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

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

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

5. 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

6. 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

7. 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

8. 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 .

9. 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

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

11. 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

12. 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

13. 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。

14. 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

15. 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

16. One Possible Solution
Increase the number of central modules

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

18. 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

19. 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 −𝑁

20. 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?

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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

32. Example: Alternating Path and Color Exchange

33. 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 −𝑁
Can the hardware complexity be further reduced?

34. 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

35. Recursive Decomposition
A 16×16 Benes Network

36. Recursive Decomposition
A 16×16 Benes Network

37. Recursive Decomposition
A 16×16 Benes Network

38. 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

39. 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