## 2 System Model and Dynamic Optimization Problem
### Data Queue Model  
To represent the dynamic of the backlogged data in each VU, we define the data queue, denoted as \(Q_{n}(t)\), for member \(n\) in the platoon in time slot \(t\). The service rate is defined as the number of served data packets at time slot \(t\), which is denoted as \(\mu_{n}(t)\). The arrival rate is defined as the number of arrived data packets at time slot \(t\), which is denoted as \(a_{n}(t)\). Then, the length of queue \(Q_{n}(t)\) is updated as shown in the following equation:  
\[Q_{n}\left(t+1\right)=\max\left[Q_{n}(t)-\mu_{n}(t),\ 0\right]+a_{n}(t). \tag{20}\]  
In this paper, it is assumed that the data arrivals of VUs are mutually independent and the data arrival of member \(n\) in the platoon is a Poisson process with a parameter that equals \(a_{n}\).  
At the beginning of each time slot, each platoon leader makes a decision regarding the channel allocation to its members. Let \(x_{nk}\) denote the variable of the channel allocation matrix \(\mathbf{X}\), which is defined as:  
\[x_{nk}(t)=\left\{\begin{array}{ll}1,\ \text{if channel $k$ is allocated to member $n$ at time slot $t$,}\\ 0,\ \text{otherwise.}\end{array}\right. \tag{21}\]  
In each slot \(t\), the state of each member in the platoon is defined as active or inactive, depending on whether a channel has been allocated to the member. If the member is active, \(x_{nk}(t)=1,\ \exists k\in\mathcal{K}\) and if the member is inactive, \(x_{nk}(t)=0,\ \forall k\in\mathcal{K}\). For member \(n\) in the platoon, depending on the relationship between the channel allocation in slot \(t-1\), the decision space in slot \(t\) can be classified into five options, represented by \(\mathbb{S}=\{D_{1},\ D_{2},\ D_{3},\ D_{4},\ D_{5}\}\):  
(1) \(D_{1}\): \(x_{nk}\left(t-1\right)=1,\ x_{nk}(t)=1,\ \exists k\in\mathcal{K}\), use the same channel as in the slot \(t-1\);  
(2) \(D_{2}\): \(x_{  
The service rate \(\mu_{n}(t)\) of member \(n\) in slot \(t\) is dependent on the channel allocation, the availability variable \(\delta_{nk}(t)\), and the non-outage variable \(\xi_{nk}(t)\), which can be derived as follows:  
\[\mu_{n}(t)=\left\{\begin{array}{ll}\sum\limits_{k\in\mathcal{K}}x_{nk}\left(t -1\right)\delta_{nk}(t)\xi_{nk}(t),&\text{if }D(t)=D_{1},\\ \sum\limits_{k\in\mathcal{K}}x_{nk}(t)\delta_{nk}(t)\xi_{nk}(t),&\text{if }D(t)=D_{2},\\ 0,&\text{if }D(t)=D_{3},\\ \sum\limits_{k\in\mathcal{K}}x_{nk}(t)\delta_{nk}(t)\xi_{nk}(t),&\text{if }D(t)=D_{4},\\ 0,&\text{if }D(t)=D_{5}.\end{array}\right. \tag{22}\]  
As represented in Equations (14) and (19), the availability probability and the non-outage probability depend on the locations of the MBS, the platoons, and the MUs. Because the network is dynamic, the channel allocation has to be updated. We define the re-allocation as the allocated channel that differs from that in the prior slot; this includes the decisions \(D_{2}\) and \(D_{4}\). Let \(r(t)\) denote the indicator of re-allocation of a platoon, which is defined as follows:  
\[r(t)=\left\{\begin{array}{ll}1,&\text{if }D(t)=D_{2},\ \exists n\in\mathcal{N}, \\ 1,&\text{if }D(t)=D_{4},\ \exists n\in\mathcal{N},\\ 0,&\text{otherwise}.\end{array}\right. \tag{23}\]  
The resource re-allocation results in additional signal overhead and latency [22; 23]. First, if the communication is active in the prior slot and then switches to another channel, the current transmission needs to be terminated. Second, synchronization between the transmitter and receiver is required if there is re-allocation. To decrease the additional signal overhead and latency in the network, the execution of re-allocation is considered a cost. Therefore, the rate of re-allocation is limited and is below a certain threshold, denoted as \(\gamma\), in our channel allocation optimization problem.