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Published byTurner Muscat Modified over 2 years ago

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The strength of routing Schemes

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

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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

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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.

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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

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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)

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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)

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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?

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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)

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

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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.

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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)).

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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

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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

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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

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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

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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

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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?

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

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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)

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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)

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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)

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H n -ln n

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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!

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21n 21n sources destinations

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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

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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.

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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

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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

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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.

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Histogram - Inet, Zipf

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Inet, Zipf

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

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