## 3 Problem Formulation and Solution
Due to the complex form of the denominator, the whole function is non-convex in (10a). We introduce variable \(\tilde{R}_{ij}\) and its constraint, so that the objective function (11a) is convex now. Note that the constraint of \(\tilde{R}_{ij}\) is also a relaxed version, and it is easy to prove that constraint (11d) turns into the tight constraint when the problem (11) obtains the optimal solution, which means that \(\tilde{R}_{ij}\) is equal to \(R_{ij}\) when the optimal problem is solved. Problem (11) is still not a convex problem due to constraints (11c) and (11d) being neither concave nor convex. For (11c), we can make the left-hand side linearized via the first-order Taylor expansion, as shown in (12). \[\min_{\mathbf{w},\theta,\mathbf{v},\mathbf{z}} \sum_{k=1}^{K}\sum_{j=1}^{N}p_{k}\Bigg{(}\frac{z_{0k}L_{k}}{R_{0}}+ \sum_{i=1}^{M}q_{ij}\frac{z_{ik}L_{k}}{R_{ij}(\mathbf{w}_{ij},\theta_{ij})} \Bigg{)}, \tag{10a}\]2  
where \(\theta_{ij}^{[\ell]}\) represents the [PERSON] expansion point. We now consider transforming inequality (11d) into a convex constraint via DC method. The basic idea of DC method is, first, to transform the original problem into the difference of two convex functions, and, then, to apply an approximation method to eliminate the non-convexity. For inequality (11d), we introduce a strongly convex function, which is added to and then taken away from the right-hand side of the original inequality, as shown in inequality (13). Inspired by [21], we choose a quadratic form as the introduced strongly convex function. We write inequality (11d) as  
\[ \begin{split}\tilde{R}_{ij}\leq&\text{log}_{2}\! \left(1+\frac{|\mathbf{h}_{ij}^{\mathrm{H}}\mathbf{w}_{ij}|^{2}}{\sum\limits_ {m\in\mathcal{M},m\
eq i}\sum\limits_{n\in\mathcal{A}_{m}\
eq j}|\mathbf{h}_{ mj}^{\mathrm{H}}\mathbf{w}_{mn}|^{2}+\sigma^{2}}\right)+\xi_{ij}(||\mathbf{w}||^{2}+ ||\boldsymbol{\theta}||^{2})\\ &-\xi_{ij}(||\mathbf{w}||^{2}+||\boldsymbol{\theta}||^{2}), \hskip 14.226378 pt\forall i\in\mathcal{M},\forall j\in\mathcal{N}.\end{split}  \tag{13}\]  
In inequality (13), \(\xi_{ij}\) is a positive constant we set. It is proved in [21] that, if \(\xi_{ij}\) is set to be positive and large enough, function \(\xi_{ij}(||\mathbf{w}||^{2}+||\boldsymbol{\theta}||^{2})\) will have strong convexity. Although the form of \(\tilde{R}_{ij}\) is non-convex and non-concave, the sum of \(\tilde{R}_{ij}\) and the introduced function would be a convex function, as long as the convexity of the introduced function is strong enough. In this way, a form of difference between two convex functions is constructed. Let  
\[ \begin{split} f_{ij}(\mathbf{w},\boldsymbol{\theta})=& \text{log}_{2}\!\left(1+\frac{|\mathbf{h}_{ij}^{\mathrm{H}}\mathbf{w}_{ij}|^{2 }}{\sum\limits_{m\
eq i}\sum\limits_{n\in\mathcal{A}_{m},n\
eq j}|\mathbf{h}_{ mj}^{\mathrm{H}}\mathbf{w}_{mn}|^{2}+\sigma^{2}}\right)+\xi_{ij}(||\mathbf{w}||^{2}+|| \boldsymbol{\theta}||^{2})\\ =& R_{ij}(\mathbf{w},\boldsymbol{\theta})+\xi_{ij}(|| \mathbf{w}||^{2}+||\boldsymbol{\theta}||^{2}),\end{split}  \tag{14}\]  
then constraint (13) is changed into  
\[\tilde{R}_{ij}\leq f_{ij}(\mathbf{w},\boldsymbol{\theta})-\xi_{ij}(||\mathbf{ w}||^{2}+||\boldsymbol{\theta}||^{2}),\hskip 14.226378 pt\forall i\in \mathcal{M},\forall j\in\mathcal{N}. \tag{15}\]  
Constraint (15) is in the form of the difference of two convex functions. To actually eliminate the non-convexity, we need to apply an approximation method, as it is noted earlier. The factor leading constraint (15) to be non-convex is \(f_{ij}(\mathbf{w},\boldsymbol{\theta})\) because it is a convex function on the right-hand side of \"\(\leq\)\". We tighten constraint (15) by replacing \(f_{ij}(\mathbf{w},\boldsymbol{\theta})\) with its lower-bound Taylor linearization. The first-order Taylor expansion for \(f_{ij}(\mathbf{w}_{ij},\boldsymbol{\theta}_{ij})\) can change the original formula into a convex constraint.