Approximation algorithms for broadcasting in duty cycled wireless sensor networks

Abstract

Broadcast is a fundamental operation in wireless sensor networks (WSNs). Given a source node with a packet to broadcast, the aim is to propagate the packet to all nodes in a collision free manner whilst incurring minimum latency. This problem, called minimum latency broadcast scheduling (MLBS), has been studied extensively in wireless ad-hoc networks whereby nodes remain on all the time, and has been shown to be NP-hard. However, only a few studies have addressed this problem in the context of duty-cycled WSNs. In these WSNs, nodes do not wake-up simultaneously, and hence, not all neighbors of a transmitting node will receive a broadcast packet at the same time. Unfortunately, the problem remains NP-hard and multiple transmissions may be necessary due to different wake-up times. Henceforth, this paper considers MLBS in duty cycled WSNs and presents two approximation algorithms, BS-1 and BS-2, that produce a maximum latency of at most $(\Delta -1) TH$ and $13TH$ respectively. Here, $\Delta$ is the maximum degree of nodes, $T$ denotes the number of time slots in a scheduling period, and $H$ is the broadcast latency lower bound obtained from the shortest path algorithm. We evaluated our algorithms under different network configurations and confirmed that the latencies achieved by our algorithms are much lower than existing schemes. In particular, compared to OTAB, the best broadcast scheduling algorithm to date, the broadcast latency and transmission times achieved by BS-1 is at least $\frac{1}{17}$ and $\frac{2}{5}$ that of OTAB respectively.

Appendices

Appendix 1

1.1 Correctness analysis of BS-1

Theorem 1

Algorithm BS-1 provides a correct and collision-free broadcast schedule.

Proof

We prove the correctness for this theorem by contradiction. We assume node $v$ in layer $i$ cannot be scheduled to receive the message collision-free because there are two or more parallel transmissions to node $v$ at the same time. Assume that node $v$’s parent $P(v)$ and one of $v$’s neighbor $u$ are scheduled to transmit at $t$. Furthermore, we consider two different cases. In the first case, node $P(v)$ is scheduled before node $u$. According to constraint 4) of Algorithm 2, if node $P(v)$ selects time $t$ as $P(v)$’s transmission time, node $u$ will not choose $t$ again, because the reception time of its neighbor $v$ has been already set to $t$. Secondly, assume that node $u$ is scheduled before node $P(v)$. According to constraint 3) of Algorithm 2, after node $u$ selects time $t$ as its transmission time, node $P(v)$ will not choose $t$ as its transmission time since node $P(v)$’s children $v$ will hear node $u$’s transmission at $t$. This is contradictory to our assumption. So node $v$ will receive the message collision-free.

We now prove BS-1 produces a schedule with 100 % reachability. We also prove this by contradiction. Assume there exists a node $v$ that has not received a broadcast message, but all of node $v$’s neighbors are covered by BS-1. Recall that node $v$ must be assigned with a parent node from its one-hop neighbors, and $v$ can receive the message collision-free from its parent node as proven above. Therefore, this case is contradictory to our assumption. Hence, this theorem holds true for all nodes in the network. $\square$

1.2 Approximation ratio analysis of BS-1

Lemma 1

Consider a parent $u$ of nodes in $L_i$. Suppose that node $u$ ’s transmission to layer $L_i$ is delayed because doing so will cause a collision at one or more of its one hop neighbors, denoted by set $W$ , with transmissions among its two-hop neighbors. Then, the following is true:

• 1. Node $w$ is not in $C(u,i)$ , where $w \in W$.

• 2. For node $u$ , there are at most $\Delta -2$ nodes among its two-hop neighbors that interfere with $u$.

Proof

We prove the correctness of the first property by contradiction. Assume that node $w$ is in $C(u,i)$, and there is a node $z$ that interferes with $u$ at $w$. According to the construction of $T_{b}$, node $z$ must have one child node in layer $i$; that is, $z$ must have one child which has the same active time slot with nodes in $C(u,i)$, and is scheduled to transmit to layer $i$ before node $u$. Recall that the order in which nodes’ parent nodes are selected (line 7 in Algorithm 1) and the order in which the transmissions are scheduled (line $5$ in Algorithm 3) are the same. That is, if node $z$ is scheduled to transmit to layer $i$ before node $u$, node $w$ must be node $z$’s child, i.e., $w \in C(z,i)$. This contradicts our assumption that node $w$ is in $C(u,i)$.

According to the first property of Lemma 1, when node $u$ is scheduled to transmit to nodes in layer $i$, none of $u$’s children in $C(u,i)$ hear a message from other nodes; That is, $I_{1}(u)= \emptyset$ as per Algorithm 2. Hence, node $u$’s transmission time is only delayed by the reception time of its one-hop neighbors as per Algorithm 2, i.e., $I_2 (u)$. Recall that node $u$ has at most $\Delta$ one-hop neighbors, in which one is $u$’s parent node and at least one is $u$’s child. Henceforth, node $u$ is adjacent to at most $\Delta -2$ interfering nodes that have been assigned with reception time when node $u$ is considered, i.e., $|I_{2}(u)| \le \Delta -2$. That is, node $u$’s transmission to layer $i$ is interfered by at most $\Delta -2$ nodes. $\square$

Lemma 2

Consider a parent $u$ of nodes in layer $L_i$ . Let time $tr(u,i)$ be its scheduled transmission time. Then, $tr(u,i) \le \ rec(u) + (\Delta -1)T$.

Proof

According to Lemma 1, $I_1(u) =\emptyset$ and $|I_2(U)| \le \Delta -2$. Recall that in Algorithm 2, $tr(u,i) = \text {min} {\left\{ t | t \,\text {mod}\,T =\tau _{i}, t> rec(u)\, \text {and} \,t \notin I_1(u) \cup I_2(u) \right\} }$, therefore $tr(u,i) \le \ rec(u) + (\Delta -2)T+T$, where $(\Delta -2)T$ accounts for the delay incurred by transmissions from $\Delta -2$ nodes to their children in layer $i$ in the worst case. $\square$

Lemma 3

Denote by $r_{i}$ the maximum reception time of nodes from layer 0 to $i$ . Then, for layer $i$ , where $i>0$ , we have $r_{i} \le (i-1)(\Delta -1)T$.

Proof

We prove this lemma by induction. For layer 1, it holds true because the reception time of nodes in layer 1 is 0. We now prove it is also true for layer $i$, where $i>1$. By Lemma 2, we get the maximum transmission time of nodes from layer 0 to $i-1$ is bounded by $r_{i-1} + (\Delta -1)T$. Since nodes in layer $i$ get the message from the upper layers, $r_{i}$ is bounded by the maximum transmission time of nodes from layer 0 to $i-1$, i.e., $r_{i} \le r_{i-1} + (\Delta -1)T \le (i-1)\Delta -1)T$. Therefore, it is also correct for layer $i$. $\square$

Theorem 2

BS-1 is an $(\Delta -1 )T$ -approximate solution for the MLBSDC problem.

Proof

By Lemma 3, the maximum reception time of nodes in $T_{b}$ is $r_{l}$, and its maximum value is $(l-1)(\Delta -1)T$. That is, BS-1 needs to take at most $(l-1)(\Delta -1)T +1$ time slots to finish the broadcast, where 1 accounts for time slot 0. Recall that $l \le H$, and therefore, in the worst case, the broadcast latency for BS-1 is $(l-1)(\Delta -1)T +1 \le (H-1)(\Delta -1)T+1 < H(\Delta -1)T$, which proves the theorem. $\square$

Theorem 3

The total number of transmissions scheduled by BS-1 do not exceed $|V|-1$.

Proof

Recall that the transmission of BS-1 is scheduled layer by layer in a top-down manner and each parent node is only allowed to transmit once to its children in a given layer. Therefore, for each layer $i$, it requires at most $|L_{i}|$ transmissions to inform all nodes in $L_{i}$. Hence, the number of transmissions performed by BS-1 is bounded by $\sum _{i=1}^{l} |L_{i}| = |V|-1$. $\square$

Theorem 4

The time complexity of BS-1 is $O(n^2)$.

Proof

We first show that the broadcast tree (Algorithm 1) can be determined in time $O(n^2)$ before proving the broadcast scheduler (Algorithm 3) of BS-1 has a running time of $O(n^2)$.

First, the SPT can be constructed in time $O(n^2)$ [6]. Then, BS-1 needs to iterate through $|E|$ edges to find the number of covered nodes for all nodes in the SPT. So it will take at most $O(|E|) \le O(n^2)$ time. To find the parent node with the most covered nodes in a given layer, it takes at most time $O(n)$. For $n$ nodes, this step needs at most time of $O(n^2)$. Overall, these steps finish in time $O(n^2)$.

The major step in the broadcast scheduler of BS-1 is the ordering for each layer (line 5 in Algorithm 3). This step takes at most time $O(n_i^2)$ for each layer $i$, where $n_i$ is the number of nodes in layer $i$. To sum up all layers, this step requires time $\sum O(n_i^{2}) \le O(n^2)$. Then, for BS-1, each node in each layer will iterate at most $n$ times to determine the minimum transmission time (Algorithm 2). For all layers, it takes at most time $\sum O(n_i\times n) \le O(n^2)$. Hence the total time complexity of BS-1 is $O(n^2)$. $\square$

Appendix 2

1.1 Correctness analysis of BS-2

Theorem 5

BS-2 yields a correct and collision-free broadcast schedule.

Proof

Recall that BS-2 has two phases. This means we only need to prove that the reception of all nodes in each phase is collision free. In each phase, BS-2 is conducted layer by layer with the same scheduling constraints as BS-1. Consequently, by the proof of Theorem 1, BS-2 also yields a collision free schedule. Furthermore, as mentioned above, all dominators and their parent nodes in BS-2 form a CDS. Therefore, all nodes can receive the broadcast message from this CDS collision-free. $\square$

1.2 Approximation ratio analysis of BS-2

Lemma 4

Consider a parent $u \in M$ of nodes in $L_i \cap X$ . Suppose that in Phase 1, node $u$ ’s transmission to nodes in layer $i$ is delayed because doing so will cause a collision at one or more of its one-hop neighbors, denoted by set $W$ , with transmissions from nodes in $M$ among its two-hop neighbors. Then the following is true.

• 1. Node $w$ is not in $C(u,i)$ , where $w \in W$.

• 2. For node $u$ , there are at most three nodes in $M$ among its two-hop neighbors that interfere with $u$.

Proof

Similar to Lemma 1, the order in which the parent of nodes in $U$ is chosen (line 13 in Algorithm 4) and the order in which the transmissions are scheduled (line $6$ in Algorithm 5) are the same. That is, if there is node $z \in M$ among $u$’s two-hop range that is scheduled to transmit to layer $i$ before node $u$, node $w$ should be in $C(z,i)$, not in $C(u,i)$. Hence, node $w$ must be not in $C(u,i)$.

According to the first property of Lemma 4, when node $u$ is scheduled to transmit to nodes in layer $i$, none of $u$’s children in $C(u,i)$ hear a message from other nodes in $M$ among $u$’s two-hop range, i.e., $I_{1}(u)=\emptyset$ as per Algorithm 2. Hence, node $u$’s transmission time is only delayed by the reception time of its one-hop neighbors that receive the message from nodes in $M$ among $u$’s two-hop range. Recall that in $T_{b}$, the children (respectively, parent nodes) of nodes in $M$ are selected from dominators, i.e., $U$, and each node can be adjacent to at most five dominators, see [9]. Hence, there are at most $5-1$ dominators among $u$’s one-hop range that are assigned with reception time when $u$ is considered, where 1 accounts for at least one child of node $u$. Except $u$’s parent node that is also a dominator, node $u$ is adjacent to at most three interfering dominators that have been assigned with reception time when node $u$ is considered, i.e., $|I_2(u)| \le 3$. In the worst case, three dominators have three different parent nodes in $M$, hence there are at most three nodes in $M$ among $u$’s two-hop neighbors that interfere with $u$. $\square$

Lemma 5

Consider a dominator $v \in U$ that is a parent of nodes in $L_i \cap X$ . Suppose that in Phase 1, node $v$ had to defer its transmission to nodes in layer $i$ which otherwise would cause a collision at one or more of its one-hop neighbors, denoted by set $W$ , with transmissions from dominators in $U$ . Then the following is true.

• 1. Node $w$ is not in $C(v,i)$ , where $w \in W$.

• 2. For node $v$ , there are at most eight dominators in $U$ among its two-hop neighbors that interfere with $v$.

Proof

Similar to Lemma 1 and 4, the order in which the parent nodes of nodes in $M$ are selected (line 19 in Algorithm 4) and the order in which the transmissions are scheduled (line $7$ in Algorithm 5) are the same. That is, if there is dominator $z \in U$ among $v$’s two-hop range that is scheduled to transmit to layer $i$ before node $v$, node $w$ should be in $C(z,i)$, not in $C(v,i)$.

None of $v$’s children in $C(v,i)$ hear a message from other dominators in $U$ among $v$’s two-hop range, when node $v$ is scheduled to transmit to nodes in layer $i$, based on the first property. Hence, dominators among $u$’s two-hop range only collide with $u$ at its one-hop neighbors that are not its children. Recall that in $T_{b}$, the children (respectively, parent nodes) of dominators in $U$ are selected from nodes in $M$, and only non-leaf nodes in $M$ are scheduled in Phase 1. Moreover, the number of dominators among $v$’s two-hop range does not exceed 19, see [2]. Let $z \in L_{j \le i}$ be a dominator among $v$’s two-hop range that is scheduled to transmit to its children $C(z,i)$ in Phase 1 before $v$. Since nodes in $C(z,i)$ belong to $M$, they must have at least one child that is also a dominator among $v$’s two-hop range in a lower layer than $i$, or else nodes in $C(z,i)$ are the leaf nodes in $T_{b}$, and $z$’s transmission will be scheduled in Phase 2, not in Phase 1. Dominators in lower layers than $i$ will not interfere with $v$, so for each $z$, there is at least one dominator among $v$’s two-hop range that does not interfere with $v$. Thus, at most half of dominators among $v$’s two-hop range are scheduled to transmit to layer $i$ before $v$, i.e., $\left\lfloor \frac{19}{2}\right\rfloor =9$. Excluding one dominator that is the parent node of $P(v)$ and also among $v$’s two-hop range, node $v$’s transmission to layer $i$ is interfered by at most eight dominators among its two-hop range. $\square$

Lemma 6

Denote by $r_{i}$ the maximum reception time of nodes from layer 0 to $i$ in Phase 1. Then, for layer $i$ , where $0< i < l$ , we have $r_{i} \le (i-1) 13T$.

Proof

We prove this lemma by induction. For layer 1, it holds true because the reception time of nodes in layer 1 is 0. We now prove it is also true for layer $i$, where $1< i < l$. For each layer $i$ in Phase 1, the transmissions by parent nodes of nodes in $U_{i}$ are scheduled before parent nodes of nodes in $M_{i} \cap X$. Let node $u$ be a parent node of nodes in $U_{i}$. Suppose that schedule the transmissions of node $u$ after $r_{i-1}$, this means, only transmissions from parent nodes to nodes in $U_{i}$ will interfere with $u$’s transmission, because all nodes from layer 0 to $i-1$ receive the message by $r_{i-1}$ and the transmitters of nodes in $M_{i}$ are scheduled after node $u$. By Lemma 4, we get node $u$’s children will not hear the message from other parent nodes of nodes in $U_{i}$, and there are at most three parent nodes of nodes in $U_{i}$ that are scheduled to transmit before $u$. Therefore, as per Algorithm 2, $|I_{u}| \le 3$ and the maximum transmission time of node $u$ is $r_{i-1}+4T$, i.e., $tr1(u, i) \le r_{i-1}+|I(u)|T +T$.

Since all parent nodes of nodes in $U_{i}$ transmit by $r_{i-1}+4T$, suppose a parent node $v \in U$ of nodes in $M_{i} \cap X$ is scheduled after time $r_{i-1}+4T$. It means, only transmissions from parent nodes of nodes in $M_{i} \cap X$ interfere with $v$’s transmission, because all nodes in $U_{i}$ and layer 0 to $i-1$ receive the message by $r_{i-1}+4T$. By Lemma 5, we get node $v$’s children will not hear the message from other parent nodes of nodes in $M_{i} \cap X$, and there are at most eight parent nodes of nodes in $M_{i} \cap X$ that are scheduled to transmit before $v$. Therefore, as per Algorithm 2, $|I_{v}| \le 8$ and the maximum transmission time of node $v$ is $r_{i-1}+13T$, i.e., $tr1(v, i) \le r_{i-1}+4T+ |I(v)|+T$. Therefore, all nodes in layer $i$ will receive the message by $r_{i-1}+13T$ in Phase 1, i.e., $r_{i} \le r_{i-1} + 13T \le (i-1)13T$. Hence, it is also correct for layer $i$. $\square$

Lemma 7

In Phase 1, all nodes receive the broadcast message by $(l-2)13T+4T$.

Proof

By Lemma 6, the maximum reception time for nodes from layer 0 to $l-1$ is $(l-2)13T$. For layer $l$, only nodes in $U_{l}$ are scheduled to receive from their parent nodes in Phase 1, and thus if their parent nodes are scheduled after $(l-2)13T$, their transmissions are delayed by at most $4T$ time for the same reason as in Lemma 6. Hence, in Phase 1, all nodes will receive the message by $(l-2)13T+4T$. $\square$

Lemma 8

Consider a node $u$ that is a member of $M_{i} \cap Y$ , where $0<i<l$ . Then, $rec(u) \le (l-2)13T+24T$ in Phase 2.

Proof

After time $(l-2)13T+4T$, node $u$’s transmission can only be corrupted by transmissions from dominators in Phase 2, because all nodes in $X$ receive the message by $(l-2)13T+4T$ and only dominators are allowed to transmit in Phase 2. Therefore in Phase 2, node $u$ must receive the message collision-free from its parent node $v\in U$ if node $v$ avoids transmitting the message at the time when other dominators among node $v$’s two hops’ range transmit to node $u$. Similar to Lemma 1, 4 and 5, the order in which the parent nodes are selected (line 19 in Algorithm 4) and the order in which the transmissions are scheduled (line 19 in Algorithm 5) are the same. Hence, $u$ will not hear a message from other nodes after time $(l-2)13T+4T$ when $u$ is scheduled. Recall that the size of dominators in a radius two circle does not exceed 19, and thus the size of dominators that interfere node $u$’s transmission at $u$’s one-hop neighbors is not over 19, i.e., $|I(u)| \le 19$ as per Algorithm 2. Hence $rec(u) \le (l-2)13T+4T+19T+T$. $\square$

Theorem 6

BS-2 provides a $13T$ -approximate solution for the latency.

Proof

By Lemma 8, we get all nodes receive the message by $(l-2)13T+24T$. That is, BS-2 takes at most $(l-2)13T+24T +1$ time slots to finish the broadcast. Recall that $l \le H$, and thus, in the worst case, the broadcast latency for BS-2 is $(l-2)13T+24T +1 \le (H-2)13T+24T +1 \le 13TH -2T+1 \le 13TH$. $\square$

Theorem 7

BS-2 is a $4(T+3)$ -approximate solution in terms of number of transmissions.

Proof

Recall that only dominators and non-leaf nodes in $M$ transmit the message in Phase 1. The number of dominators transmitting the message in Phase 1 does not exceed $|U|-1$. The value of 1 accounts for one dominator that is located in the last layer of Phase 1 and does not retransmit the message. Node $s$ does not have a parent node and each non-leaf node in $M$ must have at least one dominator as their child, thus the number of non-leaf nodes in $M$ transmitting the message in Phase 1 does not exceed $|U|-1$. Consequently, the total number of nodes transmitting in Phase 1 does not exceed $2(|U|-1)$.

For Phase 2, node $s$ does not need to transmit the message, so there are at most $(|U|-1)$ dominators transmitting the message. Recall that the parent nodes of nodes in $M_{i}$ are chosen from dominators in $U_{j \le i}$. That is, a given dominator $u \in U_{i}$ that can be a parent of node $v$ in $M_{j \ge i}$. Owing to $Lat(u,v) \le T$, and dominator $u$ only needs to transmit once to its children in the same layer, we get for any dominator $u$, the number of transmissions to children nodes is at most $T+1$ times. Hence, the total number of transmissions in Phase 2 do not exceed $(T+1)(|U|-1)$. Therefore, the total number of transmissions scheduled by BS-2 do not exceed $(T+3)(|U|-1)$, i.e., $2(|U|-1)+(T+1)(|U|-1)$. Recall that the size of $U$ does not exceed $4opt+1$ [36], where $opt$ denotes the minimum number of transmissions, BS-2 is thus a $(T+3)(4opt+1-1)=4(T+3)opt$ solution. $\square$

Theorem 8

The time complexity of BS-2 is $O(n^2)$.

Proof

We first show the broadcast tree (Algorithm 4) can be determined in time $O(n^2)$, then prove the broadcast scheduler (Algorithm 5) of BS-2 finishes in time $O(n^2)$.

First, the SPT can be constructed in time $O(n^2)$ [6]. Then, BS-2 needs to iterate through $|E|$ edges to find MIS. So it will take at most $O(|E|) \le O(n^2)$ time. Similar to Theorem 4, to determine the parent–children relationship, BS-2 needs at most time $O(n^2)$. In total, these steps are bounded by $O(n^2)$.

In Phase 1 of Algorithm 5, for the ordering step in each layer (line 6 and 7 in Algorithm 5), it takes at most time $O(n_i^2)$ for each layer $i$, where $n_i$ is the number of nodes in layer $i$. To sum up all layers, this step requires time $\sum O(n_i^{2}) \le O(n^2)$. Then, for BS-2, each node in each layer will iterate at most $n$ times to determine the minimum transmission time (Algorithm 2). For all layers, it takes at most time $\sum O(n_i\times n) \le O(n^2)$. Similar to Phase 1, in Phase 2, it also takes at most time $O(n^2)$. Hence the total time complexity of BS-2 is $O(n^2)$. $\square$