Let us consider the neighbourhood of a single RSU operating in non-saturation regime deployed on a bidirectional road segment as shown in Figure 3. According to the location to RSU, the road segment is divided into multiple regions [2]. In each region (*Rg*_{
i
}) within the RSU coverage area vehicles have different pay-load transmission rates according to their distance to the RSU. Each vehicle can transmit frames from data classes AC*k, k* = 0*..* 3, in either CCH or SCH. We assume that there are *d*_{
c
}data combinations on CCH and *d*_{
s
}data combinations on SCH. Data combination *d*_{
x
}on channel *x* ∈ (*c, s*) is characterized with an array of [*c*(*x, d*, 0) *c*(*x, d*, 1) *c*(*x, d*, 2) *c*(*x, d*, 3)]. The index value *c*(*x, d, k*) = 1 denotes the presence of data class *k*, while the value of *c*(*x, d, k*)= 0 denotes its absence. Packets from data class *k* in channel *x* arrive to the vehicle according to a Poisson process with rate *λ*_{
x,k
}. Time unit in our model is one backoff slot. For data class *k* within channel *x* at each region (*Rg*_{
r
}), we assume variable frame size of *ld*_{x,k,r}slots which includes payload, MAC header, and physical header. The PGF for frame size within the transmission range *L* of the RSU is

L{d}_{x,k}\left(z\right)=\sum _{r=1}^{R{g}_{max}}\frac{{l}_{r}}{L}{z}^{l{d}_{x,k,r}}

(1)

where *l*_{
r
}is the length of the region *Rg*_{
r
}. Duration of the SIFS period in slots will be denoted as *sifs*. We assume that RTS/CTS transmission scheme is used. Duration of RTS, CTS, and ACK frames expressed in slots will be denoted as *rts, cts*, and *ack*, respectively. We model the channel errors through Bit Error Rate (*ber*). In each region, the probability that a frame will not be corrupted by noise is denoted as {\delta}_{k,r}={\left(1-ber\right)}^{rt{s}_{b}+ct{s}_{b}+l{d}_{b,x,k,r}+ac{k}_{b}} where subscript *b* denotes values expressed in bits. Also, we consider OBU devices equipped with a single networking interface only. Each interface switches between CCH and SCH during the sync interval, as shown in Figure 4. Duration of synchronization interval is *SI* = 100 ms and duration of guard intervals is *grd* = 5 ms each, as specified in the standard [13]. Duration of CCH is (*SI* - 2grd)*SO*. We assume that duty cycle of CCH vs. SCH is *SO* = 0.5. For this value, backoff process should be completed within (*SI*/2) ms in each channel in order to avoid handover for another RSU [3]. The handover probability in each channel interval for any packets from data class *k* is

h{d}_{x,k}=\frac{(SI/2)v}{L}SO(1-{e}^{(SI/2){\lambda}_{x,k})}\le 0.0004

(2)

where *v* is the mean speed of the vehicle. The handover probability is neglected because of its small value and we assume that the backoff process will be completed in the vicinity of a single RSU.

We assume that vehicle's buffer has infinite length and use an *M*/*G*/1 queuing model. The PGF for successful packet transmission time is

S{t}_{x,k}\left(z\right)={z}^{rts+cts+3sifs+ack}\sum _{r=1}^{R{g}_{max}}\frac{{l}_{r}}{L}{z}^{l{d}_{x,k,r}}

(3)

In case of collision of RTS packets, activity on the medium has the PGF of *Ct*(*z*) = z^{rts+cts+sifs}.

### 3.1 Distribution of vehicles

The number of vehicles in each lane of road segment follows a Poisson distribution. Let *F* be the vehicle flow which corresponds to the number of vehicles that pass a fixed roadside point per unit time. Also, let λ_{
d
}be vehicle density, i.e., the number of vehicles per unit distance in one direction along the road segment. From the traffic flow theory [4], mean speed, flow, and density are related through is

Figure 5 shows the relationships among these parameters. The solid portion of the curves represents stable or free-flow state. This state holds until the vehicle density reaches a critical value. The peak of the curves is the maximum rate of flow or the capacity of road segment. Beyond the critical density, some breakdown locations appear on the road segment, which lead to the formation of some queues of vehicles. This state is called unstable or forced flow and is shown by the dashed portion of the curves. If the density increases, the traffic reaches the jam state, at which vehicles have to completely stop [7].

Moreover, Greenshield's model [6] captures the dependency between speed and density by assuming a liner relationship as follows:

v={v}_{f}\left(1-{\lambda}_{d}\u2215{\lambda}_{d,\mathsf{\text{jam}}}\right)

(5)

where *v*_{
f
}is the free-flow speed corresponding to the maximum desired speed (usually taken as the road's speed limit). λ_{d,jam}is the maximum allowable traffic density (jam density). Then, the maximum number of vehicles, *vh*_{max}, in each lane of the road segment within the transmission range *L* of the RSU can be calculated as

v{h}_{max}=L{\lambda}_{d,\mathsf{\text{jam}}}

(6)

The probability that there are *n* vehicles in each lane of the road segment is given by

P\left(n\right)=\frac{{e}^{-L{\lambda}_{d}\frac{{\left(L{\lambda}_{d}\right)}^{n}}{n!}}}{{\sum}_{i=0}^{v{h}_{max}}{e}^{-L{\lambda}_{d}\frac{{\left(L{\lambda}_{d}\right)}^{i}}{i!}}}

(7)

### 3.2 Model of IEEE 802.11p features

Let us now briefly review AIFSN-based prioritization, as shown in Figure 6. If the medium gets busy during the backoff countdown, the backoff counter will be frozen until the medium becomes idle again for the uninterrupted duration of *aifs*_{x,k}= *sifs* + AIFSN_{x,k}· *σ* where *σ* is the duration of the slot time and *x* ∈ (*c*,*s*) denotes the channel type. We model this action through *k* freezing counters [14]. They need to be re-started from the beginning if the medium becomes busy at any time during the countdown. Since access is synchronized with the end of previous transmission, no access is possible during *AIFS*_{x,3}. Figure 7 presents the Markov chain for freezing countdown, the initial values of freezing counters are set to *B*_{x,k}= *AIFSN*_{x,k}- *AIFSN*_{x,3}, *k* = 0.. 3.

When the backoff count reaches zero, the transmission starts. If a collision occurs, backoff procedure has to be repeated but with increased contention window size. There are *m*_{x,k}+ 1 backoff phases, starting from the phase 0, with increasing values of contention window. Frame can be re-transmitted up to *R* times, but window size will grow only until phase *m*_{x,k}as indicated in Tables 1 and 2. The size of contention window for channel *x* ∈ (*c, s*) and *AC* = *k* in the *i* th backoff stage (*i* = 0.. *m*_{x,k}) has the value of

{W}_{x,k,i}=\left\{\begin{array}{cc}\hfill {2}^{i}\left(CW{min}_{x,k}+1\right)-1,\hfill & \hfill 0\le i\le {m}_{x,k}\hfill \\ \hfill {2}^{{m}_{k,L}}\left(CW{min}_{x,k}+1\right)-1,\hfill & \hfill R\ge i>{m}_{x,k}\hfill \end{array}\right.

(8)

As shown in Figure 6, duration of periods where data classes *k* and higher can access channel *x* are denoted as *A*_{x,k}and their maximum durations are

{A}_{x,k,max}=\left\{\begin{array}{cc}\hfill AIFS{N}_{x,k-1}-AIFS{N}_{x,k},\hfill & \hfill k=1..3\hfill \\ \hfill {W}_{x,0,max},\hfill & \hfill k=0\hfill \end{array}\right.

(9)

Because of internal collisions of frames from different data classes within the vehicle itself, probability that a frame has completed the backoff process (we will denote it as virtual access probability, *θ*_{x,d,k}) might be different from access probability *τ*_{x,d,k}:

\begin{array}{c}{\tau}_{x,d,3}={\theta}_{x,d,3}\\ {\tau}_{x,d,k}={\theta}_{x,d,k}\prod _{l=k+1}^{3}\left(1-c\left(x,d,k\right){\theta}_{x,d,l}\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}k=0..2\end{array}

(10)

Let the probability that a backoff period is the last in the target channel be {\mathcal{X}}_{c}=\frac{1}{cch} and {\mathcal{X}}_{s}=\frac{1}{sch} for CCH and SCH, respectively. Probability that the medium will be idle during a single backoff slot in the target channel interval *A*_{x,k}is

{f}_{x,k}=\left(1-{\mathcal{X}}_{k}\right)\prod _{t=1}^{{t}_{x}}\prod _{l=k}^{3}{\left(1-c\left(x,d,k\right){\tau}_{x,d,l}\right)}^{{n}_{x,d,l}}

(11)

where *n*_{x,d,k}is the total number of vehicles from channel *x* ∈ *c, s* with data combination *d*, transmitting frames from data class *k*. However, because of virtual collisions, what the stream *c*(*x, d, k*) = 1 actually 'sees' is that no data class within the vehicle has finished the backoff process and that there is no access by any other vehicle. This value might be different from *f*_{x,k}, since it observes absolute access of data classes outside the vehicle and virtual access of classes inside the vehicle as

{\zeta}_{x,d,k}={f}_{x,k}\frac{{\prod}_{l=k+1}^{3}\left(1-c\left(x,d,k\right){\theta}_{x,d,l}\right)}{{\prod}_{l=k+1}^{3}\left(1-c\left(x,d,k\right){\tau}_{x,d,l}\right)}

(12)

Backoff counter for stream *c*(*x, d, k*) = 1, *x* ∈ (*c, s*), can be decremented during time periods *A*_{x,l}, *l* = 0..*k*, but unfortunately, with probabilities that depend on the index of the period. The overall probability of backoff counter decrement for *AC*_{
i
}is

\begin{array}{ll}\hfill {g}_{x,d,0}& =\frac{{\zeta}_{x,d,0}}{1-{\theta}_{x,d,0}}\phantom{\rule{2em}{0ex}}\\ \hfill {g}_{x,d,i}& =\frac{\left(1-{\zeta}_{x,d,i}^{{A}_{x,i,max}}\right){\zeta}_{x,d,i}}{1-{\theta}_{x,d,i}}+\frac{{\zeta}_{x,d,i}^{{A}_{x,i,max}}\left({\zeta}_{x,d,i}-1\right)}{1-{\theta}_{x,d,i}},i=1..3\phantom{\rule{2em}{0ex}}\end{array}

(13)

The probability that a transmission for AC*k* in channel *x* does not experience a virtual or real collision is {\gamma}_{{n}_{x,d,k}}={g}_{x,d,k}. The overall collision probability per lane for data combination *d* with data class *k* in channel *x* can be calculated as

E\left[co{l}_{x,d,k}\right]=1-\sum _{n=0}^{v{h}_{max}}P\left(n\right){\gamma}_{{n}_{x,d,k}}

(14)

### 3.3 PGF for the frame service time

The basic building block of the packet service time PGF is the transfer PGF for the pass through the freezing counter block which has loopback when the value of backoff counter is non-zero, and skips loopback when the backoff counter is zero. Note that transfer PGFs through the block are different if the block has to be traversed vertically from higher backoff state or laterally from the same backoff state but from higher value of backoff counter.

Let us first consider vertical traversal through the freezing counter, in which case we need to derive the probabilities *Pfa*_{l} that freezing counters of vehicles with data classes *k* < *l* will be restarted because of successful transmission on the medium in the period *A*_{x,l}, *l* = 1, 2, 3.

Pf{a}_{x,l}=\sum _{d=1}^{{d}_{x}}\sum _{i=1}^{3}\frac{{n}_{x,d,i}{\tau}_{x,d,i{{\prod}_{i=l}^{3}\left(1-{\tau}_{x,d,i}\right)}^{{n}_{x,d,i}}}}{\left(1-{\tau}_{x,d,i}\right)}

(15)

The probability that freezing counter will be restarted because of a collision is *Pfc*_{x,l}= 1 - *f*_{x,l}- *Pfa*_{x,l}.

We also need to derive duration of successful access to the medium during period *A*_{x,l}, *l* = 0.. 3 as

S{w}_{x,l}\left(z\right)=\frac{{\sum}_{d=1}^{{d}_{x}}{\sum}_{i=1}^{3}{n}_{x,d,i}{\tau}_{x,d,i}S{t}_{x,k}\left(z\right)}{{\sum}_{d=1}^{{d}_{x}}{\sum}_{i=1}^{3}{n}_{x,d,i}{\tau}_{x,d,i}}

Transfer PGFs *Bfnl*_{x,k}(*z*) for the zero value of backoff counter can be derived by noting that PGF is recursive because of the presence of a loop and has the form *Bfnl*_{x,k}(*z*) = *Bfnln*_{x,k}(*z*)/*Bfnld*_{x,k}(*z*) where

\begin{array}{ll}\hfill Bfnl{n}_{x,k}\left(z\right)& ={z}^{aif{s}_{x,3}}\prod _{i=k+1}^{3}{\left({f}_{x,i}z\right)}^{{A}_{x,i,max}}\phantom{\rule{2em}{0ex}}\\ \hfill Bfnl{d}_{x,k}\left(z\right)& =1-\sum _{i=k+1}^{3}\left(Pf{a}_{x,i}S{w}_{x,i}\left(z\right)+Pf{c}_{x,i}Ct\left(z\right)\right)\phantom{\rule{2em}{0ex}}\\ \sum _{n=0}^{{A}_{x,i,max}-1}{\left(z{f}_{x,i}\right)}^{n}z\prod _{l=i+1}^{3}{\left(z{f}_{x,l}\right)}^{{A}_{x,l,max}}\phantom{\rule{2em}{0ex}}\end{array}

except for the highest traffic class where *Bfnl*_{x,3}(*z*) = *z*.

In order to calculate vertical transfer PGFs for non-zero values of backoff counter, we need the probabilities that the backoff count will be suppressed because of successful transmission or collision. Probability that the backoff count for traffic class *k* will be suppressed in the period *A*_{x,l}, *l* = 0..*k*, because of successful transmission is

\begin{array}{ll}\hfill Pb{s}_{x,d,k,l}& =\sum _{i=l}^{3}\frac{{n}_{x,d,i}{\tau}_{x,d,i}{\prod}_{i=1}^{3}{\left(1-{\tau}_{x,d,i}\right)}^{{n}_{x,d,i}}}{\left(1-{\tau}_{x,d,k}\right)\left(1-{\tau}_{x,d,i}\right)}\phantom{\rule{2em}{0ex}}\\ -\frac{{\tau}_{x,d,k}}{{\left(1-{\tau}_{x,d,k}\right)}^{2}}\prod _{i=l}^{3}{\left(1-{\tau}_{x,d,i}\right)}^{{n}_{x,d,i}}\phantom{\rule{2em}{0ex}}\end{array}

(16)

The corresponding probabilities that the backoff count will be suppressed because of the collision on the medium are Pb{c}_{x,d,k,l}=1-\frac{{f}_{x,l}}{1-{\tau}_{x,d,k}}-Pb{s}_{x,d,k,l}.

Then, the vertical transfer PGF for non-zero value of backoff counter gets the form *Bfl*_{x,d,k}(*z*) = *α*/*β* where \alpha =c\left(x,d,k\right)\frac{{f}_{x,k}}{1-{\tau}_{x,t,k}}zBfn{l}_{x,k}\left(z\right) and *β* = 1 - *zBfnl*_{x,k}(*z*)(*Pbs*_{x,d,k}*Sw*_{x,k}(*z*) + *Pbc*_{x,d,k}*Ct*(*z*)).

Lateral transfer PGFs which connect the backoff freezing blocks, *Bfs*_{x,d,k}(*z*), can be calculated as defined in [3].

The PGF for the duration of the backoff phase *i* for data class *k* on channel *x* is

{B}_{x,d,k,i}\left(z\right)=\frac{Bfn{l}_{x,k}\left(z\right)}{{W}_{x,k,i}}+\frac{Bf{l}_{x,d,k}\left(z\right)}{{W}_{x,k,i}}\sum _{l=1}^{{W}_{x,k,i}-1}Bf{s}_{x,d,k}{\left(z\right)}^{l}

(17)

The probability that the OBU's buffer is empty at arbitrary time is *π*_{x,d,k,0}= 1-*ρ*_{x,d,k}, where *ρ*_{x,d,k}denotes offered load from data combination *d* with class *k* in channel *x*. The entire Markov chain for *AC*_{
k
}is shown in Figure 8. Markov chain has the same form for both channels since activity of the other channel is equivalent to channel busy state on the current channel. By considering Figure 8, the PGF for the backoff time for data combination *d* with data class *AC*_{
k
}in channel *x* becomes

\begin{array}{ll}\hfill Bo{f}_{x,d,k}\left(z\right)& =\sum _{r=1}^{R{g}_{max}}\frac{{l}_{r}}{L}\left(\sum _{i=1}^{{m}_{x,k}+1}\left(\prod _{j=0}^{i-1}{B}_{x,d,k,j}\left(z\right)\right)\right.\phantom{\rule{2em}{0ex}}\\ \cdot {\left(1-{\delta}_{k,r}{\gamma}_{x,d,k}\right)}^{i-1}Ct{\left(z\right)}^{i-1}{\delta}_{k,r}{\gamma}_{x,d,k}\phantom{\rule{2em}{0ex}}\\ +\sum _{i={m}_{x,k}+1}^{R}\left(\prod _{j=0}^{{m}_{x,k}}{B}_{x,d,k,j}\left(z\right)\right){B}_{x,d,k,{m}_{x,k}}{\left(z\right)}^{i-{m}_{x,k}}\phantom{\rule{2em}{0ex}}\\ \cdot {\left(1-{\delta}_{k,r}{\gamma}_{x,d,k}\right)}^{i}Ct{\left(z\right)}^{i}{\delta}_{k,r}{\gamma}_{x,d,k}\phantom{\rule{2em}{0ex}}\\ \left(\right)close=")">+\left(\prod _{j=0}^{{m}_{x,k}}{B}_{x,d,k,j}\left(z\right)\right){B}_{x,d,k,{m}_{x,k}}{\left(z\right)}^{R-{m}_{x,k}}& \phantom{\rule{2em}{0ex}}\end{array}\n \n \n close=")">\n \n \u22c5\n \n \n \n (\n \n 1\n -\n \n \n \delta \n \n \n k\n ,\n r\n \n \n \n \n \gamma \n \n \n x\n ,\n d\n ,\n k\n \n \n \n )\n \n \n \n R\n +\n 1\n \n \n C\n t\n \n \n \n (\n \n z\n \n )\n \n \n \n R\n +\n 1\n \n \n \n \n \n \n

(18)

When the buffer is found empty upon a successful packet transmission, the vehicle can proactively undertake zeroth backoff, go to the idle state, and then attempt to transmit the next arriving packet upon waiting only for *AIFS*_{x,k}. However, if this attempt is not successful, the entire backoff process (beginning from phase 1) has to be performed; the PGF for the duration of this backoff process, *Bzof*_{x,d,k}(*s*), is similar to the one in (18) but with a difference in the first line:

\begin{array}{ll}\hfill Bzo{f}_{x,d,k}\left(z\right)& =\sum _{r=1}^{R{g}_{max}}\frac{{l}_{r}}{L}\left({\gamma}_{x,d,k{\delta}_{k,r}}+\sum _{i=2}^{{m}_{x,k}+1}\left(\prod _{j=0}^{i-1}{B}_{x,d,k,j}\left(z\right)\right)\right.\phantom{\rule{2em}{0ex}}\\ \cdot {\left(1-{\delta}_{k,r}{\gamma}_{x,d,k}\right)}^{i-1}Ct{\left(z\right)}^{i-1}{\delta}_{k,r}{\gamma}_{x,d,k}\phantom{\rule{2em}{0ex}}\\ +\sum _{i={m}_{x,k}+1}^{R}\left(\prod _{j=0}^{{m}_{x,k}}{B}_{x,d,k,j}\left(z\right)\right){B}_{x,d,k,{m}_{x,k}}{\left(z\right)}^{i-{m}_{x,k}}\phantom{\rule{2em}{0ex}}\\ \cdot {\left(1-{\delta}_{k,r}{\gamma}_{x,d,k}\right)}^{i}Ct{\left(z\right)}^{i}{\delta}_{k,r}{\gamma}_{x,d,k}\phantom{\rule{2em}{0ex}}\\ \left(\right)close=")">+\left(\prod _{j=0}^{{m}_{x,k}}{B}_{x,d,k,j}\left(z\right)\right){B}_{x,d,k,{m}_{x,k}}{\left(z\right)}^{R-{m}_{x,k}}& \phantom{\rule{2em}{0ex}}\end{array}\n \n \n close=")">\n \n \u22c5\n \n \n \n (\n \n 1\n -\n \n \n \delta \n \n \n k\n ,\n r\n \n \n \n \n \gamma \n \n \n x\n ,\n d\n ,\n k\n \n \n \n )\n \n \n \n R\n +\n 1\n \n \n C\n t\n \n \n \n (\n \n z\n \n )\n \n \n \n R\n +\n 1\n \n \n \n \n \n \n

(19)

At this point, these expressions can be used to calculate extended backoff times *Bofe*_{c,d,k}(*z*) and *Bzofe*_{c,d,k}(*z*), as defined in [3].

Markov chain presents a random process with stationary distribution *y*_{x,d,k,i,j,b}, where *x* ∈ (*c*,*s*), *d* = 1..*d*_{
x
}denotes vehicle type; *k* = 0.. 3 denotes the data class, *i* = 0.. *m*_{
k
}denotes the index of the backoff phase, *j* = *0*.. *W*_{x,k,i}- 1 denotes the value of backoff counter, and *b* = 0.. *B*_{
k
}denotes the value of the freezing counter.

In order to model behavior of vehicle after the successful transmission we need to note that standard requires vehicle to perform backoff with *W*_{x,k,0}immediately after successful transmission even if the vehicle's buffer is empty. If vehicle's buffer is still empty after this backoff count, vehicle enters idle state represented by *P*_{
idle
}in Figure 8.

The total idle state probability is calculated by finding the distance between two accesses by the same vehicle. In order to obtain this result we use Laplace-Stieltjes Transform (LST) of backoff time PGF which is obtained as Bof{e}_{x,d,k}^{*}\left(s\right)=Bof{e}_{x,d,k}\left({e}^{-s}\right). If we introduce probability of no frame arrivals during zeroth backoff on channel *x* as {v}_{x,d,k,0}^{0} and LST of backoff process without the zeroth backoff phase as Bzof{e}_{x,d,k}^{*}\left(s\right) the distance between two transmissions becomes

\begin{array}{c}{D}_{x,d,k}^{*}(s)={\displaystyle \sum _{r=1}^{R{g}_{\mathrm{max}}}\frac{{l}_{r}}{L}(S{t}_{x,k}({e}^{-s})\{{\pi}_{x,d,k,0}{B}_{x,d,k,0}({e}^{-s})}\\ {v}_{x,d,k,0}^{0}{R}_{x,k}^{*}(s)[{\varphi}_{i}{I}_{x}^{*}(s)({\displaystyle \prod _{l=k+1}^{3}{f}_{x,l}^{{A}_{x,l,\mathrm{max}}}{\delta}_{k,r}}\\ +\left(1-{\displaystyle \prod _{l=k+1}^{3}{f}_{x,l}^{{A}_{x,l,\mathrm{max}}}}{\delta}_{k,r}\right)Bof{e}_{x,d,k}^{*}(s))+{\varphi}_{a}({f}_{x,0}^{aif{s}_{x,k}}{\delta}_{k,r}\\ +(1-{f}_{x,0}^{aif{s}_{x,k}}{\delta}_{k,r})Bof{e}_{x,d,k}^{*}(s))]{e}^{-saif{s}_{x,k}}\\ +{\pi}_{x,d,k,0}{B}_{x,d,k,0}({e}^{-s})(1-{v}_{x,d,k,0}^{0})({\gamma}_{x,d,k}{\delta}_{k,r}\\ +(1-{\gamma}_{x,d,k}{\delta}_{k,r})Bzof{e}_{x,d,k}^{*}(s))\\ +(1-{\pi}_{x,d,k,0})Bof{e}_{x,d,k}^{*}(s)\})\end{array}

(20)

where {R}_{x,k}^{*}\left(s\right)=\frac{{\lambda}_{x,k}}{{\lambda}_{x,k}+s} is the LST for exponential distribution of the residual frame inter-arrival time and {I}_{x}^{*}\left(s\right) denotes LST of time needed to synchronize with the beginning of CCH or SCH period [3]. *ϕ*_{
a
}is the probability that the target channel is active when the vehicle exits idle state and *ϕ*_{
i
}is the probability that the opposite channel (or guard time) is active. Then, the LST for the total active time between two successive access point, i.e., frame service time is

\begin{array}{c}T{t}_{x,d,k}^{*}(s)={\displaystyle \sum _{r=1}^{R{g}_{\mathrm{max}}}\frac{{l}_{r}}{L}}(S{t}_{x,k}({e}^{-s})\{{\pi}_{x,d,k,0}{B}_{x,d,k,0}({e}^{-s})\\ {v}_{x,d,k,0}^{0}[{\varphi}_{i}{I}_{x}^{*}(s)({\displaystyle \prod _{l=k+1}^{3}{f}_{x,l}^{{A}_{x,l,\mathrm{max}}}}{\delta}_{k,r}\\ +\left(1-{\displaystyle \prod _{l=k+1}^{3}{f}_{x,l}^{{A}_{x,l,\mathrm{max}}}{\delta}_{k,r}}\right)Bof{e}_{x,d,k}^{*}(s))+{\varphi}_{a}({f}_{x,0}^{aif{s}_{x,k}}{\delta}_{k,r}\\ +(1-{f}_{x,0}^{aif{s}_{x,k}}{\delta}_{k,r})Bof{e}_{x,d,k}^{*}(S))]{e}^{-saif{s}_{x,k}}\\ +{\pi}_{x,d,k,0}{B}_{x,d,k,0}({e}^{-s})(1-{v}_{x,d,k,0}^{0})({\gamma}_{x,d,k}{\delta}_{k,r}\\ +(1-{\gamma}_{x,d,k}{\delta}_{k,r})Bzof{e}_{x,d,k}^{*}(s))\\ +(1-{\pi}_{x,d,k,0})Bof{e}_{x,d,k}^{*}(s)\})\end{array}

(21)

and its first two moments are {\left(\right)close="|">\overline{T{t}_{x,d,k}}=\frac{dT{t}_{x,d,k}^{*}\left(s\right)}{ds}}_{}s=0\n and {\left(\right)close="|">\overline{T{t}_{x,d,k}^{\left(2\right)}}=\frac{{d}^{2}T{t}_{x,d,k}^{*}\left(s\right)}{d{s}^{2}}}_{}\n \n s\n =\n 0\n \n. Offered load on the channel now becomes {\rho}_{x,d,k}={\lambda}_{x,k}\overline{T{t}_{x,d,k}}.

The probability that the vehicle is idle can be calculated as {P}_{x,d,k,idle}=1-\frac{\overline{T{t}_{x,d,k}}}{\overline{{D}_{x,d,k}}}. Now we can form the equation for normalization condition as defined in [3].

### 3.4 Throughput and waiting time for *AC*_{
k
}in channel *x*

Normalized throughput per vehicle for data combination *d* with data class *k* in channel *x* can be obtained

as

Th{r}_{{n}_{x,d,k}}=\frac{\overline{l{d}_{x,k}}}{\overline{{D}_{x,d,k}}}

(22)

The overall normalized throughput per lane can be calculated as

E\left[Th{r}_{x,d,k}\right]=\sum _{n=0}^{v{h}_{max}}P\left(n\right)Th{r}_{{n}_{x,d,k}}

(23)

The PGF for the number of frame arrivals during frame service time is {A}_{x,d,k}\left(z\right)=T{t}_{x,d,k}^{*}\left({\lambda}_{x,k}\left(1-z\right)\right), and the PGF for the number of frames left after the departing frame in the vehicle buffer is [15]

{\prod}_{x,d,k}\left(z\right)=\frac{{\pi}_{x,d,k,0}\left(1-z\right){A}_{x,d,k}\left(z\right)}{{A}_{x,d,k}\left(z\right)-z}

(24)

In order to calculate the response time, we observe that number of frames left in the buffer after the departing frame is equal to the number of frames which arrived while the frame was buffered and serviced. If we denote the LST of the response time for *AC*_{
k
}in channel *x* as {W}_{x,d,k}^{*}\left(s\right), last statement can be written as {\prod}_{x,d,k}\left(z\right)={W}_{x,d,k}^{*}\left({\lambda}_{x,k}-z{\lambda}_{x,k}\right)T{t}_{x,d,k}^{*}\left({\lambda}_{x,k}-z{\lambda}_{x,k}\right). Because of PASTA property of M/G/1 systems, ∏_{x,d,k}(*z*) also presents probability distribution of buffer occupancy at arbitrary time. After substituting *s* = λ_{x,k}- *z* λ_{x,k}, we obtain the LST of the waiting time as {W}_{x,d,k}^{*}\left(s\right)=\frac{s\left(1-{\rho}_{x,d,k}\right)}{s-{\lambda}_{x,k}+{\lambda}_{x,k}T{t}_{x,d,k}^{*}\left(s\right)}, and its average value as \overline{{W}_{{n}_{x,d,k}}}=\frac{{\lambda}_{x,k}\overline{T{t}_{k}^{\left(2\right)}}}{2\left(1-{\rho}_{x,d,k}\right)}. Similar to (23), the overall waiting time per lane can be calculated as

E\left[\overline{{W}_{x,d,k}}\right]=\sum _{n=0}^{v{h}_{max}}P\left(n\right)\overline{{W}_{{n}_{x,d,k}}}

(25)