3D UAV placement and user association in software-defined cellular networks

Abstract

With the onset of unexpected or temporary problems resulting in degraded user performance, the flexibility and elasticity requirements of future cellular networks may not be fully satisfied by fixed ground base stations. A promising solution for this deficiency is to establish drone cells, which are formed by quickly deploying unmanned aerial vehicles (UAVs) equipped with base stations. Consequently, a UAV placement and user association algorithm for future software-defined cellular networks (SDCN) is proposed in this study. In consideration of the optimal three-dimensional placement of UAVs and the optimal drone cell users’ associations, a utility maximization problem is formulated by utilizing a global view of the SDCN controller. Following mathematical manipulation, the intractable multidimensional problem is transformed into a two-phase algorithm involving the optimal UAV placement altitude-to-radius ratio and the optimal two-dimensional (2D) drone cell horizontal coverage combined with user association. Simulation results indicate the superiority of the proposed algorithm, which increases the average throughput and average utility of all users compared with random, center and 2D UAV placement schemes. By deploying the new design, the maximum average throughput gain can reach up to $36.4\%$.

Appendices

Appendix A

Proof

Define the midpoint of the domain as $\mu _j(0)$, and denote the lower bound of the domain as LA and the upper bound as UA. Then, we have

$$\begin{aligned} \mu _j(0)=\frac{1}{2}(LA+UA). \end{aligned}$$

(19)

Bring $\mu _j(0)$ into Eq. (14). If $\frac{dF(\mu _j(0))}{d\mu _j(0)}=0$, then $\mu _j(0)$ is the root of Eq. (14).

Otherwise, if $\frac{dF(LA)}{dLA}\cdot \frac{dF(\mu _j(0))}{d\mu _j(0)}<0$, then $\mu _j^*\in (LA, \mu _j(0))$, and we set $\mu _j(1)=LA$, $UA_1=\mu _j(0)$; if $\frac{dF(\mu _j(0))}{d\mu _j(0)}\cdot \frac{dF(UA)}{dUA}<0$, then $\mu _j^*\in (\mu _j(0), UA)$, and we set $\mu _j(1)=\mu _j(0)$, $UA_1=UA$.

Note that $[LA_1, UA_1]$ is a new domain of $\mu _j$, and the length of the new domain is half that of the original one. Through this analogy, we obtain a series of domains according to

$$\begin{aligned}{}[LA, UA]\supset [LA_1, UA_1]\supset \ldots \supset [LA_n, UA_n]. \end{aligned}$$

(20)

Since each new domain is half the length of the previous one, the length of $[LA_n, UA_n]$ is

$$\begin{aligned} UA_n-LA_n=\frac{1}{2^n}(UA-LA). \end{aligned}$$

(21)

As $n\rightarrow \infty$, the length of the domain $UA_n-LA_n\rightarrow 0$.

Finally, we can conclude that the upper bound and the lower bound will converge to the point $\mu _j^*$ as $n\rightarrow \infty$.

Appendix B

Proof

Due the fact that $f_{convex}$ is differentiable and strictly convex, $f_{convex}$ must satisfy the first order condition [26, 27],

$$\begin{aligned} &f_{convex}\big (V(n+1)\big )\!\ge \! f_{convex}\big (V(n)\big )\nonumber \\ &\quad +\,\nabla f_{convex}\big (V(n)\big )\big (V(n\!+\!1)\!-\!V(n)\big )^{T}. \end{aligned}$$

(22)

To prove the monotonicity, we first assume that $V(n+1)\ge V(n)$. We have

$$\begin{aligned} f\big (V(n+1)\big )&= {} f_{concave}\big (V(n+1)\big ) +f_{convex}\big (V(n+1)\big )\nonumber \\ & \ge\, {} f_{concave}\big (V(n+1)\big )+f_{convex}\big (V(n)\big )\nonumber \\&\quad +\,\!\nabla f_{convex}\big (V(n)\big )\big (V(n\!+\!1)\!-\!V(n)\big )^{T}. \end{aligned}$$

(23)

Based on Eq. (18), we have

$$\begin{aligned}&f_{concave}\big (V(n+1)\big )+\nabla f_{convex}\big (V(n) \big )*\big (V(n+1)\big )^T\nonumber \\& \ge\, {} f_{concave}\big ((V(n))\big )+\nabla f_{convex}\big (V(n)\big )*\big (V(n)\big )^Ts, \end{aligned}$$

(24)

and after transposition, we have

$$\begin{aligned} & f_{concave}\big (V(n+1)\big )\ge f_{concave}\big ((V(n))\big )\nonumber \\ &-\nabla f_{convex}\big (V(n)\big )*\Big (\big (V(n+1)-V(n)\big )\Big )^T. \end{aligned}$$

(25)

Combining Eqs. (22) with (20) yields

$$\begin{aligned} f\big (V(n+1)\big )\ge f_{concave}\big (V(n)\big )+f_{convex}\big (V(n)\big ), \end{aligned}$$

(26)

after which we have

$$\begin{aligned} f\big (V(n+1)\big )\ge f\big (V(n)\big ). \end{aligned}$$

(27)

Thus, Eq. (17) is strictly monotonically increasing on the generated sequence $\{V(n)\}$.

If ${\mathcal {S}}_V\ne \varnothing$, it is easy to verify that ${\mathcal {S}}_V$ is closed and bounded from the constraints of (17). As shown in Remark 7 in [26], we can show that as $n\rightarrow \infty$, $||V_{n+1}-V_{n}||\rightarrow 0$, and $\{V(n+1)\}$ converges (the limit points of $\{V(n)\}$ are expressed as $\overset{\infty }{V}$).

Finally, we can conclude that the sequence of {$V(n+1)$} from the CCCP converges to $\overset{\infty }{V}$.

Appendix C

Proof of Eq. 16

Note that $F(V)=f_{concave}(V)+f_{convex}(V)$ and we define $f_{concave}(V)\triangleq L_1(r_j)$ and $f_{convex}(V)\triangleq -L_2(o_{ij})$ in the paper. To proof Eq. (), we just need to prove 16$f_{convex}(V)$ is convex and $f_{concave}(V)$ is concave. To do this, the proof is divided into two parts.

Part 1 (Prove that $F_{convex}(V)$ is convex):

Choose any $o_{ij}$, where $o_{ij}\in$dom${{\mathcal {S}}}_V$, and dom${{\mathcal {S}}}_V$ is convex (i.e., an interval).

Then, we have

$$\begin{aligned} {f_{convex}(o_{ij})}&{=-\sum _i\sum _j{\mathrm{U}}(\sum _i o_{ij})}\nonumber \\&{=-\sum _i\sum _j\log (o_{ij}).} \end{aligned}$$

(28)

The derivative of Eq. (2) is

$$\begin{aligned} {\nabla ^2 f_{convex}(o_{ij})}&{=(-\frac{1}{\sum _i\sum _j(o_{ij})})'}\nonumber \\&{=\frac{1}{\sum _i\sum _j(o_{ij})^2}\ge 0,} \end{aligned}$$

(29)

where $\nabla ^2 f_{convex}(o_{ij})$ is the second derivative of $f_{convex}(o_{ij})$.

Based on the second-order conditions [27], i.e. $\nabla ^2 f_{convex}(o_{ij})\ge 0$, we conclude that $f_{convex}(V)$ is convex.

Part 2 (Prove that $F_{concave}(V)$ is concave):

Choose any $r_{j}$, where $r_{j}\in$dom${{\mathcal {S}}}_V$, and dom${{\mathcal {S}}}_V$ is convex (i.e., an interval).

Then, we have

$$\begin{aligned} {f_{concave}(r_j)}&{=\sum _i\sum _j{\log }(B)+\sum _i\sum _j{\log } {\Bigg (}{\mathrm{{log}}}{\bigg (}\frac{P_{{T_D},j}}{\sigma ^2}{\bigg )}}\nonumber \\&\quad {-2{\log }{\bigg (}\frac{4\pi f_c}{c}{\bigg )}-2{\mathrm{{log}}}{\Big (}\sqrt{(1+\mu _j^{*2})r_j^2} {\Big )}}\nonumber \\&\quad {-\frac{1}{10}P_{\mu _j^*}\eta _{Los}-\frac{1}{10}{\Big (}1 -P_{\mu _j^*}{\Big )}\eta _{NLos}{\Bigg )}.} \end{aligned}$$

(30)

The derivative of Eq. (3) is

$$\begin{aligned} {\nabla ^2 f_{concave}(r_{j})=}&{-\frac{1}{\sum _jr^2_{j}{\Big (}\log {\Big (}\sqrt{(1+\mu _j^{*2})r_j^2}{\Big )}{\Big )}^2}}\nonumber \\&{-\frac{1}{\sum _jr_j^2\log {\Big (}\sqrt{(1+\mu _j^{*2})r_j^2}{\Big )}}.} \end{aligned}$$

(31)

We numerically construct the graph of Eq. (31) based on the interval domain $d_{ij}^h\le r_j\le (F(\mu _j^*))^{\frac{1}{2}}$ in Fig. 12 for the high-rise urban environment, since we can not judge whether the Hessian is positive or not directly.

Fig. 12

The function of $\nabla ^2 f_{concave}(r_{j})$ versus $r_j$

The results in Fig. 12 indicate that the second-order conditions [27] are satisfied, i.e. $\nabla ^2 f_{concave}(r_{j})\le 0$, which means that $F_{concave}(V)$ is concave.

Appendix D

Proof that $L_2(o_{ij})$ is differentiable with respect to $o_{ij}$

From “Appendix C”, we know that function $f_{convex}=-L_2(o_{ij})$ is differentiable at any point $o_{ij}$ in dom${{\mathcal {S}}}_V$. For simplicity, $f_{o_{ij}}$ is used to represent the value of $L_2(o_{ij})$. Thus, the increments $\varDelta o_{ij}$ and $\varDelta f_{o_{ij}}$ exist and satisfy

$$\begin{aligned} {\lim _{\varDelta o_{ij}\rightarrow 0}\frac{\varDelta f_{o_{ij}}}{\varDelta o_{ij}}=\nabla f_{convex}(o_{ij}),} \end{aligned}$$

(32)

where $\varDelta$ is the incremental symbol and $\nabla$ is the derivative symbol.

Theorem 1

The sufficient and necessary condition for the existence of the limitAof the functionf(x) is that$f(x)=A+\alpha$, where$\alpha$is infinitesimal [31].

In our problem, Theorem 1 can be expressed as: The sufficient and necessary condition for the existence of the limit $\overset{\infty }{o}_{ij}$ of function $f_{convex}(o_{ij})$ is that $f_{convex}(o_{ij})=\overset{\infty }{o}_{ij}+\alpha$, where $\alpha$ is infinitesimal. Thus, Eq. (32) can be rewritten as

$$\begin{aligned} {\varDelta f_{o_{ij}}=\nabla f_{convex}(o_{ij})\varDelta o_{ij}+\alpha \varDelta o_{ij}.} \end{aligned}$$

(33)

Since $\alpha \varDelta o_{ij}=0(\varDelta o_{ij})$, and $\nabla f_{convex}(o_{ij})$ is independent of $\varDelta o_{ij}$, Eq. (33) is equal to

$$\begin{aligned} {\varDelta f_{o_{ij}}=\nabla f_{convex}(o_{ij})\varDelta o_{ij}+0(\varDelta o_{ij}).} \end{aligned}$$

(34)

Thus, we conclude that $f_{convex}$ is differentiable at any $o_{ij}$ in dom${{\mathcal {S}}}_V$, which means that $L_2(o_{ij})$ is differentiable with respect to $o_{ij}$.