# Low-Latency Short-Packet Transmission over a Large Spatial Scale
## 3 Optimal Relay Deployment and Resource Allocation
### Properties of the Optimal Resource Allocation  
Problem (18) is challenging due to its nonlinear constraint Equation (18b). Therefore, we relax the problem's constraints. In particular, we present the following theorem. **Theorem 1**. : _If \(\ln(1+\gamma_{i})-\frac{k\ln 2}{n_{i}}>\max\left\{\frac{1}{k\ln 2},\frac{1}{n_{i} }\right\}\), the \(\epsilon\) in Problem (18) achieves its minimum when all the \(\gamma_{i}\) are greater than a constant \(\gamma_{0}\) and the sequence \(n_{i}\) and \(\gamma_{i}\) is in reverse order, i.e., for any \(n_{i}\geq n_{j}\), we have \(\gamma_{i}\leq\gamma_{j}\)._  
Proof of Theorem 1.: We use the notation \(\varphi(\gamma,n)\) to denote \(Q(\mathcal{G}(\gamma,n))\). For any \(n\) and any \(\gamma_{1}\), \(\gamma_{2}\) satisfying the constraints of Problem (18) and \(\gamma_{1}>\gamma_{2}\), let \(\chi(n)=\varphi(n,\gamma_{2})-\varphi(n,\gamma_{1})\). We first prove that \(\chi(n)\) is a decreasing function in its domain. We present the derivative of the function \(\chi(n)\) as:  
\[\chi^{{}^{\prime}}(n)=\frac{1}{\sqrt{2\pi}}\left(-\exp\left(-\frac{\mathcal{ G}_{2}^{2}}{2}\right)\frac{\partial\mathcal{G}_{2}}{\partial n}+\exp\left(- \frac{\mathcal{G}_{1}^{2}}{2}\right)\frac{\partial\mathcal{G}_{1}}{\partial n }\right). \tag{19}\]  
We next show \(\chi^{{}^{\prime}}(n)<0\), which is equivalent to the following inequation by substituting \(\mathcal{G}_{n}^{{}^{\prime}}\) and \(\mathcal{G}_{\gamma}^{{}^{\prime}}\) in Appendix A.  
\[\exp\left(\frac{\mathcal{G}_{2}^{2}-\mathcal{G}_{1}^{2}}{2}\right)<\frac{\ln( 1+\gamma_{2})+\frac{k\ln 2}{n}}{\ln(1+\gamma_{1})+\frac{k\ln 2}{n}}\frac{\gamma_{2}+1}{ \gamma_{1}+1}\sqrt{\frac{\gamma_{1}}{\gamma_{2}}\frac{\gamma_{1}+2}{\gamma_{2 }+2}}. \tag{20}\]  
Because \(\gamma_{1}>\gamma_{2}\), we derive Equation (21) and Equation (22). \[\frac{\ln(1+\gamma_{2})}{\ln(1+\gamma_{1})}<\frac{\ln(1+\gamma_{2})+\frac{k\ln 2 }{n}}{\ln(1+\gamma_{1})+\frac{k\ln 2}{n}}, \tag{21}\]\[\frac{\gamma_{2}+1}{\gamma_{1}+1}\sqrt{\frac{\gamma_{1}}{\gamma_{2}}\frac{\gamma_{ 1}+2}{\gamma_{2}+2}}>1. \tag{22}\]  
To prove Equation (20), we let \(\zeta(\gamma)=\exp\Bigl{(}-\frac{\mathcal{G}^{2}(\gamma,n)}{2}\Bigr{)}\ln(1+\gamma)\). By calculating its derivative, we obtain:  
\[\zeta^{{}^{\prime}}(\gamma)=\exp\Bigl{(}-\frac{\mathcal{G}^{2}}{2}\Bigr{)} \Bigl{(}\frac{1}{1+\gamma}-\mathcal{G}\mathcal{G}^{{}^{\prime}}_{\gamma}\ln(1+ \gamma)\Bigr{)}. \tag{23}\]  
Because \(\ln(1+\gamma)-\frac{k\ln 2}{n}>\frac{1}{n}\), we have:  
\[\mathcal{G}\mathcal{G}^{{}^{\prime}}_{\gamma}\ln(1+\gamma) >\biggl{(}\ln(1+\gamma)-\frac{k\ln 2}{n}\biggr{)}n\biggl{(}\frac{( \gamma+1)^{2}}{\gamma(\gamma+2)}\biggr{)}^{\frac{1}{2}}\frac{\ln(1+\gamma)}{ \sqrt{\gamma(\gamma+2)}} \tag{24}\] \[>\frac{\gamma+1}{\gamma(\gamma+2)}\ln(1+\gamma)\] \[>\frac{1}{\gamma+1}.\]  
The last step holds when \(\gamma>\gamma_{0}\), where \(\gamma_{0}\) is the solution of the function \(\ln(1+\gamma)=\frac{\gamma(\gamma+2)}{(\gamma+1)^{2}}\) and \(\gamma_{0}\approx 1.22\). Now, we proved that \(\zeta(\gamma)\) decreases when \(\gamma>\gamma_{0}\). Thus, \(\zeta(\gamma_{1})<\zeta(\gamma_{2})\), which leads to:  
\[\exp\Biggl{(}\frac{\mathcal{G}_{2}^{2}-\mathcal{G}_{1}^{2}}{2}\Biggr{)}<\frac {\ln(1+\gamma_{2})}{\ln(1+\gamma_{1})}. \tag{25}\]  
Combining Equations (21), (22), and (25), we can assert that Equation (20) holds, which gives us \(\chi^{{}^{\prime}}(n)<0\). Now, let \(n_{1}>n_{2}\). We have \(\chi(n_{1})<\chi(n_{2})\), i.e.,  
\[\varphi(n_{1},\gamma_{2})+\varphi(n_{2},\gamma_{1})<\varphi(n_{1},\gamma_{1}) +\varphi(n_{1},\gamma_{1}), \tag{26}\]  
in which \(\gamma_{1}>\gamma_{2}\). According to Equation (26), it is sufficient to show that when \(n_{i}\) and \(\gamma_{i}\) are in reverse order, the error probability is smaller, which completes the proof. We consider the condition \(\ln(1+\gamma_{i})-\frac{k\ln 2}{n_{i}}>\max\Bigl{\{}\frac{1}{k\ln 2},\frac{1}{n_{i}} \Bigr{\}}\) in Theorem 1 to ensure the error probability is within a reasonable range.