## 4 System Model
### Markov Chain Model  
As an example, assume the UORA trigger frame with nine UORA RUs. Because the 20-MHz IEEE 802.11 ax trigger frame can have nine RUs at most, this is a reasonable situation. Figure 8 shows the Markov chain model of the 9-RU UORA procedure. If the UORA counter of an STA is smaller than the number of UORA RUs in the received trigger frame, the STA can transmit its frame via the UL OFDMA transmission opportunity. If the initial random back-off number the STA obtained is 0 or smaller than the number of UORA RUs in the trigger frame, the STA can transmit its frame just after receiving the trigger frame with UORA RU(s). Based on Figure 8, each probability branch can be described as below. \[ \begin{array}{llll}P\{0,k|i,l\}=(1-p)/W_{0}&k\in(0,W_{0}-1)&l\in(0,9)&i\in(0,m)\\ P\{i,k|i,k+9\}=1&k\in(0,W_{i}-10)&i\in(0,m)\\ P\{i,k|i-1,l\}=p/W_{i}&k\in(0,W_{i}-1)&l\in(0,9)&i\in(0,m)\\ P\{m,k|m,l\}=p/W_{m}&k\in(0,W_{m}-1)&l\in(0,9)\end{array}  \tag{3}\]  
where \(m\) is the required number of retransmissions to reach the maximum UORA CW size. The maximum UORA CW size is called OCWmax and the initial size of UORA CW is called OCWmin. Once the OCW reaches OCWmax for successive retransmission attempts, the OCW, \(W_{i}\) in (3), shall remain at the value of OCWmax, \(W_{m}\) in (3), until the OCW is reset. The probability \(p\) means the probability of transmission failure when an STA transmits its frame via the UORA procedure. If we assume that the transmission is always successful unless there are collisions, \(p\) can be interpreted as the probability of UORA collision. From the Markov chain model and probability branches, we can derive the stationary distribution of the Markov chain. From the state (0,0), the following can be derived:  
\[b_{0,0}=\frac{(1-p)}{W_{0}}\sum_{i=0}^{m}\sum_{k=0}^{r}b_{i,k} \tag{4}\]  
Figure 7: V2X UORA channel access procedure with dedicated resource unit(s) (RUs). For each state (0,\(k\)) where \(k>0\), we can derive the following expression. \[ \begin{array}{l}b_{0,k}=b_{0,k+r}+\frac{(1-p)}{W_{0}}\sum\limits_{i=0}^{m} \sum\limits_{k=0}^{r}b_{i,k}\\ =b_{0,k+2r}+\frac{(1-p)}{W_{0}}\sum\limits_{i=0}^{m}\sum\limits_{k=0}^{r}b_{i, k}\\ =(\left\lfloor\frac{W_{0}-1-k}{r}\right\rfloor+1)\{\frac{(1-p)}{W_{0}}\sum \limits_{i=0}^{m}\sum\limits_{k=0}^{r}b_{i,k}\}\\ =(\left\lfloor\frac{W_{0}-1-k}{r}\right\rfloor+1)b_{0,0}\end{array}  \tag{5}\]  
For \(0<i<m\), \(k>0\), the expression for each \(b_{i,k}\) can be derived as below. \[ \begin{array}{l}b_{i,k}=b_{i,k+r}+p\sum\limits_{k=0}^{r}\frac{b_{i-1,k}}{W_{ i-1}}=(\left\lfloor\frac{W_{i}-1-k}{r}\right\rfloor+1)p\sum\limits_{k=0}^{r} \frac{b_{i-1,k}}{W_{i-1}}\\ b_{m,k}=p(\sum\limits_{k=0}^{r}\frac{b_{m,k}}{W_{m}}+\sum\limits_{k=0}^{r} \frac{b_{m-1,k}}{W_{m-1}})\end{array}  \tag{6}\]  
For \(b_{i,0}\) with \(i>0\),  
\[b_{i,0}=\frac{p}{W_{i-1}}\sum\limits_{k=0}^{r}b_{i-1,k} \tag{7}\]  
From Equations (5) and (7), the following expression can be derived. \[ \begin{array}{l}b_{0,k}=(\left\lfloor\frac{W_{0}-1-k}{r}\right\rfloor+1)\{ \frac{1-p}{p}(\sum\limits_{i=0}^{m}\frac{p}{W_{i}}\sum\limits_{k=0}^{r}b_{i, k})\}\\ =(\left\lfloor\frac{W_{0}-1-k}{r}\right\rfloor+1)(\frac{1-p}{p})(\sum\limits_{ i=0}^{m-1}b_{i+1,0}+\frac{p}{W_{m}}\sum\limits_{k=0}^{r}b_{m,k})\end{array}  \tag{8}\]  
If we define the virtual state term \(b_{m+1,0}\) for Equation (7), Equation (8) can be re-expressed as follows. \[b_{0,k}=(\left\lfloor\frac{W_{0}-1-k}{r}\right\rfloor+1)(\frac{1-p}{p})(\sum \limits_{i=0}^{m}b_{i+1,0}) \tag{9}\]  
In a similar way, the following equation can be derived from Equations (6) and (7). \[b_{i,k}=(\left\lfloor\frac{W_{i}-1-k}{r}\right\rfloor+1)b_{i,0} \tag{10}\]  
From the steady-state probability rule and Equations (7)-(10), we obtain  
\[