simple bicriteria linear objective.However,consider the situation where
we have a choice between two solutions,each with the same total cost,and each with an identical sum of
s
1
n
+s
2
n
.As s
1
n
¡s
2
n
increases,the probability that at least one path survives also increases.Thus we might
also seek to maximize s
1
n
¡s
2
n
as part of the secondary objective.With this in mind,we can replace (4a)
with the following:
minimize
X
(i;j)2A
c
ij
¡
x
1
ij
+x
2
ij
¢
¡®
¡
s
1
n
+s
2
n
¢
¡¯
¡
s
1
n
¡s
2
n
¢
:
The requirement that ®
¡
s
1
n
+s
2
n
¢
+¯
¡
s
1
n
¡s
2
n
¢
< 1 implies that ¯ < 1 ¡®.One can then tune ¯ relative
to ® to experiment in ¯nding the best set of parameters that encourages maximumreliable path pairs from
among alternative optimal solutions.2
3.2 Algorithmic Strategies
A relaxation of RTPD can be formed by replacing (4g) with a linear constraint that forms a necessary
condition for (4g) to hold true.Noting that s
1
n
and s
2
n
are both nonnegative,(4g) implies that
s
1
n
+s
2
n
¸ ¿:
This constraint is akin to underestimating the constraint (4g) with a linear function that intersects (4g) at
s
1
n
= 0 and at s
2
n
= 0.However,with the symmetry condition that s
1
n
¸ s
2
n
,a stronger valid inequality
would underestimate this function with a linear constraint that intersects (4g) at the points where s
1
n
= s
2
n
and where s
2
n
= 0.The former point can be calculated as s
1
n
= s
2
n
= 1 ¡
p
1 ¡¿,while the latter point is
given by s
1
n
= ¿;s
2
n
= 0.These points give us the linear constraint
(1 ¡
p
1 ¡¿)s
1
n
+(
p
1 ¡¿ ¡1 +¿)s
2
n
¸ ¿(1 ¡
p
1 ¡¿):(5)
We state this constraint as a necessary condition within the mixedinteger relaxation program,and solve
it to optimality.For the remainder of this paper,de¯ne the relaxed mixedinteger program
RTPD to be
given by model (4),with (4g) replaced with (5).We now investigate several approaches for solving problem
RTPD.Suppose (^s
1
n
;^s
2
n
) solves
RTPD.If in fact we have that ^s
1
n
+^s
2
n
¡^s
1
n
^s
2
n
¸ ¿,then this solution must
be optimal to RTPD itself.Otherwise,we state that this solution is infeasible with respect to our nonlinear
constraint,which we hereafter call a nonlinearinfeasible solution.To eliminate this infeasibility,we employ
a branchandbound algorithm.
8
Suppose that we have restricted s
1
n
to lie within some interval [a;b],where 1¡
p
1 ¡¿ · a < b · 1.De¯ne
¹
b = minfb;¿g.We generate the chord of the function (4g) that underestimates the function on the interval
s
1
n
2 [a;
¹
b] and touches (4g) exactly at s
1
n
= a and s
1
n
=
¹
b.This inequality is of the form
µ
¿ ¡a
1 ¡a
¡
¿ ¡
¹
b
1 ¡
¹
b
¶
s
1
n
+
¡
¹
b ¡a
¢
s
2
n
¸
¹
b
¿ ¡a
1 ¡a
¡a
¿ ¡
¹
b
1 ¡
¹
b
:(6)
Suppose that a nonlinearinfeasible solution with values (^s
1
n
;^s
2
n
) optimizes
RTPD over the interval (a;b).
Since the linear approximation of (4g) is exact at a and
¹
b,we have that a < ^s
1
n
<
¹
b.We split the current
interval on s
1
n
into two segments:[a;^s
1
n
] and [^s
1
n
;b].In the ¯rst interval,the linear chord that underestimates
(4g) is improved by intersecting that function at endpoints a and ^s
1
n
.The second interval is approached
similarly.Both intervals are then recursively solved in this fashion.Observe that in each of these intervals,
the previous solution can no longer be feasible,and will not be regenerated.Also observe that a lower bound
for the optimal objective function value is obtained for an interval each time it is solved.If this lower bound
is not less than the best solution found so far,then this interval search is fathomed.
Figure 3 illustrates this technique.In this ¯gure,¿ = 0:8 and the feasible region lies below the line s
1
n
= s
2
n
and above the curve s
1
n
+s
2
n
¡s
1
n
s
2
n
= 0:8.The relaxed feasible region used for
RTPD is bounded above by
s
1
n
= s
2
n
and below by the dashed line 0:5s
1
n
+0:2s
2
n
= 0:4.The solution (^s
1
n
;^s
2
n
) does not satisfy RTPD,but
it could optimize
RTPD.The ¯rst iteration of the branchandbound algorithm will partition the original
search area into regions I and II,and search those two regions in a recursive manner.
2
n
s
1
n
s
0
0.10.20.30.40.50.60.70.80.9
1
00.10.20.30.40.50.60.70.80.91
21
nn
ss=
(
)
21
ˆ
,
ˆ
nn
ss
−
−
1
1
1
ˆ
1
ˆ
,
ˆ
n
n
n
s
s
s
τ
III
Figure 3:Illustration of First Iteration of Vertical BranchandBound Algorithm
9
In the process described by Figure 3,we branch on s
1
n
,where ^s
1
n
2 [a
1
;b
1
],and create search intervals
in which s
1
n
belongs to [a
1
;^s
1
n
] or [^s
1
n
;b
1
].On the other hand,we could also choose to branch on s
2
n
,where
^s
2
n
2 [a
2
;b
2
] and create search intervals in which s
2
n
belongs to [a
2
;^s
2
n
] or [^s
2
n
;b
2
].Finally,we could branch
in a nonaxial manner by identifying some point (c
1
;c
2
) on the constraint (4g),where c
1
2 [a
1
;b
1
] and
c
2
= [(¿ ¡c
1
) =(1 ¡c
1
)] 2 [a
2
;b
2
],and creating search intervals in which s
q
n
belongs to [a
q
;c
q
] and [c
q
;b
q
]
for q = 1;2.
A wellknown key to accelerating the convergence of branchandbound algorithms on minimization prob
lems is to partition active nodes in a manner that forces the minimum objective of the two new subproblems
to be as large as possible.While a strong branching algorithmis computationally prohibitive (due to the fact
that each subproblem is a mixedinteger program),we can encourage this desired behavior by maximizing
the minimum Euclidean distance from
¡
^s
1
n
;^s
2
n
¢
to either of the new partitioning planes in the
¡
s
1
n
;s
2
n
¢
space.
For this discussion let
¹
b
2
= min
©
b
2
;1 ¡
p
1 ¡¿
ª
,where
¹
b
2
represents the greatest allowable value of s
2
n
in
the search area that also intersects (4g).For the case in which we branch on s
1
n
,called a vertical partition,
the minimum distance from
¡
^s
1
n
;^s
2
n
¢
to a partitioning plane is given by
D
V
=
¡
¹
b
1
¡ ^s
1
n
¢
³
¿¡^s
1
n
1¡^s
1
n
¡ ^s
2
n
´
r
¡
¹
b
1
¡ ^s
1
n
¢
2
+
³
¹
b
2
¡
¿¡^s
1
n
1¡^s
1
n
´
2
:(7)
For the case in which we branch on s
2
n
,called a horizontal partition,the minimum distance is given by
D
H
=
¡
a
2
¡ ^s
2
n
¢
³
¿¡^s
2
n
1¡^s
2
n
¡ ^s
1
n
´
r
(a
2
¡ ^s
2
n
)
2
+
³
a
1
¡
¿¡^s
2
n
1¡^s
2
n
´
2
:(8)
Hence,one partitioning policy would branch on s
1
n
if D
V
¸ D
H
,and would otherwise branch on s
2
n
.
Note that a\perfect"partition would choose a partitioning for which the distances from
¡
^s
1
n
;^s
2
n
¢
to either
partitioning plane are equal,thereby maximizing the minimum distance to either partitioned region.Recall
that we can choose a point (c
1
;c
2
) that lies on (4g) such that c
1
2 (a
1
;
¹
b
1
),and partition the feasible
region into areas where s
1
n
2 [a
1
;c
1
],and where s
1
n
2 [c
1
;b
1
].We can employ a simple binary search to
locate the point (c
1
;c
2
) on (4g) that yields this equidistant partitioning.A chordal constraint intersecting
(a
1
;[(¿ ¡a
1
)=(1 ¡a
1
)]) and (c
1
;c
2
) would be imposed for the former region,while a chordal constraint
intersecting (c
1
;c
2
) and (b
1
;[(¿ ¡b
1
)=(1 ¡b
1
)]) would be imposed for the latter region.We refer to this
partitioning scheme as diagonal branching.
Remark 2.
The ReformulationLinearization Technique (RLT) for continuousvariables nonlinear pro
gramming problems (Sherali and Adams 1999,Sherali and Tuncbilek 1992),can be applied to this problem
as an alternative to our branching scheme.When applied to the rectangular region tightly surrounding the
relaxation for our problem in which a
1
· s
1
n
· b
1
and [(¿ ¡b
1
) = (1 ¡b
1
)] · s
2
n
· [(¿ ¡a
1
) = (1 ¡a
1
)],RLT
10
replaces the s
1
n
s
2
n
term in (4g) with a variable w,and introduces inequalities based on the upper and lower
bounds on s
1
n
and s
2
n
.There exist four inequalities that we can obtain from these bounds:
¡
b
1
¡s
1
n
¢
µ
¿ ¡a
1
1 ¡a
1
¡s
2
n
¶
¸ 0 (9a)
¡
a
1
¡s
1
n
¢
µ
¿ ¡b
1
1 ¡b
1
¡s
2
n
¶
¸ 0 (9b)
¡
b
1
¡s
1
n
¢
µ
¿ ¡b
1
1 ¡b
1
¡s
2
n
¶
· 0 (9c)
¡
a
1
¡s
1
n
¢
µ
¿ ¡a
1
1 ¡a
1
¡s
2
n
¶
· 0:(9d)
Replacing the s
1
n
s
2
n
terms with w in equations (4g) and (9),we obtain the following relaxed linear con
straints:
w · s
1
n
+s
2
n
¡¿ (10a)
w ¸
¿ ¡a
1
1 ¡a
1
s
1
n
+b
1
s
2
n
¡b
1
¿ ¡a
1
1 ¡a
1
(10b)
w ¸
¿ ¡b
1
1 ¡b
1
s
1
n
+a
1
s
2
n
¡a
1
¿ ¡b
1
1 ¡b
1
(10c)
w ·
¿ ¡b
1
1 ¡b
1
s
1
n
+b
1
s
2
n
¡b
1
¿ ¡b
1
1 ¡b
1
(10d)
w ·
¿ ¡a
1
1 ¡a
1
s
1
n
+a
1
s
2
n
¡a
1
¿ ¡a
1
1 ¡a
1
:(10e)
Using FourierMotzkin elimination,the combinations of (10b) and (10d),(10c) and (10d),(10b) and (10e),
and (10c) and (10e) state the original constraints a
1
· s
1
n
· b
1
and [(¿ ¡b
1
) = (1 ¡b
1
)] · s
2
n
· [(¿ ¡a
1
) =(1 ¡a
1
)].
The combination of (10a) and (10b) produces the following inequality:
s
2
n
¸
¿ ¡1
(1 ¡a
1
) (1 ¡b
1
)
s
1
n
+
¿ ¡a
1
¿ ¡b
1
¿ +a
1
b
1
(1 ¡a
1
) (1 ¡b
1
)
;(11)
where (11) is in slopeintercept form,and is identical to (6).Combining (10a) and (10c) reveals the same
inequality.As a result,the RLT procedure of branching based on s
1
n
is identical to our vertical partition
scheme,while the RLTbranching on s
2
n
is identical to our horizontal partition scheme.However,the maximin
diagonal branching scheme is a problemspeci¯c strategy that does not appear to be readily obtainable from
the RLT process.2
3.3 Model Enhancements
As a precursor to this discussion,note that we can compute the most reliable path from each node i 2 N
to node n,or from node 1 to each node i 2 N,by using a simple variation of Dijkstra's algorithm.Using
this information,we can establish an upper bound on the probability of successfully reaching node i 2 N
from node 1,which we denote as u
i
.Also,given lower bounds a
q
on the minimum path probability for paths
q = 1 and 2 derived from the foregoing partitioning scheme,we derive a lowestpermissible probability of
reaching node i 2 N so that we can reach node n with probability at least a
q
,which we denote as`
q
i
.(This
11
i
`
1
i
`
2
i
u
i
v
1
i
v
2
i
m
i
A
0.81
0
1.00
4.17
4.17
1.00
B
0.64
0
0.80
1.79
1.79
0.80
C
0.72
0
0.60
2.50
2.50
0.60
D
0.56
0
0.70
2.38
2.38
0.70
E
0.45
0
0.56
1.11
1.11
0.36
F
0.58
0
0.64
1.43
1.43
0.42
G
0.51
0
0.63
1.25
1.25
0.30
H
0.41
0
0.50
1.00
1.00
0.24
Table 1:Initial Values for Pruning and Tightening
value is obtained by dividing a
q
by the most reliable path from node i to node n.) Observe that we can
possibly perform some preprocessing by noting that if u
i
p
ij
<`
q
j
for any (i;j) 2 A,then we cannot feasibly
visit node j via arc (i;j) on path q and can therefore ¯x x
q
ij
= 0.
Directed acyclic graphs provide another opportunity for preprocessing.For these graphs,we compute the
least reliable path from node 1 to each node i 2 N and from each node i 2 N to node n.(This task is
strongly NPhard for general graphs.) Denote m
i
as the smallest possible probability of reaching node i 2 N
fromnode 1,and v
q
i
as the highestpermissible probability of reaching node i 2 N such that it is still possible
to reach node n with probability no more than b
q
,where b
q
is the largest permissible path reliability for path
q for q = 1;2.This latter value is obtained by dividing b
q
by the least reliable path from node i to node
n.Similar preprocessing logic holds in this case for directed acyclic graphs:for all (i;j) 2 A,if m
i
p
ij
> v
q
j
,
then we ¯x x
q
ij
= 0.
For the example shown in Figure 1,for ¿ = 0:65,we have that 0:408 · s
1
n
· 1,and 0 · s
2
n
· 1,and
compute the values for`
q
i
,u
i
,v
q
i
,and m
i
as shown in Table 1.Immediately we see`
1
C
> u
C
and u
D
p
DF
<`
1
F
,
so we can set x
1
AC
= x
1
CE
= x
1
CF
= x
1
CG
= 0 and x
1
DF
= 0.No arcs may be removed for the second path.
Once a variable is ¯xed to zero by this process,this ¯xing remains valid for the remaining exploration
of that branch of the branchandbound tree,since the child nodes contain intervals on s
1
n
and s
2
n
that are
subsets of the parent interval.We de¯ne A
q
as the set of all arcs (i;j) 2 A such that x
q
ij
has not been ¯xed
to zero,and will henceforth tailor our methods to arcs in this set.The coe±cient tightening and lifting
techniques that we discuss below is only valid after the graph has been pruned as described above.
Next,note that (4e) essentially employs a\bigM"value of 1 as a coe±cient to the
¡
1 ¡x
q
ij
¢
term,in order
to disable the constraint when x
q
ij
= 0.However,we may tighten constraint (4e) by reducing the coe±cient
of this term and by investigating certain constraintlifting techniques.We begin by rewriting constraint (4e)
as
s
q
j
· p
ij
s
q
i
+°
q
ij
(1 ¡x
q
ij
) 8q = 1;2;8(i;j) 2 A
q
;(12)
12
where we have previously set °
q
ij
= 1.For the case in which arc (i;j) is used in path q,the value of °
q
ij
is not
signi¯cant.However,if x
q
ij
= 0,then for this particular (i;j) 2 A
q
and q = 1;2,(12) becomes the inequality
s
q
j
· p
ij
s
q
i
+°
q
ij
8q = 1;2;8(i;j) 2 A
q
:(13)
In order to tighten (13),we seek the smallest valid value for °
q
ij
,i.e.,we minimize °
q
ij
such that °
q
ij
¸ s
q
j
¡p
ij
s
q
i
for any feasible choice of s
q
i
and s
q
j
.Noting that the maximum value of s
q
j
is u
j
and the minimum value of
s
q
i
is`
q
i
,we restate (4e) as:
s
q
j
· p
ij
s
q
i
+(u
j
¡p
ij
`
q
i
) (1 ¡x
q
ij
) 8q = 1;2;8(i;j) 2 A
q
:(14)
We can perform lifting for this constraint by considering the consequences on the upper bound of s
q
j
when
a path to j that does not include arc (i;j) is selected.If x
q
ij
= 0,constraint (14) sets an upper bound on s
q
j
to be at least u
j
.However,if path q uses arc (k;j) 2 A
q
;k 6= i,then the upper bound on s
q
j
can be reduced
to u
k
p
kj
.Similarly,if path q uses arc (i;k) 2 A
q
;k 6= j,then the lower bound on s
q
i
can be increased to
`
q
k
=p
ik
.With this information,we can lift (14) as shown below:
s
q
j
· p
ij
s
q
i
+(u
j
¡p
ij
`
q
i
) (1 ¡x
q
ij
) ¡
X
k2RS(j);k6=i
(u
j
¡p
kj
u
k
) x
kj
¡
X
k2FS(i);k6=j
µ
`
q
k
p
ik
¡`
q
i
¶
p
ij
x
ik
8q = 1;2;8(i;j) 2 A
q
:(15)
4 Robust NonIndependent Paths
Although we have previously limited our discussion only to arcindependent paths,it may be preferable to
share arcs on both paths,while incurring the risk that a single shared arc failure may ruin both paths.We
discuss the formulation and algorithmadjustments required to solve the sharedarcs variation of our problem
(RTPS) in this section.
First,we de¯ne z
ij
as a binary variable equal to 1 if arc (i;j) 2 A is shared by both paths and equal to 0
otherwise.We revise our de¯nition of x
q
ij
to equal 1 if and only if arc (i;j) 2 A is used only by path q = 1
or 2 (and not by the other path).Hence,if z
ij
= 1,this implies that both x
1
ij
and x
2
ij
will equal 0.
Using these new variable de¯nitions,we adjust the objective function as follows:
minimize
X
(i;j)2A
c
ij
¡
x
1
ij
+x
2
ij
+2z
ij
¢
(16)
13
The °ow balance and capacity constraints now become
X
j2FS(1)
x
q
1j
+z
1j
= 1 8q = 1;2 (17a)
X
j2FS(i)
x
q
ij
+z
ij
=
X
h2RS(i)
x
q
hi
+z
ij
8i 2 f2;:::;n ¡1g;q = 1;2 (17b)
x
1
ij
+x
2
ij
+z
ij
· 1 8(i;j) 2 A (17c)
x
q
ij
2 f0;1g 8q = 1;2;8(i;j) 2 A;z
ij
2 f0;1g 8(i;j) 2 A:(17d)
The next challenge is in stating the probability threshold constraint.The probability that at least one
path remains operational is given by [probability that all shared arcs remain operational] £ (1 ¡ [probability
that both paths fail due to nonshared arc failures]).We further specify our de¯nition of s
q
i
in this case to
denote the probability that path q = 1;2 does not fail en route to node i 2 N due to a nonshared arc failure.
Hence,(4e) and (4f) become
s
q
j
· p
ij
s
q
i
+(1 ¡x
q
ij
) 8q = 1;2;8(i;j) 2 A (18a)
s
q
j
· s
q
i
+(1 ¡z
ij
) 8q = 1;2;8(i;j) 2 A (18b)
s
1
n
¸ s
2
n
(18c)
s
1
1
= s
2
1
= 1;(18d)
where we retain the symmetrybreaking constraint (18c).
Now s
1
n
+s
2
n
¡s
1
n
s
2
n
represents the probability that all the nonshared arcs on at least one path remain
operational.To assess the probability that a shared arc fails,we then retrace through a path and compute
the likelihood that this path fails due to sharedarc failure.(We choose path 1 arbitrarily;the sharedarc
failure probability is by de¯nition equal for either path.) Toward this end,de¯ne S
i
to be the probability
that at least one path of nonshared arcs remains operational,and that the shared arcs on path 1 remain
operational en route to node i 2 N (where S
i
is once again irrelevant if path 1 does not visit node i).We
enforce this mechanism as follows:
S
1
· s
1
n
+s
2
n
¡s
1
n
s
2
n
(19a)
S
j
· p
ij
S
i
+(1 ¡z
ij
) 8(i;j) 2 A (19b)
S
j
· S
i
+
¡
1 ¡x
1
ij
¢
8(i;j) 2 A (19c)
S
n
¸ ¿:(19d)
For the remainder of this paper,de¯ne RTPS to be given by (16),(17),(18),and (19).Again,we seek
an appropriate relaxation of (19a),by controlling the values of s
1
n
and s
2
n
.A conic partitioning scheme
proves di±cult here due to the fact that (19a) is neither a convex nor a concave constraint.Instead,we can
implement the continuousvariables RLT scheme.The equivalent of (10) for RTPS is as follows,where we
14
replace w = s
1
n
s
2
n
,and where s
1
n
2 [a
1
;b
1
] and s
2
n
2 [a
2
;b
2
].
w · s
1
n
+s
2
n
¡S
1
(20a)
w ¸ b
2
s
1
n
+b
1
s
2
n
¡b
1
b
2
(20b)
w ¸ a
2
s
1
n
+a
1
s
2
n
¡a
1
a
2
(20c)
w · a
2
s
1
n
+b
1
s
2
n
¡b
1
a
2
(20d)
w · b
2
s
1
n
+a
1
s
2
n
¡a
1
b
2
:(20e)
Note that (20d) and (20e) are not necessary to include in our formulation,since these constraints represent
upper bounds on w.Since (19a) will force w = s
1
n
s
2
n
to be as small as possible,only the lower binding
constraints (20b) and (20c) are necessary to retain.The projection of (20a),(20b),and (20c) onto the
¡
s
1
n
;s
2
n
;S
1
¢
space is de¯ned by the following inequalities:
S
1
· (1 ¡b
2
)s
1
n
+(1 ¡b
1
)s
2
n
+b
1
b
2
(21a)
S
1
· (1 ¡a
2
)s
1
n
+(1 ¡a
1
)s
2
n
+a
1
a
2
:(21b)
De¯ne
RTPS as the relaxation of RTPS,with (19a) replaced by (21).Also,de¯ne ¾
n
as the product of the
reliabilities of all shared arcs,i.e.,¾
n
=
Q
(i;j)2A
p
z
ij
ij
.A solution (^x;^z;^s;
^
S) to
RTPS also satis¯es RTPS if
^¾
n
¡
^s
1
n
+ ^s
2
n
¡ ^s
1
n
^s
2
n
¢
¸ ¿.
The procedure for solving RTPS is similar to that described in Section 3.2.The primary di®erence is
that instead of maintaining bounds for either s
1
n
or s
2
n
,we record bounds on both variables.We initially
solve
RTPS over s
1
n
2 [1 ¡
p
1 ¡¿;1] = [a
1
;b
1
] and s
2
n
2 [0;1] = [a
2
;b
2
].If the solution is not feasible to
RTPS,initialize a list of active subproblems with the intervals as described above,and with the interval
optimal solution of (^s
1
n
;^s
2
n
).If we are branching on s
1
n
,the active subproblems are divided into two search
intervals,the ¯rst with intervals s
1
n
2 [a
1
;^s
1
n
] and s
2
n
2 [a
2
;b
2
],and the second with intervals:s
1
n
2 [^s
1
n
;b
1
]
and s
2
n
2 [a
2
;b
2
].We can take advantage of the symmetry constraint to tighten the upper bound on s
2
n
in
the ¯rst interval to [a
2
;^s
1
n
].On the other hand,if we choose to branch on s
2
n
,the two search intervals created
are s
1
n
2 [a
1
;b
1
];s
2
n
2 [a
2
;^s
2
n
] and s
1
n
2 [a
1
;b
1
];s
2
n
2 [^s
2
n
;b
2
].
As would be expected,model tightening and preprocessing is also less straightforward for RTPS.Initially,
in light of our new de¯nition of s
q
i
,we recognize that u
i
should be equal to 1 for any i 2 N,since we could
choose to share all arcs on the path to 1.Calculating`
q
i
as a
q
(the minimum feasible value of s
q
n
) divided
by the most reliable path to n from i (which is now 1 since all remaining arcs may be shared),we obtain
`
q
i
= a
q
8q = 1;2;8i 2 N.Hence,while our previous pruning rules for u
i
and`
q
i
still apply to RTPS,we
must ¯rst be able to tighten the bounds on s
q
i
.
15
In order to accomplish this boundtightening,we examine conditions under which certain z
ij
variables
can be set to zero.If p
ij
< ¿,then z
ij
= 0 in any feasible solution,and hence we set z
ij
= 0 8(i;j) 2 A
where p
ij
< ¿.(This is equivalent to de¯ning L
i
= ¿ and U
i
= 1 as the lower and upper bounds on S
i
,
respectively,and then pruning z
ij
if U
i
p
ij
< L
i
.) In the example given by Figure 1,if ¿ = 0:65,then we ¯x
z
AC
= z
CE
= z
CG
= z
DF
= 0.As before,de¯ne A
q
as the set of all arcs (i;j) 2 A such that x
q
ij
has not
been ¯xed to zero by pruning and preprocessing,and de¯ne A
z
as the set of all arcs (i;j) 2 A such that z
ij
has not been ¯xed to zero.
With this information,we may be able to improve the bounds for u
i
and`
q
i
.Previously it was necessary to
set the upper bound u
i
= 1,representing the possibility that all arcs on the path to node i could be shared.
After pruning arcs to obtain A
z
,however,the situation can arise where at least one arc on any path from
the origin to a node k 2 N cannot be shared;that is,any path to node k uses at least one arc (i;j) 2 A
q
,
where (i;j) 62 A
z
.This is the case for node C in Figure 1.Any path from the origin node to node C must
use (A;C) 2 A
q
where (i;j) 62 A
z
,implying that u
C
= 0:6.Similarly,if at least one arc cannot be shared
on any path from a node k to the destination node,then`
q
k
is again calculated as a
q
divided by the most
reliable path to n from k.This path will involve at least one nonshared arc,and will have a probability less
than 1.For directed,acyclic graphs,we can still employ the same pruning rules regarding m
i
and v
q
i
for
RTPS as we did on RTPD.
One ¯nal pruning step is then taken with respect to the z
ij
variables.Note that if u
i
<`
q
j
for any
(i;j) 2 A;q = 1;2,then no feasible solution exists that shares the arc (i;j).Otherwise,path q would arrive
at node i with s
q
i
· u
i
,and if (i;j) is shared,we would have s
q
j
· u
i
<`
q
j
.We use this information to prune
the graph again and update A
z
accordingly.
We have observed that the upper and lower bounds on S
i
are L
i
= ¿ and U
i
= 1 8i 2 N.(The concepts
of m
i
and v
q
i
for s
q
n
do not have equivalent representations for S
i
without a nontrivial upper bound on S
n
.)
By analysis similar to that presented in Section 3.3,we can tighten the coe±cients on (1¡x
q
ij
) and (1¡z
ij
)
in (18a),(18b),(19b),and (19c) as shown below:
s
q
j
· p
ij
s
q
i
+(u
j
¡p
ij
`
q
i
) (1 ¡x
q
ij
) 8q = 1;2;8(i;j) 2 A
q
(22a)
s
q
j
· s
q
i
+(u
j
¡`
q
i
) (1 ¡z
ij
) 8q = 1;2;8(i;j) 2 A
z
(22b)
S
j
· p
ij
S
i
+(1 ¡p
ij
¿) (1 ¡z
ij
) 8(i;j) 2 A
z
(22c)
S
j
· S
i
+(1 ¡¿)
¡
1 ¡x
1
ij
¢
8(i;j) 2 A
1
:(22d)
Again,it is important to note that this coe±cient tightening and the lifting described below is only valid
after the graph has been fully pruned according to the foregoing procedures.We use A
1
in (22d) since
a
1
¸ a
2
and path 1 will thus experience more pruning,making jA
1
j · jA
2
j.
16
Constraint (22a) can be lifted precisely as in (15),resulting in the following lifted inequality:
s
q
j
· p
ij
s
q
i
+(u
j
¡p
ij
`
q
i
) (1 ¡x
q
ij
) ¡
X
k2RS(j);k6=i
(u
j
¡p
kj
u
k
) x
kj
¡
X
k2FS(i);k6=j
µ
`
q
k
p
ik
¡`
q
i
¶
p
ij
x
ik
8q = 1;2;8(i;j) 2 A
q
:(23)
For constraint (22b),if path q uses the shared arc (k;j) 2 A;k 6= i,then the upper bound on s
q
j
can be
reduced to u
k
,and if path q uses the shared arc (i;k) 2 A;k 6= j,then the lower bound on s
q
i
can be increased
to`
q
k
.The resulting lifted inequality is given by:
s
q
j
· s
q
i
+(u
j
¡`
q
i
) (1 ¡z
ij
) ¡
X
k2RS(j);k6=i
(u
j
¡u
k
) z
kj
¡
X
k2FS(i);k6=j
(`
q
k
¡`
q
i
) z
ik
8q = 1;2;8(i;j) 2 A
z
:(24)
For constraint (22c),if any path uses the shared arc (k;j) 2 A;k 6= i,then the upper bound on S
j
can be
reduced to p
kj
,and if any path uses the shared arc (i;k) 2 A;k 6= j,then the lower bound on S
i
can be
increased to ¿=p
ik
,yielding the lifted inequality:
S
j
· p
ij
S
i
+(1 ¡p
ij
¿) (1 ¡z
ij
) ¡
X
k2RS(j);k6=i
(1 ¡p
kj
) z
kj
¡
X
k2FS(i);k6=j
µ
1
p
ik
¡1
¶
¿p
ij
z
ik
8(i;j) 2 A
z
:(25)
Finally,for constraint (22d),if either path uses the nonshared arc (k;j) 2 A;k 6= i,then there is no net
e®ect on the upper bound on S
j
,and if either path uses the nonshared arc (i;k) 2 A;k 6= j,then again,
there is no net e®ect on the lower bound on S
i
.Therefore,we will not perform lifting on (22d).
5 Computational Results
In this section,we evaluate the computational e±cacy of our various strategies for solving RTPD and
RTPS.First,we consider the task of ¯nding the most reliable path pair in
RTPD from among multiple
optimal solutions.Second,we will compare the di®erent pruning,coe±cient tightening,and lifting strategies
discussed for
RTPD.Our third comparison will attempt to determine the most e®ective partitioning strategy
for RTPD out of those discussed in Section 3.2.Finally,we will examine pruning,tightening,and lifting for
RTPS,and then use the best strategy from this experiment to compare partitioning strategies for RTPS.
All computations were done on a 500 Mhz Sun Blade 100 running Solaris version 5.8 with 1.5 GB of installed
memory.All computational times listed are in CPU seconds.Linear and integer programming problems
were solved using CPLEX 8.1.
Problem Set Generation.
Three problem sets were necessary for the di®erent computational compar
isons.We created Set 1,a test set of 40 directed,acyclic networks,to examine the impact of secondary
17
objective criteria in ¯nding mostreliable alternative optimal solutions to
RTPD instances.Four subsets of
instances were generated,one for each combination of n = 50 or 75,and arc density of 30% or 70%.To
generate a graph with roughly d% arc density,for each possible (i;j) node pair,i < j,a random number was
generated with a uniform distribution between 0 and 1,hence prohibiting the generation of directed cycles.
An arc was generated between node i and node j if and only if the randomly generated number was not
more than d%.We prohibited the generation of an arc connecting node 1 directly to node n.Ten instances
were generated for each of the four subsets,for a total of 40 instances.For each of these instances,each arc
cost was generated according to a uniform distribution of the integers 1;:::;10.We created instances in this
manner with the density and node requirements until ten instances were discovered in each group that had
multiple optimal values of s
1
n
+s
2
n
¡s
1
n
s
2
n
and satis¯ed
RTPD.
Next,we created Set 2 instances to compare the pruning,lifting,and coe±cient tightening methods for
RTPD and
RTPS.We generated 20 directed,acyclic graphs for each combination of total nodes and arc
densities,where a graph could have 25,50,75,or 100 nodes,and could have an arc density of 20%,50%,
or 80%,for a total of 240 instances.The arc density was enforced as above,as was the exclusion of any arc
(1;n) from A.Arc costs were assigned by generating a random number with a uniform distribution between
0 and 100.
To compare the di®erent partitioning strategies discussed in Section 3.2,we generated Set 3 instances in a
similar manner as the Set 2 instances.This set consists of 20 directed,acyclic graphs for each combination
of total nodes and arc densities,for combinations of n = 25;50,or 75,and arc densities of 20%,50%,or
80%,for a total of 180 instances.Arc density requirements were met as for previous problem sets and arc
costs again were uniformly distributed real numbers between 0 and 100.Problem instances were generated
in this manner until 20 problems instances in each set were found that solved
RTPD,but that returned a
nonlinearinfeasible solution.We used the same problem instances to compare computational times for the
two partitioning methods for RTPS as described in Section 4,noting that occasionally problems requiring
branching for RTPD will not require branching for RTPS.
Remark 3.
Due to the inequalities in our problem formulation,the solver may return a value of ¹s
1
n
,
for example,which is lower than the actual (nonshared) path reliability ^s
1
n
found by plugging in values
of the x
1
variables,to allow the solution to\¯t"into the [a
1
;b
1
] range.Except in the case of the ¯rst
computational experiment where the coe±cients of s
1
n
and s
2
n
were set explicitly,the remaining experiments
used a coe±cient of 0.01 on s
1
n
,s
2
n
,and S
n
(when present) to help encourage maximum values of these
variables.When comparing values of path reliability among the methods,it is important to recalculate ^s
1
n
,
^s
2
n
,and
^
S
n
,based on the probabilities of the arcs actually used,rather than relying on the returned values.
It is on these\true"values,^s
1
n
,^s
2
n
,and
^
S
n
,that branching decisions should be made,instead of on the
values returned by the solver,in order to avoid possibly regenerating the same solution.2
18
Maximizing Reliability as a Secondary Objective for
RTPD.
We ¯rst investigated the impact of
di®erent values of ® and ¯ on ¯nding the best reliability pair among alternative optimal solutions satisfying
RTPD.Four di®erent sets of values for ® and ¯ were tested.Parameter Combination 1 (PC1) used ® = 0:01
and ¯ = 0,PC2 used ® = 0:3 and ¯ = 0:15,PC3 used ® = 0:3 and ¯ = 0:3,and PC4 used ® = 0:3 and
¯ = 0:65.
Out of the 40 Set 1 problem instances tested by varying the values of ® and ¯,the four parameter
combinations above found alternative optimal solutions with di®erent total reliability values (given by s
1
n
+
s
2
n
¡ s
1
n
s
2
n
) for 21 of the instances.On each of these 21 instances,PC2 found the best such alternative
optimal solution out of all combinations each time.As there was no clear advantage from using any of the
parameter combinations insofar as computational times,we conclude that for graphs having integer costs,
using ® = 0:3 and ¯ = 0:15 is the most e®ective of our parameter sets for ¯nding an optimal solution with
the highest total reliability.
Comparison of Pruning and Lifting Strategies for
RTPD.
We next compared the solution times
for
RTPD on Set 2 instances when pruning,coe±cient tightening,and lifting strategies were implemented.
Each instance was solved by four methods.Method I solved
RTPD with no pruning or lifting.Method
II solved
RTPD after preprocessing and pruning the graph using information about u
i
and`
q
i
;8i 2 N.
Method III solved
RTPD after pruning as in Method II and used the coe±cient tightening on the (1 ¡x
ij
)
term as suggested in (14).Method IV expanded upon Methods II and III by using the pruning,coe±cient
tightening,and lifting strategy described in (15).
Table 2 shows the average and standard deviations of the computational times by each method for each
instance group.As the problem instances get larger,Method III proves to be the most e±cient solution
method,with Method II being just slightly less e®ective.Note that for the problem instances of size 100,
density 50%,two outliers in Method I caused the average solution time to be signi¯cantly larger than that
for the problem instances of the same size but higher density.This is re°ected in the standard deviation.
The results from this comparison beg the question of whether or not there exists a threshold value of the
lifting coe±cients over which it becomes e®ective to use lifting along with pruning and coe±cient tightening.
We ran comparisons allowing only those lifting coe±cients over a certain value to be included in the s
q
j
constraints;this conditional coe±cient lifting never provided a consistently faster solving time than Method
III,regardless of the selected threshold.
Comparison of Partitioning Strategies for RTPD.
In our next experiment,we compared the solution
times for ¯ve di®erent partitioning strategies when solving RTPD on Set 3 instances.Strategy I always
used horizontal partitioning and Strategy II always used vertical partitioning.Strategy III used vertical
or horizontal partitioning to move as far as possible from the nearest relaxed feasible point,according to
equations (7) and (8).Strategy IV used vertical or horizontal partitioning to move as close as possible to the
nearest relaxed feasible point,and Strategy V chose a partitioning point where the distances to either relaxed
19
Size
Method
20% Density
50% Density
80% Density
Standard
Standard
Standard
Average
Deviation
Average
Deviation
Average
Deviation
25
I
0.08
0.03
0.80
1.38
2.42
4.38
25
II
0.04
0.02
0.15
0.20
0.42
0.73
25
III
0.04
0.02
0.13
0.12
0.28
0.22
25
IV
0.05
0.02
0.17
0.22
0.43
0.25
50
I
0.50
0.32
48.13
176.63
40.89
85.48
50
II
0.15
0.05
1.71
4.05
3.23
3.88
50
III
0.14
0.05
0.62
0.36
2.38
2.10
50
IV
0.17
0.08
1.47
1.04
6.19
3.96
75
I
5.12
7.10
43.54
74.04
100.87
190.72
75
II
0.39
0.22
7.07
14.38
7.79
6.87
75
III
0.38
0.23
6.90
15.17
7.83
7.92
75
IV
0.52
0.32
13.52
23.84
27.63
22.91
100
I
46.00
99.81
1392.36
4308.57
681.43
1589.70
100
II
2.56
7.40
16.30
36.99
39.32
67.69
100
III
1.30
2.16
12.43
29.47
35.02
62.24
100
IV
1.72
1.60
25.15
29.74
79.53
52.39
Table 2:Comparison of Computation Times for Pruning and Lifting Strategies for
RTPD
feasible region were equal and partitioned the feasible region from that point horizontally.We used pruning
with coe±cient tightening on the (1 ¡x
q
ij
) term as recommended by the previous experiment (Method III),
and stopped the branchandbound procedure when ¿ ¡
¡
s
1
n
+s
2
n
¡s
1
n
s
2
n
¢
<",where"= 0:00001.
Table 3 provides information about the average computation time per instance group,as well as the
standard deviation of those times.While these average times are all roughly equal for small RTPDinstances,
Strategy V becomes dominant as the problem size and density increases.Hence,we recommend Strategy V
in conjunction with pruning and coe±cient tightening for solving RTPD instances.
Comparison of Pruning Strategies for
RTPS.
Similar to the comparison for
RTPD,we compared
pruning,coe±cient tightening,and lifting strategies for
RTPS on Set 2 instances.Four methods were used
to solve each problem instance.Method I used no pruning or lifting,while Method II used pruning only.
Method III solved
RTPS using pruning and the coe±cient tightening described in (22),and Method IV
used pruning,coe±cient tightening,and lifting,as illustrated in (23),(24),and (25).
Table 4 demonstrates that the computational times for solving
RTPS exhibit signi¯cantly di®erent be
havior than for
RTPD.In this case,Method II had a faster average solution time,implying that coe±cient
tightening is not providing an advantage for
RTPS.This is perhaps due to the limited degree of tightening
that can be performed when arcs are allowed to be shared.Note that one problem instance in the size 50,
50% density group failed to solve within the three hour time limit,as did two instances in the size 100,50%
density group,and one in the size 100,80% density group.The associated standard deviations re°ect these
extremes.
For further clari¯cation on the extent to which coe±cient tightening might provide a computational advan
tage over using Method II,we ran an additional test comparing computational times on four new methods,
20
Size
Strategy
20% Density
50% Density
80% Density
Standard
Standard
Standard
Average
Deviation
Average
Deviation
Average
Deviation
25
I
0.09
0.03
0.22
0.14
0.53
0.39
25
II
0.10
0.03
0.27
0.18
0.66
0.41
25
III
0.10
0.03
0.22
0.13
0.53
0.39
25
IV
0.10
0.04
0.27
0.19
0.66
0.40
25
V
0.08
0.03
0.21
0.13
0.51
0.36
50
I
0.25
0.08
2.06
2.66
6.19
7.23
50
II
0.33
0.14
3.07
3.83
9.09
11.01
50
III
0.26
0.08
2.04
2.63
6.10
6.88
50
IV
0.34
0.15
3.07
3.83
9.22
11.24
50
V
0.25
0.08
1.93
2.46
6.34
6.98
75
I
0.81
0.31
6.44
8.07
31.10
46.62
75
II
0.98
0.46
9.62
9.29
34.09
40.10
75
III
0.83
0.33
6.45
8.10
31.40
46.79
75
IV
0.95
0.48
9.64
9.28
34.11
40.14
75
V
0.78
0.30
6.28
8.06
27.34
33.39
Table 3:Comparison of Computation Times for Partitioning Strategies for RTPD
Size
Method
20% Density
50% Density
80% Density
Standard
Standard
Standard
Average
Deviation
Average
Deviation
Average
Deviation
25
I
0.14
0.08
3.69
8.34
8.79
16.43
25
II
0.06
0.03
0.45
0.66
2.20
5.24
25
III
0.06
0.03
0.49
0.81
2.28
5.04
25
IV
0.06
0.04
0.57
0.82
3.03
5.23
50
I
2.38
3.16
590.66
2405.15
139.54
228.16
50
II
0.27
0.14
5.87
9.40
19.10
38.16
50
III
0.28
0.16
6.52
11.80
18.41
33.61
50
IV
0.35
0.23
11.85
25.67
43.47
71.31
75
I
109.14
245.28
527.01
999.91
653.00
891.22
75
II
1.71
1.95
40.56
105.54
95.07
166.02
75
III
1.78
2.37
61.70
141.92
124.32
150.31
75
IV
2.75
3.61
127.38
248.19
355.73
469.97
100
I
414.62
924.04
2301.08
3244.42
2065.12
3180.68
100
II
16.53
36.84
43.58
60.35
186.86
442.14
100
III
15.99
35.45
94.49
151.49
199.04
336.03
100
IV
16.83
26.67
160.46
177.57
503.40
677.47
Table 4:Comparison of Computation Times for Pruning Strategies for
RTPS
21
Size
Method
20% Density
50% Density
80% Density
Standard
Standard
Standard
Average
Deviation
Average
Deviation
Average
Deviation
25
II
0.06
0.03
0.45
0.66
2.20
5.24
25
III
a
0.06
0.04
0.66
1.37
2.30
5.29
25
III
b
0.05
0.03
0.42
0.60
2.38
5.85
25
III
c
0.06
0.04
0.64
1.06
2.23
5.65
25
III
d
0.06
0.04
0.44
0.60
2.21
5.14
50
II
0.27
0.14
5.87
9.40
19.10
38.16
50
III
a
0.25
0.12
3.41
3.27
10.94
13.78
50
III
b
0.27
0.14
8.97
17.02
17.72
28.70
50
III
c
0.27
0.14
4.62
6.15
21.49
34.95
50
III
d
0.27
0.14
4.74
4.42
27.08
50.37
75
II
1.71
1.95
40.56
105.54
95.07
166.02
75
III
a
1.92
3.22
35.08
119.24
110.96
190.64
75
III
b
1.73
2.25
42.34
129.03
80.65
141.04
75
III
c
2.45
3.99
44.07
113.35
72.03
109.60
75
III
d
1.53
1.75
63.48
177.89
142.33
247.07
100
II
16.53
36.84
43.58
60.35
186.86
442.14
100
III
a
8.74
15.03
74.64
89.72
140.10
285.20
100
III
b
9.23
19.34
89.51
141.94
181.57
336.20
100
III
c
15.37
32.09
92.30
135.46
201.53
343.49
100
III
d
9.52
16.54
56.80
88.70
170.24
342.77
Table 5:Comparison of Individual Coe±cient Tightening Strategies for
RTPS
and compared them to Method II.Method III
a
combined pruning with only the coe±cient tightening strat
egy in (22a),while Method III
b
combined pruning with only the tightening in strategy (22b),Method III
c
used (22c),and Method III
d
used (22d).The computational times for this comparison are shown in Table
5.Method III
a
appears to be slightly more e±cient than the other limited tightening methods,or using
pruning only with no tightening.We also ran experiments coupling Method III
a
with each of the other three
tightening strategies,but none of these improved the execution time required to solve
RTPS.
Comparison of Partitioning Strategies for RTPS.
As a ¯nal experiment,we compared the compu
tational times for partitioning on RTPS instances,branching on either s
1
n
(Strategy I) or on s
2
n
(Strategy
II).For Strategy III,we branched on whichever interval [a
i
;b
i
] was the largest,which,due to the symmetry
constraint will always initially branch on s
2
n
.We used Set 3 for this experiment as done for RTPD,stopping
when ¿ ¡
¡
¾
n
¡
s
1
n
+s
2
n
¡s
1
n
s
2
n
¢¢
<",where"= 0:0005.The results of this experiment are displayed in
Table 6,and demonstrate that Strategies I and III clearly dominate Strategy II.This could be due to the
fact that s
1
n
¸ s
2
n
by our antisymmetry restriction,and thus determining minimum values for s
1
n
guides the
branchandbound process toward feasible solutions more quickly by implying tighter bounds on s
2
n
as well.
6 Conclusions
In this paper we have developed several strategies for approaching the problemof ¯nding two paths between
a source and destination node wherein the probability that at least one path survives must be greater than
or equal to some threshold.We considered both the problem in which the two paths must be arcdisjoint
and where arc sharing is allowed,and proved that both these problems are strongly NPhard.Due to the
22
Size
Strategy
20% Density
50% Density
80% Density
Standard
Standard
Standard
Average
Deviation
Average
Deviation
Average
Deviation
25
I
0.35
0.26
1.40
0.93
4.04
5.00
25
II
0.36
0.26
3.17
4.87
20.59
72.76
25
III
0.34
0.27
1.45
1.22
16.84
61.05
50
I
2.48
4.48
29.67
52.05
74.70
106.04
50
II
2.28
2.77
52.68
81.57
130.11
160.69
50
III
1.89
2.36
31.70
55.21
59.75
82.88
75
I
11.16
16.79
86.77
96.38
647.27
750.75
75
II
32.98
83.77
251.67
555.30
1769.44
2403.85
75
III
18.70
40.54
88.91
113.88
516.31
623.46
Table 6:Comparison of Partitioning Strategies for RTPS
nonlinear nature of the feasible region for these problems,we built an integer programming formulation
around relaxed versions of these problems,accompanied by a nonlinear branchandbound algorithm to
provably identify an optimal solution.
In order to improve the solvability of RTPD,we introduced methods for pruning the graph and tightening
the inequality constraints in the formulation.We also examined various partitioning strategies to prevent
a nonlinearinfeasible solution from recurring during future iterations.Our computational results demon
strate that a combination of graph preprocessing,coe±cient tightening,and diagonal branching strategies
can signi¯cantly reduce the computational e®ort required to solve RTPD instances.We adapted the for
mulation and tightening techniques of RTPD to the sharedarcs version of our problem,RTPS.Our test
results suggest that graph preprocessing with limited coe±cient tightening,along with a Reformulation
LinearizationTechniquebased branching process is the most e®ective algorithm for solving RTPS of the
alternatives that we proposed.
Future studies along these lines will investigate extensions and applications of the RTPD and RTPS
problems.For instance,one can consider a robust kshortest path problem,instead of the speci¯c case in
which k = 2 here.Additionally,vertexindependent routing problems are typically much harder to solve,
even without the robustness constraint imposed in this paper.While some of the strategies used in this
paper would apply to the robust vertexindependent routing problem,their e±cacy in that setting is an
open question.Finally,one might explore conditions (either due to topology or input data) in which these
problems become polynomiallysolvable.This is an important consideration in practical settings,where
certain topologies might exist that render large RTPD and RTPS instances tractable.
Appendix
Proof of Proposition 2.
De¯ne DRTPS the same as DRTPD,with the exception that arcs may be
shared between the two paths.By the same proof as in Proposition 1,we have that DRTPS belongs
to the class NP.We show that DRTPS is NPcomplete by transformation from 3SAT,using the same
graphtopology transformation used in Proposition 1.This time,Type I arcs that connect x
j
i
to y
j
i
,or
x
j
i
23
to
y
j
i
,have a cost of 0 and a reliability of (3M
3
V + (i ¡ 1)M + j ¡ 1)=(3M
3
V + (i ¡ 1)M + j),and all
Type II arcs have a cost of 1 and a reliability of 1.All Type III arcs have a cost of 0 and a probability
of 1,except for arcs (y
j
i
;w
i
O
) and (
y
j
i
;w
j
O
) for i = 1;:::;V,j = 1;:::;M,which have a reliability of
[(j + 1)=(j + 2)][(3M
3
V + (i ¡ 1)M + j)=(3M
3
V + (i ¡ 1)M + j ¡ 1)],and for the arc connecting u
1
to
w
1
I
,which has a reliability of 1/2.The cost limit is set to · = V (M + 1),and our reliability goal is
¿ = 3M
2
=(3M
2
+1) +1=(3M
2
+1)(M +2).
First,suppose that the 3SAT instance is a yesinstance.Then we use the same solution as prescribed
in the proof of Proposition 1,wherein one path uses a procession of Type III and Type I arcs such that
the Type I arcs correspond to variable assignments that satisfy each clause,and the second path uses
only Type I and Type II arcs corresponding to the opposite of the variable assignments.Note that the
¯rst path initially traverses arcs with probability 1/2,1,(3M
3
V + (i ¡ 1)M)=(3M
3
V + (i ¡ 1)M + 1),
(2=3)(3M
3
V + (i ¡ 1)M + 1)=(3M
3
V + (i ¡ 1)M) for some i = 1;:::;V,the product of which is 1/3.
Following this,the path encounters reliabilities of 1,(3M
3
V +(i ¡1)M +1)=(3M
3
V +(i ¡1)M +2),and
(3=4)(3M
3
V +(i ¡1)M +2)=(3M
3
V +(i ¡1)M +1) for some i = 1;:::;V,such that the total reliability
of the path once it reaches w
2
O
is 1/4.This pattern continues such that the overall reliability of the path is
1=(M +2),with a cost of zero.The second path encounters V M Type I arcs and V (M +1) Type II arcs,
for a total path cost of V (M +1).Hence,the cost threshold is met.Since Type I arcs have probabilities
(3M
3
V )=(3M
3
V +1),(3M
3
V +1)=(3M
3
V +2),:::,and (3M
3
V +MV ¡1)=(3M
3
V +MV ),these terms
collapse,and the reliability of the path is given by 3M
3
V=(3M
3
V +MV ) = 3M
2
=(3M
2
+1).Since these
paths do not share any arcs,the probability that at least one path survives is given by the sum of the two
reliabilities minus their product,i.e.,
3M
2
3M
2
+1
+
1
M +2
¡
µ
3M
2
3M
2
+1
¶µ
1
M +2
¶
=
3M
2
3M
2
+1
+
1
(3M
2
+1)(M +2)
= ¿:
Hence,the reliability threshold is satis¯ed,and the foregoing solution veri¯es that the transformed DRTPS
instance is a yesinstance.
Now,suppose that the DRTPS instance is a yesinstance.First,let us show that only one path can utilize
arc (u
1
;w
1
I
) or an arc of the form(y
j
i
;w
j
O
) or (
y
j
i
;w
j
O
).By contradiction,suppose that both paths use an arc of
this form.The highestpossible reliability arc of this formis [(M+1)=(M+2)][(3M
3
V +M)=(3M
3
V +M¡1)]
(which occurs only if there exists an arc (y
M
1
;w
M
O
) or (
y
M
1
;w
M
O
) in the network).Hence,the probability that
at least one of these paths remains intact is not more than
2
µ
M +1
M +2
¶µ
3M
3
V +M
3M
3
V +M ¡1
¶
¡
·µ
M +1
M +2
¶µ
3M
3
V +M
3M
3
V +M ¡1
¶¸
2
;(26)
which is possible only if both paths use only arcs with a reliability of 1 except for an arc with an imperfect
reliability of [(M+1)=(M+2)][(3M
3
V +M)=(3M
3
V +M¡1)],and if the paths do not share the arc with
24
this imperfect reliability.This expression simpli¯es to
µ
M +1
M +2
¶µ
3M
3
V +M
3M
3
V +M ¡1
¶·
(M +3)(3M
3
V +M ¡1) ¡(M +1)
(M +2)(3M
3
V +M ¡1)
¸
<
µ
M +1
M +2
¶µ
3M
3
V +M
3M
3
V +M ¡1
¶µ
M +3
M +2
¶
;(27)
where the inequality is due to removal of the (positive) (M+1)=[(M+2)(3M
3
V +M¡1)] term on the left
handside.We now show that for V ¸ 3 and M ¸ 3,the righthandside of (27) is less than 3M
2
=(3M
2
+1),
which is less than ¿.Observe that
3M
2
3M
2
+1
¡
µ
M +1
M +2
¶µ
3M
3
V +M
3M
3
V +M ¡1
¶µ
M +3
M +2
¶
=
6M
5
V ¡12M
4
V ¡9M
3
V ¡3M
4
¡10M
3
¡16M
2
¡3M
(3M
2
+1)(3M
3
V +M ¡1)(M +2)
2
:(28)
Since the denominator of (28) is positive,it is su±cient to show that the expression given by (26) is strictly
less than ¿ by showing that the numerator of (28) is nonnegative.Let f(M;V ) = 6M
5
V ¡12M
4
V ¡9M
3
V ¡
3M
4
¡10M
3
¡16M
2
¡3M.Noting that f(3;3) = 63 > 0,we can show that f(M;V ) > 0 for any M ¸ 3
and V ¸ 3 by demonstrating that @f(M;V )=@M ¸ 0 and @f(M;V )=@V ¸ 0.Using the facts that M=3 ¸ 0
and V=3 ¸ 0,we have that
@f(M;V )
@M
= 30M
4
V ¡48M
3
V ¡27M
2
V ¡12M
3
¡30M
2
¡32M ¡3
¸ 30M
4
V ¡16M
4
V ¡3M
4
V ¡
4
3
M
4
V ¡
10
9
M
4
V ¡
32
81
M
4
V ¡
1
81
M
4
V
=
220
27
M
4
V > 0;and
@f(M;V )
@V
= 6M
5
¡12M
4
¡9M
3
¸ 6M
5
¡4M
5
¡M
5
= M
5
> 0:
Since the foregoing analysis shows that it is impossible to visit wnodes in both paths,at least one path
must use only Type I and Type II arcs.Such a path that does not use wnodes must move from u
i
to u
i+1
,
for i = 1;:::;M,along x
i
 and y
i
nodes,or
x
i
 and
y
i
nodes,and thus incurs a cost of V (M+1) and a path
reliability of 3M
3
V=(3M
3
V +MV ) = 3M
2
=(3M
2
+1).
Thus,the second path must have used only zerocost arcs.This path uses arc (u
1
;w
1
I
),traverses from w
j
I
to w
j
O
for each j = 1;:::;M either via arcs (w
j
I
;x
j
i
),(x
j
i
;y
j
i
),and (y
j
i
;w
j
O
) for some v
i
belonging to C
j
,or
by (w
j
I
;x
j
i
),(x
j
i
;y
j
i
),and (y
j
i
;w
j
O
) for some
v
i
belonging to C
j
,and ends with (w
M
O
;u
V +1
).Any such path
uses a cost of zero and a reliability of 1=(M +2).The maximum possible probability that at least one of
these paths remains operational is achieved if neither path shares imperfectreliability arcs,and is computed
25
as
3M
2
3M
2
+1
+
1
M +2
¡
µ
3M
2
3M
2
+1
¶µ
1
M +2
¶
=
3M
2
3M
2
+1
+
1
(3M
2
+1)(M +2)
= ¿:
Therefore,the DRTPS solution must have achieved this upper bound by not sharing any arcs between the
two paths.Since these arcs were not shared,we have now reduced this case to the same scenario as in the
proof of Proposition 1.Observe that all numerical data is polynomially bounded in terms of M and V,and
thus DRTPS is strongly NPcomplete,i.e.,RTPS is strongly NPhard.2
Acknowledgements.
The authors gratefully acknowledge the support of the O±ce of Naval Research
under Grant Number N000140310510.
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