On $M^{X}/G(M/H)/1$ retrial system with vacation: service helpline performance measurement

Abstract

This paper analyzes an unreliable $M^{X}/G(M/H)/1$ retrial system with vacation. We present closed-form expressions for the important performance indicators of the system, and derive the optimal vacation policies for minimizing the average waiting time of orbiting customers. The performance metrics relevant for helpline services are developed. Numerical experiments are conducted to examine the effect of vacation policy on the queue length and busy period of the system.

Appendix

Notations	Meaning
$\mathcal {P}$	Vacation policy
$M^{X}/G/1$	Batch-arrival, general service-time queue
$M^{X}/G(M/H)/1$	Unreliable batch-arrival, general service and repair-time after Markovian breakdown
$\mathcal {F}_j$	Maximum allowable vacations to be taken is j
$\mathcal {S}$	Single-vacation policy, i.e., $j=1$
$\mathcal {M}$	Multiple-vacation policy, i.e., $j\rightarrow \infty$
$Q^{\mathcal {P}}(t)$	Transient number of callers both in service and in orbit under $\mathcal {P}$ and at t
$O^{\mathcal {P}}(t)$	Transient number of callers in orbit under $\mathcal {P}$ and at t
$I^{\mathcal {P}}(t)$	Transient number of callers in service under $\mathcal {P}$ and at t
$Q^{\mathcal {F}_j}(z)$	p.g.f of the system size at arbitrary epoch
$V^i_{-}(t)$	Elapsed $i^{\text {th}}$ vacation time at t
$X^i_{-}(t)$	Elapsed $i^{\text {th}}$ service time at t
$R^i_{-}(t)$	Elapsed $i^{\text {th}}$ repair time at t
$C(t)(=\nu , 0,1,2)$	State of the server at t, i.e., vacation, idle, busy or repair
$L^{(k)}$	k-busy period for the $M^X/G(M/H)/1$ retrial system without vacation
$L^{\mathcal {P}}$	Busy period of $M^X/G(M/H)/1$ under $\mathcal {P}$ and $L^{\mathcal {P}}=L_0^{\mathcal {P}}+L_w^{\mathcal {P}}+L_r^{\mathcal {P}}$, where
$L_0^{\mathcal {P}}$	Server-waiting time for the next busy epoch under $L^{\mathcal {P}}$
$L_1^{\mathcal {P}}$	Generalized working length under $\mathcal {P}$ ($L_1^{\mathcal {P}}=L_w^{\mathcal {P}}+L_r^{\mathcal {P}}$)
$L_w^{\mathcal {P}}$	Value-adding or working length under $L^{\mathcal {P}}$
$L_r^{\mathcal {P}}$	Repair length during $L^{\mathcal {P}}$
$L_0^{(k)}$	Server-waiting time for the next busy epoch under $L^{(k)}$
$L_1^{(k)}$	Generalized working length under k-busy period ($L_1^{(k)}=L_w^{(k)}+L_r^{(k)}$)
$L_w^{(k)}$	Value-adding or working length under $L^{(k)}$
$L_r^{(k)}$	Repair length during $L^{(k)}$
I(t)	Number of services completed at time t
$\omega (t)$	The elapsed service time of the work at time t

Proof of Lemma 1

(i) The first part is to compute the p.g.f left by a departing caller. We divide our proofs into two cases.

Case 1 Given that $Q_n=0$, the server goes into vacation immediately at $S_n$ because it finds that the orbit is empty. Two events may happen, $\{Q_n=0,\overline{\nu }=0\}$ and $\{Q_n=0,\overline{\nu }>0\}$. On the event $\{Q_n=0,\overline{\nu }=0\}$, we must have $Q_{n+1}=X-1+M_{n+1}$. On the other hand, for the event $\{Q_n=0,\overline{\nu }=i\}$ for $i\ge 1$, we have

$$\begin{aligned} Q_{n+1}= & {} \left\{ \begin{array}{ll} i+M_{n+1}-1 &{}\quad \text{ w.p }\,\frac{\theta }{\lambda +\theta } \\ i+M_{n+1}-1+X &{}\quad \text{ w.p }\,\frac{\lambda }{\lambda +\theta }. \end{array} \right. \\ P\{Q_{n+1}= & {} k|Q_n=0,\overline{\nu }=i\} {=} \left\{ \begin{array}{ll} \sum _{l=1}^{k+1}c_l m_{n+1,k-l+1} &{}\quad \text{ if }\,i=0 \\ \frac{\lambda }{\lambda +\theta }\sum _{j=1}^{k-i+1}m_{n+1,k-j-i+1}c_j {+}\frac{\theta }{\lambda +\theta }m_{n+1,k-i+1}&{} \quad \text{ if }\,i\!\ge \! 1. \end{array} \right. \end{aligned}$$

Thus, we have

$$\begin{aligned} p_{0k}&=\sum _{i=0}^{\infty }P\{Q_{n+1}=k|Q_n=0,\overline{\nu }=i\} P\{\overline{\nu }=i|Q_n=0\}\nonumber \\&=\widetilde{V}(\lambda )^j\sum _{l=1}^{k+1}c_l m_{n+1,k-l+1}\nonumber \\&\quad +\sum _{i=1}^{k+1}\left\{ \frac{\lambda }{\lambda +\theta } \sum _{j=1}^{k-i+1}m_{n+1,k-j-i+1}c_j+\frac{\theta }{\lambda +\theta } m_{n+1,k-i+1}\right\} \overline{\nu }_i. \end{aligned}$$

(16)

Case 2 For $Q_n\ge 1$, then we have similar argument. There are two types of customers that competes when $Q_n\ge 1$. One is the arrival of a new batch of customer, while the other is the random customer from the orbit. On $\{Q_n=i\}$, where $i\ge 1$, we have

$$\begin{aligned} Q_{n+1} = \left\{ \begin{array}{ll} i+M_{n+1}+X-1 &{} \quad \text{ w.p }\,\frac{\lambda }{\lambda +\theta } \\ i+M_{n+1}-1 &{} \quad \text{ w.p }\,\frac{\theta }{\lambda +\theta }. \end{array} \right. \end{aligned}$$

Thus, we have for $k\ge i$,

$$\begin{aligned} p_{ik}=\frac{\lambda }{\lambda +\theta } \sum _{j=1}^{k-i+1} c_j m_{n+1,k-i+1-j} +\frac{\theta }{\lambda +\theta }m_{n+1,k-i+1}. \end{aligned}$$

(17)

We want to compute the steady state distribution of the system size immediately after departure epoches. Since $\{Q_n:n\in \mathbf N \}$ constitutes a Markov chain, we can compute its steady state by using $\pi \mathbf P =\pi$ and $\pi 1=1$. $\mathbf P =[p_{ij}]$ is the transition matrix of the embedded Markov Chain $\{Q_n\}$ and $\pi =(\pi _0,\pi _1 \pi _2\ldots )$, where $\pi _k$ is the long run fraction when a departing customer leaves behind k customers in the system. Let $Q^{\mathcal {F}_j}_{+}(z)$ be the p.g.f of the system size that is left behind by a departing customer, then $Q^{\mathcal {F}_j}_{+}(z)=\sum _{k=0}^{\infty }\pi _kz^k$. Using (2), (3) and definition of $Q^{\mathcal {F}_j}_{+}(z)$, we have after some tedious algebraic manipulation,

$$\begin{aligned} Q^{\mathcal {F}_j}_{+}(z)=\pi _0 \frac{\{(1-\zeta _j(z))(\lambda X(z)+\theta )+\theta \widetilde{V}(\lambda )^j(1-X(z))\}\widetilde{H}(\lambda -\lambda X(z))}{(\lambda X(z)+\theta )\widetilde{H}(\lambda -\lambda X(z))-(\lambda +\theta )z}. \end{aligned}$$

(18)

From the normalization condition, $Q^{\mathcal {F}_j}_{+}(1)=1$, we can find the constant $\pi _0$. Using (18) and L’Hospital rule, we obtain

$$\begin{aligned}&\lim _{z\rightarrow 1^-}\Pi (z)=\pi _0\left\{ \frac{\lambda \gamma _1\left[ \frac{1-\widetilde{V}(\lambda )^j}{1-\widetilde{V}(\lambda )}{} { EV}+\frac{\theta V(\lambda )^j}{\lambda (\lambda +\theta )}\right] }{1-\kappa (\theta )}\right\} =1\\&\Leftrightarrow \pi _0= \frac{1-\kappa (\theta )}{\lambda \gamma _1\left[ \frac{1-\widetilde{V}(\lambda )^j}{1-\widetilde{V}(\lambda )}{} { EV}+\frac{\theta V(\lambda )^j}{\lambda (\lambda +\theta )}\right] }. \end{aligned}$$

Using (1) and (18), we obtain the p.g.f of the queue length left by a departing customer is given in the main paper.

(ii) (Sufficiency) The sufficiency of the ergodicity result can be shown using Pake’s Lemma which is the statement as follows. If we have $Q(x)\ge 0$, and for all x, there exist $\epsilon >0$ such that $E(Q_{t+1}-Q_t|Q_t=x)\le -\epsilon <0$ for all x except on a finite set C, then $\{Q_n\}$ is positive recurrent. To show sufficiency of our ergodicity result given any i and $\kappa (\theta )<1$, we choose $\epsilon =\frac{1}{2}(1-\kappa (\theta ))>0$. Due to the recursive structure of the embedded sequence $\{Q_n:n\in \mathbf N \}$, we apply Pake’s lemma as follows. The mean drift $y_i=E[Q_{n+1}-Q_n|Q_n=i]$ is calculated as follows.

$$\begin{aligned} y_i= & {} \frac{\theta }{\theta +\lambda } E[M_{n+1}-1]+\frac{\lambda }{\theta +\lambda }E [X+M_{n+1}-1]\\= & {} EM_{n+1}-1+\frac{\lambda }{\theta +\lambda }\gamma _1=\lambda \gamma _1 EH-1+\frac{\lambda }{\theta +\lambda }\gamma _1. \end{aligned}$$

Then, we have $y_i=-2\epsilon$, implying $y_i<-\epsilon$ for all states except for a finite number of states. Therefore, $\kappa (\theta )<1$ is sufficient for the embedded chain to be ergodic. The necessary condition readily follows from Kaplan’s condition [see Senott et al. (1983)], namely $y_i<\infty$ for all $i\ge 0$ and there exists $i_0\in \mathbf Z ^+$ such that $y_i\ge 0$ for all $i\ge i_0$. $\square$

Proof of Theorem 1

We define the generating functions

$$\begin{aligned}&\omega _{\nu ,i}(z,x)=\sum _{k=0}^{\infty } \omega _{\nu ,k,i}(x)z^k;\omega _{0}(z)=\sum _{k=0}^{\infty } \omega _{0,k}(x)z^k;\\&\omega _{1}(z,x)=\sum _{k=0}^{\infty } \omega _{1,k}(x)z^k; \omega _{0}(z,y,x)=\sum _{k=0}^{\infty } \omega _{2,k}(x,y)z^k. \end{aligned}$$

For notational parsimony, we omit the superscript $\mathcal {F}_j$ in the following discussion. Using the technique of supplementary variables, we obtain the following system of equations for $x\ge 0, y\ge 0, k\ge 0$:

$$\begin{aligned}&\displaystyle \left[ \frac{d}{dx}+\lambda +\overline{\nu }(x)\right] \omega _{\nu ,k,i}(x)=\lambda \sum _{l=1}^kc_j\omega _{\nu ,k-l,i}(x),\quad 1\le i\le j\end{aligned}$$

(19)

$$\begin{aligned}&\displaystyle \left[ \frac{d}{dx}+\lambda +\overline{b}(x)+\alpha \right] \omega _{1,k}(x)=\lambda \sum _{l=1}^k c_j\omega _{1,k-l}(x)+\int _0^{\infty }\omega _{2,k}(x,y)\overline{r}(y)dy\end{aligned}$$

(20)

$$\begin{aligned}&\displaystyle \left[ \frac{d}{dx}+\lambda +\overline{r}(x)\right] \omega _{2,k}(x,y)=\lambda \sum _{l=1}^k c_j\omega _{2,k-l}(x,y). \end{aligned}$$

(21)

For $k\ge 1$, we have

$$\begin{aligned}&\displaystyle (\lambda +\theta )\omega _{0,k}=\int _0^{\infty } \omega _{1,k}(x)\overline{b}(x)dx+\sum _{i=1}^j\int _0^{\infty } \omega _{\nu ,k,i}(x)\overline{\nu }(x)dx \end{aligned}$$

(22)

$$\begin{aligned}&\displaystyle \lambda \omega _{0,0}=\int _0^{\infty }\overline{\nu }(x)\omega _{\nu ,0,j}(x)dx. \end{aligned}$$

(23)

These equations are to be solved under the boundary conditions:

$$\begin{aligned}&\displaystyle \omega _{\nu ,0,1}(0)=\int _{0}^{\infty }\omega _{1,0}(x)\overline{b}(x)dx \end{aligned}$$

(24)

$$\begin{aligned}&\displaystyle \omega _{\nu ,0,i}(0)=\int _{0}^{\infty }\omega _{\nu ,0,i-1}(x)\overline{\nu }(x)dx, 2\le i\le j \end{aligned}$$

(25)

$$\begin{aligned}&\displaystyle \omega _{\nu ,k,i}(0)=0, \quad k\ge 1, 1\le i\le j \end{aligned}$$

(26)

$$\begin{aligned}&\displaystyle \omega _{1,k}(0)=\lambda \sum _{j=1}^{k+1}c_j \omega _{0,k-j+1}+\theta \omega _{0,k+1}, \quad k\ge 0 \end{aligned}$$

(27)

$$\begin{aligned}&\displaystyle \omega _{2,k}(x,0)=\alpha \omega _{1,k}(x),\quad k\ge 0. \end{aligned}$$

(28)

The normalizing condition is

$$\begin{aligned} \sum _{k=0}^{\infty }\left[ \sum _{i=1}^{j} \int _0^{\infty } \omega _{\nu ,k,i}(x) dx +\omega _{0,k}+\int _0^{\infty }\omega _{1,k}(x)dx +\int _0^{\infty }\int _0^{\infty }\omega _{2,k}(x,y)dydx\right] =1. \end{aligned}$$

(29)

Using generating functions, we can express system of equations to be

$$\begin{aligned}&\left[ \frac{d}{dx}+\lambda -\lambda X(z)+\overline{\nu }(x)\right] \omega _{\nu ,i}(z,x)=0, \quad 1\le i\le j \end{aligned}$$

(30)

$$\begin{aligned}&\left[ \frac{d}{dx}+\lambda -\lambda X(z)+\overline{b}(x)+\alpha \right] \omega _{1}(z,x)=\int _0^{\infty }\omega _{2}(z,y,x)\overline{r}(y)dy \end{aligned}$$

(31)

$$\begin{aligned}&\left[ \frac{d}{dy}+\lambda -\lambda X(z)+\overline{r}(y)\right] \omega _{2}(z,y,x)=0. \end{aligned}$$

(32)

$$\begin{aligned}&\sum _{i=1}^j \omega _{\nu ,0,i}(0)+(\lambda +\theta )\omega _{0}(z)-\theta \omega _{0,0}\nonumber \\&\quad =\int _0^{\infty }\omega _1(z,x)\overline{b}(x)dx+\sum _{i=1}^j\int _0^{\infty }\overline{\nu }(x)\omega _{\nu ,i}(z,x)dx. \end{aligned}$$

(33)

The boundary conditions can be expressed as

$$\begin{aligned}&\displaystyle \omega _{\nu ,i}(z,0)=\omega _{\nu ,0,i}(0), \quad 1\le i\le j\end{aligned}$$

(34)

$$\begin{aligned}&\displaystyle z\omega _{1}(z,0)=(\lambda X(z)+\theta )\omega _0(z)-\theta \omega _{0,0}\end{aligned}$$

(35)

$$\begin{aligned}&\displaystyle \omega _2(z,0,x)=\alpha \omega _1(z,x). \end{aligned}$$

(36)

To show $\omega ^{\mathcal {F}_j}_{\nu ,i}(z,x)$, (30) implies that

$$\begin{aligned} \omega _{\nu ,i}(z,x)=\omega _{\nu ,i}(z,0)[1-V(x)]e^{-(\lambda -\lambda X(z))x}. \end{aligned}$$

Next, (26), we have $\omega _{\nu ,i}(z,0)=\omega _{\nu ,0,i}(0)$. Using this, we have $\omega _{\nu ,0,i}(x)=\omega _{i}(0,x)=\omega _{\nu ,0,i}(0)[1-V(x)]e^{-\lambda x}$. Combining it with (25), we have $\omega _{\nu ,0,i}(0)=\omega _{\nu ,0,i-1}(0)\widetilde{V}(\lambda )$. Recursively, we obtain for $1\le i\le j$,

$$\begin{aligned} \omega _{\nu ,i}(z,x)=\omega _{\nu ,0,1}(0)\widetilde{V}(\lambda )^{i-1}[1-V(x)] e^{-(\lambda -\lambda X(z))x}. \end{aligned}$$

(37)

In particular, we get

$$\begin{aligned} \sum _{i=0}^j \omega _{\nu ,0,i}(0)=g_j(\lambda ) \omega _{\nu ,0,1}(0); \lambda \omega _{0,0}=\omega _{\nu ,0,1}(0)\widetilde{V}(\lambda )^j. \end{aligned}$$

(38)

Next, (32) implies

$$\begin{aligned} \omega _{2}(z,y,x)=\omega _2(z,0,x)[1-R(y)]e^{-(\lambda -\lambda X(z))y}. \end{aligned}$$

(39)

Using both (39) and (36), we get the expression in (6). Substituting (39) into (31), we obtain the following:

$$\begin{aligned} \left[ \frac{d}{dx}+\lambda -\lambda X(z)+\overline{b}(x)+\alpha -\alpha \widetilde{R}(\lambda -\lambda X(z))\right] \omega _{1}(z,x)&=0. \end{aligned}$$

(40)

From (40), we get $\omega _1(z,x)=\omega _1(z,0)[1-B(x)]e^{(\lambda -\lambda X(z)+\alpha -\alpha \widetilde{R}(\lambda -\lambda X(z)))x}$. We substitute $\omega _1(z,x)$,(35), (37), and (38) into (33), we get

$$\begin{aligned} \omega _{\nu ,0,1}(0)g_j(\lambda )+(\lambda +\theta )\omega _0(z)&=\frac{\theta }{\lambda }\widetilde{V}(\lambda )^j\omega _{\nu ,0,1}(0)\nonumber \\&\quad +\frac{1}{z}[(\lambda X(z)+\theta )\omega _0(z)-\frac{\theta }{\lambda } \widetilde{V}(\lambda )^j\omega _{\nu ,0,1}(0)] \end{aligned}$$

(41)

After some re-arranging, we obtain

$$\begin{aligned} \omega ^{\mathcal {F}_j}_0(z)&=\omega _{\nu ,0,1}(0)\frac{ \{\frac{\theta \widetilde{V}(\lambda )^j}{\lambda }[\widetilde{H} (\lambda -\lambda X(z))-z]+z[1-\widetilde{V}(\lambda -\lambda X(z))]g_j(\lambda )\}}{[(\lambda X(z)+\theta )\widetilde{H} (\lambda -\lambda X(z))-(\lambda +\theta )z]} \end{aligned}$$

(42)

Next using (35) and (42), we have

$$\begin{aligned} \omega _1(z,x)&=\omega _{\nu ,0,1}(0)\left( \frac{[1-\widetilde{V}(\lambda -\lambda X(z))](\lambda X(z)+\theta )g_j(y) +\theta \widetilde{V}(\lambda )^j [1-X(z)]}{(\lambda X(z)+\theta )\widetilde{H}(\lambda X(z)+\theta )-(\lambda +\theta )z}\right) \nonumber \\&\quad \times [1-B(x)]e^{(\lambda -\lambda X(z)+\alpha -\alpha \widetilde{R}(\lambda -\lambda X(z)))x}. \end{aligned}$$

(43)

Integrating (43) w.r.t x, we obtain

$$\begin{aligned} \omega _1(z)&=\omega _{\nu ,0,1}(0)\left( \frac{[1-\widetilde{V}(\lambda -\lambda X(z))](\lambda X(z)+\theta )g_j(y) +\theta \widetilde{V}(\lambda )^j [1-X(z)]}{(\lambda X(z)+\theta )\widetilde{H}(\lambda X(z)+\theta )-(\lambda +\theta )z}\right) \nonumber \\&\quad \times \frac{1-\widetilde{H} (\lambda -\lambda X(z))}{\lambda -\lambda X(z)}. \end{aligned}$$

(44)

Thus, we can obtain $\omega _2(z)=\alpha \omega _1(z)\frac{1-\widetilde{R}(\lambda -\lambda X(z))}{\lambda -\lambda X(z)}$. From the normalizing condition in (29), we have $\sum _{i=1}^j\omega _{\nu ,i}(1)+\omega _{0}(1)+\omega _1(1)+\omega _2(1)=1$, and after some tedious algebra, we get $\omega _{\nu ,0,1}(0)=C_j(\theta )$. $\square$

Proof of Corollary 1

We only show (i) as the rest are similar. From Theorem 1, we consider $\omega ^{\mathcal {F}_j}_{\nu }(1)=\lim _{z\rightarrow 1^-}C_j(\theta )g_j(\lambda )\left( \frac{1-\widetilde{V}(\lambda -\lambda X(z))}{\lambda -\lambda X(z)}\right)$ and applying L’Hospital rule, we obtain $\omega ^{\mathcal {F}_j}_{\nu }(1)=C_j(\theta )g_j(\lambda ){ EV} =\frac{[1-\kappa (\theta )]g_j(\lambda ){ EV}}{g_j(\lambda ){ EV}+\frac{\theta \widetilde{V}(\lambda )^j}{\lambda (\lambda +\theta )}}< 1$. It is ready to verify that $\omega ^{\mathcal {F}_j}_{\nu }(1)+\omega ^{\mathcal {F}_j}_0(1)+\omega ^{\mathcal {F}_j}_1(1)+\omega ^{\mathcal {F}_j}_2(1)=1$. $\square$

Proof of Corollary 2

Let $O^{\mathcal {F}_j}(z)$ and $Q^{\mathcal {F}_j}(z)$ be the p.g.f for the number of callers in orbit and system respectively. In order to prove the results for (i) and (ii), they follow easily from the fact that $O^{\mathcal {F}_j}(z)=\omega ^{\mathcal {F}_j}_{\nu }(z)+\omega ^{\mathcal {F}_j}_0(z)+\omega ^{\mathcal {F}_j}_1(z)+\omega ^{\mathcal {F}_j}_2(z)$ and $Q^{\mathcal {F}_j}(z)=\omega ^{\mathcal {F}_j}_{\nu }(z)+\omega ^{\mathcal {F}_j}_0(z)+z(\omega ^{\mathcal {F}_j}_1(z)+\omega ^{\mathcal {F}_j}_2(z))$. $\square$

Proof of Theorem 2

We want to show that the p.g.f for the number of customers in the orbit can be written as the sum of three random variables. Let M be the orbit size of the system based on constant retrial policy. First, we observe that Atencia et al. (2008, Theorem 3) has shown that $N_R=N_0+M$. Finally the p.g.f for $N^{\mathcal {F}_j}(z)$ allows us to conclude that $N_{\nu }^{\mathcal {F}_j}=N^{\mathcal {F}_j}+N_0+M$. $\square$

Proof of Corollary 7

We have $EC^{\mathcal {P}}_{\nu }=EL^{\mathcal {P}}+E\zeta _{\nu }^{\mathcal {P}}$ and using results in Theorem 4 and Corollary 6, we have the required result. $\square$

Proof of Lemma 2

To show Lemma 2, we consider the following. Denote $m_n(s)=P\{M=n,\tau _s>H\}$. For any $k\ge 1$, we have $\pi _{n,1}^{(k)}(s)=m_{n-k+1}(s)$ and for $i\ge 2$,

$$\begin{aligned} \pi _{n,i}^{(k)}(s)\!=\!\sum _{j=1}^n \pi _{n,i-1}^{(k)}(s)\sum _{l=1}^{n-j+1}\frac{\lambda c_l}{s+\lambda +\theta }m_{n-j-l+1}(s)\!+\!\sum _{j=1}^{n+1}\pi ^{(k)}_{j,i-1}(s) \frac{\theta }{s+\lambda +\theta }m_{n-j+1}(s). \end{aligned}$$

Finally, the proof is completed by showing that $\sum _{n=0}^{\infty }m_n(s)z^n=\widetilde{H}(s+\lambda -\lambda X(z))$. $\square$

For ease of exposition, we define $F(s,z,y,x)=(s+\lambda +\theta )-\frac{y}{z}(\lambda X(z)+\theta )\widetilde{H}(s+\lambda -\lambda X(z))$.

1.1 Proof of Theorem 3

The proof of Theorem 3 requires two further lemmas, i.e., Lemma 5 and Lemma 6. These results allow us to compute the working and server-waiting length of the k-busy period. Finally, we obtain $EL^{(k)}$ which agrees with Atencia et al. (2008) who use the technique of supplementary variables.

Lemma 5

The generating function $\varphi _0^{(k)}(s,z,y)$ satisfies the functional equation

$$\begin{aligned} F(s,z,y,x)\varphi _0^{(k)}(s,z,y)=yz^{k-1}\widetilde{H}(s+\lambda -\lambda X(z)). \end{aligned}$$

(45)

Proof

Using the fact that $s\varphi ^{(k)}_{0ni}(s)=P\{L^{(k)}>\tau _s,C(\tau _s)=0,Q(\tau _s)=n,I(\tau _s)=i\}=\frac{s}{s+\lambda +\theta }\pi _{ni}^{(k)}(s).$ Thus, we have $\varphi ^{(k)}_{0ni}(s)= \frac{\pi _{ni}^{(k)}(s)}{s+\lambda +\theta }$. The result follows from (9) since $\varphi _{0}^{(k)}(s,z,y)=\frac{f^{(k)}(s,z,y)}{s+\lambda +\theta }$. $\square$

Lemma 6

The generating function $\varphi _1^{(k)}(s,z,y,x)$ satisfies the functional equation

$$\begin{aligned} \varphi _1^{(k)}(s,z,y,x)&=(1-H(x))e^{-(s+\lambda -\lambda X(z))x}\nonumber \\&\quad \times \left[ z^{k-1}+\frac{1}{z}(\lambda X(z)+\theta )\varphi _0^{(k)}(s,z,y)\right] . \end{aligned}$$

(46)

In addition, we have

$$\begin{aligned} \int _0^{\infty } \varphi _1^{(k)}(s,z,y,x)dx&=\frac{1-\widetilde{H}(s+\lambda -\lambda X(z))}{s+\lambda -\lambda X(z)}\nonumber \\&\quad \times \left[ z^{k-1}+\frac{1}{z}(\lambda X(z)+\theta )\varphi _0^{(k)}(s,z,y)\right] . \end{aligned}$$

(47)

Proof

Observe that $s\varphi _{1ni}^{(k)}(s,x)=P\{L^{(k)}>\tau _s,C(\tau _s)=1,\omega (\tau _s)\in (x,x+dx),Q(\tau _s)=n,I(\tau _s)=i\}$. Following the arguments in Falin and Templeton (1997) or Artalejo et al. (2002), we obtain

$$\begin{aligned} \varphi _{1n0}^{(k)}(s,x)&=[1-H(x)]e^{-sx}\sum _{r=1}^{n-k+1}e^{-\lambda x}\frac{(\lambda x)^r}{r!}P\{X_1+\cdots +X_r=n-k+1\}\\ \varphi ^{(k)}_{1ni}(s,x)&=[1-H(x)]e^{-sx}\\&\quad \times \left\{ \sum _{j=1}^n\pi _{ji}^{(k)}(s)\sum _{l=1}^{n-j+1}\frac{\lambda c_l}{s+\lambda +\theta }\sum _{r=1}^{n-j+1-l}e^{-\lambda x}\frac{(\lambda x)^r}{r!}P\left( \sum _{t=1}^r X_t{=}n-j{+}1{-}l\right) \right. \\&\quad \left. +\sum _{j=1}^{n+1}\pi _{ji}^{(k)}(s)\frac{\theta }{s+\lambda +\theta }\sum _{r=1}^{n-j+1}e^{-\lambda x}\frac{(\lambda x)^r}{r!}P\left( \sum _{t=1}^r X_t=n-j+1\right) \right\} . \end{aligned}$$

The result follows from the definition of $\varphi _1^{(k)}(s,z,y,x)$ after tedious algebraic manipulations. Finally, (47) follows from integrating (46) w.r.t x. $\square$

Proof of Theorem 3

In order to compute $L_0^{(k)}$, we need to compute $\lim _{z\rightarrow 1^-}\varphi (0,z,1)$. Let $s=0, y=1$ into (45), we obtain

$$\begin{aligned} \left[ (\lambda +\theta )-\frac{1}{z}(\lambda X(z)+\theta )\widetilde{H}(\lambda -\lambda X(z))\right] \varphi _0^{(k)}(0,z,1)=z^{k-1}\widetilde{H}(\lambda -\lambda X(z)). \end{aligned}$$

Differentiating the above equation w.r.t z and letting $z\rightarrow 1$, we obtain the desired result for $L^{(k)}_0$. The expected generalized working length of the k-busy period is given by $\lim _{z\rightarrow 1-}\int _0^{\infty }\varphi _1^{(k)}(0,z,1,x)dx$. Note that $\lim _{z\rightarrow 1^-}\frac{1-\widetilde{H}(\lambda -\lambda X(z))}{\lambda -\lambda X(z)}=EH$. $\square$

Proof of Corollary 4

The proof is immediate from Theorem 3 and the fact that $EH=EB(1+\alpha ER)$. $\square$

1.2 Proof of Theorem 4

To prove Theorem 4, we shall begin with a preliminary lemma.

Lemma 7

The generating functions $\psi _0(s,z,y)$ and $\psi _1(s,z,y,x)$ satisfy the functional equations

$$\begin{aligned} F(s,z,y,x)\psi _0(s,z,y)&=\frac{y}{z}\left\{ \zeta _j(z)\left( \frac{\lambda X(z)+\theta }{\lambda +\theta }\right) +\frac{\theta }{\lambda +\theta } \widetilde{V}(\lambda )^j[X(z)-1]\right\} \\&\quad \times \widetilde{H}(s+\lambda -\lambda X(z)). \end{aligned}$$

$$\begin{aligned} \psi _1(s,z,y,x)&=(1-H(x))e^{-(s+\lambda -\lambda X(z))x}\frac{1}{z}\nonumber \\&\quad \times \left( \left\{ \zeta _j(z)\left( \frac{\lambda X(z)+\theta }{\lambda +\theta }\right) +\frac{\theta }{\lambda +\theta }\widetilde{V}(\lambda )^j[X(z)-1]\right\} \nonumber \right. \\&\quad \left. +\,\,(\lambda X(z)+\theta )\psi _0(s,z,y)\right) . \end{aligned}$$

(48)

Proof

The results follow from applying Lemmas 5 and 6.

Proof of Theorem 4

From Lemma 48, we let $s=0, y=1$, we have

$$\begin{aligned}&\left[ (\lambda +\theta )-\frac{1}{z}(\lambda X(z)+\theta )\widetilde{H}(\lambda -\lambda X(z))\right] \varphi _0(0,z,1)\\&\quad =\frac{\zeta (z)}{z}\left( \frac{\lambda X(z)+\theta }{\lambda +\theta }\right) \widetilde{H}(\lambda -\lambda X(z)). \end{aligned}$$

Differentiate the above equation w.r.t z and let z approach 1, we obtain $L_0^{\mathcal {F}_j}$. Combining with (1), $L_1^{\mathcal {F}_j}$ is obtained by using $\lim _{z\rightarrow 1^-}\int _0^{\infty }\varphi (0,z,1,x)dx$. Finally, $EL^{\mathcal {M}}=\varphi (0,1,1)+\lim _{z\rightarrow 1^-}\int _0^{\infty }\varphi (0,z,1,x)dx$.

Proof of Lemma 4

To see this, we apply Theorem 4 and after some re-arranging, we have

$$\begin{aligned} { EL}^{\mathcal {F}_j}=-\frac{1}{\lambda +\theta }+\left[ \widetilde{V}(\lambda )^j \left( \frac{\theta }{\lambda (\lambda +\theta )} -\frac{{ EV}}{1-\widetilde{V}(\lambda )}\right) + \frac{{ EV}}{1-\widetilde{V}(\lambda )}\right] \frac{\kappa (\theta )}{1-\kappa (\theta )}. \end{aligned}$$

Given that $\widetilde{V}(\lambda )\ge 0$, $\widetilde{V}(\lambda )^j$ is always increasing function in j. It is easy to see that the sign of $\frac{\theta }{\lambda (\lambda +\theta )}-\frac{{ EV}}{1-\widetilde{V}(\lambda )}$ determines the if $EL^{\mathcal {F}_j}$ is increasing or decreasing. $\square$