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Cooperative Relaying & Power Allocation Strategies in Sensor Networks Jingqiao Zhang Oct. 27, 2005

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2 ECSE 6962: Wireless Ad Hoc and Sensor Networks Outline Introduction Relay Channel Strategies: Decode-and-Forward (DF), Amplify-and-Forward (AF), Compress-and-Forward (CF) Power Allocation Strategies in DF & AF Two-hop Relay Networks Minimum Energy Accumulative Routing Relay based routing scheme in sensor networks

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3 ECSE 6962: Wireless Ad Hoc and Sensor Networks Outline Introduction Relay Channel Strategies: Decode-and-Forward (DF), Amplify-and-Forward (AF), Compress-and-Forward (CF) Power Allocation Strategies in DF & AF Two-hop Relay Networks Minimum Energy Accumulative Routing Relay based routing scheme in multi-hop networks

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4 ECSE 6962: Wireless Ad Hoc and Sensor Networks Information Theory Point to point communication Network Information Theory Multiple Access Channel (M A C), Broadcasting Channel (BC), Relay Channel X1 X2 Y Y1 Y2 X Relay Y X R1R1 R2R2 R1R1 R2R2 Binary Symmetric Broadcast Channel The capacity is known for the special physically degraded relay channel -- p(y,y1|x,x1) = p(y1|x,x1)p(y|y1,x1) Y1:X1 BC MAC

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5 ECSE 6962: Wireless Ad Hoc and Sensor Networks Relay Strategies The transmission is divided into to two periods 1 st period: the source node broadcasts its signal. -- BC 2 nd period: relay nodes transmits the source’s information. -- MAC Regenerative Decode-and-Forward (DF) Receive Decode Re-encode Transmit Non-regenerative Amplify-and-Forward (AF) Receive Amplify Transmit Compress-and-Forward (CF) Receive Quantize/Compress Transmit Relay D S

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6 ECSE 6962: Wireless Ad Hoc and Sensor Networks Outline Introduction Relay Channel Strategies: DF, AF, CF Power Allocation Strategies in DF & AF Two-hop Relay Networks Minimum Energy Accumulative Routing Relay based routing scheme in multi-hop networks

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7 ECSE 6962: Wireless Ad Hoc and Sensor Networks Channel Model A Shared Channel Y = m (√ m X m ) + Z node 0 node M M M−1M−1 X0X0 Y Orthogonal channels Y = BX + Z B = diag( √ 0, √ 1, …, √ M ) X = [ X 0, X 1, …, X M ] Z: a vector of AWGN ~ N(0,1) √1√1 √2√2 √m√m √M√M √0√0 Y 1 = √ 1 X 0 + Z 1 √1√1 √2√2 √m√m √M√M Y 1 X 1 Y M X M X 1 = √b 1 Y 1 (AF)

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8 ECSE 6962: Wireless Ad Hoc and Sensor Networks Objective Notations Power allocation: p=[p 0, p 1, …, p M ]; p 0 is the source power, and p m is the power at relay node m. r : the expected communication rate from the source to the destination The objective is to find the optimal power allocation p that supports a reliable communication at rate r between the source and the destination. subject to Channel capacity ¸ r

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9 ECSE 6962: Wireless Ad Hoc and Sensor Networks An Overview of Conclusions Orthogonal ChannelsShared Channels DF Bandwidth: the larger, the better Power: At most one relay node at the “best” position is used AF Bandwidth: an optimal bandwidth Power: water-filling allocation Bandwidth: an optimal bandwidth Power: Maximum Ratio Combining (powers /equivalent channel gains) According to the two recent papers of Maric, et. al

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10 ECSE 6962: Wireless Ad Hoc and Sensor Networks DF: Constraints Relay node: A( p 0 ): the subset of relay nodes that decode the transmission from s For i 2 A( p 0 ), we have W log(1+ i p 0 /W ) ¸ r, Destination node: 2W I DF (p/W) ¸ r I DF (p/W): the maximum mutual information I DF (p/W) between the channel input X and output Y Relay ds

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11 ECSE 6962: Wireless Ad Hoc and Sensor Networks DF: Problem Formulation Optimization Problem Subject to log(1+x) ¼ x

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12 ECSE 6962: Wireless Ad Hoc and Sensor Networks DF: Problem Simplification Subject to Subject to Theorem: The wideband DF relay problem admits an optimal solution in which at most one relay node transmits.

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13 ECSE 6962: Wireless Ad Hoc and Sensor Networks DF: Optimal Power Allocation Relay node k is used ( p k > 0 ) only if k > 0 and k > 0. The optimal solution is to choose from the set U={k| k > 0 and k > 0 }, and find the one minimizing the total power

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14 ECSE 6962: Wireless Ad Hoc and Sensor Networks AF Receive Amplify Transmit Y m = √ 1 X 0 + Z m, and X m = √b m Y m The communication rate r is no greater than the maximum mutual information 2W I AF (p/W) between the channel input X and output Y Unlike DF, the capacity is not a increasing function of the bandwidth. To see this, No benefit from the relay transmission

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15 ECSE 6962: Wireless Ad Hoc and Sensor Networks AF: Problem Formulation Define P = p/W (power per dimension) Two sub-problems Decide the best set of relay node {1, 2, …, K} For a given the set {1, 2, …, K}, find the optimal power and bandwidth Subject to Lagrangian:

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16 ECSE 6962: Wireless Ad Hoc and Sensor Networks AF: Optimal Power Allocation Theorem 3: The AF relay problem has an optimum solution in which the optimum bandwidth W *, rate r and the total transmit power p * have a linear relationship Subproblem-1 is not solved completely. Actually, the author solves the problem for a given source power P 0 : where is the constant such that

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17 ECSE 6962: Wireless Ad Hoc and Sensor Networks The Effect of the Source Power The best choice of relay nodes significantly depends on the transmit power at the source.

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18 ECSE 6962: Wireless Ad Hoc and Sensor Networks Outline Introduction Relay Channel Strategies: DF, AF, CF Power Allocation Strategies in DF & AF Two-hop Relay Networks Minimum Energy Accumulative Routing Relay based routing scheme in multi-hop networks

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19 ECSE 6962: Wireless Ad Hoc and Sensor Networks Minimum Energy Accumulative Routing Traditional multi-hop Model (TM) Point-to-point communication in each hop Accumulative Routing (AR) Store the partially overheard packet (leakage) The leakage contributes to the final reception of the packet. For example, several copies of the same packet can be decoded by maximum ratio combining (MRC). v1v1 v4v4 v2v2 v3v3

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20 ECSE 6962: Wireless Ad Hoc and Sensor Networks TM: Point-to-Point Transmission TM: Point-to-point transmission (without relay): p i ¸ H/g i,j Total tx power: i p i = H/g i,i+1 vivi vjvj g i,j / d i,j pipi p j r = p i g i,j ¸ H p 1 = H / g 1,2 p 2 = H / g 2,3 p 3 = H / g 3,4 v1v1 v4v4 v2v2 v3v3

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21 ECSE 6962: Wireless Ad Hoc and Sensor Networks TR: Relay-based Routing Relay transmission: the previous transmission of the same packet can be overheard and stored by the latter node. E.g., before v i v i+1, the leakage from v 1, v 2, …, v i-1 is: l i+1 = vj 2 {v1, …, vi-1} p j g j,i+1 Total tx power: i p i = (H – l i+1 )/g i,i+1 v1v1 v2v2 v3v3 v4v4 p 2 = (H – p 1 g 1,3 ) / g 2,3 p 1 = H / g 1,2 p 3 = (H – p 1 g 1,4 – p 2 g 2,4 ) / g 3,4

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22 ECSE 6962: Wireless Ad Hoc and Sensor Networks An Example Assume the power attenuation exponent = 2. In simulation, 30% energy saving is achieved sd r H H sd r H 0.5 H overheard Total energy = 2H Total energy = 1.5H

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23 ECSE 6962: Wireless Ad Hoc and Sensor Networks Problem Formulation To find a feasible transmission schedule S = [(v 1, p 1 ), …, (v w, p w ))] S is an ordered list of node-power pair Feasible: v 1 = s, v w = d relay: decode – transmit A general AR routing: j=1 i-1 p j g j,i ¸ H, 8 i > 1 A k-Relay case: j=i-k i-1 p j g j,i ¸ H, 8 i > 1 Minimum Energy Accumulative Routing: MEAR (V, s, d) For a source-destination pair (s, d), to look for a feasible transmission schedule S = [(v 1, p 1 ), …, (v w, p w ))], such that the total tx energy E(S), i=1 w p i is minimized

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24 ECSE 6962: Wireless Ad Hoc and Sensor Networks Complexity of the Problem Theorem 1: The MEAR(V, s, t) problem is NP-complete for a general graph with arbitrary link gains and a cap on the transmission energy a node can spend on one packet. D-MEAR (V,s,t) problem: Given (V,s,t), is there a feasible tx schedule S for (s,t) such that E(S) · P?

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25 ECSE 6962: Wireless Ad Hoc and Sensor Networks Shortest Path Heuristic (SPH) Algorithm Shortest path algorithm point-to-point communication in each hop p i = H/g i,i+1 as the weight of edge (i,i+1) Theorem 2: In a general graph model, E(OPT) ∈ o( E(SPH) ) in the worst case SPH be the solution from the shortest path heuristic, and OPT be the optimal solution. Proof: consider a special case where H =1, g i,j =(|i - j|+ ) -1 SPH: from node v 1 to v n directly. E(SPH) = n+ OPT:

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26 ECSE 6962: Wireless Ad Hoc and Sensor Networks The Structure of Optimal Schedules Theorem 3: A MEAR(V, s, t) problem always has an optimal schedule that is a wavepath. Wavepath: A schedule S is a wavepath iff v i v j ( 8 i j ) ---- no loop in the route j=1 i-1 p j g j,i = H ---- wavepath property p 2 = (H – p 1 g 1,3 ) / g 2,3 p 1 = H / g 1,2 p 3 = (H – p 1 g 1,4 – p 2 g 2,4 ) / g 3,4 (3,4), (2,3), (1,2), (u,v): cost (or distance) from node u to node v

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27 ECSE 6962: Wireless Ad Hoc and Sensor Networks RPAR Algorithm Initialization RPAR generate a tree T rooted as s Calculate the cost (distance) from s to any other node v : e(v) = H / g s,v At each iteration, Find a new node u that is closest to the source node. Add it to the tree T. Update the cost (distance) from the rest node v to the root s. (u, v) e(v)e(v) e(u)e(u) s u v

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28 ECSE 6962: Wireless Ad Hoc and Sensor Networks Comparison (OPT, RPAR, SPH) A number of nodes are randomly distributed in a 1000m £ 1000m region. Attenuation exponent = 2 RPAR is very close to the optimal solution. E(RPAR) / E(OPT) < 1.1 SPH RPAR

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29 ECSE 6962: Wireless Ad Hoc and Sensor Networks Effects of Node Density

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30 ECSE 6962: Wireless Ad Hoc and Sensor Networks Implementation Issues - 1 Identify each received packet before MRC Reliable header & partially overhead payload Adopt a strong modulation/coding of the packet header Store all the partially received packets the same source address and SN. Combine various copies and decode the packet Maximum Ratio Combining v1v1 v4v4 v2v2 v3v3

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31 ECSE 6962: Wireless Ad Hoc and Sensor Networks Implementation Issues - 2 Wireless channel: RTS/CTS handshake IEEE like MAC protocol RTS/CTS collision avoidance mechanism should be modified to prevent interference at overhearing nodes E.g., let RTS/CTS cover the all routing area -- only suitable for a low-load network

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32 ECSE 6962: Wireless Ad Hoc and Sensor Networks Conclusions Communication performance can be improved by the cooperation of intermediate relay nodes In a two-hop network, DF: At most one relay node at the “best” position assists the communication AF: There exist an optimal bandwidth. Optimal power allocation is like water-filling (orthogonal channels) or CMR (shared channels) In a multi-hop network, Energy saving can be achieved by a relay-based routing algorithm A heuristic algorithm achieves the sub-optimal solution in polynomial time.

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33 ECSE 6962: Wireless Ad Hoc and Sensor Networks Future Work What if both DF and AF modes are allowed? What will be the optimal power allocation? Is there any analytical result on the RPAR, since seemingly we can transform its operation into a Trellis structure? Is it possible to find the near-optimal set of participating nodes distributedly? If yes, the distributed power allocation can be easily obtianed according to the wavepath property.

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34 ECSE 6962: Wireless Ad Hoc and Sensor Networks References 1. G. Kramer, M. Gastpar, and P. Gupta, Cooperative Strategies and Capacity Theorems for Relay Networks, IEEE Transactions on Information Theory, vol. 51, no. 9, pp – 3063, Sept I. Maric and D. Yates, Bandwidth and Power Allocation for Coopertive Strategies in Gaussian Relay Network, In Proc. of Asilomar, Nov Monterey, CA. (invited) 3. I. Maric and D. Yates, Forwarding Strategies for Gaussian Parallel-Relay Networks, Intl. Symposium on Information Theory, pp. 269, June J. Chen, L. Jia, X. Liu, G. Noubir, and R. Sundaram, Minimum Energy Accumulative Routing in Wireless Networks, Infocom 2005, vol. 3, pp – 1886, March T. M. Cover and J.A. Thomas, Elements of Information Theory, Wiley, 1991

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35 ECSE 6962: Wireless Ad Hoc and Sensor Networks Thank you!

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