The strength of routing Schemes

Main issues Eliminating the buzz: Are there real differences between forwarding schemes: OSPF vs. MPLS? Can we quantify them?

Outline Define packet forwarding paradigms: –Vanilla IP, OSPF, MPLS, general bifurcation Compare their relative strength: –upper and lower bounds on performance ratio A centralized heuristic for vanilla IP forwarding –control is centralized anyway –achieves good performance

Packet forwarding in practice Vanilla IP –forward all packets destined to some addr. to a selected shortest path OSPF –like above, but allow equal splitting when multiple shortest paths exist MPLS –pre-select routes for flows.

Forwarding Modelling Network as a graph : G(V,E), |V|=n, |E|=m. N v – the set of neighbors of node v. c e >0– the capacity of link e E D={d i,j } – the demand matrix Routing assignment: R: V 4 [0..1], φ u,v (i,j) is the relative amount of (i,j)-flow that is routed from a node u to a neighbor v. 1. For all u,i,j V: Σ v N u φ u,v (i,j)=1 2. For all u,i,j,v V, v N u : φ u,v (i,j)=0

source invariance A routing assignment R is source invariant if it does not depend on the source: φ u,v (i 1,j) = φ u,v (i 2,j) φ u,v (j)

Unrestricted Splitable Routing (US-R) Restricted Splitable Routing (RS-R) –Split over at most L outgoing links –Special case: Unsplittable flow problem (RS-R 1 ) Standard IP Forwarding (IP-R) –Source invariant RS-R 1 OSPF Routing (OSPF-R) –Source invariant routing assignments splitting flow evenly among next hops. Routing Paradigms u,j V, v N u : φ u,v (j)=1 u,j,v,v V, if φ u,v (j)>0 and φ u,v (j)>0 then φ u,v (j)= φ u,v (j)

How packets are splitted? Option 1 (basic): packet sprinkler –each packet chooses next hop with prob. φ u,v (j) –may cause reordering hurts performance. Option 2 (flow-cached): hashing –each flow is hashed to next hop with prob. φ u,v (j) –may not result in splitting at desired ratios –can we afford double hashing/buckets at core?

Performance Measures Decide on an allocation matrix –say use max-min fairness Min Congestion –congestion factor (CF) = link flow / link capacity –hard constraint: congestion 1, –soft constraints minimize the penalty Max Flow (MF)

Hardness Result IP-R is NP even for a single destination!

Hardness Result node i has demand a i node x is connected to dest with capacity B node y is connected to dest with capacity a i -B 123 n … xy dest nodes 1,2,…,n are connected to nodes x and y with infinite capacity Equiv. to subset sum: The partition can be made if the max cong. = 1.

Comparison between paradigms Lower Bound on ratio: Example that shows the ratio is at least as high as (f(n)) Upper Bound on ratio: Show that a ratio of, at least, O(g(n)) can always be achieved. If f(n)=g(n) the bound is tight (g(n)).

IP-R vs RS-R 1 and OSPF-R Lower bound Ω(N) –IP-R: single path CF=N –RS-R 1 : separate routes CF=1 –OSPF-R: divide equally CF=1 Upper bound O(N) –IP-R can use the highest flow of RS-R 1 /OSPF-R 123n … x dest … n

IP-R vs RS-R 1 and OSPF-R Lower bound Ω(N) –IP-R: single path CF=N, MF=1 –RS-R 1 : separate routes CF=1, MF=N –OSPF-R: divide equally CF=1, MF=N Upper bound O(N) –IP-R can use the highest flow of RS-R 1 /OSPF-R 123n … x dest … n

N flows, each carry a unit demand OSPF-R –use single path thruput is 1 –use two paths thruput is 2 –use more - still limited by 2 (due to the first split) RS-R 1 can do N Lower bound Ω(N) N N-2 N-3 1 OSPF-R vs. RS-R 1 Max Flow (basic) N N

N flows, each carry a unit demand OSPF-R –to max. throughput must split the flows –max thruput is log N –given log* N stages: max thruput is 2 RS-R 1 can do N Lower bound Ω(N) N N-2 N-3 1 OSPF-R vs. RS-R 1 Max Flow (flow-cached) N N N N-2 N-3

N flows, each carry a unit demand OSPF-R –use single path CF=N –use two paths CF=N/2 on the down link RS-R 1 can do CF=1 Lower bound Ω(N) N N-2 N-3 1 OSPF-R vs. RS-R 1 Congestion Factor (both cases) N N

What do we have thus far? IP-R vs. RS-R 1 and OSPF-R (N) in both criteria. OSPF-R vs. RS-R 1 O(N) in all criteria and cases. But, we sometime used fairly complex topologies! What if topologies are simple? or very simple?

A Simple Topology SD L wlog, the link capacities are C 1 C 2 C L

OSPF-R –c l non-decreasing use all links from l * and above. –throughput is given by: (L- l * +1) c l* = C/ ln L OSPF-R vs. RS-R 1 Max Flow (basic)

OSPF-R –c l non-decreasing use all links from l * and above. –throughput is given by: (L- l * +1) c l* = C/ ln L OSPF-R vs. RS-R 1 Max Flow (basic)

OSPF-R –c l non-decreasing use all links from l * and above. –throughput is given by: (L- l * +1) c l* = C/ ln L RS-R 1 can achieve C Lower bound Ω(log L) We can also show that for any capacity allocation OSPF-R can achieve, at least, C/ ln L, hence (log L) OSPF-R vs. RS-R 1 Max Flow (basic)

H n -ln n

A centralized heuristic for vanilla IP forwarding Aim: improve performance of centrally controlled IP networks. Why centralized? –networks are centrally controlled anyway: IPNC. Static weight setting sucks!

21n 21n sources destinations

A centralized heuristic for vanilla IP forwarding Aim: improve performance of centrally controlled IP networks. Why centralized? –networks are centrally controlled anyway: IPNC. Static weight setting sucks! dynamic link weights adjustment

Link Weights The family of exponential weights: Proved to perform well by [AAP93] for related problems. [Fortz, Thorup,2000] used a piece-wise linear approximation of it. control the routing sensitivity to load.

Algorithm Input: network topology & demand matrix Output: forwarding tables 1. sort flows 2. initialize link weights 3. for every flow in sort order 4.route flow along SP with IP constraint 5.adjust weights

Simulation Setting –Two types of random networks: Flat & Inet –demand d i,j {1,2,3…}. D = Σ i,j d i,j –Demand matrices Destinations – uniformly chosen Sources – uniformly or Zipf-like chosen (param.=.5) –Link capacities – all 1 –Infinite bandwidth requirements –Three heuristics: rand, sort, dest –α=β/D, β=0,1,20,100,D

Total Flow When =D (Max Sensitivity) the flow increase by 30-50% All other cases, the total flow is almost the same. Even =1 improved performance significantly with almost no penalty in added flow.

Histogram - Inet, Zipf

Inet, Zipf

Summary At least, in theory OSPF cannot compete with MPLS abilities. In practice vanilla IP may be enough if you have central control.