Energy-Efficient Computing Offloading Based on Multi-UAV Dispatch via NOMA in Emergency Communication Networks

Abstract

Unmanned aerial vehicles (UAVs) have become an inevitable choice due to their advantages of flexible movement and rapid deployment for ensuring emergency communication services. The endurance of UAVs may be limited due to the inability to charge in time in emergency communication scenarios. In this paper, we present a non-orthogonal multiple access (NOMA) based multi-UAV dispatching mobile edge computing (MEC) offloading for emergency communication networks, where each UAV is equipped with MEC server to provide computing services for terrestrial users, and NOMA technology is used to increase spectrum utilization and reduce task processing energy consumption. The main goal is to maximize the energy efficiency of UAV-MEC systems by jointly optimizing computing resources, power control, and UAV dispatch in this paper. The system energy efficiency (SEE) optimization problem is complex and non-convex, to solve this, we first adopt the Dinkelbach method to transform the original problem into an equivalent problem. Then, we decompose the equivalent problem into a resource allocation subproblem based on a given UAV dispatching strategy and UAV dispatching subproblem for a given resource allocation strategy. To address the resource allocation problem, we derive closed-form solutions for computing resources and power allocation with the Lagrangian dual method. Next, we propose a three-stage UAV energy-efficient dispatching scheme based on the global K-Means (GKM) algorithm to optimize the dispatching of UAVs. Numerical results demonstrate the effectiveness of the proposed scheme.

Appendices

Appendix 1: The Derivation of Eq. (4)

From (3), for $m = 1$, we obtain

$$O_{1u} [k] = \tau B\log_{2} \left( {1 + \frac{{p_{1u} [k]g_{1u} [k]}}{{\sigma^{2} }}} \right).$$

(47)

For $m = 2$, we have

$$O_{2u} [k] = \tau B\log_{2} \left( {1 + \frac{{p_{2u} [k]g_{2u} [k]}}{{p_{1u} [k]g_{1u} [k] + \sigma^{2} }}} \right),$$

(48)

by analogy, for $m = M_{u}$, we have

$$O_{{M_{u} u}} [k] = \tau B\log_{2} \left( {1 + \frac{{p_{{M_{u} u}} [k]g_{{M_{u} u}} [k]}}{{\sum\nolimits_{i = 1}^{{M_{u} - 1}} {p_{iu} [k]g_{iu} [k]} + \sigma^{2} }}} \right).$$

(49)

Expand $O_{u} [k] = \sum\limits_{m = 1}^{{M_{u} }} {O_{mu} [k]}$, we can obtain

$$O_{u} [k] = O_{1u} [k] + O_{2u} [k] + \cdots + O_{{M_{u} u}} [k].$$

(50)

Substitute (46), (47), and (48) into (49), we have

$$\begin{aligned} O_{u} [k] & = \tau B\left[ {\log_{2} \left( {1 + \frac{{p_{1u} [k]g_{1u} [k]}}{{\sigma^{2} }}} \right)} \right. \\ & \quad \log_{2} \left( {1 + \frac{{p_{2u} [k]g_{2u} [k]}}{{p_{1u} [k]g_{1u} [k] + \sigma^{2} }}} \right) \\ & \left. {\quad + \cdots + \log_{2} \left( {1 + \frac{{p_{{M_{u} u}} [k]g_{{M_{u} u}} [k]}}{{\sum\nolimits_{i = 1}^{{M_{u} - 1}} {p_{iu} [k]g_{iu} [k]} + \sigma^{2} }}} \right)} \right]. \\ \end{aligned}$$

(51)

According to the properties of logarithmic functions, the total throughput of users to UAV u at the kth time slot can be expressed as

$$\begin{aligned} O_{u} [k] & = \tau B\log_{2} \left( {\frac{{\sigma^{2} + p_{1u} [k]g_{1u} [k]}}{{\sigma^{2} }}} \right. \\ & \quad \cdot \frac{{\sigma^{2} + p_{1u} [k]g_{1u} [k] + p_{2u} [k]g_{2u} [k]}}{{p_{1u} [k]g_{1u} [k] + \sigma^{2} }} \\& \quad \left. { \cdot \cdots \cdot \frac{{\sigma^{2} + \sum\nolimits_{i = 1}^{{M_{u} - 1}} {p_{iu} [k]g_{iu} [k]} + p_{{M_{u} u}} [k]g_{{M_{u} u}} [k]}}{{\sum\nolimits_{i = 1}^{{M_{u} - 1}} {p_{iu} [k]g_{iu} [k]} + \sigma^{2} }}} \right) \, \\ & = \tau B\log_{2} \left( {\frac{{\sigma^{2} + \sum\nolimits_{i = 1}^{{M_{u} - 1}} {p_{iu} [k]g_{iu} [k]} + p_{{M_{u} u}} [k]g_{{M_{u} u}} [k]}}{{\sigma^{2} }}} \right) \\ & = \tau B\log_{2} \left( {1 + \sum\nolimits_{i = 1}^{{M_{u} }} {\frac{{p_{iu} [k]g_{iu} [k]}}{{\sigma^{2} }}} } \right) \\ \end{aligned}$$

(52)

The Eq. (4) is achieved.

Appendix 2: Prove Lemma 1

For $\min \left\{ {\tau B\log_{2} \left( {1 + \sum\limits_{m = 1}^{{M_{u} }} {\frac{{p_{mu} [k]g_{mu} [k]}}{{\sigma^{2} }}} } \right),\sum\limits_{m = 1}^{{M_{u} }} {\frac{{\tau f_{mu} [k]}}{{C_{u} }}} } \right\}$, there are three cases:

• (i) $\tau B\log_{2} \left( {1 + \sum\limits_{m = 1}^{{M_{u} }} {\frac{{p_{mu} [k]g_{mu} [k]}}{{\sigma^{2} }}} } \right) > \sum\limits_{m = 1}^{{M_{u} }} {\frac{{\tau f_{mu} [k]}}{{C_{u} }}} ;$

• (ii) $\tau B\log_{2} \left( {1 + \sum\limits_{m = 1}^{{M_{u} }} {\frac{{p_{mu} [k]g_{mu} [k]}}{{\sigma^{2} }}} } \right) < \sum\limits_{m = 1}^{{M_{u} }} {\frac{{\tau f_{mu} [k]}}{{C_{u} }}} ;$

• (iii) $\tau B\log_{2} \left( {1 + \sum\limits_{m = 1}^{{M_{u} }} {\frac{{p_{mu} [k]g_{mu} [k]}}{{\sigma^{2} }}} } \right) = \sum\limits_{m = 1}^{{M_{u} }} {\frac{{\tau f_{mu} [k]}}{{C_{u} }}} .$

Next, we would use contradiction analysis to prove that (iii) holds. We assume that there is another set of solutions besides the optimal solution. Let $\Lambda$ be the objective function of problem P3-2, then we can define the corresponding objective functions for the solution $\{ p_{mu} [k]^{*} ,f_{mu} [k]^{*} \}$ and $\{ p_{mu} [k]]^{\prime},f_{mu} [k]^{\prime}\}$ as $\Lambda^{*}$ and $\Lambda^{\prime}$, respectively. Suppose (i) holds, and $f_{mu} [k]^{*} = f_{mu} [k]{\prime}$, $p_{mu} [k]^{*} \ge p_{mu} [k]{\prime}$, and $\sum\nolimits_{u = 1}^{{\widehat{U}}} {\sum\nolimits_{m = 1}^{{M_{u} }} {p_{mu} [k]^{*} } } > \sum\nolimits_{u = 1}^{{\widehat{U}}} {\sum\nolimits_{m = 1}^{{M_{u} }} {p_{mu} [k]{\prime} } }$. Obviously, the objective function $\Lambda$ decreases with $p_{mu} [k]$ increasing, thus we have $\Lambda^{*} < \Lambda^{\prime}$, the case (i) is not satisfied.

Similarly, suppose (ii) holds, and $p_{mu} [k]^{*} = p_{mu} [k]{\prime}$, $f_{mu} [k]^{*} \ge f_{mu} [k]{\prime}$. It is obvious that the objective function decreases with increasing, thus we have $\Lambda^{*} < \Lambda{\prime}$, the case (ii) is not satisfied.

Based on the above analysis, case (iii) holds, and Lemma 1 is proved.