Transcript Buffer Sizing: Going from Small To Tiny

Reducing the Buffer Size
in Backbone Routers
Yashar Ganjali
High Performance Networking Group
Stanford University
February 23, 2005
[email protected]
http://www.stanford.edu/~yganjali
Motivation

Problem
–
–
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Possible solution
–

Internet traffic is doubled every year
Disparity between traffic and router growth
(space, power, cost)
All-optical networking
Consequences
–
–
Large capacity  large traffic
Very small buffers
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Outline of the Talk
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Buffer sizes in today’s Internet
From huge to small (Guido’s results)
–

2-3 orders of magnitude reduction
From small to tiny
–
Constant buffer sizes?
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Backbone Router Buffers
Source
Router
Destination
C
2T

Universally applied rule-of-thumb
–
A router needs a buffer size: B=2TxC

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2T is the two-way propagation delay
C is the capacity of the bottleneck link

Known to the inventors of TCP
Mandated in backbone routers
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Appears in RFPs and IETF architectural guidelines
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Review: TCP Congestion Control
Only W packets
may be outstanding
Rule for adjusting W
–
–
W ← W+1/W
W ← W/2
If an ACK is received:
If a packet is lost:
Source
Dest
Window size
Wmax
Wmax
2
t
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Multiplexing Effect in the Core
W
B
0
Buffer Size
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Probability
Distribution
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Backbone router buffers

It turns out that
–
–
The rule of thumb is wrong for a core routers
today
2T  C
Required buffer is
instead of 2T  C
n
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Required Buffer Size
2T  C
n
Simulation
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Impact on Router Design

10Gb/s linecard with 200,000 x 56kb/s flows
–
Rule-of-thumb: Buffer = 2.5Gbits

–
Becomes: Buffer = 6Mbits
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Requires external, slow DRAM
Can use on-chip, fast SRAM
Completion time halved for short-flows
40Gb/s linecard with 40,000 x 1Mb/s flows
–
–
Rule-of-thumb: Buffer = 10Gbits
Becomes: Buffer = 50Mbits
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How small can buffers be?
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Imagine you want to build an all-optical
router for a backbone network…
…and you can build a few dozen packets in
delay lines.
Conventional wisdom: It’s a routing problem
(hence deflection routing, burst-switching,
etc.)
Our belief: First, think about congestion
control.
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TCP with ALMOST No Buffers
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Utilization
of bottleneck link = 75%
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Problem Solved?
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75% utilization with only one unit of buffering
More flows  Less buffer
Therefore, one unit of buffering is enough
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TCP Throughput withSmall Buffers
TCP Throughput vs. Number of Flows
0.9
0.8
Throughput
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
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200
400
600
Number of Flows
800
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1000
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TCP Reno Performance
Buffer Size = 10; Load = 80%
1
0.9
Throughput
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
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1000
2000
3000
4000
Bottleneck Capacity Mbps
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5000
6000
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Two Concurrent TCP Flows
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Simplified Model
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Flow 1 sends W packets during each RTT
Bottleneck capacity = C packets per RTT
Example: C = 15, W = 5
1 RTT
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Flow 2 sends two consecutive packets during
each RTT
Drop probability is increased with W
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Simplified Model (Cont’d)
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W(t+1) = p(t)x[W(t)/2] + [1-p(t)]x[W(t)+1]
But, p grows linearly with W
E[W] = O(C ½)
Link utilization = W/C
As C increases, link utilization goes to zero.
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Snow model!!!
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Q&A
Q. What happens if flow 2 never sends any
consecutive packets?
A. No packet drops unless utilization = 100%.
Q. How much space we need between the two
packets?
A. At least the size of a packet.
Q. What if we have more than two flows?
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Per-flow Queueing
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Let us assume we have a queue for each
flow; and
Server those queues in a round robin
manner.
Does this solve the problem?
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Per-flow Buffering
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Per-Flow Buffering
Temporarily Idle
Time t

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Time t + RTT
Flow 3, does not have a packet at time t; Flows 1
and 2 do.
At time t+RTT we will see a drop.
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Ideal Solution

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If packets are spaced out perfectly; and
The starting times of flows are chosen randomly;
We only need a small buffer for contention
resolution.
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Randomization
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Mimic an M/M/1
queue
Under low load,
queue size is
small with high
Probability
Loss can be
bounded
r
M/M/1
EX 
1
Buffer B
Packet
Loss
b
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1 r
P X  k   r k
X
P(Q > b)
r
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TCP Pacing
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Current TCP:
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Send packets when ACK received.
Paced TCP:
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Send one packet every W/RTT time units.
Update W, and RTT similar to TCP
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CWND: Reno vs. Paced TCP
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TCP Reno: Throughput vs. Buffer Size
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Paced TCP: Throughput vs. Buffer Size
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Early Results
Congested core router with 10 packet buffers.
Average offered load = 80%
RTT = 100ms; each flow limited to 2.5Mb/s
source
source
>10Gb/s
>10Gb/s
router
10Gb/s
server
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What We Know
Arbitrary
Injection
Process
If Poisson Process
with load < 1
Theory
Any rate > 0
need unbounded
buffers
Need buffer
size of approx:
O(logD + logW)
i.e. 20-30 pkts
Complete
Centralized
Control
Experiment
TCP Pacing:
Results as good
or better than for
Poisson
Constant fraction
throughput with
constant buffers
[Leighton]
D=#of hops
W=window size
[Goel 2004]
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RENO
PACING
Limited Congestion Window
Limited Window
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Unlimited Window
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Slow Access Links
Congested core router with 10 packet buffers.
RTT = 100ms; each flow limited to 2.5Mb/s
source
source
5Mb/s
5Mb/s
router
10Gb/s
server
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Conclusion
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We can reduce 1,000,000 packet buffers to
10,000 today.
We can “probably” reduce to 10-20 packet
buffers:
–
–
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With many small flows, no change needed.
With some large flows, need pacing in the access
routers or at the edge devices.
Need more work!
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Extra Slides
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Pathological Example
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Flow 1: S1  D; Load = 50%
Flow 2: S2  D; Load = 50%
S1
D
1
2
3
4
S2
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If S1 sends a packet at time t, S2 cannot
send any packets at time t, and t+1.
To achieve 100% throughput we need at
least one unit of buffering.
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