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Experiment And Experiment And Analysis ofAnalysis ofDynamic TCP Dynamic TCP AcknowledgementAcknowledgement

Daeseob Lim

Sam Lai

Wing-Ho Gordon Wong

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What is the main problem of dynamic TCP acknowledgment? To ACK, or not to ACK: that is the

question. When a packet arrives to the receiver,

there are two choices:ACK immediately.Wait and ACK later, such that you may have

the chance to acknowledge multiple packets with just one ACK.

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What’s the difference?

ACK immediately: low latency : time elapses between the packet

arrival and the ACK for this packet is send. No. of ACK packet increase.

Wait and ACK later:High latencySmall amount of ACK packet generated

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What consider to be the best solution? Low no. of acknowledge Low aggregate acknowledgement latency

for all packets

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Aggregate latency example:

The total ACK latency is 30ms in this case.

Packet A Arrive Packet B Arrive Send single ACK for both A and B

10ms 10ms 10ms

ACK latency for B is 10ms

ACK latency for A is 20ms

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Naïve Solutions

Send ACK immediately for each packet?Low or even no ACK latency, but this will

generate too much ACK packets Send one ACK for all the end of all

transmits?Only one ACK is neededHowever, high latencyDon’t know which one is the last packetWhat if the link is unreliable?

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An Online Randomized Algorithm

Dynamic TCP acknowledgement and other stories about e/(e-1)By Anna R. Kalin, Claire Kenyon, Dana

Randall Able to achieve a competitive ratio of 1.58

compare to the optimal solution Competitive ratio = performance of the

algorithm / performance of the optimal solution

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Detail about this algorithm

P(t, t`) be the set of packets arrive between time t and t`

There exists z such that z is between 0 and 1 inclusively

Distributed function to produced z. (randomized factor)

Suppose that ith acknowledgement happens at time i and the next one happen at time ti+1

(con’t)

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Detail about the algorithm

By the algorithm, we should locate Ti+1 such that ti <= Ti+1 <= ti+1 and P(ti, Ti+1)(Ti+1 - ti+1) = z

If we do that, z unit of latency cost will be saved by sending a single additional ACK at Ti+1.

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Before applying the algorithm

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After the algorithm

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Why this works?

The rectangle is guarantee to have area of at least 1

By sending 1 additional ACK, the acknowledgement cost increase by 1, but the latency cost decreases by at least 1.

The new sequence is at least as good as the original one.

More detail proof in the paper. Dynamic TCP acknowledgement and other stories

about e/(e-1)

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Contribution of this research

Implement a randomized online algorithm about delayed ACK into Linux kernel

Compare real performance of the randomized algorithm and the current TCP implementation

Observe its superiority in terms of cost Analyze its inability in terms of throughput

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ACK for data packet

Data packet

Receiver

Data packetReceiver

ACK packet

Immediate Immediate ACKACK

Delayed Delayed ACKACK

Data packet

Receiver

ACK packet

Schedule a timer

Timer expired

≈ 40ms

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Interval of Delayed-ACK Timer

Determined by some factorsMinimum/maximum interval by kernel constantsEstimated RTT

Restrictions by RFC 2581The maximum is 500ms.Acknowledge at least every second segment.Acknowledge out-of-order data immediately.

In most cases, ≈ 40ms (~ 200ms)

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Implementation of TCP on Linux

Need to send an Need to send an immediate ACK?immediate ACK?

tcp_rcv_established()

__tcp_ack_snd_check()

tcp_send_ack() tcp_send_delayed_ack()

Received a data packet from IP-layer

YesNo,

Then why not ‘Delayed ACK’?

The point to hack kernel

codes !!

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Hacking protocol stack

Cost > Random Cost > Random value ?value ?

Send additional ACK !

tcp_send_delayed_ack()

Choose a random value

Yes

No

Scale to threshold value

Cost = Unacked data size * Elapsed time since last ACK

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Generating random number

Generate random numbers in advance, store them into kernel codes, and select a number sequentially

y = ex/(e-1)

1X

Y

0.599, 0.761, 0.232, 0.378, 0.619, 0.997,

….

….

unsigned rand_numbers[1000] = {0.599, 0.761, 0.232,0.378, 0.619, 0.997,….….

};

….

number = rand_numbers[index++];

Generate numbers with off-line program

Select random number in the array

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Test Environment

Client

Router

Server

ModifiedKernel

+NetworkSniffer

(Ethereal)

NetworkEmulator

(ns2)

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Competitive Ratio Experiment The server sends out 100 packets to the

client at random time spacing at most 70ms apart.

The competitive ratio is calculated for each cost ratio starting from 0.05 to 0.95 stepping by 0.05, then 0.900 to 0.995 stepping by 0.005.

Run on simulated networks having bandwidth of 100Mbps and RTT of 2ms and 100ms for both versions of TCP.

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Overall Competitive Ratioon 2ms Network

1

10

100

1000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Cost Ratio

Com

petit

ive

Rat

io

Original TCPNew TCP

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Overall Competitive Ratioon 100ms Network

1

10

100

1000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Cost Ratio

Com

petit

ive

Rat

io

Original TCPNew TCP

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Blowup Competitive Ratioon 2ms Network

1

1.2

1.4

1.6

1.8

2

2.2

2.4

0.9 0.92 0.94 0.96 0.98 1Cost Ratio

Com

petit

ive

Rat

io

Original TCPNew TCP

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Blowup Competitive Ratioon 100ms Network

1

1.2

1.4

1.6

1.8

2

2.2

2.4

0.9 0.92 0.94 0.96 0.98 1Cost Ratio

Com

petit

ive

Rat

io

Original TCPNew TCP

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Analysis For new TCP, overall the competitive ratio

is within 1.58 except for borderline cases.Small cost ratio:

Expensive latency cost Overhead from network sniffer Overhead from new TCP

Large cost ratio: Expensive acknowledgement cost Original TCP acknowledgements Possibility of additional acknowledgement

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Analysis

For original TCP, the competitive ratio starts out extremely high, then converges rapidly with the new TCP. Eventually, it starts to increase, but at a slower rate.Favors delay acknowledgement even when

latency cost is high.Always acknowledge within 200ms or every 2

packet full of data even when acknowledgement cost is high.

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Streaming Data Experiment

The client sends out a request to the server asking for data of a certain size to be sent.

The server replies with the data. The client measures the total duration to

determine throughput. Run on simulated networks having

bandwidth of 100Mbps and RTT of 2ms and 100ms for both versions of TCP.

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Streaming Data Result

RTTOriginal

TCP

New TCP

Throughput Speedup

2ms 6.926Mbps 6.873Mbps 0.992

100ms 0.577Mbps 0.577Mbps 1

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Analysis The new TCP can not outperform the

original TCP in terms of throughput. Intuitively, you can imagine, if the incoming

traffic is regular and data keeps pouring in, to optimize throughput, you’d want to delay ack as long as possible.

The new TCP can not do delay ack longer than the original TCP.

Random scale down of the threshold for sending an additional acknowledgement.

Our implementation induces little overhead.

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Conclusion

Prove the randomized algorithm can achieve the competitive ratio of 1.58 in most cases.

Our implementation achieves better competitive ratio comparing to the original TCP in most cases.

Low overhead implementation. Can not improve network performance in

terms of throughput.