Optimal Load-Balancing Isaac Keslassy (Technion, Israel), Cheng-Shang Chang (National Tsing Hua University, Taiwan), Nick McKeown (Stanford University, U.S.A.), Duan-Shin Lee (National Tsing Hua University, Taiwan)

Router Designer Wishlist 1. Mesh Switch: avoid switch reconfiguration and complex scheduling algorithms. Practical for optics (AWGR) % Throughput: router guaranteed to be stable under any admissible traffic matrix 3. Minimum Linecard Complexity  Minimize maximum rate at which packets arrive to/depart from any input/output.  Buffering Speed  Processing Speed

In Out R R R R R R ? ? ? ? ? ? ? ? ? R R R R R R R R R R R R R Naive Mesh with 100% Throughput Output Write Speed = NR

Output-Queued Mesh 1. Mesh % throughput 3. … but output write speed = NR

R In Out R R R R R R/N If Traffic Is Uniform R R

100% Throughput: Non-Uniform Traffic Matrices R In Out R R R R R R/N R R R R R R R R R ?

Out R R R R/N Load-Balanced Router Load-balancing meshForwarding mesh In Out R R R R/N R R R Theorem: 100% Throughput [Val. 82, CLJ 01, K. et al. 03]

Out R R R R/N In R R R R/N Load-balancing mesh Forwarding mesh Load-Balanced Router

Out R R R R/N In R R R R/N Load-balancing mesh Forwarding mesh Load-Balanced Router

1. Mesh → 2 meshes % throughput 3. Node speed? ? ≈

Out R R R R/N In R R R R/N One linecard In Out In Out Combining the Two Meshes

In Out In Out In Out In Out R In Out In Out In Out In Out R 2R/N A Single Combined Mesh

Matrix for the Combined Mesh  Combined mesh matrix:  The combined mesh matrix gets 100% throughput

Node Speed for Combined Mesh Max input/output read/write speed = 2R In Out In Out In Out In Out R In Out In Out In Out In Out R 2R/N

Combined Mesh 1. Single Mesh % Throughput 3. Max Node Speed = 2R Question: is 2R optimal? Any better architecture? ?

Can Another Architecture Do Better?  Criteria: Given all architectures with 100% throughput, minimize the maximum node processing speed.

Other Mesh Architectures We Consider  Any number of stages (e.g., 3 stages, 4 stages….)  Any mesh architecture (e.g., ring)  Any link capacities (e.g., non-uniform mesh)  Any packet routing algorithm (e.g., adaptive algorithm) Any mesh and any routing.

Example 1: Add A Third Mesh? In Out R R R R/N R R R In Out R R R R/N In Out R R R R/N 1 st stage 2 nd stage3 rd stage Combine the 3 meshes  Max speed = 3R (instead of 2R )

Example 2: Use a Non-Uniform Mesh  This is actually a ring!

Example 2: Unidirectional Ring  Assume that each node sends all traffic to itself.  Then each packet goes through N nodes.  To get 100% throughput, each node needs to run N times faster i N  Max speed = NR (instead of 2R )

At First Glance…  … it seems that the uniform mesh is optimal with 2R !  Why:  All links have the same capacity,  And it is perfectly symmetric.  However…. uniform mesh is NOT optimal!

Why Uniform Mesh is Not Optimal  Links between two different nodes used for spreading and forwarding  Same-node links only used for forwarding, not spreading  need less capacity.  Example: packet from node 1 to node 2. No point in sending it from node 1 to node 1 before forwarding to node 2! 12

Main Result Slightly Non- Uniform Mesh Slightly better than 2R

However…  The result is actually good for the load-balanced router with uniform mesh.  The uniform mesh is optimal as N → 1  In other words, asymptotically with N, the load- balanced router is at least as good as any other mesh architecture with any other routing algorithm. The load-balanced router satisfies the wishlist goals.

Generalization: Load-Balanced Network 1 3 … N  Hotnets III, Nov. 2004:  Zhang-Shen and McKeown  Kodialam, Lakshman and Sengupta  Two steps: 1. Uniform spreading of incoming packets (independently of destination) 2. Forwarding to destination A uniform load-balanced backbone  guarantees 100% throughput for any traffic matrix  is at least as good as any other backbone design

Thank you.