# Virtual Private Network Layout

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9 Δεκ 2013 (πριν από 5 χρόνια και 1 μήνα)

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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

-
(
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!