CS558 Syllabus & Progress

Modeling TCP Throughput (Padhye)

Goal
- Find an expression (relationship) for Throughput of TCP
- This expression will expose how the throughput of a TCP connection will depend on a number of factors.
Throughput
- Throughput is the total amount of data transmitted by a source
- Average Throughput is the average amount of data transmitted by a source per time unit
- Example:
Modeling the TCP protocol
- In this webpage, we will analyze the behavior of the TCP protocol
- This material is taken from the research paper by Padhye (1998): click here
- ... which is part of his PhD thesis: click here

Overview of TCP operation

Recall that TCP operates in 2 phases:

Recall also that TCP Reno is the most prevalent TCP version

TCP Reno includes the Fast Recovery mechanism (See: click here )

Typical Reno CWND behavior:

Notes:

Initially, there is a transient period that corresponds to the slow start phase where the CWND (and throughput) is very chaotic
After the transient period, the CWND (and throughput) because periodic
This period corresponds to the congestion avoidance phase
This is also the steady state period...

Facts:

The analysis of performance of TCP (of any protocol for that matter) is very difficult for transient periods
In other words: analyzing the performance of TCP during the Slow Start phase is very difficult
On the other hand, TCP Reno will not remain in the Slow Start phase for a long period of time !

Therefore:

The contribution to the total performance/throughput of TCP by the slow start periods is very small
This contribution to TCP throughput can generally be ignored for long (file) transfers (which is the main interest of the study)

So in the analysis of the TCP performance, we ignore the Slow Start phase and only consider TCP's performance during the Congestion Avoidance phase

Simplifying assumptions
- To make the analysis tractable, the author made some simplifying assumptions:

Assumption 1: rounds with constant RTT

The analysis first abstracts the continuous operation of TCP into a series of rounds:

Notes:

Assume the congestion window CWND = W

At the start of the round, the sender sends W packets

Note:

Unlike actual TCP that works with a unit of one byte, Padhye's analysis works with a unit of "one packet".
So the window W in Padhye's analysis is actually represents W * MSS bytes

The end of the round of packet transmissions is when:
The round is ended because the next round will be started !!!
Because when the sender receives the ACK of the first packet, the sender can send the first packet in the next round.
And by the time that the sender finishes sending the "new" first packet, the second ACK packet is received by the sender, which allows it to send the second packet in the new round.... And so on...
The length of one round of transmissions is approximately RTT seconds
To simplify the analysis, Padhye assumed that RTT is always the same value (constant)

Assumption 2: "integral" increase of TCP's congestion window (when there are no errors)

Assumption:

Furthermore, the congestion window size is assumed to change as follows:

Note:

TCP still updates CWND (= W) in a piecemeal fashion
After receiving a new ACK packet, the congestion window is always increased by the constant value 1/W
(In actual TCP, the increase is larger for later increments - because W is itself increased)

After receiving all W ACKs, the congestion window CWND (which was equal to W at the start of the round) is increased to W+1

Delayed ACK

Some TCP receivers use delayed ACK

Delayed ACK is a technique used to reduce the number of ACK messages:

An ACK message is sent:

after b packets (typically b = 2) are received, or
after a certain amount of time has expired (otherwise the sender will time out !)

The following figure shows how the congestion window size of the sender is changed when delayed ACK are used:
1. The figure shows an exmaple where the receiver sends back an ACK for every TWO messages received
2. You can see that the sender will receive half as many NEW ACKs
3. After one round of messages, sender's CWND is only increased to W + 1/2
4. In general, if the receiver sends back an ACK for every b messages received, in ONE round, the sender's CWND is increased to W + 1/b

Conclusion:

CWND = W at the start of a round, and
there are no packet loss in the next round,

then

CWND = W + 1/b at the start of the next round
where b is the number packets received before an ACK message is sent

Assumption 3: one loss, all loss packet loss model

The following assumption on packet loss is made to make the analysis simpler:

Graphically:

Notes:

Suppose that packet 6 in a round is lost

Then:

all packets following packets 6 in the same round will also be lost
Note:

Effect of this assumption:

After a packet loss in a round, the sender will not receive any more ACKs for packets in the same round
(This help simplify the analysis)

Note:

This does not mean that the sender will not receive any more ACKs
The next opportunity that the sender will receive an ACK is for the first packet in the next transmission round
In the figure, you can see that the sender will receive duplicate ACKs when the new packets 11 and 12 are ACKed by the receiver.

Steady State Behavior

Typical Reno "saw tooth" CWND plot:
Note:

Detailed description of one "saw tooth" period:

Notes:

A new "saw tooth" is started by the packet loss in the previous "saw tooth"
After the packet loss, TCP will use a new CWND
There will be a number of packet transmission rounds where there are no packet losses.
Then the final round of the saw tooth will be a transmission round with some packet losses
The pattern repeats itself again....

Receovering from packet losses in TCP

Recall that TCP can recover from packets losses in 2 ways:

Triple Duplicate ACK (fast retransmit/recovery) - See: click here
Recovery Action:
Time out
When will a time out event occur:
Recovery Action:

Padhye was the first to analyse the window change in both cases....
(Earlier works only studied triple duplicate ACK receovery)
We will now enter the Mathematical part of Padhye's paper...