multi-terabit ip lookup using parallel bidirectional pipelines author: weirong jiang, viktor k....

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Multi-Terabit IP Lookup Using Parallel Bidirectional Pipelines Author: Weirong Jiang, Viktor K. Prasanna Publisher: May 2008 CF '08: Proceedings of the 2008 conference on Com puting frontiers ACM Presenter: Yu-Ping Chiang Date: 2008/09/16

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Multi-Terabit IP LookupUsing Parallel Bidirectional Pipelines

Author: Weirong Jiang, Viktor K. Prasanna Publisher: May 2008 CF '08: Proceedings of the 2008

conference on Computing frontiers ACM Presenter: Yu-Ping ChiangDate: 2008/09/16

Outline

Overview Front End Back End

Memory Balancing Trie Partitioning Subtrie-to-Pipeline Mapping Node-to-Stage Mapping

Performance

Front End

Receive packets Dispatch packets to pipelines:

cache hit / miss set delay

Back End

Process packets Output retrieved next-hop information:

using delay to retrieve

output information

Outline

Overview Front End Back End

Memory Balancing Trie Partitioning Subtrie-to-Pipeline Mapping Node-to-Stage Mapping

Performance

Trie Partitioning

Initial stride (I) following section: I=12

I=2

Subtrie-to-Pipeline Mapping Problem formulation

Algorithm – O(KP) K = # of triesP = # of pipelines

Performance

Node-to-Stage Mapping

Problem formulation

Constraint: ancestor mapped preceding to child. Main idea:

two subtries mapped onto different direction. two same trie level nodes mapped onto differe

nt stages.

Inversion: Methods:

largest leaf least height largest leaf per height least average depth per leaf

(use in following section)

Inversion Factor (IFR) (in following section: 4~8)

O(HN) Node fields:

Distance to child Memory address of child

H = # of pipeline stagesN = total # of trie nodes

Outline

Overview Front End Back End

Memory Balancing Trie Partitioning Subtrie-to-Pipeline Mapping Node-to-Stage Mapping

Performance

Performance Memory: 1.8 MB

(13+5)*2^13*25*4=14.75Mb=1.8MB 18 KB/stage

18.75 G packets / sec 7.5 PPC*2.5GHz=18.75G packets/sec

Power consumption: 0.2 W / IP lookup 0.008*25=0.2W