1. Introduction

• STM problem: sum-throughput maximization problem [6,9,13,14,15],
• EEM problem: energy efficiency maximization problem [5,10,12], and
• ECm problem: energy consumption minimization problem [7,11,16,17].

• Different from the general SWIPT, in this paper, we propose a new DT-SWIPT mechanism, which allows wireless stations that finish information transmission to share their excess energy to other wireless stations that have not yet transmitted, thereby reducing energy consumption in the network.
• Ensure that the SNR value of the received signal of the HAP can meet the SNR threshold so that the HAP can decode the signals from wireless stations.
• The optimization model, named DaTA, is proposed to obtain the best solution for the ECm problem. In addition, the DaTA-H scheme is also proposed so that an approximate solution can be obtained in a reasonable time.
• Two scenarios, special scenarios and general scenarios are conducted in the simulation to verify the advantages of the proposed schemes. We compare the DaTA and the DaTA-H schemes proposed in this paper against the STM, EEM, and ECm models of the related works. From the simulation results, it can be concluded that the DaTA scheme has better performance in the energy consumption and the energy efficiency in both general and special scenarios. In special scenarios, when the SNR threshold reaches a certain value, the energy consumption and the energy efficiency of the DaTA-H scheme are also better than those of the related works.

2. Preliminaries

3. DT-SWIPT Assisted Time Allocation for the ECm Problem

3.1. Problem Formulation

According to the HTT protocol, at the beginning of a period, the HAP will transfer energy to all wireless stations. The power used by the HAP is denoted as $P_0$. There is an upper limit, $P_{max}$, for the transmission power. The constraint of $P_0$ is shown in Eq. (1).

$$0\le P_0\le P_{max}.$$

(1)

In the ET phase, the channel gain between the HAP ($s_0$) and $s_i$ is denoted as $h_i$, which can be expressed in Eq. (2).

$$h_i=c_0{d}_{0,i}^{-\lambda },$$

(2)

where $d_{0,i}$ is the distance between the HAP ($s_0$) and $s_i$. Other parameters, such as antenna gain, antenna height of the HAP, etc., are neglected in the paper and termed $c_0$. For simplicity, $c_0$ is assumed to be a constant in the paper [23]. $\lambda$ is the path loss exponent [25].

With the channel gain in Eq. (2), when the HAP transfers energy in the ET phase, the received signal power of $s_i$ is denoted as $P_{0,i}^r$, which can be expressed in Eq. (3).

$$P_{0,i}^r=P_0{h}_i.$$

(3)

By Eq. (3), the energy that $s_i$ can obtain in the ET phase is denoted as $E_i$, which can be expressed in Eq. (4).

$$E_i={\eta }_i{P}_{0,i}^r{\tau }_0,$$

(4)

where ${\eta }_i$ is the energy conversion efficiency of $s_i$ when receiving energy during energy harvesting, and $0\le {\eta }_i\le 1$. ${\tau }_0$ is the duration of the ET phase.

In addition to the energy required to transmit information to the HAP, wireless stations also need the energy for basic operations, such as receiving beacons, modulating information, and maintaining basic circuits. Therefore, among the obtained energy $E_i$, the energy of $s_i$ will have a proportion of ${\rho }_i$ for transmission, and the remaining $1-{\rho }_i$ energy will be used for basic operations, where $0\le {\rho }_i\le 1$. Thus, through Eq. (4), the transmission power of $s_i$, denoted as $P_i$, can be expressed as Eq. (5).

$$P_i={\displaystyle \frac{{\rho }_i{E}_i}{{\tau }_i}}.$$

(5)

According to the DT-SWIPT scheme, $s_i$ completes the information transmission during ${\alpha }_i{\tau }_i$, and then shares the energy to other wireless stations during the remaining time, that is, $(1-{\alpha }_i){\tau }_i$. Let $g_{i,j}$ be the channel gain coefficient from $s_i$ to $s_j$ (assume that $s_i$ is transmitted earlier than $s_j$), which can be expressed in Eq. (6).

$$g_{i,j}=c_i{d}_{i,j}^{-\lambda },$$

(6)

where $d_{i,j}$ is the distance between $s_i$ and $s_j$. $c_i$ is a constant and stands for the other parameters, such as the antenna gain, antenna height of $s_i$, like $c_0$ in Eq. (2).

Therefore, with the channel gain coefficient $g_{i,j}$, the additional energy shared from $s_i$ and obtained by $s_j$ is denoted as $E_{i,j}^{add}$, which can be expressed as Eq. (7).

$$E_{i,j}^{add}={\eta }_j{P}_i{g}_{i,j}(1-{\alpha }_i){\tau }_i.$$

(7)

By TDMA protocol, wireless stations transmit information to the HAP sequentially. Therefore, each wireless station receives a different amount of energy shared by previously transmitted wireless stations. For this situation, let $rank\left(i\right)$ be the transmission order of $s_i$, and $ran{k}^{-1}\left(k\right)$ be the index of the wireless station whose transmission order is k. That is, if $s_i$ is the kth transmitting wireless station, then $rank\left(i\right)=k$ and $ran{k}^{-1}\left(k\right)=i$. If $s_i$ is the mth transmitting wireless station ($1<k<m\le n$), then $s_i$ will obtain the energy shared by $s_{ran{k}^{-1}\left(1\right)},\dots ,s_{ran{k}^{-1}\left(k\right)},\dots ,s_{ran{k}^{-1}(m-1)}$. Therefore, the total additional energy obtained by $s_i$, which is shared from the previously transmitted wireless stations, is denoted as $E_i^{add}$, which can be expressed in Eq. (8).

$$E_i^{add}=\sum _{j=1}^{rank\left(i\right)-1}{E}_{ran{k}^{-1}\left(j\right),i}^{add}.$$

(8)

Besides the energy obtained from the HAP, $s_i$ also obtains additional energy shared by previously transmitted wireless stations. Therefore, the total energy obtained by $s_i$ is denoted as $E_i^{total}$, which can be expressed in Eq. (9).

$$E_i^{total}=E_i+E_i^{add}.$$

(9)

Since the first transmitting wireless station does not obtain additional energy, the wireless station with $rank\left(i\right)=1$ does not need to calculate the additional energy. As a result, the transmission power of $s_i$ will be replaced from Eq. (5) to Eq. (10).

$$P_i=\{\begin{array}{cc}{\displaystyle \frac{{\rho }_i{E}_i^{total}}{{\tau }_i}},\hfill & rank\left(i\right)=2,\dots ,n,\hfill \\ {\displaystyle \frac{{\rho }_i{E}_i}{{\tau }_i}},\hfill & rank\left(i\right)=1.\hfill \end{array}$$

(10)

Let $g_{i,0}$ be the channel gain coefficient from $s_i$ to the HAP ($s_0$), which can be expressed in Eq. (11).

$$g_{i,0}=c_i{d}_{i,0}^{-\lambda }.$$

(11)

While $s_i$ starts to transmit information to the HAP, with the channel gain coefficient $g_{i,0}$, the received signal strength of the HAP from $s_i$ is denoted as $P_{i,0}^r$, which can be expressed in Eq. (12).

$$P_{i,0}^r=P_i{g}_{i,0}.$$

(12)

To ensure that the HAP can decode the signal transmitted from $s_i$, the ratio of the received signal power $P_{i,0}^r$ to the noise power N must be larger than or equal to a certain threshold, which is denoted as $SN{R}_{HAP}^{th}$. The relationship is shown in Eq. (13).

$${\displaystyle \frac{P_{i,0}^r}{N}}\ge SN{R}_{HAP}^{th}.$$

(13)

According to the Shannon’s Capacity Theorem [26], the channel capacity of $s_i$, which is denoted as $C_i$, can be expressed in Eq. (14).

$$C_i=B{log}_2(1+{\displaystyle \frac{P_{i,0}^r}{N}}).$$

(14)

Obviously, $C_i$ will be affected by $P_{i,0}^r$ and the bandwidth B. According to Eq. (14), the throughput of $s_i$ during the period T is denoted as $R_i$, which can be expressed in Eq. (15).

$$R_i={\displaystyle \frac{{\alpha }_i{\tau }_i}{T}}{C}_i.$$

(15)

Since the work performed by each wireless station may be different, $s_i$ has its own throughput demand, $R_i^{th}$. The throughput of $s_i$ must be larger than or equal to $R_i^{th}$. The relationship between the $R_i^{th}$ and $R_i$ is shown in Eq. (16).

$$R_i\ge R_i^{th}.$$

(16)

This paper aims to minimize the energy consumption of the network under the conditions of Eqs. (13) and (16), for i from 1 to n. Since the energy source of wireless stations is harvested from the HAP, which can be considered as the energy consumption of the network. The energy consumption of the network is denoted as $E_c$, which can be expressed in Eq. (17).

$$E_c=P_0{\tau }_0.$$

(17)

From [20], it shows that when $P_0=P_{max}$, the energy consumption of the network will be the lowest. Therefore, let $P_0$ be $P_{max}$. By Eq. (17), the goal will be to minimize the WET time, that is, ${\tau }_0$.

Based on the goal and constraints, a non-linear programming model for the D-ECm problem can be formulated in Model 1.

Optimization Model 1 (D-ECm Problem).

In Model 1, $C_1$ denotes the SNR constraint. That is, the ratio of the received signal power $P_{i,0}^r$ to the noise power N must be greater than $SN{R}_{HAP}^{th}$ to ensure that the HAP can decode the signal transmitted from $s_i$. $C_2$ denotes the throughput constraint, which requires that the throughput of $s_i$ must meet its throughput demand $R_i^{th}$. $C_3$ denotes that the transmission time of the HAP and $s_i$ can not be longer than T and should be larger than 0. $C_4$ denotes that the sum of the transmission time of the HAP and wireless stations can not be longer than T. $C_5$ denotes that the DT-SWIPT ratio of $s_i$ during the transmission phase should be smaller than 1.

Since Model 1 is a nonlinear optimization problem, we use an SQP (Sequential Quadratic Programming) method [27] to obtain the optimal solution.

3.2. SQP-Based Algorithm for Finding the Optimal Solution of the D-ECm Problem

SQP [27] is an iterative method for solving constrained nonlinear optimization problems. SQP obtains the optimal solution by solving a sequence of QP subproblems. The standard form of the QP problem is formulated in Model 2. In Model 2, $\nabla f\left(\overline{x}\right)$ and ${[\nabla f\left(\overline{x}\right)]}^{Tr}$ represent the gradients of the function $f\left(\overline{x}\right)$ and the transpose matrix of $\nabla f\left(\overline{x}\right)$, respectively. M and N represent the numbers of constraints of the equations and inequalities, respectively. The QP problem can get the corresponding optimal solution w according to the input $\overline{x}$. We can then take $\overline{x^{\prime }}=\overline{x}+w$ into the original QP problem to form a new QP problem and obtain another optimal solution $w^{\prime }$. Therefore, w can be regarded as the iterative step in SQP. When $\overline{x^{\prime }}-\overline{x}$ approaches 0, the corresponding $\overline{x}$ is the optimal solution of the SQP problem.

Optimization Model 2 (Standard Form of the QP Problem).

To satisfy the standard form of the QP problem, we turn the objective function of Model 1 into minimize ${\tau }_0^2$, and let $X=[{\tau }_0,{\tau }_1,\dots ,{\tau }_n,{\alpha }_1,\dots ,{\alpha }_{n-1}]$, where $X\in {\mathbb{R}}^{2n+}$ and ${\mathbb{R}}^{+}$ represents the set of non-negative real numbers. As a result, we rewrite Model 1 as Model 3.

Optimization Model 3 (Reduced D-ECm Problem).

Model 3 can be converted into a QP problem, called QP-ECm, shown in Model 4. In Model 4, $w^k$ is the optimal solution corresponding to $X^k$, $w^k\in {\mathbb{R}}^{2n+}$. k is an iteration counter of the SQP.

Optimization Model 4 (QP-ECm Problem).

To solve the D-ECm problem, a SQP-based algorithm is proposed, called DaTA (DT-SWIPT assisted Time Allocation), which is shown in Algorithm 1. In line 2, we convert the ECm problem into a QP-ECm problem. In line 3, set the iterative counter k and select an initial point $X^0$ that meets all constraints. In line 5, solve the QP-ECm problem according to $X^k$ and get the optimal solution $w^k$. In lines 6 and 7, increase the iterative counter k, and let $X^k=X^{k-1}+w^{k-1}$. If $X^k-X^{k-1}$ approaches 0, return the optimal solution $X^{k-1}$; otherwise, return to line 5 to solve a new QP problem again.

In this section, we have formulated a nonlinear optimization problem, called D-ECm, and found the optimal solution through a SQP-based algorithm, named DaTA. However, it costs much time to find the optimal solution. However, to our best knowledge, there is no heuristic method proposed in the literature to find a plausible solution. Therefore, we propose a heuristic method, called DaTA-H, and describe it in the following section.

Algorithm 1: DaTA: a SQP-based Algorithm for the D-ECm Problem

4. DaTA-H: The Heuristic Method

4.1. Problem Formulation of the Simplifying D-ECm Problem

In Section 3.1, we can know that the transmission time of $s_i$, ${\tau }_i$, is related to ${\tau }_0$ through Eqs. (4) and (10). To simplify the D-ECm problem, we only take ${\tau }_0$ into account. The remaining time, $T-{\tau }_0$, is evenly distributed to all wireless stations. Therefore, in DaTA-H, the transmission time of $s_i$, i.e. ${\tau }_i$, can be expressed as Eq. (18).

$${\tau }_i={\displaystyle \frac{T-{\tau }_0}{n}},i=1,\dots ,n.$$

(18)

By Eq. (18), we can simplify the variables such as ${\tau }_1,\dots ,{\tau }_n$ into a function containing ${\tau }_0$.

In addition, we further simplify ${\alpha }_1,\dots ,{\alpha }_{n-1}$ by means of changing the transmission order. According to Eqs. 2,  3, and 4, the distance $d_{0,i}$ between the HAP ($s_0$) and $s_i$ will directly affect the energy obtained by $s_i$. The wireless station closest to the HAP can obtain the most energy, while the wireless station farthest from the HAP harvests the least. Therefore, we rearrange the order of the wireless stations according to $d_{0,i}$. The wireless station closest to the HAP is marked as $s_1$, the next is $s_2$, and so on. The farthest one is $s_n$. In DaTA-H, we let the wireless station that obtains the most energy from the HAP (i.e. $s_1$) be the first one to transmit information and share the energy to other stations for the longest time. The wireless station that obtains the second most energy (i.e. $s_2$) is the second one to transmit information and share energy for the second-longest time, and so on. The farthest wireless station (i.e. $s_n$) is the last one to transmit information and needs not to share energy. In summary, the time ratio ${\alpha }_i$ of the DT-SWIPT can be allocated by an arithmetic sequence, which can be expressed as Eq. (19).

$${\alpha }_i=\alpha +i*{\displaystyle \frac{{\alpha }_n-\alpha }{n}},i=1,\dots ,n,$$

(19)

where $\alpha$ is a base value of the DT-SWIPT. Since $\alpha$ behaves differently due to the different SNR thresholds of the HAP, we assume that $\alpha$ is known in advance. Section 5 will discuss the impact of $\alpha$ on the performance.

According to the above summary, we can rewrite the D-ECm problem as an S-ECm (Simplified D-ECm) problem and is shown in Model 5.

Optimization Model 5 (S-ECm: Simplified D-ECm Problem).

Model 5 is still a non-linear programming problem due to $C_{19}$. Since the variable to be optimized is only ${\tau }_0$, we can use a linear search method to iteratively find an approximate solution of ${\tau }_0$.

4.2. DaTA-H Algorithm for S-ECm

The DaTA-H algorithm is shown in Algorithm 2. In addition to the inputs used in Algorithm 1, the inputs also include the $\alpha$ and the step size $\delta$. Algorithm 2 is divided into two steps: the initialization step and the iterative step.    

Algorithm 2: DaTA-H Algorithm for the S-ECm Problem

In the Initialization Step, because we need to find an approximate solution of ${\tau }_0$, we first set a variable k to count the number of iterations (line 2) and set the WET time ${\tau }_0^k$ in line 3.

 Lemma 1.

Let ${\tau }_0^{th}$ be the minimum of the WET time and $\theta ={\displaystyle \frac{SN{R}_{HAP}^{th}*N}{g_{1,0}{\rho }_1{\eta }_1{h}_1{P}_0+\frac{SN{R}_{HAP}^{th}*N}{n}}}$. Then ${\tau }_0^{th}$ can be obtained as follows.

$${\tau }_0^{th}={\displaystyle \frac{T}{n}}*\theta .$$

Proof. 

To simplify the D-ECm problem, we have proposed a simplified model, S-ECm. There we have assumed that the wireless station closest to the HAP is $s_1$, followed by $s_2$, and so on. Therefore, the minimum WET time should at least satisfy the energy requirement of $s_1$. As a result, ${\tau }_0^{th}$ can be regarded as the WET time required for $s_1$. By $C_{18}$, we can obtain Eq. (20).

$$P_{1,0}^r\ge SN{R}_{HAP}^{th}*N.$$

(20)

By Eq. (12), Eq. (20) can be rewritten as below.

$$P_1\ge SN{R}_{HAP}^{th}*{\displaystyle \frac{N}{g_{1,0}}}.$$

(21)

Since $s_1$ is the first wireless station to transmit information to the HAP (i.e. $rank\left(1\right)=1$), the energy obtained by $s_1$ is totally from the HAP, no additional energy from any other wireless station. By Eq. (10), Eq. (21) is turned as follows.

$$E_1\ge SN{R}_{HAP}^{th}*{\displaystyle \frac{N{\tau }_1}{g_{1,0}{\rho }_1}}.$$

(22)

Furthermore, by Eqs. (3) and (4), Eq. (22) is turned as follows.

$${\tau }_0\ge SN{R}_{HAP}^{th}*{\displaystyle \frac{N{\tau }_1}{g_{1,0}{\rho }_1{\eta }_1{h}_1{P}_0}}.$$

(23)

${\tau }_0^{th}$ is the minimum value of ${\tau }_0$. By Eq. (18), ${\tau }_1={\displaystyle \frac{T-{\tau }_0^{th}}{n}}$. As a result, we have the following equation.

$${\tau }_0^{th}={\displaystyle \frac{SN{R}_{HAP}^{th}*N*T}{n*g_{1,0}{\rho }_1{\eta }_1{h}_1{P}_0+SN{R}_{HAP}^{th}*N}}.$$

(24)

Since $SN{R}_{HAP}^{th}$, N, n , $g_{1,0}$, ${\rho }_1$, ${\eta }_1$, $h_1$, and $P_0$ are all known parameters, let

$\theta ={\displaystyle \frac{SN{R}_{HAP}^{th}*N}{g_{1,0}{\rho }_1{\eta }_1{h}_1{P}_0+\frac{SN{R}_{HAP}^{th}*N}{n}}}$. We simplify Eq. (24) as below.

$${\tau }_0^{th}={\displaystyle \frac{T}{n}}*\theta ,$$

which concludes the proof of the lemma. □

According to Lemma 1, we can know that if $s_1$ satisfies $C_{18}$, ${\tau }_0^k$ must be equal to ${\tau }_0^{th}$. To avoid ${\tau }_0^k$ starting from 0 and causing too many iterations, therefore, in the initialization step, we let ${\tau }_0^k$ iterates from ${\tau }_0^{th}$. According to Eq. (18), line 4 is to evenly set ${\tau }_i^k$ for all wireless stations. Based on $\alpha$, line 5 sets the DT-SWIPT time ratio ${\alpha }_i$. Therefore, we can derive $P_{i,0}^r$ and $R_i$, for $i=1,\dots ,n$.

In the Iterative Step, we first check whether all wireless stations satisfy the constraints of $C_{18}$ and $C_{19}$ based on the derived $P_{i,0}^r$ and $R_i$ in the initialization step (line 7). If one of the constraints is not satisfied, the iterative step starts. Firstly, line 8 increases the iteration index k. Line 9 increases ${\tau }_0^k$ by adding a step size $\delta$. According to Eq. (18), line 10 sets a new transmission time for all wireless stations. Line 11 recalculates the $P_{i,0}^r$ and $R_i$, $i=1,\dots ,n$. Finally, go back to line 7 to recheck whether the constraints of $C_{18}$ and $C_{19}$ are satisfied for all wireless stations or not.

In the iterative step, there is an upper limit for the number of iterations. That is, the WET time ${\tau }_0^k$ must be smaller than or equal to T after several iterations. Therefore, we derive the upper bound of the number of iterations by Lemma 2.

 Lemma 2.

In Algorithm 2, the number of iterations in the iterative step is not greater than ${log}_{1+\delta }n-{log}_{1+\delta }\theta$.

Proof. 

Since the increase of the ${\tau }_0^k$ must be smaller than or equal to T, it bounds the number of iterations in the iterative step. In other words, we can have the following inequality.

$${\tau }_0^k={\tau }_0^{th}{(1+\delta )}^k\le T.$$

According to Lemma 1, ${\displaystyle \frac{T}{n}}\theta {(1+\delta )}^k\le T$. As a result,

$$k\le ({log}_{1+\delta }n-{log}_{1+\delta }\theta ),$$

which concludes the proof of the lemma. □

Furthermore, the time complexity of Algorithm 2 is analyzed in Theorem 1.

 Theorem 1.

The time complexity of Algorithm 2 is $O(nlogn)$.

Proof. 

In the Initialization Step, the time to calculate ${\tau }_0^k$, ${\tau }_i^k$, $P_{i,0}^r$, and $R_i$ is $O\left(1\right)$ and the numbers of calculations of $P_{i,0}^r$ and $R_i$ are respectively n. Therefore, the time complexity of the Initialization Step is $O\left(n\right)$. In the Iterative Step, according to Lemma 2, the number of iterations of the while loop is ${log}_{1+\delta }n-{log}_{1+\delta }\theta$. The time to calculate lines 8 and 9 is respectively $O\left(1\right)$ and that to calculate lines 10 and 11 is n, respectively. Therefore, the time complexity of the Iterative Step is $O(nlogn)$. As a result, the time complexity of Algorithm 2 is $O(nlogn)$, which concludes the theorem. □

In this section, we have simplified the S-ECm problem and proposed a heuristic method, called DaTA-H. Based on the result of Lemma 1, we avoid ${\tau }_0^k$ iterating from 0 to cause too many iterations. We use a linear search method to find an approximate solution of ${\tau }_0$. In addition, we also prove that the time complexity of Algorithm 2 is $O(nlogn)$) in Theorem 1 based on Lemma 2. In the following section, We will compare the efficiency of DaTA and DaTA-H and show the advantages against the related work.

5. Performance Evaluations

6. Conclusions

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