Interference-aware lifetime maximization with joint routing and charging in wireless sensor networks

Abstract

Radio Frequency based Wireless power transfer (RF-WPT) has become increasingly popular in recent years. Its ability to harvest Radio Frequency (RF) energy enables a novel approach to charge low-power wireless devices, resulting in benefits to product design, usability, and reliability in wireless sensor networks (WSN). It is however known that RF-WPT also introduces interference to wireless communications, leading to poor data throughput. The joint routing and charging remains a challenging job in RF energy harvesting networks. In this paper, we take initial steps towards understanding both data routing and charger scheduling in WSNs. We propose a smart interference-aware scheduling to maximize network lifetime and avoid potential data loss caused by charging interference. Then, we theoretically prove the optimality of the proposed design, i.e., $1 - \frac{\phi }{W}$, where W is an arbitrary positive integer and $\phi$ is determined by network properties. The evaluation indicates that the proposed design can guarantee 99% optimality and significantly improve network lifetime in WSNs.

Appendices

Prove of Theorem 1

Proof

Suppose P is the optimal solution of problem (OR-C), which consists of the maximum network lifetime T, the optimal charger’s travel path, sojourn/travel durations $U(x_l)$ and $U(y_l)$, data flow functions $g_{ij}(t)$, $g_{i0}(t)$, $g^s_i(t)$ and $g^r_i(t)$. Based on solution P, we can construct a solution $P^*$ with lifetime $T^*$ as follows. First, we keep the charger’s travel path and sojourn/travel durations the same as solution P. Then we show that both of these solutions achieve the same network lifetime:

$$\begin{aligned} T^* = \sum _{l\in L} [U(x_l) + U(\nu _l)] = \sum _{l\in L} [U(x_l) + U(y_l)] = T \end{aligned}$$

Next, we construct data flow functions of solution $P^*$ as described in Sect. 3.2. And we need to show that $P^*$ is a feasible solution of problem (OR-D). Specifically, we need to prove that solution $P^*$ meets constraints Eqs. (5), (9)–(15). Here, we focus on constraint Eq. (13) while others are similar and thus omitted to conserve space. The proofs are based on 3 different cases described in Sect. 3.2.

Considering Case 1, for sensor i, we have:

$$\begin{aligned}&\sum _{k\in N}^{k\ne i}f_{ki}(\nu _l) + g_i \\&\quad = \sum _{k\in N}^{k\ne i} \frac{\int _{t_l+U(x_l)}^{t_{l+1}}g_{ki}(t)dt}{U(\nu _l)} + \frac{\int _{t_l+U(x_l)}^{t_{l+1}}g_i dt}{U(\nu _l)}\\&\quad = \frac{\int _{t_l+U(x_l)}^{t_{l+1}}[ \sum _{k\in N}^{k\ne i} g_{ki}(t) + g_i]dt}{U(\nu _l)} \\&\quad = \frac{\int _{t_l+U(x_l)}^{t_{l+1}}[ \sum _{j\in N}^{j\ne i} g_{ij}(t) + g_{i0}(t) - g_i^r(t) ]dt}{U(\nu _l)} \\&\quad = \sum _{j\in N}^{j\ne i}f_{ij}(\nu _l)+f_{i0}(\nu _l)-f_i^r(\nu _l) \end{aligned}$$

Cases 2 and 3 can be proved in the same way. Since all the constraints of problem (OR-D) are satisfied by solution $P^*$, $P^*$ is a feasible solution to problem (OR-D).

Finally, we need to show that $P^*$ is the optimal solution of problem (OR-D). Suppose $\overline{P}$ is the optimal solution of problem (OR-D) with the maximum network lifetime $\overline{T} > T$. Based on $\overline{P}$, we can construct a solution of problem (OR-C) with lifetime $\overline{T}$, which contradicts with the fact that T is the maximum network lifetime.

Since we have proved that $T^* = T$, the optimal solution of problem (OR-D) can achieve the same maximum network lifetime as problem (OR-C), which concludes the proof. $\square$

Proof of Theorem 2

The proof of Theorem 2 is based on the following lemmas.

Lemma 1

For a given feasible charger travel path and a sojourn point$x_l$, we have$U(\nu _l) \ge \lambda _l U(x_l)$.

Proof

Since both $g^l_{max}$ and $g_{max}$ are positive parameters, $\lambda _l > 0$ holds. If $U(x_l) = 0$, $U(\nu _l) \ge \lambda _l U(x_l) = 0$ holds. Here, we emphasize on proving the $U(x_l) > 0$ case. Based on Eqs. (12), (14) and (15), we can derive:

$$\begin{aligned} g_i U(x_l) = f_i^s(x_l) U(x_l) = f_i^r(\nu _l) U(\nu _l) \le g_{max} U(\nu _l) \end{aligned}$$

(22)

Hence, $g_{max} U(\nu _l) \ge g_i U(x_l)$ holds for all sensor $i \in N_l$. It holds for the sensor with the maximum data generation rate in $N_l$, thus $g_{max} U(\nu _l) \ge g^l_{max} U(x_l)$, which concludes the proof. Note that this lemma also holds for the time dependent continuous formulation, we omit the proof to conserve space. $\square$

Lemma 2

Considering a data routing with full sensor participation, denote$\pi _i$the minimum energy consumption rate required to forward aunit of data from sensorito the sink, and the correspondingrouting path is$F_i$. Suppose each sensor has a unit of datageneration rate, then, the combination of$\sum _{i\in N}F_i$is the minimum energy routing with the minimum total energy consumptionrate$\sum _{i\in N}\pi _i$.

Proof

It is apparent that $\sum _{i\in N}F_i$ is a routing scheme of the sensor network. Next, we prove $\sum _{i\in N}F_i$ is the minimum energy routing using contradictions. Suppose $M^{*}$ is a routing scheme that consumes less energy than $\sum _{i\in N}\pi _i$. Thus at least one routing path $F^{*}_i$ in $M^{*}$ consumes less energy than $\pi _i$, which contradicts with the fact that $F_i$ is the minimum energy routing path. $\square$

Now, we begin to prove Theorem 2.

Proof

During $[t_l+U(x_l),t_{l+1}]$, the sensory data generated by sensor i is consisted of two parts: $g^s_i(x_l)U(x_l)$ and $g_i U(\nu _l)$. The former is generated and stored during $[t_l,t_l+U(x_l)]$, and the latter is newly generated during $[t_l+U(x_l),t_{l+1}]$. Suppose the minimum energy routing is adopted during $[t_l,t_{l+1}]$, based on Lemma 2, the total energy consumption is:

$$\begin{aligned}&\sum _{i\in N}\pi _i [g^s_i(x_l) U(x_l) + g_i U(\nu _l)] \\&\quad = \sum _{i\in N}\pi _i g^s_i(x_l) U(x_l) + \sum _{i\in N}\pi _i g_i U(\nu _l) \\&\quad = \sum _{i\in N_l}\pi _i g_i U(x_l) + \sum _{i\in N}\pi _i g_i U(\nu _l) \\&\quad \ge \sum _{i\in N_l}\pi _i g_i U(x_l) + \sum _{i\in N}\pi _i g_i \lambda _l U(x_l) \end{aligned}$$

When $U(\nu _l) = \lambda _l U(x_l)$, the total energy consumption during $[t_l+U(x_l),t_{l+1}]$ is minimized, which concludes the proof. $\square$

Proof of Proposition 1

Proof

The proofs of Relations 2 and 3 are apparent. After we solve problem (MIN-E), $e_{il} = \varepsilon _{il} U(x_l) + \mu _{il} U(\nu _l)$ can be easily obtained. Here, we focus on Relation 1. We assume that the minimum energy routing during $[t_l, t_{l+1}]$ is unique (the proof of non-unique situation is similar). To prove the proposition, we only need to prove that the total energy consumption $\sum _{i\in N}e_{il}$ is minimal.

Based on Lemma 2, we can derive $\sum _{i\in N} \eta _i = \sum _{i\in N}\pi _i g_i$ and

$$\begin{aligned} \sum _{i\in N} \mu _{il} \lambda _l U(x_l) = \sum _{i\in N_l}\pi _i g_i U(x_l) + \sum _{i\in N}\pi _i g_i \lambda _l U(x_l) \end{aligned}$$

(23)

Then, we have:

$$\begin{aligned} \begin{aligned}&\sum _{i\in N}e_{il} \\&\quad = \sum _{i\in N}\varepsilon _{il} U(x_l) + \sum _{i\in N} \mu _{il} \lambda _l U(x_l) + \sum _{i\in N}[U(\nu _l) - \lambda _l U(x_l)] \eta _i \\&\quad = \sum _{i\in N}\varepsilon _{il} U(x_l) + \sum _{i\in N_l}\pi _i g_i U(x_l) + \sum _{i\in N}\pi _i g_i \lambda _l U(x_l) \\&\qquad + \sum _{i\in N}[U(\nu _l) - \lambda _l U(x_l)] \pi _i g_i \\&\quad = \sum _{i\in N}\varepsilon _{il} U(x_l) + \sum _{i\in N_l}\pi _i g_i U(x_l) + \sum _{i\in N}\pi _i g_i U(\nu _l) \end{aligned} \end{aligned}$$

(24)

where $\sum _{i\in N}\varepsilon _{il} U(x_l)$ is the minimum energy consumption during $[t_l,t_l+U(x_l)]$ and the successive two polynomials together represent the minimum energy routing during $[t_l+U(x_l),t_{l+1}]$, which concludes the proof. $\square$

Proof of theorem 3

The proof is based on the following lemmas.

Lemma 3

A feasible solution of problem (LP-T) is also a feasiblesolution of problem (RLX).

Proof

Suppose P is a feasible solution of problem (LP-T) that consists of $U(x_l)$ and $U(\nu _l)$. Since energy consumption results $\eta _i$, $\varepsilon _{il}$ and $\mu _{il}$ are all obtained by solving problems (MIN-B) and (MIN-E), and data routing constraints Eqs. (11)–(15) are naturally satisfied by P. Let the charger travel a single TSP path (with each sensor visited once) and constraint Eq. (18) is equivalent to constraint Eq. (16). Therefore, constraints Eqs. (10)–(16) are all satisfied by solution P. Thus P is a feasible solution to problem (RLX), which concludes the proof. $\square$

Lemma 4

In terms of problem (RLX), as long as the total sojourndurations$\sum _{l\in N}U(x_l)$and$\sum _{l\in N}U(\nu _l)$at each sensor’s location remains the same, the network lifetime will remain unchanged regardless of the charger’s travel path.

Proof

Since the energy constraint Eq. (9) is relaxed to Eq. (16) in problem (RLX), the above lemma can be easily proved by analyzing sensor’s energy profiles at each location, which only relates to the duration spent at sojourn/virtual points. We omit the proof here to conserve space. $\square$

Lemma 5

Suppose$P^*$is an optimal solution of problem (RLX), which consists of$\varepsilon ^*_{il}$, $U^*(x_l)$, $\mu ^*_{il}$, $U^*(\nu _l)$, $\eta ^*_i$, $H^*_i$, $\tau ^*_i$, $T^*_0$and the maximum network lifetime

$$\begin{aligned} T^* = \sum _{l\in N} [U^*(x_l) + U^*(\nu _l)]. \end{aligned}$$

We can always construct a solution of problem (LP-T) with the network lifetime$T^{'} \ge T^*$.

Proof

Based on Lemma 4, we can regulate the charger’s travel path to a single TSP path ($L = N$), and then problem (RLX) can be solved by CPLEX. The resulted solution is denoted by $P^*$.

Next, we construct a solution $\overline{P}$ as follows: we keep the charger’s travel path, sojourn and travel durations unchanged while data routing is altered to the minimum energy routing. Specifically, $\overline{P}$ consists of $\varepsilon _{il}$, $U^*(x_l)$, $\mu _{il}$, $U^*(\nu _l)$, $\eta _i$, $H_i$, $\tau _i$, $T_0$ and $\overline{T}$ (note that $\overline{T} = T^*$). Suppose the total energy consumption of solution $P^*$ and $\overline{P}$ during the operational interval are $\omega ^*$ and $\omega$, respectively. To prove Lemma 5, we need to prove: (i) $\omega ^* \ge \omega$; (ii) $T^*_0 \ge T_0$; (iii) For $P^*$, Eq. (10) reaches equality.

• (i) Based on the definition of $\omega ^*$ and $\omega$, we have:

  $$\begin{aligned} \omega ^* = \sum _{i\in N}\sum _{l\in N} e^*_{il} \quad \hbox {and} \quad \omega = \sum _{i\in N}\sum _{l\in N} e_{il} \end{aligned}$$

  where

  $$\begin{aligned} e^*_{il} = (\varepsilon ^*_{il} + \mu ^*_{il} \lambda _l) U^*(x_l) + [U^*(\nu _l) - \lambda _l U^*(x_l)] \eta ^*_i \end{aligned}$$

  and

  $$\begin{aligned} e_{il} = (\varepsilon _{il} + \mu _{il} \lambda _l) U^*(x_l) + [U^*(\nu _l) - \lambda _l U^*(x_l)] \eta _i \end{aligned}$$

  Solution $\overline{P}$ adopts the minimum energy routing during the whole network lifetime. Therefore, for any $l\in L$, we have $\sum _{i\in N}\varepsilon _{il} \le \sum _{i\in N}\varepsilon ^*_{il}$, $\sum _{i\in N}\mu _{il} \le \sum _{i\in N}\mu ^*_{il}$ and $\sum _{i\in N}\eta _i \le \sum _{i\in N}\eta ^*_i$. Thus, we have

  $$\begin{aligned} \omega ^* - \omega&= \sum _{l\in N} \bigg\{ \bigg(\sum _{i\in N}\varepsilon ^*_{il} - \sum _{i\in N}\varepsilon _{il}\bigg) U^*(x_l) \\&+ \bigg(\sum _{i\in N}\mu ^*_{il} - \sum _{i\in N}\mu _{il}\bigg)\lambda _l U^*(x_l) \\&+ \bigg(\sum _{i\in N}\eta ^*_i - \sum _{i\in N}\eta _i\bigg) [U^*(\nu _l) - \lambda _l U^*(x_l)] \} \ge 0 \end{aligned}$$

• (ii) Based on energy conservation, we have:

  $$\begin{aligned} \omega ^* + N e_0 T^*_0 = N h_0 + \varpi _0\sum _{i\in N}\tau ^*_i + \varpi \sum _{i\in N}U^*(x_i) \end{aligned}$$

  (25)

  and

  $$\begin{aligned} \omega + N e_0 T_0 = N h_0 + \varpi _0\sum _{i\in N}\tau _i + \varpi \sum _{i\in N}U^*(x_i) \end{aligned}$$

  (26)

  Let Eq. (25) minus Eq. (26), then we can obtain

  $$\begin{aligned} (\varpi _0 - N e_0) \bigg(\sum _{i\in N}\tau ^*_i - \sum _{i\in N}\tau _i\bigg) = \omega ^* - \omega \ge 0 \end{aligned}$$

  The charging rate during the initial time $\varpi _0$ is larger than the total energy consumption rate, i.e., $\varpi _0 - N e_0 > 0$ holds (otherwise, sensor batteries may deplete before the beginning of the operational interval $T_0$). Thus we obtain $\sum _{i\in N}\tau ^*_i \ge \sum _{i\in N}\tau _i$. Based on $T_0 = t_{TL} + \sum _{i\in N} \tau _i$, we can derive $T^*_0 \ge T_0$.

• (iii) This can be explained intuitively. Suppose the equality is not reached, which means that part of the total energy E is unallocated. Then, we can always find a method to reallocate the unallocated energy and obtain a larger T, which contradicts the fact that $T^*$ is the maximum network lifetime. Thus we have:

  $$\begin{aligned} N h_0 + \varpi _0\sum _{i\in N}\tau ^*_i + \varpi \sum _{l\in N}U^*(x_l) = E \end{aligned}$$

The above equation is based on the fact that we are in full charge of energy allocation of the sensor network. To prove Lemma 5, we first need to prove that $\overline{P}$ is a feasible solution of problem (LP-T). Since $H_i$ is an intermediate parameter that can be removed by reformulation and $U^*(\nu _l) \ge \lambda _l U^*(x_l)$ holds, we only need to prove that Eq. (10) is satisfied by $\overline{P}$. Based on (i) and (ii), we can derive:

$$\begin{aligned} \begin{aligned}&N h_0 + \varpi _0\sum _{i\in N}\tau _i + \varpi \sum _{l\in N}U^*(x_l) = \omega + N e_0 T_0 \\&\quad \le \omega ^* + N e_0 T^*_0 = N h_0 + \varpi _0\sum _{i\in N}\tau ^*_i + \varpi \sum _{l\in N}U^*(x_l) = E \end{aligned} \end{aligned}$$

Thus $\overline{P}$ is feasible to problem (LP-T). The above equation also shows that a part of energy is unallocated in solution $\overline{P}$. Based on (iii), we can always construct a solution of problem (LP-T) with longer network lifetime $T^{'} \ge \overline{T} = T^*$, which concludes the proof. $\square$

Finally, we prove Theorem 3.

Proof

Suppose $P^*$ is an optimal solution of problem (RLX) with the maximum network lifetime $T^*$. Based on Lemma 5, we can always construct a solution of problem (LP-T), e.g., $\overline{P}$, with a network lifetime $\overline{T} \ge T^*$. Meanwhile, based on Lemma 3, solution $\overline{P}$ is also feasible to problem (RLX). And the maximum network lifetime of problem (RLX) is $T^*$. Hence, solution $\overline{P}$ is also the optimal solution of problem (RLX) with the network lifetime $\overline{T} = T^*$, which concludes the proof. $\square$

Proof of Theorem 4

Proof

As to $W = 1$, problem (LP-W) equals to problem (LP-T). Here, we emphasize on the $W > 1$ situation. Let $U^w(x_l) = W U(x_l)$ and $U^w(\nu _l) = W U(\nu _l)$, then the objective function of problem (LP-W) becomes

$$\begin{aligned} T = \sum _{l\in N} [U^w(x_l) + U^w(\nu _l)] \end{aligned}$$

(27)

And constraint Eq. (19) is converted to:

$$\begin{aligned} \sum _{l\in N} e^w_{il} - \varpi U^w(x_i) = H_i \end{aligned}$$

(28)

where

$$\begin{aligned} e^w_{il} = \varepsilon _{il} U^w(x_l) + \mu _{il} \lambda _l U^w(x_l) + (U^w(\nu _l) - \lambda _l U^w(x_l)) \eta _i \end{aligned}$$

Similarly, constraint Eq. (20) is converted to:

$$\begin{aligned} N h_0 + \varpi _0\sum _{i\in N} \tau _i + \varpi \sum _{l\in N}U^w(x_l) \le E \end{aligned}$$

(29)

Then, based on the new objective function Eq. (27) and constraints Eqs. (28), (29), problem (LP-W) can be equivalently transformed to problem (LP-T), which concludes the proof. $\square$