# A Delay-Aware and Reliable Data Aggregation for Cyber-Physical Sensing
## 4 Design of the TSR-DARDA Scheme
### Performance Analysis  
**Theorem 1**. : _Consider the network depth is \(k\), Let \(h_{k}\) denote the total number of subsets in tier \(T_{k}\). Assume the maximum number of retransmission is \(\mu\) and nodes in hotspots need \(\tau\) time unit to forward the aggregated information to sink. The time required for all nodes to finish a single transmission is \(D=h_{k}(1+\mu)+\sum\limits_{i=1}^{k-2}(1+\mu)+\tau\)._  
Proof. : Consider the network depth is \(k\), which represents the number of tiers. From the Section 4.2, we know that \(H(k,m)\) is the subset of nodes in \(T_{k}\) that can be scheduled simultaneously. Let \(h_{d}\) denote the total number of subsets in each tier \(T_{d}\) such that \(\cup_{m=1}^{h_{d}}H(d,m)=T_{d}\). Clearly, all nodes in \(T_{d}\) can finish a single transmission in \(h_{d}\) time slots. Based on the assumption that nodes transmission in different tiers do not interfere with each other, the subset of nodes \(H(d,m)\) in different tiers could transmit data simultaneously. That is nodes in \(T_{d}\)\(T_{d}\) can start their transmissions simultaneously with nodes in outer tier \(T_{d+1}\), but the number of nodes in \(T_{d}\) is more than that in \(T_{d+1}\). Assume the  
Figure 7: Optimization of the proposed TSR-DARDA scheme. maximum number of retransmissions is \(\mu\). Therefore, the time needed for all nodes not in hotspots to finish data aggregation and transmission to collecting nodes in hotspots is \(h_{k}(1+\mu)+\sum\limits_{i=1}^{k-2}(1+\mu)\) when the total number of tier is \(k\), so, we can obtain the time required for all nodes to finish a single transmission as follows:  
\[D=h_{k}(1+\mu)+\sum\limits_{i=1}^{k-2}(1+\mu)+\tau \tag{23}\]  
**Theorem 2**: _Let \(D\) and \(D^{\Delta}\) respectively represent the time required for all nodes to finish a single transmission under the cases of network partition with the optimized TSR-DARDA scheme and other schemes. It is \(D^{\Delta}-D>0\)._  
**Proof. ** The width set are denoted by \(\{d_{k},d_{k-1},...,d_{1}\}\) partitioned according to the optimized TSR-DARDA scheme from the out layer to inner layer to sink, And \(\left\{d_{k},d_{k-1}^{\Delta},d_{k-2}^{\Delta}...,d_{1}\right\}\) denote the other partition case which has the identical widths of \(d_{k}\) and \(d_{1}=r\). \(\square\)  
When \(k=4\), the width set of surface circular partition are respectively \(\{d_{4},d_{3},d_{2},d_{1}\}\) with the TSR-DARDA scheme and \(\{d_{4},d^{\Delta}{}_{3},d^{\Delta}{}_{2},d_{1}\}\) without the TSR-DARDA scheme. Because the outmost tier \(T_{4}\) with the identical circular width under the two partition cases, the time required for nodes in the tier to finish data aggregation is identical and equals to \(h_{4}(1+\mu)\). For the TSR-DARDA scheme, it needs \(D=h_{d}(1+\mu)+(1+\mu)+(1+\mu)+\tau\) for all nodes to finish a single transmission to a sink. For other partition schemes:  
1. If \(d^{\Delta}{}_{3}>d_{3}\), there is \(d^{\Delta}{}_{2}<d_{2}\). At the time point \(h_{4}(1+\mu)+(1+\mu)\), the tier \(T_{4}\) has forwarded data aggregation to collecting nodes in tier \(T_{3}\) and \(T_{3}\) has not finished its own data aggregation. And more time \(\Lambda\) is needed to finish its own data aggregation. Therefore, all nodes to finish a single transmission to sink needs \(D^{\Delta}=h_{d}(1+\mu)+(1+\mu)+\Lambda+(1+\mu)+\tau\) time slots. It is obviously \(D^{\Delta}-D>0\). 2. If \(d^{\Delta}{}_{3}<d_{3}\), there is \(d^{\Delta}{}_{2}>d_{2}\). At the time point \(h_{4}(1+\mu)+(1+\mu)\), \(T_{3}\) has finished its own data aggregation. And, at the time point \(h_{4}(1+\mu)+(1+\mu)+(1+\mu)\), tier \(T_{2}\) has not finished its own data aggregation and needs more time \(\Gamma\) to finish data collection to collecting nodes in the tier. Hence, the total time required to finish a single transmission for all nodes to sink is \(D^{\Delta}=h_{d}(1+\mu)+(1+\mu)+(1+\mu)+\Gamma+\tau\). It is obviously \(D^{\Delta}-D>0\). By an analogous method, when \(k=5\), \(6\), \(7\),..., they holds the same conclusion for other partition schemes.