1 Delay Efficient Sleep Scheduling in Wireless Sensor Networks Gang Lu, Narayanan Sadagopan, and Bhaskar Krishnamachari IEEE INFOCOM, Miami, FL, March /6/2005 Hong-Shi Wang

2 Contents  Introduction  Problem Definition –Delay Efficient Sleep Scheduling (DESS) –Average Delay Efficient Sleep Scheduling (ADESS)  Analysis –NP-Completeness –Optimal Assignment on Specific Topologies  Heuristic Approaches –Centralized –Local –Randomization –Concentric Ring for the Grid topology  Performance Evaluation  Conclusions

3 Sleep Latency  Largest source of energy consumption is keeping the radio on (even if idle). Particularly wasteful in low-data-rate applications.  Solution: regular duty-cycled sleep-wakeup cycles. E.g. S-MAC  Another Problem: increased latency time

4 Special Case Solution: D-MAC Gang Lu, Bhaskar Krishnamachari and Cauligi Raghavendra, "An Adaptive Energy-Efficient and Low- Latency MAC for Data Gathering in Sensor Networks," IEEE WMAN Staggered sleep wake cycles minimize latency for one-way data gathering.

5 General Problem Formulation  Each node is assigned one slot out of k to be an active reception slot which is advertised to all neighbors that may have to transmit to it.  Nodes sleep on all other slots unless they have a packet to transmit.  Assume low traffic so that only sleep latency is dominant and there is low interference/contention.  The per-hop sleep delay is the difference between reception slots of neighboring nodes  Data between any pair of nodes is routed on lowest-delay path between them (arbitrary communication patterns possible)  Goal: assign slots to nodes to minimize the worst case end to end delay (delay diameter)

6 Illustration

7 Problem Definition  Let G = (V,E) be an arbitrary graph.  Let k be the parameter that dictates the duty cycling requirements.  Assigning a slot s [0 … k-1] to a node i schedules i to wake up only at slot s.  Let f : V →[0 … k-1] be a slot assignment function that assigns a slot to every node in the graph.

8 Problem Definition  For a given f, let d f (i, j ) be the delay in transmitting data from i to j where (i, j ) E:  From the definition above, it also follows that:  Delay on a path P under a slot assignment f is defined as

9 Problem Definition  All to all Communication –In this scenario, every pair of sensors is equally likely to communicate. –Hence, it is desirable to assign slots to the nodes such that no two nodes incur arbitrarily long delays in communications.  Weighted Communication –In this scenario, the frequency of communication between a pair of sensor is not the same across all pairs. –This may happen in the case of a hierarchical network structure.

10 All to all Communication  Definition 1 – Delay diameter (D f ) For a given graph G = (V, E), number of slots k and slot assignment function f : V → [0 … k-1], the delay diameter is defined as, where P f (i, j) is the delay along the shortest delay path between nodes i and j under the given slot assignment function f.  Definition 2 – Delay Efficient Sleep Scheduling (DESS): Given a graph G = (V, E) and the number of slots k, find an assignment function f : V → [0 … k - 1] that minimizes the delay diameter i.e.

11 Illustration

12 Weighted Communication  Definition 3 – Delay diameter (D f ) For a given graph G = (V, E), number of slots k and slot assignment function f : V → [0 … k-1] and weights w(i, j) ≧ 0, the average delay diameter is defined as, where P f (i, j) is the delay along the shortest delay path between nodes i and j under the given slot assignment function f.  Definition 4 – Average Delay Efficient Sleep Scheduling (ADESS): Given a graph G = (V, E) and the number of slots k, weight w(i, j) ≧ 0, find an assignment function f : V → [0 … k - 1] that minimizes the delay diameter i.e.

13 Analysis  Definition 5 – DESS(G, k, f, △ ): Given a graph G = (V,E), number of slots k, a positive number △ and a slot assignment function f : V → [0 … k-1], is D f ≦ △.  Definition 6 – ADESS(G, k, f, w, △ ) : Given a graph G = (V,E), number of slots k, a positive number △, a slot assignment function f : V →[0 … k-1], and positive weights w ij for all i,j V, is D f avg ≦ △.

14 NP-Completeness  Theorem 1: DESS(G, k, f, △ ) is NP-Complete  Proof : To Prove that DESS(G, k, f, △ ) is NP-complete, we show a polynomial time reduction from 3-CNF-SAT to DESS(G,2,f ’,4). Consider a 3-CNF formula F consisting of n clauses and m literals i.e. F= c 1 ∩ c 2 ∩ …c n, where each ci = x i1 ∪ x i2 ∪ xi3 and. For non-triviality, we assume that a clause does not contain a literal and its complement(as such a clause is trivially satisfiable).

15 NP-Completeness  Given a 3-CNF formula F, construct a graph G = (V,E) are follows: –S V –For each variable x i : X i, X i1 (representing x i ), and X i2 (representing x i ) V –For each clause c i : C i V. – –If X i appear in c j, (X i1, C j ) E. If X i appears in c j, (X i2, C j ) E.

16 NP-Completeness  The diameter of G is 4. Consider the following slot assignment function f‘ : –f’ (S) = 1 i.e. S wakes up only at slot 1. – –f ’(X i1 ) = 0 iff x i is true, else f ‘(x i1 ) = 1. Moreover, f ’(X i1 ) = f ‘(X i2 ) = 1.  Since k = 2, d f ’ (i, j) = d f ’ (j, i) = 1 iff f ‘(i) ≠ f ’(j). If f ’(i) = f ’(j), then df ’(i, j) = df ‘(j, i) = 2.  This reduction can be computed in polynomial time. We will now show that a formula F is satisfiable iff D f’ ≦ 4 in G.

17 NP-Completeness

18 NP-Completeness  If the formula F is satisfiable, for every clause c i, at least one literal x j is true. Thus for every node C i in G, there exists a node X jk (k = 1 or k = 2) such that f ‘(X jk ) = 0. Thus, we can make the following observations about the delays along the paths from various nodes to S: –  Thus, for any given pair of nodes a and b, the maximum delay incurred on a path from a → S → b is at most 4. Hence, D f’ ≦ 4 in G.

19 NP-Completeness  If the formula F is not satisfiable, there exists at least one clause c i such that none of its literals are true. Thus, d f ‘(C i,X jk ) = d f ’(X jk,C i ) = 2, for all (C i, X jk ) E.  Now, let y l be a literal that appears in c i. Consider a path from C i to the node X lp (where X lp is the node that represents the complement of y l.). Every path from C i will reach a vertex X jk (such that the corresponding variable x jk appears in c i ) for which f ‘(x jk ) = 1. This first hop will incur a delay of 2. From X jk, one can either go to S (f ‘(s) = 1) or C j (f ‘(C j ) = 1) or X j (f ‘(X j ) = 1). This hop also incurs a delay of 2. At least one more edge has to be traversed to reach node X lp, which has a delay of at least 1.  Thus, there exists 2 nodes C i and X lp such that the shortest delay path between them has a delay of at least 5. Thus, D f ‘ > 4.

20 NP-Completeness

21 Optimal Assignment on Specific Topologies  Optimal Assignment on a Tree : Theorem 2 : Consider a tree T = (V, E). Let the number of slots be k. Let the diameter of T(in hops) be h. Then for every slot assignment f : V → [0,…k - 1], D f ≧ hk / 2.

22 Optimal Assignment on Tree  Proof : Consider a path between two nodes p to q having x hops. Since T is a tree, this is the only path between p and q. Consider an arbitrary slot assignment function f : V → [0,…k - 1]. Thus,  This is true for each pair of nodes including a and b. Thus, for every slot assignment function f, D f ≧ hk / 2, where h is the diameter of T.

23 Optimal Assignment on Tree  Based on theorem 2, the following assignment function f will minimize the delay diameter of the Tree T = (V, E) whose hop diameter is h (from a to b): Just use 2 slot values, 0 and.  Let d f (a) = 0. Adjacent vertices are assigned different slots (similar to a chess board pattern). In this case.. Hence, which tightly matches the lower bound on delay diameter of T.

24 Optimal Assignment on Specific Topologies  Optimal Assignment on a Tree : Theorem 3 : Consider n = mk nodes 0, 1, …. mk –1 arranged on a ring in the clockwise direction. The optimal slot assignment function f is specified as follows : f (0) = 0.

25 Optimal Assignment on a Ring  Proof : We will refer to such an f as the sequential slot assignment as it assigns a sequentially increasing slot (modulo k) to the nodes around the ring. We prove theorem 3 by contradiction.  For k = 2, it is easy to show that assigning 2 adjacent nodes the same slot incurs a dealy of 2 in both directions on that link, while a sequential assignment will yield a delay of 1 in either direction. Hen, we focus on the case where k ≧ 3. For a sequential slot assignment f, it is easy to show that the delay diameter is given by  Assume that exists a slot assignment function f ‘, such that D f ’ < D f.In the rest of the proof, we will focus on the delay in the ring due to f ‘.

26 Optimal Assignment on a Ring

27 Optimal Assignment on a Ring  Consider a block of m links on the ring from node 0 to node m. Since we assumed that D f ’ < m(k-1), the shortest delay path from node 0 to node m (and vice versa) must lie completely within the block.  The alternative path has m(k-1) links each incurring a delay of at least 1 (if this alternative path is the shortest delay path, it contradicts our assumption that D f ’ < m(k-1)).This is true for every block of m links on the ring.

28 Optimal Assignment on a Ring , let d i1 be the delay in block i from node (i-1)m to im, while d i2 be the delay in block i form node im to (i-1)m. We claim that d min = min i,j {d ij } < 2m. This can again be proved by contradiction as follows:  Consider a path from node 0 to node (k-1)m / 2. There are two possibilities. –0 → m → 2m … → (k-1)m / 2. The delay along this path is at least (k-1)*d min / 2. –0 → mk – m … → (k-1)m / 2. The delay along this path is at least (k+1)*d min / 2.  Thus, if d min ≧ 2m, it contradicts the assumption that D f ’ < m(k-1). Moreover, since each block has m links, each incurring a dealy of at least 1,

29 Optimal Assignment on a Ring

30 Optimal Assignment on a Ring  Let d min = m + x, where x [0, m). Consider the block that has the lowest delay d min. Without loss of generality, label the starting and ending node in this block as mk – m and 0. Consider a path from node 0 to node mk – m – x. There are two possibilities. –0 → mk – m → … → mk – m – x. Delay along this path is at least mk – d min + x = m(k-1), which contradicts our assumption about D f ’ < m(k-1). –0 → m → 2m … → mk – m – x. Delay along this path is given by:  This again contradicts our assumption that D f ’ < m(k-1).

31 Optimal Assignment on a Ring

32 Algorithm - Centralized Assign slot 0 to all nodes in G d = D(G)//delay diameter for i ← 1 to n//number of iterations for each node s in the network for k 1 ← 0 to k – 1// total slots ok ← slot(s) slot(s) ← k 1 md ← D(G) if d min < d then d ← d min minslot ← k1 if d min == d then minslot ← k 1 with 50% probability minslot ← ok with 50% probability slot(s) ← minslot

33 Algorithm Local-Neighbor Each node s get the slots of its direct neighbor N(s) mind ← MAX_VALUE for k 1 ← 0 to k – 1 // total slots slots(s) ← k 1 fd(s,t) ← delay from s to t in N(s) bd(s,t) ← delay from t in N(s) to s maxd ← max(fd, bd) if maxd < mind then mind ← maxd minslot ← k 1 slot(s) ← minslot

34 Algorithm Local-DV Each node s calculate DV tables FDV, BDV Get the FDV, BDV of its direct neighbor N(s) mind ← MAX_VALUE for k 1 ← 0 to k – 1 // total slots slot(s) ← k update FDV, BDV maxd ← max(FDV, BDV) if maxd < mind then mind ← maxd minslot ← k 1 slot(s) ← minslot

35 Randomization  The simplest slot assignment is to just randomly choose a slot for each node once.  In a dense network where a node has a large number of neighbors (where multiple path are available for any pair of nodes), there is a high probability that assignment may lead to a short delay path. This decentralized random slot assignment named “Random-Average”.  The randomized slot assignment can also be done in a centralized manner. This centralized version named “Random-Minimum strategy” After a certain number of iterations of choosing random slots for all the nodes, this strategy choose the assignment that gives the minimum delay diameter and then deploys the slot assignment in the network.

36 Concentric Ring for the Grid topology  Concentric ring allocation for a grid of 4 x 4 nodes with k = 5. The dotted lines illustrate the concentric rings at each level.

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43 Multi-Schedule Solutions  If each node is allowed to adopt multiple schedules, then can find much more efficient solutions: Grid: delay diameter of at most d + 8k (create four cascading schedules at each node, one for each direction) Tree: delay diameter of at most d+4k (create two schedules at each node, one for each direction) On general graphs can obtain a O( (d + k)log n) approximation for the delay diameter

44 Conclusion  Summary –This paper addressed and proved that DESS problem is a NPC problem. –Provided optimal solution for specific topologies (tree and ring) –For arbitrary topologies, them proposed several heuristics and evaluated them through simulations.  Future Work –Techniques to compute good lower bounds on the optimal delay diameter for an arbitrary graph. –Good distributed heuristics for the DESS problem –In-depth analysis and algorithms for the weighted communication average delay problem (ADESS)