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Scheduling for information gathering on sensor network

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Abstract

We investigate a unique wireless sensor network scheduling problem in which all nodes in a cluster send exactly one packet to a designated sink node in an effort to minimize transmission time. However, node transmissions must be sufficiently isolated either in time or in space to avoid collisions. The problem is formulated and solved via graph representation. We prove that an optimal transmission schedule can be obtained efficiently through a pipeline-like schedule when the underlying topology is either line or tree. The minimum time required for a line or tree topology with \(n\) nodes is \(3(n - 2)\). We further prove that our scheduling problem is NP-hard for general graphs. We propose a heuristic algorithm for general graphs. Our heuristic tries to schedule as many independent segments as possible to increase the degree of parallel transmissions. This algorithm is compared to an RTS/CTS based distributed algorithm. Preliminary simulated results indicate that our heuristic algorithm outperforms the RTS/CTS based distributed algorithm (up to 30%) and exhibits stable behavior.

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Correspondence to Hongsik Choi.

Appendix

Appendix

1.1 Proof of Lemma 1

This is a proof of correctness of the algorithm PIPE. We will show that the schedule meet all the constraints of TAPL from (1) to (4). First, the schedule of LPIPE satisfies the constraint (1), meaning no two messages can be sent by one node at the same time, i.e., \(f(v_i,\,m_j) \ne f(v_i,\, m_k)\) iff \(j \ne k\) for all \(v_i \in V\). If not, i.e., \(f(v_i,\,m_j) = f(v_i,\, m_k)\), \( t_{(i-1)\%3 + 3*l_1} = t_{(i-1)\%3 + 3*l_2} \) for some \(l_1\) and \(l_2\) such that \(j = i + l_1,\, k = i+ l_2\) i by the statement 3 of LPIPE. This is possible only if \(l_1 = l_2\). That implies \(j = k\), which contradicts the assumption that \(j \ne k\).

Second, the schedule generated by LPIPE satisfies constraint (2), meaning no nodes within two hops of node \(i\) can transmit any messages when node \(i\) transmit a message. The proof is simple with a contradiction. If not, \(f(v_i,\,m_*) = <$> <$>f(v_j,\, m_*)\), i.e., \(t_{(i-1)\%3 + 3k} = t_{(j-1)\%3 + 3k'}\). It is possible only if \(k = k'\) and \((i-1)\%3 = (j-1)\%3\), which implies \((i-j)\%3 = 0\). In other words, \(|i-j | \geq 3\) which contradicts the assumption, \(| i- j | \leq 2\).

Third, the schedule generated by LPIPE satisfies constraint (3) that ensures no node can transfer a message before it actually receives the message. For a given message, \(m_j\), it will be sent by \(v_i, 1 \leq i \leq j\), at time \((i-1)\%3 + 3k\) where \(i+k = j\) by statement 3 of PIPE. It is easy to verify that \(f(v_i, m_j) = (i-1) \%3 + 3k < <$> <$> (i-2)\%3 + 3(k+1) = f(v_{i-1},\,m_j)\) since \((i-1)\%3 < <$> <$>(i-2)\%3 +3\). Fourth, the schedule satisfies constraint (4), namely the existing path constraint. More formally, we have to show that \(f(v_i,\, m_j) \ne null\) for all \(1\leq i \leq n-1,\, <$> <$>i \leq j \leq n-1\). For a given message, \(m_j\), originating at \(v_j\), we need to show that it will be relayed by nodes between \(v_j\) and \(v_0\) with different time slots. Note that \( j = i+k\) for \(0 \leq i \leq j\). From statement 3 of LPIPE, we know that \(m_{i+k}\) is sent by \(v_i\) at time \((i-1)\%3 + 3*k\), which implies \(f(v_i,\,m_j) \ne null\) for all \(1 \leq i \leq j\).

1.2 Proof of Lemma 2

Since statement 1 of LPIPE will be executed \(n-3\) times and statement 3 will assign 3 time slot, i.e., \(k,\,k+1,\,k+2\), for each execution of statement 1, it will use at most \(3(n-3)\) time slots. Statements 4, 5, and 6 will use one time slot per statement. Therefore, LPIPE will generate a schedule with \(3(n-2)\) time slots, i.e, \(K \leq 3(n-2)\).

1.3 Proof of Lemma 3

The DGCP problem asks for the chromatic number of the given graph. The upper bound of the chromatic number is the maximum degree, \(\Delta + 1\) and the trivial lower bound is the maximum clique size that the graph contains. For simplicity, let’s call nodes in \(\{ v_{i,k},\, k \leq i \leq m \}\) the nodes in column \(K\). With the edge set of \(\{(v_{i,k},\,v_{j,k})\), \(i \ne j\), \(1 \leq k \leq m-1 \}\), the induced subgraph of \(G_m\) with the nodes in column \(K\) is a complete graph with size of \(m-k+1\).

The edge set of \(\{(v_{i,k},\,v_{j,k+1}), 0 < k < m, k < i \leq m,<$> <$> k +1 \leq j \leq m \}\) defines the relationship between the nodes in column \(K\) and nodes in column \(k+1\). It makes all nodes in column \(K\) to be connected to all nodes in column \(k+1\). With the fact that the nodes in each column induced subgraph is a complete graph, this makes an induced subgraph with nodes in column \(K\) and \(k+1\) makes a complete graph of size \(m - k + 1 + m - (k+1) +1 = 2m -2k +1\). This can be applied recursively for the nodes in column \(k+1\) and column \(k+2\).

The edge set of \(\{ (v_{i,k},\, v_{j,k+2}),\, 1 \leq k \leq m-2,\, k \leq i \leq m,<$> <$>\, k+2 \leq j \leq m \}\) defines the relationship between nodes in column \(K\) and the nodes in column \(k+2\). It makes edges between all nodes in column \(K\) and all nodes in column \(k+2\). With the results of the relation among nodes in column \(K\) and the relation between nodes in neighboring columns, it makes the induced subgraph with the nodes in three consecutive neighboring columns, column \(k, k+1i,\) and \(k+2\), to be a complete graph with size of \(m-k-1 <$> <$>+ m- (k+1) -1 + m-(k+2) + 1 =<$> <$> 3(m-k)\). It has a maximum size when \(k = 1\). Therefore, the maximum degree of the graph \(G_m\), \(\Delta(G_d)\), is \( 3(m-1)-1\). Furthermore, the maximum clique in the graph, \(G_d\) is the induced subgraph with the nodes in columns 1, 2, and 3 with size of \(3(m-1)\). We now have an upperbound and a lowerbound of the chromatic number for graph, \(G_d\), \(3(m-1) <$> <$>\leq \chi(G_d) \leq \Delta(G_d) +1 = 3(m-1)\). In other words, \(\chi(G_d)<$> <$> = 3(m-1)\).

1.4 Proof of Lemma 4

Construct an instance of TAPL, \(L(V,\, E)\) from the instance of DGCP, \(G_d = (V_d,\, E_d)\) as follow. Let \(V(L) = \{ v_i,\, 0 \leq<$> <$> i \leq n-1 \}\). Let \(E(L) = \{(v_i,\, v_{i-1})\ \hbox{for all}\ v_i \in V(G_d) \}\). Let \(M = \{m_i\ \hbox{for all}\ v_i \in V(L)/v_0 \}\). Now we have an instance of \(L(V,E)\) with \(n\) nodes from \(v_0\) to \(v_{n-1}\).

Intuitively, the edge set \(\{(v_{i,k},\,v_{j,k}), i \ne j, 1 \leq k \leq<$> <$> m-1 \} \) implies constraint (1) of TAPL. The edge set \(\{(v_{i,k},<$> <$>\,v_{j,k+1}), 0 < k < m, k < i \leq m, <$> <$>k +1 \leq j \leq m \}\) and \(\{ (v_{i,k},<$> <$>\, v_{j,k+2}),\, 1 \leq k \leq m-2,\, k \leq i \leq<$> <$> m,\, k+2 \leq j \leq m \}\) corresponds to constraint (2) of TAPL.

Let us show that one can get the solution of DGCP from that of TAPL. Let the solution of TAPL be the function \(f: V \times M \rightarrow \{0,1,2, \ldots, K \} \). Note that function \(f\) satisfies constraints (1) and (2) of TAPL by definition. Let the color of \(v_{i,j}\) be \(c(v_{i,j})\). Now, let \(c(v_{i,j}) = f(v_j,\, m_i)\) for \( 0 < <$> <$> j \leq m,\, j \leq i \leq m\). This assignment gives the solution of DGCP and we will show that this is the proper coloring. Note that \(c(u) \ne c(v)\) if \((u,v) \in E\) for proper coloring. We will show this by contradiction. If the coloring using TAPL solution is not proper, there is at least one vertex pair, \(v_{a,b}\), \(v_{c,d}\), such that \(c(v_{a,b}) = c(v_{c,d})\) and \((v_{a,b},\, v_{c,d}) \in E(G_m)\). By definition of the edge set of DGCP, we have three cases to verify.

First, if \((v_{a,b}, v_{c,d}) \in \{(v_{i,k},v_{j,k}), i \ne j, 1 \leq k \leq m-1 \} \), then \(b = d\). \(c(v_{a,b}) = c(v_{c,d})\) means \(c(v_{a,b}) = c(v_{c,b})\) and it implies \(f(v_b,m_a) = f(v_b, m_c)\) which is contradictary to the fact that \(f\) is the solution of TAPL (violate constraint (1)).

Second, if \( (v_{a,b}, v_{c,d}) \in \{(v_{i,k},\, v_{j,k+1}), 0 < k < m, k < i \leq<$> <$> m,\, k +1 \leq j \leq m \},\, |i-k| = 1\). \(c(v_{a,b}) = c(v_{c,d})\) means \(c(v_{a,b}) = c(v_{c,b+1})\) (or \(c(v_{a,b}) = c(v_{c,b-1})\) and it implies \(f(v_b,m_a) = f(v_{b+1}, m_c)\) (or \(f(v_b,m_a) = f(v_{b-1}, m_c))\), respectively, which is contradictory to the fact that \(f\) is the solution of TAPL (violate constraint (2)).

Third, if \((v_{a,b}, v_{c,d}) \in \{(v_{i,k},v_{j,k+2}), 0 < k < m-1,<$> <$> k < i \leq m, k +2 \leq j \leq m \}, |i-k| = 2\). \(c(v_a,b) = c(v_c,d)\) means \(c(v_{a,b}) = c(v_{a,b+2})\) (or \(c(v_{a,b}) = c(v_{a,b-2}))\) and it␣implies \(f(v_b,m_a) = f(v_{b+2}, m_c)\) (or \(f(v_b,m_a) = <$> <$>f(v_{b-2}, <$> <$>m_c))\), respectively, which contradicts the fact that \(f\) is the solution of TAPL (violate constraint (2)). Therefore, the coloring of \(c\) is the proper coloring.

1.5 Proof of Lemma 5

By Lemma 4, we know that DGCP, \(G_{n-1}\) can be colored with \(3(n-2)\) colors if TAPL can be scheduled with \(3(n-2)\) time slots. We also know the lower bound of DGCP, \(G_{n-1}\) is \(3(n-2)\) by Lemma 3. From these two facts we know that the lower bound of TAPL with \(n\) nodes is the number of colors for DALP with \(G_{n-1}\). Therefore, we have \(\chi(G_d) = 3(n-2) \leq K \) of TAPL with \(n\) nodes.

1.6 Proof of Lemma 7

Since statement 2 of TPIPE use the scheduling result of LPIPE with the same number of nodes, it should generate the schedule with \(3(n-2)\) time slots, i.e. \(K \leq 3(n-2)\).

1.7 Proof of Lemma 8

Note that line topology is a special case of tree topology. TAPT is a general version of TAPL. Therefore, lowerbound for the TAPT can not be smaller than that of TAPL which forces the lower bound of TAPT to be \(3(n-2)\).

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Choi, H., Wang, J. & Hughes, E.A. Scheduling for information gathering on sensor network. Wireless Netw 15, 127–140 (2009). https://doi.org/10.1007/s11276-007-0050-9

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