EECS 122:
Introduction to Computer Networks
Computer Science Division
Department of Electrical Engineering and Computer Sciences
University of California, Berkeley
Berkeley, CA 94720-1776
Katz, Stoica F04

Outline


 Exponential averaging and its applications
Retransmission timeout (RTO) computation
Little’s Theorem (revisited)
Katz, Stoica F04 2

Exponential Averaging


Let X
1
, X
2
, …, X
N be a series of measurements
Then average value A i computed as after the i -th measurement is
A i
= a A i-1
+ (1 a
) X i where a is a constant between 0 and 1


Note that (assuming A
0
A i
= 0):
= a N-1
(1 a
) X
1
+ a N-2
(1 a
) X
2
+ (1 -
What is the role of a
? What does a control? a
) X
N
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Exponential Averaging: Example
 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



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)
Katz, Stoica F04 6

Moving Window Average


Let X
1
, X
2
, …, X
N measurements be a series of
Then average value A i after the measurement is computed as i -th
A i
= ( X i-1
+ X i-2
+ … + X i-w
)/ W
 where W is the window size
How do exponential averaging and moving window averaging compare?
Katz, Stoica F04 7

Outline
 Exponential averaging and its applications

 Retransmission timeout (RTO) computation
Little’s Theorem (revisited)
Katz, Stoica F04 8

RTT
1
1
Timeout
RTT
1
Timeout
1
Timeout too long  inefficiency
Timeout too short  duplicate packets
Katz, Stoica F04 9

 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
Katz, Stoica F04 10

•

•
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 ?
Katz, Stoica F04 11

 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?
Katz, Stoica F04 12

Outline
 Exponential averaging and its applications


Retransmission timeout (RTO) computation
Little’s Theorem (revisited)
Katz, Stoica F04 13

Little’s Theorem



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 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
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 x(t)
2
Receiver time
T
What is the system occupancy, i.e., average number of packets in transit between 1 and 2 ?
Katz, Stoica F04 15

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
Sender x(t)
S= area
T
Average occupancy = S/T
2
Receiver time
Katz, Stoica F04 16

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
Sender
P S(N-1)
S(N) d(N-1)
2
Receiver x(t)
S= area time
T
S = S(1) + S(2) + … + S(N) = P*(d(1) + d(2) + … + d(N))
Katz, Stoica F04 17

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
Sender
P S(N-1)
S(N) d(N-1)
2
Receiver x(t)
S= area time
Average occupancy
T
S/T = (P*(d(1) + d(2) + … + d(N)))/T
= ((P*N)/T) * ((d(1) + d(2) + … + d(N))/N)
Average arrival time Average delay
Katz, Stoica F04 18

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
Sender a(i) S(N-1)
S(N) d(N-1)
2
Receiver x(t)
S= area
T
Average occupancy = (average arrival rate) x (average delay) time
Katz, Stoica F04 19