Multipath Routing Algorithms for Congestion Minimization Ron Banner

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

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.

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.

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

How much is gained by optimal flow distribution? u Experiment: Generated random networks that include 10, 000 Waxman topologies & 10, 000 power-law topologies. u 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? ur(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. Power law Waxman The full potential of multipath routing is far from having been exploited. . .

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.

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 .

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

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

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}

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

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).

Approximation Scheme u Goal: reduce the number of variables to be polynomial in |V| and u Scaling: u Apply the linear program for the new instance. |E| instead of L. · The new instance relaxes the original instance. · Hence, congestion is not worse than the optimum. 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). u

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.

Minimizing Congestion while Routing Along at Most K Different Paths. ü /K- integral flows that minimize congestion ü An optimal /K- integral flow is a 2 -APX scheme. üComputing optimal /K- integral flows.

/K- integral flows that minimize congestion u Minimize the network congestion factor such that: l ina ig Or nt con ai r t s The demand is routed along at most K paths. =3 fe=0 Ne w co ns tr ain t The flow over each path is a multiple of /K. =3 fe=0 fe=2 fe=1 K=2

An optimal /K- integral flow is a 2 -apx scheme u Theorem: The minimum congestion of a /K-integral flow is at most solution. · Proof twice the congestion of the optimal 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.

Computing optimal /K-integral flows u The network congestion factor of each /Kintegral 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

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 E. ; otherwise repeat the process with a smaller Output the flow that transfers flow units and has the smallest

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.

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

Questions?

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

Pseudo-Polynomial solution

Nodal flow conservation constraint for each v V{s, t} v

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? u The partition problem: Given an ordered set of elements a 1, a 2 , …, a 2 n 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 2 i-1, a 2 i for 1≤i≤n such that ∑a A’ s(a)=∑a AA’ s(a)? S(a ) u S(a 1) S(a 3) 5 2 n-1 S T S(a 2) S(a 4) S(a 6) S(a 2 n) All link capacities are 1. u 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 2 i-1, a 2 i for 1≤i≤n and ∑a A’ s(a)=∑a AA’ s(a). u

The end-to-end delay restriction is intractable <= u u u There is a a subset A’ A such that A’ contains exactly one element of a 2 i-1, a 2 i for 1≤i≤n and ∑a A’ s(a)=∑a AA’ 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 AA’ 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). => u u u 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 2 i-1, a 2 i 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’=Ap 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).

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? u The single source unsplittable flow problem: Given a network G with a source s, targets t 1, t 2 , …, tk and corresponding demands D 1, D 2 , …, Dk , is there an assignment of traffic to paths such that for each 1≤i≤k demand Di is routed over a single path without violating the capacity constraints? u S t 1 t 2 D 1 u Dk tk T Claim: There exists a path flow that transfers = D 1+ D 2 +…+ Dk 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 , …, Dk to paths such that Di, 1≤i≤k is routed over a single path without violating the capacity constraints Ü There is exactly one path from S to ti for each 1≤i≤k. Hence, there are exactly K paths from S to T that carry a positive flows. Þ There is at least one path from S to ti for each 1≤i≤k. However, since there at most K paths there is exactly one path from S to ti for each 1≤i≤k.