techniques for fast packet buffers sundar iyer, ramana rao, nick mckeown (sundaes,ramana,...

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Techniques for

Fast Packet Buffers

Sundar Iyer, Ramana Rao, Nick McKeown(sundaes,ramana, nickm)@stanford.eduDepartments of Electrical Engineering & Computer Science, Stanford University

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Problem Statement RedefinedMotivation:

To design an extremely high speed packet buffer architecture with fast access time and large size.

This talk:Is about the analysis of one such well known approach.

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Characteristics of Packet Buffer Architectures

• The total throughput needed is at least 2(Ingress Rate)

• Size of Buffer is at least R * RTT

• The buffers have one or more FIFOs

• The sequence in which the FIFOs are accessed is determined by an arbiter and is unknown apriori

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Memory Hierarchy of Packet Buffer

ArrivingPackets

DepartingPackets

Large DRAM memory with access time T’

Ingress SRAM Egress SRAM cache of FIFO heads

1

Q

1

Q

1

Q

b cells

R RArbiter

b cells b cells

Write Access Read Access Time = T= 2T’ Time = T = 2T’

Memory Management Algorithm

cache of FIFO tails

grants

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System Design ParametersMain Parameters

– SRAM Size– Latency faced by a cell

System Parameters– I/O Bandwidth– Number of addresses

• Use single address on every DRAM• Use different addresses on every DRAM

– Use/Non Use of DRAM Burst Mode– (non) Existence of Bank conflicts

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Today’s Talk…

Optimize Main Parameters– Minimize latency at cost of SRAM size – (Necessity and Sufficiency)

…… (later) Minimize SRAM size at cost of Latency

Assumptions on system parameters• No speedup on I/O

– I/O = 2R• Simple address architecture

– Use single address from every DRAM

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More Assumptions ..

• We shall assume that we have only cells of size “C” which arrive in the system

• No use of DRAM Burst Mode

• No bank conflicts

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Symmetry Argument

• The analysis and working of the ingress and egress buffer architectures are similar

• We shall analyze only the egress buffer architecture

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A Bad Case for the Queues …1

t = 0 t = 1 t = 2 t = 3

t = 4 t = 5 t = 6 t = 7

w

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A Bad Case for the Queues … 2

t = 8 t = 9 t = 10 t = 11

t = 12 t = 13 t = 14 … t = 17

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Observation

• There exists some value of “w” for which the buffer does not overflow

• w = qb is one such sufficient value• Threshold value “Ti” governs “w”.

wTib -1

Q

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Definitions• Occupancy

– This is the number of cells in the SRAM for a particular queue

• Active Queue– An active queue is one which has an

occupancy less than the threshold and has cells in the DRAM present for it

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One More Definition • Deficit

– This is defined as the difference between the threshold ‘T’ and the occupancy of an active queue.

– For a queue which is not active the deficit is zero

occupancy

b -1 deficit

Ti

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Can we Bound the Maximum Value of the Deficit?

• Define f(i,q)– The maximum deficit that a set of “i”

queues can have in a system of “q” queues

• We are interested in f(1,q)

• f(q,q) < qb …. trivially

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Largest Deficit Queue First

Recurrence Equations

• f(2,q) >= f(1,q) –b + [f(1,q) –b]• f(3,q) >= f(2,q) –b + [f(2,q) –b]/2• f(4,q) >= f(3,q) –b + [f(3,q) –b]/3• ……• f(q,q) >= f(q-1,q) –b + [f(q-1,q) –b]/(q-

1)

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Dirty Math..• qb > f(q,q) … trivially >= [f(q-1,q) –b] + [f(q-1,q) –b]/(q-1) >= f(q-1,q)(q/q-1) – b(q/q-1)

>= {f(q-2,q)(q-1/q-2) –b(q-1/q-2)}(q/q-1) – b[q/q-1]

>= f(q-2,q)q/q-2 –bq/q-2 –bq/q-1 >= f(q-3,q)q/q-3 –bq/q-3 –bq/q-2 - bq/q-1 ….. >= f(1,q) q/1 – bq sigma [1/i]• This gives, f(1,q) <= b[1 + ln q]

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Results

• If the MMA services the queue,– with the largest deficit &– has a simple address architecture – and no I/O speedup

• then– A latency of zero can be guaranteed when the – width of the SRAM is b[1 + lnq] + b = b [2 +

ln q]– And the size of SRAM is [2 + lnq]qb

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Necessity Traffic Pattern – b=2, q=8

t = 0 t = 8 t = 8 +8/2 t = 8 + 8/2 + 8/4

w w w w

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Necessity Analysis … 1

• In 1st iteration – q(b-1/b) queues with deficit 1

• In 2 nd iteration– q(b-1/b)2 queues with deficit 2

• In xth iteration– q(b-1/b)x = 1 queues with deficit x

• X = log (b/b-1) q = ln q/ ln (1 +1/b-1) ; (Use ln (1+x) = x) = ln q(b-1)

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Necessity Analysis ….2

• In xth iteration– We can delete another “b”– Deficit is x + b = ln q(b-1) + b = b[ 1 + ln q(b-1)/b] = approx b [1 + lnq]

• Width of SRAM = b [2 + lnq] • Size of SRAM = qb[2 + lnq]

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A Dose of Reality• Typical values

– “b” is typically <= 10– q = Np, where

• N = # of ports (for VOQ)• p = number of classes per port

• Implementations– VOQ

• N = 32, p = 1, q = 25, b = 23, SRAM = 700 kb

– Diffserv• N = 32, p = 16, q = 29, b = 23, SRAM = 17 Mb

– Intserv• Lets not think about it!

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Future Work

• Discussion on trading off latency for SRAM size

• Analysis of other parameters– Relaxing I/O, address constraints

• Implementation Pain

• …. Still a long way to go