1 How Many Packets Can We Encode? - An Analysis of Practical Wireless Network Coding Jerry Le, John...

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How Many Packets Can We Encode?- An Analysis of Practical Wireless

Network Coding

Jerry Le, John C.S. Lui, Dah Ming Chiu

Chinese University of Hong Kong

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Outline

• Introduction and Problem Formulation

• The Physical Upper Bound on how many we can encode

• How many can we encode under Random-Access

• Bounding the throughput gain

• Conclusion

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Introduction and Problem Formulation• Quick review of the Practical Coding System (COPE [1])

[1] S. Katti, et al. XORs in the Air: Practical Wireless Network Coding. Sigcomm 2006.

1

4

5

3

2

Two flows: 1→5→2 3→5→4

P1

P1 P2

P2

Encoded broadcast:

P1⊕P2

decode P2 decode P1

1 23

Two flows: 1→3→2 2→3→1

P1 P2

Encoded broadcast:

P1⊕P2

decode P2 decode P1

One transmission (slot) saved!

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Introduction and Problem Formulation• Quick review of the Practical Coding System (COPE [1])

[1] S. Katti, et al. XORs in the Air: Practical Wireless Network Coding. Sigcomm 2006.

1

3

52 4

P1

P3

P4P2P1⊕P2 ⊕P3 ⊕P4

decode P3

decode P2decode P4

decode P1

- more complicated scenario...- 4 bi-directional flows

Three transmissions saved!

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Introduction and Problem Formulation• Question 1 on COPE:

encode 2 at once

encode 4 at once

encode 6 at once

Q1. Is there a limit on how many we can encode? (Assuming everything else is optimal)

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1 23P1 P2

Encoded broadcast:

P1⊕P2

decode P2 decode P1

Wait a minute...what if...

1 23P1 P2

P1

?

Nothing to encode with...? The order is important!

Q2. How many can we encode under random access?

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1 23P1 P2

Encoded broadcast:

P1⊕P2

decode P2 decode P1

Q3. What is the limit of throughput gain in general topology?

- “Non-coding” uses 4 slots- “Coding” uses 3 slots- Throughput gain?

Of course, 4/3!

But how about this...?

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Outline



• How many can we encode under Random Access


• Conclusion

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The “Physical” Upper Bound• Coding Structure

- 1 coding node

- n coding flows

coding structurewith 2 coding flows coding structure

with 4 coding flows

Assuming scheduling (ordering), routing is optimal

encode at most 2 at onceencode at most 4 at once

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The “Physical” Upper Bound• Coding Structure

- 1 coding node

- n coding flows, need to find out the upper bound of n

• Two main constraints for coding to happen

- opportunistic overhearing

- two-hop relaying 1

4

5

3

2

P1

P1 P2

P2

Encoded broadcast:

P1⊕P2

decode P2 decode P1

Main result: n is indeed upper bounded, due to the geometrical constraints.

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The “Physical” Upper Bound

• Main results - in 2D networks, n ≤ O[(r/δ)2] - in 3D networks, n ≤ O[(r/δ)3] - r: successful transmission range -δ: “distance gap” between successful transmission and unsuccessful transmission

r

δ

r/(r+δ) 0.6 0.7 0.8 0.9

n* 3.39 3.95 4.88 6.97

Numerical example

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Outline





• Conclusion

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Upper Bound under Random Access

1

4

5

3

2

P1

P1 P2

P2

Encoded broadcast:

P1⊕P2

The order is important...

P1

P2

Node 1 sendsNode 5’s bufferNode 3 sends

Node 5 sends

P1⊕P2

We are lucky...this works just fine

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1

4

5

3

2

P1

P1 P2

P2

The order is important...

P1P2

Node 1 sendsNode 5’s bufferNode 5 sends

Node 3 sends

P1P2 P1

Node 5 sendsP2

No coding happens...

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1

4

5

3

2

A more complex example- a randomly generated transmission sequence: 1,1,3,5,1,3,5,5,3,5

Node 5’s buffer

- 10 slots used to send 3 packets for each flow- optimum case is 9 - worst case is 12

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Upper Bound under Random AccessKey intuition- higher packet diversity (in coding node’s buffer) results in higher encoding number

What can affect packet diversity?- traffic intensity- buffer size- random access “Equal Access”

“K-Priority” (bottleneck gets higher priority)

“Wait-for-X” (hold transmission until having something to encode)

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Upper Bound under Random AccessMain results

1) “Equal Access” is particularly suitable for coding - In saturation, symmetric case, the average encoding number at each transmission is: n*M/(M+1)

Example:n=4Avg. encoding number vs. buffer size

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2) Enough buffer size is essential for coding

Example:n=4Avg. encoding number vs. buffer size

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2) Enough buffer size is essential for coding

Example:n=4Throughput vs. buffer size

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3) Surprisingly, Wait-for-X is not suitable

- Equal Access + Enough buffer is already near optimal- Too much waiting can only bring more packet loss

The simpleX=1, K=1 pairhas the lowest loss...

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Outline





• Conclusion

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Bounding the Throughput Gain

Remember this?...

- Throughput gain = T* of coding / T* or non-coding

- What affect throughput? (Routing, Scheduling)

- The best routing for coding may not be optimal for non-coding!

- The same is true for scheduling!

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Lemma 1. Only considering the “same routing” case canguarantee that we get an upper bound of throughput gainfor the general case.

**

*

*

*

**

cnc

cc

ncnc

cc

nc

c

RT

RT

RT

RT

T

TG

optimal routing for coding

optimal routing for non-coding

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Lemma 2. For a general network, the throughput gain is bounded by the throughput gain in one of its coding structures.

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Theorem. 1) For a general network, the throughput gain is bounded by 2n/(n+1), where n is the number of coding flows in one of its coding structures.2) The bound can be approximated as 2n/(n+M(M+1)) when using equal access. (M is buffer size)

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Outline





• Conclusion

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Conclusion

• A performance analysis of COPE

• Focus: the “encoding number”

• Physical upper bound on encoding number

• Performance under Random Access

• Bound in general network

1 How Many Packets Can We Encode? - An Analysis of Practical Wireless Network Coding Jerry Le, John...

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