High Speed TCP

Effect of very high bandwidth

Number of packets carried by a link:

Network link:
Propagation delay on a 300 Km link = 300,000/300,000,000 sec = 10^-3 sec
The 300 Km link carries (B × 10^-3) bits
Suppose each packet consists of 10,000 bits (~ 1 Kbytes), then a 300 Km link carries approximately (B × 10^-7) packets
Number of packets carried by a 300 Km link:

Effect of a high speed link:

Effect of very high speed links on a congested router

Effect of a congested router:

The queue of a congested router is eventually overflow
The overflowed queue will drop newly arriving packets

Effect of high speed link:

Notes:

Even when the TCP (sender) times out and set CWND = 1, the arrival rate into the congested route will not decrease significantly
The congested router will take a very long time to clear its buffer
TCP will experience multiple consecutive timeout
Consequently: the SSThreh become very small

Example:

Note:

When the SSThresh has become very small, TCP will take a very long time to acquire additional bandwidth for its transmission use !!!
(Because TCP is in congestion avoidance mode and increases CWND by 1 (one) after every RTT sec)

Some pre-requisites...

CWND update in TCP:

1 new ACK: CWND = CWND + ------ CWND 1 Triple Duplicate ACKs: CWND = CWND - --- CWND 2

Throughput of TCP: (using the first order approximation of the Padhye's result: (see: click here )

------- 1 / 3 B(p) = ----- \ / ----- RTT \/ 2bp

For the case where b = 1 (no delayed ACK), we have:

------ ------- 1 / 3 1 / 1.5 B(p) = ----- \ / ---- = ----- \ / ----- RTT \/ 2p RTT \/ p 1 1.2 B(p) = ----- * ------- (packets per sec) . . . . . . . . . . . . (1) RTT ---- \/ p

Why can TCP not achieve Gbps throughput....

The TCP throughput formula (Equation (1)) can tell us that TCP cannot achieve gigabit per second throughput

Assume that:

RTT = 0.1 sec Packetsize = 1500 bytes

The TCP throughput is equal to: (See: click here)

1 1.2 B(p) = ----- * ------- (packets per sec) RTT ---- \/ p 1 1.2 = ----- * ------- (packets per sec) 0.1 ---- \/ p 1 1.2 = ----- * ------- * 1500 (bytes per sec) 0.1 ---- \/ p 12 * 1500 * 8 = --------------- bps ---- \/ p 144,000 = --------- bps ---- \/ p

To achieve, says 10 Gbps throughput, the packet loss probability p must be equal to:

144,000 B(p) = --------- = 10,000,000,000 ---- \/ p
---- 144,000 ==> \/ p = --------------- = 0.0000144 10,000,000,000
<==> p = 0.00000000020736 = 2 x 10^-10 (in other words: 1 packet loss in every 5,000,000,000 packets !)

Existing communication networks cannot achieve 1 packet loss in every 5,000,000,000 packets sent

Clarification:

Fiber optic has error rate of 1 error in 10¹⁰ bits
Probab of bit error in fiber = 10⁻¹⁰
A packet of 1500 bytes has 8 * 1500 = 12000 bits

Every bit have be received correctly for the packet to be received correctly

Therefore:

1 − p = (1 − 10⁻¹⁰)¹²⁰⁰⁰ ~= 1 − 12000 × 10⁻¹⁰

Packet loss probability for a 1500 byte packet in fiber:

p = 12000 × 10⁻¹⁰ = 1.2 × 10⁻⁶

p is much larger than 2 × 10⁻¹⁰

Conclusion:

Another way to explain why TCP cannot achieve Gbps throughput....

Classic TCP:

1 Increase: W = W + --- (when new ACK is received) W Decrease: W = W - 0.5 * W (when packet loss is detected by 3 dup ACKs)
1 1.2 Avg TCP Throughput = ----- * ------- (in # packets/sec) RTT --- \/ p p = packet loss probability

The average CWND used by (classic) TCP:

Throughput when using CWND = k (packets):

What value of CWND used by TCP will yield the average TCP throughput:

1 1.2 Avg TCP Throughput = ----- * ------- (in # packets/sec) RTT --- \/ p

Conclusion: 1.2 Avg TCP CWND = ------- (in # packets) --- \/ p

Another way to explain why TCP cannot achieve Gbps throughput.... continued....

Relationship:

1.2 Avg. CWND = ------- (in # packets) --- \/ p

Some explicit values:

Packet Drop Rate P Congestion Window W ------------------ ------------------- 10^-2 12 10^-3 38 10^-4 120 10^-5 379 10^-6 1200 10^-7 3795 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.

Observation:

Packet loss probability on fiber optic links is in the order of p = 10^-7
From the table above, we see that the (standard) TCP protocol will avg. CNWD = 3795
This CWND value is too small not allow TCP to transmit at Giga bits per sec

How to achieve Gbps throughput with TCP
- Modify TCP:
- Note:

A generalized TCP congestion control scheme

A more general update algorithm for TCP's CWND is:

a new ACK: CWND = CWND + ------ CWND 3 dup ACK: CWND = CWND − b * CWND

The TCP throughput is then (approximately): (see: click here )

------------ 1 / a (2 - b) B(p) = ----- \ / ---------- (packets/sec) RTT \/ 2bp

Notes:

No proof here - it's in the paper: (see: click here )
The parameter b has a different meaning then the b in Padhye's paper.
The parameter b in Padhye's paper relates to delayed ACKs
The parameter b here is the multiplicative decrease factor !!!
(I am using the authors' notation so you can read their paper along side with the class notes with more ease...)

High Speed TCP research literature...
- The first attempt to make TCP operate at Gigabit speed is the work:
  This proposal builds on TCP Reno and is going through the Internet standardization process.
- Another commerically avaialble product is "FAST-TCP" (from CalTech):
  - "FAST TCP: Motivation, Architecture, ALgorithms and Performance" click here
- A friend of mine is also in the business of designing high speed TCP protocols:
  - "CUBIC: A New TCP-Friendly High Speed TCP variant" click here
- And there are many other protocols... too many to cover :-)
- We will therefore cover the first one - that will give you an idea what the other are doing....