# ATWin: An Improved and Detailed Startup Model of TTP/C
## 4 Formal Demonstration of the Validity of the ATWin Model
### The Ability of Contention Eliminating
All receiving nodes in set \(S_{l}(n)\) compulsorily discard the first valid frame and set the contention status flag to be true, even if no contention occurs in the \(n\)-_th_ round of startup. All nodes with the true contention status flag should start to reenter the listening state at the end of ATW. For any node \(j\in\mathsf{S}_{l}(n)\), it must also belong to \(\mathsf{S}_{l}(n+1)\). This means that if node \(j\) does not obtain authentication to send frames in the \(n\)-th round of startup, it still cannot obtain the sending authentication in the (\(n\)+\(1\))-th startup round. Proof. : As shown in Figure (b)b, node \(i\) (\(i\in\mathsf{S}_{l}(n)\)) obtains authentication to send frames and to perform its ATWin at the time \(T^{i}_{listen}(n)\). Then it enters the PRP phase and arrives at the decision time point \(T^{i}_{start}(n)\) after the PRP. Node \(i\) needs to perform a TDMA round after sending frames successfully as well as perform clique detection in the PSP phase. In this process, the TDMA round will firstly take \((n-1)\Delta_{slot}\) time, and the clique detection operation takes \(\Delta_{PSP}\) time. Node \(i\) then receives the CB error and waits for \(\Delta^{k}_{startup}\) to try the cold-start again. If the receiving node \(j\) detects the contention within the ATW, it must restart the listening timeout timer immediately at \(T^{j}_{start}(n)\), which is the end of the ATW. Since there must be a head node in the (\(n\)+\(1\))-_th_ round, a node with the earliest timeout time of the priority timer will start first. It is discussed whether the node may belong to the set \(\mathsf{S}_{l}(n)\) as follows. We assume that the head node \(hd(n+1)\) in the (\(n\)+\(1\))-_th_ of startup belongs to set \(\mathsf{S}_{l}(n)\). For any node \(j\in\mathsf{S}_{l}(n)\), Formula (18) must hold (see Appendix C for detailed derivation). \[T^{j}_{listen}(n+1)-TR^{i}_{j}(n+1)\geq\Delta_{slot}+\Delta_{frame} \tag{18}\]  
For any node in set \(\mathsf{S}_{l}(n)\), its timeout time of the priority timer is greater than that of any node in \(\mathsf{S}_{l}(n)\). The node in \(\mathsf{S}_{l}(n)\) does not obtain authentication to send frames before the nodes in set \(\mathsf{S}_{l}(n)\); thus, the assumption is valid. As shown in Figure (b)b, node \(j\) inevitably receives a cold-start frame from node \(i\) at \(TR^{i}_{j}(n+1)\) before \(T^{j}_{listen}(n+1)\). The node in set \(\mathsf{S}_{l}(n)\) is still the receiving node in the (\(n\)+\(1\))-_th_ round of startup because of the timeout value of the listening timeout timer. **Theorem 4**. : _In the n-th round of startup, if node \(i(i\in\mathrm{S}_{z}(n))\) starts and enters into the listening state during the contention, it must belong to \(\mathrm{S}_{l}(n+1)\)._  
Proof. : All nodes in set \(\mathrm{S}_{z}(n)\) are inactive before the head node \(hd(n)\) node sends frames. If the node \(i\) of set \(\mathrm{S}_{z}(n)\) enters the listening state before \(\left( \begin{array}{c}{Thd(n)}\\ {T}_{listen}^{hd(n)}(n)+\Delta_{prop}\end{array} \right)\), it inevitably receives a cold-start frame from the head node in the _n-th_ round. Meanwhile, if the node happens to receive a cold-start frame, according to ATwin, it must discard the frame and set the contention status flag. If it does not receive a valid frame, it still needs to set the contention status flag. If it does not detect channel activity, it can be equivalently considered to belong to the set \(\mathrm{S}_{l}(n+1)\) in the _(n+1)-th_ round. If the node, starting for the first time, detects channel activity, it can be determined that the contention occurs in the first round. Thus, it enters the listening state again at the end of the ATW. Equivalently, the node belongs to \(\mathrm{S}_{l}(n)\), known from Theorem 3. This node does not obtain the sending authentication before \(\mathrm{S}_{l}(n)\). **Theorem 5**. : _For any node \(i\) in set \(\mathrm{S}_{l}(n)\), if its number ID \(N_{i}\) is not the minimum number in set \(\mathrm{S}_{l}(n)\), the node must belong to \(\mathrm{S}_{l}(n+1)\)._  
Proof.