Senatore--ORPtalk_To..

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August 2, 2013 – Flexible Networks Design 2013, Toronto
The Online Replacement Path Problem
Marco Senatore, University of Rome “Tor Vergata”
With David Adjiashvili (ETH) and Gianpaolo Oriolo (University of Rome “Tor Vergata”)
Introduction The Online Replacement Path Problem Robust Length vs. Nominal Length k-Hop ORP Policy-ORP and Canadian Traveller’s Problem
Motivation
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University of Rome “Tor Vergata”
Motivation
Collaboration with the Public Trasport (PT) Agency of Rome: historical and on-line
data.
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Motivation
data.
PT in Rome is very unreliable!!
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Motivation
data.
On the other hand, customers want to take reliable journeys.
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Motivation
data.
We want to be robust with respect to missing connections.
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Motivation
data.
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Motivation
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Motivation
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Motivation
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Introduction
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Introduction
Given an edge-weighted graph G = (V , E , `), ` : E → R+ , a source s ∈ V , a
destination t ∈ V and an s-t path P.
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Introduction
Assume at most one edge e ∈ E can fail.
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Introduction
Routing Mechanism (RM) wants to route a package along P without knowing the
identity of the failed edge.
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Introduction
RM can discovers the identity of e while traversing P.
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Introduction
t
s
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Introduction
t
s
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Introduction
t
s
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Introduction
t
s
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Introduction
t
s
!
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Introduction
t
s
As soon as the RM discovers the failure, it is allowed to take a detour to reach t.
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Introduction
t
s
As soon as the RM discovers the failure, it is allowed to take a detour to reach t.
Goal: Provide an s-t path to the RM, such that the worst-case total travel time is
minimized.
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Introduction
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Introduction
Different variants of the problem can be defined.
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Introduction
RM wants to route a package along P without knowing the identity of the failed
edge.
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Introduction
edge.
The RM discovers the identity of the failed edge e as soon as:
s
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t
Introduction
edge.
1
it tries to cross it → Online Replacement Path (ORP);
s
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t
Introduction
edge.
1
⇓
s
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t
Introduction
edge.
1
2
it reaches a node that is at most k hops away from e on P → k-Hop ORP;
s
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t
Introduction
edge.
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2
k=2
⇓
s
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t
Introduction
edge.
1
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3
it reaches a node that is at distance at most R from the closer endpoint of e on P →
Radius ORP;
s
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t
Introduction
edge.
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2
3
Radius ORP;
R=6
⇓
s
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2
5
t
Introduction
edge.
1
2
3
4
Radius ORP;
it reaches a node of P that is at most k hops away from e on G → Strong k-Hop
ORP
s
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t
Introduction
edge.
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Radius ORP;
ORP
k=2
⇓
s
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t
Introduction
edge.
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Radius ORP;
ORP
k=2
⇓
s
t
As soon as RM is aware of the failed edge, it follows the shortest path in the
residual graph.
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Outline
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Introduction
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3
Robust Length vs. Nominal Length
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k-Hop ORP
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Policy-ORP and Canadian Traveller’s Problem
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Outline
1
Introduction
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3
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k-Hop ORP
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Policy-ORP and Canadian Traveller’s Problem
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Some Notations
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Some Notations
For each v ∈ V ,
1
2
Pv ,t is the set of v -t paths;
π(v ) := shortest v -t path length.
For e ∈ E and u ∈ V :
1
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let Qu−e be some fixed shortest u-t path in G \ e;
π −e (u) = `(Qu−e ).
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Definition (Detour)
Given a path P ∈ Pv ,t and an edge e = uu 0 ∈ E , the detour P −e is:
0
the walk P[v , u] ⊕ Qu−uu if uu 0 ∈ P;
P otherwise.
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Definition (Detour)
0
P otherwise.
t
v
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Definition (Detour)
0
P otherwise.
t
v
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Definition (Detour)
0
P otherwise.
t
v
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Definition (Detour)
0
P otherwise.
t
v
Definition (Robust length)
The robust length of the v -t path P is
Val(P) = max `(P −e ).
e∈E
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Definition (Detour)
0
P otherwise.
t
v
e∈E
For each v ∈ V , we define πrob (v ) as the minimum of Val(P) over all P ∈ Pv ,t .
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Definition (Detour)
0
P otherwise.
t
v
e∈E
For each v ∈ V , we define πrob (v ) as the minimum of Val(P) over all P ∈ Pv ,t .
Find: an s-t path with minimum robust length.
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Val(P) is finite iff every v ∈ P is 2-connected to t.
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Consider Pu .
u
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t
Consider Pu .
v
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u
t
Consider Pu . Then Val(vu ⊕ Pu ) = max{`(vu) + Val(Pu ), π −vu (v )}.
v
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u
t
v
u
t
[Monotonicity]:Val(P) ≥ Val(P[x, t]), whenever x ∈ V (P).
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v
u
t
1000
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v
u
t
1000
Lemma (Weak optimality principle)
Let Pv ∈ Pv ,t be an optimal path from v , u ∈ V (Pv ) and Pu ∈ Pu,t be an optimal path
from u. Then the path Pv0 = Pv [v , u] ⊕ Pu satisfies Valk (Pv0 ) = Valk (Pv ), namely it is
also optimal from v .
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We devise a label setting algorithm.
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Lemma
Let U ⊂ V , with t ∈ U, be a set of nodes for which πrob is known, and let vu ∈ δ(U):
vu =
max{`(zw ) + πrob (w ), π −zw (z)}.
arg min
zw ∈E :w ∈U,z6∈U
Then πrob (v ) = Val(vu ⊕ Pu ), where Pu is any optimal (robust) u-t path .
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Lemma
vu =
arg min
Theorem
Given an instance of ORP, the values πrob and the corresponding paths can be computed
in time O(m + n log n) in undirected graphs, and O(mn + n2 log n) in directed graphs.
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Lemma
vu =
arg min
Theorem
This complexity bound builds upon a result by Proietti et al. of ’99 providing an
O(mα(m, n)) algorithm to compute π −e (u), ∀e = uu 0 ∈ E .
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Lemma
vu =
arg min
Theorem
This complexity bound builds upon a result by Proietti et al. of ’99 providing an
O(mα(m, n)) algorithm to compute π −e (u), ∀e = uu 0 ∈ E .
The same complexity bound is claimed in Bar-Noy and Schieber ’91, ”The Canadian
Traveller Problem”.
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In some applications faults occur rarely, hence it is preferred to have the cost of the
chosen path as low as possible.
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At the same time we could have a budget B on the robust length.
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We aim at solving:
P ∗ :=
arg min
`(P).
P∈Ps,t : Val(P)≤B
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We aim at solving:
P ∗ :=
arg min
`(P).
We can solve this problem with a simple modification of Dijkstra’s algorithm.
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We aim at solving:
P ∗ :=
arg min
`(P).
We can solve this problem with a simple modification of Dijkstra’s algorithm.
Algorithm
1:
2:
3:
4:
5:
6:
7:
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S = ∅; S̄ = V
d(s) = 0; d(u) = ∞ ∀u ∈ V \ s
while t ∈
/ S do
Find u = arg minz∈S̄ d(z)
S = S + u S̄ = S̄ − u
for q ∈ N(u) \ S do
if d(u) + `(uq) ≤ d(q) and d(u) + π −uq (u) ≤ B then
d(q) = d(u) + `(uq)
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`(P ∗ )
πrob (s) := minimum s-t robust length
π(s) := shortest s-t length
π(s)
B
πrob (s)
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`(P ∗ )
π(s)
B
πrob (s)
The number of steps is O(m).
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`(P ∗ )
π(s)
B
πrob (s)
The number of steps is O(m).
The profile can be found in O(m ∗ (m + n log n)).
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k-Hop ORP
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k-Hop ORP
The RM is informed about the failure of an edge e ∈ E as soon as it reaches a node
that is at most k hops away from e on P.
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k-Hop ORP
The RM is informed about the failure of an edge e ∈ E as soon as it reaches a node
that is at most k hops away from e on P.
k=2
⇓
s
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t
k-Hop ORP
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k-Hop ORP
Consider a path P and an edge e ∈ E (P). Denote by v (P, e) the first node u of P
such that e is visible from u.
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k-Hop ORP
Definition (Detour)
Given a path P ∈ Pu,t and an edge e ∈ E , the detour P −e is:
the walk P[u, v (P, e)] ⊕ Qv−e
(P,e) if e ∈ P;
P otherwise.
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k-Hop ORP
Definition (Detour)
(P,e) if e ∈ P;
P otherwise.
Definition (k-Robust length)
The k-robust length of the u-t path P is
Valk (P) = max `(P −e ).
e∈E
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k-Hop ORP
Definition (Detour)
(P,e) if e ∈ P;
P otherwise.
e∈E
Find: an s-t path with minimum k-robust length.
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k-Hop ORP
Definition (Detour)
(P,e) if e ∈ P;
P otherwise.
e∈E
Find: an s-t path with minimum k-robust length.
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k-Hop ORP
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k-Hop ORP
More challenging than ORP: the k-robust length is not monotonic!
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k-Hop ORP
k=2
u
t
20
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k-Hop ORP
k=2
u
t
20
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k-Hop ORP
5
k=2
v
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u
t
20
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k-Hop ORP
5
k=2
v
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u
t
20
Property: If P ∈ Pu,t is such that Valk (P) = π −e (u) for some e ∈ P visible from u,
then P is an optimal path from u.
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k-Hop ORP
5
k=2
v
20
u
t
20
then P is an optimal path from u. =⇒ Preprocessing step!
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k-Hop ORP
5
k=2
v
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u
t
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then P is an optimal path from u. =⇒ Preprocessing step!
Weak optimality principle still holds for k-Hop ORP
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k-Hop ORP and Radius ORP
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Theorem
Given an instance of k-Hop ORP we can compute the v -t path with minimum k-robust
length for any v ∈ V in time O(m + n log n) in undirected graphs, and O(mn + n2 log n)
in directed graphs.
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Theorem
Given an instance of k-Hop ORP we can compute the v -t path with minimum k-robust
length for any v ∈ V in time O(m + n log n) in undirected graphs, and O(mn + n2 log n)
in directed graphs.
Our algorithm for k-Hop ORP solves Radius ORP as well.
R=6
⇓
s
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2
5
t
Strong k-Hop ORP
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Strong k-Hop ORP
k=2
⇓
s
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t
Strong k-Hop ORP
k=2
⇓
s
t
Theorem
For any > 0 it is NP-hard to approximate Strong 1-Hop ORP within a factor of 3 − in
undirected graphs. In directed graphs it is strongly NP-hard to decide if there exists a
path with finite robust length.
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Strong k-Hop ORP
k=2
⇓
s
t
Theorem
For any > 0 it is NP-hard to approximate Strong 1-Hop ORP within a factor of 3 − in
undirected graphs. In directed graphs it is strongly NP-hard to decide if there exists a
path with finite robust length.
Any shortest path is a 3 approximation for undirected graphs.
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Policy-ORP and Canadian Traveller’s Problem (Ongoing work)
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Question: Can we do better if not forced to follow a path? Can it be advantageous
to explore the graph?
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A policy is a function φ : (u ∈ V , W ⊆ V , F ⊆ E ) → v ∈ N(u), where:
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3
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u is the current node;
W is the set of vertices visited before arriving to u;
F is the set of failed edges discovered along the way (|F | ≤ 1).
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1
2
3
The robust value of a policy φ is the maximum cost we pay over all possible failure
scenarios.
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1
2
3
scenarios.
Policy-ORP is to find a policy with minimum robust value.
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1
2
3
scenarios.
Lemma
A solution to ORP from vertex s returns an optimal policy with respect to the initial
state (s, {s}, ∅).
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to explore the graph? NO!
1
2
3
scenarios.
Lemma
state (s, {s}, ∅).
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1
2
3
scenarios.
Lemma
state (s, {s}, ∅).
Other variants of ORP?
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1
2
3
scenarios.
Lemma
state (s, {s}, ∅).
Other variants of ORP?
Close problem: In ’91 Yannakakis and Papadimitriou defined the Canadian
Traveller’s Problem, Bar-Noy and Schieber ’91
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A Game Theory Perspective (Ongoing work)
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Given G = (V , E , `), ` : E → R+ , a source s ∈ V and a destination t ∈ V .
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We consider two players, a path builder and an interdictor:
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1
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The path builder wants to route a package from s to t as quickly as possible.
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1
2
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The interdictor wants to increase the s-t shortest distance as much as possible by
removing an edge from the network.
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1
2
The strategies of the interdictor are edges e ∈ E ; if she moves first, she needs to
solve the Most Vital Arc Problem. Let z ∗ (MVA) be the optimal value.
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1
2
The strategies of the path builder are s-t paths; if she moves first, she needs to
solve the ORP Problem. Let z ∗ (ORP) be the optimal value.
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1
2
The strategies of the path builder are s-t paths; if she moves first, she needs to
solve the ORP Problem. Let z ∗ (ORP) be the optimal value.
Lemma
Let P and e be an s-t path and an edge of E , respectively. Then
Val(P) ≥ z ∗ (ORP) ≥ z ∗ (MVA) ≥ π −e (s).
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ORP Game
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ORP Game
We call the ORP Game, that game in which both the path builder and the
interdictor communicate their strategies at the same time.
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ORP Game
Theorem
Let P and e be optimal solutions to the ORP and MVA instances on G = (V , E ). Then
(P, e) is a pure NE of the ORP Game if and only if Val(P) = π −e (s). Moreover, in this
case, Val(P) = z ∗ (ORP) = z ∗ (MVA) = π −e (s).
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ORP Game
Theorem
We can compute a pure NE or certify that no pure NE exists in O(m + n log n) for
undirected grpahs, in O(mn + n2 log n) for directed graph.
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ORP Game
Theorem
In the case where z ∗ (ORP) 6= z ∗ (MVA), the ORP Game will still admit a NE in
mixed strategy.
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ORP Game
Theorem
In the case where z ∗ (ORP) 6= z ∗ (MVA), the ORP Game will still admit a NE in
mixed strategy.
Question: Is it possible to find a mixed NE in polynomial time?
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Stochastic ORP
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Stochastic ORP
In this variant of ORP:
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Stochastic ORP
1
2
3
4
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at most one edge can fail;
for each e ∈ E , p(e) ∈ [0, 1] is the probability that e will be the unique edge to fail;
p(∅) ∈ [0, 1] is the
P probability that no edge fails;
it holds p(∅) + e∈E p(e) = 1.
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Stochastic ORP
1
2
3
4
p(∅) ∈ [0, 1] is the
Definition
Given a node v ∈ V , the expected robust length of the v -t path P is
i
X
X h
0
EVal(P) = (p(∅) +
p(uu 0 ))`(P) +
p(uu 0 )(`(P[v , u]) + π −uu (u)) .
uu 0 ∈P
/
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uu 0 ∈P
Stochastic ORP
1
2
3
4
p(∅) ∈ [0, 1] is the
Definition
i
X
X h
0
EVal(P) = (p(∅) +
p(uu 0 ))`(P) +
p(uu 0 )(`(P[v , u]) + π −uu (u)) .
uu 0 ∈P
/
uu 0 ∈P
Stochastic ORP is to find, for some v ∈ V , a path P minimizing EVal(P) over all
paths P ∈ Pv ,t .
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Stochastic ORP
1
2
3
4
p(∅) ∈ [0, 1] is the
Definition
i
X
X h
0
EVal(P) = (p(∅) +
p(uu 0 ))`(P) +
p(uu 0 )(`(P[v , u]) + π −uu (u)) .
uu 0 ∈P
/
uu 0 ∈P
paths P ∈ Pv ,t .
Stochastic ORP is NP-hard by reduction from the Hamiltonian Path problem.
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Stochastic ORP
1
2
3
4
p(∅) ∈ [0, 1] is the
Definition
i
X
X h
0
EVal(P) = (p(∅) +
p(uu 0 ))`(P) +
p(uu 0 )(`(P[v , u]) + π −uu (u)) .
uu 0 ∈P
/
uu 0 ∈P
paths P ∈ Pv ,t .
Stochastic ORP is NP-hard by reduction from the Hamiltonian Path problem.
Open question: Is it possible to find a mixed NE in polynomial time?
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Questions?
Thank You...
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Senatore--ORPtalk_To..

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