Transcript The Gaussian Nature of TCP Mark Shifrin Supervisor: Dr. Isaac Keslassy M.Sc Seminar

Faculty of Electrical Engineering
The Gaussian Nature of TCP
Mark Shifrin
Supervisor: Dr. Isaac Keslassy
M.Sc Seminar
Large Network Presentation
Characterizing
traffic is essential to understanding
the behavior of large Internet networks.
Network is large and complex – contains millions of flows.
 The interaction and mutual impact is difficult to understand.

2
Large Network Related Challenges –What do we pursue?

Ability to analyze problems:
 Congestions
 High

packet losses
Ability to make planning decisions:
 Bottleneck
capacity decision
 Buffer
size
 RTT (Round Trip Time) distribution
3
Obstacles and Difficulties
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
Today: no existing efficient large network
model because of numerous problems.
Problems:
 Many TCP feedback behaviors (congestion
avoidance, slow-start, …)
 Complex interactions among flows
 Complex queueing analysis
 Many topologies
 Many protocols (TCP, UDP, …)
 Complex interactions between flows and
ACKs
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Reducing the Network into Basic Components
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Bottlenecks dictate the behavior of the flows.
Assumption: The network can be subdivided into
basic dumbbell topologies with a single forward
bottleneck.
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Previous work – rule-of-thumb
Source
Router
Destination
C
RTT

Universally applied rule-of-thumb for TCP flows:
 A router needs a buffer size: B  RTT  C

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RTT is the two-way propagation delay
C is capacity of bottleneck link
Context
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Mandated in backbone and edge routers.
Appears in RFPs and IETF architectural guidelines.
Usually referenced in [Villamizar and Song, ’94]
Already known by inventors of TCP [Van Jacobson, ‘88]
Has major consequences for router design
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Previous Work – Stanford Model
[Appenzeller et al., ’04 | McKeown and Wischik, ’05 | McKeown et
al., ‘06]
 Assumption 1: TCP Flows modeled as i.i.d.
 total window W has Gaussian distribution
 Assumption 2: Queue is the only variable part
 queue has Gaussian distribution: Q=W-CONST
 use smaller buffer than in rule of thumb

We also consider small buffers. But, with small buffers:
 Buffers lose Gaussian property
 Link variations cannot be neglected
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Previous Work – Total Arrival is Poisson Distributed
Assumption: M/M/1/K model of buffer with
TCP [Avrachenkov et al., ‘01]
 Using the model of [Padhye et al., ’98] for
the M/M/1/K analysis.
 We show that the Poisson model is far
from reality.

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What is new in our analysis?

Dividing the Dumbbell Topology into components
(Lines).
 Finding
separately the pdf of the traffic of each
component we can describe the statistics of the entire
network.
 We show the impact of one component on another.

We find a new model of TCP congestion window
(cwnd) distribution, considering correlated losses.
 We


use it to find the pdf of the traffic on the Lines.
We prove that distributions of the packets on the
Lines are Gaussian.
We find Transmission rate, Queue (Q) distribution
and the packet loss.
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Contents of the Dumbbell Topology
Line L2
Line L5
Line L3
Line L4
Line L1
Q
Line L6
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Contents of the Dumbbell Topology
Line L5
Line L2
Line L3
Line L4
Line L1
Q
Line L6
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Outline
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Models of single objects
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Transmission rate and packet loss derivation
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Arrival rate of single flow
Total arrival rate
Q distribution
Packet loss
Gaussian models of Lines



Finding the model of cwnd .
Correlated losses
Packet Loss Event derivation
Model of li1
L1 model , Lindeberg Central Limit Theorem
Models of other lines and of W.
Conclusions
12
Model Development Flow
li1 pdf
total arrival rate
cwnd
Q pdf
packet loss
l1i
arrival rates
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TCP Newreno Congestion Control (Reminder)
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Slow Start w → w*2
Congestion Avoidance: w → w+1
Fast Retransmit and Fast Recovery: w → w/2

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Treats multiple losses in the same window
Timeout (TO): w →1
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CWND Simple Markov Chain Solution

Simple Markov Chain.
 Assumption:
Timeout (TO) probability is small
comparatively to the halving.

Model 1: Independent losses [Altman et al., 04]
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Model 2: Packet loss event derivation
w=n/2
w=n
pe(n)*n
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p: packet loss probability. Original transition was given by p*n.
We cannot use p because the losses are correlated.
probability of the Loss Event for window size n is equal to
pe(n)n
Assumption: In window of size n with losses , each packet in the
same window has equal probability to experience the packet
loss event.
Intuition: pe is the effective packet loss, which is independent
for every packet.
Fast Retransmit and Fast Recovery able to recover almost in
the same way no matter how many packets are lost in the
window.
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Correlated Losses
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Packet losses are strongly correlated: given
one packet was lost in a TCP window, the
probability is higher for others in the same
window to be lost.
Assumption: For a given window size n,
given at least one lost packet, the
distribution of the number of lost packets
is independent of the network condition!
(No actual impact of p, RTT distribution,
Buffer size)
Denote this distribution as Pr(l(n)=k), for
k=1…n
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Packet Loss Burst Size Distribution
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Packet Loss Burst Size Distribution
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Expected packet loss burst size
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Pr(l(n)=l) = Pr(l lost packets for w=n|loss event
happened)
E[l(n)]- expected value of the size of the bursty
loss, given at least one packet was lost, for
window size n.
At every loss event in window of size n, E[l(n)]
packets fall in average.
Every loss event is effectively treated as a
single loss (approximation)
pe – effective packet loss probability.
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What is this pe actually?

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Theorem: pe(n)/p =1/ E[l(n)], i.e. independent of
p for a given n.
Proof outline:
 For
arbitrary lost packet B denote AB as P(some
packets lost in window containing B)
 p=P(B|AB)*p(AB)+ P(B|ABC)*p(ABC)
0
 P(B|AB)=E[l(n)]/n
 p(AB)=pe(n)*n
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Conclusion: pe(n)=p/ E[l(n)]
Intuition: E[l(n)] losses treated as a single loss!
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pe/p - simulation results
500 flows traced by NS2 tools, event by event.
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MC Transitions (Approach 1 & 2)

Approach 1:
 Pr(w(t+1)=w(t)/2
| w(t)=n)=n*pe(n)
 Pr(TO)=const (known)
 Pr(w(t+1)=w(t)+1 | w(t)=n)=1-Pr(TO)Pr(w(t+1)=w(t)/2 | w=n).

Approach 2:
 Pr(w(t+1)=w(t)/2

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| w(t)=n)=Ph1+Ph2
Then Ph1 is the probability of loss in the first half of the
window and Ph2 is the probability of loss in the second half of
the window.
Improvement: introduce Slow Start in the model
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Results ( pdf,low p)
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Results (cdf, low p)
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Results (pdf, medium p)
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Results (cdf, medium p)
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Results (pdf, high p)
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Results (cdf, high p)
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Model of li1
li1 pdf
cwnd
packet loss
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li1 distribution – uniform vs. bursty
source
li
li2
B
1
li
li5
3
li4
dest.
li6
rtti=tp1i+tp2i+tp3i+tp4i+tp5i+tp6i
• Approach 1: Uniform packet distribution. l1(t)i = wi(t)*
tpi1/rtti.
• Approach 2: Bursty Packet Distribution
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Bursty Transmission and li1 model
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Assumption: All packets in a flow move in a
single burst
 Transmission and queueing times of the
entire burst negligible comparatively to
propagation latencies
Conclusion 1: All packets belonging to an
arbitrary flow i are present almost always on
the same link (l1i,l2i,l3i,l4i,l5i,l6i).
Conclusion 2: The probability of burst of flow i
being present on the certain link is equal to
the ratio of its propagation latency to the total
rtti.
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li1 probability density function
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We compare the measured results vs. the models 1
and 2
Reminder of the model 1: l1(t)i = wi(t)* tpi1/rtti.
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Model results li1 , logarithmic scale
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Outline

Models of single objects

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


Transmission rate and packet loss derivation


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Arrival rate of single flow
Total arrival rate
Q distribution
Packet loss
Gaussian Models of Lines
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Finding the model of cwnd .
Correlated losses
Packet Loss Event derivation
Model of li1
L1 model , Lindeberg Central Limit Theorem
Models of other lines and of W.
Conclusions
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Model Development Flow
li1 pdf
cwnd
total arrival rate
l1i arrival rates
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Rate Transmission Derivation
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The objective is to find the pdf of the number
of packets sent on some link i, ri, in a time
unit δt.
Assumptions:
 The
rate on each one of the links in L1 is
statistically independent
 We assume that the transmissions are bursty.
 We assume that the rate is proportional to the
distribution of l1i and to the ratio δt/tp1i.

tp1i is the latency on the corresponding link.
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Arrival rate of a single flow
source
B
li
2
li1
li1 *δt/tp1i
δt
tp1i
• δt is the same for all flows.
• We find the arrival distribution for every flow in δt msec.
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Rate PDF on the single line
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
The distribution is the same, but with different
parameters.
For finding the total rate Lindeberg Central Limit
Theorem is needed.
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Total Rate
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Theorem:
The proof is using the Lindeberg condition.
 The
idea: Central Limit Theorem holds if the share
of each flow comparatively to the sum is negligible
as the number of the flows grows.
 Argument for the proof: cwnd is limited by
maximum value  so does the l1i and ri.

Another alternative: Using a Poisson arrival
rate - too small variance (result of the bursty
transmission not reflected)
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Rate Model - Results
Probability
Total arrival rate – number of packets per δt
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Model Development Flow
li1 pdf
total arrival rate
cwnd
Q pdf
packet loss
l1i
arrival rates
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Q pdf

Two possible ways to find the Q
distribution:
 Running
MC simulation using samples of R
 G/D/1/K analysis using R for arrivals [TranGia and Ahmadi, ‘88]
Both ways yield almost the same results.
 Packet loss p is derived from the Q
distribution.
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Q pdf - results
Probability
Q state – 0 to 585 packets
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Fixed Point Solution: p=f(p)
li1 pdf
total arrival rate
cwnd
Q pdf
packet loss
l1i
arrival rates
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Packet Loss Rate
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Solution by gradient algorithm
1.
2.
3.
4.
5.
6.
7.
Start with approximate p
Find the cwnd and all l1i
Find all rates of all flows
Find the total rate and Q pdf
Find new p
Make a correction according to the error
weighted by step size and go to 2.
Till the error small enough – fixed point
found
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Packet loss - results
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Model gives about 10%-25% of discrepancy.
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Case 1: Measured: p=2.7%, Model: p=3%
Case 2: Measured: p=0.8%, Model: p=0.98%
Case 3: Measured: p=1.4%, Model: p=1.82%
Case 4: Measured: p=0.452%, Model: p=0.56%
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Outline

Models of single objects





Transmission rate and packet loss derivation





Arrival rate of single flow
Total arrival rate
Q distribution
Packet loss
Gaussian Models of Lines



Finding the model of cwnd .
Correlated losses
Packet Loss Event derivation
Model of li1
L1 model , Lindeberg Central Limit Theorem
Models of other lines and of W.
Conclusions
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L1 (reminder)
Line L5
Line L3
Line L1
Q
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Lindeberg Central Limit Theorem
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All li1 are statistically independent.
The distribution is the same but with different
parameters – because of different RTT –
simple Central Limit Theorem won’t fit.
Theorem:
Conclusion : Using the model for cwnd and
then for li1 we are able to find the L1
distribution.
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L1 Model Results
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What else can we know from our model?
Line L2
Line L5
Line L3
Line L4
Line L1
Q
Line L6
•Stanford Model: Gaussian Queue, Gaussian W, constant traffic on the
links
•Our model: non-Gaussian Queue, Gaussian W, Gaussian traffic on the
links.
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Packets on other lines
client
cwnd
li
1
li2
B
bcwnd
li
li5
3
li4
li6
server
bcwnd
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Model Development Flow
cwnd
W
Total
l i2
L2
l i4
L4
l i5
L5
l i6
L6
bcwnd
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cwnd on the Lines
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We know the pdf for the cwnd – the window at
the source.
This window travels the lines as the single burst
The burst which passes through the Q loses
some packets.
How can we know the distribution of the window
of packets after it passes the Q?
Suggestion: Using l(n) distribution for window of
size n.
Reminder: Pr(l(n)=l) = Pr(l lost packets for w=n|loss
event happened)
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Packet Loss Burst Size Distribution (Reminder)
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Markov Chain of bcwnd
1-Ploss
Ploss*P(l(n)=1)
w=1
w=2
w=3
w=n-1
w=n
Ploss*P(l(n)=n-3)
Ploss*P(l(n)=n-2)
Ploss*P(l(n)=n-1)
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Back cwnd algorithm (bcwnd)
The fall probability for window of size n:
Ploss= pe(n)*n. (TO is small and omitted)
 For each n from 1 to 64 update:

 Pr(bcwnd=n)=P(cwnd=n)-P(cwnd=n)*
 For
Ploss
each I, i=0 to n-1 update:
P(bcwnd=i)=P(cwnd=i)+ P(cwnd=n)* Ploss*P(l(n)=n-i)

The update done from low to the highest
state.
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bcwnd and cwnd comparison
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Models for L2, L4,L5,L6
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Typical case: tp1i, tp2i, tp4i, tp5i, tp6i, all have a
different distribution
Approach 1: Use model of L1 with cwnd and
scale by ratio of means of tp.
 ratio
of tp gives unreliable result for both the expected
value and variance

Approach 2: Use bcwnd for the model for l2i, l4i,
l 5i , l6i .
model is exactly the same as for l1i, but bcwnd
instead of cwnd used.
 Corresponding latencies for each line used.
 Use The Lindeberg Central Limit Theorem, exactly in
the same way as for L1!
 The
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L2 Model – result demonstration
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Topology – Reminder of the Lines
Line L2
Line L5
Line L3
Line L4
Line L1
Q
Line L6
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The Gaussian distributions
L4
L2
L1
L6
L5
W
NS2 simulation of 500 flows with all different time propagations
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Are these lines really Gaussian?
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W – sum of all cwnd

Two approaches to find W.
model – is known. All TCP connections are i.i.d
– classical Central Limit Theorem is applicable.
 Counting by summing all the components:
W=L1+L2+L3+L4+L5+L6+Q.
 cwnd
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L3 and Q – are non Gaussian components.
cwndi of different flows have some small
correlation.
L1,L2,L4,L5,L6 - all Gaussian. L1 – almost the
perfect Gaussian.
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What Size of Buffer do We Need?
Case 1
Buffer
154
772
3862
Formula
p (%) Avg cwnd
C*RTT/sqrt(N)/17.5
0.8
21
C*RTT/sqrt(N)/3.5
0.664
24.5
C*RTT/sqrt(N)/0.7
0.45
28.5
Case 2
Buffer
154
450
1802
Formula
C*RTT/sqrt(N)/8
C*RTT/sqrt(N)/2
C*RTT/sqrt(N)/0.5
p (%) Avg cwnd
3.26
8
2.98
8.8
2.01
11.5
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Outline

Models of single objects





Transmission rate and packet loss derivation





Arrival rate of single flow
Total arrival rate
Q distribution
Packet loss
Gaussian Models of Lines



Finding the model of cwnd .
Correlated losses
Packet Loss Event derivation
Model of li1
L1 model , Lindeberg Central Limit Theorem
Models of other lines and of W.
Conclusions
72
Conclusions..

We know the probabilistic models of
different components of the network
 Single
flows
 General components

We can plan routers or networks now – by
deciding the C and the B, given some
topology and packet loss limit.
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Thanks are going to:
Dr. I. Keslassy for guidance and
inspiration
 Lab Staff for relentless support:

 Hai
Vortman, Yoram Yihyie, Yoram Or-Chen,
Alex
Students for some simulations.
 Listeners in this class for the patience..

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