- The Throughput analysis
from the previous discussion is good when
TCP recovers through
Triple Duplicate ACKs
- $64,000 question:
- How often does that happen ?
- They ran an experiment
for an hour
and obtain the following result:
Meaning of the table:
- Each row
in the table represents
one TCP connection
(between a "Sender" and
a "Receiver")
- The Loss Indic.
contains the
total
number of loss indications
in the experiment for each connection
- The column labeled
TD contains
the
number of loss indications that are
Triple Duplicate ACKs
- The column labeled TO contains the number of loss indications that are Timeouts
- Each row
in the table represents
one TCP connection
(between a "Sender" and
a "Receiver")
- Conclusion from the experiment:
-
The number of
Timeouts
is very significant
- In fact, most of loss indications are time outs !
-
The number of
Timeouts
is very significant
- Consequence:
- We cannot
omit
time out
in the analysis
of TCP performance
- It also means that existing analysis that omitted timeouts is flawed or at least, incomplete for understanding the behaviour of TCP.
- We cannot
omit
time out
in the analysis
of TCP performance
- During the discussion of the
TCP congestion protocol,
we did not discuss
the details of the
setting of TCP Time out timer
- We must do this before we can
understand
the TCP analysis
- The TCP Timeout Timer:
- To determine when a
packet has time out,
TCP estimates
the RTT from the
ACK messages
of the TCP connection
(I won't go into the details of the estimation formula, other than tell you that it uses a moving average formula based on the observed RTT)
- TCP maintains a
TO timeout
estimate
(that is based on the measured RTT
to the destination)
- TCP adjusts its TO
estimate when a new ACK is received.
- When a timeout
occurs, TCP will assume that the
network is congested
In response, TCP will assume that RTT has increased significantly
- Actions performed by
TCP
in response to a Timeout event:
- Retransmit the packet (ofcourse :-))
- Set CWND = 1 (this is
slow start)
-
Double
the TO timeout
value for the connection:
TO = 2 x TO
- Retransmit the packet (ofcourse :-))
- To determine when a
packet has time out,
TCP estimates
the RTT from the
ACK messages
of the TCP connection
- Example:
- NOTE:
- TCP can go through
multiple consecutive
Timeouts
- Each time TCP times out, the timeout estimate TO is doubled !!!
- TCP can go through
multiple consecutive
Timeouts
- Example:
- The following figure show a
plot of the value of
CWND
in TCP-Reno:
In general, TCP Reno goes the following transmission cycle with Triple Dupl. ACKs and time out:
- Let's call the current cycle
i and the
previous
cylce is
i-1
- Suppose the previous cycle
i-1
ended
in a time out
with
a congestion window size
CWND = Wi-1
- Then at the
start of
the current cycle
i ,
we have:
- SSThreshi = Wi-1/2.
- CWNDi = 1
- TCP Reno will operate in
slow start mode
until:
- CWND > Wi-1/2
Then TCP Reno will enter the congestion avoidance phase...
- During congestion avoidance,
TCP will detect packet loss
using:
- Triple Duplicate ACKs when the congestion is not severe or
- Timeout when the congestion is severe
The cylce always ends with a time out packet loss detection (because we will be back to CWND = 1)
There may a few TD periods before a Timeout occurs...
- Also,
TCP Reno may timeout several times
before the slow start is
successful
Note:
- When multiple timeouts
occurs,
the TO value is
increased.
- But the value of
SSThreshi+1
of the next cycle
is equal to CWNDi/2
independent from
the number of timeout events.
- Example:
- Suppose the current cycle ended
with
CWNDi
= Wi
(this is the value of CWND
timeout
occured).
- Then the next TCP Reno cycle will use:
- SSThreshi+1 = Wi/2.
- Suppose the current cycle ended
with
CWNDi
= Wi
(this is the value of CWND
timeout
occured).
(If TCP would reduce CWND for each timeout event, it may reduce SSThresh = 1....
This will make TCP Reno will transmit very very slowly !!!)
- When multiple timeouts
occurs,
the TO value is
increased.
- After the time out, the whole thing repeats itself from the start...
- Let's call the current cycle
i and the
previous
cylce is
i-1
- Ignore the (short) slow start period
- Consider the diagram:
- There is a very short period
where TCP Reno performs
the slow start
- Simplifying assumption:
- We will simplify
the short slow start period
at the start
of the TCP Reno cycle.
- By replacing it with a "fast" start where the CWND = SSThresh !!!
- We will simplify
the short slow start period
at the start
of the TCP Reno cycle.
- Consider the diagram:
- So the simplified
Reno cycle is as follows:
(This assumption is very very helpful and compromises very little accuracy)
- Consider the operation of
TCP Reno:
Observation:
- TCP Reno operates in two phases:
- First, there is a phase
where TCP Reno
uses Triple Duplicate ACKs
to receover from packet losses
- Then this is followed by a phase where TCP Reno uses Time outs to receover from packet losses
- First, there is a phase
where TCP Reno
uses Triple Duplicate ACKs
to receover from packet losses
- TCP Reno operates in two phases:
- Notation:
- TD =
a consecutive interval
where TCP Reno recovers
using only Triple Duplicate ACKs
Example:
- In the above figure, the ZiTD period is a TD period
- TO =
a consecutive interval
where TCP Reno recovers
using only Timeouts
Example:
- In the above figure, the ZiTO period is a TO period
- TD =
a consecutive interval
where TCP Reno recovers
using only Triple Duplicate ACKs
- Long term progression of TCP Reno
- Consider the detailed operation
of the TCP Reno protocol:
Notice that:
- TCP Reno will alternate
continuously between:
- TD period
- TO period
- TCP Reno will alternate
continuously between:
- Long term behavior of TCP Reno:
- Consider the detailed operation
of the TCP Reno protocol:
- Definitions and notations based on the long term behavior of
TCP Reno:
- TDi =
the ith
TD period.
- TOi =
the ith
TO period.
- ZiTD
= the duration
of TDi
- ZiTO
= the duration
of TOi
- Si
=
the duration of
a "Reno cycle"
(I made up this name :-))
= ZiTD + ZiTO
- Mi = number of packets transmitted in a "Reno cycle" (in Si)
- TDi =
the ith
TD period.
- NOTE:
- The interval
ZiTD
consists of a
random number
of
Triple Duplicate (TD) Periods
(For definition of TD Period, see: click here)
- The interval
ZiTD
consists of a
random number
of
Triple Duplicate (TD) Periods
- Initial throughput expression:
E[M] TCP Throughput = ------ ...... (1) E[S]Where:
- E[M] = the average number of packets sent in one TCP Reno cycle.
- E[S] = the average duration of one TCP Reno cycle.
- We need some
random variables
to compute
the throughput of
TCP Reno.
- The following figure
helps to illustrate
the meaning of
4 of these
random variables
(a second figure below will illustrate the remaining 2 random variables)
Recall that the interval ZiTD consists of a random number of Triple Duplicate Periods (for definition of TD Period, see: click here)
- ni
= the number of
TD periods
in
ZiTD
( ni is a random variable)
- Aij
= duration
of the jth
TD period
in
ZiTD
( Aij is also a random variable)
- Wij
= the congestion window size (CWND)
at the end
of the jth
TD periods
in
ZiTD
( Wij is a random variable)
- Ri
= the number of
retransmissions
in
ZiTO
( Ri is a random variable)
- ni
= the number of
TD periods
in
ZiTD
- These 2 random variables
were used to study the
throughput
of TCP Reno using
only Triple Duplicate ACK
recovery.
They are re-introduced to study the throughput of TCP Reno using both Triple Duplicate ACK and Timeout
Yj and Aj:
Notes:
- Yj
= the
number of packets
transmitted in the
jth
TD period
- Aj = the duration of the jth TD period
- Yj
= the
number of packets
transmitted in the
jth
TD period
- Reno throughput
(Equation (1)):
E[M] TCP Throughput = ------ ...... (1) E[S]Re-write E[M] and E[S] using the random variables defined above....
- To re-write
the E[M] (average number of packets sent)
, we
count
the
number of packets sent
in
one Reno Cylce
(This diagram will help you count):
- E[n] =
average number of
TD periods
in one Reno cycle.
- E[Y] =
average number of
packets transmitted in
one
TD period
NOTE: we have derived E[Y] before: click here - E[R] = average number of retransmissions in one TO period
E[M] = E[Yi1] + E[Yi2] + ... + E[Yi,ni] + 1 + 1 + ... + 1 | | | | +------------------------------+ +---------------+ sent in the TD periods sent in TO periods = E[Yi1] + E[Yi2] + ... + E[Yi,ni] + E[R]
Each of the random variable Yik has the same expected value Therefore: E[M] = E[n] * E[Y] + E[R] . . . . . . . . . . . . . . . . (2)Where:
- E[n] =
average number of
TD periods
in one Reno cycle.
- To re-write
the E[S] (average duration of a Reno cycle)
, we
"count"
the
duration of each
sub-period
in
one Reno Cylce
(This diagram will help you "count"):
- E[n] =
average number of
TD periods
in one Reno cycle.
- E[A] =
average number of
packets transmitted in
one
TD period
NOTE: we have derived E[A] before: click here - E[Z] = average duration of the timeout recovery period in one Reno cycle.
E[S] = E[Ai1] + E[Ai2] + ... + E[Ai,ni] + E[ ZiTO] | | | | +------------------------------+ +--------+ lengths of the TD periods length of the TO period
Each of the random variable Aik has the same expected value Therefore: E[S] = E[n] * E[A] + E[Z] . . . . . . . . . . . . . . . . (3)Where:
- E[n] =
average number of
TD periods
in one Reno cycle.
- We can now write a more detailed equation for
TCP Reno throughput:
E[M] TCP Throughput = ------ ............. (1) E[S]
Improved expression: E[n] * E[Y] + E[R] TCP Throughput = ----------------------- ................(4) E[n] * E[A] + E[Z]There are 5 unknowns in Equation (4)....
But 2 unknowns have been computed previously...
(There's still a lots of work ahead :-))
- We have already computed
two
of the 5 unknowns
- We have computed
E[Y] and
E[A]
when
we studied the
Reno's throughput
when the loss indications
were
only
Triple Duplicate ACKs
- Some important result are
summarized below for
convenience:
1 - p E[Y] = ------ + E[W] (See: click here) p2 E[A] = ( --- E[W] + 1) * RTT (See: click here) bThe expression for E[W] is:
---------------------- / 2+b / 8(1-p) 2+b 2 E[W] = ----- + \ / -------- + ( ----- ) (See: click here) 3b \/ 3bp 3b
- Therefore:
---------------------- / 1-p 2+b / 8(1-p) 2+b 2 E[Y] = ----- + ----- + \ / -------- + ( ----- ) . . . . . . . . (5) p 3b \/ 3bp 3b
-- ------------------------- --- | 2+b / 2 b (1-p) 2+b 2 | E[A] = RTT * | --- + \ / ----------- + ( ----- ) + 1 | . . . (6) | 6 \/ 3p 6 | -- ---
- We still need to find expressions for:
- E[n] =
average number of
TD periods in
one Reno cycle.
- E[R] =
average number of
retransmissions
in one TO period.
- E[Z] = average duration of the timeout recovery period in one Reno cycle.
- E[n] =
average number of
TD periods in
one Reno cycle.