# Low-Latency Short-Packet Transmission over a Large Spatial Scale
## 3 Optimal Relay Deployment and Resource Allocation
### Properties of the Optimal Resource Allocation
Noticing Equation (6) holds when \(\epsilon_{i}\approx 0\), we force channel capacity \(C_{i}\) to be relatively larger than the coding rate \(\frac{k}{n_{i}}\). If \(C_{i}-\frac{k}{n_{i}}\approx 0\), we obtain \(\epsilon_{i}\approx 0.5\) from Equation (5). For those \(\epsilon_{i}\) being rather large, we can simply remove relay \(R_{i}\), which does not affect the overall error probability. (Thus, 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 the rest of the paper.) By this means, we further shed some light on allocating the blocklength and power to the relays in Theorem 1. When the overall error probability is minimized, the blocklength allocated to a relay should be a decreasing function of its transmission power. According to Theorem 1, we present the following corollary. **Corollary 1**. : _Problem (18) is equivalent to the following optimization problem:_  
\[\exp\left(\frac{\mathcal{G}_{2}^{2}-\mathcal{G}_{1}^{2}}{2}\right)<\frac{\ln( 1+\gamma_{2})+\frac{k\ln 2}{n}}{\ln(1+\gamma_{1})+\frac{k\ln 2}{n}}\frac{\gamma_{2}+1}{ \gamma_{1}+1}\sqrt{\frac{\gamma_{1}}{\gamma_{2}}\frac{\gamma_{1}+2}{\gamma_{2 }+2}}. \tag{20}\]0 \[\exp\left(\frac{\mathcal{G}_{2}^{2}-\mathcal{G}_{1}^{2}}{2}\right)<\frac{\ln( 1+\gamma_{2})+\frac{k\ln 2}{n}}{\ln(1+\gamma_{1})+\frac{k\ln 2}{n}}\frac{\gamma_{2}+1}{ \gamma_{1}+1}\sqrt{\frac{\gamma_{1}}{\gamma_{2}}\frac{\gamma_{1}+2}{\gamma_{2 }+2}}. \tag{20}\]1 (27b) \[\exp\left(\frac{\mathcal{G}_{2}^{2}-\mathcal{G}_{1}^{2}}{2}\right)<\frac{\ln( 1+\gamma_{2})+\frac{k\ln 2}{n}}{\ln(1+\gamma_{1})+\frac{k\ln 2}{n}}\frac{\gamma_{2}+1}{ \gamma_{1}+1}\sqrt{\frac{\gamma_{1}}{\gamma_{2}}\frac{\gamma_{1}+2}{\gamma_{2 }+2}}. \tag{20}\]2 (27c) \[\exp\left(\frac{\mathcal{G}_{2}^{2}-\mathcal{G}_{1}^{2}}{2}\right)<\frac{\ln( 1+\gamma_{2})+\frac{k\ln 2}{n}}{\ln(1+\gamma_{1})+\frac{k\ln 2}{n}}\frac{\gamma_{2}+1}{ \gamma_{1}+1}\sqrt{\frac{\gamma_{1}}{\gamma_{2}}\frac{\gamma_{1}+2}{\gamma_{2 }+2}}. \tag{20}\]3 (27d) \[\exp\left(\frac{\mathcal{G}_{2}^{2}-\mathcal{G}_{1}^{2}}{2}\right)<\frac{\ln( 1+\gamma_{2})+\frac{k\ln 2}{n}}{\ln(1+\gamma_{1})+\frac{k\ln 2}{n}}\frac{\gamma_{2}+1}{ \gamma_{1}+1}\sqrt{\frac{\gamma_{1}}{\gamma_{2}}\frac{\gamma_{1}+2}{\gamma_{2 }+2}}. \tag{20}\]4  
Proof of Corollary 1.: We start the proof by observing that each pair \((n_{t},\gamma_{t})\) is independent in Problem (18) of the others. If we exchange the subscript of \((n_{i},\gamma_{i})\) and \((n_{i},\gamma_{j})\) for \(i\
eq j\), the objective \(\epsilon\) does not change. Thus, we can assume \(n_{1}\geq n_{2}\geq\cdots\geq n_{L}\). Then, from Theorem 1, we know that Problem (18) achieves the minimum value when \(\gamma_{1}\leq\gamma_{2}\leq\cdots\leq\gamma_{L}\). We denote by \(e_{1}^{*}\) and \(e_{2}^{*}\) the optimal value of Problem (18) and Problem (27), respectively. Because the domain of Problem (27) is a part of the domain of Problem (18), \(e_{1}^{*}\leq e_{2}^{*}\). Moreover, \(e_{1}^{*}\) is achieved when \(\gamma_{1}\leq\gamma_{2}\leq\cdots\leq\gamma_{L}\) from Theorem 1, which satisfies (27d) and (27e). Therefore, the optimal solution of Problem (18) falls into the domain of Problem (27). This gives \(e_{2}^{*}\leq e_{1}^{*}\). Thus, \(e_{1}^{*}=e_{2}^{*}\), which completes the proof.