Tree-based object tracking without mobility statistics in wireless sensor networks

Abstract

Object tracking in wireless sensor networks is to track mobile objects by scattered sensors. These sensors are typically organized into a tree to deliver report messages upon detecting object’s move. Existing tree construction algorithms all require a mobility profile that characterizes the movement statistics of the target object. Mobility profiles are generally obtained based on historical running traces. The contribution of this work is twofold. We first show that the problem of finding an optimal message report tree that requires the least amount of report messages is NP-hard. We then propose analytic estimates of mobility profiles based on Markov-chain model. This profiling replaces an otherwise experimental process that generates and analyzes running traces. Simulation results show that the analytic profiling works well and can replace costly statistical profiling without noticeable performance degradation.

Appendix

We prove Theorem 1 by reducing the Exact Cover by 3-sets (X3C) problem [4] to the BCMRT problem. Given a set $Q= \{q_{1}, q_{2}, \ldots, q_{3k}\}$ and a collection $\chi = \{X_{1}, X_{2}, \ldots, X_{p}\}$ of 3-element subsets of Q, the X3C problem asks whether there exists a subcollection $\chi^{\prime}\subseteq \chi$ such that every element of Q occurs in exactly one member of χ′.

For an instance of the X3C problem, we construct a mobility profile M as in Fig. 16, in which v _i and s _j are for q _i and X _j , respectively. There is another vertex r connected to every s _j with an edge of weight L ₁. There is an edge (v _i , s _j ) of weight L ₂ whenever q _i ∈ X _j . All v _i ’s are connected by unit-weight edges to form a clique. L ₁ and L ₂ are large enough so that these edges are enforced to be in the MRT if it exists. We claim that there exists an exact cover of the X3C problem if and only if the BCMRT problem has a solution with cost

$$ c^*\,=\,pL_1\,+\,3k(L_2)\,+\,3(3p\,-\,3k)L_2\,+\,3k(2\,\times\,2\,+\,(3k\,-\,3)\,\times\,4)/2\,=\,pL_1\,+\,(9p\,-\,6k)L_2\,+\,3k(6k\,-\,4). $$

Fig. 16

The mobility profile corresponding to an instance of the X3C problem. Edges in the clique are not shown

For the convenience of discussion, we divide the node set into three subsets: V = {v _i |∀ i}, S = {s _j |∀ j}, and R = {r}. Let E ₁ = {(r, s _j )|1 ≤ j ≤ p}, E ₂ = {(s _j , v _i )|q _i ∈ X _j } and E ₃ = {(v _i , v _j )|1 ≤ i < j ≤ 3k} be three types of node pairs that are connected by edges in M. Since an exact cover does not exist if $\exists v_i\,\in\,V, \forall s_j\,\in\,S: (s_j, v_i)\,\notin\,E_2,$ we preclude such trivial cases in the rest of the proof. (the only-if part) Suppose that there exists an exact cover χ′. We show that there exists a spanning tree T with cost c*. The edge set of T consists of all edges in E ₁ and edges (s _j , v _i ) for all X _j ∈ χ′ and q _i ∈ X _j . Since χ′ is an exact cover, T is a spanning tree. The cost of T consists of three distinct parts.

• Cost associated with E ₁: This part of cost is pL ₁.

• Cost associated with E ₂: There are 3p node pairs in E ₂, 3k of which are included as edges in T and each of the others has a distance of 3 (with common ancestor r). The total cost is therefore 3kL ₂ + 3(3p − 3k)L ₂.

• Cost associated with E ₃: Each v _i ∈ V shares with other two nodes in V a common parent s _j ∈ S and with each of the remaining 3k − 3 nodes in V the common parent r. Each of the former node pairs has a distance of 2 while the distance of the latter is 4. The cost is therefore 3k(2 × 2 + (3k − 3) × 4)/2 = 3k(6k − 4).

Clearly, the total cost of T is c*. (the if part) M has p + 3k + 1 nodes, and therefore any spanning tree has exactly p + 3k edges. The weights L ₁ and L ₂ are set large enough such that if the OMRT has cost c* it must contain all the p edges in E ₁ and 3k edges in E ₂, and therefore no edge in E ₃ is in the OMRT. Suppose that T* is the OMRT of cost c*. To our aim, we assign L ₁ = (9p − 6k)L ₂ + 3k(6k − 4). If any edge (r, s _j ) is not in T*, the cost is larger than (p + 2)L ₁ > c* since the distance from r to s _j is at least three. Since all edges in E ₁ are in T*, no node in V can be connected to more than one node in S. Otherwise it forms a cycle.

For convenience, we root T* at r. By the above discussion, all nodes in S are at level one. Next, we set L ₂ = 9k and claim that all nodes in V are at level two, which also means that no edge in E ₃ can be an edge of T*. Suppose by contradiction that there exists a node v in level 3. We shall show that there is a way to reduce the cost of T*, which contradicts the optimality of T*. Let v _i be the parent of v and s _j the parent of v _i . Also assume that (v, s) is an edge in E ₂. We replace edge (v _i , v) with (s, v) and make all children of v, if any, the children of v _i . The cost between v and all nodes in S is reduced by at least L ₂, while the cost between v and all nodes in V is increased by at most 3(3k − 1) since the new distance from v to v _i is 4.

We have shown that the entire E ₁ is in T* and the other tree edges are from E ₂. The cost in the cuts (R, S) and (S, V) is therefore pL ₁ + (9p − 6k)L ₂ no matter which of the 3k edges in E ₂ are chosen. For each pair of nodes in V, the update cost is 2 if the node pair is connected to the same parent and 4 otherwise. Since each s _j can have at most three children, the cost is lower bounded by 3k(6k − 4), and it happens only when k nodes in S have exactly three children and others have none. In this case there is an exact cover for the X3C problem.