1. 1 Distributed End-to-End Bandwidth Allocation in Ad Hoc Network Zhijun Cai, Mi Lu, Xiaodong Wang 指導教授：石貴平 報告學生：莊宗翰 報告日期： 2002/09/10

2. 2 Outline Introduction Analysis of Previous Work Bandwidth Allocation Scheme Simulation Results Conclusions

3. 3 Introduction End-to-end bandwidth allocation scheme –Topology-transparent scheduling technology Reduce control overhead. –Code distribution method Avoid hidden terminal problem. –Utilize the global resource information along the route Improve performance.

4. 4 Analysis of Previous Work (1) QoS Routing in Ad Hoc Networks [1] –C. Lin and J. Liu –IEEE Journal on Selected Areas in Communications, vol. 17, no. 8, pp. 1426-1438, Aug. 1999. An On-demand QoS Routing Protocol for Mobile Ad Hoc Networks [16] –Chunhung Richard Lin and Chungching Liu –Proc. of Globalcom’00, 2000, vol. 3, pp. 1783-1787.

5. 5 Analysis of Previous Work (2) Drawback 1: –The control subframe is composed of N control slots Each node is assigned one unique control slot. –Significant control overhead Topology-transparent spatial re-use technology. Reduce the length of the control subframe. Topology-transparent spatial re-use technology. Reduce the length of the control subframe.

6. 6 Analysis of Previous Work (3) Drawback 2: A D BC use {1,3} free{2,4,5,6,7,8,9} use {2,4}use {1,6,8} use {3,9} Free_slot={5,7} ？？？

7. 7 Analysis of Previous Work (4) Drawback 3: ADBC {1,2,3,4} {1,2,5,6} {5,6} Case 1: {1,2} {5,6} X Case 2: {3,4} {1,2} {5,6} Previous Work: Utilize only local resource information. Our Work: Utilize the global resource information. Previous Work: Utilize only local resource information. Our Work: Utilize the global resource information.

8. 8 Bandwidth Allocation Scheme Control Subframe Structure Native Code Distribution Method Proposed Algorithm On-demand Bandwidth-Guaranteed Routing

9. 9 Control Subframe Structure Topology-transparent spatial re-use technology –Reduce the length of the control subframe. Modified Galois field topology transparent broadcast scheduling algorithm The control frame with p×q control slots. (p,q are determined by N and D) [ref. 20] Overhead reduce gain g=N/(p×q) The control frame with p×q control slots. (p,q are determined by N and D) [ref. 20] Overhead reduce gain g=N/(p×q)

10. 10 Native Code Distribution Method (1) ABC ID A =1 ID B =2 ID C =3 NC A = null NC B = null NC C = null Native Code Native Codes (NCs) 12 3 4 … 214

11. 11 Native Code Distribution Method (2) DE ID A =3 ID B =2 ID C =1 ID D =4 ID E =5 ABC Control packet: ID, NC, MNN, NNCS MNN: Minimum ID of its Neighbors whose need NCs NNCS: Neighbors have utilized NCs Set Control packet: ID, NC, MNN, NNCS MNN: Minimum ID of its Neighbors whose need NCs NNCS: Neighbors have utilized NCs Set MNNc=1 MNN B =1MNN D =1

12. 12 Native Code Distribution Method (3) Any two nodes within 2-hop distance can not share the same code. The number of NCs (Native Code) is no less than the max number of 2-hop neighbor. Any two nodes within 2-hop distance can not set their NCs at the same time. If one link is broken, no node need to update its NC; while if one link is created, at most two nodes may need to update their NCs.

13. 13 Proposed Algorithm (1) I i-1 I i+2 IiIi I i+1 FSL(I i-1 ) FSL(I i ) FSL(I i+1 ) FSL(I i+2 ) LSL(i-1) LSL(i) LSL(i+1) FSL (Free Slot List) LSL (Link Slot List): LSL(i)=FSL(I i )∩FSL(I i+1 ) → CF (Conflict Free): neither in LSL (i-1) nor in LSL (i+1) → CE (Conflict Existence): LSL － CF FSL (Free Slot List) LSL (Link Slot List): LSL(i)=FSL(I i )∩FSL(I i+1 ) → CF (Conflict Free): neither in LSL (i-1) nor in LSL (i+1) → CE (Conflict Existence): LSL － CF AB LSL={1,4,6} FSL B ={1,2,4,5,6,8}FSL A ={1,3,4,6,9} Send: {2,5} Receive: {7,8} Example:

14. 14 Proposed Algorithm (2) NCE(i) : the number of CE slots in LSL(i) NASL(i) : the number of slots in ASL(i) NCE(i) : the number of CE slots in LSL(i) NASL(i) : the number of slots in ASL(i) For a CE slot, its SV(Stability Value) is defined as follows: LSV(i) is the min value of all the SVs of CE slots in LSL(i) I i-1 I i+2 IiIi I i+1 LSL(i-1) LSL(i) LSL(i+1)

15. 15 Proposed Algorithm (3) ABCD link 1 link2 link3 LSL1={ } LSL2={ } LSL3={ } ASL1={ } ASL2={ } ASL3={ } 12 12 13 3 ASL (Available Slot List): ASL(i) ∩ ASL(i+1)=NULL

16. 16 Proposed Algorithm (4) ABCD link 1 link2 link3 LSL1={ } LSL2={ } LSL3={ } ASL1={ } ASL2={ } ASL3={ } 12 12 13 3 1.Select min number of ASL 2.Select min number of LSL 3.Select min of LSV 4.Randomly select 1.Select min number of ASL 2.Select min number of LSL 3.Select min of LSV 4.Randomly select Select slots with min SV 2 X X

17. 17 Proposed Algorithm (5) ABCD link 1 link2 link3 LSL1={ } LSL2={ } LSL3={ } ASL1={ } ASL2={ } ASL3={ } 1 1 13 3 1.Select min number of ASL 2.Select min number of LSL 3.Select min of LSV 4.Randomly select 1.Select min number of ASL 2.Select min number of LSL 3.Select min of LSV 4.Randomly select 2 X X X 1

18. 18 Proposed Algorithm (7) ABCD link 1 link2 link3 LSL1={ } LSL2={ } LSL3={ } ASL1={ } ASL2={ } ASL3={ } 33 3 1.Select min number of ASL 2.Select min number of LSL 3.Select min of LSV 4.Randomly select 1.Select min number of ASL 2.Select min number of LSL 3.Select min of LSV 4.Randomly select 2 X 1 X

19. 19 Proposed Algorithm (8) ABCD link 1 link2 link3 LSL1={ } LSL2={ } LSL3={ } ASL1={ } ASL2={ } ASL3={ } 3 1.Select min number of ASL 2.Select min number of LSL 3.Select min of LSV 4.Randomly select 1.Select min number of ASL 2.Select min number of LSL 3.Select min of LSV 4.Randomly select 21 Bandwidth =2

20. 20 On-demand Bandwidth-Guaranteed Routing AFBCED Route REQ: ID, FSL, route list Route Response

21. 21 Simulation Results Number of nodes: 50 Number of data slots per frame: 20 Average number of neighbors for a node: 6 Bandwidth requirement: 2

22. 22 Simulation Results (Cont.)

23. 23 Conclusion An efficient end-to-end distributed bandwidth allocation scheme. Utilize the topology-transparent scheduling technology. An efficient orthogonal code distribution scheme. Utilize the global resource information along the route.

24. 24 END Thank you !!

25. 25 Example (1) AFBCED link 1 link2 link3 link4 link5 LSL1={3,4,8,11,14,15,16,17,18,19,20} LSL2={4,14,17,18,19} LSL3={14,17,18} LSL4={1,2,3,7,10,11,14,16,18} LSL5={4,7,11,12,13,19,20}

26. 26 Example (2) AFBCED link 1 link2 link3 link4 link5 LSL1={4,14,17,18,19} LSL2={4,14,17,18,19} LSL3={14,17,18} LSL4={7,11,14,18} LSL5={7,11} ASL1={3,8,11,15,16,20} ASL2={} ASL3={} ASL4={1,2,3,10,16} ASL5={4,12,13,19,20} NASL1=6 NASL2=0 NASL3=0 NASL4=5 NASL5=5 NCE1=5 NCE2=5 NCE3=3 NCE4=4 NCE5=2

27. 27 Example (3) AFBCED link 1 link2 link3 link4 link5 LSL1={4,14,17,18,19} LSL2={4,14,17,18,19} LSL3={14,17,18} LSL4={7,11,14,18} LSL5={7,11} ASL1={3,8,11,15,16,20} ASL2={} ASL3={} ASL4={1,2,3,10,16} ASL5={4,12,13,19,20} NASL1=6 NASL2=0 NASL3=0 NASL4=5 NASL5=5 NCE1=5 NCE2=5 NCE3=3 NCE4=4 NCE5=2 min SV X X

28. 28 Example (4) AFBCED link 1 link2 link3 link4 link5 LSL1={4,14,17,18,19} LSL2={4,17,18,19} LSL3={17,18} LSL4={7,11,18} LSL5={7,11} ASL1={3,8,11,15,16,20} ASL2={} ASL3={14} ASL4={1,2,3,10,16} ASL5={4,12,13,19,20} NASL1=6 NASL2=0 NASL3=1 NASL4=5 NASL5=5 NCE1=5 NCE2=4 NCE3=2 NCE4=3 NCE5=2

29. 29 Example (5) AFBCED link 1 link2 link3 link4 link5 LSL1={14,17,19} LSL2={19} LSL3={} LSL4={7,11,18} LSL5={7,11} ASL1={3,8,11,15,16,20} ASL2={4,18} ASL3={14,17} ASL4={1,2,3,10,16} ASL5={4,12,13,19,20} NASL1=6 NASL2=2 NASL3=2 NASL4=5 NASL5=5 NCE1=3 NCE2=1 NCE3=0 NCE4=3 NCE5=2 Route bandwidth

30. 30 Background (1) The channel structure Orthogonal code –Any two nodes having common neighbors cannot share the same code.

31. 31 Background (2) I0I0 IrIr I1I1 I2I2 Route: R(I 0 →I r )={ I 0, I 1, I 2, I r } SIL(I 0, I 1 ) SIL(I 1, I 2 ) SIL(I 2, I r ) Slot Index List B(I 0, I 1 ) B(I 1, I 2 ) B(I 2, I r ) Bandwidth SIL={1,2,3,4} SIL={5,6} SIL={4,7,8} B=4 B=2 B=3 Route Bandwidth: B( R(I 0 →I r ) )=min (0 ≦ i ≦ r-1) {B( I i, I i+1 )} Route Bandwidth: B( R(I 0 →I r ) )=min (0 ≦ i ≦ r-1) {B( I i, I i+1 )} FSL(I 0 ) FSL(I 1 ) FSL(I 2 ) FSL(I r ) Free Slot List