TCP Modelling

Prentice HallHigh Performance TCP/IP Networking, Hassan-Jain

TCP Modelling


Objectives

Appreciate mathematical modeling of TCP for performance optimization of TCP/IP networks

Understand the fundamental relationship between packet loss probability and TCP performance

Investigate the Period Model and the Detailed Packet Loss Models


Contents

Motivation Essentials TCP Features Investigate TCP Mathematical Models

Periodic Model Detailed Packet Loss Model


Motivation


Why Model TCP Mathematically?

Options: Experimental Observation Computer Simulation Mathematical Modeling


Motivation for TCP Modeling

The sheer scale of the TCP operating environment Models in general are required to gain a deeper understanding of

TCP dynamics Uncertainties exist in the TCP operating environment

Uncertainties can be modeled as stochastic processes to drive TCP responses

To quantify metrics that define system performance To drive the design of TCP-friendly algorithms for

multimedia applications TCP is the most widely used transport layer protocol


Essentials of

TCP Modeling


TCP Essentials

Every TCP model must captureWindow Dynamics (internal and deterministic)Packet Loss Process (external and uncertain)

Mainly Reno flavours are modelled Triple duplicate ACK

window = window / 2TCP state = Congestion Avoidance

Packet Losswindow = 1TCP state = slow start


Standard Modeling Assumption

X(t) = W(t)/RTT

where, X(t) is TCP Throughput,

W(t) is Window Size,

RTT is Round Trip Time.

What does X(t) = W(t)/RTT implicity assume?


TCP Window Dynamics

Slow Start: Exponential Increase Congestion Avoidance: Linear Increase Exponential Back off: Multiplicative Decrease


Packet Loss Process

Packet loss triggers window decrease Packet loss is uncertain This uncertainty is typically modeled as a

stochastic processE.g. probability p of losing a packet


TCP Models


Gallery of TCP Models

Periodic Model Detailed Packet Loss Model Stochastic model Control system model Network system model


Periodic Model

Goal

To find an expression for the average TCP Throughput as a function of packet loss probability.

with TCP in Steady State Periodic Window Size Constant Packet Loss Probability

X(P) = number of packets per period / period duration


Steady State / Periodic Model

time (RTT)

W/2

W

loss occurs

period

Simplest possible model W is the maximum

supportable window size (then loss occurs)

TCP window starts at w/2 grows to W, then halves, repeat forever …

W(t) packets transmitted each RTT

W(t+1) = W(t) + 1 each round until a loss occurs

Why not model slow start?


T

Packets Transmitted Per Period

To find W as a function of P.

# packets sent per “period” = base * average height = (T / RTT) * (W/2 + W) / 2

= W/2 * (W/2 + W) / 2 = 1/P

W/2 * (W/2 + W) / 2 = 1/P => W = root(8/3P)


Inverse Square Root P Law X(P)

X(P) is the average sending rate of the TCP source

X(P) = (number of packets sent per period ) / (period duration)

= (1/P) / (RTT * W/2) , since W – W/2 = W/2

= (1/RTT) * (3/2P)^(1/2)

where P is the packet loss probability, W = root(8/3P).


Inverse Square P Law Question

If a TCP connection has an average RTT of 200ms, and packet loss probability 0.05, what is the average throughput?


Inverse Square P Law Answer

RTT=0.2, P=0.05.

Using the Inverse Square P Law

X(P) = (1/P) / root(3/2P)

X(0.05) = (1/0.2)*(root(3/2x0.05))

= 27.4 packets/second


Detailed Packet Loss Model

Goal

To find an expression for the average TCP Throughput as a function of packet loss probability.

with random packet loss events and considering

triple duplicate ACKs time outs receiver window limits

E[X(P)] = number of packets per period / period duration


Triple Duplicate ACK (TDA)Goal: E[X(P)] = E[Y] / E[A]

E[Y] is the expected number of packets per TDA periodE[A] is the expected TDA period duration


ith TDA PeriodApproach

Find expressions for Yi and Ai (ith TDA period).

Wi - Window size of ith TDA periodRi - Number of roundsΒi - Number of packets in final roundαi - Number of the packet lost in the ith period

Yi - number of packets transmitted Ai - period duration

Find E[Y] and E[A] as a function of P.


TDA’s


TDAs Question

If a TCP connection has an average RTT of 200ms, and packet loss probability 0.05, what is the average throughput?


TDAs Answer

RTT=0.2, P=0.05.

Using the TDA expression

X(0.05) = 26.69 packets/second


Timeouts


TimeoutsApproach

Find an expressions for E[X(P)] = E[M] / E[S].

Si = ZiTD + Zi

TO (ith period duration)

ZiTD - TDAs duration, losses recovered by TDA

ZiTO - Timeouts duration, loss results in timeout


Timeouts


Limits on the Receiver Window Size


Complete Model Question

If a TCP connection has an average RTT of 200ms, and packet loss probability 0.05, if the receiver limits the its window size to Wm = 20 packets and the TCP sender has a timeouts value of To = 500ms, what is the average throughput?


Complete Model Answer

RTT=0.2, P=0.05, Wm=20 packets, To=500ms.

Using the TDA expression

X(0.05) = 24.73 packets/second


Comparison Against ns-2 trans-Atlantic TCP Connection

Very Low

Low

High

Moderate


Results

Against a simulated trans-Atlantic TCP simulation connection

Inverse Square P Law

If moderately low packet loss probability, then good estimates are provided

If very low receiver window limits send rate and overestimates are provided

Otherwise, overestimates the send rate, due to a lack of accounting for timeouts

Suitable for small packet loss probability and no receiver window limits.


Results

Complete Detailed Packet Loss Model

Provides a good approximation for any probability loss or enforced receiver window size limit.

Documents

TCP Modelling