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

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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

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Outline

Introduction Analysis of Previous Work Bandwidth Allocation Scheme Simulation Results Conclusions

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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.

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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.

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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.

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Drawback 2:

A

D

B C

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}

？？？？？？

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Drawback 3:

A DB C{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.

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Bandwidth Allocation Scheme

Control Subframe Structure Native Code Distribution Method Proposed Algorithm On-demand Bandwidth-Guaranteed Routing

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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)

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Native Code Distribution Method (1)

A B CIDA=1 IDB=2 IDC=3

NCA= null NCB= null NCC= null Native Code

Native Codes (NCs)Native Codes (NCs)

1 2 34 …

2 1 4

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D E

IDA=3 IDB=2 IDC=1 IDD=4 IDE=5

A B C

Control packet: ID, NC, MNN, NNCSMNN: Minimum ID of its Neighbors whose need NCsNNCS: Neighbors have utilized NCs Set

Control packet: ID, NC, MNN, NNCSMNN: Minimum ID of its Neighbors whose need NCsNNCS: Neighbors have utilized NCs Set

MNNc=1MNNB=1 MNND=1

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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.

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Proposed Algorithm (1)

Ii-1 Ii+2Ii Ii+1

FSL(Ii-1) FSL(Ii) FSL(Ii+1) FSL(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

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

A BLSL={1,4,6}

FSLB={1,2,4,5,6,8}FSLA={1,3,4,6,9}

Send: {2,5}Receive: {7,8}

Example:

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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)

Ii-1 Ii+2Ii Ii+1LSL(i-1) LSL(i) LSL(i+1)

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A B C Dlink 1 link2 link3

LSL1={ }

LSL2={ }

LSL3={ }

ASL1={ }

ASL2={ }

ASL3={ }

1 2

1 2

1 3

3

ASL (Available Slot List): ASL(i) ∩ ASL(i+1)=NULLASL (Available Slot List): ASL(i) ∩ ASL(i+1)=NULL

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LSL1={ }

LSL2={ }

LSL3={ }

ASL1={ }

ASL2={ }

ASL3={ }

1 2

1 2

1 3

3

1. Select min number of ASL2. Select min number of LSL3. Select min of LSV4. Randomly select


Select slots with min SV

2 X

X

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LSL1={ }

LSL2={ }

LSL3={ }

ASL1={ }

ASL2={ }

ASL3={ }

1

1

1 3

3



2 X

X

X1

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LSL1={ }

LSL2={ }

LSL3={ }

ASL1={ }

ASL2={ }

ASL3={ } 33

3



2

X1 X

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LSL1={ }

LSL2={ }

LSL3={ }

ASL1={ }

ASL2={ }

ASL3={ }3



2 1

Bandwidth =2

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On-demand Bandwidth-Guaranteed Routing

A FB C ED

Route REQ: ID, FSL, route list Route Response

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Simulation Results

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

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Simulation Results (Cont.)

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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.

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END

Thank you !!

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Example (1)

A FB C EDlink 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}

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Example (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}

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

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Example (3)


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

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Example (4)


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

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Example (5)


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

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Background (1)

The channel structure

Orthogonal code– Any two nodes having common neighbors cannot share

the same code.

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Background (2)

I0 IrI1 I2

Route: R(I0→Ir)={ I0, I1, I2, Ir }Route: R(I0→Ir)={ I0, I1, I2, Ir }

SIL(I0, I1) SIL(I1, I2) SIL(I2, Ir) Slot Index List

B(I0, I1) B(I1, I2) B(I2, Ir) Bandwidth

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)}

Route Bandwidth:B( R(I0→Ir) )=min(0 i r-1)≦ ≦ {B( Ii, Ii+1)}

FSL(I0) FSL(I1) FSL(I2) FSL(Ir) Free Slot List

Documents

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

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