On energy efficiency of geographic opportunistic routing in lossy multihop wireless networks

Abstract

Geographic opportunistic routing (GOR) is an emerging technique that can improve energy efficiency in lossy multihop wireless networks. GOR makes local routing decision by using nodes’ location information, and exploits the broadcast nature and spatial diversity of the wireless medium to improve the packet forwarding reliability. In this paper, our goal is to fully understand the principles and tradeoffs in GOR, thus provide insightful analysis and guidance to the design of more efficient routing protocols in multihop wireless networks. We propose a local metric, one-hop energy efficiency (OEE), to balance the packet advancement, reliability and energy consumption in GOR. We identify and prove important properties about GOR on selecting and prioritizing the forwarding candidates in order to maximize the expected packet advancement. Leveraging the proved properties, we then propose two localized candidate selection algorithms with O(N ³) running time to determine the forwarding candidate set that maximizes OEE, where N is the number of available next-hop neighbors. Through extensive simulations, we show that GOR applying OEE achieves better energy efficiency than the existing geographic routing and blind opportunistic routing schemes under different node densities and packet sizes.

Appendix 1

1.1 Proof of Lemma 5.3

Proof

We begin by proving the following useful result. Let $\mathcal{A}$ be an ordered node set $\langle a_1,a_2, \ldots ,a_M\rangle$ ³ with M ($1\leq M<\infty$) nodes, in which the nodes are ordered as $d_{{_{{a_{1} }} }} > d_{{a_{2} }} > \cdots > d_{{a_{M} }}$. Let $\mathcal{B}$, $\langle b_1,b_2,\ldots,b_N\rangle$ with N (0 ≤ N < M) nodes, be a subset of $\mathcal{A}$ such that $d_{{_{{b_{1} }} }} > d_{{b_{2} }} > \cdots > d_{{b_{N} }}$ and b _N  = a _M . For any node $q\notin\mathcal{A}$ with $d_{q} \le d_{{a_{M} }}$, we have the following inequality

$$ \begin{aligned} \Updelta A&:={\rm EPA}(\langle{\mathcal A},q\rangle)-{\rm EPA}(\langle q,{\mathcal A}\rangle)\\ \geq\Updelta B&:={\rm EPA}(\langle{\mathcal B},q\rangle)-{\rm EPA}(\langle q,{\mathcal B}\rangle) \end{aligned} $$

(32)

The equality holds only when $d_{{a_{1} }} = d_{{a_{2} }} = \cdots = d_{{a_{M} }} = d_{q}$.

As $\mathcal{B}\subset \mathcal{A}$, we can represent $\mathcal{A}$ as

$$ {\mathcal A} = \langle{\mathcal A}_{1},b_{1},{\mathcal A}_{2},b_{2},\ldots,{\mathcal A}_{N},b_{N}\rangle $$

(33)

where $\mathcal{A}_{i}$ (1 ≤ i ≤ N) is an ordered node set and can be ∅. Assume $|\mathcal{A}_{i}|=L_{i}$. Denote the kth element in $\mathcal{A}_{i}$ as $\mathcal{A}_{{i_{k} }}$. According to the definition of EPA of an ordered node set in Eq. (3), we have

$$ {\rm EPA}({\mathcal A}_{i}) = \sum_{k=1}^{L_i}d_{A_{i_{k}}}p_{A_{i_{k}}}\cdot \prod_{n=0}^{k-1}(1-p_{A_{i_{n}}}) $$

(34)

where $p_{{A_{{i_{0} }} }} : = 0$.

For convenience, we handle the ordered set $\mathcal{A}_{i}$ as a new node by giving the following definition.

$$ p_{{\mathcal A}_{i}} = 1 - \prod_{n=0}^{L_i}(1-p_{A_{i_{n}}}) \quad {\rm and}\quad \,d_{{\mathcal A}_{i}} = \frac{{\rm EPA}({\mathcal A}_{i})}{p_{{\mathcal A}_{i}}} $$

(35)

If $\mathcal{A}_{i}=\emptyset$, we can choose a positive real number for ${d_{\mathcal{A}_{i}}}$, s.t. ${d_{b_{i-1}}\geq d_{\mathcal{A}_{i}}\geq d_{b_{i}}}$, and ${p_{\mathcal{A}_{i}}=0}$. $d_{{b_{0} }}$ is chosen as a positive real number that makes the inequality ${d_{b_{0}}\geq d_{\mathcal{A}_{1}} \geq d_{b_{1}}}$ hold.

From Eq. (34) and (35), we can easily derive that

$$d_{A_{i_{1}}} \geq d_{{\mathcal A}_{i}} \geq d_{A_{i_{L_i}}},\;{\hbox{then}}\; d_{b_{i-1}}\geq d_{{\mathcal A}_{i}}\geq d_{b_{i}} \forall 1\leq i\leq N $$

(36)

Now EPA of $\mathcal{A}$ is represented as

$$ {\rm EPA}({\mathcal A}) = \sum_{k=1}^{N}[d_{{\mathcal A}_{k}}p_{{\mathcal A}_{k}} + d_{b_{k}}p_{b_{k}}(1-p_{{\mathcal A}_{k}})] \cdot\prod_{n=0}^{k-1}(1-p_{{\mathcal A}_{n}})(1-p_{b_{n}}) $$

(37)

where ${p_{\mathcal{A}_{0}}:=0}$ and $p_{{b_{0} }} : = 0$.

Define ${d_{\mathcal{A}_{N+1}}:=d_q}$ and ${p_{\mathcal{A}_{N+1}}:=p_q}$, then

$$ \Updelta A = \sum_{k=1}^{N}[p_{{\mathcal A}_{k}}(d_{{\mathcal A}_{k}}-d_{{\mathcal A}_{N+1}}) + (1-p_{{\mathcal A}_{k}})p_{b_{k}}(d_{b_{k}}-d_{{\mathcal A}_{N+1}})]\\ \cdot p_{{\mathcal A}_{N+1}}\prod_{n=0}^{k-1}(1-p_{{\mathcal A}_{n}})(1-p_{b_{n}}) $$

(38)

$$ \Updelta B =\sum_{k=1}^{N}p_{b_{k}}p_{{\mathcal A}_{N+1}}(d_{b_{k}}-d_{{\mathcal A}_{N+1}})\prod_{n=0}^{k-1}(1-p_{b_{n}}) $$

(39)

So

$$ \begin{aligned} \Updelta A - \Updelta B &= \sum\limits_{{k = 1}}^{N} {[(d_{{\mathcal{A}_{k} }} - d_{{b_{k} }} ) + (1 - p_{{b_{k} }} )(d_{{b_{k} }} - d_{{\mathcal{A}_{{k + 1}} }} )]} \\ \quad \cdot \left\{ {p_{{\mathcal{A}_{{N + 1}} }} \left[ {1 - \prod\limits_{{n = 0}}^{k} {(1 - p_{{\mathcal{A}_{n} }} )} } \right]\left[ {\prod\limits_{{n = 0}}^{{k - 1}} {(1 - p_{{b_{n} }} )} } \right]} \right\} \ge 0 \\ \end{aligned} $$

(40)

As ${\mathcal{A}\supset\mathcal{B}, \exists p_{\mathcal{A}_{n}}\neq 0 (1\leq n\leq N)}$. We also have ${d_{\mathcal{A}_{k}}\geq d_{b_{k}}}$ and ${d_{b_{k}}\geq d_{\mathcal{A}_{k+1}}}$. So inequality (40) holds, that is, inequality (32) holds.

Now we are ready to prove the containing property of $\mathcal{F}^{*}_r$’s. Recall that for arbitrary finite node number, $N=|\mathcal{C}|\geq 1$, we want to prove $\forall \mathcal{F}^{*}_{r-1}, \exists$ an $\mathcal{F}^{*}_{r}$, s.t. $\mathcal{F}^{*}_{r-1}\subset\mathcal{F}^{*}_{r}, \forall 1\leq r\leq N$. For each r, denote one feasible $\mathcal{F}^{*}_{r}=\langle r_{1},r_{2},\ldots,r_{r}\rangle$ and $d_{{r_{1} }} \ge d_{{r_{2} }} \ge \cdots \ge d_{{r_{r} }}$. We prove this Lemma by induction.

First, for arbitrary N, when r = 1, as $\mathcal{F}^{*}_{0}=\emptyset$, and $\mathcal{F}^{*}_{1}\neq\emptyset$, it is obvious that the containing property holds.

Second, when r = 2, implicating N ≥ 2, we want to prove $\forall \mathcal{F}^{*}_{1}, \exists \mathcal{F}^{*}_{2}$, s.t. $\mathcal{F}^{*}_{1}\subset\mathcal{F}^{*}_{2}$. Assume one feasible $\mathcal{F}^{*}_{1}=\langle 1_{1}\rangle$. We know node 1₁ achieves the maximum EPA among the N nodes. If we assume for any feasible $\mathcal{F}^{*}_{2}$, $\mathcal{F}^{*}_{1}\nsubseteq\mathcal{F}^{*}_{2}$, that is $1_{1}\notin\mathcal{F}^{*}_{2}$, then we can always replace node 2₂ as 1₁ to achieve at least an equal or larger EPA than EPA($\mathcal{F}^{*}_{2}$). Then the new node set containing node 1₁ should be one feasible node set achieving maximum EPA by selecting 2 nodes from the N nodes, thus the assumption is wrong, then $\exists \mathcal{F}^{*}_{2}$, s.t. $\mathcal{F}^{*}_{1}\subset\mathcal{F}^{*}_{2}$. That is, the containing property holds when r = 2.

Then, we assume $\forall \mathcal{F}^{*}_{m-1}, \exists$ an $\mathcal{F}^{*}_{m}$, s.t. $\mathcal{F}^{*}_{m-1}\subset\mathcal{F}^{*}_{m}$, when r = m (m ≥ 2), and we want to prove $\forall \mathcal{F}^{*}_{m}, \exists$ an $\mathcal{F}^{*}_{m+1}$, s.t. $\mathcal{F}^{*}_{m}\subset\mathcal{F}^{*}_{m+1}$, when r = m + 1, implicating N ≥ m + 1.

We will prove under the following two cases, the containing property holds.

Case I: $(m+1)_{1}=m_{i}, i\in \{2, 3, \ldots , m\}$.

Let node (m + 1) _j (2 ≤ j ≤ m + 1) be the first node in $\mathcal{F}^{*}_{m+1}$ but not in $\mathcal{F}^{*}_{m}$. We can always find such a node because $\langle (m+1)_{2},\ldots,(m+1)_{m+1}\rangle$ contains more nodes than $\langle m_{i},\ldots,m_{m}\rangle$. We know that EPA($\mathcal{F}^{*}_{m}$) is the largest EPA achieved by selecting m nodes, then

$$ {\rm EPA}({\mathcal F}^{*}_{m}) \geq {\rm EPA}({\mathcal F}^{*}_{m+1} \setminus \{(m+1)_{j}\}) $$

(41)

where $\mathcal{F}^{*}_{m+1}\setminus\{(m+1)_{j}\}$ is the ordered subset obtained by excluding (m + 1) _j from $\mathcal{F}^{*}_{m+1}$.

From the definition of EPA in Eq. (3), we have

$$ \begin{aligned} &{\rm EPA}(\langle (m+1)_{j},{\mathcal F}^{*}_{m}\rangle)\\ &\geq {\rm EPA}(\langle (m+1)_{j},{\mathcal F}^{*}_{m+1} \setminus \{(m+1)_{j}\}\rangle) \end{aligned} $$

(42)

Assume (m + 1) _j−1 = m _l (i ≤ l ≤ m), we have

$$ \begin{aligned} &\Updelta 1:={\rm EPA}(\langle m_{1},\ldots,m_{l},(m+1)_{j},m_{l+1},\ldots,m_m\rangle)\\ &\quad -{\rm EPA}(\langle (m+1)_{j},{\mathcal F}^{*}_{m}\rangle)\\ &={\rm EPA}(\langle m_{1},\ldots,m_{l},(m+1)_{j}\rangle)\\ &\quad -{\rm EPA}(\langle (m+1)_{j},m_{1},\ldots,m_{l}\rangle) \end{aligned} $$

(43)

$$ \begin{aligned} \Updelta 2:={\rm EPA}({\mathcal F}^{*}_{m+1})\\ -{\rm EPA}(\langle (m+1)_{j},{\mathcal F}^{*}_{m+1}\setminus \{(m+1)_{j}\}\rangle)\\ ={\rm EPA}(\langle(m+1)_{1},\ldots,(m+1)_{j-1},(m+1)_{j}\rangle)\\ -{\rm EPA}(\langle (m+1)_{j},(m+1)_{1},\ldots,(m+1)_{j-1}\rangle) \end{aligned} $$

(44)

According to the definition of (m + 1) _j and the assumption that (m + 1)₁ = m _i , we have $\langle (m+1)_{1},\ldots,(m+1)_{j-1}\rangle\subset\langle m_{1},\ldots,m_{l}\rangle$. Then according to inequality (32), we have $\Updelta 1 \geq\Updelta 2$, and combining with inequality (42), we get

$$ {\rm EPA}(\langle m_{1},\ldots,m_{l},(m+1)_{j},m_{l+1},\ldots,m_m\rangle)\geq{\rm EPA}({\mathcal F}^{*}_{m+1}) $$

(45)

According to the definition of $\mathcal{F}^{*}_{m+1}$, we have

$$ {\rm EPA}(\langle m_{1},\ldots,m_{l},(m+1)_{j},m_{l+1},\ldots,m_m\rangle)\leq{\rm EPA}({\mathcal F}^{*}_{m+1}) $$

(46)

From inequalities (45) and (46), we have

$$ {\rm EPA}(\langle m_{1},\ldots,m_{l},(m+1)_{j},m_{l+1},\ldots,m_m\rangle)={\rm EPA}({\mathcal F}^{*}_{m+1}) $$

(47)

We find the set $\langle m_{1},\ldots,m_{l},(m+1)_{j},m_{l+1},\ldots,m_m\rangle$ achieving the maximum EPA by using m + 1 nodes, and this set contains $\mathcal{F}^{*}_{m}=\langle m_{1},\ldots,m_{l},m_{l+1},\ldots,m_m\rangle$.

In the above proof, we implicitly assume N ≥ m + i. However, it’s easy to get that node (m + 1)₁ can not be m _i when N < m + i. Because according to the Relay priority rule in Proposition 5.2, when we assume (m + 1)₁ = m _i , we implicitly exclude nodes m ₁,…, m _i−1 as the forwarding candidates. The number of nodes that can be chosen is N − (i − 1) < (m + i) − (i − 1) = m + 1, then we do not have enough nodes to choose for $\mathcal{F}^{*}_{m+1}$.

Case II: (m + 1)₁ ≠ m _i , ∀ 2 ≤ i ≤ m

In this case, we have two sub-cases:

• 1) (m + 1)₁ ≠ m ₁

Then $(m+1)_{1}\notin \mathcal{F}^{*}_{m}$. Assume for any feasible $\mathcal{F}^{*}_{m}, \langle (m+1)_2,\ldots,(m+1)_{m+1}\rangle\neq\mathcal{F}^{*}_{m}$, we can replace $\langle (m+1)_2,\ldots,(m+1)_{m+1}\rangle$ in $\mathcal{F}^{*}_{m+1}$ by $\mathcal{F}^{*}_{m}$ to get a sequence $\langle (m+1)_1,\mathcal{F}^{*}_{m}\rangle$ which achieves at least an equal or larger EPA than EPA($\mathcal{F}^{*}_{m+1}$). So $\exists \mathcal{F}^{*}_{m+1}$, s.t. $\mathcal{F}^{*}_{m}\subset \mathcal{F}^{*}_{m+1}$.

• 2) (m + 1)₁ = m ₁

To find $\langle (m+1)_2,\ldots,(m+1)_{m+1}\rangle$ is equivalent to selecting m nodes achieving maximum EPA from the remained node set $\mathcal{C}_{m_{1}}\setminus\{m_{1}\}$ in which all the nodes have advancements larger than $d_{{m_{1} }}$. As $\mathcal{F}^{*}_{m}\setminus\{m_{1}\}$ is the node set that achieves the maximum EPA by selecting m − 1 nodes in $\mathcal{C}_{m_{1}}\setminus\{m_{1}\}$, by the induction assumption, we have $\mathcal{F}^{*}_{m}\setminus\{m_{1}\}\subset\langle (m+1)_2,\ldots,(m+1)_{m+1}\rangle$, then $\mathcal{F}^{*}_{m}\subset\mathcal{F}^{*}_{m+1}$.

From the induction above, we know for arbitrary finite node number, N ≥ 1, we have ∀ $\mathcal{F}^{*}_{r-1}, \exists \mathcal{F}^{*}_{r}$ s.t. $\mathcal{F}^{*}_{r-1}\subset\mathcal{F}^{*}_{r}, \forall 1\leq r\leq N$.□