## 2 System Model
### Computation Model
\begin{cases}C_{\text{rem}}^{\text{MEC}}(t)-R_ {k}^{\text{MEC}}(t),&C_{\text{rem}}^{\text{MEC}}(t)\geq R_{k}^{\text{MEC}}(t), \\ 0,&C_{\text{rem}}^{\text{MEC}}(t)<R_{k}^{\text{MEC}}(t).\end{cases}  \tag{8}\]  
In time slot \(t\), at each task completion, \(C_{\text{rem}}^{\text{MEC}}(t)\) decreases by the corresponding data volume. This process continues until either all tasks in the queue are completed or \(C_{\text{rem}}^{\text{MEC}}(t)\) is fully exhausted. If the total data volume of the task queue is less than or equal to \(C_{\text{max}}^{\text{MEC}}\), it indicates that the MEC server can process all tasks within the time slot \(t\), resulting in the task queue being cleared. Conversely, if the total data volume of the task queue exceeds \(C_{\text{max}}^{\text{MEC}}\), it means that the MEC server cannot process all tasks within the time slot \(t\), leading to only a subset of tasks being processed, with the remaining tasks deferred to the next time slot. Therefore, for task \(k\) with remaining data volume \(R_{k}^{\text{MEC}}(t)\), the amount of data processed by the \(k\)-th task in the MEC server in time slot \(t\) is  
\[P_{k}^{\text{MEC}}(t)=\min\{R_{k}^{\text{MEC}}(t),C_{\text{rem}}^{\text{MEC}}(t)\}. \tag{9}\]  
Let \(N_{\text{MEC}}(t)\) denote the number of tasks that can be processed by the MEC in time slot \(t\). So, the total task data of the \(i\)-th task vehicle processed by the MEC server in time slot \(t\) is  
\[D_{i}^{\text{MEC}}(t)=\sum_{k\in\mathcal{I}_{i}^{\text{MEC}}(t)}P_{k}^{\text{ MEC}}(t), \tag{10}\]  
where \(\mathcal{I}_{i}^{\text{MEC}}(t)=\{k\mid N_{k}^{\text{MEC}}=i,k\in[1,N_{\text{ MEC}}(t)]\}\) denotes the set of tasks that belong to the \(i\)-th task vehicle and are processed by the MEC server in time slot \(t\). The service vehicles are the same as the MEC server, so the total task data of the \(i\)-th task vehicle processed by the \(n\)-th service vehicle in time slot \(t\) is  
\[D_{i,n}(t)=\sum_{k\in\mathcal{I}_{i,n}(t)}P_{k,n}(t), \tag{11}\]  
where \(\mathcal{I}_{i,n}(t)=\{k\mid N_{k,n}=i,k\in[1,N_{n}(t)]\}\) represents the set of tasks that belong to the \(i\)-th task vehicle and are processed by the the \(n\)-th service vehicle in time slot \(t\), \(N_{n}(t)\) is the number of tasks that can be processed by the \(n\)-th service vehicle in time slot \(t\), and \(P_{k,n}(t)\) is the amount of data processed by task \(k\) in the the \(n\)-th service vehicle in time slot \(t\). So, the amount of data that belong to the \(i\)-th task vehicle's task computed by the service nodes in time slot \(t\) is given by  
\[D_{i}^{\text{off}}(t)=D_{i}^{\text{MEC}}(t)+\sum_{n=1}^{S}D_{i,n}(t). \tag{12}\]  
As mentioned above, \(D_{i}^{\text{rem}}(t)\) is the remaining data amount of the task at the head of the task queue for the \(i\)-th task vehicle in time slot \(t\). If \(D_{i}^{\text{rem}}(t)=0\), the task at the head of the queue is completed and is removed from the task queue. Then, the remaining data amount of the task at the head of the task queue for the \(i\)-th task vehicle in time slot \((t+1)\) is  
\[D_{i}^{\text{rem}}(t+1)=\min\Bigl{\{}0,D_{i}^{\text{rem}}(t)-D_{i}^{\text{ loc}}(t)-D_{i}^{\text{off}}(t)\Bigr{\}}. \tag{13}\]