SCIENCE CHINA Information Sciences, Volume 63 , Issue 2 : 122302(2020) https://doi.org/10.1007/s11432-019-9936-y

Joint utility optimization for wireless sensor networks with energy harvesting and cooperation

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  • ReceivedMar 26, 2019
  • AcceptedJul 17, 2019
  • PublishedJan 16, 2020



This work was supported in part by National Science and Technology Major Project of China (Grant No. 2018ZX03001008-002), in part by Natural Science Foundation of Jiangsu Province (Grant No. BK20180011), and in part by National Natural Science Foundation of China (Grant Nos. 61571120, 61871122).



Proof of Theorem th1

The Lyapunov function 15 consists of two parts, the data stream queue and the energy queue. We separately handle these two parts in order to obtain the upper bound of the Lyapunov drift 16.

For one thing, we discuss the energy queue part. By introducing $\theta_n$, squaring both sides of 12 and rearranging it, we then have \begin{align} &\frac{1}{2}{{\left[ {{e}_{n}}(t+1)-{{\theta }_{n}} \right]}^{2}} -\frac{1}{2}{{\left[ {{e}_{n}}(t)-{{\theta }_{n}} \right]}^{2}} \\ & \le {{\left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{\left( {{P}_{nm}}( t )+{{\varepsilon}_{nm}}( t ) \right)} \right]}^{2}} +{{\left[ \sum\limits_{k\in \mathcal{N}_{n}^{(\rm in)}}{{{\varepsilon}_{kn}}( t )}+{{h}_{n}}( t ) \right]}^{2}} \\ & - \left[ {{e_n}(t) - {\theta _n}} \right]\left[ {\sum\limits_{m \in \mathcal{{\cal N}}_n^{(\rm out)}} {{P_{nm}}( t )} - {h_n}( t )} \right] -\left[ {{e}_{n}}(t)-{{\theta }_{n}} \right] \left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{{{\varepsilon}_{nm}}( t )-\sum\limits_{k\in \mathcal{N}_{n}^{(\rm in)}}{{{\widehat{\varepsilon}}_{kn}}( t )}} \right]. \tag{33} \end{align} Since ${{P}_{nm}}(~t~)$, ${{\varepsilon}_{nm}}(~t~)$, ${{h}_{n}}(~t~)$ have upper bounds ${P}_{\rm~max}$, ${\varepsilon}_{\rm~max}$, ${h}_{\rm~max}$ respectively, we can obtain that \begin{equation} {{\left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{\left( {{P}_{nm}}( t )+{{\varepsilon}_{nm}}( t ) \right)} \right]}^{2}}+{{\left[ \sum\limits_{k\in \mathcal{N}_{n}^{(\rm in)}}{{{\widehat{\varepsilon}}_{kn}}( t )}+{{h}_{n}}( t ) \right]}^{2}} \le {{\left[ {{D}_{\max }}\left( {{P}_{\max }}+{{\varepsilon}_{\max }} \right) \right]}^{2}}+{{\left[ {{D}_{\max }}{{\varepsilon}_{\max }}+{{h}_{\max}} \right]}^{2}}. \tag{34}\end{equation}

Let $B_1$ denote ${{[~{{D}_{\max~}}(~{{P}_{\max~}}+{{\varepsilon}_{\max~}}~)~]}^{2}}+{{[~{{D}_{\max~}}{{\varepsilon}_{\max~}}+{{h}_{\max}}~]}^{2}}$. Hence, we sum both sides of 34 over $n$ and it can be recast as \begin{align} \frac{1}{2}\sum\limits_{n=1}^{N}{{{\left[ {{e}_{n}}(t+1)-{{\theta }_{n}} \right]}^{2}}}-\frac{1}{2}\sum\limits_{n=1}^{N}{{{\left[ {{e}_{n}}(t)-{{\theta }_{n}} \right]}^{2}}}\le & N{{B}_{1}} -\sum\limits_{n=1}^{N}{\left[ {{e}_{n}}(t)-{{\theta }_{n}} \right]}\left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{{{P}_{nm}}( t )-{{h}_{n}}( t )} \right] \\ & -\sum\limits_{n=1}^{N}{\sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{{{\varepsilon}_{nm}}( t )\left[ {{\widehat{\varepsilon}}_{n}}( t )-{{\eta }_{nm}}{{\widehat{\varepsilon}}_{m}}(t) \right]}}. \tag{35} \end{align} We define ${{\widehat{E}}_{n}}(t)={{e}_{n}}(~t~)-{{\theta~}_{n}}$. And due to the fact that $0\le~{{\eta~}_{kn}}\le~1$, we have $(~1-{{\eta~}_{nm}}~){{\widehat{E}}_{n}}(~t~)\ge~-(~1-{{\eta~}_{nm}}~){{\theta~}_{n}}$. Let $B_2$ denote ${{B}_{2}}=N{{B}_{1}}+\sum\nolimits_{n=1}^{N}{\sum\nolimits_{m=1}^{N}{(~1-{{\eta~}_{nm}}~)}}{{\varepsilon}_{\max~}}$, and thus 35 can be rewritten as \begin{align} \frac{1}{2}\sum\limits_{n=1}^{N}{{{\left[ {{e}_{n}}(t+1)-{{\theta }_{n}} \right]}^{2}}}-\frac{1}{2}\sum\limits_{n=1}^{N}{{{\left[ {{e}_{n}}(t)-{{\theta }_{n}} \right]}^{2}}}\le & {{B}_{2}} -\sum\limits_{n=1}^{N}{{{\widehat{E}}_{n}}( t )}\left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{{{P}_{nm}}( t )-{{h}_{n}}( t )} \right] \\ &-\sum\limits_{n=1}^{N}{\sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{{{\eta }_{nm}}{{\varepsilon}_{nm}}( t )}}\big[ {{\widehat{E}}_{n}}( t )-{{\widehat{E}}_{m}}(t) \big]. \tag{36} \end{align} For another, the data queue part is studied and proved. Since the inequality $(~{{[~f(x)~]}^{+}}~)\le~{{f}^{2}(x)}$ holds for any $x\in~\mathbb{R}$, Eq. 9 can be converted into \begin{align} {{\big[ q_{n}^{(d)}(t+1) \big]}^{2}} -{{\big[ q_{n}^{(d)}(t) \big]}^{2}}\le & -2q_{n}^{(d)}(t)\left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{\lambda _{nm}^{( d )}(t)}-\sum\limits_{k\in \mathcal{N}_{n}^{(\rm in)}}{\lambda _{kn}^{( d )}(t)}-R_{n}^{(d)} \right] \\ &+{{\left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{\lambda _{nm}^{( d )}(t)} \right]}^{2}}+{{\left[ \sum\limits_{k\in \mathcal{N}_{n}^{(\rm in)}}{\lambda _{kn}^{( d )}(t)}+R_{n}^{(d)} \right]}^{2}}. \tag{37} \end{align} Rearrange 37 similarly to how we deal with that in the energy queue part, and it can be written as \begin{equation} \frac{1}{2}{{\big[ q_{n}^{(d)}(t+1) \big]}^{2}}-\frac{1}{2}{{\big[ q_{n}^{(d)}(t) \big]}^{2}}\le {{B}_{3}} -q_{n}^{(d)}(t)\left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{\lambda _{nm}^{( d )}(t)}-\sum\limits_{k\in \mathcal{N}_{n}^{(\rm in)}}{\lambda _{kn}^{( d )}(t)}-R_{n}^{(d)} \right], \tag{38}\end{equation} where ${{B}_{3}}=\frac{3}{2}D_{\max~}^{2}C~_{\max~}^{2}+R_{\max~}^{2}$.

Hence, we sum 38 over all $(n,d)$, and then obtain the following bound combined with 35. \begin{align} L({\boldsymbol Q}(t+1))-L({\boldsymbol Q}(t)) \le & B -\sum\limits_{n,d}{q_{n}^{(d)}}(t)\left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{\lambda _{nm}^{( d )}(t)}-\sum\limits_{k\in \mathcal{N}_{n}^{(\rm in)}}{\lambda _{kn}^{( d )}(t)}-R_{n}^{(d)} \right] \\ & -\sum\limits_{n=1}^{N}{{{{\hat{E}}}_{n}}( t )}\left[ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{{{P}_{nm}}( t )-{{h}_{n}}( t )} \right] -\sum\limits_{n=1}^{N}{\sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{{{\eta }_{nm}}{{\varepsilon}_{nm}}( t )}}[ {{{\hat{E}}}_{n}}( t )-{{{\hat{E}}}_{m}}(t) ], \tag{39} \end{align} where \begin{align} B & ={{N}^{2}}{{B}_{3}}+{{B}_{2}} \\ &={{N}^{2}}\left( \frac{3}{2}D_{\max }^{2}C _{\max }^{2}+R_{\max }^{2} \right) + N \big\{ {{[ {{D}_{\max }}( {{P}_{\max }}+{{\varepsilon}_{\max }} )]}^{2}}+{{[ {{D}_{\max }}{{\varepsilon}_{\max }}+{{h}_{\max}} ]}^{2}} \big\} +\sum\limits_{n=1}^{N}{\sum\limits_{m=1}^{N}{( 1-{{\eta }_{nm}} )}}{{\varepsilon}_{\max }}. \tag{40} \end{align} According to 16, we take expectations over the network state ${\boldsymbol~Q}(t)$ on both sides of 39, and then obtain 18.

Theorem th1 is thus proved.

Proof of Theorem th2

First, we use mathematical induction (MI) to prove the boundary constraint of $q_{n}^{(d)}(t)$ in 29 and $e_{n}(t)$ in 30.

As for $q_{n}^{(d)}(t)$, it is clear to see that Eq. 29 holds for $t=0$, because $q_{n}^{(d)}(0)=0$ for all $n,d~\in~\mathcal{N}$. Then, we assume that at time slot $t$, $q_{n}^{(d)}(t)$ obeys $q_{n}^{(d)}(t)\le~\beta~V+{{R}_{\max~}}$, $\forall~n,d~\in~\mathcal{N}$. Hence, two cases are considered in the next time slot $t+1$ classified by whether to receive data stream.

If node $n$ does not receive any data stream from other nodes through links at $t$, then it comes to $~q_{n}^{(d)}(t+1)\le~\beta~V+{{R}_{\max~}}$ referring to 9.

If node $n$ receives data stream from node $k$, with $~q_{k}^{(d)}(t)\le~\beta~V+{{R}_{\max~}}$, then it comes to \begin{equation} q_{n}^{(d)}(t) \le q_{k}^{(d)}(t)-\tau \le (\beta V+{{R}_{\max }})-({{D}_{\max }}{{C}_{\max }}+{{R}_{\max }}) \le \beta V. \tag{41}\end{equation} Since the transmitted data cannot exceed ${{R}_{\max~}}$, we have $q_{n}^{(d)}(t+1)\le~\beta~V+{{R}_{\max~}}$.

If node $n$ receives data packets from the outside of the network, we have $q_{n}^{(d)}(t)\le~Vf'(~0~)=\beta~V$ and get the same result as the above case.

Hence, $~q_{n}^{(d)}(t+1)\le~\beta~V+{{R}_{\max~}}$ holds for any $n,d~\in~\mathcal{N}$, and thus 29 is proved.

As for $e_{n}(t)$, since we have made the assumption that there is no energy in the initial state, we can get $e_{n}(t)=0$ for any $n~\in~\mathcal{N}$. For node $n$, we assume $e_{n}^{{}}(t)\le~{{\theta~}_{n~}}+{{h}_{\max~}}+{{D}_{\max~}}{{\varepsilon}_{\max~}}{{\eta~}_{\max~}}$ at time slot $t$.

If $e_{n}^{{}}(t)~<~{{\theta~}_{n}}$, $e_{n}(t+1)$ holds due to the limit of the input energy, no more than ${{h}_{\max~}}+{{D}_{\max~}}{{\varepsilon}_{\max~}}{{\eta~}_{\max~}}$.

If $e_{n}^{{}}(t)\ge~{{\theta~}_{n}}$, node $n$ does not harvest energy but may receive energy from other nodes. As an example, $e_{m}(t)~\le~{{\theta~}_{m}+h_{\max}+{{D}_{\max~}}{{\varepsilon}_{\max~}}{{\eta~}_{\max~}}}$ and energy is transferred from node $m$ to node $n$, which implies that \begin{align} e_{n}^{{}}(t) & < e_{m}^{{}}(t)-{{\theta }_{m}}-\frac{\zeta }{{{\eta }_{nm}}}+{{\theta }_{n}} \\ & \le e_{m}^{{}}(t)+{{\theta }_{n}}-\zeta \\ & \le ({{\theta }_{m}+h_{\max}+{{D}_{\max }}{{\varepsilon}_{\max }}{{\eta }_{\max }}})+{{\theta }_{n}} -({{\theta }_{\max}} +{{D}_{\max }}{{\varepsilon}_{\max }}{{\eta }_{\max }}) \\ & \le {{\theta }_{n}}+ h_{\max}. \tag{42} \end{align} Since the transferred energy cannot surpass ${{D}_{\max~}}{{\varepsilon}_{\max~}}{{\eta~}_{\max~}}$, we then have $e_{n}(t+1)~<~{{\theta~}_{n}+h_{\max}+{{D}_{\max~}}{{\varepsilon}_{\max~}}{{\eta~}_{\max~}}}$.

Hence, $e_{n}(t+1)~<~{{\theta~}_{n}+h_{\max}+{{D}_{\max~}}{{\varepsilon}_{\max~}}{{\eta~}_{\max~}}}$ holds for any $n~\in~\mathcal{N}$, and thus 30 is proved.

Therefore, the boundary constraint of $q_{n}^{(d)}(t)$ and $e_{n}(t)$ are provided in 29 and 30.

Next we prove 31 by contradiction when the data transmission or energy transfer action is in progress, that is $P_{nm}>0$ or $e_{\max}>0$.

In the power control step of the EDPR algorithm, we strive to choose the optimal $\varepsilon_{nm}^{*}(~t~)$ and $P_{nm}^{*}(~t~)$ to solve the energy-availability problem 28 as \begin{align} G({{\varepsilon}_{nm}}(t),{{P}_{nm}}(t))=& \sum\limits_{n}{\left\{ \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{{{\varepsilon}_{nm}}(t)}\big[ {{\eta }_{nm}}\big( {{{\hat{E}}}_{n}}(t)-{{{\hat{E}}}_{m}}(t) \big)-\zeta \big] \right.} \\ &+\sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{C_{nm}^{{}}(t)}\left[ q_{n}^{(d)}(t)-q_{m}^{(d)}(t)-\tau \right] \left. +{{{\hat{E}}}_{n}}(t)\sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{{{P}_{nm}}(t)} \right\}. \tag{43} \end{align} If there exist the optimal power allocation ${{\boldsymbol\varepsilon}^{*}}$ and the transferred energy ${{{\boldsymbol~P}}^{*}}$ that maximize 43, it is sure that they satisfy $P_{nm}^{*}(t)>0$ and ${\varepsilon}_{nm}^{*}(t)>0$ on link $(n,m)$. We now create a new solution $({{\boldsymbol\varepsilon}^{*}},{{{\boldsymbol~P}}^{*}})$ to the problem, by setting ${{P}_{nm}}^{*}(t)=0$, $\forall~k\ne~n$, $\forall~m$ and ${{\varepsilon}_{nm}}(~t~)$, $\forall{n,m}$ remains the same. Then, we have \begin{equation} G({{{\boldsymbol P}}^{*}})-G({\boldsymbol P})=\sum\limits_{n}\left\{ {{{\hat{E}}}_{n}}( t ){{{\boldsymbol P}}^{*}} + \sum\limits_{m\in \mathcal{N}_{n}^{(\rm out)}}{\left[ {{C}_{nm}}\left( {{{\boldsymbol P}}^{*}} \right)-{{C}_{nm}}\left( {\boldsymbol P} \right) \right]}{{\widetilde{W}}_{( n,m )}}(t) \right\}. \tag{44}\end{equation}

Contrary to Theorem th2 we assume $e_{n}^{{}}(t)<{{P}_{\max~}}+{{\varepsilon}_{\max~}}$ in such case that node $n$ cannot transfer energy to other nodes with energy transferred weight ${{W}_{(~n,m~)}}<0$. Combined with 22, then ${{\widehat{E}}_{n}}(~t~)<\delta~(~\beta~V+{{R}_{\max~}}~)$. In addition, we observe it from 29 that the backlog of all data queue is no more than $\beta~V+{{R}_{\max~}}$, so the backlog weight of data stream over link $(~n,m~)$ satisfies $\widetilde{W}_{(n,m)}^{{}}(t)\le~\beta~V-{{D}_{\max~}}{{C}_{\max~}}$ at any time slot. Hence, the above problem can be simplified into \begin{align} G(P_{nm}^{*}(t))-G({{P}_{nm}}(t))&={{C}_{nm}}P_{nm}^{*}(t)\text{ }\widetilde{W}_{( n,m )}^{{}}(t)+{{\hat{E}}_{n}}( t )P_{nm}^{*}( t ) \\ &\le\left( \beta V-{{D}_{\max }}{{C}_{\max }} \right)\delta P_{nm}^{*}(t)-\delta \left( \beta V+{{R}_{\max }} \right)P_{nm}^{*}(t) \\ &=\delta P_{nm}^{*}(t)\left({{D}_{\max }}{{C}_{\max }}+{{R}_{\max }}\right) \\ &<0. \tag{45} \end{align} This implies that ${{{\boldsymbol~P}}^{*}}$ is not the optimal solution of the problem 44 since it conflicts with our assumption. The proof of the optimal transferred energy $\varepsilon_{nm}^{*}(t)$ can be performed similarly. Hence, $e_{n}^{{}}(t)\ge~{{P}_{\max~}}+{{\varepsilon}_{\max~}}$, namely 31, holds whether node $n$ transfers any non-negative energy or allocates any non-zero power over its subsequent nodes. What's more, it turns out that all the power allocation and energy transfer actions are feasible, and the “energy-availability" constraint is indeed redundant.

Therefore, part (a) of Theorem th2 is proved. And the proof of part (b) is similar to the proof of Theorem 5.1 in [26], so we omit it here for brevity.


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  • Figure 1

    (Color online) (a) The energy harvesting and data receiving process of a source node; (b) the topology of a WSN.


    Algorithm 1 EDPR algorithm with optimal utility at time slot $t$

    Require:Initialize $V$ and obtain the corresponding ${\theta~}_n$ (${n~\in~\mathcal{N}}$) for nodes according to 22;

    Output:Calculate the optimal amount of data transmission ${R_n^{(~d~)}(t)}$, the energy transfer ${{\varepsilon}_{nm}}(~t~)$, and the power allocation ${{P}_{nm}}(t)$.

    Determine ${{h}_{n}}(t)$ for node $n$ through energy management;

    Determine $R_{n}^{(~d~)}(t)$ according to 23 through data transmission;

    if the link from node $n$ to node $m$ exists then

    Calculate the weight ${{W}_{n}}(t)$ and $~{{\widetilde{W}}_{(~n,m~)}}(t)$ according to 26 and 27 through power control and routing scheduling;

    end if

    Obtain ${{\varepsilon}_{nm}(t)}$, $P_{nm}(t)$, $\lambda~_{nm}^{{d}}(~t~)$ based on the value of $e_{n}(t)$, $\theta_{n}$, $W_{n}(t)$;

    Update $q_{n}^{(~d~)}(t)$ and $e_n(t)$ according to 9 and 12.