## 3 Grant-Free Scheduling
### Collisions with Other UEs  
First, we focus on the probability \(P_{c}\) that a packet is not correctly received due to the collisions of its \(K\) replicas with the packets transmitted by other UEs. To this end, we consider UL transmissions and \(N\) UEs within a single cell with aperiodic traffic. Packets are generated by each UE following a Poisson distribution with exponential inter-arrival time. The average packet inter-arrival time is equal to \(1/\lambda\), where \(\lambda\) is the average number of packets generated per second. We consider the transmission of small packets with a size of 32 bytes [22], and we assume without loss of generality that each packet requires only one RB. Figure 1: Illustration of the Fifth Generation New Radio (5G NR) resource grid map: Transmission of a data packet with four repetitions and a random selection of Resource Blocks (RBs) per slot. The probability \(P_{g}\) that one or more packets are generated for a UE in a time period \(T_{slot}\) is equal to:  
\[P_{g}=1-\exp(-T_{slot}\cdot\lambda) \tag{2}\]  
We define \(R_{i}\) as the set of UEs for which a new packet could be generated in a slot \(s_{i}\) (the slot has a time duration equal to \(T_{slot}\)). Here, \(n_{i}\) is the number of UEs that do have a new packet to transmit in \(s_{i}\). This \(n_{i}\) can then take any value between 0 and the cardinality of \(R_{i}\). The probability \(P_{tx}(n_{i},R_{i})\) that \(n_{i}\) UEs from the set \(R_{i}\) of UEs have new packets to be transmitted in \(s_{i}\) with duration \(T_{slot}\) is equal to:  
\[P_{tx}(n_{i},R_{i})=\left( \begin{array}{c}|R_{i}|\\ n_{i}\end{array} \right)\cdot P_{g}{}^{n_{i}}\cdot\left(1-P_{g}\right)^{|R_{i}| -n_{i}} \tag{3}\]  
where \(|R_{i}|\) represents the number of elements or the cardinality of the set \(R_{i}\). A packet will not be successfully delivered to the BS if all its \(K\) replicas collide with the trans-missions of other UEs. A UE has an active transmission in \(s_{i}\) if it generated a new data packet in the previous slots \(s_{i-(K-1)}\),..., \(s_{i-1}\), and \(s_{i}\). If this is the case, then the UE would be transmitting one of the \(K\) replicas in \(s_{i}\). We denote as \(n_{i}^{act}\) the number of UEs with active transmissions in \(s_{i}\). The probability \(P_{nrc}(n_{i}^{act},U)\) that \(n_{i}^{act}\) UEs do not collide with a given UE is equal to the probability that they do not select the same RB at a given slot for their next transmission as the UE under study. \(P_{nrc}(n_{i}^{act},U)\) is given by:  
\[P_{nrc}(n_{i}^{act},U)=\left( \begin{array}{c}U-1\\ U\end{array} \right)^{n_{i}^{act}} \tag{4}\]  
Equations (2)-(4) are necessary to compute the probability \(P_{c}\) that a packet is not correctly received at the BS due to the collision of all its \(K\) replicas with the transmissions of other UEs. To compute \(P_{c}\), let us consider the case of a particular UE\({}_{1}\) that has to transmit the \(K\) replicas of a packet in slots \(s_{i}\), \(s_{i+1}\),..., \(s_{i+K-1}\). For the sake of clarity, we consider an example with \(K=4\), and \(s_{i}\) corresponding to \(s_{3}\). \(P_{c}\) is then equal to the probability of collision of the 4 replicas transmitted in \(s_{3}\), \(s_{4}\), \(s_{5}\), and \(s_{6}\), which is represented by \(P_{rc}(s_{3},s_{4},s_{5},s_{6})\):  
\[P_{c}=P_{rc}(s_{3},s_{4},s_{5},s_{6}) \tag{5}\]  
To determine \(P_{rc}(s_{3},s_{4},s_{5},s_{6})\), we first study the probability \(P_{rc}(s_{3})\) that the replica of the packet transmitted in \(s_{3}\) collides with a transmission from any other UE. \(P_{rc}(s_{3})\) is given by the probability that one or more UEs (in addition to UE\({}_{1}\)) have an active transmission in \(s_{3}\) (i.e., \(n_{3}^{act}\geq 1\)), and that one or more of the \(n_{3}^{act}\) UEs select the same RB as UE\({}_{1}\) for their transmission. \(n_{3}^{act}\) is equal to \(n_{0}+n_{1}+n_{2}+n_{3}\), and the probability \(P_{rc}(s_{3})\) has to consider all possible combinations of \(n_{0}\), \(n_{1}\), \(n_{2}\) and \(n_{3}\) that result in \(n_{3}^{act}\geq 1\).