## 3 Grant-Free Scheduling
### Collisions with Other UEs
The probability \(P(n_{3}^{act}\geq 1)\) can then be expressed as:  
\[P(n_{3}^{act}\geq 1)=\sum_{n_{0}=n_{0}^{min}}^{n_{max}^{max}}\left\{P_{tx}(n_{0},R_{0})\cdot\sum_{n_{1}=n_{1}^{min}}^{n_{max}^{max}}\left(P_{tx}(n_{1},R_{1}) \cdot\sum_{n_{2}=n_{2}^{min}}^{n_{max}^{max}}\left[P_{tx}(n_{2},R_{2})\cdot \sum_{n_{3}=n_{3}^{min}}^{n_{max}^{max}}P_{tx}(n_{3},R_{3})\right]\right\}\right\} \tag{6}\]  
where \(n_{i}^{max}\) and \(n_{i}^{min}\) represent the maximum and minimum possible values of \(n_{i}\) in each slot, and are equal to:  
\[n_{i}^{max}=|R_{i}|,\ \forall i\leq 3 \tag{7}\]  
\[n_{i}^{min}=\left\{ \begin{array}{ll}1&\text{if}\ i=3\ \&\ |R_{i}|=N-1\\ 0&\text{otherwise}\end{array} \right.,\ i\leq 3 \tag{8}\]  
where \(R_{i}\) is the set of UEs that could have a new packet to be transmitted in \(s_{i}\). \(R_{i}\) is equal to the total number of UEs (\(N\)) minus UE\({}_{1}\) and all active UEs in the slot previous to \(s_{i}\). The cardinality of \(R_{i}\) is then equal to:  
\[|R_{i}|=N-1-\sum_{j=\max\{i-3,0\}}^{i-1}n_{j},\ i\leq 3 \tag{9}\]It should be noted that \(n_{i}^{min}\) is equal to 0 or 1 in order to guarantee that \(n_{3}^{act}\) is equal to or higher than one. \(n_{i}^{act}\) can be expressed as:  
\[n_{i}^{act}=\sum\
olimits_{j=\max\{i-3,0\}}^{i}n_{j},\ i\leq 3 \tag{10}\]  
To achieve finally the expression of \(P_{rx}(s_{3})\), we need to incorporate to the expression of \(P(n_{3}^{act}\geq 1)\) in (6) the probability that one or more of the \(n_{4}^{act}\) UEs select the same RB as UE\({}_{1}\) for their transmissions. This probability is equal to \(1-P_{nrc}(n_{3}^{act},U)\). \(P_{rx}(s_{3})\) is then calculated as:  
\[P_{rx}(s_{3})=\sum\limits_{n_{0}=n_{0}^{min}=0}^{n_{0}^{max}}\left\{P_{tx}(n_ {0},R_{0})\cdot\sum\limits_{n_{1}=n_{1}^{min}}^{n_{max}^{max}}\left\{P_{tx}(n_ {1},R_{1})\cdot\sum\limits_{n_{2}=n_{2}^{min}}^{n_{max}^{max}}\left[P_{tx}(n_ {2},R_{2})\cdot\sum\limits_{n_{3}=n_{3}^{min}}^{n_{max}^{max}}\left\{P_{rx}(n_ {3},R_{3})\cdot(1-P_{nrc}(n_{3}^{act},U))\right\}\right]\right\}\right\} \tag{11}\]  
The probability of collision of the replica transmitted in \(s_{4}\) depends upon the number \(n_{4}^{act}\) of UEs with active transmissions in \(s_{4}\). This \(n_{4}^{act}\) depends on the number \(n_{1}\), \(n_{2}\), \(n_{3}\) and \(n_{4}\) of UEs that have new packets to transmit in \(s_{1}\), \(s_{2}\), \(s_{3}\), and \(s_{4}\), respectively. The probability that UEs have new packets to transmit in \(s_{1}\), \(s_{2}\), and \(s_{3}\) is already included in (11) (\(P_{tx}(n_{1},R_{1})\), \(P_{tx}(n_{2},R_{2})\), and \(P_{tx}(n_{3},R_{3})\) respectively). In this context, \(P_{rx}(s_{3})\) and \(P_{rx}(s_{4})\) are not independent, and they must be calculated jointly. We then compute the joint probability \(P_{rx}(s_{3},s_{4})\) that the replicas transmitted in \(s_{3}\) and \(s_{4}\) collide with transmissions from other UEs. Computing \(P_{rx}(s_{3},s_{4})\) only requires including in (11) the probability that there are UEs with new packets to be transmitted in \(s_{4}\) (i.e., \(P_{tx}(n_{4},R_{4})\)), and the probability that one or more of the active \(n_{4}^{act}\) UEs in \(s_{4}\) select the same RB for their transmission than UE\({}_{1}\). \(P_{rx}(s_{3},s_{4})\) can then be expressed as:  
\[ \begin{array}{l}P_{rc}(s_{3},s_{4})=\sum\limits_{n_{0}=n_{0}^{min}}^{n_{0}^ {max}}\left\{P_{tx}(n_{0},R_{0})\cdot\sum\limits_{n_{1}=n_{1}^{min}}^{n_{max}^ {max}}\left\langle P_{tx}(n_{1},R_{1})\cdot\sum\limits_{n_{2}=n_{2}^{min}}^{n_{ max}^{max}}\left[P_{tx}(n_{2},R_{2})\cdot\right.\right.\\ \left.\left.\sum\limits_{n_{3}=n_{3}^{min}}^{n_{3}^{max}}\left\langle\left\{P _{tx}(n_{3},R_{3})\cdot(1-P_{nrc}(n_{3}^{act},U))\right\}\cdot\sum\limits_{n_ {4}=n_{4}^{min}}^{n_{max}^{max}}\left\{P_{tx}(n_{4},R_{4})\cdot(1-P_{nrc}(n_{ 4}^{act},U))\right\}\right\rangle\right\}\right\rangle\right\}\right\}\end{array}