Conditional Waiting Time Analysis in Tandem Polling Queues

We consider a system of tandem polling queues with two stations $\left(j={}1,2\right)$, and two types of customer class $\left(i={}1,2\right)$ $\left(\text{see Figure }\ref{fig:mesh3.1}\right)$. Customers of type $i$ arrive to their respective queue at station 1 according to Poisson process with parameter $\lambda_{i}$ and have service time with exponential distribution having parameter $\mu_{i1}$ and $\mu_{i2}$ at station 1 and 2 respectively. We define $\tau_{ij}$ $\left(={}1/\mu_{ij}\right)$ as the mean service time of customer i at station j. Each customer first gets served at station 1 and then at station 2 before exiting the system. At each station, the server uses a cyclic discipline with exhaustive service to serve customers of type 1 and type 2 alternatively. We assume that the buffers are infinite and that switchover time is negligible at each station. The state of the system at time $t$ is defined by $\Big{(}\,L_{11}\left(t\right),L_{21}\left(t\right),S_{1}\left(t\right),L_{12}\left(t\right),L_{22}\left(t\right),S_{2}\left(t\right)\,\Big{)}$ where $L_{ij}\left(t\right)$ is the number of customers of type $i$ at station $j$ at time $t$ and $S_{j}(t)$ is the queue being served at station $j$ at time $t$. Also, let $\boldsymbol{L^{a}}={}\left(\,L_{11}\left(0\right),L_{21}\left(0\right),L_{12}\left(0\right),L_{22}\left(0\right)\,\right)$ denote the queue length seen by an arriving customer at $t={}0$.

The traffic intensity $\rho_{ij}$ at queue i of station j is defined as $\rho_{ij}={}\lambda_{i}\tau_{ij}$ and the total traffic intensity at station j, $\rho_{j}$, is defined as $\rho_{j}={}\sum_{i={}1}^{2}\rho_{ij}$. Then, from Takagi et al. [15], for the system to be stable, it is required that $\rho_{j}<$ 1 for each $j$. We assume this condition holds for our system. Let $W_{ij}^{a}$ denote the conditional waiting times for arriving type $i$ customer at station $j$ seeing a system with queue lengths $\boldsymbol{L^{a}}$ and let $W_{i}^{a}$ denote the conditional waiting times for arriving type $i$ customer in the system seeing a system with queue lengths $\boldsymbol{L^{a}}$. Then, $\displaystyle\mathop{\mathbb{E}}\left[W_{i}^{a}\right]={}\displaystyle\mathop{\mathbb{E}}\left[W_{i1}^{a}\right]+\displaystyle\mathop{\mathbb{E}}\left[W_{i2}^{a}\right]$ and our goal in this paper is to determine $\displaystyle\mathop{\mathbb{E}}\left[W_{i}^{a}\right]$ for $i={}1,2$.

To determine $\displaystyle\mathop{\mathbb{E}}\left[W_{i}^{a}|m\right]$ for an arriving customer in the system, we consider different sub-scenarios $\left(n={}\text{1, 2, \ldots}\right)$ for each of the scenarios $m$. A sub-scenario is defined as sequence of events required for an arriving tagged customer to exit the system from station 2. Then, to determine $\displaystyle\mathop{\mathbb{E}}\left[W_{i}^{a}|m\right]$, we first determine the mean conditional waiting times $\displaystyle\mathop{\mathbb{E}}\Big{[}W_{i}^{a}|m,n\Big{]}$ for each sub-scenarios under scenario $m$, and then determine the conditional expected waiting times $\displaystyle\mathop{\mathbb{E}}\left[W_{i}^{a}|m\right]$ using Equation $\left(\ref{eq:3.1}\right)$ below$\colon$

$$\displaystyle\mathop{\mathbb{E}}\left[W_{i}^{a}|m\right]={}\sum_{s={}1}^{\infty}\displaystyle\mathop{\mathbb{E}}\left[W_{i}^{a}|m,n={}s\right]\Pr\left(n={}s|m\right)$$

(1)

where $\Pr\left(n={}s|m\right)$ is the probability of sub-scenarios under scenario $m$. The details of the analysis of these sub-scenarios are provided in Section 5. However, it is useful to make an important observation regarding this analysis here. The events that constitute the various sub-scenarios for each scenarios are summarized in Table 1. We simplify our analysis by leveraging the occurrence of repeating events across sub-scenarios. For instance, in sub-scenarios $n\geq 5$ for scenario $m={}1$, the sequence of events begin to repeat. Therefore, we use the analysis of events in sub-scenario $n={}4$ to analyze the events in the sub-scenarios $n\geq 5$. Additionally, for different sub-scenarios under scenarios $m={}2,3,$ and 4, either portions of sequence of event are similar to the sequence of events observed for sub-scenarios under scenario $m={}1$ or have some additional events than the sequence of events observed for sub-scenarios under scenario $m={}1$ $\left(\text{see highlights in Table $\ref{T:3.1}$}\right)$. We use this information to simplify our analysis of scenarios $m={}2,3,$ and 4.

We first determine the conditional waiting times and probabilities for each sub-scenarios and then, calculate the waiting time $\displaystyle\mathop{\mathbb{E}}\left[W_{1}^{a}|m={}1\right]$ using Equation $\left(\ref{eq:3.7}\right)$.

$$\displaystyle\mathop{\mathbb{E}}\left[W_{1}^{a}|m={}1\right]={}\sum_{s={}1}^{\infty}\displaystyle\mathop{\mathbb{E}}\left[W_{1}^{a}|m={}1,n={}s\right]\Pr\left(n={}s|m={}1\right)$$

(7)

From the perspective of the tagged customer arriving at station 1 at $t={}0$, the sub-scenario $\left(m={}1,n={}1\right)$ corresponds to a sequence of two events, A and B, that needs to occur before the tagged customer departs from station 2 $\left(\text{see Figure \ref{fig:mesh3.11}}\right)$. In the event A, for some $K_{1}>L_{11}^{a}$, exactly $K_{1}$ customers are served at station 1 before station 2 server switches to serve queue 2 at $t={}t_{B}$. As $K_{1}\geq L_{11}^{a}+1$, the tagged customer gets served at station 1 and moves to station 2 before the server at station 2 switches to queue 2. In the event B, the tagged customer and the $L_{11}^{a}$ customers ahead of it are served at station 2 in the first cycle of station 2. It is important to note that at no instance until the service of tagged customer at station 2 is queue 1 at station 2 empty. The probability of this sub-scenario $\displaystyle\mathop{\mathbb{P}}\Big{(}n={}1|m={}1\Big{)}$ that $L_{11}^{a}+1$ type 1 customers are served at station 1 before queue 1 at station 2 becomes empty is given by Equation $\left(\ref{eq:3.8}\right)$ and is determined using Result 4.

$$\displaystyle\mathop{\mathbb{P}}\Big{(}n={}1|m={}1\Big{)}={}\displaystyle\mathop{\mathbb{P}}\Big{(}K_{1}>L_{11}^{a}\Big{)}$$

(8)

For sub-scenario $\left(m={}1,n={}1\right)$, $\displaystyle\mathop{\mathbb{E}}\Big{[}W_{1}^{a}|m={}1,n={}1\Big{]}$ is the expected time needed to serve the $L_{12}^{a}$ customers at station 2 at $t={}0$ and the $L_{11}^{a}+1$ customers that moved to station 2 after being served at station 1. For sub-scenario $\left(m={}1,n={}1\right)$, we write $\displaystyle\mathop{\mathbb{E}}\Big{[}W_{1}^{a}|m={}1,n={}1\Big{]}$ as

$$\displaystyle\mathop{\mathbb{E}}\Big{[}W_{1}^{a}|m={}1,n={}1\Big{]}={}\displaystyle\mathop{\mathbb{E}}\Big{[}h_{12}\Big{(}L_{11}^{a}+L_{12}^{a}+1\Big{)}\Big{]}$$

(9)

Next, we conduct a similar analysis for sub-scenarios $n={}2,3,4$ for scenario $m={}1$.

From the perspective of the tagged customer arriving at station 1 at $t={}0$, the sub-scenario $\left(m={}1,n={}2\right)$ corresponds to a sequence of 4 events: A’, C, D, and $\text{E}_{1}$ before the tagged customer exits the system as shown in Figure 12. In event A’, exactly $K_{1}$ customers, where $K_{1}\leq L_{11}^{a}$, are served at station 1 before station 2 server switches to serve queue 2 at $t={}t_{C}$ after serving $L_{12}^{a}+K_{1}$ type 1 customers. The time $T_{A}$ it takes for server at station 2 to serve $L_{12}^{a}+K_{1}$ customers of queue 1 is $h_{12}\Big{(}L_{12}^{a}+K_{1}\Big{)}$. The probability $\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event A'}\Big{)}$ is the complement of $\displaystyle\mathop{\mathbb{P}}\Big{(}n={}1|m={}1\Big{)}$.

In event C, all customers of queue 2 at station 2 at $t={}t_{C}$ are served before the tagged customer is served at station 1. The time $T_{C}$ it takes for the server at station 2 to serve $L_{22}\left(t_{C}\right)$ customers of queue 2 is $h_{22}\left(L_{22}\left(t_{C}\right)\right)$, where $L_{22}\left(t_{C}\right)={}L_{22}^{a}$. The number of customers ahead of the tagged customer at $t={}t_{C}$ is

$$L_{11}^{{}^{\prime}}\left(t_{C}\right)={}L_{11}^{a}-K_{1}$$

(10)

where the conditional probability of $K_{1}$ is given by Equation $\left(\ref{eq:3.12}\right)$ that uses $\displaystyle\mathop{\mathbb{P}}\Big{(}K_{1}={}u\Big{)}$ obtained using Result 4.

$$\displaystyle\mathop{\mathbb{P}}\Big{(}K_{1}={}k|K_{1}\leq L_{11}^{a}\Big{)}={}\frac{\displaystyle\mathop{\mathbb{P}}\Big{(}K_{1}={}v\Big{)}}{\sum_{k=0}^{{}L_{11}^{a}}\displaystyle\mathop{\mathbb{P}}\Big{(}K_{1}={}k\Big{)}}$$

(11)

The probability $\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event C}\Big{)}$ that $L_{22}\left(t_{C}\right)$ customers of queue 2 at station 2 are served before $L_{11}^{{}^{\prime}}\left(t_{C}\right)+1$ customer gets served at station 1 is given by Equation $\left(\ref{eq:3.12}\right)$ and is determined using Result 6.

	$\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event C}\Big{)}$	$\displaystyle={}\mathop{\mathbb{P}}\Big{(}h_{22}\Big{(}L_{22}\left(t_{C}\right)\Big{)}<{}h_{11}\left(L_{11}^{{}^{\prime}}\left(t_{C}\right)+1\right)\Big{)}$		(12)
		$\displaystyle={}\sum_{k={}1}^{L_{11}^{a}}\mathop{\mathbb{P}}\Big{(}h_{22}\Big{(}L_{22}\left(t_{C}\right)\Big{)}<{}h_{11}\left(L_{11}^{a}-k+1\right)\Big{)}\mathop{\mathbb{P}}\Big{(}K_{1}={}k|K_{1}\leq L_{11}^{a}\Big{)}$

In the events D and $\text{E}_{1}$, the tagged customer first gets served at station 1 after the service of $L_{11}^{{}^{\prime}}\left(t_{D}\right)$ customers ahead of it and then moves to station 2 at $t={}t_{E_{1}}$, where it eventually gets served and exits the system. It is important to note that as queue 1 is being served at station 1 during event D, $L_{22}\left(t\right)$ remains 0 for $t_{D}\leq t\leq t_{E_{1}}$ and hence, the server at station 2 does not switch to serve queue 2 once it empties queue 1. Thus, $\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event D}\Big{)}\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event $\text{E}_{1}$}\Big{)}={}1$.

Let $V_{1}$ be the number of customers that got served in queue 1 at station 1 during event C. As the tagged customer did not get served at station 1 during event C, the number of customers ahead of the tagged customer at $t={}t_{D}$ is equal to $L_{11}^{{}^{\prime}}\left(t_{D}\right)={}{}L_{11}^{a}-K_{1}-V_{1}$ where the conditional probability distribution of $V_{1}$ is

$$\displaystyle\mathop{\mathbb{P}}\Big{(}V_{1}={}v|V_{1}\leq L_{11}^{a}-K_{1}\Big{)}={}\frac{\displaystyle\mathop{\mathbb{P}}\Big{(}V_{1}={}v\Big{)}}{\sum_{v=0}^{{}L_{11}^{a}-K_{1}}\displaystyle\mathop{\mathbb{P}}\Big{(}V_{1}={}v\Big{)}}$$

(13)

and

$$\displaystyle\mathop{\mathbb{P}}\Big{(}V_{1}={}v\Big{)}=\Big{(}1/v!\Big{)}e^{-\mu_{11}\left(T_{C}\right)}\Big{(}\mu_{11}\left(T_{C}\right)\Big{)}^{v}$$

(14)

Note that the number of type 1 customers at station 2 at $t={}t_{D}$, $L_{12}\left(t_{D}\right)$, is $V_{1}$. At $t={}t_{D}$, we know that $L_{11}^{{}^{\prime}}\left(t_{D}\right)$ customers are ahead of the tagged customer at station 1 and $L_{12}\left(t_{D}\right)$ customers are queued at station 2. Let $\displaystyle\mathop{\mathbb{E}}\Big{[}W_{L_{11}^{{}^{\prime}}\left(t_{D}\right),L_{12}\left(t_{D}\right)}\Big{]}$ be the mean waiting time of the tagged customer for events D and $\text{E}_{1}$. We can determine $\displaystyle\mathop{\mathbb{E}}\Big{[}W_{L_{11}^{{}^{\prime}}\left(t_{D}\right),L_{12}\left(t_{D}\right)}\Big{]}$ using Result 5.

Then, based on this analysis, for sub-scenario $\left(m={}1,n={}2\right)$, $\displaystyle\mathop{\mathbb{E}}\Big{[}W_{1}^{a}|m={}1,n={}2\Big{]}$ is equal to the sum of the mean times for events A’, C, D, and $\text{E}_{1}$ and is given by Equation $\left(\ref{eq:3.15}\right)$.

$\displaystyle\mathop{\mathbb{E}}\Big{[}W_{1}^{a}|m={}1,n={}2\Big{]}={}\mathop{\mathbb{E}}\Big{[}h_{12}\Big{(}L_{12}^{a}+K_{1}\Big{)}\Big{]}+\mathop{\mathbb{E}}\Big{[}h_{22}\Big{(}L_{22}\left(t_{c}\right)\Big{)}\Big{]}+\mathop{\mathbb{E}}\Big{[}W_{L_{11}^{{}^{\prime}}\left(t_{D}\right),L_{12}\left(t_{D}\right)}\Big{]}$

(15)

The probability of this sub-scenario $\displaystyle\mathop{\mathbb{P}}\Big{(}n={}2|m={}1\Big{)}$ is given by Equation $\left(\ref{eq:3.16}\right)$.

$$\displaystyle\mathop{\mathbb{P}}\Big{(}n={}2|m={}1\Big{)}=\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event A'}\Big{)}\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event C}\Big{)}$$

(16)

From the perspective of the tagged customer arriving at station 1 at $t={}0$, the sub-scenario $\left(m={}1,n={}3\right)$ corresponds to a sequence of 4 events: A’, C’, $\text{F}_{1}$, and $\text{E}_{2}$ as shown in Figure 13. In event A’, exactly $K_{1}$ $\Big{(}\text{for $K_{1}\leq L_{11}^{a}$}\Big{)}$ customers are served at station 1 in time $h_{12}\Big{(}L_{12}^{a}+K_{1}\Big{)}$ before station 2 server switches to serve queue 2 at $t={}t_{C^{\prime}}$ after serving $L_{12}^{a}+K_{1}$ type 1 customers.

In the event C’, the tagged customer and the $L_{11}^{\prime}\left(t_{C^{\prime}}\right)$ customers ahead of it at $t={}t_{C}$ are served at station 1 before $L_{22}\left(t_{C^{\prime}}\right)$ customers of queue 2 at station 2 are served. The time $T_{C^{\prime}}$ it takes for server at station 1 to serve $L_{11}^{{}^{\prime}}\left(t_{C^{\prime}}\right)+1$ customers of queue 1 is $h_{11}\left(L_{11}^{{}^{\prime}}\left(t_{C^{\prime}}\right)+1\right)$, where $L_{11}^{{}^{\prime}}\left(t_{C^{\prime}}\right)$ is given by Equation $\left(\ref{eq:3.10}\right)$. The probability $\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event C'}\Big{)}$ that $L_{11}^{{}^{\prime}}\left(t_{C^{\prime}}\right)+1$ are served at station 1 before $L_{22}\left(t_{C^{\prime}}\right)$ customers are served at station 2 is given by Equation $\left(\ref{eq:3.17}\right)$ and is determined using Result 6.

$\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event C'}\Big{)}={}\mathop{\mathbb{P}}\Big{(}h_{11}\left(L_{11}^{{}^{\prime}}\left(t_{C^{\prime}}\right)+1\right)<h_{22}\Big{(}L_{22}\left(t_{C^{\prime}}\right)\Big{)}\Big{)}$

(17)

In the event $F_{1}$, all customers of queue 2 at station 2 at $t={}t_{F_{1}}$ are served before queue 1 at station 1 empties. Let $S_{1}\left(t\right)$ be the Poisson random variable denoting number of arrivals of type 1 customers at station 1 in time $t$. During the event A’ and C’, external $S_{1}\left(T_{A^{\prime}}+T_{C^{\prime}}\right)$ type 1 customers arrive at station 1. As the tagged customer and all the customers ahead of it at station 1 has moved to station 2 at the end of event C’, the length of queue 1 at station 1 at $t={}t_{F_{1}}$ is given by Equation $\left(\ref{eq:3.18}\right)$.

$$L_{11}\left(t_{F_{1}}\right)={}S_{1}\left(T_{A^{\prime}}+T_{C^{\prime}}\right)$$

(18)

Let $V_{2}$ be the number of customers that got served in queue 2 at station 2 during event C’. As $L_{22}\left(t_{F_{1}}\right)\neq 0$, the upper bound on $V_{2}$ is $L_{22}^{a}-1$. Thus, $L_{22}\left(t_{F_{1}}\right)$ can be written as

$$L_{22}\left(t_{F_{1}}\right)={}L_{22}^{a}-V_{2}$$

(19)

where the conditional probability distribution of $V_{2}$ is

$$\displaystyle\mathop{\mathbb{P}}\Big{(}V_{2}={}v|V_{2}\leq L_{22}^{a}-1\Big{)}={}\frac{\displaystyle\mathop{\mathbb{P}}\Big{(}V_{2}={}v\Big{)}}{\sum_{v=0}^{{}L_{22}^{a}-1}\displaystyle\mathop{\mathbb{P}}\Big{(}V_{2}={}v\Big{)}}$$

and

$$\displaystyle\mathop{\mathbb{P}}\Big{(}V_{2}={}v\Big{)}=\Big{(}1/v!\Big{)}e^{-\mu_{22}\left(T_{C^{\prime}}\right)}\Big{(}\mu_{22}\left(T_{C^{\prime}}\right)\Big{)}^{v}$$

The time $T_{F_{1}}$ it takes for the server at station 2 to serve $L_{22}\left(t_{F_{1}}\right)$ customers of queue 2 is $h_{22}\left(L_{22}\left(t_{F_{1}}\right)\right)$. The probability $\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event $\text{F}_{1}$}\Big{)}$ that $L_{22}\left(t_{F_{1}}\right)$ customers are served at station 2 before queue 1 at station 1 empties with $L_{11}\left(t_{F_{1}}\right)$ customers at $t={}t_{F_{1}}$ is given by Equation $\left(\ref{eq:3.20}\right)$ and is determined using Result 7.

$\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event $\text{F}_{1}$}\Big{)}={}\mathop{\mathbb{P}}\Big{(}h_{22}\left(L_{22}\left(t_{F_{1}}\right)\right)<g_{11}\left(L_{11}\left(t_{F_{1}}\right)\right)\Big{)}$

(20)

In the event $\text{E}_{2}$, the tagged customer and the $L_{12}^{{}^{\prime}}\left(t_{E_{2}}\right)$ customers ahead of it are served at station 2 before the tagged customer exits the system. The number of customers $L_{12}^{{}^{\prime}}\left(t_{E_{2}}\right)$ ahead of the tagged customer at $t={}t_{E_{2}}$ is equal to $L_{11}^{{}^{\prime}}\left(t_{C^{\prime}}\right)$ as all the customers that were ahead of the tagged customer at station 1 are also ahead of it at station 2 since the server at station 2 was serving queue 2 during the events C’ and $\text{F}_{1}$. The time it takes to serve the tagged customer and customers ahead of it at station 2 after the server at station 2 begins its second cycle after the arrival of the tagged customer is $h_{12}\left(L_{12}^{{}^{\prime}}\left(t_{E_{2}}\right)+1\right)$.

Therefore, for sub-scenario $\left(m={}1,n={}3\right)$, $\displaystyle\mathop{\mathbb{E}}\Big{[}W_{1}^{a}|m={}1,n={}3\Big{]}$ is equal to the sum of mean times for the events A’, C’, $\text{F}_{1}$, and $\text{E}_{2}$ and is given by Equation $\left(\ref{eq:3.21}\right)$.

	$\displaystyle\mathop{\mathbb{E}}\Big{[}W_{1}^{a}|m={}1,n={}3\Big{]}=$	$\displaystyle{}\mathop{\mathbb{E}}\Big{[}h_{12}\Big{(}L_{12}^{a}+K_{1}\Big{)}\Big{]}+\mathop{\mathbb{E}}\Big{[}h_{11}\Big{(}L_{11}\left(t_{C^{\prime}}\right)+1\Big{)}\Big{]}+$		(21)
		$\displaystyle\mathop{\mathbb{E}}\Big{[}h_{22}\Big{(}L_{22}\left(t_{F_{1}}\right)\Big{)}\Big{]}+\mathop{\mathbb{E}}\Big{[}h_{12}\left(L_{12}^{{}^{\prime}}\left(t_{E_{2}}\right)+1\right)\Big{]}$

The probability of this sub-scenario $\displaystyle\mathop{\mathbb{P}}\Big{(}n={}3|m={}1\Big{)}$ is given by Equation $\left(\ref{eq:3.22}\right)$.

$$\displaystyle\mathop{\mathbb{P}}\Big{(}n={}3|m={}1\Big{)}=\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event A'}\Big{)}\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event C'}\Big{)}\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event $\text{F}_{1}$}\Big{)}$$

(22)

From the perspective of the tagged customer arriving at station 1 at $t={}0$, the sub-scenario $\left(m={}1,n={}4\right)$ corresponds to a sequence of 5 events: A’, C’, $\text{F'}_{1}$, G, and H as shown in Figure 14. Events A’ and C’ in sub-scenario $\left(m={}1,n={}4\right)$ are identical to events A’ and C’ in sub-scenario $\left(m={}1,n={}3\right)$.

In the event $\text{F'}_{1}$, queue 1 at station 1 with external arrivals empty before queue 2 at station 2. The time $T_{F^{\prime}_{1}}$ to empty queue 1 at station 1 with $L_{11}\left(t_{F^{\prime}_{1}}\right)$ customers is $g_{11}\Big{(}L_{11}\left(t_{F^{\prime}_{1}}\right)\Big{)}$ where $L_{11}\left(t_{F^{\prime}_{1}}\right)$ is given by Equation $\left(\ref{eq:3.19}\right)$. The probability $\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event $\text{F'}_{1}$}\Big{)}$ that queue 1 at station 1 with $L_{11}\left(t_{F^{\prime}_{1}}\right)$ customers empties before $L_{22}\left(t_{F^{\prime}_{1}}\right)$ type 2 customers are served at station 2 is given by Equation $\left(\ref{eq:3.23}\right)$ and is determined using Result 7.

$\displaystyle\mathop{\mathbb{P}}\Big{(}\text{Event $\text{F'}_{1}$}\Big{)}={}\mathop{\mathbb{P}}\Big{(}g_{11}\left(L_{11}\left(t_{F^{\prime}_{1}}\right)\right)<h_{22}\left(L_{22}\left(t_{F^{\prime}_{1}}\right)\right)\Big{)}$

(23)

In the event G, queue 2 at station 2 with additional arrivals from station 1 empties before queue 2 at station 1 with additional external arrivals. Let $V_{3}$ be the number of type 2 customers served at station 2 during events C’ and $\text{F'}_{1}$. As $L_{22}\left(t_{G}\right)\neq 0$, the upper bound on $V_{3}$ is $L_{22}\left(t_{C^{\prime}}\right)-1$ for event G to occur. The queue length $L_{22}\left(t_{G}\right)$ is given by Equation $\left(\ref{eq:3.24}\right)$.

$$L_{22}\left(t_{G}\right)={}L_{22}\left(t_{C^{\prime}}\right)-V_{3}$$

(24)

where the conditional probability distribution of $V_{3}$ is

$$\displaystyle\mathop{\mathbb{P}}\Big{(}V_{3}={}v|V_{3}\leq L_{22}\left(t_{C^{\prime}}\right)-1\Big{)}={}\frac{\displaystyle\mathop{\mathbb{P}}\Big{(}V_{3}={}v\Big{)}}{\sum_{v=0}^{{}L_{22}\left(t_{C^{\prime}}\right)-1}\displaystyle\mathop{\mathbb{P}}\Big{(}V_{3}={}v\Big{)}}$$

and

$$\displaystyle\mathop{\mathbb{P}}\Big{(}V_{3}={}v\Big{)}=\Big{(}1/v!\Big{)}e^{-\mu_{22}\left(T_{C^{\prime}}+T_{F^{\prime}_{1}}\right)}\Big{(}\mu_{22}\left(T_{C^{\prime}}+T_{F^{\prime}_{1}}\right)\Big{)}^{v}$$

The hitting time to 0, $T_{G}$, for queue 2 with $L_{22}\left(t_{G}\right)$ customers at station 2 and $L_{21}\left(t_{G}\right)$ type 2 customers at station 1 is $\phi_{\left(L_{21}\left(t_{G}\right),L_{22}\left(t_{G}\right)\right)}$ and is determined using Result 3 where $x={}2$ and $y={}1$. During the time $\left(0,t_{G}\right)$, additional $S_{2}\left(t_{G}\right)$ type 2 customers arrive at station 1 where $S_{2}\left(t_{G}\right)$ is a Poisson random with parameter $\lambda_{2}t_{G}$. We write $L_{21}\left(t_{G}\right)$ as

$$L_{21}\left(t_{G}\right)={}L_{21}^{a}+S_{2}\left(t_{G}\right)$$

(25)