CSE 291-a Interconnection Networks Lecture 7: February 7, 2007 Prof. Chung-Kuan Cheng CSE Dept, UC San Diego Winter 2007 Transcribed by Thomas Weng

Topics Circuit Switching - Definitions: Nonblocking, rearrangeable, strict. Crossbar Clos Networks

Crossbar n inputs, n outputs, n 2 switches Rearrangeable, strict and wide nonblocking If n is small, this is usually the best choice. xxx xxx xxx 1 2 n 12n

Engineering Issues 1.Physical layout (what to do with many nodes?) 2.Control (packet switching) Paper - BlackWidow: High-Radix Clos Networks, S. Scott, D. Abts, J. Kim, W.J. Dally 123n 1 2 n

Physical Layout (example) wires per row 64 horizontal wires 64 wires in between each signal 8x64 vertical wires in all 64 8x8 switches 1 8 8x8 crossbar Goes to row 1, row 2, …, row 8

Clos Network: Three Stage Clos(m,n,r) 1 n 2 n r n n x m 1 2 m r n n n

Clos Network (continued) 1.rn inputs, rn outputs 2.2rnm + mr 2 switches (this is less than r 2 n 2 ) Clos(m,n,r) is rearrangeable iff m >= n Let m = n 1.rn inputs 2rn 2 + nr 2 switches = (2n + r)rn (a crossbar is rn 2 switches) Optimal choice of n and r?

Clos Network - Proof Proof: By induction Clos(1,1,r) – you have r boxes, each box is 1 x 1 This is a crossbar, which we know is rearrangeable. 1 n 2 n r n.... r 1 2 r....

Clos Network – Proof (cont) Assume that for the case Clos(n-1, n-1, r), n>=2, the statement is true. For the case Clos(n, n, r), we use the first switch in the middle to reduce the requirement to Clos(n-1, n-1, r). 1 n r n n x m 1 n r n n

Clos Network – Proof (cont) Because n inputs, n outputs, we can always find a perfect matching. If we take out a middle box, and now have (n-1) inputs, (n-1) outputs n n n (r-1)n+1.. rxnrxn 11 22. nn PermutationOutput side (i,j) for (j) = i Each box will have n outputs Each box will have n input edges 1 n 2 n n n Each box is a node with degree=n 1 2 n n n n Bipartite graph Perfect matching

Clos Network – Strictly non- blocking 1 n 2 n r n n x (2n–1) 1 2 2n r n n n Clos Network is non-blocking in strict sense when m >= 2n-1. Each box has 2n-1 output pins Each box has 2n-1 input pins

Clos Network – Proof Proof by contradiction From i to j, we cannot make connection, e.g. from 1 to 2, we cannot make connection. Only time we can’t make a connection is if all paths are taken. Input i has taken n-1 signals, output j has taken n-1 signals. Thus, at most 2n-2 paths are taken. However, we have 2n-1 boxes for 2n-1 distinct paths between i and j. So we will always have at least one path to go through.

Clos Network – Proof (cont) 1 n 2 n r n n x (2n–1) 1 2 2n r n n n At most n-1 boxes taken from 1, and n-1 boxes taken from 2, so 2n-2 boxes are taken. n-1

Clos Network – More than three stages 1 n r n m r.... n n Clos Networks: Adding wires to reduce switches. Can we do better? Add even more wires to reduce number of switches? Yes! By increasing number of stages. Change middle stage box into another 3-stage Clos Network, this gives us 5-stage Clos Network. Can repeat this process! Replace with 3-stage Clos Network

Clos Network – More than three stages (cont) C(1) = N 2 switches (crossbar) C(3) = 6N 3/2 – 3N C(5) = 16N 4/3 – 14N + 3N 2/3 C(7) = 36N 5/4 – 46N + 20N 3/4 – 3N 1/2 C(9) = 76N 6/5 – 130N + 86N 4/5 – 26N 3/5 + 3N 2/5 (if N is huge, we want more levels)

Benes Network Start with Butterfly Network. What if we flip this, and repeat this network to the other side? This is Benes Network n n.... 2n 2x2 n inputs: (2n+1)2 n x4 switches N inputs: 2N(2log 2 N-1) switches