Delay Efficient Sleep Scheduling in Wireless Sensor Networks Gang Lu, Narayanan Sadagopan, Bhaskar Krishnamachari, Anish Goel Presented by Boangoat(Bea) Jarupan Appeared in INFOCOMM 2005

Introduction Minimize the communication latency for periodic energy-efficient radio sleep cycles A single wake up schedule: –Formulate a graph-based optimization problem –Show that the problem is NP-hard –Derive and analyze the optimal solution in special cases (tree and ring topologies) –Propose several heuristics solutions for arbitrary topologies Multiple wake up schedule

Existing works Energy efficiency is a primary goal in wireless sensor networks Avoid idle listening of the radio SMAC: – Synchronized periodic duty cycling – Sleep latency TMAC, SMAC (extension): – Adaptive listening – Increase energy, overhearing, not sufficient for long path DMAC: – Communication pattern is restricted to data gathering tree – Node wakes up to receive the packet and sleeps after its transmit the packet How can the sensors be scheduled in arbitrary networks in order to minimize the sleep latency while providing energy efficiency through periodic sleeping?

Assumptions Consider k duty cycle requirement for an application Low traffic load (no additional delay due to congestion, interference/collision) Fixed packet length slot for a single packet transmission and fixed number of slots A single wake-up schedule: –Each node is assigned one of k slot as an active slot –Node can only receive the packet in its active slot –Node can wake up in any active slot of the neighbors to transmit the packet

G = (V, E) be an arbitrary graph k = number of slots = schedule length f : V → [0…k] be a slot assignment function For a given f, let d f (i,j) be a delay in transmitting data from i to j where (i, j) is in E Notation

Two Models All to all communication –Every pair of sensors are equally likely to communicate –Assign slots so that no two sensors incur arbitrarily long delays Weighted communication –Communication between some pairs is more frequent than other pairs –Assign slots so that average delay in the network is minimized

All to all communication A  B 5 A  C 3 A  D 5 A  E 2 A  F 3 B  A 4 B  C 4 All pairsMin delay P f (i,j) Max { min delay } = D f = 5 Delay Diameter A  B 3 A  C 2 A  D 3 A  E 1 A  F 2 B  A 3 B  C 3 All pairsMin hops Max { min hops } = 3 Hop Diameter Non Symmetric Symmetric P f (i,i) is delay along the shortest delay path between i an j

Example slot assignments Different slot assignments for the same topology Delay diameter = 5 (A  B)Delay diameter = 8 (D  C) Problem : Delay Efficient Sleep Scheduling (DESS) Given G and k, find an assignment (f) that minimizes the delay diameter D f

Weighted Communication Minimize the delay across the network Average Delay Diameter –P f (i,j) is delay along shortest delay path between i, j –Weights, w(i,j) ≥ 0 – Problem : Average Delay Efficient Sleep Scheduling (ADESS) Given G and k, find an assignment (f) that minimizes the average delay diameter

Analysis of DESS & ADESS Consider the equivalent decision problems DESS(G,k,f,∆) –Given a graph G = (V,E), number of slots k, a positive ∆ and slot assignment f, Is D f ≤ ∆? (decision problem) ADESS(G,k,f,w,∆) –Given a graph G = (V,E), number of slots k, a positive ∆ and slot assignment f, and a positive weights w i,j for all i, j in V, Is D f avg ≤ ∆?

Showing NP-Completeness Find a polynomial time reduction from a known NPC problem (say p) Why ? –If “our problem” is solvable in polynomial time (say q) then “known NPC problem” is solvable in polynomial time p + q  A contradiction Known NPC problem Our problem polynomial NP Complete problems can not be solvable in polynomial time i.e., there are no efficient algorithms

NP-Completeness of DESS 3-CNF-SAT DESS ( G, k, f, ∆ ) polynomial CNF – Conjunctive Normal Form – Product of Sums F = (X 2 OR ~X 3 OR X 1 ) AND (~X 3 OR X 2 OR X 4 ) AND (~X 4 OR ~X 1 OR X 2 ) Given F, find a satisfying assignment i.e., when F evaluates to TRUE ? Polynomial time reduction to DESS(G,k,f,∆) –Given F, we construct a graph G to form a problem DESS(G,2,f’,4) where hop diameter of G = 4 –Argue that Formula F is satisfiable iff D f ≤ 4 –i.e., given a solution to 3-CNF-SAT, we can construct a solution for DESS in polynomial time

NP-Completeness of DESS For a given satisfying assignment to F, we can construct the slot assignment to G in polynomial time

Optimal Assignment on a Tree A tree T=(V,E) and number of slot = k A diameter of T (in hops) = h (from a to b) We can prove the lower bound : D f ≥ hk/2 Proof:

Optimal Assignment on a Tree How to find f that achieves the lower bound hk/2? Just use two slot values: 0 and Adjacent nodes are assigned different slot values (like a chess board format)

Optimal Assignment on a Ring If number of nodes n = mk (0 … mk-1) then optimality is achieved by a sequential slot assignment f – f(0) = 0, for all i : 1 ≤ i ≤ mk-1 : f(i) = (f(i-1)+1) mod k D f in such an assignment is m(k-1) m = 4, k = 3, n = 12 D f = 4(3-1) = If n = mk + t then also we can find a similar assignment –Proof is involved

Heuristics : Centralized (d min )

Heuristics : Local-Neighbor

Heuristics : Local-DV

Heuristics : Randomization Random-Average: –Randomly chose the slot assignment for each node –High probability to find a short delay path in a dense network Random-Minimum: –Centralized version where the minimum delay diameter is found from multiple iterations of random slot assignment for all nodes

Heuristics: Concentric Ring for the Grid Topology Grid of 4x4 with k = 5 The outer most ring is given a sequential assignment going clockwise direction start at 0. For other rings, the slot assignment is chosen as best delay diameter for that ring.

Evaluation on Grids k = 15 Grid size = 9x9 Concentric ring performs the best while local scheme performs the worst

Evaluation on random topology area 10x10 area 3x33 Total # of sensors deployed = 100 Random-Avg performs worse in random graph due to bottlenecks problem

Evaluation on random topology k=10, 100 nodes in 10X10 area Random-Min with N=50,100 in 10X10 area As R increases, the delay diameter decreases

Multi-schedule Solutions A node can be active in more than one slot Tree networks (TREE-MULTI-SYNCH) –Delay diameter is less than d+4k (d is hop diameter) –Previously, we had lower bound –Independent of network size Grid networks (GRID-MULTI-SYNCH) –Consider a packet from X at (i,j) to Y at (p,q) –Shortest path distance: d = (p – i) + (q – j) –Delay diameter (i.e., total latency) is less than d+8k

Multi-schedule Solutions General networks –Decompose given network into a set of spanning trees –d G (u,v) = shortest path distance between u and v in G –c is a constant –Given G, there exist a collection S of spanning trees such that for all nodes u,v in G, –For a given X,Y we find a tree that minimizes d T (X,Y) and then use TREE-MULTI-SYNCH

Conclusions Formulated problems DESS, ADESS Proved their hardness Designed some heuristics (-) How good are heuristics when compared to optimal ? (-) Theoretical bounds on the performance of heuristics (-) Incorporate traffic constraints: interference, congestion etc.