On 1-Persistent Retransmission in Grant-free Access for URLLC Service

In this paper, we study the effect of 1-pR on the collision process of GF access, under the condition that the group of UEs in the coverage of a BS is relatively fixed. Though there are three strategies for GF access, we model the Reactive strategy [3] in this paper. We define the number of UEs that try to send packets as the attempt rate. We show that the 1-pR makes the attempt rate seen by the packet in retransmission larger than that seen by the packet in the first transmission. As a result, the successful probability of retransmission is lower than that of the first transmission. Taking this observation into consideration, we derive the distribution of the user-plane delay of URLLC packets, which includes the delay incurred by queueing process. We demonstrate that whether to include the effect of 1-pR and queueing process in the analysis would have a significant impact on the prediction accuracy of delay distribution. We further show that our analytical result can be used to predict the maximum number of UEs that can be supported, if other parameters, such as the number of available RBs and the traffic rate are known.

In summary, the contributions of this paper are as follows:

1. 1.

   We study the impact of the 1-pR of GF access on user-plane latency in the case where the group of UEs in the coverage of a BS is relatively fixed.

2. 2.

   Different from previous works [5, 7, 8, 9, 10], we take into account the queueing process of packets in our model and show that it has a significant influence on delay performance.

3. 3.

   Our result can be used to predict the number of UEs that can be supported when the number of RBs is known.

The rest of this paper is organized as follows. Section II presents the system model and our assumptions. Section III analyzes the impact of 1-pR on the attempt rates seen by the packets in the first transmission and retransmission. Taking into account the queueing process of packets, we derive the distribution of user-plane delay in section IV, based on which section V demonstrates the impact of 1-pR and queueing process on delay performance and system design. Section VI concludes this paper.

To derive the attempt rates, we need to study the competition process that an HOL packet will experience before it is successfully received by the BS. After a packet becomes an HOL packet, it experiences the first transmission lasting for 1 TTI, as Fig. 2 shows. After that, it waits 3 TTIs in the buffer. If the transmission fails, it will undergo a retransmission in the 5-th TTIs after the first transmission. This process continues until this packet is received by the BS successfully.

As Fig. 3 plots, the transmission process of an HOL packet can be described by a Markov chain, where $F$ and $R_{i}$ are the state of the first transmission and the state of the $i$-th retransmission, $W_{F}$ and $W_{R_{i}}$ are the waiting states after the first transmission and the $i$-th retransmission, and $p_{F}$ and $p_{R_{i}}$ are the successful probabilities of the first transmission and the $i$-th retransmission, respectively. Clearly, the duration times of $F$ and $R_{i}$ are 1 TTI, and those of $W_{F}$ and $W_{R_{i}}$ are 3 TTIs. Let $\pi_{F}$, $\pi_{W_{F}}$, $\pi_{R_{i}}$, and $\pi_{W_{R_{i}}}$ be the limiting probabilities of states $F$, $W_{F}$, $R_{i}$, and $W_{R_{i}}$. According to Fig. 3, we have the following equilibrium equations:

Solving them together with the normalization condition

we have the limiting probabilities as follows

Let $\tilde{f}$ and $\tilde{r}_{i}$ be the time-average probabilities that the UE stays at states $F$ and $R_{i}$ during the time interval when the UE is busy with HOL-packet transmission. From (2), we can easily obtain $\tilde{f}$ and $\tilde{r}_{i}$ as follows

and

It follows that the probability that a UE is non-empty, denoted by $\alpha$, is given by [16]

When UE $\mathcal{A}$ transmits the HOL packet for the first time, the average number of non-empty UEs seen by UE $\mathcal{A}$ is $(N-1)\alpha$. Also, the probability that a UE is busy in transmission is $\tilde{f}+\sum_{i=1}^{\infty}\tilde{r}_{i}$. The attempt rate seen by the packet in the first transmission is thus given by

Intuitively, the attempt rate seen by the packet in the $i$-th retransmission $G_{R_{i}}$ depends on the attempt rate seen by the packet in the $(i-1)$-th retransmission $G_{R_{i-1}}$. In the following, we first derive $G_{R_{1}}$ from $G_{F}$, and then calculate $G_{R_{i}}$ from $G_{R_{i-1}}$, where $i=2,3,\cdots$.

$G_{R_{1}}$ consists of two parts. Let $g_{1}$ be the average number of UEs that do not compete for the same RB with UE $\mathcal{A}$ in $\mathcal{A}$’s first transmission and send packets in $\mathcal{A}$’s first retransmission, and $g_{2}$ be the average number of UEs that fail in the competition with UE $\mathcal{A}$ in $\mathcal{A}$’s first transmission. Clearly, $G_{R_{1}}=g_{1}+g_{2}$.

Before calculating $g_{1}$ and $g_{2}$, we need to know the condition probability $\operatorname{Pr}\left\{S_{k}\mid H\right\}$, where $S_{k}$ is the event that $k$ UEs compete for the same RB with $\mathcal{A}$, and $H$ is the event that $\mathcal{A}$ fails in the first transmission of HOL packet. According to the Bayes’ theorem

When $N$ is sufficiently large, it is easy to show that the number of UEs sending packets in a TTI obeys the Poisson distribution, that is,

According to Ref. [17], the probability that $\mathcal{A}$ fails in the first transmission under the condition that $k$ UEs compete for the same RB with $\mathcal{A}$ is given by

where $\rho=\bar{P}/\sigma^{2}$ is mean received signal-to-noise ratio (SNR). Also, $\operatorname{Pr}\{H\}$ can be determined via the total probability formula as follows

where we use the condition that $N$ is sufficiently large for the second step. We verify via simulation that the difference between the second and the third terms of (9) is less than $10^{-76}$, when $N\geq 20$ and $B=20$. Substituting (7) through (9) into (6), we have

Consider the term $G_{F}/B$ in (10). As Eq. (4) shows, $G_{F}\rightarrow\hat{\lambda}/(4\tilde{f})$ when $N$ is sufficiently large. Also, the network must keep the successful probability of packet transmission very high to meet the strict delay requirement. This implies that $\tilde{f}$ in (3) is approximately equal to 1/4. It follows that $G_{F}$ is almost equal to $\hat{\lambda}$. According to assumptions A3 and A5, $G_{F}/B$ will be very small, such that $e^{-G_{F}/B}\rightarrow 1-G_{F}/B$. Our simulation verifies that $G_{F}=0.0286$ and the difference $e^{-G_{F}/B}-(1-G_{F}/B)$ is less than $10^{-6}$ when $B\geq 20$ and $N=20$. Therefore, (10) can be rewrote as follows

From (11), we can find that

1. 1)

   When $k=0$,

   $$\operatorname{Pr}\left\{S_{0}\mid H\right\}=\frac{1-e^{-\frac{\mu}{\rho}}}{1-e^{-\frac{\mu}{\rho}}\left(1-\frac{\mu}{\mu+1}\frac{G_{F}}{B}\right)}.$$

2. 2)

   When $k=1$,

   $$\operatorname{Pr}\left\{S_{1}\mid H\right\}=\frac{\frac{G_{F}}{B}\left(1-\frac{e^{-\frac{\mu}{\rho}}}{\mu+1}\right)}{1-e^{-\frac{\mu}{\rho}}\left(1-\frac{\mu}{\mu+1}\frac{G_{F}}{B}\right)}.$$

3. 3)

   When $k\geq 2$, $\operatorname{Pr}\left\{S_{k}\mid H\right\}$ becomes the high-order infinitesimal of $G_{F}/B$, indicating that the failure of $\mathcal{A}$ in the first transmission is induced by the collision with more than one UE with negligible probability.

In other words, when $B$ is sufficiently large, $\operatorname{Pr}\left\{S_{k}\mid H\right\}\rightarrow 0$ for $k\geq 2$.

Let $s$ be the probability that a UE competes for the same RB with UE $\mathcal{A}$ in $\mathcal{A}$’s first transmission, given that $\mathcal{A}$ fails in the first transmission. Using (11), we have

It follows that $(N-1)(1-s)$ is the average number of UEs that do not compete for the same RB with UE $\mathcal{A}$ in $\mathcal{A}$’s first transmission. Each of these UEs will send a packet when $\mathcal{A}$ makes the first retransmission with probability $\alpha/4$. Thus, we have

As (11) indicates, $\mathcal{A}$ collides with at most 1 UE if it fails in the first transmission. If $\mathcal{A}$ collides with a UE and both of them fail, $\mathcal{A}$ will see this UE make a retransmission 4 TTIs later. In this case, this UE will contribute to $g_{2}$, which is the average number of UEs that fail in the competition with UE $\mathcal{A}$ in $\mathcal{A}$’s first transmission. Let $p_{1}$ be the probability that a UE fails in the competition with $\mathcal{A}$ when $\mathcal{A}$ sends a packet, given that $\mathcal{A}$ fails in the first transmission. According to Appendix A, we have

Combining (13) with the fact that the failure of $\mathcal{A}$ in the first transmission is induced by the collision with at most one UE yields $g_{2}=0\times\operatorname{Pr}\left\{S_{0}\mid H\right\}+1\times p_{1}\operatorname{Pr}\left\{S_{1}\mid H\right\}$. As a result, the attempt rate seen by $\mathcal{A}$ in the first retransmission is equal to

Following the similar argument, we can derive $G_{R_{i}}$ as follows

where $i=2,3,\cdots$.

To clearly demonstrate the relationship between $G_{F}$ and $G_{R_{i}}$, we consider the ideal wireless channel, where there is no noise [18]. In this case, the mean SNR $\rho=\bar{P}/\sigma^{2}\rightarrow\infty$ and $\mu/\rho\rightarrow 0$, and thus

where

In the case of ideal channel, Eq. (16) shows that $G_{R_{1}}^{*}=G_{R_{2}}^{*}=\cdots=G_{R_{\infty}}^{*}\approx G_{F}^{*}+p_{1}^{*}$, where $p_{1}^{*}$ is actually the reduction of $g_{2}$ and is contributed by the UE who fails in the competition with $\mathcal{A}$ in the last transmission. This result is formally stated by the following theorem.

Note that if the SINR threshold $\mu$ satisfies $0<\mu\leq 1$, the difference $p_{1}^{*}=0$, which means the UE in collision with $\mathcal{A}$ in the last transmission succeeds. This can be interpreted as follows. The ideal channel has no additive noise, i.e., $\sigma^{2}=0$. In this case, the SINRs of $\mathcal{A}$ and the UE in collision with $\mathcal{A}$ are reciprocal of each other, according to (1). It follows that given $\mathcal{A}$ fails, i.e., its SINR is less than 1, the SINR of the other UE must be larger than 1 and thus succeed in the competition.

As Eqs. (14) and (15) exhibit, the relationship between $G_{F}$ and $G_{R_{i}}$ in general cases is quite complicated. We thus use numerical results and simulations to demonstrate that Theorem 1 approximately holds. For a channel in general cases, it is necessary to keep the mean received SNR $\rho$ much larger than the SINR threshold $\mu$, i.e., $\mu/\rho\ll 1$, such that very high reliability of URLLC service can be guaranteed. Under this condition, Fig. 4 plots $G_{R_{i}}$ as a function of retransmission times $i$, where the UE population is $N=100$, the number of RBs is $B=48$, the average power of additive noise is $\sigma^{2}=-112$dBm, and the mean received power of packets is $\bar{P}=-60\mathrm{dBm}$ [19]. Note that $G_{R_{0}}$ in Fig. 4 stands for $G_{F}$, as we mark. From the result, we can find that the difference among the $G_{R_{i}}$s is much smaller than that between $G_{R_{i}}$ and $G_{F}$, except when $0<\mu\leq 1$.

Section III demonstrates that all the $G_{R_{i}}$s are almost equal. This indicates that it is reasonable to adopt the following simplification:
(*) All the attempt rates seen by the packet in different rounds of retransmission are all equal,
when we solve the service time, such that the derivation can be largely simplified. Accordingly, the Markov chain in Fig. 3 reduces to a four-state Markov chain in Fig. 5, where there is a first-transmission state $\widehat{F}$, a waiting state after the first transmission $\widehat{W}_{F}$, a retransmission state $\widehat{R}$, and a waiting state after the retransmission $\widehat{W}_{R}$. In the simplified model, the attempt rates seen by the packet in the first transmission and the packet in retransmissions are denoted as $\widehat{G}_{F}$ and as $\widehat{G}_{R}$, respectively. In Fig. 5, $\widehat{p}_{F}$ and $\widehat{p}_{R}$ denote the successful probabilities of the first transmission and the retransmissions, respectively.

According to the simplified Markov chain in Fig. 5, we can obtain the stationary probabilities of $\widehat{F}$, $\widehat{W}_{F}$, $\widehat{R}$, and $\widehat{W}_{R}$ as follows

and

from which we further obtain the non-empty probability

Accordingly, we have

and

Following the argument that the packet can successfully transmitted if the received SINR is larger than the preset threshold $\mu$, we drive $\widehat{p}_{F}$ and $\widehat{p}_{R}$ according to (17) and (18) as follows

and

In this case of ideal channel where $\mu/\rho\rightarrow 0$, (18), (19) and (20) can be simplified as follows

and

These two equations indicate that $p_{R}^{*}$ is less than $p_{F}^{*}$ by a constant, which is the consequence of (16).

As Fig. 5 shows, a UE may need several attempts for a successful transmission. The service time is 4 TTIs if the first transmission is successful with probability $p_{F}$, and $4+4i$ if the packet is sent successfully until the $i$-th retransmission with probability $\left(1-\hat{p}_{F}\right)\hat{p}_{R}\left(1-\hat{p}_{R}\right)^{i-1}$. It follows that the distribution of service time $X$ is given by

where $i=1,2,3\cdots$.

As Fig. 6 plots, the user-plane delay of packet $\mathcal{a}$ consists of the following parts. After arriving at the UE, $\mathcal{a}$ will experience an alignment delay, denoted by $A$, which is the time interval from the arrival to the start of the next TTI. After that, $\mathcal{a}$ enters the queue and experiences a waiting time, denoted by $W$, which includes the following two components. First, $\mathcal{a}$ has to wait a residual service time, denoted by $R$, if $\mathcal{a}$ sees a packet in service when it enters the queue. Second, $\mathcal{a}$ may need to wait the service completion of the packets waiting in the queue before $\mathcal{a}$. After $\mathcal{a}$ becomes an HOL packet, it will be successfully decoded by the BS after $X-2$ TTIs, called transmission time and denoted by $Z$. Thus, the user-plane delay is given by

The packet can be removed from the buffer of UE, only when the UE finishes processing an ACK from the BS. Let $V=W+X$ be the sojourn time that the packet spends in the queue. Let $d(t)$, $w(t)$ and $v(t)$ be the probability density functions (PDFs) of the user-plane delay, the waiting time and the sojourn time, the Laplace transforms of which are respectively denoted by $D(s)$, $W(s)$, and $V(s)$. According to (22), $D(s)=A(s)W(s)X(s)e^{2s}=A(s)V(s)e^{2s}$.

According to the P-K formula of M/G/1 queue [22], $V(s)$ is given by

where $\bar{X}\triangleq E[X]=4\left(1+\hat{p}_{R}-\hat{p}_{F}\right)/\hat{p}_{R},$ and $X(s)$ is the Laplace transform of (21)

Also, since the arrival process of packets at the UE is a Poisson process, $A$ is a uniformly distributed random variable in the range between 0 and 1 TTI, the Laplace transform of which is given by

Substituting (23), (24), and (25) into $D(s)$

We can obtain $d(t)$ through the inversion transform of (26), and finally the delay distribution $D(t)$.

As sections III and IV discuss, the 1-pR will cause the attempt rate seen by the packet in retransmissions larger than that in the first transmission. It follows that the 1-pR will lower the successful probability of packet retransmission, which will enlarge the distribution tail of user-plane delay. This implies that whether to consider the effect of 1-pR in the analysis would have a significant impact on the accuracy of the analytical results.

To better verify this point, we first derive the distribution of user-plane delay without considering the effect of 1-pR for comparison. Ignoring the effect of 1-pR is equivalent to set $\hat{p}_{R}=\hat{p}_{F}=\tilde{p}$ in Fig. 3. Following the similar derivation process in section IV, we have

and

where

is the Laplace transform of the PDF of the service time when the effect of 1-pR is ignored.

Using (26) and (27), Fig. 7 compares the complementary cumulative distribution functions (CCDFs) of user-plane delay when the effect of 1-pR is or is not considered in the analysis. The input traffic rates to a UE in Fig. 7(a) and 7(b) are 5 and 10 packets/s. As Fig. 7 plots, the curves are roughly stair-like, and the $i$-th step describes the delay distribution of packets that are transmitted $i$ times before success. Since the 1-pR is considered in the derivation of (26), our analytical result $D$ agrees with the simulation result very well. As a comparison, $\widetilde{D}$ in (27) that is obtained by assuming $\hat{p}_{R}=\hat{p}_{F}$ has a remarkable error on the second step. This is attributed to the reason that the assumption $\hat{p}_{R}=\hat{p}_{F}$ overestimates the successful probability of retransmission, which leads to the curve of $\widetilde{D}$ is lower than the simulation curve in the second step. This clearly indicates that taking into consideration the effect of 1-pR in delay analysis is necessary.

Though the arrive rate of URLLC traffic is very low, it is still necessary to consider the effect of queueing process on the user-plane delay. We demonstrate this point by comparing the delay performance with and without considering the queueing process. Let $D_{nq}$ be the user-plane delay when the arriving packet is assumed to see the UE with empty queue. The user-plane delay in this case consists of the alignment delay $A$ and the transmission delay $Z$, i.e., $D_{nq}=A+Z=A+X-2$. Let $d_{nq}(t)$ be the PDF of $D_{nq}$. Using (24) and (25), we have

where $i=1,2,3\cdots$. From (28), we can obtain $D_{nq}(t)$.

We compare $D_{nq}$, $D$, and simulation in Fig. 8, where the input traffic rates to a UE are 5 and 10 packets/s. A packet will have to wait several TTIs in the buffer before transmission if it sees a nonempty queue when it arrives at the UE. Therefore, the stair-like curves of the simulation result and $D$ are sloped in each step, as Fig. 8 plots. However, if we assume that the arriving packet always sees an empty UE, the user-plane delay will not take some specific values. For example, the user-plane delay in this case will not take values like 4 and 5, as (28) shows. Accordingly, in Fig. 8, there includes a horizontal line segment in each step of the curve of $D_{nq}$, e.g., the horizontal line segment from the 3rd TTI to the 6th TTI in the first step. This makes the curve of $D_{nq}$ remarkably lower than that of the simulation and $D$ at each step.

We further observe the impact of queueing process on user-plane delay under different input traffic rates to a UE, while fixing the aggregate input traffic rate $\hat{\lambda}=200$ packets/s. It is very interesting to see from Fig. 9 that the curve of small $N$ and relatively large $\lambda$ is higher than that of large $N$ and relatively small $\lambda$ in each step. This clearly indicates that the input traffic rate has a greater impact on the delay performance than the UE population, because an arriving packet in the case with relatively large $\lambda$ and small $N$ will have more chance to see a non-empty UE and hence suffer a larger queueing delay. In particular, the access network in the case with $\lambda=10$ packets/s and $N=20$ already cannot meet the reliability requirement that the user-plane delay of $>99.999\%$ packets should be less than 1ms. This suggests that, due to the impact of queueing process, we should consider the traffic rate to each UE $\lambda$, not just the aggregate rate $\hat{\lambda}$, in the system design.

Compared to the simulation, the analytical result can obtain the outage probability very fast, such that it can be used to assist the design of networks. In practice, the BS is able to know if there is a UE that newly joins or departs from its coverage, via Dual Active Protocol Stack [23]. Hence, a possible application of analytical results is to assign a proper number of RBs, denoted by $B^{*}$, to the URLLC traffic, according to the channel state, the number of UEs, and the input traffic rate to a UE . However, if the analytical model is not accurate enough, it will exert a negative influence on system configuration, as we demonstrate below.

The outage probability, denoted by $P_{o}$, is defined as the probability that the user-plane delay of a packet is larger than a preset threshold. According to the 3GPP standard [2], $P_{o}$, must be less than $10^{-5},$ given that the delay threshold is 1ms. Clearly, the outage probability can be easily calculated from the PDF of user-plane delay. For example, we can find the outage probability using $d(t)$:

Assuming that $\sigma^{2}$, $\bar{P}$, $N$, and $\lambda$ are known, we find the proper number of RBs $B^{*}$ via a simple method as follows. Initially, select a $B$ and calculate $P_{o}$. If $P_{o}>10^{-5}$, increase $B$ by a step size $b<B$ and recalculate $P_{o}$. Repeat the process until $P_{o}<10^{-5}$. We plot $P_{o}$ as a function of $B$, and find the intersection of the curve with the line $P_{o}=10^{-5}$. The rounding up of the abscissa of the intersection point is the $B^{*}$ to be found. If the initial $P_{o}$ is small than $10^{-5}$, we can find $B^{*}$ in the similar way. Table I compares the $B^{*}$ predicted by simulation, $D$, $\tilde{D}$, and $D_{nq}$, where $\lambda=5$ packets/s and $N$ changes from 40 to 100. It can be seen that $D$ can provide an exact prediction, while $\widetilde{D}$ and $D_{nq}$ lead to a remarkably underestimated $B^{*}$. Clearly, the underestimation of $B^{*}$ will lead to a bad consequence that the reliability cannot be guaranteed. For example, when $N=60$, the $B^{*}$s predicted by $\widetilde{D}$ and $D_{nq}$ are 29 and 39, respectively. If we employ these predicted $B^{*}$s, our simulation shows that the outage probabilities would be $2.4\times 10^{-5}$ and $1.7\times 10^{-5}$, respectively, which are higher than $10^{-5}$.

Note that we also consider the case where $\lambda=10$ packets/s. We fail to find a $B^{*}$ such that the network can support, no matter what value $N$ takes and whether we use the simulation or the analytical model. This is because a considerable portion of packets will experience the queueing process. In this case, some additional mechanism should be taken to eliminate the queue delay, such as 4 stop-and-wait (SAW) channel discussed in [4].

Our model is developed under the assumption that the group of UEs in the coverage of a BS is fixed. However, the UE in reality may move from one cell to another. We thus verify the effectiveness of our model in a dynamic scenario, where the UE group changes slowly.

To do that, we make the following assumptions. We assume the radius of a cell is 250m [24], and the moving speed of a UE is 30km/h, meaning that a UE will leave a cell with probability $p_{l}=2.3\times 10^{-6}$ in each TTI. Let $N_{t}$ be the number of UEs in TTI $t$ and $\bar{N}$ be the average number of UEs in a cell. Suppose that the average number of UEs moving into a cell is $\left(2\bar{N}-N_{t}\right)p_{l}$, such that $N_{t}$ fluctuates around $\bar{N}$. At last, we assume that the channel condition keeps unchanged.

To utilize the analytical results in such a dynamic scenario, we calculate the $B^{*}$s for all possible $N$s beforehand and store them in a table similar to Table I. Once the BS detects the variation of $N_{t}$, it adjusts $B^{*}$ according to the table, such that spectrum utilization can be enhanced.

We perform simulation according to the above description. Table II gives the outage probability when the BS adjusts the number of RBs according to $N_{t}$, using $D$, $\widetilde{D}$ and $D_{nq}$, and confirms that the adjustment based on $D$ can ensure an outage probability of less than $10^{-5}$ in the dynamic situation.