multipath routing algorithms for congestion minimization ron banner and ariel orda department of...
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TRANSCRIPT
Multipath Routing Algorithms for
Congestion Minimization
Ron Banner and Ariel Orda
Department of Electrical Engineering Technion- Israel Institute of Technology
Introduction
Traditional routing schemes route all traffic along a single ldquooptimalrdquo path
Traffic is always routed over a single path High congestion Waste of network resources
Multipath Routing split the traffic among several paths in order to ease congestion
Multipath Routing
Multipath routing can be fundamentally more efficient than the traditional approach
It can significantly reduce congestion in ldquohot spotsrdquo
As congested links result in high variance it provides steady and smooth data streams
Previous work mainly focused on heuristics
Equal Cost MultiPath (ECMP) Equal Distribution of traffic along multiple shortest paths The shortest path and equal partition limitations considerably
reduce load balancing capabilities
OSPF-OMP Allows splitting traffic among paths unevenly Heuristic traffic distribution scheme that often results in an
inefficient flow distribution
Proportionally split traffic among several ldquowidestrdquo paths that are disjoint wrt bottlenecks [Nelakuditi et al
1xxx] Again Heuristic and evaluated by way of simulations
How much is gained by optimal flow distribution
0
005
01
015
02
025
03
035
04
045
L 11L 12L 13L 14L 15L 16L 17L 18L 19L 2L
Length Restriction (L is the length of the shorest path)
r (L
)
Experiment Generated random networks that include 10000 Waxman topologies amp 10000 power-law topologies
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by OMP
Power law
Waxman
How much is gained by optimal flow distribution
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by ECMP
The full potential of multipath routing is far from having been
exploited
0
01
02
03
04
05
1 11 12 13 14 15 16 17 18 19 2Length Restriction
r(L
)
Waxman
Power law
Problem formulation
Goals Minimize network congestion Cope with constraints
Path Length distribute traffic only among paths with satisfactory quality (length)
Number of paths Too many paths per destination pose major complication amp considerable overhead
Performance Objective network congestion factor
Minimizing
Eg [RFC 2702] [xxx]
No link becomes over-utilized
More room for future traffic growth
e
e Ee
fmax
c
Computational Intractability
Minimizing the network congestion factor under path length restrictions is NP- hard Proof
Minimizing the network congestion factor while routing traffic along at most K paths is NP-hard Proof
Minimizing Network Congestion Under length Restrictions
Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Introduction
Traditional routing schemes route all traffic along a single ldquooptimalrdquo path
Traffic is always routed over a single path High congestion Waste of network resources
Multipath Routing split the traffic among several paths in order to ease congestion
Multipath Routing
Multipath routing can be fundamentally more efficient than the traditional approach
It can significantly reduce congestion in ldquohot spotsrdquo
As congested links result in high variance it provides steady and smooth data streams
Previous work mainly focused on heuristics
Equal Cost MultiPath (ECMP) Equal Distribution of traffic along multiple shortest paths The shortest path and equal partition limitations considerably
reduce load balancing capabilities
OSPF-OMP Allows splitting traffic among paths unevenly Heuristic traffic distribution scheme that often results in an
inefficient flow distribution
Proportionally split traffic among several ldquowidestrdquo paths that are disjoint wrt bottlenecks [Nelakuditi et al
1xxx] Again Heuristic and evaluated by way of simulations
How much is gained by optimal flow distribution
0
005
01
015
02
025
03
035
04
045
L 11L 12L 13L 14L 15L 16L 17L 18L 19L 2L
Length Restriction (L is the length of the shorest path)
r (L
)
Experiment Generated random networks that include 10000 Waxman topologies amp 10000 power-law topologies
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by OMP
Power law
Waxman
How much is gained by optimal flow distribution
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by ECMP
The full potential of multipath routing is far from having been
exploited
0
01
02
03
04
05
1 11 12 13 14 15 16 17 18 19 2Length Restriction
r(L
)
Waxman
Power law
Problem formulation
Goals Minimize network congestion Cope with constraints
Path Length distribute traffic only among paths with satisfactory quality (length)
Number of paths Too many paths per destination pose major complication amp considerable overhead
Performance Objective network congestion factor
Minimizing
Eg [RFC 2702] [xxx]
No link becomes over-utilized
More room for future traffic growth
e
e Ee
fmax
c
Computational Intractability
Minimizing the network congestion factor under path length restrictions is NP- hard Proof
Minimizing the network congestion factor while routing traffic along at most K paths is NP-hard Proof
Minimizing Network Congestion Under length Restrictions
Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Multipath Routing
Multipath routing can be fundamentally more efficient than the traditional approach
It can significantly reduce congestion in ldquohot spotsrdquo
As congested links result in high variance it provides steady and smooth data streams
Previous work mainly focused on heuristics
Equal Cost MultiPath (ECMP) Equal Distribution of traffic along multiple shortest paths The shortest path and equal partition limitations considerably
reduce load balancing capabilities
OSPF-OMP Allows splitting traffic among paths unevenly Heuristic traffic distribution scheme that often results in an
inefficient flow distribution
Proportionally split traffic among several ldquowidestrdquo paths that are disjoint wrt bottlenecks [Nelakuditi et al
1xxx] Again Heuristic and evaluated by way of simulations
How much is gained by optimal flow distribution
0
005
01
015
02
025
03
035
04
045
L 11L 12L 13L 14L 15L 16L 17L 18L 19L 2L
Length Restriction (L is the length of the shorest path)
r (L
)
Experiment Generated random networks that include 10000 Waxman topologies amp 10000 power-law topologies
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by OMP
Power law
Waxman
How much is gained by optimal flow distribution
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by ECMP
The full potential of multipath routing is far from having been
exploited
0
01
02
03
04
05
1 11 12 13 14 15 16 17 18 19 2Length Restriction
r(L
)
Waxman
Power law
Problem formulation
Goals Minimize network congestion Cope with constraints
Path Length distribute traffic only among paths with satisfactory quality (length)
Number of paths Too many paths per destination pose major complication amp considerable overhead
Performance Objective network congestion factor
Minimizing
Eg [RFC 2702] [xxx]
No link becomes over-utilized
More room for future traffic growth
e
e Ee
fmax
c
Computational Intractability
Minimizing the network congestion factor under path length restrictions is NP- hard Proof
Minimizing the network congestion factor while routing traffic along at most K paths is NP-hard Proof
Minimizing Network Congestion Under length Restrictions
Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Previous work mainly focused on heuristics
Equal Cost MultiPath (ECMP) Equal Distribution of traffic along multiple shortest paths The shortest path and equal partition limitations considerably
reduce load balancing capabilities
OSPF-OMP Allows splitting traffic among paths unevenly Heuristic traffic distribution scheme that often results in an
inefficient flow distribution
Proportionally split traffic among several ldquowidestrdquo paths that are disjoint wrt bottlenecks [Nelakuditi et al
1xxx] Again Heuristic and evaluated by way of simulations
How much is gained by optimal flow distribution
0
005
01
015
02
025
03
035
04
045
L 11L 12L 13L 14L 15L 16L 17L 18L 19L 2L
Length Restriction (L is the length of the shorest path)
r (L
)
Experiment Generated random networks that include 10000 Waxman topologies amp 10000 power-law topologies
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by OMP
Power law
Waxman
How much is gained by optimal flow distribution
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by ECMP
The full potential of multipath routing is far from having been
exploited
0
01
02
03
04
05
1 11 12 13 14 15 16 17 18 19 2Length Restriction
r(L
)
Waxman
Power law
Problem formulation
Goals Minimize network congestion Cope with constraints
Path Length distribute traffic only among paths with satisfactory quality (length)
Number of paths Too many paths per destination pose major complication amp considerable overhead
Performance Objective network congestion factor
Minimizing
Eg [RFC 2702] [xxx]
No link becomes over-utilized
More room for future traffic growth
e
e Ee
fmax
c
Computational Intractability
Minimizing the network congestion factor under path length restrictions is NP- hard Proof
Minimizing the network congestion factor while routing traffic along at most K paths is NP-hard Proof
Minimizing Network Congestion Under length Restrictions
Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
How much is gained by optimal flow distribution
0
005
01
015
02
025
03
035
04
045
L 11L 12L 13L 14L 15L 16L 17L 18L 19L 2L
Length Restriction (L is the length of the shorest path)
r (L
)
Experiment Generated random networks that include 10000 Waxman topologies amp 10000 power-law topologies
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by OMP
Power law
Waxman
How much is gained by optimal flow distribution
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by ECMP
The full potential of multipath routing is far from having been
exploited
0
01
02
03
04
05
1 11 12 13 14 15 16 17 18 19 2Length Restriction
r(L
)
Waxman
Power law
Problem formulation
Goals Minimize network congestion Cope with constraints
Path Length distribute traffic only among paths with satisfactory quality (length)
Number of paths Too many paths per destination pose major complication amp considerable overhead
Performance Objective network congestion factor
Minimizing
Eg [RFC 2702] [xxx]
No link becomes over-utilized
More room for future traffic growth
e
e Ee
fmax
c
Computational Intractability
Minimizing the network congestion factor under path length restrictions is NP- hard Proof
Minimizing the network congestion factor while routing traffic along at most K paths is NP-hard Proof
Minimizing Network Congestion Under length Restrictions
Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
How much is gained by optimal flow distribution
r(L= the ratio between the congestion of an optimal assignment of traffic to paths (with a length restriction L) to the congestion produced by ECMP
The full potential of multipath routing is far from having been
exploited
0
01
02
03
04
05
1 11 12 13 14 15 16 17 18 19 2Length Restriction
r(L
)
Waxman
Power law
Problem formulation
Goals Minimize network congestion Cope with constraints
Path Length distribute traffic only among paths with satisfactory quality (length)
Number of paths Too many paths per destination pose major complication amp considerable overhead
Performance Objective network congestion factor
Minimizing
Eg [RFC 2702] [xxx]
No link becomes over-utilized
More room for future traffic growth
e
e Ee
fmax
c
Computational Intractability
Minimizing the network congestion factor under path length restrictions is NP- hard Proof
Minimizing the network congestion factor while routing traffic along at most K paths is NP-hard Proof
Minimizing Network Congestion Under length Restrictions
Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Problem formulation
Goals Minimize network congestion Cope with constraints
Path Length distribute traffic only among paths with satisfactory quality (length)
Number of paths Too many paths per destination pose major complication amp considerable overhead
Performance Objective network congestion factor
Minimizing
Eg [RFC 2702] [xxx]
No link becomes over-utilized
More room for future traffic growth
e
e Ee
fmax
c
Computational Intractability
Minimizing the network congestion factor under path length restrictions is NP- hard Proof
Minimizing the network congestion factor while routing traffic along at most K paths is NP-hard Proof
Minimizing Network Congestion Under length Restrictions
Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Computational Intractability
Minimizing the network congestion factor under path length restrictions is NP- hard Proof
Minimizing the network congestion factor while routing traffic along at most K paths is NP-hard Proof
Minimizing Network Congestion Under length Restrictions
Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Minimizing Network Congestion Under length Restrictions
Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Pseudo-Polynomial Solution
= the total flow along e=(uv) that has been routed from s to u through paths with a total length of
ef
u vw
0 0ef 1 3ef
3 0ef
2 7ef
4 0ef
0 0ef 1 0ef
3 3ef
2 0ef
4 7ef
2el
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Pseudo-Polynomial Solution (Linear Program)
Objective function Minimize α
Constraints α is the network congestion factor ie for each eE
Nodal flow conservation constraint ie for each vVst
0
Le e
e e
f f
c c
Out( ) In( )
ele e
e v e v
f f
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Pseudo-Polynomial Solution (Linear Program)
Constraints (cont) Demand constraint
The complete linear program
0
Out( )se
e
f
( ) ( )
( ) ( )
0
( )
0
Minimize
0 0
0 1
0 0
0
e
e
le e
e O v e I v
le e
e O s e I s
ee O s
L
e e
e e
e
s t
f f v V s t L
f f L
f
f c e E
f e E L l
f
0
0
e E L
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Pseudo-Polynomial solution
The linear program can be solved within time complexity that is polynomial in the number of variables
Therefore the complexity incurred by solving the linear program is polynomial in L
Indeed the number of variables is O(|E|L) ef
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Approximation Scheme
Goal reduce the number of variables to be polynomial in |V| and |E| instead of L
Scaling
Apply the linear program for the new instance The new instance relaxes the original instance Hence congestion is not worse than the optimum
Convert each non-simple path into a simple path Accumulating error for a path (N-1) New path length is at most L+ N=L∙(1+є)
L L L= where e
e
ll
N
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Extensions
Multi-commodity It is straightforward to extend the linear program to the
multi-commodity case
End-to-End Reliability Constraints
Multipath Routing has increased vulnerability to failures
A failure in each path causes the entire transmission to fail
The problem Minimize congestion under end-to-end reliability constraints Our approximation scheme can be modified to solve this
problem
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Minimizing Congestion while Routing Along at Most K Different Paths
K- integral flows that minimize congestion An optimal K- integral flow is a 2-APX schemeComputing optimal K- integral flows
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
K- integral flows that minimize congestion
Minimize the network congestion factor such that
=3
K=2
fe=2
fe=1
The demand is routed along at most K paths
The flow over each path is a multiple of K
New constraintOriginal constr
aint
fe=0 =3
K=2
fe=0
3
2ef
3
2ef
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
An optimal K- integral flow is a 2-apx scheme
Theorem The minimum congestion of a K-
integral flow is at most twice the congestion of the optimal solution Proof
Each K- integral flow satisfies the requirement to ship the demand on at most K paths
Corollary minimizing the congestion while restricting the flow to be integral in K is a 2-approximation scheme for the original problem
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Computing optimal K-integral flows
The network congestion factor of each K-integral flow belongs to
The flow over each link is integral in K and is at most Hence for each eE it holds that
Thus for each eE it holds that
In particular
Sufficient to find the K-integral flow that has the minimum network congestion factor in
0e
n e E n KK c
0 ef n n KK
max 0 e
e Ee e
fn e E n K
c K c
0 e
e e
fn n K
c K c
0 e
n e E n KK c
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Computing optimal K-integral flows (cont)
Goal Find a K-integral flow that has the minimum network
congestion factor in
Solution
A Multiply all link capacities by a factor of
B Round down the capacity of each link to a multiply of K
Since the flow must be K-integral such a rounding has no affect
C Apply a maximum flow algorithm
Since all capacities are integral in K the algorithm returns a K-
integral flow
D If the K-integral flow fails to transfer flow units repeat the process
with a larger otherwise repeat the process with a smaller
E Output the flow that transfers flow units and has the smallest
0e
A n e E n KK c
A
e E max
e e
e ee e
f ff c
c c
A A
A
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Computing optimal K-integral flows
Since the set A is polynomial the complexity of the solution is polynomial
Thus we established a polynomial algorithm that admits at most K paths and has a network congestion factor that is at most twice larger than the optimum
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Future Work
A unifying scheme that bounds the number of paths AND the length of each path
Distributed implementation of both algorithms
Heuristic schemes with lower complexity
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Questions
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Proof (Sketch)
Step 1 Find a flow that minimizes congestion while routing traffic along K paths
1f
2f
3f
e EMax e
e
f
c
Step 2 Double the flow over each path
12 f
22 f
32 f
E
e
22Max e
e
f
c
2
Step 3 Round down the flow over each path to a multiple of K
1f 2f
3f
e
EMax 2
2 e
e
f
c
By construction is a K-integral flow
In step 3 the total flow is reduced by at most units There are K paths each
ldquoloosesrdquo at most K units
Hence transfers at least flow units
f
f
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Pseudo-Polynomial solution
( ) ( )
0 0ele e
e O v e I v
f f v V s t L
( ) ( )
0 1ele e
e O s e I s
f f L
0
( )e
e O s
f
Minimize
s t
0
1 L
ee
fc
e E
0ef
0
0ef
0 ee E L l
0e E L
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
Nodal flow conservation constraint
v
for each vVst
( ) ( )
ele e
e Out v e In v
f f
2 1ef
3 1ef
4el
3el
6 2ef
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
The end-to-end delay restriction is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t such that if path p transfers a positive amount of flow then D(p)leD
The partition problem Given an ordered set of elements a1 a2 hellip a2n that constitute a set A with a size s(a)+ for each a A is there a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen such that sumaArsquo s(a)=sumaAArsquo s(a)
All link capacities are 1 Claim It is possible to transfer 2 flow units over paths whose end-to-end
delays are not larger than frac12sumaA s(a) iff there is a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for 1leilen and sum
aArsquo s(a)=sum
aAArsquo s(a)
S(a1) S(a3) S(a5) S(a2n-1)
S T
S(a2) S(a4) S(a6) S(a2n)
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
The end-to-end delay restriction is intractable
lt=lt= There is a a subset ArsquoA such that Arsquo contains exactly one element of a2i-1 a2i for
1leilen and sumaArsquo
s(a)=sumaAArsquo
s(a) The selection of the links that correspond to the elements of Arsquo and the zero
delay links that connect these links constitutes a path p Path p is disjoint to the path that the complement subset AArsquo defines Since all capacities are equal to 1 we have two disjoint paths that can transfer
together 2 units of flow The end-to-end delay of each path is frac12sumaA s(a)
=gt=gt There is a path flow that transfers two flow units over paths that are not larger
than frac12sumaA s(a) Let p be a path that carries a positive flow by construction p contains exactly
one element of a2i-1 a2i for 1leilen Since all the links have one unit of capacity p can transfer at most 1 flow unit Therefore there exists a path prsquo that is disjoint to p that transfers a positive
flow by construction prsquo=Ap Hence D(p) lefrac12sumaA s(a) and D(prsquo) lefrac12sumaA s(a) Therefore since D(p)+ D(prsquo)=sumaA s(a) it follows that sum
ap s(a)=sumaprsquo
s(a)=frac12sumaA
s(a)
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-
The restriction on the number of paths is intractable
A special case of our problem Is there a path flow that transfers flow units from s to t over at most K paths
The single source unsplittable flow problem Given a network G with a source s targets t1 t2 hellip tk and corresponding demands D1 D2 hellip Dk is there an assignment of traffic to paths such that for each 1leilek demand Di is routed over a single path without violating the capacity constraints
Claim There exists a path flow that transfers = D1+ D2 +hellip+ Dk flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D1 D2 hellip Dk to paths such that Di 1leilek is routed over a single path without violating the capacity constraints There is exactly one path from S to ti for each 1leilek Hence there are exactly K paths from S to T
that carry a positive flows There is at least one path from S to ti for each 1leilek However since there are at most K paths
there is exactly one path from S to ti for each 1leilek
S
t1 t2 tk
TD1
D2 Dk
- Multipath Routing Algorithms for Congestion Minimization
- Introduction
- Multipath Routing
- Previous work mainly focused on heuristics
- How much is gained by optimal flow distribution
- Slide 6
- Problem formulation
- Computational Intractability
- Minimizing Network Congestion Under length Restrictions
- Pseudo-Polynomial Solution
- Pseudo-Polynomial Solution (Linear Program)
- Slide 12
- Pseudo-Polynomial solution
- Approximation Scheme
- Extensions
- Minimizing Congestion while Routing Along at Most K Different Paths
- K- integral flows that minimize congestion
- An optimal K- integral flow is a 2-apx scheme
- Computing optimal K-integral flows
- Computing optimal K-integral flows (cont)
- Slide 21
- Future Work
- Questions
- Proof (Sketch)
- Slide 25
- Nodal flow conservation constraint
- The end-to-end delay restriction is intractable
- Slide 28
- The restriction on the number of paths is intractable
-