1. 1 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

2. 2 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

3. 3 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 P1P1 P1P1 P2P2 P2P2 Encoded broadcast: P 1 ⊕ P 2 decode P 2 decode P 1 123 Two flows: 1→3→2 2→3→1 P1P1 P2P2 Encoded broadcast: P 1 ⊕ P 2 decode P 2 decode P 1 One transmission (slot) saved!

4. 4 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 3 524 P1P1 P3P3 P4P4 P2P2 P 1 ⊕ P 2 ⊕ P 3 ⊕ P 4 decode P 3 decode P 2 decode P 4 decode P 1 - more complicated scenario... - 4 bi-directional flows Three transmissions saved!

5. 5 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)

6. 6 Introduction and Problem Formulation Question 2 on COPE: 123 P1P1 P2P2 Encoded broadcast: P 1 ⊕ P 2 decode P 2 decode P 1 Wait a minute...what if... 123 P1P1 P2P2 P1P1 ? Nothing to encode with...? The order is important! Q2. How many can we encode under random access?

7. 7 Introduction and Problem Formulation Question 3 on COPE: 123 P1P1 P2P2 Encoded broadcast: P 1 ⊕ P 2 decode P 2 decode P 1 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...?

8. 8 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

9. 9 The “Physical” Upper Bound Coding Structure - 1 coding node - n coding flows coding structure with 2 coding flows coding structure with 4 coding flows Assuming scheduling (ordering), routing is optimal encode at most 2 at once encode at most 4 at once

10. 10 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 P1P1 P1P1 P2P2 P2P2 Encoded broadcast: P 1 ⊕ P 2 decode P 2 decode P 1 Main result: n is indeed upper bounded, due to the geometrical constraints.

11. 11 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) 0.60.70.80.9 n* 3.393.954.886.97 Numerical example

12. 12 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

13. 13 Upper Bound under Random Access 1 4 5 3 2 P1P1 P1P1 P2P2 P2P2 Encoded broadcast: P 1 ⊕ P 2 The order is important... P1P1 P2P2 Node 1 sends Node 5’s buffer Node 3 sends Node 5 sends P1⊕P2P1⊕P2 We are lucky...this works just fine

14. 14 Upper Bound under Random Access 1 4 5 3 2 P1P1 P1P1 P2P2 P2P2 The order is important... P1P1 P2P2 Node 1 sends Node 5’s buffer Node 5 sends Node 3 sends P1P1 P2P2 P1P1 Node 5 sends P2P2 No coding happens...

15. 15 Upper Bound under Random Access 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

16. 16 Upper Bound under Random Access Key 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)

17. 17 Upper Bound under Random Access Main 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=4 Avg. encoding number vs. buffer size

18. 18 Upper Bound under Random Access Main results 2) Enough buffer size is essential for coding Example: n=4 Avg. encoding number vs. buffer size

19. 19 Upper Bound under Random Access Main results 2) Enough buffer size is essential for coding Example: n=4 Throughput vs. buffer size

20. 20 Upper Bound under Random Access Main results 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 simple X=1, K=1 pair has the lowest loss...

21. 21 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

22. 22 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!

23. 23 Bounding the Throughput Gain Lemma 1. Only considering the “same routing” case can guarantee that we get an upper bound of throughput gain for the general case. optimal routing for coding optimal routing for non-coding

24. 24 Bounding the Throughput Gain Lemma 2. For a general network, the throughput gain is bounded by the throughput gain in one of its coding structures.

25. 25 Bounding the Throughput Gain 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)

26. 26 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

27. 27 Conclusion A performance analysis of COPE Focus: the “encoding number” Physical upper bound on encoding number Performance under Random Access Bound in general network