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Distributed End-to-End Bandwidth Allocation in Ad Hoc Network

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

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Distributed End-to-End Bandwidth Allocation in Ad Hoc Network

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

  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. 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. 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.

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

  7. Analysis of Previous Work (4) • Drawback 3: A B C D {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.

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

  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)

  10. Native Code Distribution Method (1) Native Codes (NCs) … 4 1 2 3 A B C IDA=1 IDB=2 IDC=3 NCA= null NCB= null NCC= null Native Code 2 1 4

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

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

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

  15. Proposed Algorithm (3) link 1 link2 link3 A B C D ASL1={ } ASL2={ } ASL3={ } LSL1={ } LSL2={ } LSL3={ } 1 2 1 2 3 1 3 ASL (Available Slot List): ASL(i) ∩ ASL(i+1)=NULL

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

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

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

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

  20. On-demand Bandwidth-Guaranteed Routing Route Response Route REQ: ID, FSL, route list A B C D E F

  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. Simulation Results (Cont.)

  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. END Thank you !!

  25. Example (1) link 1 link2 link3 link4 link5 A B C D E F 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. Example (2) link 1 link2 link3 link4 link5 A B C D E F 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 LSL1={4,14,17,18,19} LSL2={4,14,17,18,19} LSL3={14,17,18} LSL4={7,11,14,18} LSL5={7,11}

  27. Example (3) link 1 link2 link3 link4 link5 A B C D E F 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 LSL1={4,14,17,18,19} LSL2={4,14,17,18,19} LSL3={14,17,18} LSL4={7,11,14,18} LSL5={7,11} X min SV X

  28. Example (4) link 1 link2 link3 link4 link5 A B C D E F 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 LSL1={4,14,17,18,19} LSL2={4,17,18,19} LSL3={17,18} LSL4={7,11,18} LSL5={7,11}

  29. Example (5) link 1 link2 link3 link4 link5 A B C D E F 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 LSL1={14,17,19} LSL2={19} LSL3={} LSL4={7,11,18} LSL5={7,11} Route bandwidth

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

  31. Background (2) Route: R(I0→Ir)={ I0, I1, I2, Ir } B(I0, I1) B(I1, I2) B(I2, Ir) Bandwidth SIL(I0, I1) SIL(I1, I2) SIL(I2, Ir) Slot Index List FSL(I0) FSL(I1) FSL(I2) FSL(Ir) Free Slot List I0 I1 I2 Ir SIL={1,2,3,4} SIL={5,6} SIL={4,7,8} B=4 B=2 B=3 Route Bandwidth: B( R(I0→Ir) )=min(0≦i≦r-1) {B( Ii, Ii+1)}

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