Transcript 13-Analysis.ppt

EECS 122:
Introduction to Computer Networks
Transport Analysis
Computer Science Division
Department of Electrical Engineering and Computer Sciences
University of California, Berkeley
Berkeley, CA 94720-1776
Katz, Stoica F04
Outline
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Exponential averaging and its applications
Retransmission timeout (RTO) computation
Little’s Theorem (revisited)
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Exponential Averaging
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Let X1, X2, …, XN be a series of measurements
Then average value Ai after the i-th measurement is
computed as
Ai = a Ai-1 + (1 - a ) Xi
where a is a constant between 0 and 1
Note that (assuming A0 = 0):
Ai = a N-1 (1 - a) X1 + a N-2 (1 - a) X2 + (1 - a) XN
What is the role of a? What does a control?
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Exponential Averaging: Example
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X constant; A converges to X
X
A
1 2 3
4 5 6 7 8 9 10
…
iteration
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Exponential Averaging: Example

Maintaining queue size in RED
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Exponential Averaging: Example
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RTT estimation in TCP
Measure RTT for each packet/ACK pair
Compute average of RTT as
- EstRTT = a x EstRTT + (1 - a ) x RTT
- a is 0.9 (or 0.8)
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Moving Window Average

Let X1, X2, …, XN be a series of
measurements
Then average value Ai after the i-th
measurement is computed as
Ai = (Xi-1 + Xi-2 + … + Xi-w)/W
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where W is the window size
How do exponential averaging and moving
window averaging compare?
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Outline



Exponential averaging and its applications
Retransmission timeout (RTO) computation
Little’s Theorem (revisited)
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Retransmission Timeout (RTO)
Computation: The Problem
1
1
Timeout
RTT
RTT
1
Timeout
1
Timeout too long 
inefficiency
Timeout too short 
duplicate packets
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RTO Computation: Original Algorithm
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Measure RTT for each packet/ACK pair, then
perform
1. EstRTT = a x EstRTT + (1 - a ) x RTT,
where a is 0.9 (or 0.8)
2. RTO = 2 x EstRTT
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RTO Computation: Jacobson/Karels
Algorithm
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Measure RTT for each packet/ACK pair, then perform
1. Err = RTT - EstRTT
2. EstRTT = EstRTT + (1 – a ) x Err
(Note: equivalent to EstRTT = a x EstRTT + (1 - a ) x RTT)
3. DevRTT = (1 - b ) x DevRTT + b x |Err|
4. RTO = EstRTT + 4 DevRTT
where a = 0.9 and b = 1/8
•
•
DevRTT represents the mean of the deviation (like
standard deviation) of the RTT
Why do we need DevRTT?
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RTO Computation: Karn/Partridge
Algorithm
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Add the following two considerations to
Jacobson/Karels algorithm
- EstRTT is updated only when an ACK is received
before the timeout expires. Why?
- If a packet timeouts, double EstRTT. Why?
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Outline



Exponential averaging and its applications
Retransmission timeout (RTO) computation
Little’s Theorem (revisited)
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Little’s Theorem
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Assume a system (e.g., a queue) at which packets
arrive at rate a(t)
Let d(i) be the delay of packet i , i.e., time packet i
spends in the system
What is the average number of packets in the
system?
d(i) = delay of packet i
a(t) – arrival rate
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system
Intuition:
- Assume arrival rate is a = 1 packet per second and the delay
of each packet is s = 5 seconds
- What is the average number of packets in the system?
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Little’s Theorem
1
2
Latest bit seen
by time t
d(i) = delay of packet i
x(t) = number of packets
in transit (in the system)
at time t
Sender
Receiver
x(t)
time
T
What is the system occupancy, i.e., average
number of packets in transit between 1 and 2 ?
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Little’s Theorem
1
2
Latest bit seen
by time t
d(i) = delay of packet i
x(t) = number of packets
in transit (in the system)
at time t
Sender
Receiver
x(t)
S= area
time
T
Average occupancy = S/T
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Little’s Theorem
1
2
Latest bit seen
by time t
d(i) = delay of packet i
x(t) = number of packets
in transit (in the system)
at time t
Sender
Receiver
S(N)
S(N-1)
P
d(N-1)
x(t)
S= area
time
T
S = S(1) + S(2) + … + S(N) = P*(d(1) + d(2) + … + d(N))
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Little’s Theorem
1
2
Latest bit seen
by time t
d(i) = delay of packet i
x(t) = number of packets
in transit (in the system)
at time t
Sender
Receiver
S(N)
S(N-1)
P
d(N-1)
x(t)
S= area
time
T
Average
S/T = (P*(d(1) + d(2) + … + d(N)))/T
occupancy
= ((P*N)/T) * ((d(1) + d(2) + … + d(N))/N)
Average arrival time
Average delay
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Little’s Theorem
1
Latest bit seen
by time t
d(i) = delay of packet i
x(t) = number of packets
in transit (in the system)
at time t
2
Sender
Receiver
S(N)
a(i)
S(N-1)
d(N-1)
x(t)
S= area
time
T
Average occupancy = (average arrival rate) x (average delay)
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