1. Multipath Routing Algorithms for Congestion Minimization Ron Banner and Ariel Orda Department of Electrical Engineering Technion- Israel Institute of Technology

2. Introduction u Traditional routing schemes route all traffic along a single “optimal” path u Traffic is always routed over a single path  High congestion  Waste of network resources. u Multipath Routing split the traffic among several paths in order to ease congestion.

3. Multipath Routing u Multipath routing can be fundamentally more efficient than the traditional approach. u It can significantly reduce congestion in “hot spots”. u As congested links result in high variance, it provides steady and smooth data streams.

4. Previous work mainly focused on heuristics u Equal Cost MultiPath (ECMP): Equal Distribution of traffic along multiple shortest paths  The shortest path and equal partition limitations considerably reduce load balancing capabilities. u OSPF-OMP: Allows splitting traffic among paths unevenly.  Heuristic traffic distribution scheme that often results in an inefficient flow distribution. u Proportionally split traffic among several “widest” paths that are disjoint w.r.t. bottlenecks [Nelakuditi et al., 1xxx]  Again: Heuristic and evaluated by way of simulations.

5. How much is gained by optimal flow distribution? u Experiment : Generated random networks that include 10,000 Waxman topologies & 10,000 power-law topologies.   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

6. 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... Waxman Power law

7. Problem formulation u 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 & considerable overhead. u Performance Objective: network congestion factor  Minimizing  E.g.: [RFC 2702], [xxx].  No link becomes over-utilized.  More room for future traffic growth.

8. Computational Intractability u Minimizing the network congestion factor under path length restrictions is NP- hard.  Proof. u Minimizing the network congestion factor while routing traffic along at most K paths is NP-hard.  Proof.

9. Minimizing Network Congestion Under length Restrictions Pseudo-Polynomial Algorithm є- Optimal Approximation Scheme Extensions

10. Pseudo-Polynomial Solution  = the total flow along e=(u,v) that has been routed from s to u through paths with a total length of u vw

11. Pseudo-Polynomial Solution (Linear Program) u Objective function:  Minimize α u Constraints:  α is the network congestion factor i.e., for each e  E  Nodal flow conservation constraint i.e., for each v  V\{s,t}

12. Pseudo-Polynomial Solution (Linear Program) u Constraints (cont.):  Demand constraint: u The complete linear program:

13. Pseudo-Polynomial solution u The linear program can be solved within time complexity that is polynomial in the number of variables. u Therefore, the complexity incurred by solving the linear program is polynomial in L.  Indeed, the number of variables is O(|E|·L).

14. Approximation Scheme u Goal: reduce the number of variables to be polynomial in |V| and |E| instead of L. u Scaling: u Apply the linear program for the new instance.  The new instance relaxes the original instance.  Hence, congestion is not worse than the optimum. u Convert each non-simple path into a simple path. u Accumulating error for a path: (N-1)· . u New path length is at most: L+ N·  =L∙(1+є).

15. Extensions u Multi-commodity  It is straightforward to extend the linear program to the multi-commodity case. u 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.

16. 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.

17.  /K- integral flows that minimize congestion u Minimize the network congestion factor such that:  = 3 K=2 f e =2 f e =1 The demand  is routed along at most K paths. The flow over each path is a multiple of  /K. New constraint Original constraint f e =0  = 3 K=2 f e =0

18. An optimal  /K- integral flow is a 2-apx scheme u Theorem: The minimum congestion of a  /K-integral flow is at most twice the congestion of the optimal solution.  Proof u Each  /K- integral flow satisfies the requirement to ship the demand  on at most K paths. u Corollary: minimizing the congestion while restricting the flow to be integral in  /K is a 2-approximation scheme for the original problem.

19. Computing optimal  /K-integral flows u 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 e  E it holds that  Thus, for each e  E it holds that  In particular, u Sufficient to find the  /K-integral flow that has the minimum network congestion factor in

20. 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

21. Computing optimal  /K-integral flows u Since the set A is polynomial the complexity of the solution is polynomial. u 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.

22. Future Work u A unifying scheme that bounds the number of paths AND the length of each path. u Distributed implementation of both algorithms. u Heuristic schemes with lower complexity.

23. Questions?

24. Proof (Sketch) u Step 1: Find a flow that minimizes congestion while routing traffic along K paths u Step 2: Double the flow over each path u Step 3: Round down the flow over each path to a multiple of  /K. u By construction, is a  /K- integral flow. u In step 3 the total flow is reduced by at most  units.  There are K paths, each “looses” at most  /K units. u Hence, transfers at least flow units.

25. Pseudo-Polynomial solution

26. Nodal flow conservation constraint v for each v  V\{s,t}

27. 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) ≤ D?  The partition problem: Given an ordered set of elements a 1, a 2,…, a 2n that constitute a set A with a size s(a)  + for each a  A, is there a subset A’  A such that A’ contains exactly one element of a 2i-1, a 2i for 1 ≤ i ≤ n such that ∑ a  A’ s(a)=∑ a  A\A’ s(a)? u 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 ½∑ a  A s(a) iff there is a subset A’  A such that A’ contains exactly one element of a 2i-1, a 2i for 1 ≤ i ≤ n and ∑ a  A’ s(a)=∑ a  A\A’ s(a). S(a 1 ) S(a 3 ) S(a 5 ) S(a 2n-1 ) S T S(a 2 )S(a 4 ) S(a 6 ) S(a 2n )

28. The end-to-end delay restriction is intractable <=  There is a a subset A’  A such that A’ contains exactly one element of a 2i-1, a 2i for 1≤i≤n and ∑ a  A’ s(a)=∑ a  A\A’ s(a).  The selection of the links that correspond to the elements of A’ and the zero delay links that connect these links constitutes a path p.  Path p is disjoint to the path that the complement subset A\A’ 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 ½∑ a  A s(a).=>  There is a path flow that transfers two flow units over paths that are not larger than ½∑ a  A s(a).  Let p be a path that carries a positive flow; by construction, p contains exactly one element of a 2i-1, a 2i for 1≤i≤n.  Since all the links have one unit of capacity p can transfer at most 1 flow unit.  Therefore, there exists a path p’ that is disjoint to p that transfers a positive flow; by construction, p’=A\p  Hence, D(p) ≤ ½∑ a  A s(a) and D(p’) ≤ ½∑ a  A s(a).  Therefore, since D(p)+ D(p’)=∑ a  A s(a) it follows that ∑ a  p s(a)=∑ a  p’ s(a)=½∑ a  A s(a).

29. The restriction on the number of paths is intractable u 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 t 1, t 2,…, t k and corresponding demands D 1, D 2,…, D k, is there an assignment of traffic to paths such that for each 1 ≤ i ≤ k demand D i is routed over a single path without violating the capacity constraints?  Claim: There exists a path flow that transfers  = D 1 + D 2 +…+ D k flow units from S to T over at most K paths iff it is possible to find an assignment of the demands D 1, D 2,…, D k to paths such that D i, 1 ≤ i ≤ k is routed over a single path without violating the capacity constraints  There is exactly one path from S to t i for each 1 ≤ i ≤ k. Hence, there are exactly K paths from S to T that carry a positive flows.  There is at least one path from S to t i for each 1 ≤ i ≤ k. However, since there are at most K paths there is exactly one path from S to t i for each 1 ≤ i ≤ k. S t1t1 t2t2 tktk T D1D1 D2D2 DkDk