## 4 Resource Allocation Design for Delay Minimization  
Problem \(P0\) is non-convex due to the binary constraints, the coupling effect between variables and the min-max form of the objective function. Additionally, it is a mixed integer non-linear programming (MINLP) which is NP-hard. Thus, it is a challenge to find its solution. In order to solve the optimization problem, a low-complexity iterative algorithm is proposed. In the problem \(P0\), the objective function, Equations (16)-(18) and (22) are non-convex. Next, we solve them in turn. First, for the Equation (22), it is a binary constraint, which can be equivalent to  
\[0\leq\mu_{k}\leq 1,\ \forall k\in\mathcal{K}, \tag{23}\] \[\mu_{k}(\mu_{k}-1)\geq 0,\ \forall k\in\mathcal{K}. \tag{24}\]  
Note that optimization variable \(u_{k}\) is a continuous value between \(0\) and \(1\). However, Equation (24) is a non-convex. In order to deal with the Equation (24), we reformulate problem \(P0\) as  
\[P1: \min_{\mathbf{\mu},\mathbf{\
u},\mathbf{p},\mathbf{f},\mathbf{\phi}}\max_{k\in \mathcal{K}}\mu_{k}(L\frac{c_{k}D_{k}}{f_{k}}+\frac{d}{v_{k}})-\eta\sum_{k\in \mathcal{K}}\mu_{k}(\mu_{k}-1), \tag{25}\] \[s.t. \ \eqref{eq:P0}-\eqref{eq:P0},\eqref{eq:P0}\]  
where \(\eta\) is usually a constant and represents a penalty factor for penalizing the objective function that any \(\mu_{k}\) is not equal to \(0\) or \(1\)[24]. In order to solve the min-max form of the objective function, the variable \(T\) is introduced. Therefore, the problem \(P1\) can be reformulated as  
\[P2:\min_{\mathbf{\mu},\mathbf{\
u},\mathbf{\sigma},\mathbf{f},\mathbf{\phi},T}T, \tag{26}\] \[s.t. \mu_{k}(L\frac{c_{k}D_{k}}{f_{k}}+\frac{z}{\
u_{k}})-\eta\sum_{k\in \mathcal{K}}\mu_{k}(\mu_{k}-1)\leq T,\] (27) \[(\ref{eq:P1})-(\ref{eq:P2}), \tag{23}\]  
Note that although we introduce the variable \(T\) to solve the min-max form of the objective function, it also brings a new non-convex Equation (27). Due to non-convex Equations (16)-(18) and (27), the problem \(P2\) is still difficult to solve. For Equation (16), its non-convexity comes from the form \(x/y\) of the second term. By replacing the variable \(p_{k}\) with a new variable \(q_{k}=\sqrt{p}_{k}\), the Equation (16) can be rewritten as  
\[L\frac{\alpha_{k}}{2}f_{k}2c_{k}D_{k}+\frac{dq_{k}^{2}}{\
u_{k}} \leq e_{k\max}. \tag{28}\]  
It is obvious that the Equation (28) is convex. Thus, the optimization problem \(P2\) can be rewritten as  
\[P3:\min_{\mathbf{\mu},\mathbf{\
u},\mathbf{\sigma},\mathbf{f},\mathbf{\phi},T}T, \tag{29}\] \[s.t. \sigma_{h,k}^{2}(-\frac{1}{\log(1-\delta_{0})}\frac{\hat{\gamma}_ {k}\sigma^{2}}{q_{k}^{2}\sigma_{h,k}^{2}}-2)\leq\Big{|}\hat{h}_{k}\Big{|}^{2},\] (30) \[\mu_{k}(\mu_{k}-1)\geq 0,\ \forall k\in\mathcal{K}. \tag{24}\]0  
Note that the problem \(P3\) is a non-convex quadratically constrained quadratic program (QCQP). By introducing variables \(\kappa\) and \(\mathbf{\Lambda}\), the problem \(P3\) can be rewritten as  
\[\mu_{k}(\mu_{k}-1)\geq 0,\ \forall k\in\mathcal{K}. \tag{24}\]1 \[\mu_{k}(\mu_{k}-1)\geq 0,\ \forall k\in\mathcal{K}. \tag{24}\]2 (32) \[\mu_{k}(\mu_{k}-1)\geq 0,\ \forall k\in\mathcal{K}. \tag{24}\]3 (33) \[\mu_{k}(\mu_{k}-1)\geq 0,\ \forall k\in\mathcal{K}. \tag{24}\]4 (34) \[\mu_{k}(\mu_{k}-1)\geq 0,\ \forall k\in\mathcal{K}. \tag{24}\]5  
where \(\mathbf{\tilde{\phi}}=\left[ \begin{array}{c}\mathbf{\phi}\\ \kappa\end{array} \right]\), \(\Lambda=\mathbf{\tilde{\phi}}\mathbf{\tilde{\phi}}^{H}\), \(\mathbf{\hat{\Xi}}_{k}=\left[ \begin{array}{cc}\hat{\mathbf{\alpha}}_{k}^{H}&\hat{ \mathbf{\alpha}}_{k}\hat{h}_{dk}^{H}\\ \hat{h}_{dk}\hat{h}_{k}^{H}&0\end{array} \right]\), and \(\text{Rank}(\mathbf{\Lambda})\) represents the rank of matrix \(\mathbf{\Lambda}\). Due to non-convex Equations (27), (32) and (34), problem \(P4\) still cannot be solved directly. For the Equation (34), SDR can be invoked to relax it.