State-dependent M/G/1 queueing systems

Abstract

We consider a state-dependent $M_{n}$/$G_{n}$/1 queueing system with both finite and infinite buffer sizes. We allow the arrival rate of customers to depend on the number of people in the system. Service times are also state dependent and service rates can be modified at both arrivals and departures of customers. We show that the steady-state solution of this system at arbitrary times can be derived using the supplementary variable method, and that the system’s state at arrival epochs can be analyzed using an embedded Markov chain. For the system with infinite buffer size, we first obtain an expression for the steady-state distribution of the number of customers in the system at both arbitrary and arrival times. Then, we derive the average service time of a customer observed at both arbitrary times and arrival epochs. We show that our state-dependent queueing system is equivalent to a Markovian birth-and-death process. This equivalency demonstrates our main insight that the $M_{n}$/$G_{n}$/1 system can be decomposed at any given state as a Markovian queue. Thus, many of the existing results for systems modeled as an M / M / 1 queue can be carried through to the much more practical M / G / 1 model with state-dependent arrival and service rates. Then, we extend the results to the $M_{n}$/$G_{n}$/1 queueing systems with finite buffer size.

Appendix: Proofs

Proof

(Proof of Observation 2) Note that the auxiliary queue is defined as an $M_{n}$/$M_{n}$/1 queue with arrival rate $\lambda _{\kappa +i}$ and service rate $\mu _{\kappa +i}$ when there are i customers in this queue. Therefore, the steady-state distribution of the number of people in this queue can be obtained using (5) and (6). Using (5), the probability of having $\kappa +i$ people in the original queue, $P(\kappa +i)$, is

$$\begin{aligned} P(\kappa +i)= & {} \frac{\lambda _0 P(0)}{\lambda _{\kappa +i}}\prod \limits ^{\kappa +i-1}_{j=0} \frac{\lambda _{j+1}}{ \mu _{j+1}}=\frac{\lambda _0 P(0)}{\lambda _{\kappa +i}}\left( \prod \limits ^{\kappa -1}_{j=0} \frac{\lambda _{j+1}}{ \mu _{j+1}}\right) \left( \prod \limits ^{\kappa +i-1}_{j=k} \frac{\lambda _{j+1}}{ \mu _{j+1}}\right) \nonumber \\= & {} \left( \frac{\lambda _0 P(0)}{\lambda _{\kappa }P^{A}(0)}\prod \limits ^{\kappa -1}_{j=0} \frac{\lambda _{j+1}}{ \mu _{j+1}}\right) \left( \frac{\lambda _{\kappa } P^{A}(0)}{\lambda _{\kappa +i}}\prod \limits ^{i-1}_{j=0} \frac{\lambda _{\kappa +j+1}}{ \mu _{\kappa +j+1}}\right) {,} \end{aligned}$$

(55)

where $P^{A}(0)$ denotes the probability of having no customers in the auxiliary queue.

We next prove that the first term in (55) is equal to $\left( 1-F( \kappa -1)\right)$. Substituting $P^{A}(0)$ given in (6) into the first term in (55), we get

$$\begin{aligned}&\left( \frac{\lambda _0 P(0)}{\lambda _{\kappa }P^{A}(0)}\prod \limits ^{\kappa -1}_{j=0} \frac{\lambda _{j+1}}{ \mu _{j+1}}\right) =\left( \frac{\lambda _0 P(0)}{\lambda _{\kappa }}\prod \limits ^{\kappa -1}_{j=0} \frac{\lambda _{j+1}}{ \mu _{j+1}}\right) \left( 1+\sum _{i=1}^{\infty }\frac{\lambda _{\kappa }}{\lambda _{\kappa +i}}\prod \limits ^{i-1}_{j=0} \frac{\lambda _{\kappa +j+1}}{ \mu _{\kappa +j+1}}\right) \\&\quad =\left( \frac{\lambda _0 P(0)}{\lambda _{\kappa }}\prod \limits ^{\kappa -1}_{j=0} \frac{\lambda _{j+1}}{ \mu _{j+1}}\right) \left( 1+\sum _{i=\kappa +1}^{\infty }\frac{\lambda _{\kappa }}{\lambda _{i}}\prod \limits ^{i-1}_{j=\kappa } \frac{\lambda _{j+1}}{ \mu _{j+1}}\right) \\&\quad =\left( \frac{\lambda _0 P(0)}{\lambda _{\kappa }}\prod \limits ^{\kappa -1}_{j=0} \frac{\lambda _{j+1}}{ \mu _{j+1}}+\sum _{i=\kappa +1}^{\infty }\frac{\lambda _{0}P(0)}{\lambda _{i}}\prod \limits ^{i-1}_{j=0} \frac{\lambda _{j+1}}{ \mu _{j+1}}\right) \\&\quad =\sum _{i=\kappa }^{\infty }\frac{\lambda _{0}P(0)}{\lambda _{i}}\prod \limits ^{i-1}_{j=0} \frac{\lambda _{j+1}}{ \mu _{j+1}}=\left( 1-F( \kappa -1)\right) . \end{aligned}$$

Comparing the second term in (55) with (5), we find that the former is the probability of having i people in the auxiliary queue, $P^{A}(i)$. Therefore, (55) is equivalent to (8);

$$\begin{aligned} P(\kappa +i)=\left( 1-F( \kappa -1)\right) P^{A}(i). \end{aligned}$$

(56)

$\square$

1.1 Supplementary result for proof of Theorem 2

Before we prove Theorem 2, we prove the following lemma. Let $p_{j}(\eta )$ denote the steady-state density of the residual service time of the customer in service when there are j customers in the system, assuming that such steady-state density exists. Then,

Lemma 4

$$\begin{aligned}&\int _{\eta }^{\infty }e^{- \lambda _1 u}p^\prime _{1}(u)du = \int _{\eta }^{\infty }e^{- \lambda _1 u}\left( \lambda _1 p_{1}(u) - \lambda _0 P(0) b_1(u) - p_{2}(0) b_1 (u)\right) du , \end{aligned}$$

(57)

$$\begin{aligned}&\int _{\eta }^{\infty }e^{- \lambda _j u}p^\prime _{j} (u)du = \int _{\eta }^{\infty }e^{- \lambda _j u}\left( \lambda _j p_{j}(u) - \alpha _{j}p_{j-1}\left( \alpha _{j}u \right) \lambda _{j-1} -p_{j+1}(0) b_j (u)\right) du.\nonumber \\ \end{aligned}$$

(58)

Proof

(Proof of Lemma 4)

As we assume that $b_{j}(\cdot )$ are all absolutely continuous, a sufficient condition for $p_{j}(\eta )$ to exist is that the $M_{n}$/$G_{n}$/1 system is stable. Dividing both sides of (12) and (13) by dt and rearranging them we get

$$\begin{aligned} \frac{p_{1}(\eta ,t)-p_{1}(\eta -dt,t+dt)}{dt}= & {} p_{1}(\eta ,t)\lambda _1 -p_{2}(0,t)b_1(\eta ) - \lambda _0 p_{0}(t) b_1(\eta )+\frac{o(dt)}{dt},\\ \frac{p_{j}(\eta ,t)-p_{j}(\eta -dt,t+dt)}{dt}= & {} p_{j}(\eta ,t)\lambda _j -p_{j+1}(0,t)b_j(\eta ) \\&-\alpha _{j}p_{j-1}\left( \eta \alpha _{j},t\right) \lambda _{j-1}+\frac{o(dt)}{dt}. \end{aligned}$$

Taking the limit as $dt\rightarrow 0$, we get

$$\begin{aligned}&p^\prime _{1}(\eta ,t) = \lambda _1 p_{1}(\eta ,t) - \lambda _0 p_{0}(t) b_1(\eta ) - p_{2}(0,t) b_1 (\eta ),\\&p^\prime _{j} (\eta ,t) = \lambda _j p_{j}(\eta ,t) - \alpha _{j}p_{j-1}\left( \alpha _{j}\eta ,t\right) \lambda _{j-1} -p_{j+1}(0,t) b_j (\eta ). \end{aligned}$$

Assuming that the steady-state exists and taking the limit $t\rightarrow \infty$, we get:

$$\begin{aligned}&p^\prime _{1}(\eta ) = \lambda _1 p_{1}(\eta ) - \lambda _0 P(0) b_1(\eta ) - p_{2}(0) b_1 (\eta ) , \end{aligned}$$

(59)

$$\begin{aligned}&p^\prime _{j} (\eta ) = \lambda _j p_{j}(\eta ) - \alpha _{j}p_{j-1}\left( \alpha _{j}\eta \right) \lambda _{j-1} -p_{j+1}(0) b_j (\eta ). \end{aligned}$$

(60)

Multiplying both sides of (59) and (60) by $e^{- \lambda _j u}$ and taking the integral, we get (57) and (58).$\square$

Proof

(Proof of Theorem 2)

We note that $p_{j}(\infty )=0$, $p_{j}(0)\ge 0$ for $j>0$, and $P(0)>0$ because we assume that the system is stable. Also, $\int _{0}^{\infty }p_{j}(\eta )d\eta =P(j)$ and similarly from (1), $\int _{0}^{\infty }\alpha _{j} p_{j-1}(\alpha _{j} \eta )d\eta =P(j-1)$. Then, by setting $\eta =\lambda _{j}=0$ in Lemma 4, from (57) and (58) we get

$$\begin{aligned}&-p_{1}(0)= \lambda _{1}P(1)-\lambda _{0}P(0)-p_{2}(0), \\&-p_{j}(0)= \lambda _{j}P(j)-\lambda _{j-1}P(j-1)-p_{j+1}(0). \end{aligned}$$

Therefore, for $j\ge 1$

$$\begin{aligned}&\lambda _{j-1}P(j-1)-p_{j}(0)= \lambda _{j}P(j)-p_{j+1}(0). \end{aligned}$$

Note that $\lambda _{j-1}P(j-1)-p_{j}(0)$ is independent of j and must go to zero in the limit when $j\rightarrow \infty$ if the system is stable (see, for example, [18]). Therefore,

$$\begin{aligned} p_{j+1}(0)= \lambda _{j}P(j), \quad j\ge 0. \end{aligned}$$

(61)

Note that (61) can be explained using level crossing as well. Considering level j (number of customers in the system), $\lambda _{j}P(j)$ is the rate of up-crossing this level and $p(j+1, 0)$ is the rate of down-crossing (i.e., the rate at which a departure leaves j customers behind).

Recalling that $h_{j}(\cdot )$ denotes the steady-state density of the residual service time observed by an arrival that sees j customers in the system, we have

$$\begin{aligned} h_{j}(\eta )=\frac{p_{j}(\eta )}{\int _{0}^{\infty }p_{j}(r)dr}=\frac{p_{j}(\eta )}{P(j)}. \end{aligned}$$

(62)

Therefore,

$$\begin{aligned} \tilde{h}_{j}(s)=\frac{\int _{0}^{\infty }e^{-sr}p_{j}(r)dr}{P(j)}. \end{aligned}$$

(63)

Substituting (61) in (59) and (60), and multiplying both sides by $e^{-\lambda _{j}\eta }$, we get, after some algebra,

$$\begin{aligned}&e^{-\lambda _{1}\eta } (p^\prime _{1} (\eta ) - \lambda _1 p_{1}(\eta )) = - b_1 (\eta ) e^{-\lambda _1 \eta }(\lambda _{0} P(0) + \lambda _1 P(1)), \end{aligned}$$

(64)

$$\begin{aligned}&e^{-\lambda _{j}\eta } (p^\prime _{j} (\eta ) - \lambda _j p_{j}(\eta )) =-e^{- \lambda _j \eta } \left( \alpha _{j}p_{j-1}\left( \alpha _{j}\eta \right) \lambda _{j-1} + \lambda _j P(j) b_j (\eta )\right) . \end{aligned}$$

(65)

Considering (63), Lemma 4, and recalling $b_{j}(\infty )=0$ for $j\ge 1$, we get

$$\begin{aligned}&p_{1} (\eta ) = e^{\lambda _1 \eta }\int _{\eta }^{\infty }\left( b_1 (u) e^{-\lambda _1 u}(\lambda _{0} P(0) + \lambda _1 P(1))\right) du,\\&p_{j} (\eta )=e^{\lambda _j \eta }\int _{\eta }^{\infty }\left( e^{- \lambda _j u} \left( \alpha _{j}p_{j-1}\left( \alpha _{j}u \right) \lambda _{j-1} + \lambda _j P(j) b_j (u)\right) \right) du. \end{aligned}$$

Substituting (61) in the above equations, we get for $\eta =0$

$$\begin{aligned}&\lambda _{0}P(0) = (\lambda _{0} P(0) + \lambda _1 P(1)) \int _{0}^{\infty }\left( b_1 (u) e^{-\lambda _1 u}\right) du,\\&\lambda _{j-1}P(j-1)=\int _{0}^{\infty }\left( e^{- \lambda _j u} \left( \alpha _{j}p_{j-1}\left( u \alpha _{j}\right) \lambda _{j-1} + \lambda _j P(j) b_j (u)\right) \right) du. \end{aligned}$$

Solving the integrals on the right-hand sides of the above equations, we get

$$\begin{aligned}&\lambda _{0}P(0) = (\lambda _{0} P(0) + \lambda _1 P(1))\tilde{b}_{1}\left( \lambda _{1}\right) , \\&\lambda _{j-1}P(j-1)=\lambda _{j-1}P(j-1)\tilde{h}_{j-1}\left( \frac{\lambda _{j}}{\alpha _{j}}\right) +\lambda _{j}P(j)\tilde{b}_{j}\left( \lambda _{j}\right) {,} \end{aligned}$$

because from (63) we have

$$\begin{aligned} \int _{0}^{\infty }e^{- \lambda _j u} \left( \alpha _{j}p_{j-1}\left( u \alpha _{j}\right) \right) du=P(j-1)\tilde{h}_{j-1}\Big (\frac{\lambda _{j}}{\alpha _{j}}\Big ). \end{aligned}$$

(66)

Finally, with $\tilde{h}_{0}(\cdot )=\tilde{b}_{1}(\cdot )$ and $\alpha _{1}=1$, after some algebra and using induction we get for each $j\ge 1$

$$\begin{aligned} P(i) = \frac{\lambda _0 P(0)}{\lambda _i}\prod \limits ^{i-1}_{j=0} \frac{1-\tilde{h}_{j}\left( \frac{\lambda _{j+1}}{\alpha _{j+1}}\right) }{ \tilde{b}_{j+1}(\lambda _{j+1})}. \end{aligned}$$

Standard arguments establish the rest of the theorem.$\square$

Proof

(Proof of Theorem 3)

Setting $\eta =0$ in Lemma 4, from (57) and (58) we get (similar to (66))

$$\begin{aligned}&P(1)s\tilde{h}_{1}(s)-p_{1}(0)=\lambda _{1}P(1)\tilde{h}_{1}(s)-\tilde{b}_{1}(s)\left( p_{2}(0)+\lambda _{0}P(0)\right) , \end{aligned}$$

(67)

$$\begin{aligned}&P(j)s\tilde{h}_{j}(s)-p_{j}(0)=\lambda _{j}P(j)\tilde{h}_{j}(s)-\lambda _{j-1}P(j-1)\tilde{h}_{j-1}\left( \frac{s}{\alpha _{j}}\right) -\tilde{b}_{j}(s)p_{j+1}(0). \nonumber \\ \end{aligned}$$

(68)

Substituting (61) in (67) and (68), we get

$$\begin{aligned} P(1)s\tilde{h}_{1}(s)-\lambda _{0}P(0)= & {} \lambda _{1}P(1)\tilde{h}_{1}(s)-\tilde{b}_{1}(s)\left( \lambda _{1}P(1)+\lambda _{0}P(0)\right) , \end{aligned}$$

(69)

$$\begin{aligned} P(j)s\tilde{h}_{j}(s)-\lambda _{j-1}P(j-1)= & {} \lambda _{j}P(j)\tilde{h}_{j}(s)-\lambda _{j-1}P(j-1)\tilde{h}_{j-1}\nonumber \\&\qquad \times \, \left( \frac{s}{\alpha _{j}}\right) -\tilde{b}_{j}(s)\lambda _{j}P(j). \end{aligned}$$

(70)

Now considering (69) we get

$$\begin{aligned} \tilde{h}_{1}(s)=\frac{\left( \lambda _{0}P(0)-\tilde{b}_{1}(s)\left( \lambda _{1}P(1)+\lambda _{0}P(0)\right) \right) }{\left( P(1)s-\lambda _{1}P(1)\right) }. \end{aligned}$$

Substituting $\lambda _{1}P(1)$ from (16) and recalling that $\tilde{h}_{0}(\cdot )=\tilde{b}_{1}(\cdot )$, we get

$$\begin{aligned} \tilde{h}_{1}(s)=\left( \frac{\lambda _{1}}{s-\lambda _{1}}\right) \left( \frac{\tilde{b}_{1}\left( \lambda _{1}\right) }{1-\tilde{b}_{1}\left( \frac{\lambda _{1}}{\alpha _{1}}\right) }\left( 1-\tilde{b}_{1}(s)\right) -\tilde{b}_{1}(s)\right) {.} \end{aligned}$$

Considering that $\tilde{h}_{0}(\cdot )=\tilde{b}_{1}(\cdot )$ and $\alpha _{1}=1$, we get

$$\begin{aligned} \tilde{h}_{1}(s)\,=\,\frac{\lambda _1}{s-\lambda _1}\left[ \tilde{b}_{1}(\lambda _1)\frac{1-\tilde{h}_{0}\Big (\frac{s}{\alpha _{1}}\Big )}{1-\tilde{h}_{0}\Big (\frac{\lambda _1}{\alpha _{1}}\Big )}-\tilde{b}_{1}(s)\right] . \end{aligned}$$

(71)

Similarly, we can obtain $\tilde{h}_{j}(s)$ for $j>1$ using (70):

$$\begin{aligned} \tilde{h}_{j}(s)\,=\,\frac{\lambda _j}{s-\lambda _j}\left[ \tilde{b}_{j}(\lambda _j)\frac{1-\tilde{h}_{j-1}\Big (\frac{s}{\alpha _{j}}\Big )}{1-\tilde{h}_{j-1}\Big (\frac{\lambda _j}{\alpha _{j}}\Big )}-\tilde{b}_{j}(s)\right] , \quad j\ge 1. \end{aligned}$$

(72)

$\square$

Proof

(Proof of Observation 3) From Theorem 2 we have

$$\begin{aligned} P\left( i\right)= & {} \frac{\lambda _{0}P\left( 0\right) }{\lambda _{i}}\prod _{j=0}^{i-1}\frac{1-\tilde{h}_{j}\left( \frac{\lambda _{j+1}}{\alpha _{j+1}} \right) }{\tilde{b}_{j+1}\left( \lambda _{j+1}\right) } \\= & {} \lambda _{i-1}\frac{1-\tilde{h}_{i-1}\left( \frac{\lambda _{i}}{\alpha _{i}}\right) }{\lambda _{i}\tilde{b}_{i}\left( \lambda _{i}\right) }P\left( i-1\right) . \end{aligned}$$

Multiplying both sides by $\lambda _{i-1}$ and comparing the result with (3), we get $\hat{\mu }_{i}$ as in (19).$\square$

Proof

(Proof of Observation 4) Consider the auxiliary queue. We define a continuous time MP for the auxiliary queue similar to the one expressed in (12) and (13) with states $(j,\eta )$, where j is the number of jobs in the auxiliary queue, while $\eta$ denotes the remaining service time. We define $g_{j}(\eta ,t)$ as the probability that there are j jobs in the auxiliary queue, and the remaining service time is $\eta$ at time t. Therefore,

$$\begin{aligned} g_{1}(\eta -dt,t+dt)= & {} g_{1}(\eta ,t)(1-\lambda _{\kappa +1}dt)+g_{2}(0,t)b_{\kappa +1}(\eta )dt\nonumber \\&+\,g_{t}(0,0)\lambda _{\kappa }b_{0}^{A}(\eta )dt, \end{aligned}$$

(73)

$$\begin{aligned} g_{j}(\eta -dt,t+dt)= & {} g_{j}(\eta ,t)(1-\lambda _{\kappa +j}dt)+g_{j-1}(\eta ,t)\lambda _{\kappa +j-1}dt\nonumber \\&+\,g_{j+1}(0,t)b_{\kappa +j}(\eta )dt,\qquad j\ge 2, \end{aligned}$$

(74)

where $b_{0}^{A}(\cdot )$ is the steady-state service time density of a job that finds 0 jobs in this queue.

Now consider the original system defined in (12) and (13). Given that there are more than $\kappa -1$ customers in the system, (12) and (13) reduced to the same equations as given in (73) and (74), where $b_{0}^{A}(\cdot )$ is the equilibrium service time densities of customers that find $\kappa$ customers in the original system (Step 2 of the definition of the auxiliary queue). Thus, the steady-state distribution of the number of jobs in the auxiliary queue, $P^{A}(i)$, is identical to the steady-state distribution of the number of customers in the original queue given that there are more than $\kappa$ customers in the system, $P(\kappa +i)$.$\square$

Proof

(Proof of Corollary 1)

The proof is similar to the proof of Theorem 2.2.2 in [18]. van Doorn and Regterschot [11] define the adapted LAA as the LAA conditioning on the state of the system, and show that under the adapted LAA PASTA holds. We next establish that the adapted LAA holds in our setting. Let X(t) denote the state of the system, and $N_{s}$ be the Poisson process that generates the future arrivals when the state of the system is $X(t)=s$. Then, for every s we have that $\{N_{s}(t+u)-N_{s}(t), u\ge 0\}$ and X(t) are independent and the adaptive LAA holds. Therefore, Theorem 1 in [11] holds.$\square$

Proof

(Proof of Theorem 4)

We first obtain the transition probabilities $p_{0k}$. Consider $p_{00}$, the probability that the next departing customer leaves no customer behind given that the last departing customer left no customer behind, $P\left( M_{n+1}=0|M_{n}=0\right)$. This probability is equal to the probability of having no arrivals during the next service time, i.e., it is $\tilde{b}_{1}(\lambda _{1})$ (for example, [9], p. 171). Therefore,

$$\begin{aligned} p_{00}=\tilde{b}_{1}(\lambda _{1}). \end{aligned}$$

(75)

Next consider $p_{01}$, the probability that the next departing customer leaves one customer behind given that the last departing customer left no customer behind, $P\left( M_{n+1}=1|M_{n}=0\right)$. This probability is equal to the probability of one arrival during the next service time. The only sample path that would lead to this event is that there is exactly one arrival during the sojourn time of the departing customer. Because this sojourn time is identical to the service time of this customer, the probability $p_{01}$ is equal to the probability that (i) a customer arrives after the service time starts, $\left( 1-\tilde{b}_{1}(\lambda _{1})\right)$ and (ii) no customers arrive during the remaining service time observed by this arrival. Noting that this arrival sees one customer in the system upon the arrival, the equilibrium remaining service time observed by this arrival is $h_{1}(\cdot )$. Considering that the rate of the residual service time is modified by $\alpha _{2}$ after this customer joins the queue, the probability that no customers arrive during the remaining service time observed by this arrival is $\tilde{h}_{1}(\frac{\lambda _{2}}{\alpha _{2}})$. Therefore,

$$\begin{aligned} p_{01}=\left( 1-\tilde{b}_{1}(\lambda _{1})\right) \tilde{h}_{1}\left( \frac{\lambda _{2}}{\alpha _{2}}\right) . \end{aligned}$$

(76)

With similar logic, the probability that the next departing customer leaves k customers behind given that the last departing customer left no customer behind, $P\left( M_{n+1}=k|M_{n}=0\right)$, is equal to the probability of k arrivals during the next service time. This probability is equal to the probability that a customer arrives after the first service time starts, $\left( 1-\tilde{b}_{1}(\lambda _{1})\right)$, followed by a customer arrival during the remaining service times of all arrivals that see $i<k$ customers in the system, each with probability $\left( 1-\tilde{h}_{i}\Big (\frac{\lambda _{i+1}}{\alpha _{i+1}}\Big )\right)$, and no arrival during the remaining service time once there are k customers in the system, $\tilde{h}_{k}\Big (\frac{\lambda _{k+1}}{\alpha _{k+1}}\Big )$. Recalling our definition $\tilde{h}_{0}(\cdot )=\tilde{b}_{1}(\cdot )$, we have

$$\begin{aligned} p_{0k}=\tilde{h}_{k}\left( \frac{\lambda _{k+1}}{\alpha _{k+1}}\right) \prod _{i=0}^{k-1}\left( 1-\tilde{h}_{i}\Big (\frac{\lambda _{i+1}}{\alpha _{i+1}}\Big )\right) . \end{aligned}$$

(77)

Using the transition probabilities in (30) derived in the body of the paper and (77), we next derive the steady-state probabilities of the embedded MC. Considering (25), we need to derive the steady-state distribution of the number of customers in the system using the embedded MC. Note that the steady-state probability that a departure leaves k customers behind satisfies $P_{d}(k)=\sum _{j=0}^{\infty }P_{d}(j)p_{jk}$.

First consider $P_{d}(0)$:

$$\begin{aligned}&P_{d}(0)=P_{d}(0)p_{00}+P_{d}(1)p_{10}=P_{d}(0)\tilde{b}_{1}(\lambda _{1})+P_{d}(1)\tilde{b}_{1}(\lambda _{1}),\nonumber \\\Rightarrow & {} P_{d}(1)=P_{d}(0)\frac{1-\tilde{b}_{1}(\lambda _{1})}{\tilde{b}_{1}(\lambda _{1})}. \end{aligned}$$

(78)

Next consider $P_{d}(1)$:

$$\begin{aligned} P_{d}(1)= & {} P_{d}(0)p_{01}+P_{d}(1)p_{11}+P_{d}(2)p_{21}\nonumber \\= & {} P_{d}(0)\left( 1-\tilde{b}_{1}(\lambda _{1})\right) \tilde{h}_{1}\left( \frac{\lambda _{2}}{\alpha _{2}}\right) +P_{d}(1)\left( 1-\tilde{b}_{1}(\lambda _{1})\right) \tilde{h}_{1}\left( \frac{\lambda _{2}}{\alpha _{2}}\right) +P_{d}(2)\tilde{b}_{2}(\lambda _{2}).\nonumber \\ \end{aligned}$$

(79)

Substituting $P_{d}(1)$ from (78) into (79), we get

$$\begin{aligned} P_{d}(2) = P_{d}(0)\prod \limits ^{1}_{j=0} \frac{1-\tilde{h}_j\left( \frac{\lambda _{j+1}}{\alpha _{j+1}} \right) }{ \tilde{b}_{j+1}(\lambda _{j+1})}. \end{aligned}$$

(80)

The rest of the probabilities can be obtained similarly.$\square$

Proof

(Proof of Corollary 2)

Substituting (19) in (3) we get

$$\begin{aligned} \lambda _{i}P_{F}\left( i\right) =\frac{\lambda _{i+1}\tilde{b}_{i+1}\left( \lambda _{i+1}\right) }{1- \tilde{h}_{i}\left( \frac{\lambda _{i+1}}{\alpha _{i+1}}\right) }P_{F}\left( i+1\right) . \end{aligned}$$

(81)

Therefore,

$$\begin{aligned} P_{F}\left( i+1\right) =\frac{\lambda _{i}}{\lambda _{i+1}}\frac{1- \tilde{h}_{i}\left( \frac{\lambda _{i+1}}{\alpha _{i+1}}\right) }{\tilde{b}_{i+1}\left( \lambda _{i+1}\right) }P_{F}\left( i\right) , \end{aligned}$$

(82)

which after some algebra leads to (44).$\square$

Proof

(Proof of Lemma 1) (35) is equivalent to

$$\begin{aligned} P(i)=\frac{\bar{P}^{a}(i)\sum _{k=0}^{\infty }\lambda _{k}P(k)}{\lambda _{i}}. \end{aligned}$$

(83)

Using (83) we get P(0):

$$\begin{aligned} P(0)=\frac{\bar{P}^{a}(0)\sum _{k=1}^{\infty }\frac{\lambda _{k}}{\lambda _{0}}P(k)}{1-\bar{P}^{a}(0)}. \end{aligned}$$

(84)

Using (83) and (84), we obtain P(i) as the function of P(0) given in (36).$\square$

Proof

(Proof of Observation 6)

The proof is based on Theorem 4 and similar to the proof of Observation 3.$\square$

Proof

(Proof of Lemma 2)

Noting that $1/\mu _{b}=-\frac{d\tilde{h}_{\kappa }(s)}{ds}|_{s=0}$, by taking the derivative of (18) with respect to s and setting $s=0$, we get

$$\begin{aligned} \frac{1}{\mu _{b}}=-\frac{d\tilde{h}_{k}(s)}{ds}|_{s=0}= & {} \frac{\lambda _{k}}{\left( s-\lambda _{k}\right) ^{2}}\left[ \tilde{b}_{k}(\lambda _{k})\frac{1-\tilde{h}_{k-1}\Big (\frac{s}{\alpha _{k}}\Big )}{1-\tilde{h}_{k-1}\Big (\frac{\lambda _{k}}{\alpha _{k}}\Big )}-\tilde{b_{k}}(s)\right] |_{s=0}\\&-\,\frac{\lambda _{k}}{\left( s-\lambda _{k}\right) } \left[ -\frac{\tilde{b}_{k}(\lambda _{k})}{1-\tilde{h}_{k-1}\Big (\frac{\lambda _{k}}{\alpha _{k}}\Big )}\frac{d\tilde{h}_{k-1}\Big (\frac{s}{\alpha _{k}}\Big )}{ds}-\frac{d\tilde{b}_{k}(s)}{ds}\right] |_{s=0}\\= & {} -\,\frac{1}{\lambda _{k}}+\frac{1}{\mu _{k}}-\left[ \frac{\tilde{b}_{k}(\lambda _{k})}{1-\tilde{h}_{k-1}\Big (\frac{\lambda _{k}}{\alpha _{k}}\Big )}\frac{d\tilde{h}_{k-1}\Big (\frac{s}{\alpha _{k}}\Big )}{ds}|_{s=0}\right] .\\ \end{aligned}$$

We prove the result by induction. For $k=1$ we have $\alpha _{1}=1$ so that

$$\begin{aligned} \frac{1}{\mu _{b}}=-\frac{1}{\lambda _{1}}+\frac{1}{\mu _{1}}+\left[ \frac{\tilde{b}_{1}(\lambda _{1})}{1-\tilde{b}_{1}(\frac{\lambda _{1}}{\alpha _{1}})}\frac{d\tilde{b}_{1}\Big (\frac{s}{\alpha _{1}}\Big )}{ds}|_{s=0}\right] =-\frac{1}{\lambda _{1}}+\frac{1}{\mu _{1}}-\left[ \frac{\tilde{b}_{1}(\lambda _{1})}{1-\tilde{b}_{1}(\frac{\lambda _{1}}{\alpha _{1}})}\frac{1}{\mu _{1}}\right] ,\\ \end{aligned}$$

which with $\tilde{h}_{0}(\cdot )=\tilde{b}_{1}(\cdot )$ is equivalent to (39) for $k=1$. Now suppose (39) holds for $\kappa =m-1$, i.e.,

$$\begin{aligned} -\frac{d\tilde{h}_{m-1}(s)}{ds}|_{s=0}= & {} \frac{1}{\mu _{m-1}}-\frac{1}{\lambda _{m-1}}+\sum _{j=1}^{m-2}\prod _{i=j}^{m-2}\frac{1}{\alpha _{i+1}}\left( \frac{1}{\mu _{j}}-\frac{1}{\lambda _{j}}\right) \frac{\tilde{b}_{i+1}(\lambda _{i+1})}{1-\tilde{h}_{i}\big (\frac{\lambda _{i+1}}{\alpha _{i+1}}\big )}\nonumber \\&-\,\frac{1}{\mu _{1}}\prod _{i=1}^{m-2}\frac{1}{\alpha _{i}}\frac{\tilde{b}_{i+1}(\lambda _{i+1})}{1-\tilde{h}_{i}\big (\frac{\lambda _{i+1}}{\alpha _{i+1}}\big )}. \end{aligned}$$

(85)

We next show that it holds for $k=m$.

$$\begin{aligned} \frac{1}{\mu _{b}}= & {} -\frac{1}{\lambda _{m}}+\frac{1}{\mu _{m}}+\left[ \frac{\tilde{b}_{m}(\lambda _{m})}{1-\tilde{h}_{m-1}\big (\frac{\lambda _{m}}{\alpha _{m}}\big )}\frac{d\tilde{h}_{m-1}\Big (\frac{s}{\alpha _{m}}\Big )}{ds}|_{s=0}\right] \nonumber \\= & {} -\frac{1}{\lambda _{m}}+\frac{1}{\mu _{m}}+\left[ \frac{\tilde{b}_{m} (\lambda _{m})}{1-\tilde{h}_{m-1}\big (\frac{\lambda _{m}}{\alpha _{m}}\big )}\frac{1}{\alpha _{m}}\frac{d\tilde{h}_{m-1}(s)}{ds}|_{s=0}\right] . \end{aligned}$$

(86)

Substituting (85) in (86), we get (39) for $\kappa =m$, which completes the proof.$\square$

Proof

(Proof of Theorem 5) Based on Observation 4, we have

$$\begin{aligned}&P(k)=\left( 1-F(k-1)\right) P^{A}(0). \end{aligned}$$

Substituting P(k) and $F( k-1)$ from (16), we get

$$\begin{aligned} \frac{\lambda _0 P(0)}{\lambda _k}\prod \limits ^{k-1}_{i=0} \frac{1-\tilde{h}_i\left( \frac{\lambda _{i+1}}{\alpha _{i+1}}\right) }{ \tilde{b}_{i+1}(\lambda _{i+1})}=\left( 1-\sum _{j=0}^{k-1}\frac{\lambda _0 P(0)}{\lambda _j}\prod \limits ^{j-1}_{i=0} \frac{1-\tilde{h}_i\left( \frac{\lambda _{i+1}}{\alpha _{i+1}}\right) }{ \tilde{b}_{i+1}(\lambda _{i+1})}\right) (1-\rho _{b}), \end{aligned}$$

resulting in (40) after some algebra.$\square$

Proof

(Proof of Lemma 3) Let $N^{a}$ denote the number of customers in the system seen by an arrival. Then,

$$\begin{aligned} \bar{P}_{F}^{a}(i)= & {} P\left( N^{a}=i|{\textit{accepted}} \right) P\left( {\textit{accepted}} \right) +P\left( N^{a}=i|{\textit{not accepted}} \right) \nonumber \\&\qquad P\left( {\textit{not accepted}} \right) . \end{aligned}$$

(87)

By a level crossing argument for states $i<K$, the frequency of transitions from state i to state $i+1$ is equal to the frequency of transitions from state $i+1$ to state i. Therefore, the probability that an arriving customer observes i people in the system given that she is accepted to the queue, $P\left( N^{a}=i|{\textit{accepted}} \right)$, is identical to the probability that a departing customer leaves i people behind, $\bar{P}_{F}^{d}(i)$. Moreover, for states $i<K$, $P\left( N^{a}=i|{\textit{not accepted}} \right)$ is zero. Therefore,

$$\begin{aligned} \bar{P}_{F}^{a}(i)= & {} \bar{P}_{F}^{d}(i)\left( 1-\bar{P}_{F}^{a}(K) \right) +\left( 0\right) \bar{P}_{F}^{a}(K)\\= & {} \bar{P}_{F}^{d}(i)\left( 1-\bar{P}_{F}^{a}(K) \right) , \quad i=0,\ldots ,K-1. \end{aligned}$$

$\square$