# Multi-Objective Optimization for EE-SE Tradeoff in Space-Air-Ground Internet of Things Networks
## 3 Problem Solution
### Sub-Problem 3: UAV Position Deployment  
After solving the sub-channel selection and power control, we need to consider the UAV position deployment sub-problem. According to sub-problems 1 and 2, we have \(\mathbf{A}\), \(\mathbf{P}_{I}\), and \(\mathbf{P}_{U}\). On this basis, the objective function of \(\mathcal{P}_{2}\) has been fixed. Like sub-problem 1,we consider maximizing the device transmission rate, so the UAV position deployment sub-problem can be described as  
\[ \begin{split}\mathcal{SP}_{6}&:\quad\max_{\mathbf{D}} \sum_{i}r_{iu}\\ \text{s.t.}&\quad C6,C7,C8.\end{split}  \tag{32}\]  
Obviously, \(\mathcal{SP}_{6}\) is a non-convex problem. In order to solve this problem, we introduce a new variable \(\Theta=\{\theta_{iu}\}\). \(\mathcal{SP}_{6}\) can be redescribed as  
\[ \begin{split}\mathcal{SP}_{7}&:\quad\max_{\mathbf{D}, \Theta}\sum_{i}r_{iu}\\ \text{s.t.}&\quad C6,C7,C8,\\ &\quad C10:\theta_{iu}=\arctan\left(\frac{h_{\text{U}}}{\|\mathbf{ d}_{u}^{\text{U}}-\mathbf{d}_{i}^{\text{I}}\|}\right),\quad\forall i,u.\end{split}  \tag{33}\]  
Then, to address the non-affine constraint C9 [26], we can relax the problem into  
\[ \begin{split}\mathcal{SP}_{8}&:\quad\max_{\mathbf{D}, \Theta}\sum_{i}r_{iu}\\ \text{s.t.}&\quad C6,C7,C8,\\ &\quad C10^{\prime}:\theta_{iu}\leq\arctan\left(\frac{h_{\text{U} }}{\|\mathbf{d}_{u}^{\text{U}}-\mathbf{d}_{i}^{\text{I}}\|}\right),\quad \forall i,u,\end{split}  \tag{34}\]  
where C6, C7, C8, C9\({}^{\prime}\) and objective function are non-convex. Using SCA algorithm for constraint C8, the lower bounder of \(\|\mathbf{d}_{m}^{\text{U}}-\mathbf{d}_{n}^{\text{U}}\|^{2}\) at \(\mathbf{d}_{m}^{\text{U}0}\) and \(\mathbf{d}_{n}^{\text{U}0}\) can be expressed as  
\[ \begin{split}\|\mathbf{d}_{m}^{\text{U}}-\mathbf{d}_{n}^{\text{U} }\|^{2}&\geq-\|\mathbf{d}_{m}^{\text{U}0}-\mathbf{d}_{n}^{\text{ U}0}\|^{2}\\ &+2\Big{(}\mathbf{d}_{m}^{\text{U}0}-\mathbf{d}_{n}^{\text{U}0} \Big{)}^{\text{T}}\Big{(}\mathbf{d}_{m}^{\text{U}}-\mathbf{d}_{n}^{\text{U}} \Big{)}.\end{split}  \tag{35}\]  
In addition, it is also necessary to convert non-convex constraints C6, C7 and objective function into convex constraints. Denote \(\sqrt{\|\mathbf{d}_{u}^{\text{U}}-\mathbf{d}_{i}^{\text{I}}\|^{2}+h_{\text{U} }^{2}}\) as \(x\) and \((1+\phi e^{-\sigma(\theta_{iu}-\phi)})\) as \(y\). We can konw that C6 and C7 are convex for variables \(x\) and \(y\), as shown below. **Proof. ** Obviously, both \(x\) and \(y\) are greater than 0. Under normal circumstances, \(\eta_{\text{NLOS}}\) is much larger than \(\eta_{\text{LOS}}\). Therefore, \(g_{iu}\) can be expressed as  
\[ \begin{split} g_{iu}&=10^{-2\log\left(\frac{4\pi r_ {iu}}{r}\right)+\frac{P_{iu}(p_{\text{NLOS}}-\eta_{\text{LOS}})}{20}-\frac{p_{ \text{NLOS}}}{10}}\\ &=A_{1}\frac{A_{2}}{x^{2}}10^{\frac{A_{3}}{y}},\end{split}  \tag{36}\]  
where \(A_{1}\), \(A_{2}\), and \(A_{3}\) are constants greater than 0. Therefore, \(r_{iu}\) can be expressed as  
\[ \begin{split} r_{iu}&=a_{iu}B\log_{2}\left(1+\frac{ p_{i}S_{iu}}{BN_{0}}\right)\\ &=a_{iu}B\log_{2}\left(1+\frac{p_{i}A_{1}\frac{A_{2}}{x^{2}}10^{ \frac{A_{3}}{y}}}{BN_{0}}\right)\\ &=a_{iu}B\log_{2}\left(1+\frac{C_{1}}{x^{2}}10^{\frac{C_{2}}{y}} \right),\end{split}  \tag{37}\]where \(C_{1}\) and \(C_{2}\) are constants greater than 0. Then, calculate the Hessian matrix \(\mathbf{H}\) of data transmission rate \(r_{iu}\)  
\[\mathbf{H}= \begin{bmatrix}\frac{\partial_{t}^{2}r_{iu}}{\partial x^{2}}&\frac{ \partial_{t}^{2}r_{iu}}{\partial x\partial y}\\ \frac{\partial_{t}^{2}r_{iu}}{\partial y\partial x}&\frac{\partial_{t}^{2}r_{ li}}{\partial y^{2}}\end{bmatrix}