## 5 Analysis  
In this section, we first present an analysis on the carryover epidemic model and then describe the analytical results on the number of time slots required for the entire network infection. In the following Lemma, we first derive the cumulative number of infected nodes when the carryover epidemic model is used. **Lemma 1**. : _In carryover epidemic model, the cumulative number of infected nodes from the 0 th time slot to the \(\tau\)th time slot is \(I_{\tau}-q_{\tau}\)._  
Proof. : Recall that \(I_{t+1}-I_{t}+q_{t}\) is the infection quota in the \(t+1\)th time slot (\(t\geq 1\)) in carryover epidemic model, where \(I_{1}-I_{0}\) is the infection quota in the first time slot and \(q_{t}\) is the infection quota deficit in the \(t\)th time slot. The number of infected nodes in a time slot is the difference between the infection quota and the quota deficit in a time slot. Hence, the cumulative number of infected nodes from the 0 th time slot to the \(\tau\)th time slot is given by:  
\[I_{0}+I_{1}-I_{0}-q_{1}+\sum_{t=1}^{\tau-1}I_{t+1}-I_{t}+q_{t}-q_{t+1}=I_{\tau }-q_{\tau}\]  
\(\square\)  
If the entire network infection is completed in the \(\tau\)th time slot,  
By the definition of discrete Poisson distribution, \(\text{Pr}(X=j)\) is given by:  
\[\text{Pr}(X=j)=\frac{e^{\mu}\mu^{j-1}}{(j-1)!}\]  
and \(\text{E}[X]=\mu\). Then the probability that the entire network infection is completed at most the \(\tau\)th time slot is calculated as \(\sum_{j=1}^{\tau}\text{Pr}(X=j)\). Thus, 1-\(\sum_{j=1}^{\tau}\text{Pr}(X=j)\) is the probability that more than \(\tau\) time slots are consumed to infect all nodes in the network. Finally, we derive the Chernoff bounds on the number of time slots required for the entire network infection. **Lemma 2**. : _If \(j>\tau\) and \(\mu\leq j(\frac{\tau}{j-\tau})\ln\frac{j}{\tau}\), then_  
\[\text{Pr}(X\geq j)\leq\frac{\tau^{j}}{j^{j}}e^{\mu(\frac{j}{\tau}-1)}.\]  
Proof. : For any \(y>0\), the Markov's inequality is given by:  
\[\text{Pr}(X\geq j) = \text{Pr}(\text{e}^{yX}\geq\text{e}^{yj})\] \[\leq \frac{\text{E}[\text{e}^{yX}]}{\text{e}^{yj}}\]  
Since \(E[\text{e}^{yX}]=\text{e}^{\mu(\text{e}^{y}-1)}\) is the moment generating function of the Poisson discrete random variable \(X\), we have  
\[\text{Pr}(X\geq j)\leq\text{e}^{\mu(\text{e}^{y}-1)-yj}\]  
Given that \(j>\tau\) and \(\mu\leq j(\frac{\tau}{j-\tau})\ln\frac{j}{\tau}\), by selecting \(y=\ln(\frac{j}{\tau})>0\), we have  
\[\text{Pr}(X\geq j)\leq\frac{\tau^{j}}{j^{j}}\text{e}^{\mu(\frac{j}{\tau}-1)}\leq 1\]  
We investigate how \(j\) affects the Chernoff bound of Pr(\(X\geq j\)) when \(\tau=10\). As shown in Figure 1, the Chernoff bound of Pr(\(X\geq j\)) decreases as \(j\) increases. From this observation, as \(j\) is away from \(\tau\), it decays the Chernoff bound on the probability that the number of time slots required for the entire network infection is at least \(j\). Given a fixed value of \(j\), the Chernoff bound of Pr(\(X\geq j\)) increases as \(\mu\) increases. This means that the Chernoff bound of Pr(\(X\geq j\)) rises as an average number of slots required for the entire network infection is close to \(\tau\). **Lemma 3**. : _If \(j<\tau+1\) and \(\mu\geq j(\frac{\tau+1}{j-\tau-1})\ln(\frac{j}{\tau+1})\), then_  
\[\text{Pr}(X\leq j)\leq\frac{(\tau+1)^{j}}{j^{j}}\text{e}^{\mu(\frac{j}{\tau+1}- 1)}.\]  
Proof. : For any \(y<0\), the [PERSON]'s inequality is given by:  
\[\text{Pr}(X\leq j) = \text{Pr}(\text{e}^{yX}\geq\text{e}^{yj})\] \[\leq \frac{\text{E}[\text{e}^{yX}]}{\text{e}^{yj}}\]  
Since \(E[\text{e}^{yX}]=\text{e}^{\mu(\text{e}^{y}-1)}\) is the moment generating function of the Poisson discrete random variable \(X\), we have  
\[\text{Pr}(X=j)=\frac{e^{\mu}\mu^{j-1}}{(j-1)!}\]0  
Given that \(j<\tau+1\) and \(\mu\geq j(\frac{\tau+1}{j-\tau-1})\ln(\frac{j}{\tau+1})\), by selecting \(y=\ln(\frac{j}{\tau+1})<0\), we have  
\[\text{Pr}(X=j)=\frac{e^{\mu}\mu^{j-1}}{(j-1)!}\]1  
We investigate how \(j\) affects the Chernoff bound of Pr(\(X\leq j\)) when \(\tau=10\).