Resource allocation algorithm with worst case delay guarantees in energy harvesting body area networks

Abstract

In Body Area Networks, for energy harvesting-powered, managing the renewable energy to provide delay-sensitive services is of significant importance. In this paper, we propose an online algorithm to allocate resources, i.e., energy and channel, to maximize the user utility while guaranteeing the worst-case delay. To this end, we first formulate a user utility optimization problem, characterizing the stochastic nature of energy harvesting and energy consumption. Furthermore, a priori knowledge of any this processes is not required. Using Lyapunov optimization techniques, we decompose the problem into four sub-problems, i.e., battery management, collecting rate control, transmission power allocation and dropping rate control. Low-complexity online resource allocation algorithm is proposed to address these problems for user utility maximization, while we further analyze the performance of the online algorithm, in terms of the upper bounds of queues, required battery capacity, and the optimality of the proposed algorithm. Simulation results verify our analysis and the efficacy of the proposed algorithm.

Appendices

Appendix A: PROOF OF LEMMA 1

For any time slot t, r_n(t) represents the data that collects on this time slot. We show that all of this data departs on or before time t + I^max. Using the method of contradiction, we assume this is not true. It must be that W_n(t) > 0 for all ρ ∈{t + 1,...,t + I^max}. From Eq. 13, we have:

$$\begin{array}{ll} &M_{n}(\rho + 1) = max[M_{n}(\rho)+\epsilon - \eta_{n}(\rho)-d_{n}(\rho), 0], \\ &M_{n}(\rho + 1)\geq M_{n}(\rho)+\epsilon - \eta_{n}(\rho)-d_{n}(\rho). \end{array} $$

Summing the above over ρ ∈{t + 1,...,t + I^max} yields:

$$M_{n}(t+I^{max} \!+ 1)-M_{n}(t + 1) \!\geq \epsilon I^{max} -\! \sum\limits^{t+I^{max}}_{\rho = t + 1}[\eta_{n}(\rho)+d_{n}(\rho)]. $$

Due to M_n(t + I^max + 1) ≤ M^max and M_n(t) ≥ 0, we have:

$$ \epsilon I^{max} - M^{max} \leq \sum\limits_{\rho = t + 1}^{t+I^{max}}[\eta_{n}(\rho)+d_{n}(\rho)], $$

(30)

$$ \sum\limits_{\rho = t + 1}^{t+I^{max}}[\eta_{n}(\rho)+d_{n}(\rho)] < W_{n}(t + 1)\leq W^{max}. $$

(31)

Combining (30) and (31), we have:

$$\epsilon I^{max} -M^{max}< W^{max}. $$

Therefore:

$$I^{max} < (W^{max}+M^{max})/ \epsilon. $$

This contradicts the Eq. 14, thus the Eq. 14 is proved.

Appendix B: PROOF OF THEOREM 1

By squaring both sides of data queue, i.e., Eq. 6, we can get Eq. 32. Similarly, by squaring both sides of energy queue and virtual queue, we have Eqs. 33 and 34. Substituting $X_{n}(t) = 1_{(W_{n}(t)>0)}$ and $Y_{n}(t) = 1_{(W_{n}(t) = 0)}$ into Eq. 13. Rearranging the equation, we have Eq. 21.

$$ \begin{array}{ll} \!\frac{1}{2}&\![(W_{n}(t\,+\,1))^{2}\,-\,(W_{n}(t)^{2})] \\ \!\leq\! & \!\frac{1}{2}[(r_{n}(t))^{2}\,+\,(\eta_{n}(t)\,+\,d_{n}(t))^{2} \,+\,2W_{n}(t)(r_{n}(t)\,-\,\eta_{n}(t)\,-\,d_{n}(t))] \\ \!\leq\! & \!\frac{(r_{n}(t))^{2}+(\eta_{n}(t)+d_{n}(t))^{2}}{2}\,+\,W_{n}(t)(r_{n}(t)-\eta_{n}(t)-d_{n}(t), \end{array} $$

(32)

$$ \begin{array}{ll} &\frac{1}{2}[(K_{n}(t + 1)-{\Omega})^{2}-(K_{n}(t)-{\Omega})^{2}] \\ & \leq\frac{1}{2}[(h_{n}(t))^{2}+(b_{n}(t))^{2}-2\hat{K}_{n}(t)(h_{n}(t)-b_{n}(t))]\\ &\leq \frac{(h_{n}(t))^{2}+(b_{n}(t))^{2}}{2}-\hat{K}_{n}(t)(h_{n}(t)-Cr_{n}(t)-a_{n}(t)), \end{array} $$

(33)

$$ \begin{array}{ll} \!\frac{1}{2}&\![(M_{n}(t + 1))^{2}-(M_{n}(t))^{2}] \\ \!\leq & \!\!\frac{1}{2} [max[\epsilon^{2},(\eta_{n}^{max}\,+\,d_{n}^{max})^{2}] \,+\, 2M_{n}(t)[X_{n}(t)(\epsilon\,-\,\eta_{n}(t))\,-\,d_{n}(t)\,-\,Y_{n}(t)\eta_{n}^{max}]]\\ \!\leq & \!\frac{max[\epsilon^{2},(\eta_{n}^{max}+d_{n}^{max})^{2}]}{2} + M_{n}(t)[X_{n}(t)(\epsilon-\eta_{n}(t))-d_{n}(t)-Y_{n}(t)\eta_{n}^{max}]. \end{array} $$

(34)

Appendix C: PROOF OF THEOREM 2

Due to that the data collecting and data transmission of sensor n need to consume energy, if K_n(t) < b_max, sensor n does not collect data, i.e., r_n(t) = 0. $r^{*}_{n}(t)$ from Eq. 22 determines the collecting rate in time slots t. The user utility function O(r_n(t)) is concave, so, $O^{^{\prime }-1}(r_{n}(t))$ and r_n(t) are negative correlation. According to Eq. 22, sensor n does not collect data if

$$ \frac{W_{n}(t)+C\hat{K}_{n}(t)}{V} \geq \nu \geq O^{\prime}(0). $$

(35)

To make sure sensor n not collect data when K_n(t) < b_max, we can set Ω as

$$ {\Omega} = \frac{V\nu}{C} + b^{max}. $$

(36)

Appendix D: PROOF OF THEOREM 3

We use induction to prove (25). At t = 0, Eq. 25 holds. Then, we assume that Eq. 25 holds in time slot t, we prove it holds in time slot t + 1.

1)If sensor n does not collect data, then we can get W_n(t + 1) ≤ W_n(t) ≤ Vν + r^max.

2)If sensor n collects data with rate $r^{*}_{n}(t)$, we have $VO^{\prime }(r^{*}_{n}(t)) = W_{n}(t)+\hat {K}_{n}(t)C = W_{n}(t)-C(K_{n}(t)-{\Omega })$ and we know C(K_n(t) −Ω) ≤ 0, so $W_{n}(t)\leq VO^{\prime }(r^{*}_{n}(t))$. ν is the maximum slope of the user utility function, W_n(t) ≤ Vν. Since $r^{*}_{n}(t)\leq r^{max}$, we have W_n(t + 1) ≤ W_n(t) + r^max ≤ Vν + r^max.

Thus, Eq. 25 is proved.

Appendix E: PROOF OF THEOREM 4

The proof of the M^max bound: If M_n(t) ≤ Vνβ, then M_n(t + 1) ≤ Vνβ + 𝜖 = M^max. Else, the DRC will choose d_n(t) = d^max, so that the queue cannot increase on the next slot, i.e., M_n(t + 1) ≤ M^max. Thus THEOREM 4 is proved.

Appendix F: PROOF OF THEOREM 6

We prove the Theorem 6 by comparing the Lyapunov drift of proposed resource allocation algorithm and another random algorithm. We use π to denote the random algorithm. a^π(t),r^π(t),d^π(t)andh^π(t) denote the optimal solution of the random algorithm. η^π(t) represents the transmission data rate. b^π(t) represents the total energy consumption. Due to the channel state, energy harvesting process changes in i.i.d manner across the time slots. According to Theorem 4.5 in [37], we can get the following inequalities:

$$ \mathbb{E}[\sum\limits_{n\in N}O(r^{\pi}_{n}(t))-\sum\limits_{n\in N}\beta \nu d^{\pi}_{n}(t)]\leq U^{*} + \delta, $$

(37)

$$ \mid\mathbb{E}[\sum\limits_{n\in N}(r^{\pi}_{n}(t)-\eta^{\pi}_{n}(t))]\mid \leq \rho_{1} \delta, $$

(38)

$$ \mid\mathbb{E}[\sum\limits_{n\in N}(h^{\pi}_{n}(t)-b^{\pi}_{n}(t))]\mid \leq \rho_{2}\delta, $$

(39)

where δ is a arbitrarily small parameter, and δ > 0, and ρ₁,ρ₂ are constant scalars.

In each slot time, the proposed algorithm minimizes the right hand side of the Lyapunov drift in Eq. 21. From the Eqs. 18, 19, we can get

$$ \begin{array}{ll} {\Delta}(t)-V\mathbb{E}[\overline{U}(t)]\leq &B + \mathbb{E}[D_{V}(t)|{\Theta}(t)]\\ \leq& B + \mathbb{E}[D^{\pi}_{V}(t)]\\ \leq& B + (\rho_{1} +\rho_{2}+ 1)\delta -VU^{*}, \end{array} $$

(40)

where $\overline {U}$ denotes the user utility of proposed alogorithm. U^∗ denotes the user utility of algorithm π. By setting δ to zero, we can get

$$ {\Delta}(t)-V\mathbb{E}[\overline{U}(t)] \leq B -VU^{*}. $$

(41)

Take expectations on both sides, we have

$$ \frac{L(T-1)-L(0)}{T}-\frac{1}{T}\sum\limits^{T-1}_{t = 0}\mathbb{E}[\overline{U}(t)]\leq B-VU^{*}, $$

(42)

where L(T − 1) and L(0) are finite. Let T →∞, we can get $\overline {U}\geq U^{*} - \frac {B}{V}$. Thus, Theorem 6 is proved.