The state-dependent M / G / 1 queue with orbit

Abstract

We consider a state-dependent single-server queue with orbit. This is a versatile model for the study of service systems, where the server needs a non-negligible time to retrieve waiting customers every time he completes a service. This situation arises typically when the customers are not physically present at a system, but they have a remote access to it, as in a call center station, a communication node, etc. We introduce a probabilistic approach for the performance evaluation of this queueing system, that we refer to as the queueing and Markov chain decomposition approach. Moreover, we discuss the applicability of this approach for the performance evaluation of other non-Markovian service systems with state dependencies.

Appendix

In this Appendix, we first provide several technical analytic proofs of various results of the paper. We then use the supplementary variable method to prove the main results in the paper. The proofs are quite tedious but straightforward, depending mainly on algebraic manipulations, and they are given for completeness.

1.1 Technical analytic proofs

Corollary 5.2—Proof

In light of (5.17), (5.19) yields

$$\begin{aligned} \mu _{01}=\lambda _{10} p_{10}. \end{aligned}$$

(7.1)

Now, by substituting $\mu _{1j}$ from the RHS of (5.18) into (5.20), canceling $\lambda _{0j} p_{0j}$ yields $\lambda _{1,j-1}p_{1,j-1}+\mu _{0,j+1}=\mu _{0j}+\lambda _{1j} p_{1j}$, for $j\ge 1$, which can be written as

$$\begin{aligned} \mu _{0,j+1}-\mu _{0j}=\lambda _{1j} p_{1j} -\lambda _{1,j-1} p_{1,j-1},\quad j\ge 1. \end{aligned}$$

(7.2)

Summing (7.2) for consecutive values of j and canceling equal terms yields

$$\begin{aligned} \mu _{0,j+1}-\mu _{01}=\lambda _{1j}p_{1j}-\lambda _{10}p_{10},\quad j\ge 1, \end{aligned}$$

which reduces to $\mu _{0,j+1}=\lambda _{1j}p_{1j}$, for $j\ge 1$, using (7.1). $\square$

Lemma 5.2—Proof

We have

$$\begin{aligned} E[\min (X,T_{\lambda })]= & {} \int _{0}^{\infty } \Pr [\min (X,T_{\lambda })>x] \mathrm{d}x=\int _{0}^{\infty } \Pr [X>x]\mathrm{e}^{-\lambda x} \mathrm{d}x \\= & {} \int _{0}^{\infty } \int _{x}^{\infty } \mathrm{d}F(y) \mathrm{e}^{-\lambda x}\mathrm{d}x = \int _{0}^{\infty } \int _{0}^y \mathrm{e}^{-\lambda x} \mathrm{d}x \mathrm{d}F(y) \\= & {} \int _{0}^{\infty } \frac{1-\mathrm{e}^{-\lambda y}}{\lambda } \mathrm{d}F(y)=\frac{1-\tilde{F} (\lambda )}{\lambda }. \end{aligned}$$

$\square$

Theorem 5.1—Proof

We first establish the initial conditions. The initial condition (5.40) is simply (5.21). By (5.40) and (5.34) we have that

$$\begin{aligned} \lambda _{00}p_{00}= & {} \lambda _{00}p_{00}\tilde{B}_{10}(\lambda _{10})+\mu _{01}\tilde{B}_{00}(\lambda _{10}), \end{aligned}$$

which yields (5.41). By (5.33) for $j=1$ we have

$$\begin{aligned} \mu _{01}=\mu _{11}\tilde{A}_{1}(\lambda _{01}), \end{aligned}$$

which, in combination with (5.41), yields (5.42). Also, (5.43) is identical to (5.8).

Now, solving (5.35) for $\mu _{0,j+1}$ yields (5.36). Indeed, we have that

$$\begin{aligned} \mu _{0,j+1}= & {} \frac{\mu _{1j}-\lambda _{0j}p_{0j}\tilde{B}_{1j}(\lambda _{1j})-\lambda _{1,j-1}p_{1,j-1}\tilde{S}_{1,j-1}(\lambda _{1j})}{\tilde{B} _{0j}(\lambda _{1j})} \\= & {} \frac{\mu _{1j}-(\mu _{1j}-\mu _{0j})\tilde{B}_{1j}(\lambda _{1j})-\mu _{0j} \tilde{S}_{1,j-1}(\lambda _{1j})}{\tilde{B}_{0j}(\lambda _{1j})} \\= & {} \mu _{1j}\frac{1-\tilde{B}_{1j}(\lambda _{1j})}{\tilde{B}_{0j}(\lambda _{1j})}+\mu _{0j}\frac{\tilde{B}_{1j}(\lambda _{1j})-\tilde{S} _{1,j-1}(\lambda _{1j})}{\tilde{B}_{0j}(\lambda _{1j})},\quad j\ge 1, \end{aligned}$$

where in the second equality we used equations (5.22) and (5.23). Now, (5.37) immediately follows from (5.33) for $j\ge 1$ and (5.38) is identical to (5.9). Substituting (5.1) into (5.13) and using (5.22) yields

$$\begin{aligned} a_{j}= & {} \frac{\mu _{0,j+1}\left( 1-\tilde{B}_{j}(\lambda _{1j})\right) +\lambda _{0j}p_{0j}\left( 1-\tilde{B}_{j}(\lambda _{1j})\right) }{\lambda _{1j}p_{1j}} \\= & {} \frac{\mu _{0,j+1}\left( 1-\tilde{B}_{j}(\lambda _{1j})\right) +(\mu _{1j}-\mu _{0j})\left( 1-\tilde{B}_{j}(\lambda _{1j})\right) }{ \mu _{0,j+1}} \\= & {} \left( 1-\tilde{B}_{j}(\lambda _{1j})\right) \left( 1+\frac{ \mu _{1j}-\mu _{0j}}{\mu _{0,j+1}}\right) , \end{aligned}$$

which gives the first line in (5.39). The second line follows using (5.1), (5.2) and (5.22). Indeed, we have

$$\begin{aligned} a_{j}= & {} \left( 1-\frac{\mu _{0,j+1} \tilde{B}_{0j}\left( \lambda _{1j}\right) +\left( \mu _{1j}-\mu _{0j}\right) \tilde{B}_{1j}\left( \lambda _{1j}\right) }{ \mu _{0,j+1} +\mu _{1j}-\mu _{0j}}\right) \frac{\mu _{0,j+1}+\mu _{1j}-\mu _{0j}}{ \mu _{0,j+1}} \\= & {} \left( \frac{\mu _{0,j+1} +\mu _{1j}-\mu _{0j}-\mu _{0,j+1} \tilde{B} _{0j}\left( \lambda _{1j}\right) -\left( \mu _{1j}-\mu _{0j}\right) \tilde{B} _{1j}\left( \lambda _{1j}\right) }{\mu _{0,j+1} +\mu _{1j}-\mu _{0j}}\right) \\&\times \frac{\mu _{0,j+1}+\mu _{1j}-\mu _{0j}}{\mu _{0,j+1}} \\= & {} \frac{\mu _{0,j+1} \left( 1-\tilde{B}_{0j}\left( \lambda _{1j}\right) \right) +\left( \mu _{1j}-\mu _{0j}\right) \left( 1-\tilde{B} _{1j}\left( \lambda _{1j}\right) \right) }{\mu _{0,j+1}}. \end{aligned}$$

Finally, (5.44) is identical to (5.7).$\square$

Theorem 5.2—Proof

Using (5.22) and (5.23), we deduce that

$$\begin{aligned} \mu _{1j}=\lambda _{0j}p_{0j}+\mu _{0j}=\lambda _{0j}p_{0j}+\lambda _{1,j-1}p_{1,j-1},\quad j\ge 1. \end{aligned}$$

(7.3)

Equations (5.23) and (5.33) yield

$$\begin{aligned} \lambda _{1,j-1}p_{1,j-1}=\mu _{0j}=\mu _{1j}\tilde{A}_{j}(\lambda _{0j}),\quad j\ge 1, \end{aligned}$$

so, using (7.3), we conclude that

$$\begin{aligned} \lambda _{1,j-1}p_{1,j-1}=\left( \lambda _{0j}p_{0j}+\lambda _{1,j-1}p_{1,j-1}\right) \tilde{A}_{j}(\lambda _{0j}),\quad j\ge 1. \end{aligned}$$

Solving for $p_{0j}$ yields (5.45). Next, equations (5.34) and (5.21) imply that

$$\begin{aligned} \lambda _{00}p_{00}= & {} \mu _{01}\tilde{B}_{00}\left( \lambda _{10}\right) +\lambda _{00}p_{00}\tilde{B}_{10}\left( \lambda _{10}\right) =\nu _{10} \tilde{B}_{0}\left( \lambda _{10}\right) \\= & {} \left( \lambda _{10}p_{10}+\lambda _{00}p_{00}\right) \tilde{B}_{0}\left( \lambda _{10}\right) , \end{aligned}$$

where the second and third equality follow by (5.2) and (5.24), respectively. Noting that $\mu _{01}=\lambda _{10}p_{10}$ from (5.23) and solving for $p_{10}$ yields both expressions in (5.47).

Finally, substituting (7.3) and (5.23) into the left-hand side and right-hand side of (5.35), respectively, we get

$$\begin{aligned} \lambda _{0j}p_{0j}+\lambda _{1,j-1}p_{1,j-1}= & {} \lambda _{1j}p_{1j}\tilde{B} _{0j}(\lambda _{1j})+\lambda _{0j}p_{0j}\tilde{B}_{1j}(\lambda _{1j})\nonumber \\&+\lambda _{1,j-1}p_{1,j-1}\tilde{S}_{1,j-1}(\lambda _{1j}) \nonumber \\= & {} \nu _{1j}\tilde{B}_{j}\left( \lambda _{1j}\right) +\lambda _{1,j-1}p_{1,j-1}\tilde{S}_{1,j-1}\left( \lambda _{1j}\right) ,\quad j\ge 1.\nonumber \\ \end{aligned}$$

(7.4)

Rearranging terms in the first equality yields

$$\begin{aligned}&\lambda _{1j}p_{1j}\tilde{B}_{0j}(\lambda _{1j})\\&\quad =\lambda _{0j}p_{0j}\left( 1- \tilde{B}_{1j}\left( \lambda _{1j}\right) \right) +\lambda _{1,j-1}p_{1,j-1}\left( 1-\tilde{S}_{1,j-1}\left( \lambda _{1j}\right) \right) ,\quad j\ge 1. \end{aligned}$$

Solving for $p_{1j}$ yields the first line in (5.46). Substituting (5.24) in the second line of (7.4) implies that

$$\begin{aligned}&\lambda _{0j}p_{0j}+\lambda _{1,j-1}p_{1,j-1}\nonumber \\&\quad =\left( \lambda _{1j}p_{1j}+\lambda _{0j}p_{0j}\right) \tilde{B}_{j}\left( \lambda _{1j}\right) +\lambda _{j-1}p_{1,j-1}\tilde{S}_{1,j-1}\left( \lambda _{1j}\right) ,\quad j\ge 1. \end{aligned}$$

Solving for $p_{1j}$ yields the second line in (5.46). $\square$

Corollary 5.4—Proof

We prove the result by induction. First, note that (5.40)–(5.42) imply that $\mu _{10}, \mu _{01}$ and $\mu _{11}$ are linear in $p_{00}$. Also, (5.44) shows that LSTs $\tilde{S}_{0j}(s)$ are independent of $p_{00}$, for $j\ge 1$. Since $\mu _{01}$ is linear in $p_{00}$, (5.1) for $j=1$ implies that $\nu _{10}$ is linear in $p_{00}$. Moreover, since $\nu _{10}$ and $\mu _{01}$ are linear in $p_{00}$, (5.2) implies that $\tilde{B}_{0}(s)$ is independent of $p_{00}$ and consequently $\tilde{S}_{10}(s)$ is independent of $p_{00}$. From formulas (5.45) for $j=1$ and (5.47), we have that $p_{01}$ and $p_{10}$ are linear in $p_{00}$.

We make the induction hypothesis: Assume that the rates $\mu _{01}, \mu _{02},\ldots , \mu _{0j}$, the rates $\mu _{10}, \mu _{11},\ldots , \mu _{1j}$, the rates $\nu _{10}, \nu _{11},\ldots , \nu _{1,j-1}$, the probabilities $p_{10}, p_{11},\ldots , p_{1,j-1}$, and the probabilities $p_{01}, p_{02},\ldots , p_{0j}$ are linear in $p_{00}$. Moreover, assume that the LSTs $\tilde{B}_{0}(s), \tilde{B}_{1}(s),\ldots , \tilde{B}_{j-1}(s)$ and $\tilde{S}_{10}(s), \tilde{S}_{11}(s),\ldots , \tilde{S}_{1,j-1}(s)$ are independent of $p_{00}$.

We will prove the following: (i) $\mu _{0,j+1}$ is linear in $p_{00}$, (ii) $\mu _{1,j+1}$ is linear in $p_{00}$, (iii) $\nu _{1j}$ is linear in $p_{00}$, (iv) $\tilde{B}_{j}(s)$ is independent of $p_{00}$, (v) $\tilde{S}_{1j}(s)$ is independent of $p_{00}$, (vi) $p_{1j}$ is linear in $p_{00}$ and (vii) $p_{0,j+1}$ is linear in $p_{00}$.

• (i) Since $\mu _{0j}$ and $\mu _{1j}$ are linear in $p_{00}$ and $\tilde{S}_{1,j-1}(s)$ is independent of $p_{00}$, (5.36) implies that $\mu _{0,j+1}$ is linear in $p_{00}$.

• (ii) Using that $\mu _{0,j+1}$ is linear in $p_{00}$, (5.37) gives that $\mu _{1,j+1}$ is linear in $p_{00}$.

• (iii) Since $\mu _{0,j+1}$ and $p_{0j}$ are linear in $p_{00}$, (5.1) implies that $\nu _{1j}$ is linear in $p_{00}$.

• (iv) Since $\mu _{0,j+1}, p_{0j}$ and $\nu _{1j}$ are linear in $p_{00}$, (5.2) implies that $\tilde{B}_{j}(s)$ is independent of $p_{00}$.

• (v) Using that $\mu _{1j}, \mu _{0j}$, and $\mu _{0,j+1}$ are linear in $p_{00}$, (5.39) gives that $a_j$ is independent of $p_{00}$. Then, using that $a_j, \tilde{S}_{1,j-1}(s)$ and $\tilde{B}_{j}(s)$ are independent of $p_{00}$, (5.38) implies that $\tilde{S}_{1j}(s)$ is independent of $p_{00}$.

• (vi) Since $p_{0j}$ and $p_{1,j-1}$ are linear in $p_{00}$ and $\tilde{S}_{1,j-1}(s)$ is independent of $p_{00}$, (5.46) shows that $p_{1j}$ is linear in $p_{00}$.

• (vii) (5.45), using that $p_{1j}$ is linear in $p_{00}$, yields that $p_{0,j+1}$ is linear in $p_{00}$. $\square$

Theorem 5.3—Proof

Using (5.46) in combination with (5.45) gives

$$\begin{aligned} p_{1j}=\frac{\lambda _{1,j-1}\left[ \left( 1{-}\tilde{B}_{1j}(\lambda _{1j})\right) \left( 1-\tilde{A}_{j}(\lambda _{0j})\right) {+}\tilde{A}_{j}(\lambda _{0j}) \left( 1{-}\tilde{S}_{1,j-1}(\lambda _{1j})\right) \right] }{\lambda _{1j}\tilde{B}_{0j}(\lambda _{1j})\tilde{A}_{j}(\lambda _{0j})}p_{1,j-1}. \end{aligned}$$

Thus,

$$\begin{aligned} p_{1j}=k_jp_{1,j-1},\quad j\ge 1. \end{aligned}$$

(7.5)

Also, (5.45) and (5.47) can be written as

$$\begin{aligned} p_{0j}=l_jp_{1,j-1},\quad j\ge 1, \end{aligned}$$

(7.6)

and

$$\begin{aligned} p_{10}=qp_{00}, \end{aligned}$$

(7.7)

respectively. Then,

$$\begin{aligned}&p_{0j}+p_{1j}\overset{(7.5),(7.6)}{=}(k_j+l_j)p_{1,j-1} \overset{(7.5)}{=}(k_j{+}l_j)\prod _{i=1}^{j-1}k_ip_{10} \overset{(7.7)}{=}q(k_j{+}l_j)\prod _{i=1}^{j-1}k_ip_{00},\nonumber \\&\quad j\ge 1. \end{aligned}$$

(7.8)

The normalization equation yields

Thus, the stability condition is given by (5.48) and $p_{00}$ is given by (5.49). $\square$

Theorem 5.4—Proof

Let $\tilde{P}_{ij}(s)=p_{ij}\tilde{S}_{ij}(s)$ be the Laplace transform of the steady-state density

$$\begin{aligned} p_{ij}(r)=\lim _{t\rightarrow \infty }\lim _{\mathrm{d}r\rightarrow 0^+} \frac{\Pr [C(t)=i,Q(t)=j,S(t)\in (r,r+\mathrm{d}r]]}{\mathrm{d}r}, \end{aligned}$$

i.e.,

$$\begin{aligned} \tilde{P}_{ij}(s)=\int _0^{\infty } \mathrm{e}^{-sr} p_{ij}(r)\mathrm{d}r. \end{aligned}$$

Multiplying (5.7)–(5.9) with the corresponding $p_{ij}$s yields the following equations for $\tilde{P}_{ij}(s)$:

$$\begin{aligned} \tilde{P}_{0j}(s)= & {} \frac{1}{s-\lambda }\left( \frac{\lambda p_{0j}\tilde{A}(\lambda )}{1-\tilde{A}(\lambda )}-\frac{\lambda p_{0j}}{1-\tilde{A}(\lambda )}\tilde{A}(s)\right) ,\quad j\ge 1, \end{aligned}$$

(7.9)

$$\begin{aligned} \tilde{P}_{10}(s)= & {} \frac{1}{s-\lambda }\left( \frac{\lambda p_{10} \tilde{B}(\lambda )}{1-\tilde{B}(\lambda )}-\frac{\lambda p_{10} }{1-\tilde{B}(\lambda )}\tilde{B}(s)\right) , \end{aligned}$$

(7.10)

$$\begin{aligned} \tilde{P}_{1j}(s)= & {} \frac{1}{s-\lambda }\left( \frac{\lambda p_{1j} a_j \tilde{B}(\lambda )}{1-\tilde{B}(\lambda )} + \frac{\lambda p_{1j} (1-a_j) \tilde{S}_{1,j-1}(\lambda )}{1-\tilde{S}_{1,j-1}(\lambda )}\right. \nonumber \\&\quad -\,\left. \frac{\lambda p_{1j} a_j }{1-\tilde{B}(\lambda )}\tilde{B}(s) - \frac{\lambda p_{1j} (1-a_j) }{p_{1,j-1} (1-\tilde{S}_{1,j-1}(\lambda ))}\tilde{P}_{1,j-1}(s) \right) ,\quad j\ge 1. \nonumber \\ \end{aligned}$$

(7.11)

We simplify the coefficients appearing in these equations, using (5.21)–(5.23) and (5.33)–(5.35) of the second and third QMCD system. We have that

$$\begin{aligned} \nu _{1j}= & {} \mu _{0,j+1}+\lambda p_{0j}=\lambda p_{1j}+\lambda p_{0j},\quad j\ge 1,\\ a_j= & {} \frac{\nu _{1j}(1-\tilde{B}(\lambda ))}{\lambda p_{1j}},\quad j\ge 1,\\ 1-a_j= & {} \frac{\lambda p_{1,j-1}(1-\tilde{S}_{1,j-1}(s))}{\lambda p_{1j}},\quad j\ge 1, \end{aligned}$$

so we can easily see (using (5.13), (5.21)–(5.23) and (5.33)–(5.35)) that

$$\begin{aligned} \frac{\lambda p_{0j} \tilde{A}(\lambda )}{1-\tilde{A}(\lambda )}= & {} \mu _{0j},\quad j\ge 1, \end{aligned}$$

(7.12)

$$\begin{aligned} \frac{\lambda p_{0j}}{1-\tilde{A}(\lambda )}= & {} \mu _{1j},\quad j\ge 1, \end{aligned}$$

(7.13)

$$\begin{aligned} \frac{\lambda p_{10} \tilde{B}(\lambda )}{1-\tilde{B}(\lambda )}= & {} \mu _{10}, \end{aligned}$$

(7.14)

$$\begin{aligned} \frac{\lambda p_{10}}{1-\tilde{B}(\lambda )}= & {} \nu _{10}, \end{aligned}$$

(7.15)

$$\begin{aligned} \frac{\lambda p_{1j} a_j}{1-\tilde{B}(\lambda )}= & {} \nu _{1j},\quad j\ge 1, \end{aligned}$$

(7.16)

$$\begin{aligned} \frac{\lambda p_{1j} (1-a_j) }{p_{1,j-1} (1-\tilde{S}_{1,j-1}(\lambda ))}= & {} \lambda ,\quad j\ge 1, \end{aligned}$$

(7.17)

and

$$\begin{aligned} \frac{\lambda p_{1j} a_j \tilde{B}(\lambda )}{1-\tilde{B}(\lambda )} + \frac{\lambda p_{1j} (1-a_j) \tilde{S}_{1,j-1}(\lambda )}{1-\tilde{S}_{1,j-1}(\lambda )}=\mu _{1j},\quad j\ge 1. \end{aligned}$$

(7.18)

Substituting (7.12)–(7.18) in (7.9)–(7.11) yields

$$\begin{aligned} \tilde{P}_{0j}(s)= & {} \frac{1}{s-\lambda } \left( \mu _{0j}-\mu _{1j}\tilde{A}(s)\right) ,\quad j\ge 1, \end{aligned}$$

(7.19)

$$\begin{aligned} \tilde{P}_{10}(s)= & {} \frac{1}{s-\lambda } \left( \mu _{10}-\nu _{10}\tilde{B}(s)\right) , \end{aligned}$$

(7.20)

$$\begin{aligned} \tilde{P}_{1j}(s)= & {} \frac{1}{s-\lambda } \left( \mu _{1j}-\nu _{1j}\tilde{B}(s)-\lambda \tilde{P}_{1,j-1}(s)\right) ,\, j\ge 1. \end{aligned}$$

(7.21)

We now apply a generating function approach to determine the equilibrium distribution of the number of customers in the system. To this end, we introduce the generating functions

$$\begin{aligned}&\tilde{P}_0(s,z)=\sum _{j=1}^{\infty } \tilde{P}_{0j}(s)z^j,\quad \tilde{P}_1(s,z)=\sum _{j=0}^{\infty } \tilde{P}_{1j}(s)z^j,\\&\tilde{M}_{0}(z)=\sum _{j=1}^{\infty } \mu _{0j},\quad \tilde{M}_{1}(z)=\sum _{j=0}^{\infty } \mu _{1j}. \end{aligned}$$

Multiplying (7.19) by $(s-\lambda )z^j$, and summing for $j\ge 1$, yields

$$\begin{aligned} (\lambda -s) \tilde{P}_0(s,z)=(\tilde{M}_1(z)-\mu _{10})\tilde{A}(s)-\tilde{M}_0(z). \end{aligned}$$

(7.22)

Similarly, multiplying (7.21) by $(s-\lambda )z^j$, summing for $j\ge 1$, and summing also (7.20) yields

$$\begin{aligned} (\lambda -s)\tilde{P}_1(s,z)=\sum _{j=0}^{\infty } (\mu _{0,j+1}+\lambda p_{0j})z^j\tilde{B}(s)+\lambda z \tilde{P}_1(s,z)-\tilde{M}_1(z). \end{aligned}$$

(7.23)

Note, however, that

$$\begin{aligned} \sum _{j=0}^{\infty } \mu _{0,j+1} z^j= & {} \frac{\tilde{M}_0(z)}{z},\\ \sum _{j=0}^{\infty }\lambda p_{0j} z^j= & {} \lambda p_{00}+\lambda \sum _{j=1}^{\infty } \tilde{P}_{0j}(0)z^j=\lambda p_{00}+\lambda \tilde{P}_0(0,z), \end{aligned}$$

so (7.23) assumes the form

$$\begin{aligned} (\lambda -s)\tilde{P}_1(s,z)= & {} \frac{1}{z}\tilde{M}_0(z)\tilde{B}(s)+\lambda p_{00} \tilde{B}(s)+\lambda \tilde{P}_0(0,z)\tilde{B}(s)\nonumber \\&+\,\lambda z \tilde{P}_1(s,z)-\tilde{M}_1(z). \end{aligned}$$

(7.24)

Setting $s=0$ in (7.22), $s=0$ in (7.24), $s=\lambda$ in (7.22), and $s=\lambda (1-z)$ in (7.24) yields the system

$$\begin{aligned} \lambda \tilde{P}_0(0,z)= & {} \tilde{M}_1(z)-\mu _{10}-\tilde{M}_0(z), \end{aligned}$$

(7.25)

$$\begin{aligned} \lambda (1-z) \tilde{P}_1(0,z)= & {} \frac{1}{z}\tilde{M}_0(z)+\lambda p_{00}+\lambda \tilde{P}_0(0,z)-\tilde{M}_1(z), \end{aligned}$$

(7.26)

$$\begin{aligned} \tilde{M}_1(z)\tilde{A}(\lambda )= & {} \mu _{10} \tilde{A}(\lambda )+\tilde{M}_0(z),\\ \tilde{M}_1(z)= & {} \frac{1}{z}\tilde{M}_0(z)\tilde{B}(\lambda (1-z)){+}\lambda p_{00} \tilde{B}(\lambda (1-z)){+}\lambda \tilde{P}_0(0,z)\tilde{B}(\lambda (1{-}z)),\nonumber \end{aligned}$$

(7.27)

in the unknowns $\tilde{M}_0(z), \tilde{M}_1(z), \tilde{P}_0(0,z)$, and $\tilde{P}_1(0,z)$. We can easily solve the system by using sequentially (7.27), (7.25), and (7.26), to express the unknowns $\tilde{M}_1(z), \tilde{P}_0(0,z)$, and $\tilde{P}_1(0,z)$ in terms of $\tilde{M}_0(z)$. We get

$$\begin{aligned} \tilde{M}_1(z)= & {} \lambda p_{00} +\frac{1}{\tilde{A}(\lambda )}\tilde{M}_0(z), \end{aligned}$$

(7.28)

$$\begin{aligned} \tilde{P}_0(0,z)= & {} \frac{1-\tilde{A}(\lambda )}{\lambda \tilde{A}(\lambda )}\tilde{M}_0(z), \end{aligned}$$

(7.29)

$$\begin{aligned} \tilde{P}_1(0,z)= & {} \frac{1}{\lambda z}\tilde{M}_0(z). \end{aligned}$$

(7.30)

We can then plug (7.28)–(7.30) and obtain the equation

$$\begin{aligned} \lambda p_{00} +\frac{1}{\tilde{A}(\lambda )}\tilde{M}_0(z)= \left( \frac{1}{z}\tilde{M}_0(z) +\lambda p_{00} +\frac{1-\tilde{A}(\lambda )}{\tilde{A}(\lambda )}\tilde{M}_0(z)\right) \tilde{B}(\lambda (1-z)) \end{aligned}$$

(7.31)

for $\tilde{M}_0(z)$. Solving (7.31) for $\tilde{M}_0(z)$ and substituting in (7.28)–(7.30) yields the following explicit expressions for $\tilde{M}_0(z), \tilde{M}_1(z), \tilde{P}_0(0,z)$, and $\tilde{P}_1(0,z)$:

$$\begin{aligned} \tilde{M}_0(z)= & {} \frac{\lambda \tilde{A}(\lambda )z(1-\tilde{B}(\lambda (1-z)))}{\tilde{A}(\lambda )\tilde{B}(\lambda (1-z))(1-z)-z(1-\tilde{B}(\lambda (1-z)))}p_{00},\nonumber \\ \tilde{M}_1(z)= & {} \frac{\lambda \tilde{A}(\lambda )\tilde{B}(\lambda (1-z))(1-z)}{\tilde{A}(\lambda )\tilde{B}(\lambda (1-z))(1-z)-z(1-\tilde{B}(\lambda (1-z)))}p_{00},\nonumber \\ \tilde{P}_0(0,z)= & {} \frac{(1-\tilde{A}(\lambda ))z(1-\tilde{B}(\lambda (1-z)))}{\tilde{A}(\lambda )\tilde{B}(\lambda (1-z))(1-z)-z(1-\tilde{B}(\lambda (1-z)))}p_{00}, \end{aligned}$$

(7.32)

$$\begin{aligned} \tilde{P}_1(0,z)= & {} \frac{\tilde{A}(\lambda )(1-\tilde{B}(\lambda (1-z)))}{\tilde{A}(\lambda )\tilde{B}(\lambda (1-z))(1-z)-z(1-\tilde{B}(\lambda (1-z)))}p_{00}. \end{aligned}$$

(7.33)

The generating function of the number of customers in the system in steady-state is

$$\begin{aligned} K(z)=\sum _{j=0}^{\infty } p_{0j}z^j+\sum _{j=1}^{\infty } p_{1,j-1}z^j=p_{00}+\tilde{P}_0(0,z)+z\tilde{P}_1(0,z). \end{aligned}$$

Using (7.32)–(7.33), we obtain

$$\begin{aligned} K(z)=\frac{\tilde{A}(\lambda )(1-z)\tilde{B}(\lambda (1-z))}{\tilde{A}(\lambda )\tilde{B}(\lambda (1-z)))(1-z)-z(1-\tilde{B}(\lambda (1-z)))}p_{00}. \end{aligned}$$

(7.34)

The probability of an empty system $p_{00}$ is determined from the normalization equation $K(1)=1$. Using L’Hospital’s rule, we obtain

$$\begin{aligned} p_{00}=1-\frac{\lambda (-1)\tilde{B}'(0)}{\tilde{A}(\lambda )}=1-\frac{\lambda E[B]}{\tilde{A}(\lambda )}. \end{aligned}$$

(7.35)

The system is stable if and only if $p_{00}>0$, which gives the stability condition (5.50). Substituting (7.35) in (7.34) yields (5.51). $\square$

1.2 Alternative proofs using the supplementary variable method

In the rest of the Appendix, we use the well-known supplementary variable method, introduced by Cox [8], to provide analytic proofs for the three QMCD Systems (Corollaries 5.1–5.3) which constitute the basis for the recursive schemes for the performance analysis of the model.

To this end, we introduce the following transient probabilities and densities:

$$\begin{aligned} p_{ij}^{(t)}= & {} \Pr [C(t)=i,Q(t)=j],\quad i=0,1,\quad j\ge 0, \\ p_{0j}^{(t)}(r)= & {} \lim _{\mathrm{d}r\rightarrow 0^+} \frac{ \Pr [C(t)=0,Q(t)=j,S(t)\in (r,r+\mathrm{d}r]]}{\mathrm{d}r},\quad j\ge 1, \\ p_{1j}^{(t)}(r)= & {} \lim _{\mathrm{d}r\rightarrow 0^+} \frac{\Pr [C(t)=1,Q(t)=j,S(t)\in (r,r+\mathrm{d}r]]}{\mathrm{d}r},\quad j\ge 0, \end{aligned}$$

and their steady-state counterparts

$$\begin{aligned} p_{ij}= & {} \lim _{t\rightarrow \infty }p_{ij}^{(t)},\quad i=0,1,\quad j\ge 0, \end{aligned}$$

(7.36)

$$\begin{aligned} p_{0j}(r)= & {} \lim _{t\rightarrow \infty }p_{0j}^{(t)}(r),\quad j\ge 1, \end{aligned}$$

(7.37)

$$\begin{aligned} p_{1j}(r)= & {} \lim _{t\rightarrow \infty }p_{1j}^{(t)}(r),\quad j\ge 0. \end{aligned}$$

(7.38)

A moment of reflection reveals that the analytic quantity $p_{ij}(0)$ is identical to the corresponding rate $\mu _{ij}$ of service/seeking time completions that we considered in the probabilistic derivations, for all i, j. Therefore, similarly to (5.1) and (5.2), we have in the present analytic framework the equations

$$\begin{aligned} \nu _{1j}= & {} p_{0,j+1}(0)+\lambda _{0j}p_{0j}, \end{aligned}$$

(7.39)

$$\begin{aligned} \tilde{B}_{j}\left( s\right)= & {} \frac{1}{\nu _{1j}}\left( \mu _{0,j+1} \tilde{B }_{0j}\left( s\right) +\lambda _{0j}p_{0j}\tilde{B}_{1j}\left( s\right) \right) . \end{aligned}$$

(7.40)

Moreover, we denote by $\tilde{P}_{ij}(s)$ the Laplace transform (LT) of the steady-state density $p_{ij}(r)$, i.e.,

$$\begin{aligned} \tilde{P}_{ij}(s)=\int _{0}^{\infty }\mathrm{e}^{-sr}p_{ij}(r)\mathrm{d}r=p_{ij}\tilde{S} _{ij}(s),\quad (i,j)\in \{0,1\}\times \{0,1,2,\ldots \}{\setminus }\{(0,0)\}. \end{aligned}$$

(7.41)

We consider the evolution of the continuous time Markov process $\{(C(t),Q(t),S\left( t\right) ):t\ge 0\}$ in the interval $[t,t+\mathrm{d}t]$. Then, we have Lemma 7.1.

Lemma 7.1

The steady-state probabilities $p_{0j}, j\ge 1,$ the steady-state densities $p_{ij}(0)$ and the LTs $\tilde{P}_{ij}(s), (i,j)\in \{0,1\}\times \{0,1,2,\ldots \}{\setminus }\{(0,0)\}$, satisfy the following system of equations:

$$\begin{aligned} \lambda _{00}p_{00}-p_{10}(0)= & {} 0, \end{aligned}$$

(7.42)

$$\begin{aligned} (s-\lambda _{0j})\tilde{P}_{0j}(s)= & {} p_{0j}(0)-p_{1j}(0)\tilde{A}_{j}(s),\quad j\ge 1, \end{aligned}$$

(7.43)

$$\begin{aligned} (s-\lambda _{10})\tilde{P}_{10}(s)= & {} p_{10}(0)-p_{01}(0)\tilde{B} _{00}(s)-\lambda _{00}p_{00}\tilde{B}_{10}(s), \end{aligned}$$

(7.44)

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

(7.45)

Proof

Considering the evolution of $\{(C(t),Q(t),R(t)):t\ge 0\}$ in the interval $[0,t+\mathrm{d}t]$ and conditioning on its value at time t, we have the equations

$$\begin{aligned} p_{00}^{(t+\mathrm{d}t)}= & {} p_{00}^{(t)}(1-\lambda _{00}\mathrm{d}t)+p_{10}^{(t)}(0)\mathrm{d}t+o(\mathrm{d}t), \end{aligned}$$

(7.46)

$$\begin{aligned} p_{0j}^{(t+\mathrm{d}t)}(r-\mathrm{d}t)= & {} p_{0j}^{(t)}(r)(1-\lambda _{0j}\mathrm{d}t)+p_{1j}^{(t)}(0)a_{j}(r)\mathrm{d}t+o(\mathrm{d}t),\quad j\ge 1, \end{aligned}$$

(7.47)

$$\begin{aligned} p_{10}^{(t+\mathrm{d}t)}(r-\mathrm{d}t)= & {} p_{10}^{(t)}(r)(1-\lambda _{10}\mathrm{d}t)+p_{01}^{(t)}(0)\mathrm{d}tb_{00}(r)+p_{00}^{(t)}\lambda _{00}\mathrm{d}tb_{10}(r)+o(\mathrm{d}t),\nonumber \\ \end{aligned}$$

(7.48)

$$\begin{aligned} p_{1j}^{(t+\mathrm{d}t)}(r-\mathrm{d}t)= & {} p_{1j}^{(t)}(r)(1-\lambda _{1j}\mathrm{d}t)+p_{0,j+1}^{(t)}(0)\mathrm{d}tb_{0j}(r)+p_{0j}^{(t)}\lambda _{0j}\mathrm{d}tb_{1j}(r) \nonumber \\&+\,p_{1,j-1}^{(t)}(r)\lambda _{1,j-1}\mathrm{d}t+o(\mathrm{d}t),\quad j\ge 1, \end{aligned}$$

(7.49)

for $\mathrm{d}t\rightarrow 0^+$. Taking the limits of (7.46)–(7.49) as $t\rightarrow \infty ,$ and using (7.36)–(7.38) yields

$$\begin{aligned} p_{00}= & {} p_{00}(1-\lambda _{00}\mathrm{d}t)+p_{10}(0)\mathrm{d}t+o(\mathrm{d}t), \end{aligned}$$

(7.50)

$$\begin{aligned} p_{0j}(r-\mathrm{d}t)= & {} p_{0j}(r)(1-\lambda _{0j}\mathrm{d}t)+p_{1j}(0)a_{j}(r)\mathrm{d}t+o(\mathrm{d}t),\quad j\ge 1, \end{aligned}$$

(7.51)

$$\begin{aligned} p_{10}(r-\mathrm{d}t)= & {} p_{10}(r)(1-\lambda _{10}\mathrm{d}t)+p_{01}(0)\mathrm{d}tb_{00}(r)+p_{00}\lambda _{00}\mathrm{d}tb_{10}(r)+o(\mathrm{d}t),\nonumber \\ \end{aligned}$$

(7.52)

$$\begin{aligned} p_{1j}(r-\mathrm{d}t)= & {} p_{1j}(r)(1-\lambda _{1j}\mathrm{d}t)+p_{0,j+1}(0)\mathrm{d}tb_{0j}(r)+p_{0j}\lambda _{0j}\mathrm{d}tb_{1j}(r) \nonumber \\&+\,p_{1,j-1}(r)\lambda _{1,j-1}\mathrm{d}t+o(\mathrm{d}t),\quad j\ge 1, \end{aligned}$$

(7.53)

for $\mathrm{d}t\rightarrow 0^+$. Rearranging the terms appropriately, dividing by $\mathrm{d}t$ and taking the limits as $\mathrm{d}t\rightarrow 0^+$ in (7.50)–(7.53) yields

$$\begin{aligned} p_{10}(0)= & {} \lambda _{00}p_{00}, \end{aligned}$$

(7.54)

$$\begin{aligned} p_{0j}^{\prime }(r)= & {} \lambda _{0j}p_{0j}(r)-p_{1j}(0)a_{j}(r),\quad j\ge 1, \end{aligned}$$

(7.55)

$$\begin{aligned} p_{10}^{\prime }(r)= & {} \lambda _{10}p_{10}(r)-p_{01}(0)b_{00}(r)-\lambda _{00}p_{00}b_{10}(r), \end{aligned}$$

(7.56)

$$\begin{aligned} p_{1j}^{\prime }(r)= & {} \lambda _{1j}p_{1j}(r)-p_{0,j+1}(0)b_{0j}(r)-\lambda _{0j}p_{0j}b_{1j}(r)-\lambda _{1,j-1}p_{1,j-1}(r),\quad j\ge 1.\nonumber \\ \end{aligned}$$

(7.57)

Equation (7.54) yields immediately (7.42). Also, multiplying (7.55)–(7.57) by $\mathrm{e}^{-sr}$ and integrating with respect to $s\in [0,\infty )$ yields equations (7.43)–(7.45).$\square$

Now, we can provide an analytic proof of the second and the third QMCD systems.

Lemma 7.2

The steady-state probabilities $p_{ij}$ and the steady-state densities $p_{ij}(0)$ satisfy the second QMCD system:

$$\begin{aligned} \lambda _{00}p_{00}= & {} p_{10}(0), \end{aligned}$$

(7.58)

$$\begin{aligned} \lambda _{0j}p_{0j}= & {} p_{1j}(0)-p_{0j}(0),\quad j\ge 1, \end{aligned}$$

(7.59)

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

(7.60)

Together with the LSTs $\tilde{S}_{ij}(s)$, they also satisfy the third QMCD system:

$$\begin{aligned} p_{0j}(0)= & {} p_{1j}(0)\tilde{A}_j(\lambda _{0j}),\quad j\ge 1, \end{aligned}$$

(7.61)

$$\begin{aligned} p_{10}(0)= & {} \lambda _{00} p_{00}\tilde{B}_{10}(\lambda _{10})+p_{01}(0)\tilde{ B}_{00}(\lambda _{10})=\nu _{10}\tilde{B}_0(\lambda _{10}), \end{aligned}$$

(7.62)

$$\begin{aligned} p_{1j}(0)= & {} \lambda _{0j} p_{0j}\tilde{B}_{1j}(\lambda _{1j}) +\lambda _{1,j-1} p_{1,j-1} \tilde{S}_{1,j-1}(\lambda _{1j})+p_{0,j+1}(0)\tilde{B} _{0j}(\lambda _{1j}) \nonumber \\= & {} \nu _{1j}\tilde{B}_j(\lambda _{1j})+\lambda _{1,j-1} p_{1,j-1} \tilde{S} _{1,j-1}(\lambda _{1j}),\quad j\ge 1. \end{aligned}$$

(7.63)

Proof

Equation (7.42) yields immediately (7.58). Taking $s=0$ in (7.43)–(7.45) yields

$$\begin{aligned} \lambda _{0j}p_{0j}= & {} p_{1j}(0)-p_{0j}(0),\quad j\ge 1, \end{aligned}$$

(7.64)

$$\begin{aligned} \lambda _{10}p_{10}= & {} p_{01}(0)+\lambda _{00}p_{00}-p_{10}(0), \end{aligned}$$

(7.65)

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

(7.66)

Equation (7.64) is identical to Eq. (7.59). Also, using (7.58), Eq. (7.65) yields (7.60) for $j=0$.

Using (7.59), Eq. (7.66) assumes the form

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

Iterating this equation yields

$$\begin{aligned} \lambda _{1j}p_{1j}= & {} \sum _{i=1}^j \left( p_{0,i+1}(0)-p_{0i}(0)\right) +\lambda _{10}p_{10} \\= & {} p_{0,j+1}(0)-p_{01}(0)+\lambda _{10}p_{10},\quad j\ge 1. \end{aligned}$$

Then, using (7.60) for $j=0$, we deduce (7.60) for $j\ge 1$.

Equation (7.43) for $s=\lambda _{0j}$ yields (7.61). Also, setting $s=\lambda _{10}$ in Eq. (7.44) and using (7.40) yields (7.62). Similarly, Eq. (7.45) for $s=\lambda _{1j}$ in combination with (7.40) and (7.41) yields (7.63). $\square$

Finally, we provide an analytic proof of the first QMCD system, i.e., of Corollary 5.1 with $a_j$ given by (5.13).

Corollary 5.1—Proof

Analytic proof of (5.7)–(5.9) with $a_j$ given by (5.13).

Equation (7.43) can be written as

$$\begin{aligned} \tilde{P}_{0j}(s)=\frac{1}{s-\lambda _{0j}}\left( p_{0j}(0)-p_{1j}(0)\tilde{A }_{j}(s)\right) ,\quad j\ge 1. \end{aligned}$$

Using (7.61), the above equation yields

$$\begin{aligned} \tilde{P}_{0j}(s)=\frac{1}{s-\lambda _{0j}}p_{1j}(0)\left( \tilde{A} _{j}(\lambda _{0j})-\tilde{A}_{j}(s)\right) ,\quad j\ge 1. \end{aligned}$$

Dividing by $p_{0j}$, we deduce that

which proves formula (5.7).

Equation (7.44) can be written as

$$\begin{aligned} \tilde{P}_{10}(s)=\frac{1}{s-\lambda _{10}}\left( p_{10}(0)-p_{01}(0)\tilde{B }_{00}(s)-p_{00}\lambda _{00}\tilde{B}_{10}(s)\right) . \end{aligned}$$

Using (7.62) we obtain

$$\begin{aligned} \tilde{P}_{10}(s)=\frac{1}{s-\lambda _{10}}\left[ p_{01}(0)\left( \tilde{B} _{00}(\lambda _{10})-\tilde{B}_{00}(s)\right) +p_{00}\lambda _{00}\left( \tilde{B}_{10}(\lambda _{10})-\tilde{B}_{10}(s)\right) \right] , \end{aligned}$$

and dividing by $p_{10}$ yields

which proves formula (5.8).

Similarly, Eq. (7.45) is written as

$$\begin{aligned} \tilde{P}_{1j}(s)=\frac{1}{s-\lambda _{1j}}\left( p_{1j}(0)-p_{0,j+1}(0) \tilde{B}_{0j}(s)-p_{0j}\lambda _{0j}\tilde{B}_{1j}(s)-\lambda _{1,j-1} \tilde{P}_{1,j-1}(s)\right) ,\quad j\ge 1. \end{aligned}$$

Using (7.63) and (7.41) we obtain

$$\begin{aligned} \tilde{P}_{1j}(s)= & {} \frac{1}{s-\lambda _{1j}}\left[ p_{0,j+1}(0)\left( \tilde{B}_{0j}(\lambda _{1j})-\tilde{B}_{0j}(s)\right) +\lambda _{0j}p_{0j}\left( \tilde{B}_{1j}(\lambda _{1j})-\tilde{B}_{1j}(s)\right) \right. \\&+\left. \lambda _{1,j-1}\left( \tilde{P}_{1,j-1}(\lambda _{1j})-\tilde{P} _{1,j-1}(s)\right) \right] ,\quad j\ge 1. \end{aligned}$$

Dividing by $p_{1j}$ yields

where

$$\begin{aligned} \alpha _{j}=\frac{\nu _{1j}\left( 1-\tilde{B}_{j}(\lambda _{1j})\right) }{ \lambda _{1j}p_{1j}},\quad j\ge 1, \end{aligned}$$

(7.67)

and

$$\begin{aligned}&\frac{\lambda _{1,j-1}p_{1,j-1}\left( 1-\tilde{S}_{1,j-1}(\lambda _{1j})\right) }{\lambda _{1j}p_{1j}} \\&\quad \overset{(7.63)}{=} \frac{\lambda _{1,j-1}p_{1,j-1}-p_{1j}(0)+p_{j}^{1}\tilde{B}_{j}(\lambda _{1j})}{\lambda _{1j}p_{1j}} \\&\quad \overset{(7.59)}{=} \frac{\lambda _{1,j-1}p_{1,j-1}-\lambda _{0j}p_{0j}-p_{0j}(0)+p_{j}^{1}\tilde{B} _{j}(\lambda _{1j})}{\lambda _{1j}p_{1j}} \\&\quad \overset{(7.60)}{=} \frac{-\lambda _{0j}p_{0j}+p_{j}^{1} \tilde{B}_{j}(\lambda _{1j})}{\lambda _{1j}p_{1j}} \\&\quad \overset{(7.39)}{=} \frac{p_{0,j+1}(0)-p_{j}^{1}+p_{j}^{1} \tilde{B}_{j}(\lambda _{1j})}{\lambda _{1j}p_{1j}} \\&\quad \overset{(7.60)}{=} \frac{\lambda _{1j}p_{1j}-p_{j}^{1}\left( 1-\tilde{B}_{j}(\lambda _{1j})\right) }{\lambda _{1j}p_{1j}} \\&\quad \overset{(7.67)}{=} 1-\alpha _{j},\quad j\ge 1. \end{aligned}$$

Thus, formula (5.9) has been proved with $a_j$ given by (5.13). $\square$