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