Design Principles of Scalable Switching Networks Shanghai Jiao Tong University Design Principles of Scalable Switching Networks

2×2 Switching Network Most simplest switching network Bar State Cross State

𝑁×𝑁 Switching Network 1 ? 1 2 2 3 3 4 4

𝑁×𝑁 Crossbar Implement an 𝑁×𝑁 switch by an array of 2×2 switches Connection from input 𝑖 to output 𝑗 Set the switch 𝑖,𝑗 to the bar state Set other switches along the path to the cross state 1 1 2 2 3 3 4 4

A new Connection New connections between a pair of free input and output can be freely set up without distribution of existing connections 1 1 2 2 3 3 4 4

Strictly Non-Blocking (SNB) A switch is SNB if a connection can always be set up between any idle (or free) input and output without the need to rearrange the paths of the existing connections

Performance of 𝑁×𝑁 Crossbar Crossbar Implementation: SNB Routing complexity: 𝑂 1 Hardware complexity: 𝑂 𝑁 2 1 1 2 2 3 3 4 4

Lower Bound of Hardware Requirement An 𝑁×𝑁 switch can support 𝑁! mappings Let 𝑀 be the minimal number of required crosspoints # states ≥ # mappings 2 𝑀 ≥𝑁! 𝑀≥ log 2 𝑁! ≈𝑁 log 2 𝑁 for large 𝑁 How to reach this lower bound? 1 2 N .

Scalable Design: Two-Stage Network Blocking Input 2 cannot be connected to output 2 if input 1 is already connected to output 1 1 2 3 4

Scalable Design: Dilated Two-Stage Network There are multiple links between the modules at two adjacent stages 1 2 3 4

Performance of Dilated Networks Suppose 𝑁=𝑟𝑛 Bandwidth expansion factor = 𝑛: nonblocking Number of crosspoints: 𝑛×𝑟𝑛 ×2𝑟=2 𝑁 2 =𝑂 𝑁 2 1 2 r . n n 虽然不是低复杂度的设计，但是提供了一个很好的信息：要想无阻塞，在每对输入输出模块之间至少要有n条路

Scalable Design: Three-Stage Network Clos Network by C. Clos in 1953 𝑛 1 𝑟 1 = 𝑛 3 𝑟 3 = 𝑁 for 𝑁×𝑁 switch There is exactly one link between the modules at two adjacent stages n1 × r2 r1 × r3 r2 × n3 (1) (2) (r1) (r2) (r3) Input Stage Central Stage Output Stage

Routing Constraint Contention-free routing condition: Two connections from the same input module (or to the same output module) can not share the same central module A F G B 1 2 3 4 5 6 7 8 9 H 虽然clos网络也可以实现无阻塞，但是也不是没有前提的。

Condition on 𝑟 2 𝑟 2 central modules provide 𝑟 2 paths for each pair of input module and output module It seems that 𝑟 2 should be larger than 𝑛 1 and 𝑛 3 Is this condition enough? A F G B 1 2 3 4 5 6 7 8 9 H

Example A request for connection from input 9 to output 4 is blocked 𝑆 3 ≜ set of central modules used by input module 3 = 𝐹,𝐺 𝐷 2 ≜ set of central modules used by output module 2 = 𝐻 Hint: find an central module that is not used by both the input module and the output module A F G B 1 2 3 4 5 6 7 8 9 H

One Possible Solution Increase the number of central modules

One Possible Solution Such that a new path always available for new connection

SNB iff 𝑟 2 ≥ min 𝑛 1 + 𝑛 3 −1,𝑁 Proof: 1) Trivial case: 𝑁≤ 𝑛 1 + 𝑛 3 −1 2) If 𝑁> 𝑛 1 + 𝑛 3 −1, consider the worst case where all other inputs in input module 𝑖 and all other outputs in output module 𝑗 are busy, i.e., 𝑆 𝑖 = 𝑛 1 −1, 𝐷 𝑗 = 𝑛 3 −1. It follows that 𝑆 𝑖 ∪ 𝐷 𝑗 = 𝑆 𝑖 + 𝐷 𝑗 − 𝑆 𝑖 ∩ 𝐷 𝑗 ≤ 𝑆 𝑖 + 𝐷 𝑗 = 𝑛 1 + 𝑛 3 −2. Thus, if 𝑟 2 ≥ 𝑛 1 + 𝑛 3 −1, there is at least one central module available for the request from input module 𝑖 to output module 𝑗 if there is a new request from 𝑖 to 𝑗

Hardware Complexity of SNB Clos Networks Suppose 𝑟 1 = 𝑛 1 = 𝑟 3 = 𝑛 3 = 𝑁 . To be SNB, 𝑟 2 ≥ 𝑛 1 + 𝑛 3 −1=2 𝑁 −1. In this case, the number of needed crosspoints is 2 𝑁 × 𝑁 × 𝑁 + 2 𝑁 −1 × 𝑁 × 𝑁 =4 𝑁 3 2 −𝑁

Comments on SNB Clos Networks The hardware complexity of an 𝑁×𝑁 SNB Clos network is less than that of an 𝑁×𝑁 crossbar when 𝑁>14 However, the routing process in the SNB Clos network is more complex than the crossbar Can the hardware complexity of Clos networks be further reduced?

Another Possible Approach If a new connection request is blocked, it can still be satisified if we rearrange some exsiting connections 4 1 2 3 4 1 2 3

Rearrangeably Non-Blocking (RNB) A switch is rearrangeably non-blocking if a connection can always be set up between any idle input and output, although it may be necessary to rearrange the existing connections

Paull Matrix Row 𝑖: input module 𝑖 Column 𝑗: output module 𝑗 Entry 𝑖,𝑗 : set of central modules used by the connections from input module 𝑖 to output module 𝑗 F,G,H 2 1 i r1 j r2 Output Module Input Module i j F G H 为了更好地观察这种路由限制，定义Paull矩阵

A Legitimate Paull Matrix In row 𝑖: 𝑆 𝑖 is the set of central modules appear in row 𝑖 A central module appears in row 𝑖 or 𝑆 𝑖 only once Number of employed modules: 𝑆 𝑖 ≤min 𝑛 1 , 𝑟 2 In column 𝑗: 𝐷 𝑖 is the set of central modules appear in column 𝑗 A central module appears in column 𝑗 or 𝐷 𝑗 only once Number of employed modules: 𝐷 𝑗 ≤min 𝑛 3 , 𝑟 2

Example: Blocking Let 𝑀= 𝐴,𝐵,𝐶 be the set of central modules 𝑆 2 = 𝐴,𝐵 𝐷 3 = 𝐴,𝐶 Request 6->9, input 6 is an input of input module 2 and output 9 is an output of output module 3, can not be built up since 𝑀− 𝑆 2 ∪ 𝐷 3 =𝜙 B 2 1 3 C A 4

Example: Alternating Chain 𝑆 2 does not contain 𝐵, and 𝐷 3 does not have 𝐶: find C-B chain B 2 1 3 C A 4 C 2 1 3 B A 4

Example: Setup New Connection 𝑆 2 and 𝐷 3 do not have 𝐶: assign 𝐶 to the new connection C 2 1 3 B A,C 4 A

RNB iff 𝑟 2 ≥ min 𝑛 1 , 𝑛 3 Proof: 1) Only if: trivial 2) if: similarly, we consider the worst case, where where all other inputs in input module 𝑖 and all other outputs in output module 𝑗 are busy, i.e., 𝑆 𝑖 = 𝑛 1 −1, 𝐷 𝑗 = 𝑛 3 −1. Thus, there are two cases: a) 𝑆 𝑖 ∪ 𝐷 𝑗 < 𝑟 2 , and b) 𝑆 𝑖 ∪ 𝐷 𝑗 = 𝑟 2 . As to case (1), a new connection from input module 𝑖 to output module 𝑗 can be immediately set up. For case (2), we have 𝑆 𝑖 − 𝐷 𝑗 = 𝑆 𝑖 ∪ 𝐷 𝑗 − 𝐷 𝑗 ≥ 𝑟 2 − 𝑛 3 −1 ≥ 𝑛 3 − 𝑛 3 +1=1, and 𝐷 𝑗 − 𝑆 𝑖 = 𝑆 𝑖 ∪ 𝐷 𝑗 − 𝑆 𝑖 ≥ 𝑟 2 − 𝑛 1 −1 ≥ 𝑛 1 − 𝑛 1 +1=1. It follows that: ∃ a central module, say 𝐴, in 𝑆 𝑖 but not in 𝐷 𝑗 or

RNB iff 𝑟 2 ≥ min 𝑛 1 , 𝑛 3 (con’t) ∃ a central module, say 𝐵, in 𝐷 𝑗 but not in 𝑆 𝑖 . This is visualized as follows: A B 𝑖 ′ 𝑖 𝑗 ′ 𝑗 𝑖 ′′ 𝑖 ′′′ 𝑗 ′′ 𝑖 ′ 𝑖 ′′ 𝑖 𝑖 ′′′ . 𝑗 ′ 𝑗 𝑗 ′′ A B 𝑖 already connects to all middle-stage nodes except 𝐵 𝑗 already connects to all middle-stage nodes except 𝐴

RNB iff 𝑟 2 ≥ min 𝑛 1 , 𝑛 3 (con’t) We can obtain a new legitimate Paull matrix, by rearranging the Paull matrix as follows: A B 𝑖 ′ 𝑖 𝑗 ′ 𝑗 𝑖 ′′ 𝑖 ′′′ 𝑗 ′′ B . 𝑖 ′ 𝑖 ′′ 𝑖 𝑖 ′′′ . 𝑗 ′ 𝑗 𝑗 ′′ A B

Bipartite Model Input/output module ↔ a node in 𝑋/𝑌 Connection ↔ an edge Central module ↔ a color Routing constraints: two edges share the same node can not use the same color

Example: Alternating Path and Color Exchange

Hardware Complexity of SNB Clos Networks Suppose 𝑟 1 = 𝑛 1 = 𝑟 3 = 𝑛 3 = 𝑁 . To be RNB, 𝑟 2 ≥ max 𝑛 1 , 𝑛 3 = 𝑁 . In this case, the number of needed crosspoints is 2 𝑁 × 𝑁 × 𝑁 + 𝑁 × 𝑁 × 𝑁 =3 𝑁 3 2 Recall that, the number of crosspoints needed by an SNB network is 4 𝑁 3 2 −𝑁 Can the hardware complexity be further reduced?

Benes Network The 𝑁×𝑁 network is RNB if each 𝑁 2 × 𝑁 2 network are RNB 2 × 2 𝑁 2 × 𝑁 2 . 1 2 3 4 N-1 N

Recursive Decomposition A 16×16 Benes Network

Recursive Decomposition A 16×16 Benes Network

Recursive Decomposition A 16×16 Benes Network

Complexity Let 𝑁= 2 𝑛 and 𝑓 𝑘 be number of stages of a 𝑘×𝑘 Benes network. We have 𝑓 𝑁 =𝑓 𝑁 2 +2 then 𝑓 2 𝑛 =𝑓 2 𝑛−1 +2 =𝑓 2 𝑛−2 +2×2 =⋯ =𝑓 2 𝑛−𝑗 +2×𝑗 =⋯ =𝑓 2 +2 𝑛−1 =1+2 𝑛−1 =2𝑛−1=2 log 𝑁 −1 Thus, number of 2×2 switches is 𝑂 𝑁 log 𝑁

Loop Routing Algorithm: 𝑂 𝑁 log 𝑁 Set up paths for input-output pairs (1, 4), (2, 5), (3, 6), (4, 3) (5, 7), (6, 8), (7, 1), (8, 2) 1 2 3 4 5 6 7 8