## 5 Proposed Work
### Optimizing Path Planning Strategy of DCVs  
The trajectory planning of the DCV is designed to meet certain deadline time requirements that guarantee the efficiency and usefulness of the collected data and minimize the energy-hole problem in the network. An energy-hole problem results when the energy of a sensor node has fallen below the threshold, making continuous data sensing operations difficult. The energy-hole problem can be minimized by achieving timeliness in data collection and transmission. Each DCV can be positioned at a specified location on an  
Figure 11: A working example of the proposed model using multiple DCVs. Figure 12: Maximum time usage with different numbers of DCVs from one to five. Figure 13: Number of DCVs at specific deadlines. optimal path, which is calculated using Equation (12). We employ the nearest-neighbor (NN) heuristic algorithm presented in [63] to obtain the optimal trajectory. To this end, a DCV begins from the initial location and traverses along a specified path, only stopping at designated points for data collection, and returns to its initial location after each tour. This process is repeated until all the DCVs have completed their tasks. The time taken for a DCV to accomplish its task depends on the number of regional optimal points. To optimize time and energy consumption, each DCV is assigned an average number of regional optimal points closest to them. The proposed approach deals with a multi-routed problem to determine optimal \(Q\) routes (\(R_{1},\ R_{2},\ \ldots,\ R_{Q}\)) to collaboratively visit all z optimal points, such that the longest time taken to traverse the Q routes is minimized. We note that data gathering using multiple mobile elements is the same as the multiple traveling salesman problem (TSP) [64], and by adapting the approaches presented in [11; 65], it can be shown to be an \(N\!P\)-hard problem. Thus, heuristic algorithms can be employed to achieve a near-optimal solution. To achieve the deadline time constraint and optimize energy consumption, we determine an optimal route for each DCV and assign an average number of optimal points to each DCV. Equation (13) helps to calculate the average number of optimal points assigned to each MCV. \[A_{NOP}(MCV)=\sfrac{s}{V} \tag{13}\]  
where \(A_{NOP}\) denotes the average number of optimal points assigned to each DCV, \(s\) represents the number of optimal points, and \(V\) denotes the number of DCVs. In situations in which the number of optimal points (\(s\)) is not evenly divisible by \(V\), Equation (14) can be employed. \[\sfrac{s}{V}=Q,\ \ldots,\ R(for\ V\ DCVs,\ R=Q+1\ and\ else\ Q) \tag{14}\]  
Equation (14) implies that \(R\) DCVs must traverse a \(Q+1\) number of optimal points, while the others must traverse \(Q\) number of optimal points. While \(Q\) represents the quotient of \(s/V\), \(R\) represents the remainder. For instance, if three DCVs are deployed to eight optimal points according to Equation (14), this means that \(s=8\), \(V=3\), \(Q=2\), and \(R=2\). Therefore, the allocation of eight optimal points to three DCVs will take the form of [3; 3; 2]. The number of optimal points (OPs) allocated to each DCV can be computed using a distance matrix whereby the distances between the OPs and all DCVs is represented by the following entries:  
\[d_{matrix}(DCVs,\ OP_{s})= \begin{bmatrix}d_{11}\ d_{12}\ d_{13}\ \ldots\ d_{1s}\\ d_{21}\ d_{22}\ d_{23}\ \ldots\ d_{2s}\\ d_{31}\ d_{32}\ d_{33}\ \ldots\ d_{3s}\\ \ldots\ \ldots\ \ldots\\ d_{V1}\ d_{V2}\ d_{V3}\ \ldots\ d_{Vs}\end{bmatrix}  \tag{15}\]  
where the number of DCVs is the row index, while that of the OPs is the column index. The number of OPs allocated to a DCV is determined by the minimal value of the distance matrix in the respective column. For instance, an \(OP_{2}\) is assigned to \(DCV_{2}\) if the distance between it and \(DCV_{2}\) is less than that from other DCVs. The same goes for all other OPs and the relevant DCVs. That is, if \(d(DCV_{2},OP_{2})<d(DCV_{1},\ OP_{2})\), then \(OP_{2}\to DCV_{2}\).