## 3 The Analysis
### Effect of Data Duplication
#### 3.4.1 Minimum Bandwidth Allocation  
When duplication transmission is applied, delay and error probability will be impacted relying on different duplication schemes. Then, the MCS selection and bandwidth allocation can be derived based on Theorem 1 as summarized in Corollary 1 followed by its proof.  
**Corollary 1**: _Consider a URLLC information packet with length of \(k\) bits, and it is transmitted for \(J\) times either consequently in time domain or parallelled in frequency domain under channel quality of \(P\). If the overall delay and error probability after \(J\) times duplication is constraint by \((d_{0},\epsilon_{0})\), there are multiple solutions for MCS selection \((n_{J},M_{J})\) and bandwidth allocation \(W_{J}\) as far as \(W_{J}\) is no less than_  
\[W_{J}\geq J\cdot\frac{k}{d_{0}\cdot R_{2}(P,n_{J},\sqrt[]{\epsilon_{0}},M_{J} )}. \tag{15}\]  
**Proof.** We first consider the duplication in time domain, which means the same information packet will be transmitted for \(J\) times consequently one by one using the same bandwidth. Since the maximum delay for URLLC and the length of information packet \(k\) are both short, we assume that the channel quality \(P\) as well as the coding and modulation scheme \((n_{J},M_{J})\) will be the same for several duplicated transmissions. Let \((d_{t},\epsilon_{t})\) denote the delay and error probability of each transmission, the worst case is that the last transmission is received correctly, therefore the maximum delay \(d_{0}\) and overall error probability \(\epsilon_{0}\) for \(J\) duplications in time domain will be\[d_{0} =J\cdot d_{t}, \tag{16}\] \[e_{0} =\epsilon_{0}^{J}.\]  
In order to fulfill the requirement of \((d_{0},e_{0})\), the delay and error probability of each transmission \((d_{t},\epsilon_{t})\) should be  
\[d_{t} =d_{0}/J, \tag{17}\] \[e_{t} =\sqrt[]{e_{0}}.\]  
By applying Theorem 1, the bandwidth allocation for each transmission \(W_{t}\) should be no less than  
\[W_{t}\geq\frac{k}{d_{t}\cdot R_{2}(P,n_{J},e_{t},M_{J})}=J\cdot\frac{k}{d_{0} \cdot R_{2}(P,n_{J},\sqrt[]{e_{0}},M_{J})}. \tag{18}\]  
Since duplication is made in time domain, the bandwidth will be \(W_{t}\) during \(J\) transmission.  
When considering duplication in frequency domain, the same information packet will be transmitted for \(J\) times in different bandwidth resources at the same time. Let \((d_{f},e_{f})\) denote the delay and error probability of each transmission, then the maximum delay \(d_{0}\) and overall error probability \(\epsilon_{0}\) for \(J\) duplications in frequency domain will be  
\[d_{0} =d_{f}, \tag{19}\] \[\epsilon_{0} =\epsilon_{f}^{J}.\]  
The delay and error probability of each transmission \((d_{f},e_{f})\) in time domain will be  
\[d_{f} =d_{0}, \tag{20}\] \[e_{f} =\sqrt[]{e_{0}}.\]  
Based on Theorem 1, the bandwidth allocation for each transmission, denoted \(W_{f}^{1}\), should be no less than  
\[W_{f}^{1}\geq\frac{k}{d_{f}\cdot R_{2}(P,n_{J},e_{f},M_{J})}=\frac{k}{d_{0} \cdot R_{2}(P,n_{J},\sqrt[]{e_{0}},M_{J})}. \tag{21}\]  
The whole bandwidth used for \(J\) duplications in frequency domain \(W_{f}(J)\) will be  
\[W_{f}\geq J\cdot W_{f}^{1}=J\cdot\frac{k}{d_{0}\cdot R_{2}(P,n_{J},\sqrt[]{e_{ 0}},M_{J})}. \tag{22}\]