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Nick McKeown

Spring 2012 Lecture 4 Parallelizing an OQ Switch. EE384x Packet Switch Architectures. Nick McKeown. many outputs. one output. 1. 1. 1. 1. 1. k. k. N. N. Scaling an OQ Switch. Work conserving if memory b/w >= R(N+1). Not so clear. B5. C5. B5. A5. B6. A5. B6. B6. C6. A6.

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Nick McKeown

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  1. Spring 2012 Lecture 4 Parallelizing an OQ Switch EE384x Packet Switch Architectures Nick McKeown

  2. many outputs one output 1 1 1 1 1 k k N N Scaling an OQ Switch Work conserving if memory b/w >= R(N+1) Not so clear.

  3. B5 C5 B5 A5 B6 A5 B6 B6 C6 A6 B5 Time slot = 2 Time slot = 3 Time slot = 1 A6 B A8 A8 A7 A8 A7 Parallel OQ SwitchMay not be work-conserving Constant size packets 1 A 1 2 k=3 C N=3 At most two memory operations per time slot: 1 write and 1 read

  4. Work Conserving Problem How can we design a parallel OQ work-conserving switch from slower parallel memories? • Theorem (sufficiency) • A parallel output-queued switch is work-conserving with 3N –1 memories, each able to perform at most one memory operation per time slot.

  5. Re-stating the Problem • There are K cages which can contain an infinite number of pigeons. • Assume that time is slotted, and in any one time slot • At most N pigeons can arrive and at most N can depart. • At most 1 pigeon can enter or leave a cage via a pigeon hole. • The time slot at which arriving pigeons will depart is known • For any switchWhat is the minimum K, such that all N pigeons can be immediately placed in a cage when they arrive, and can depart at the right time?

  6. DT=t DT=t+X Only one packet can enter or leave a memory at time t Only one packet can enter or leave a memory at any time DT=t+X Intuition for Theorem Time = t Only one packet can enter a memory at time t Memory

  7. Proof of Theorem When a packet arrives in a time slot it must choose a memory not chosen by • The N – 1 other packets that arrive at that timeslot. • The N other packets that depart at that timeslot. • The N - 1 other packets that can depart at the same time as this packet departs (in future). Proof By the pigeon-hole principle, the switch can be work-conserving if there are 3N –1 memories, each able to perform at most one memory operation per time slot.

  8. Memory Memory Memory Memory Memory Memory Memory Memory A5 1 A4 A3 A4 A1 A1 A5 B1 C3 C3 C1 B3 B1 K=8 C1 A Parallel Shared Memory Switch At most one operation – a write or a read per time slot A R B R C Arriving Packets Departing Packets • From theorem 1, k = 7 memories don’t suffice .. but 8 memories do

  9. Distributed Shared Memory Switch Switch Fabric Memories Memories Memories R R R Line Card 1 Line Card 2 Line Card N The central memories are distributed to the line cards and shared. Memory and line cards can be added incrementally. From theorem 1, the switch is work-conserving if we have a total of 3N –1 memories, each able to perform one operation per time slot i.e. a total memory bandwidth of  3NR.

  10. Switch bandwidth What switch bandwidth does the DSM switch need in order to be work-conserving? Theorem (sufficiency)A switch bandwidth of 4NR is sufficient for a distributed shared memory switch to be work-conserving. Proof There are a maximum of 3 memory accesses and 1 external line access per time slot.

  11. Switch Algorithm What switching algorithm allows the DSM switch to be work-conserving? • Shared bus: No algorithm needed. • Crossbar switch: Algorithm needed because only permutations are allowed. Theorem An edge coloring algorithm can switch packets for a work-conserving distributed shared memory switch ProofKönig’s theorem: Any bipartite graph with maximum degree  has an edge coloring with  colors.

  12. Summary - Switches with 100% throughput Switch Algorithm Total MemBW Switch BW Fabric # Mem. Mem. BW OQ Bus N (N+1)R N(N+1)R NR None Shared Mem. Bus 1 2NR 2NR 2NR None MWM IQ Crossbar N 2R 2NR NR 2N 3R 6NR 2NR Maximal CIOQ Cisco GSR Crossbar Time Reserve* 2N 3R 6NR 3NR PSM Bus k 3NR/k 3NR 3NR C. Sets DSM Juniper M-series N 3R 3NR 4NR Edge Color Xbar N 3R 3NR 6NR C. Sets N 4R 4NR 4NR C. Sets PPS - OQ Clos Nk 2R(N+1)/k 2N(N+1)R 4NR C. Sets Nk 4NR/k 4NR 4NR C. Sets PPS Clos None Nk 2NR/k 2NR 2NR

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