High Speed TCP

A generalized CWND update algorithm

The Average Generalized TCP Throughput is given by the formula (See: click here):

a Increase: W = W + --- W Decrease: W = W - b * W ------------ 1 / a (2 - b) Avg TCP Throughput = ----- \ / ---------- (# packets/sec) RTT \/ 2bp

We can also conclude that the Average TCP CWND is:

------------ / a (2 - b) Avg TCP CWND = \ / ---------- (# packets) \/ 2bp

The Gist of a High Speed TCP protocol

Achieving higher throughput in TCP:
Limiting circumstance:
Therefore:

Herein lies the difficulty in the design (if we can tossed out "Ordinary Reno", any speedier increase and decrease will work):

A High speed version of TCP must be "friendly" to the "normal" (Reno) TCP
In other words, when a High speed version of TCP and a "normal" (Reno) TCP compete for bandwidth over a common bottle neck link, the "normal" (Reno) TCP must be starve for bandwidth
We call this property: "TCP-friendly"

Key observation in the design of a High Speed TCP

Considerations:

TCP Reno cannot take advantage of high speed connections
However, the new High Speed TCP protocol must not overwhelm TCP Reno when operating in a low speed environment

Gist of the design of a TCP Reno-friendly High Speed TCP protocol:

The High Speed TCP protocol uses TCP Reno increase/descrease when operating over a low speed connection:

1 Increase: W = W + --- (when a new ACK is received) W Decrease: W = W - 0.5 * W (when 3 dup ACKs are received)

The High Speed TCP protocol uses faster increase/descrease when operating over a high speed connection:

a Increase: W = W + --- (when a new ACK is received) W Decrease: W = W - b * W (when 3 dup ACKs are received)

$64,000 question:

How can we tell if the connection is a high speed connection or a low speed connection ?????

Ingenious solution:

We can tell if a connection is a high speed connection or a low speed connection by the CWND value !!!!

(The "how" is explained below....)

Fact:

High speed links (optical fiber) have very low packet loss probability
Low speed links (copper) have relatively high packet loss probability

Average CWND used by classic TCP Reno:

Packet Drop Rate P Congestion Window W ------------------ ------------------- 10^-2 12 10^-3 38 (Low speed links) 10^-4 120 10^-5 379 10^-6 1200 10^-7 3795 (High speed links) 10^-8 12000 10^-9 37948 10^-10 120000 Table 2: TCP Response Function for Standard TCP. The average congestion window W in MSS-sized segments (packets) is given as a function of the packet drop rate P.

Conclussion:

From the above table, we see that:
(Assuming that MSS of TCP is approximately 1500 bytes)

Summarizes the "High Speed TCP" algorithm:

Notes:

The HighSpeed TCP will use a dynamic increase and decrease functions that depended on its current CWND value.
To avoid overwhelming TCP Reno in low speed networks:
To fully utilize the vast amount of bandwidth in high speed networks:

Dynamic window update rate in High bandwidth range

Dynamic update rate:

CWDN update algorithm for 1 ≤ CWND ≤ 38:

1 Increase: CWND = CWND + ------ CWND Decrease: CWND = CWND - 0.5 * CWND

CWDN update algorithm for CWND > 38 is dynamically changing with the value of CWND:

a(CWND) Increase: CWND = CWND + ------ CWND Decrease: CWND = CWND - b(CWND) * CWND

Example: (not scientifically correct, just to illustrate the concept of "dynamic update rate")

CWND ≤ 38

1 Increase: CWND = CWND + ------ CWND Decrease: CWND = CWND - 0.5 * CWND

CWND = 40

2 Increase: CWND = CWND + ------ CWND Decrease: CWND = CWND - 0.45 * CWND

CWND = 50

3 Increase: CWND = CWND + ------ CWND Decrease: CWND = CWND - 0.42 * CWND

Dynamic update rate:

The factors a and b changes with the current congestion window size CWND
I.e.:
(The paper uses W to denote CWND)

The CWND that yield 10 Gbps throughput

The authors assumed the following ball park (representative) values:
Throughput of TCP when using CWND = W:

CWND that yield 10 Gbps throughput:

120,000 × CWND = 1,000,000,000 ==> CWND = 1,000,000,000 / 120,000 = 83,333 ==> CWND ~= 83,000

Conclusion:

The High Speed TCP increase/decrease algorithm must allow CWND to reach CWND ~= 83,000
(using a high speed connection with packet drop probability of p = 10⁻⁷)

TCP increase/decrease algorithms that can reach CWND = 83,000 when p = 10⁻⁷

Recall the general TCP throughput and CWND formulas:

------------ / a (2 - b) 1 Avg TCP Throughput = \ / ---------- × ------- \/ 2bp RTT ------------ / a (2 - b) Avg TCP CWND = \ / ---------- \/ 2bp

CWND increase/decrease method that can reach CWND = 83,000 when p = 10⁻⁷:

------------- / a (2 - b) 83000 = \ / ---------- \/ 2b * 10⁻⁷
Or: ----------------- / a (2 - b) 83000 = \ / ---------- * 10⁷ \/ 2b

Summary:

If the packet loss probability is equal to 10⁻⁷ and the TCP CWND update algorithm uses:

a Increase: CWND = CWND + ------ CWND Decrease: CWND = CWND - b * CWND

such that a and b satisfy the relationship:

----------------- / a (2 - b) 83000 = \ / ---------- * 10⁷ \/ 2b

then:

The TCP CWND will reach an average value of CWND = 83,000
(Assuming the packet size = 1500 bytes, the average TCP throughput will be 10 Gbps)

Note:

Possible solution 1: a = 72 and b = 0.1

Verify: --------------------- / 72 (2 - 0.1) \ / ------------- * 10⁷ \/ 2*0.1 ------------------ / 72 * 1.9 = \ / ---------- * 10⁷ \/ 0.2 -------- / = \ / 684 * 10⁷ \/ = 82704.29 (approximately 83000)

Possible solution 2: a = 459 and b = 0.5:

Verify: --------------------- / 459 (2 - 0.5) \ / ------------- * 10⁷ \/ 2*0.5 ------------------ / 459 * 1.5 = \ / ---------- * 10⁷ \/ 1.0 -------- / = \ / 688.5 * 10⁷ \/ = 82975.90 (also approximately 83000)

Selecting an appropriate value for a and b

Consider the choice between:
Which CWND update method is better ?

(a = 72 and b = 0.1):

If there are no packet losses in a round of transmission (RTT = 0.1 sec), the CWND of the TCP connection will increase 72 packets
If there are packet losses resulting in 3 duplicate ACKs in a round of transmission, the CWND of TCP connection will decrease by 0.1 × CWND (a 10% decrease)

(a = 459 and b = 0.5):

If there are no packet losses in a round of transmission (RTT = 0.1 sec), the CWND of the TCP connection will increase 459 packets
If there are packet losses resulting in 3 duplicate ACKs in a round of transmission, the CWND of TCP connection will decrease by 0.5 × CWND (a 50% decrease)

Finding:
Bottomline:

Designing a(W) and b(W)

Summary of the findings:

Notes:

TCP Reno operates well in low bandwidth connections
The increase/decrease parameters of TCP Reno are:
We have found that to achieve 10 Gbps throughput over a high bandwidth connection, High Speed TCP will use as increase/decrease parameters:

High Speed TCP uses dynamic increase/decrease parameters that depends on the network reliability (or CWND):

a = a(CWND)
b = b(CWND)

$64,000 Question:

What functions do we use for a(CWND) and b(CWND) ???

Initial answer:

When High Speed TCP detects that it is operating in a low bandwidth connection (i.e., CWND ≤ 38), High Speed TCP will use:
(This will ensure TCP (Reno) friendiness)
When High Speed TCP detects that it is operating in a high bandwidth connection (i.e., CWND = 83,000), High Speed TCP will use:

Between these two extreme values:

The value of a will gradually change from 1 to 72 depending on CWND
The value of b will gradually change from 0.5 to 0.1 depending on CWND

I.e.:

Designing a(CWND) and b(CWND) using interpolation

Plot of CWND vs. Packet drop probability (i.e., a function W(p)):

So far:

We only have 2 operational points:
Between these 2 extreme (end) points, we need to find a smooth function
This can be done through interpolation

Question:

Possible interpolation function W(p):

Observation on the interpolation functions:

The red interpolation function will cause CWND to increase rapidly when p drop slightly below 10⁻³
The green interpolation function will cause CWND to increase rapidly only when p is very small (~ 10⁻⁷)

Conclusion:
Note:

The interpolation function used by HS-TCP

Consider an interpolation function of the form:

u W(p) = ----- p^v

W(p) must satisfy:

W( 10⁻⁷ ) = 83,000 ........ (1) W( 10⁻³ ) = 38 ........ (2)

We can find u and v as follows:

u ------ = 83,000 ........ (1a) (10⁻⁷)^v u ------ = 38 ........ (2a) (10⁻³)^v
Or: u = 83,000 × (10⁻⁷)^v ........ (1b) u = 38 × (10⁻³)^v ........ (2b)
==> 83,000 × (10⁻⁷)^v = 38 × (10⁻³)^v <==> log(83,000) + log(10⁻⁷)^v = log(38) + log(10⁻³)^v <==> 4.919 - 7v = 1.580 - 3v <==> 4v = 3.339
Therefore:

v = 0.835 ...... (3)

and:

u = 38 × (0.001)^0.835 = 0.12 ...... (4)

Proposed function W(p): :

0.12 W(p) = --------- . . . . . . . . . . . . . . (5) 0.835 p

Plot of W(p):
(The scale is too large :-))

The Inverse Interpolation Function p = p(W)

In the design of the HS-TCP protocol, we will need to:

So us will compute the inverse function p(W) here:

0.12 W = --------- 0.835 p 0.12 <==> p^0.835 = ------ (1/(0.835) = 1.1976) W 0.12^1.1976 <==> p = ------------ W^1.1976
0.0789 p(W) = --------- . . . . . . . . . . . . . (6) W^1.2

Designing increase/decrease parameters (a and b) for HS-TCP - Introduction

Recall that the High Speed (HS) TCP protocol uses dynamic a and b that depends on the CWND value w:
- Note: w = current congestion window (CWND)

Recall also the general TCP throughput and CWND formulas:

------------ / a (2 - b) 1 Avg TCP Throughput = \ / ---------- × ------- ........ (7) \/ 2bp RTT ------------ / a (2 - b) Avg TCP CWND = \ / ---------- (# packets) ........ (8) \/ 2bp

We write Equation (8) in a different form:

-------------- / a (2 - b) w = \ / ------------- (# packets) \/ 2 b p 2 a (2 - b) <==> w = ----------- 2 b p <==> a * (2 - b) = w² * 2 * b * p 2 w² b p <==> a = ---------- 2 - b
Or:

2 w² b(w) p(w) a(w) = ---------------- . . . . . . . . . (9) 2 - b(w)

(Because a, b, and p changes as CWND w changes)

Recall the relationship between CWND w and the packet loss probability p: (See: click here )

0.0789 p(W) = --------- . . . . . . . . . . . . . (6) W^1.2

Notice that there are two undetermined variables (degrees of freedom) in Equation (9):

2 w² b(w) p(w) a(w) = ---------------- . . . . . . . . . (9) 2 - b(w)

Namely:

a(w)
b(w)
w is not a degree of freedom because it is the parameter on which the other values depends)
p(w) is not a degree of freedom because it is already determined by the throughput formula of TCP Reno)

In other words:

From the form of Equation (9):

2 w² b(w) p(w) a(w) = ---------------- . . . . . . . . . (9) 2 - b(w)

It is obvious that we should:

Choose the value for b(w)
use Equation (3) to compute the value of a(w)....

Designing the decrease parameter b(w) for HS-TCP

Note that we cannot choose arbitrary values for b(w) because:

When w = 38 (low speed connection), HS-TCP must used a factor of b(w) = 0.5 for the decrement operation
When w = 83,000 (high speed bandwidth connection), HS-TCP must used a factor of b(w) = 0.1 for the decrement operation

Graphically:

To choose an appropriate value for b(w) , we can use interpolation....

Consider the choices:

Note:

Using the red interpolation line, TCP will decrement CWND more rapidly
(CWND decreases by approximately 0.5 or 50%)
Using the green interpolation line, TCP will decrement CWND more gradually
(CWND decreases by approximately 0.3-0.4 or 30%-40%)

The authors' choice for b(w) :

The designers of the HighSpeed TCP protocol uses the following Interpolation curve for b(w) :

log(w) - log(38) b(w) = (0.1 - 0.5) * ---------------------- + 0.5 . . . . . (10) log(83000) - log(38)

The Interpolation curve for b(w) looks like this:

My own comments....

Personally, I disgree with the choice:

The factor b is used to decrease TCP traffic
Fact:
Personally, I think the red interpolation curve is better....
(Just a gut feeling....)

(I will continue with the paper....)

Implementation

CWND update method in HS-TCP:

a(w) New ACK: w <- w + ------ w Triple Dup ACK: w <- w - b(w) * w

a(w) and b(w) depends on the current congestion window CWND = w.

Assume that the current congestion window CWND is equal to:

The factor b(w) is interpolated using the formula:

log(w) - log(38) b(w) = (0.1 - 0.5) * ---------------------- + 0.5 .......(10) log(83000) - log(38)

After computing b(w), the term a(w) is computed using Equation (9): (See: click here )

2 w² b(w) p(w) a(w) = ---------------- .......... (9) 2 - b(w)

The value p(w) is computed from w using Equation (6):

0.0789 p(w) = --------- ........ (6) w^1.2

Implementation Code of HighSpeed TCP in NS2

The implemation code of the HighSPeed TCP can be found in NS2 in the file NS-DIR/tcp/tcp.cc: click here
Look for the method "double TcpAgent::decrease_param()" to find b(w)
Look for the method "double TcpAgent::increase_param()" to find a(w)

Here is a "cleaned up" code segment of the HighSpeed TCP algorithm in NS2:

# These values are set in TCL Agent/TCP set low_window_ 38 ; # default changed on 2002/8/12. Agent/TCP set high_window_ 83000 Agent/TCP set high_p_ 0.0000001 # 10^-7 Agent/TCP set high_decrease_ 0.1 Agent/TCP set max_ssthresh_ 0
double TcpAgent::decrease_param() { double decrease; // ============================================ // These values has been computed elsewhere: // ============================================ // hstcp_.dec1 = 0.5 - log(low_window_) * (high_decrease_ - 0.5)/highLowWin; // hstcp_.dec2 = (high_decrease_ - 0.5)/highLowWin; decrease = hstcp_.dec1 + hstcp_.dec2 * log(cwnd_); // They have re-worked Equation (10) into // the form: // b = constant1 + constant2 * log(w) return decrease; }
double TcpAgent::increase_param() { double increase, decrease, p, answer; // p ranges from 1.5/W^2 at congestion window low_window_, to // high_p_ at congestion window high_window_, on a log-log scale. // The decrease factor ranges from 0.5 to high_decrease // as the window ranges from low_window to high_window, // as the log of the window. // For an efficient implementation, this would just be looked up // in a table, with the increase and decrease being a function of the // congestion window. if (cwnd_ <= low_window_) { answer = 1 / cwnd_; // Standard Reno: CWND = CWND + 1/CWND return answer; } else { // ============================================ // These values has been computed elsewhere: // ============================================ // hstcp_.p1 = log(hstcp_.low_p) // - log(low_window_) * highLowP/highLowWin; // hstcp_.p2 = highLowP/highLowWin; // hstcp_.dec1 = 0.5 - log(low_window_) // * (high_decrease_ - 0.5)/highLowWin; // hstcp_.dec2 = (high_decrease_ - 0.5)/highLowWin; p = exp(hstcp_.p1 + log(cwnd_) * hstcp_.p2); decrease = hstcp_.dec1 + log(cwnd_) * hstcp_.dec2; // decrease = b(w) increase = cwnd_ * cwnd_ * p /(1/decrease - 0.5); // (w * w * p)/(1/b - 0.5) // = (w² b p) / (1 - 0.5*b) // = (2 w² b p ) / (2 - b) answer = increase / cwnd_; return answer; } }

How to use HighSpeed TCP in NS2:
Example CWND change - Reno vs. HS-TCP:
DEMO: (Reno vs. HighSpeed TCP)
- Reno NS prog file: click here
- HighSpeed TCP NS prog file: click here