Virtual Private Network Layout

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Dec 9, 2013 (3 years and 6 months ago)

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Virtual Private Network Layout

A proof of the tree conjecture on a ring network



Leen Stougie

Eindhoven University of Technology (TUE)

&

CWI, Amsterdam


http://www.win.tue.nl/math/bs/spor/2004
-
15.pdf




































Input to the VPN problem


Undirected graph
G=(V,E)



Subset of the vertices
W
µ

V

(terminals)


Communication bounds on the terminals
b(i)
for all

i
2

W



Unit capacity costs on the edges
c(e)
for
all
e
2

E



Communication bounds and scenarios


b(i)

is bound on total of incoming and outgoing
communication of node
v

(
symmetric VPN
)


A valid demand scenario is symmetric matrix
D=(d
ik
)
ik
2

W

with
d
ii
=0

satisfying




d
ik
¸

0
8

i,k
2

W

and

k
2

W

d
ik



b(i)
8

i
2

W



D

is the set of all valid scenarios



VPN Robust optimization


Select for each pair
i,k
2

W

a path for
communication


Reserve enough capacity on the edges E


All demand in every valid communication
scenario
D
2
D

can be routed on the selected
paths


The total cost of reserving capacity is minimum


The paths are to be selected before seeing
any communication scenario

Routing variations of VPN


SPR (
Single path routing
)

For each pair
i,k
2

W

exactly one path
P
ik
µ

E



TTR (
Terminal tree routing
)

SPR with for each
i
2

W
,
[
k
2

W
P
ik

is a tree in
G



TR (
Tree routing
)

SPR with
[
i,k
2

W

P
ik

is a tree in
G



MPR (
Multi
-
path routing
)

For each pair
i,k
2

W

for each path
P

between
i

and
k
,
specify fraction of communication using
P

Relation between the variations


Lemma:


OPT(MPR)


OPT(SPR)


OPT(TTR)


OPT(TR)


Proof:

SPR is the MPR problem with the extra restriction
that all fractions must be 0 or 1.

The other inequalities are similarly trivial.

The open VPN
-
problem


Conjecture 1:

SPR
2

P (polynomially solvable)



Conjecture 2:

OPT(SPR)=OPT(TR)



Conjecture 3:

OPT(MPR)=OPT(TR)

What do we know about VPN?


TR
2

P

Kumar et al. 2002



OPT(TR)= OPT(TTR)

Gupta et al. 2001



OPT(TR)


2OPT(MPR)

Gupta et al. 2001



MPR
2

P

Erlebach and Ruegg 2004, Altin et al. 2004,
Hurkens et al. 2004


The asymmetric VPN

b
+
(v)

outgoing communication bound

b
-
(v)

incoming communication bound




TR is NP
-
hard

Gupta et al. 2001


TR
2

P if

v
2

W
b
-
(v)=

v
2

W
b
+
(v)

Italiano et al. 2002


MPR
2

P

Erlebach and Ruegg 2004, Altin et al. 2004, Hurkens et al. 2004


Constant Aprroximation ratios for SPR

Gupta et al. 2001, Eisenbrandt et al. 2005 (randomized)

Conjecture 3 is true:


If
G

is a tree (trivial)



If
G

is
K
4



If
G

is a cycle !!!!



If
G

is a 1
-
sum of graphs for which
Conjecture 3 is true

Path
-
formulation of VPN

P
ik

set of paths in
G

between
i

and
k

P

set of all paths in
G

For each path p in G we define
x
p

For all
i

and
k
2

W,

p
2

P
ik
x
p
=1



SPR:
x
p
2

{0,1}
8
p
2

P


MPR:
0


x
p



1
8
p
2

P

The capacity problem


Given selected paths: given values for
x(p)



Problem: find capacities on edges
z(e)
8

e
2

E












e
p
=1

if
e
2
P

and
0

otherwise


Dual of the capacity finding problem


Path
-
formulation of SPR


SPR: Find
x(p)

minimizing

e
2

E
c
e
z
e

Path
-
formulation of MPR


MPR: SPR with
x(p)
¸

0

i.o.
x(p)
2

{0,1}


Dual of the Path
-
formulation of MPR

Dual
-
MPR

MPR and TR


OPT(MPR)


OPT(TR)


Weak duality: any feasible
(

,

)

has



ik


OPT(MPR)



Conjecture 3:

OPT(MPR)=OPT(TR)


Conjecture 3:

OPT(TR)=Optimal solution
value of the dual of MPR

Optimal solution of TR (1)

Notation
b(U)=

v
2

U

b(v)

Take tree
T

Each
e
2

T

is cut in
T

splitting
V

in
L(e)

and
R(e)

Direct
e

to minimum of
b(L(e))

and
b(R(e))



There is a unique vertex
r

with indegree
0
, root


Cost of
T
:

e
min{b(L
e
),b(R
e
)} c(e)


The minimum cost tree with
r

as the root is the shortest
path tree from
r

in
G

w.r.t. length function
c


OPT(TR) can be found in polynomial time


Optimal solution of TR (2)

Let
d
G
(u,v)
the distance between
u

and
v

in
G

w.r.t. length
function
c


The cost of optimal tree
T

is given by


v
b(v)
d
T
(r,v)

for some root vertex
r
.



Moreover, it is bounded from below by


v
b(v)
d
G
(r,v)
.


Clearly it is bounded from above by


v
b(v)
d
T
(u,v) forall u
2
V


Compute shortest path tree rooted at
u

for all
u
2
V
and select

the one with minumum cost solves OPT(TR) in polynomial time


Conjecture 3 true for the cycle

Lemma:

If Conjecture 3 is true for any cycle with:

-

W=V

-
b(v)=1

8

v
2

V

-
|V|

is even

Then Conjecture 3 is true for any cycle


Theorem:

Conjecture 3 is true for any even cycle
with the above three properties


The even cycle (1)


Vertices
0,1,2,...,2n
-
1


Edges
e
1
,e
2
,...,e
2n


Cost of tree by deleting edge
e
k
:


(using

e
min{b(L
e
),b(R
e
)} c(e)
)




We show there exist a dual solution with
value equal to


min
e
k




The even cycle (2)


MPR
-
dual restricitions for even cycle with
b(v)=1


Only two possible paths between each pair of
vertices

The even cycle (3)

The Tool Lemma


The Tool Lemma:


-

Let
G=(V,E)

even circuit

-

b
´

1
.

-

F
µ

E, F


;



Then there exist

:E
!

R
+
,


not equal
0
, and
K

such that


support(

)
µ
F


8

f
2

F: K=C(f;

)=min
e
2

E

C(e;

)



There is a dual solution
(

,

)

with value
K

for the MPR
-
dual problem with cost function


The even cycle (4)

Part of Proof of Tool Lemma


Proof:

By induction on
|F|

-
|F|=1

(easy):
F={e
k
}


-
Take

k
=1

and

i
=0
8
i


k


-
Clearly,
min
e
2

E
C(e;

)=C(e
k
;

)=0


-
A feasible dual solution with value
0

is

e
ih
=0
,

ih
=0
8
e
2

E
8
i,h
2

V


The even cycle (5)

Part of Proof of Tool Lemma


Proof (continued):
|F|>1

-
Case (i): There is a k such that
e
k
2

F

and its
opposite edge
e
k+n
2

F

-
(
in figure read e
k
=a and e
k+n
=b
)




Choose

k
=

k+n
=1

and

i
=0
8

i


k,n+k


)
C(e;

)=n
8
e
2

E


Choose







Verify that



ij
=n

The even cycle (6)


Theorem:

Let
G=(V,E)

be an even circuit,
c: E
!
R
+

and
b(v)=1
8
v
2

V
. Then the cost of
an optimal tree solution equals the value
of an optimal dual solution.



Proof:

An inductive primal
-
dual argument
using the Tool Lemma.


(By request on the blackboard)

Postlude


OPT(MPR)=OPT(TR) for any graph?


SPR polynomially solvable for any graph?


Proof for the cycle is complicated!


Is there an easier proof for the cycle?


The crucial insight?


Complexity of the non
-
robust MPR
-
problem is also open!