Equilibrium Strategies in an M_n/M/1 Queue with Server Breakdowns and Delayed Repairs

Abstract

Ophthalmic units use sophisticated equipment to enable accurate diagnosis of refractive errors. This equipment is subject to two types of breakdowns. One is simple breakdown which can be repaired in-house, and the other is complex breakdown which requires repair by the original equipment manufacturer (OEM) and results in a delay time between the server breakdown and the start of the repair. In this paper, we model this scenario as an ${\mathrm{M}}_{\mathrm{n}}$/M/1 queuing system with two types of breakdowns and delayed repairs due to complex breakdowns, where the delay time and repair times for simple and complex breakdowns are generally distributed. We obtain the steady-state probabilities and provide the recursive formulas for the Laplace–Stieltjes transforms (LSTs) of conditional residual delay time and repair times given the system state. For the fully observable case, we derive the equilibrium joining strategies of customers who decide to join or balk based on their observation of the system state. Moreover, two numerical experiments are conducted to explore the equilibrium joining probabilities.

1. Introduction

In recent years, refractive error has been a significant public health concern worldwide. Inappropriate prevention and correction of refractive error can lead to serious consequences. By 2020, uncorrected refractive error was the primary cause of moderate to severe visual impairment and the second leading cause of blindness among people aged 50 and older; see Worku et al. [1]. Ophthalmic units rely on the equipment of phoropter and trial lens set to accurately and effectively diagnose refractive errors, such as myopia, hyperopia, and astigmatism. For patients with refractive errors and asymptomatic individuals, screening examinations, including visual acuity, are recommended every 1–2 years by an ophthalmologist or optometrist. However, sudden breakdowns of equipment are critical, as they can lead to visual impairment, economic losses and patient harm. Due to the technology complexity of equipment, some simple breakdowns, such as blown bulb and loose power cord, can be repaired in-house, while other complex breakdowns require external repair services; see Jiang et al. [2]. For the latter breakdowns, the ophthalmic unit can have a service contract with OEM, who can provide the necessary parts and technical expertise for equipment repair.

If a simple breakdown occurs, the equipment is repaired immediately. In contrast, if a complex breakdown occurs, the ophthalmic unit must stop service and promptly contact OEM. Subsequently, the OEM arranges for technicians to repair the equipment, implying a delay time between the breakdown of the server and the start of the repair. With regard to strategic patients, they will decide whether to join or balk based on their observations of the queue length and equipment’s status upon arrival. If patients choose to balk, they will seek care from another ophthalmic unit. Therefore, it is more realistic to consider the state-dependent joining probabilities of strategic patients. In this paper, we establish an ${\mathrm{M}}_{\mathrm{n}}$/M/1 queuing system with two types of breakdowns and delayed repairs due to complex breakdowns, where delay time and repair times have general distributions. In the fully observable case, we mainly analyze the equilibrium strategies of customers.

Nowadays, with the rapid development of science and technology, machines are gradually replacing human workers, such as ATMs, cloud computing, and automatic sorting equipment. Queuing systems with breakdowns have been studied in many application scenarios; see Dudin et al. [3], Ke et al. [4] and Liu et al. [5]. With economic considerations of strategic customers in a queuing system, Naor [6] first studies an M/M/1 queueing model with a linear reward–cost structure, and derives the equilibrium threshold of strategic customers in the observable case. Moreover, the unobservable case is studied by Edelson and Hildebrand [7]. Hassin [8] and Shone et al. [9] weigh up the relative merits of the observable and unobservable models in different situations. Subsequently, a large amount of research has studied the behavior of strategic customers in different queuing models, and the relevant work has been reviewed and summarized in the monographs of Hassin and Haviv [10] and Hassin [11].

Research on strategic behavior of customers in queuing systems with vacation/breakdowns has been widely studied. Burnetas and Economou [12] introduce the strategic behavior of customers in an M/M/1 vacation queuing system, and analyze the equilibrium threshold strategies under four different information levels (fully observable, almost observable, almost unobservable, and fully unobservable cases). Wang et al. [13] propose a vacation queuing system, where arriving customers who find the server is on vacation have the option of paying a fee to terminate the vacation immediately. They find that this option can effectively enhance the system throughput when the arrival rate is intermediate. Due to non-availability of repair facilities, Wang and Zhang [14] consider the delayed repairs in an M/M/1 queuing model with breakdowns, and analyze the equilibrium strategies in the fully observable and partially observable cases. Xu and Xu [15] analyze the strategic behavior of customers and social welfare in an M/M/1 queue with sever partial failures and repairs in the fully observable and fully unobservable cases. In a queuing model with catastrophes and working vacation, Wang et al. [16] analyze the customer strategies and social welfare under four different information levels.

In contrast with the Markovian queuing models, where a customer’s strategy is to join if the queue length is shorter than a ‘threshold’ upon arrival, the analysis of customers’ strategies in the non-Markovian models is more challenging, as the memoryless property no longer holds. In an M/G/1 queue, customers’ decisions to join or balk are related not only to the queue length, but also to the residual service time of the customer in service. Kerner [17] considers an ${\mathrm{M}}_{\mathrm{n}}$/G/1 queuing system and gives a recursive algorithm for calculating the equilibrium strategies, where the arrival rate depends on the queue length. Different from Kerner [17], our study takes into account the unreliable server. We have expanded the scope of our analysis to consider that the joining probabilities are influenced not merely by the queue length, but also by the server’s state. Subsequently, Economou and Manou [18] analyze the same model by a probabilistic approach. Zhang et al. [19] study strategic behavior of customers in an observable M/G/1 vacation queue with generally distributed setup time. In an almost observable ${\mathrm{M}}_{\mathrm{n}}$/G/1 queuing model with server breakdowns, Zhu et al. [20] present an auxiliary system to derive the stationary distribution and customers strategies. Recently, Xu et al. [21] studied the equilibrium strategies of customers in an M/G/1 retrial queue with setup times and server breakdowns.

In the present paper, we consider a single-server queuing system with two types of breakdowns and delayed repairs due to complex breakdowns, where arrival rate is dependent on the system state. In contrast with Wang and Zhang [14], we assume that the delay time and repair times are independently generally distributed. In the fully observable case, arriving customers observe the queue length and the server’s state, which provide relevant information about the residual delay time and repair times. Clearly, this information is essential for arriving customers to assess their sojourn time and make strategic decisions. Our main goal is to analyze the customers’ strategies of joining or balking upon arrival. The main contributions of this paper are as follows:

(1)

We extend the work of Wang and Zhang [14] to include state-dependent joining probabilities and general distributions of delay and repair times, where the joining probabilities depend on the queue length and the server’s state. Moreover, these probabilities not only reflect customers’ preferences for the system but also define their equilibrium strategies.

(2)

Given the system state, we derive recursive formulas for the LSTs of the conditional residual delay time and repair times for both simple and complex breakdowns. According to the property of LST, we obtain the corresponding conditional expectations, which are crucial to analyze equilibrium joining strategies.

(3)

The equilibrium strategies of customers are obtained. When the server is busy, there exists an equilibrium threshold; that is, customers decide to join if the queue length is shorter than this threshold. Moreover, when the server malfunctions, the equilibrium joining strategies may not be unique, as demonstrated by a numerical experiment.

The rest of the paper is structured as follows. In Section 2, we describe the basic queuing model. In Section 3, we analyze the stationary distribution and the LSTs of the conditional residual delay time and repair times for simple and complex breakdowns. In Section 4, we derive the equilibrium strategies of customers. In Section 5, two numerical examples are conducted to illustrate the equilibrium strategies. We conclude and provide some ideas for future research in Section 6.

2. Model Description

We consider an observable single-server queue with two types of breakdowns and delayed repairs due to complex breakdowns, which is described as follows. Under the first-come, first-served (FCFS) principle, customers’ arrival follows a Poisson process at rate $\lambda$, and service time is exponentially distributed at rate $\mu$. The server is subject to simple and complex breakdowns in operation. Specifically, the simple breakdowns follow a Poisson process with parameter ${\zeta }_0$, and the complex breakdowns follow a Poisson process with parameter ${\zeta }_1$. If a simple breakdown occurs, the server is repaired immediately, where the repair time for a simple breakdown has probability density function $r_0\left(x\right)$, cumulative distribution function $R_0\left(x\right)$, and finite first moment $\beta$. The corresponding LST is $R_0^{*}\left(s\right)={\int }_0^{\infty }{e}^{-sx}d{R}_0\left(x\right)$. If a complex breakdown occurs, the server cannot be repaired immediately due to non-availability of OEM. It is assumed that the delay time from a complex breakdown of the server to the start of the repair has probability density function $d\left(x\right)$, cumulative distribution function $D\left(x\right)$, and finite first moment $\alpha$. The corresponding LST is $D^{*}\left(s\right)={\int }_0^{\infty }{e}^{-sx}dD\left(x\right)$. Moreover, the repair time for a complex breakdown has probability density function $r_1\left(x\right)$, cumulative distribution function $R_1\left(x\right)$, and first moment $\theta$. The associated LST is $R_1^{*}\left(s\right)={\int }_0^{\infty }{e}^{-sx}d{R}_1\left(x\right)$. The server stops service during the delay time and repair times for both simple and complex breakdowns, and resumes work after the server is repaired.

Let random variables ($N\left(t\right),I\left(t\right)$) represent the state of the system at arbitrary time t, where $N\left(t\right)$ represents the number of customers in the system and $I\left(t\right)$ ($I\left(t\right)=0:$ repair for a simple breakdown; $I\left(t\right)=1$: work; $I\left(t\right)=2$: delayed repair; $I\left(t\right)=3$: repair for a complex breakdown) represents the server’s state. For the fully observable case, if a customer finds that there are n customers in the system and the server’s state is i upon arrival, he will join the system with probability $q_{n,i}$ and balk with probability $1-q_{n,i}$. Note that the joining probability is solely dependent on the system state and not on time t. Clearly, the effective arrival rate is ${\lambda }_{n,i}=\lambda q_{n,i}$ when the system state is $(n,i)$.

We use the residual delay time and repair times for simple and complex breakdowns as supplementary variables. Specifically, when the server’s state is 2 at time t, the residual delay time is defined as the time from one customer’s arrival to the start of the repair; when the server’s state is 0 (3) at time t, the residual repair time for a simple (complex) breakdown is defined as the time from one customer’s arrival until the server resumes work. We now define the following probabilities and notations:

• $p_t(n,1)=P\{N\left(t\right)=n,I\left(t\right)=1\}$ is the probability that there are n customers in the system and the server’s state is 1 at time t, for $n\ge 0.$
• $p_t(n,0,z)$ is the probability density that the system state is $(n,0)$ and the residual repair time for a simple breakdown is z at time t, for $n\ge 0$.
• $p_t(n,2,y)$ is the probability density that the system state is $(n,2)$ and the residual delay time is y at time t, for $n\ge 0$.
• $p_t(n,3,x)$ is the probability density that the system state is $(n,3)$ and the residual repair time for a complex breakdown is x at time t, for $n\ge 0$.
• $p(n,0,z)=\underset{t\to \infty }{lim}{p}_t(n,0,z)$ is the limiting probability that the system state is $(n,0)$ and the residual repair time for a simple breakdown is z, for $n\ge 0$.
• $p(n,2,y)=\underset{t\to \infty }{lim}{p}_t(n,2,y)$ is the limiting probability that the system state is $(n,2)$ and the residual delay time is y, for $n\ge 0$.
• $p(n,3,x)=\underset{t\to \infty }{lim}{p}_t(n,3,x)$ is the limiting probability that the system state is $(n,3)$ and the residual repair time for a complex breakdown is x, for $n\ge 0$.
• ${\pi }_{n,1}=\underset{t\to \infty }{lim}P\{N\left(t\right)=n,I\left(t\right)=1\}$ is the probability that the system state is $(n,1)$, for $n\ge 0.$
• ${\pi }_{n,i}={\int }_0^{\infty }p(n,i,x)dx$ is the probability that the system state is $(n,i)$, for $n\ge 0$ and $i=0,2,3.$
• $P^{*}(n,i,s)={\int }_0^{\infty }{e}^{-sx}p(n,i,x)dx$ is the LST of $p(n,i,x)$, for $n\ge 0$ and $i=0,2,3.$
• $Q_{n,i}^{*}\left(s\right)={\displaystyle \frac{P^{*}(n,i,s)}{{\pi }_{n,i}}}$ is the LST of the conditional residual repair time given the system $(n,i)$, for $i=0,3$ and $n\ge 0$.
• $Q_{n,2}^{*}\left(s\right)={\displaystyle \frac{P^{*}(n,2,s)}{{\pi }_{n,2}}}$ is the LST of the conditional residual delay time given the system $(n,2)$, for $n\ge 0$.
• ${\overline{r}}_{n,i}$ is the conditional expectation of residual repair time given the system $(n,i)$, for $i=0,3$, and $n\ge 0$.
• ${\overline{r}}_{n,2}$ is the conditional expectation of residual delay time given the system $(n,i)$, for $n\ge 0$.

Remark 1.

The limiting probabilities stated above exist when the system is in stationary state. In particular, it is not necessary to make an assumption about stability in our paper. Specifically, strategic customers will balk if the queue length is too long, due to the finite reward R from receiving service (R is defined in Section 4). Therefore, our model is similar to a queuing system with finite capacity. Consequently, we do not make additional stability assumptions.

3. Residual Delay Time and Repair Time

In this section, we explore the steady-state probabilities and the conditional expectations of residual delay time and repair times for the simple and complex breakdowns. We now derive the steady-state probabilities in the following lemma.

Lemma 1.

The steady-state probabilities that there are n customers in the system and the server’s state is $i(i=0,2,3)$ are given by

$$\begin{array}{cc}\hfill {\pi }_{n,i}& =\left\{\begin{array}{cc}{\displaystyle \frac{{\zeta }_0{A}_{n,0}}{{\lambda }_{n,0}}}{\pi }_{n,1},& n\ge 0,\\ {\displaystyle \frac{{\zeta }_1{A}_{n,i}}{{\lambda }_{n,i}}}{\pi }_{n,1},& n\ge 0\mathit{and}i=2,3,\end{array}\right.\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill {\pi }_{n,i}& ={\displaystyle \frac{{\lambda }_{n-1,i}{B}_{n,i}}{{\lambda }_{n,i}}}{\pi }_{n-1,i},n\ge 1\mathit{and}i=0,2,3,\hfill \end{array}$$

(2)

where

$$\begin{array}{cc}\hfill A_{0,0}& =1-R_0^{*}\left({\lambda }_{0,0}\right),\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill A_{0,2}& =1-D^{*}\left({\lambda }_{0,2}\right),\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill A_{0,3}& =D^{*}\left({\lambda }_{0,2}\right)\left[1-R_1^{*}\left({\lambda }_{0,3}\right)\right],\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill A_{n,0}& =1-R_0^{*}\left({\lambda }_{n,0}\right)+{\displaystyle \frac{\mu A_{n-1,0}\left[1-Q_{n-1,0}^{*}\left({\lambda }_{n,0}\right)\right]}{{\lambda }_{n-1,1}+{\zeta }_0{A}_{n-1,0}+{\zeta }_1{A}_{n-1,2}+{\zeta }_1{A}_{n-1,3}}},n\ge 1,\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill A_{n,2}& =1-D^{*}\left({\lambda }_{n,2}\right)+{\displaystyle \frac{\mu A_{n-1,2}\left[1-Q_{n-1,2}^{*}\left({\lambda }_{n,2}\right)\right]}{{\lambda }_{n-1,1}+{\zeta }_0{A}_{n-1,0}+{\zeta }_1{A}_{n-1,2}+{\zeta }_1{A}_{n-1,3}}},n\ge 1,\hfill \\ \hfill A_{n,3}& =D^{*}\left({\lambda }_{n,2}\right)\left[1-R_1^{*}\left({\lambda }_{n,3}\right)\right]+{\displaystyle \frac{\mu A_{n-1,2}{Q}_{n-1,2}^{*}\left({\lambda }_{n,2}\right)\left[1-R_1^{*}\left({\lambda }_{n,3}\right)\right]}{{\lambda }_{n-1,1}+{\zeta }_0{A}_{n-1,0}+{\zeta }_1{A}_{n-1,2}+{\zeta }_1{A}_{n-1,3}}}\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill & +{\displaystyle \frac{\mu A_{n-1,3}\left[1-Q_{n-1,3}^{*}\left({\lambda }_{n,3}\right)\right]}{{\lambda }_{n-1,1}+{\zeta }_0{A}_{n-1,0}+{\zeta }_1{A}_{n-1,2}+{\zeta }_1{A}_{n-1,3}}},n\ge 1,\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill B_{n,0}& =1-Q_{n-1,0}^{*}\left({\lambda }_{n,0}\right)+{\displaystyle \frac{\left({\lambda }_{n-1,1}+{\zeta }_0{A}_{n-1,0}+{\zeta }_1{A}_{n-1,2}+{\zeta }_1{A}_{n-1,3}\right)\left[1-R_0^{*}\left({\lambda }_{n,0}\right)\right]}{\mu A_{n-1,0}}},n\ge 1,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill B_{n,2}& =1-Q_{n-1,2}^{*}\left({\lambda }_{n,2}\right)+{\displaystyle \frac{\left({\lambda }_{n-1,1}+{\zeta }_0{A}_{n-1,0}+{\zeta }_1{A}_{n-1,2}+{\zeta }_1{A}_{n-1,3}\right)\left[1-D^{*}\left({\lambda }_{n,2}\right)\right]}{\mu A_{n-1,2}}},n\ge 1,\hfill \\ \hfill B_{n,3}& =1-Q_{n-1,3}^{*}\left({\lambda }_{n,3}\right)+{\displaystyle \frac{D^{*}\left({\lambda }_{n,2}\right)\left({\lambda }_{n-1,1}+{\zeta }_0{A}_{n-1,0}+{\zeta }_1{A}_{n-1,2}+{\zeta }_1{A}_{n-1,3}\right)\left[1-R_1^{*}\left({\lambda }_{n,3}\right)\right]}{\mu A_{n-1,3}}}\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill & +{\displaystyle \frac{A_{n-1,2}{Q}_{n-1,2}^{*}\left({\lambda }_{n,2}\right)\left[1-R_1^{*}\left({\lambda }_{n,3}\right)\right]}{A_{n-1,3}}},n\ge 1,\hfill \end{array}$$

(11)

and $Q_{n-1,i}^{*}\left({\lambda }_{n,i}\right)$$(n\ge 1,$$i=0,2,3)$ are given in Lemma 2.

Proof. 

Consider the system states at the instants t and $t+dt$, where $dt$ is a small time increment. If the system state is $(0,0)$ and the residual repair time for a simple breakdown is z at the instant $t+dt$, then it is necessary that at the instant t, either the system state is $(0,0)$ with residual repair time $z+dt$ and no customer arrives in the time $dt$, or a simple breakdown occurs in the time $dt$ and the repair time is z. Moreover, all other events happen with probability $o\left(dt\right)$. Hence, we have

$$p_{t+dt}(0,0,z)=p_t(0,0,z+dt)(1-{\lambda }_{0,0}dt)+{\zeta }_0{p}_t(0,1)r_0\left(z\right)dt+o\left(dt\right),$$

where $1-{\lambda }_{0,0}dt$ is the probability that no customer arrives in time $dt$ and ${\zeta }_{0}dt$ is the probability that a simple breakdown occurs in time $dt$. As $t\to \infty$ in the above equality, we have

$$p(0,0,z+dt)-p(0,0,z)={\lambda }_{0,0}p(0,0,z+dt)dt-{\zeta }_0{\pi }_{0,1}{r}_0\left(z\right)dt+o\left(dt\right).$$

Dividing both sides of the above equation by $dt$ and taking the limit $dt\to 0$, we obtain

$$p^{\prime }(0,0,z)={\lambda }_{0,0}p(0,0,z)-{\zeta }_0{\pi }_{0,1}{r}_0\left(z\right).$$

(12)

Similarly to the above case where the system state is $(0,0)$, we can obtain

$$\begin{array}{cc}\hfill & p^{\prime }(0,2,y)={\lambda }_{0,2}p(0,2,y)-{\zeta }_1{\pi }_{0,1}d\left(y\right),\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill & p^{\prime }(0,3,x)={\lambda }_{0,3}p(0,3,x)-p(0,2,0)r_1\left(x\right),\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill & p^{\prime }(n,0,z)={\lambda }_{n,0}p(n,0,z)-{\zeta }_{0}p(n,1)r_0\left(z\right)-{\lambda }_{n-1,0}p(n-1,0,z),n\ge 1,\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill & p^{\prime }(n,2,y)={\lambda }_{n,2}p(n,2,y)-{\zeta }_1{\pi }_{n,1}d\left(y\right)-{\lambda }_{n-1,2}p(n-1,2,y),n\ge 1,\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill & p^{\prime }(n,3,x)={\lambda }_{n,3}p(n,3,x)-p(n,2,0)r_1\left(x\right)-{\lambda }_{n-1,3}p(n-1,3,x),n\ge 1,\hfill \\ \hfill & ({\lambda }_{0,1}+{\zeta }_0+{\zeta }_1){\pi }_{0,1}=p(0,0,0)+p(0,3,0)+\mu {\pi }_{1,1},\hfill \\ \hfill & ({\lambda }_{n,1}+{\zeta }_0+{\zeta }_1+\mu ){\pi }_{n,1}=p(n,0,0)+p(n,3,0)+\mu {\pi }_{n+1,1}+{\lambda }_{n-1,1}{\pi }_{n-1,1},n\ge 1.\hfill \end{array}$$

(17)

Integrating both sides of Equations (12)–(17) from 0 to ∞, respectively, we have

$$\begin{array}{cc}\hfill & p(0,0,0)=-{\lambda }_{0,0}{\pi }_{0,0}+{\zeta }_0{\pi }_{0,1},\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill & p(0,2,0)=-{\lambda }_{0,2}{\pi }_{0,2}+{\zeta }_1{\pi }_{0,1},\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill & p(0,3,0)=-{\lambda }_{0,3}{\pi }_{0,3}+p(0,2,0),\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill & p(n,0,0)=-{\lambda }_{n,0}{\pi }_{n,0}+{\zeta }_0{\pi }_{n,1}+{\lambda }_{n-1,0}{\pi }_{n-1,0},n\ge 1,\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill & p(n,2,0)=-{\lambda }_{n,2}{\pi }_{n,2}+{\zeta }_1{\pi }_{n,1}+{\lambda }_{n-1,2}{\pi }_{n-1,2},n\ge 1.\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill & p(n,3,0)=-{\lambda }_{n,3}{\pi }_{n,3}+p(n,2,0)+{\lambda }_{n-1,3}{\pi }_{n-1,3},n\ge 1,\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill & {\lambda }_{n,0}{\pi }_{n,0}+{\lambda }_{n,1}{\pi }_{n,1}+{\lambda }_{n,2}{\pi }_{n,2}+{\lambda }_{n,3}{\pi }_{n,3}=\mu {\pi }_{n+1,1}.\hfill \end{array}$$

(24)

Multiplying both sides of Equation (12) by $e^{-{\lambda }_{0,0}z}$, and integrating from 0 to ∞, we have

$$p(0,0,0)={\zeta }_0{\pi }_{0,1}{R}_0^{*}\left({\lambda }_{0,0}\right).$$

(25)

Similarly, it follows Equations (13)–(17) that

$$\begin{array}{cc}\hfill p(0,2,0)& ={\zeta }_1{\pi }_{0,1}{D}^{*}\left({\lambda }_{0,2}\right),\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill p(0,3,0)& =p(0,2,0)R_1^{*}\left({\lambda }_{0,3}\right),\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill p(n,0,0)& ={\zeta }_0{\pi }_{n,1}{R}_0^{*}\left({\lambda }_{n,0}\right)+{\lambda }_{n-1,0}{\pi }_{n-1,0}{Q}_{n-1,0}^{*}\left({\lambda }_{n,0}\right),n\ge 1,\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill p(n,2,0)& ={\zeta }_1{\pi }_{n,1}{D}^{*}\left({\lambda }_{n,2}\right)+{\lambda }_{n-1,2}{\pi }_{n-1,2}{Q}_{n-1,2}^{*}\left({\lambda }_{n,2}\right),n\ge 1.\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill p(n,3,0)& =p(n,2,0)R_1^{*}\left({\lambda }_{n,3}\right)+{\lambda }_{n-1,3}{\pi }_{n-1,3}{Q}_{n-1,3}^{*}\left({\lambda }_{n,3}\right),n\ge 1,\hfill \end{array}$$

(30)

Combining Equations (18)–(30), we obtain the results after some algebraic manipulations. □

Note that $p(n,2,0)$ is the probability that the repair just begins (server’s state changes from 2 to 3) when there are n customers in the system. Similarly, for $i=0,3$, $p(n,i,0)$ is the probability that the server just resumes work (server’s state changes from i to 1) when there are n customers in the system. Using the steady-state probabilities, we analyze the LSTs of the conditional residual delay time and repair times in the following lemma.

Lemma 2.

Given the system state, the LSTs of the conditional residual delay time and repair times have the following recursive formulas:

$$\begin{array}{cc}\hfill Q_{n,0}^{*}\left(s\right)& ={\displaystyle \frac{{\lambda }_{n,0}\left[R_0^{*}\left({\lambda }_{n,0}\right)-R_0^{*}\left(s\right)\right]}{A_{n,0}\left(s-{\lambda }_{n,0}\right)}}+{\displaystyle \frac{{\lambda }_{n,0}\left[Q_{n-1,0}^{*}\left({\lambda }_{n,0}\right)-Q_{n-1,0}^{*}\left(s\right)\right]}{B_{n,0}\left(s-{\lambda }_{n,0}\right)}},n\ge 1,\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill Q_{n,2}^{*}\left(s\right)& ={\displaystyle \frac{{\lambda }_{n,2}\left[D^{*}\left({\lambda }_{n,2}\right)-D^{*}\left(s\right)\right]}{A_{n,2}\left(s-{\lambda }_{n,2}\right)}}+{\displaystyle \frac{{\lambda }_{n,2}\left[Q_{n-1,2}^{*}\left({\lambda }_{n,2}\right)-Q_{n-1,2}^{*}\left(s\right)\right]}{B_{n,2}\left(s-{\lambda }_{n,2}\right)}},n\ge 1,\hfill \\ \hfill Q_{n,3}^{*}\left(s\right)& ={\displaystyle \frac{{\lambda }_{n,3}{D}^{*}\left({\lambda }_{n,2}\right)\left[R_1^{*}\left({\lambda }_{n,3}\right)-R_1^{*}\left(s\right)\right]}{A_{n,3}\left(s-{\lambda }_{n,3}\right)}}+{\displaystyle \frac{{\lambda }_{n,3}{A}_{n-1,2}{Q}_{n-1,2}^{*}\left({\lambda }_{n,2}\right)\left[R_1^{*}\left({\lambda }_{n,3}\right)-R_1^{*}\left(s\right)\right]}{A_{n-1,3}{B}_{n,3}\left(s-{\lambda }_{n,3}\right)}}\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill & +{\displaystyle \frac{{\lambda }_{n,3}\left[Q_{n-1,3}^{*}\left({\lambda }_{n,3}\right)-Q_{n-1,3}^{*}\left(s\right)\right]}{B_{n,3}\left(s-{\lambda }_{n,3}\right)}},n\ge 1,\hfill \end{array}$$

(33)

with the initial conditions

$$\begin{array}{cc}\hfill Q_{0,0}^{*}\left(s\right)& ={\displaystyle \frac{{\lambda }_{0,0}}{s-{\lambda }_{0,0}}}{\displaystyle \frac{R_0^{*}\left({\lambda }_{0,0}\right)-R_0^{*}\left(s\right)}{1-R_0^{*}\left({\lambda }_{0,0}\right)}},\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill Q_{0,2}^{*}\left(s\right)& ={\displaystyle \frac{{\lambda }_{0,2}}{s-{\lambda }_{0,2}}}{\displaystyle \frac{D^{*}\left({\lambda }_{0,2}\right)-D^{*}\left(s\right)}{1-D^{*}\left({\lambda }_{0,2}\right)}},\hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill Q_{0,3}^{*}\left(s\right)& ={\displaystyle \frac{{\lambda }_{0,3}}{s-{\lambda }_{0,3}}}{\displaystyle \frac{R_1^{*}\left({\lambda }_{0,3}\right)-R_1^{*}\left(s\right)}{1-R_1^{*}\left({\lambda }_{0,3}\right)}}.\hfill \end{array}$$

(36)

Proof. 

Multiplying both sides of Equations (12) and (15) by $e^{-sz}$, (13) and (16) by $e^{-sy}$, (14) and (17) by $e^{-sx}$, then integrating with respect to x, y and z, respectively, we obtain

$$\begin{array}{cc}\hfill s{P}^{*}(0,0,s)-p(0,0,0)& ={\lambda }_{0,0}{P}^{*}(0,0,s)-{\zeta }_0{\pi }_{0,1}{R}_0^{*}\left(s\right),\hfill \\ \hfill s{P}^{*}(n,0,s)-p(n,0,0)& ={\lambda }_{n,0}{P}^{*}(n,0,s)-{\zeta }_0{\pi }_{n,1}{R}_0^{*}\left(s\right)-{\lambda }_{n-1,0}{P}^{*}(n-1,0,s),n\ge 1,\hfill \\ \hfill s{P}^{*}(0,2,s)-p(0,2,0)& ={\lambda }_{0,2}{P}^{*}(0,2,s)-{\zeta }_1{\pi }_{0,1}{D}^{*}\left(s\right),\hfill \\ \hfill s{P}^{*}(n,2,s)-p(n,2,0)& ={\lambda }_{n,2}{P}^{*}(n,2,s)-{\zeta }_1{\pi }_{n,1}{D}^{*}\left(s\right)-{\lambda }_{n-1,2}{P}^{*}(n-1,2,s),n\ge 1,\hfill \\ \hfill s{P}^{*}(0,3,s)-p(0,3,0)& ={\lambda }_{0,3}{P}^{*}(0,3,s)-p(0,2,0)R_1^{*}\left(s\right),\hfill \\ \hfill s{P}^{*}(n,3,s)-p(n,3,0)& ={\lambda }_{n,3}{P}^{*}(n,3,s)-p(n,2,0)R_1^{*}\left(s\right)-{\lambda }_{n-1,3}{P}^{*}(n-1,3,s),n\ge 1.\hfill \end{array}$$

Combining with Equations (1)–(2) and (25)–(30), we derive the results. □

In Lemma 1, note that the steady-state probabilities can be expressed in terms of ${\pi }_{0,1}$, but the explicit expression for ${\pi }_{0,1}$ cannot be given without any assumptions on the state space. Specifically, the state space of the system can be represented by a set $S=\left\{(n,i):i=0,1,2,3\mathrm{and}0\le n\le n_i\right\}$, where $n_i$ is the maximum value of the queue length when the server’s state is i. Consequently, the probability ${\pi }_{0,1}$ can be determined by the normalization Equation ${\displaystyle \sum _{(n,i)\in S}{\pi }_{n,i}=1}$. However, in Lemma 2, we can calculate the LSTs $Q_{n,i}^{*}\left(s\right)(i=0,2,3\mathrm{and}n\ge 0)$ recursively without any assumptions, since these results are independent of ${\pi }_{0,1}$. Therefore, we immediately obtain the conditional expectations of residual delay time and repair times.

Theorem 1.

Given that there are n customers in the system and the server’s state is i, the conditional expectations of residual delay time and repair times are

$$\begin{array}{cc}\hfill {\overline{r}}_{0,0}& ={\displaystyle \frac{\beta }{1-R_0^{*}\left({\lambda }_{0,0}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,0}}},\hfill \end{array}$$

(37)

$$\begin{array}{cc}\hfill {\overline{r}}_{n,0}& ={\displaystyle \frac{{\overline{r}}_{n-1,0}}{B_{n,0}}}+{\displaystyle \frac{\beta }{A_{n,0}}}-{\displaystyle \frac{1}{{\lambda }_{n,0}}},n\ge 1,\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill {\overline{r}}_{0,2}& ={\displaystyle \frac{\alpha }{1-D^{*}\left({\lambda }_{0,2}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,2}}},\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill {\overline{r}}_{n,2}& ={\displaystyle \frac{{\overline{r}}_{n-1,2}}{B_{n,2}}}+{\displaystyle \frac{\alpha }{A_{n,2}}}-{\displaystyle \frac{1}{{\lambda }_{n,2}}},n\ge 1,\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill {\overline{r}}_{0,3}& ={\displaystyle \frac{\theta }{1-R_1^{*}\left({\lambda }_{0,3}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,3}}},\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill {\overline{r}}_{n,3}& ={\displaystyle \frac{{\overline{r}}_{n-1,3}}{B_{n,3}}}+{\displaystyle \frac{\theta D^{*}\left({\lambda }_{n,2}\right)}{A_{n,3}}}+{\displaystyle \frac{\theta A_{n-1,2}{Q}_{n-1,2}^{*}\left({\lambda }_{n,2}\right)}{A_{n-1,3}{B}_{n,3}}}-{\displaystyle \frac{1}{{\lambda }_{n,3}}},n\ge 1.\hfill \end{array}$$

(42)

Proof. 

Taking the negative value of the derivative of Equations (31)–(36) with respect to s and setting $s=0$, the results can be obtained by using Equations (6)–(11). □

In Theorem 1, we derive the recursive formulas for the conditional expectations of the residual delay time and repair times. To enhance clarity, we outline the method for calculating these expectations as follows.

• Step 1. For $n=0$, calculate the LSTs $D^{*}\left({\lambda }_{n,2}\right)$, $R_0^{*}\left({\lambda }_{n,0}\right)$, and $R_1^{*}\left({\lambda }_{n,3}\right)$, respectively.
• Step 2. Substituting the results of Step 1 into Equations (37), (39) and (41), respectively, we obtain ${\overline{r}}_{n,i}$ for $i=0,2,3$.
• Step 3. For $n\ge 1$, calculate $D^{*}\left({\lambda }_{n,2}\right)$, $R_0^{*}\left({\lambda }_{n,0}\right)$, $R_1^{*}\left({\lambda }_{n,3}\right)$, and $Q_{n-1,i}^{*}\left({\lambda }_{n,i}\right)$ for $i=0,2,3$ by Lemma 2.
• Step 4. Calculate $A_{n,i}$ and $B_{n,i}$ for $i=0,2,3$ by Lemma 1 and the results of Step 3.
• Step 5. If $n=1$, use Theorem 1 and the results of Steps 2–4 to compute ${\overline{r}}_{1,i}$ for $i=0,2,3$; otherwise, use Theorem 1 and the results of Steps 3–5 to compute ${\overline{r}}_{n,i}$ for $i=0,2,3$.
• Step 6. Repeat Steps 3–5.

4. Equilibrium Analysis

In this section, we derive the equilibrium strategies of customers who decide whether to join or not upon arrival. If joining, each customer receives a reward of R ($R<\infty$) units after completing service. In addition, customers incur a cost of C units per unit of time as long as they sojourn in the system. In the fully observable case, customers observe the queue length and the server’s state upon arrival. Based on their observation of the system state, customers make irrevocable decisions after comparing their sojourn costs with their rewards.

To derive the equilibrium strategies, we first introduce the generalized service time, which is the period from the start of one customer’s service to the end of his service. Clearly, this time can be decomposed into two terms. One is the actual service time and the other is the interruption time due to delayed repairs and server repairs. Denote by X and $\overline{X}$ the actual service time and the generalized service time of a customer, respectively. Then, we calculate the expected value of $\overline{X}$.

Lemma 3.

The expected generalized service time of a customer is

$$\overline{\mu }\triangleq E\left[\overline{X}\right]={\displaystyle \frac{1+{\zeta }_0\beta +{\zeta }_1\alpha +{\zeta }_1\theta }{\mu }}.$$

(43)

Proof. 

Clearly, the generalized service time satisfies

$$\overline{X}=\left\{\begin{array}{cc}X,& \mathrm{if}\mathrm{no}\mathrm{breakdown}\mathrm{occurs}\mathrm{during}\mathrm{service},\\ R_0+\widehat{X},& \mathrm{if}\mathrm{a}\mathrm{simple}\mathrm{breakdown}\mathrm{occurs}\mathrm{during}\mathrm{service},\\ D+R_1+\tilde{X},& \mathrm{if}\mathrm{a}\mathrm{complex}\mathrm{breakdown}\mathrm{occurs}\mathrm{during}\mathrm{service},\end{array}\right.$$

where random variables D, $R_0$ and $R_1$ represent the delay time and repair times for a simple breakdown and a complex breakdown, respectively. Since complex breakdowns follow Poisson processes and the assumption that D and $R_1$ are independent, we conclude that $\tilde{X}$, D and $R_1$ are independent of each other. Similarly, we conclude that $\widehat{X}$ and $R_0$ are independent. Moreover, due to the memoryless property of exponential distribution of the actual service time, $\overline{X}$, $\widehat{X}$ and $\tilde{X}$ have the same distribution. Therefore, we have

$$E\left[\overline{X}\right]={\displaystyle \frac{\mu }{\mu +{\zeta }_0+{\zeta }_1}}E\left[X\right]+{\displaystyle \frac{{\zeta }_0}{\mu +{\zeta }_0+{\zeta }_1}}E\left[R_0+\widehat{X}\right]+{\displaystyle \frac{{\zeta }_1}{\mu +{\zeta }_0+{\zeta }_1}}E\left[D+R_1+\overline{X}\right],$$

which is equivalent to

$$\begin{array}{c}\hfill E\left[\overline{X}\right]=E\left[X\right]+{\displaystyle \frac{{\zeta }_0}{\mu }}E\left[R_0\right]+{\displaystyle \frac{{\zeta }_1}{\mu }}E\left[D+R_1\right]={\displaystyle \frac{1+{\zeta }_0\beta +{\zeta }_1\alpha +{\zeta }_1\theta }{\mu }}.\end{array}$$

□

We now consider customers’ strategies in different server states. First, in the case where the server’s state is 1, there exists an equilibrium threshold; that is, customers join the system when the queue length is shorter than this threshold.

Theorem 2.

When an arriving customer finds that there are n customers in the system and the server is busy, he/she joins if and only if n is shorter than the threshold $n_1$, where

$$n_1=⌊{\displaystyle \frac{\mu R}{C\left(1+{\zeta }_0\beta +{\zeta }_1\alpha +{\zeta }_1\theta \right)}}⌋.$$

Proof. 

Consider a tagged customer who finds that there are n customers in the system and the server’s state is 1 upon arrival. According to Lemma 3, the expected utility of the tagged customer is $R-C(n+1)\overline{\mu }$. Clearly, the tagged customer will join if and only if $(n+1)\overline{\mu }\le R/C$. Using Equation (43), we obtain the result. □

Then, we consider the case where the server’s state is 2. In this case, customers need to infer the residual delay time based on the information of system state, and then make an irrevocable decision whether to join or not. Assume that $Q_{n,2}^e=(q_{0,2}^e,q_{1,2}^e,\cdots ,q_{n,2}^e)$ is a vector of the equilibrium strategies, where $q_{n,2}^e$ is the equilibrium joining probability of a customer who finds that the system state is $(n,2)$ upon arrival. To calculate the equilibrium strategies when the server’s state is 2, we consider whether the system is empty or not.

Theorem 3.

When an arrival finds that the system is empty and the server’s state is 2, equilibrium strategy exists if one of the following cases occurs:

(1) 

If $\overline{\mu }+\theta +\underset{0\le q_{0,2}\le 1}{max}\left\{{\overline{r}}_{0,2}\right\}\le {\displaystyle \frac{R}{C}},$ then $q_{0,2}^e=1$ is a dominant strategy.

(2) 

If $\overline{\mu }+\theta +\underset{0\le q_{0,2}\le 1}{min}\left\{{\overline{r}}_{0,2}\right\}\ge {\displaystyle \frac{R}{C}},$ then $q_{0,2}^e=0$ is a dominant strategy.

(3) 

(i) 

If $\overline{\mu }+\theta +{\displaystyle \frac{\alpha }{1-D^{*}\left(\lambda \right)}}-{\displaystyle \frac{1}{\lambda }}\le {\displaystyle \frac{R}{C}}$, then $q_{0,2}^e=1$ is a symmetric equilibrium strategy.

(ii) 

If $\overline{\mu }+\theta +{\displaystyle \frac{{\alpha }_2}{2\alpha }}\ge {\displaystyle \frac{R}{C}}$, where ${\alpha }_2$ is the second moment of $D\left(x\right)$, then $q_{0,2}^e=0$ is a symmetric equilibrium strategy.

(iii) 

There exists at least one value $q^{*}$ ($0<q^{*}<1$) satisfying $\overline{\mu }+\theta +{\displaystyle \frac{\alpha }{1-D^{*}\left(\lambda q^{*}\right)}}-{\displaystyle \frac{1}{\lambda q^{*}}}={\displaystyle \frac{R}{C}}$, then $q_{0,2}^e=q^{*}$ is a mixed symmetric equilibrium strategy.

Proof. 

(1) For any $0\le q_{0,2}\le 1$, the customer’s expected utility $R-C\left(\overline{\mu }+\theta +{\displaystyle \frac{\alpha }{1-D^{*}\left({\lambda }_{0,2}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,2}}}\right)$ is nonnegative. Therefore, the optimal strategy of customer is joining the system regardless of other customers’ strategies.

(2) For any $0\le q_{0,2}\le 1$, the customer’s expected utility $R-C\left(\overline{\mu }+\theta +{\displaystyle \frac{\alpha }{1-D^{*}\left({\lambda }_{0,2}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,2}}}\right)$ is nonpositive. Therefore, the optimal strategy of the customer is balking the system regardless of other customers’ strategies.

(i) ((ii), respectively) If all other customers join (balk), then a customer’s benefit from joining is nonnegative (nonpositive). Therefore, joining (balking) is a symmetric equilibrium strategy in case (i) (case (ii)).

(iii) Since ${\displaystyle \frac{R}{C}}-\overline{\mu }-\theta -{\displaystyle \frac{{\alpha }_2}{2\alpha }}\ge 0$, ${\displaystyle \frac{R}{C}}-\overline{\mu }-\theta -{\displaystyle \frac{\alpha }{1-D^{*}\left(\lambda \right)}}+{\displaystyle \frac{1}{\lambda }}\le 0$, and $\overline{\mu }+\theta +{\displaystyle \frac{\alpha }{1-D^{*}\left({\lambda }_{0,2}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,2}}}$ is a continuous function with respect to $q_{0,2}$, there exists at least one $q^{*}\in (0,1)$ as defined in case (iii). Therefore, the best response of the customer is joining with probability $q^{*}$ if all other customers decide to join with probability $q^{*}$. As a result, $q^{*}$ is a mixed symmetric equilibrium. □

Theorem 4.

When an arrival finds that the system state is $(n,2)$ and the equilibrium probabilities are $Q_{n-1,2}^e$, there exists at least one equilibrium strategy if one of the following cases occurs:

(1) 

If $(n+1)\overline{\mu }+\theta +{\overline{r}}_{n,2}(Q_{n-1,2}^e,1)\le {\displaystyle \frac{R}{C}}$, then $q_{n,2}^e=1$ is an equilibrium strategy.

(2) 

If $(n+1)\overline{\mu }+\theta +{\overline{r}}_{n,2}(Q_{n-1,2}^e,0)\ge {\displaystyle \frac{R}{C}}$, then $q_{n,2}^e=0$ is an equilibrium strategy.

(3) 

There exists at least one value $q^{*}$ ($0<q^{*}<1$) satisfying $(n+1)\overline{\mu }+\theta +{\overline{r}}_{n,2}(Q_{n,2}^e,q^{*})={\displaystyle \frac{R}{C}}$, then $q_{n,2}^e=q^{*}$ is a mixed symmetric equilibrium strategy.

Finally, we consider the cases where the server’s state is i$(i=0,3)$. Similar to the above case, denote by $Q_{n,i}^e=(q_{0,i}^e,q_{1,i}^e,\cdots ,q_{n,i}^e)$ a vector of the equilibrium strategies, where $q_{n,i}^e$ is the equilibrium joining probability of a customer who finds that the system state is $(n,i)$ upon arrival. In these cases, we consider whether the system is empty or not.

Theorem 5.

When an arrival finds that the system is empty and the server’s state is 0, equilibrium strategy exists if one of the following cases occurs:

(1) 

If $\overline{\mu }+\underset{0\le q_{0,0}\le 1}{max}\left\{{\displaystyle \frac{\beta }{1-R_0^{*}\left({\lambda }_{0,0}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,0}}}\right\}\le {\displaystyle \frac{R}{C}},$ then $q_{0,0}^e=1$ is a dominant strategy.

(2) 

If $\overline{\mu }+\underset{0\le q_{0,0}\le 1}{min}\left\{{\displaystyle \frac{\beta }{1-R_0^{*}\left({\lambda }_{0,0}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,0}}}\right\}\ge {\displaystyle \frac{R}{C}},$ then $q_{0,0}^e=0$ is a dominant strategy.

(3) 

(i) 

If $\overline{\mu }+{\displaystyle \frac{\beta }{1-R_0^{*}\left(\lambda \right)}}-{\displaystyle \frac{1}{\lambda }}\le {\displaystyle \frac{R}{C}}$, then $q_{0,0}^e=1$ is an equilibrium strategy.

(ii) 

If $\overline{\mu }+{\displaystyle \frac{{\beta }_2}{2\beta }}\ge {\displaystyle \frac{R}{C}}$, where ${\beta }_2$ is the second moment of $R_0\left(x\right)$, then $q_{0,0}^e=0$ is an equilibrium strategy.

(iii) 

There exists at least one $q^{*}$ ($0<q^{*}<1$) satisfying $\overline{\mu }+{\displaystyle \frac{\beta }{1-R_0^{*}\left(\lambda q^{*}\right)}}-{\displaystyle \frac{1}{\lambda q^{*}}}={\displaystyle \frac{R}{C}}$, then $q_{0,0}^e=q^{*}$ is a mixed symmetric equilibrium strategy.

Theorem 6.

When an arrival finds that the system is empty and the server’s state is 3, equilibrium strategy exists if one of the following cases occurs:

(1) 

If $\overline{\mu }+\underset{0\le q_{0,3}\le 1}{max}\left\{{\displaystyle \frac{\theta }{1-R_1^{*}\left({\lambda }_{0,3}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,3}}}\right\}\le {\displaystyle \frac{R}{C}},$ then $q_{0,3}^e=1$ is a dominant strategy.

(2) 

If $\overline{\mu }+\underset{0\le q_{0,3}\le 1}{min}\left\{{\displaystyle \frac{\theta }{1-R_1^{*}\left({\lambda }_{0,3}\right)}}-{\displaystyle \frac{1}{{\lambda }_{0,3}}}\right\}\ge {\displaystyle \frac{R}{C}},$ then $q_{0,3}^e=0$ is a dominant strategy.

(3) 

(i) 

If $\overline{\mu }+{\displaystyle \frac{\theta }{1-R_1^{*}\left(\lambda \right)}}-{\displaystyle \frac{1}{\lambda }}\le {\displaystyle \frac{R}{C}}$, then $q_{0,3}^e=1$ is an equilibrium strategy.

(ii) 

If $\overline{\mu }+{\displaystyle \frac{{\theta }_2}{2\theta }}\ge {\displaystyle \frac{R}{C}}$, where ${\theta }_2$ is the second moment of $R_1\left(x\right)$, then $q_{0,3}^e=0$ is an equilibrium strategy.

(iii) 

There exists at least one $q^{*}$ ($0<q^{*}<1$) satisfying $\overline{\mu }+{\displaystyle \frac{\theta }{1-R_1^{*}\left(\lambda q^{*}\right)}}-{\displaystyle \frac{1}{\lambda q^{*}}}={\displaystyle \frac{R}{C}}$, then $q_{0,3}^e=q^{*}$ is a mixed symmetric equilibrium strategy.

Theorem 7.

If an arrival finds that the system state is $(n,i)$ and the equilibrium probabilities are $Q_{n-1,i}^e$ for $i=0,3$, there exists a symmetric equilibrium if one of the following cases occurs:

(1) 

If $(n+1)\overline{\mu }+{\overline{r}}_{n,i}(Q_{n-1,i}^e,1)\le {\displaystyle \frac{R}{C}}$, then $q_{n,i}^e=1$ is an equilibrium strategy.

(2) 

If $(n+1)\overline{\mu }+{\overline{r}}_{n,i}(Q_{n-1,i}^e,0)\ge {\displaystyle \frac{R}{C}}$, then $q_{n,i}^e=0$ is an equilibrium strategy.

(3) 

There exists at least one $q_i^{*}$ ($0<q^{*}<1$) satisfying $(n+1)\overline{\mu }+{\overline{r}}_{n,i}(Q_{n-1,i}^e,q^{*})={\displaystyle \frac{R}{C}}$, then $q_{n,i}^e=q_i^{*}$ is a mixed symmetric equilibrium strategy.

Note that the proofs of Theorems 4–7 are similar to the proof of Theorem 3, so we omit these proofs to make the article concise.

Theorems 2–7 provide an algorithm to identify the equilibrium strategies $q_{n,i}^e$ for $n\ge 0$ and $i=0,1,2,3$. According to Theorem 2, we can determine an equilibrium threshold $n_1$. Specifically, $q_{n,1}^e=1$ if $n<n_1$; otherwise, $q_{n,1}^e=0$. Moreover, Theorem 3 determines $q_{0,2}^e$, and Theorem 4 recursively determines $q_{n,2}^e$ for $n\ge 1$. Similarly, Theorems 5–6 determine $q_{0,0}^e$ and $q_{0,3}^e$, respectively, and Theorem 7 recursively determines $q_{n,0}^e$ and $q_{0,3}^e$ for $n\ge 1$.

In addition, there is no need to calculate the equilibrium joining probabilities when the queue length is no less than the equilibrium threshold $n_1$ as defined in Theorem 2. Specifically, if the server’s state is 0, 2 or 3 and the queue length n exceeds the threshold $n_1$, an arriving customer will balk because his expected utility is less than ${\displaystyle \frac{R}{C}}-(n+1)\overline{\mu }$, which is less than 0, as shown in Theorem 2. Consequently, the queue length never exceeds the threshold $n_1$.

5. Numerical Experiments

Thus far, we have provided recursive methods to derive the equilibrium strategies of customers in different system states. In this section, we conduct two numerical examples; one is a Markovian case, and the other is a case where the delay time has Erlangian distribution, repair time for a complex breakdown has Hyper-exponential distribution, and repair time for a simple breakdown is determinate. In the Markovian case, our results extend those of Wang and Zhang [14] (Theorem 1) and Economou and Kanta [22] (Theorem 3.1). Moreover, in the second case, we find that multiple equilibrium joining probabilities exist for some values of R.

5.1. Example 1

In the Markovian case, the delay time and repair times for both simple and complex breakdowns are assumed to be exponentially distributed. The equilibrium strategies are defined by the thresholds $\left(n_0,n_1,n_2,n_3\right)$. Specifically, when the server’s state is i upon arrival, customers join the system if the queue length is shorter than $n_i$; otherwise, customers balk.

Theorem 8.

Consider a fully observable ${\mathit{M}}_{\mathit{n}}$/M/1 queue with two types of server breakdowns and delayed repairs in the Markovian case. There exists a vector of equilibrium thresholds $\left(n_0,n_1,n_2,n_3\right)$, i.e.,

$$\begin{array}{cc}\hfill & n_0=⌊x_0⌋,\mathit{with}{x}_0={\displaystyle \frac{(R-C\beta )\mu }{C(1+{\zeta }_0\beta +{\zeta }_1\alpha +{\zeta }_1\theta )}},\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill & n_1=⌊x_1⌋,\mathit{with}{x}_1={\displaystyle \frac{R\mu }{C(1+{\zeta }_0\beta +{\zeta }_1\alpha +{\zeta }_1\theta )}},\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill & n_2=⌊x_2⌋,\mathit{with}{x}_2={\displaystyle \frac{(R-C\theta -C\alpha )\mu }{C(1+{\zeta }_0\beta +{\zeta }_1\alpha +{\zeta }_1\theta )}}.\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill & n_3=⌊x_3⌋,\mathit{with}{x}_3={\displaystyle \frac{(R-C\theta )\mu }{C(1+{\zeta }_0\beta +{\zeta }_1\alpha +{\zeta }_1\theta )}}.\hfill \end{array}$$

(47)

Proof. 

According to Theorem 2, we immediately obtain Equation (45). We now calculate the equilibrium thresholds when the server’s state is 0, 2 and 3. Using Theorem 1 and the exponential distributions of delay time and repair times, we have

$${\overline{r}}_{n,0}=\beta ,{\overline{r}}_{n,2}=\alpha ,\mathrm{and}{\overline{r}}_{n,3}=\theta ,\mathrm{for}n\ge 0.$$

Denote by $U_{n,i}$ the expected utility of customer who observes that the system state is $(n,i)$ upon arrival. By Theorems 3–7, for $n\ge 0$, we obtain

$$U_{n,0}=R-C\left(\beta +(n+1)\overline{\mu }\right),U_{n,2}=R-C\left(\alpha +\theta +(n+1)\overline{\mu }\right),\mathrm{and}{U}_{n,3}=R-C\left(\theta +(n+1)\overline{\mu }\right).$$

Therefore, the inequality $U_{n,0}\ge 0$ is equivalent to

$$\begin{array}{c}\hfill n+1\le {\displaystyle \frac{R-C\beta }{C\overline{\mu }}}.\end{array}$$

Using Equation (43), we obtain Equation (44). Similarly, calculating the inequalities $U_{n,2}\ge 0$ and $U_{n,3}\ge 0$, we obtain Equations (46) and (47), respectively. □

Clearly, the thresholds obtained in Theorem 8 are consistent with those in Wang and Zhang [14] (Theorem 1) when ${\zeta }_0=0$ and with those in Economou and Kanta [22] (Theorem 1) when ${\zeta }_1=0$. Moreover, the stationary probabilities in the Markovian case can be calculated by Lemma 1 and normalization function ${\displaystyle \sum _{i=0}^3\sum _{n=0}^{n_1}{\pi }_{n,i}=1}$.

5.2. Example 2

In this subsection, we consider a specific numerical example with Erlangian delay times, Hyper-exponential repair times for complex breakdowns, and Uniform repair times for simple breakdowns. Specifically, we set

$$d\left(x\right)=4x^2{e}^{-2x},r_0\left(x\right)={\mathbb{1}}_{\left\{x=0.5\right\}}\left(x\right),\mathrm{and}{r}_1\left(x\right)={\displaystyle \frac{1}{3}}{e}^{-x}+{\displaystyle \frac{2}{3}}{e}^{-2x},\mathrm{for}x\ge 0,$$

where ${\mathbb{1}}_{\left\{·\right\}}$ is the indicator function. Therefore, we have

$$D^{*}\left(s\right)={\displaystyle \frac{8}{{(2+s)}^3}},R_0\left(x\right)=e^{-0.5s},R_1^{*}\left(s\right)={\displaystyle \frac{1}{3+3s}}+{\displaystyle \frac{4}{6+3s}},\alpha =1.5,\beta =0.5,\mathrm{and}\theta ={\displaystyle \frac{2}{3}}.$$

Suppose that $\lambda =2$, $\mu =1$, ${\zeta }_0=1$, ${\zeta }_1=1$, and $C=1$. Hence, we have $\overline{\mu }=11/3$ according to Equation (43). Now, we consider two cases where $R=12.8\mathrm{and}R=11.71$.

In the case where $R=12.8$, the balking threshold $n_1$ is 3 when the server’s state is 1, by Theorem 2. Consequently, we only need to calculate the joining probabilities $q_{n,i}^e$ for $0\le n\le 2$, and $i=0,2,3$.

First, consider the joining probabilities $q_{0,i}^e$ for $i=0,2,3$. From Equation (39), we have

$${\overline{r}}_{0,2}={\displaystyle \frac{3{(2+2q_{0,2})}^3}{2{(2+2q_{0,2})}^3-16}}-{\displaystyle \frac{1}{2q_{0,2}}}.$$

According to Theorem 3, we obtain that $q_{0,2}^e=1$ is the dominant strategy. Similarly, from Equations (37) and (41), we have

$${\overline{r}}_{0,0}={\displaystyle \frac{1}{2(1-e^{-q_{0,0}})}}-{\displaystyle \frac{1}{2q_{0,0}}}\mathrm{and}{\overline{r}}_{0,3}={\displaystyle \frac{2(1+2q_{0,3})(2+2q_{0,3})}{2(4+6q_{0,3})}}-{\displaystyle \frac{1}{2q_{0,3}}}.$$

According to Theorems 5 and 6, we obtain that $q_{0,0}^e=1$ and $q_{0,3}^e=1$ are the dominant strategy. Subsequently, for $i=0,2,3$, substituting $q_{0,i}^e=1$ into Equations (3)–(5), (34)–(36), (37), (39) and (41), we obtain the values $A_{0,i}$, ${\overline{r}}_{0,i}$, and functions $Q_{0,i}^{*}\left(s\right)$ with respect to s.

Second, consider the joining probabilities $q_{1,i}^e$ for $i=0,2,3$. From Equations (7) and (10), we obtain $A_{1,2}$ and $B_{1,2}$, which can be expressed in terms of $q_{1,2}$. Therefore, inserting $A_{1,2}$, $B_{1,2}$, and $Q_{0,2}^{*}\left(s\right)$, we can express ${\overline{r}}_{1,2}$ in terms of $q_{1,2}$ from Equation (40). By Theorem 4, we obtain that $q_{1,2}=1$. After the probability $q_{1,2}^e=1$ is determined, we can calculate the values $A_{1,2}$, $B_{1,2}$, ${\overline{r}}_{1,2}$, and $Q_{0,2}\left({\lambda }_{1,2}\right)$. Similarly, from Equations (6), (8), (9), (11), (38) and (42), we can express $A_{1,i}$, $B_{1,i}$ and ${\overline{r}}_{1,i}$ in terms of $q_{1,i}$ for $i=0,3$. According to Theorem 7, we obtain $q_{1,0}^e=q_{1,3}^e=1$. Subsequently, for $i=0,3$, the values of $A_{1,i}$, $B_{1,i}$, ${\overline{r}}_{1,i}$ and $Q_{0,i}\left({\lambda }_{1,i}\right)$ are determined by $q_{1,i}^e=1$.

Finally, consider the joining probabilities $q_{n,i}^e$ for $n\ge 1$ and $i=0,2,3$. Repeating the second step, we obtain $q_{2,0}^e=1$, $q_{2,2}^e=0.644$ and $q_{2,3}^e=1$ from Theorems 4 and 7. In summary, when the queue length is shorter than 3 and the server is busy upon arrival, customers join the system; otherwise, customers balk. When the server’s state is 0, 2, or 3, customers’ equilibrium strategies are $Q_{3,0}^e=(1,1,1,0)$, $Q_{3,2}^e=(1,1,0.644,0)$ and $Q_{3,3}^e=(1,1,1,0)$.

In the case where $R=11.71$, the balking threshold $n_1$ is 3 when the server’s state is 1 according to Theorem 2. Therefore, we calculate the joining probabilities $q_{n,i}^e$ for $0\le n\le 2$ and $i=0,2,3$. Similar to the computational procedure of the case $R=12.8$, we derive the customers’ equilibrium strategies as $Q_{3,0}^e=(1,1,1,0)$, $Q_{3,2}^e=(1,1,0,0)$ and $Q_{3,3}^e=(1,1,0/0.642/1,0)$. As we can see, when the system state is $(2,3)$, there are three equilibrium joining probabilities. Therefore, the equilibrium joining strategy is not unique for some values of reward R.

6. Conclusions

In this paper, we analyze the equilibrium strategies of customers in a single-server queuing system with two types of server breakdowns and delayed repairs due to complex breakdowns, where the arrival rate is dependent on the system state. Specifically, one type is a simple breakdown, where the server can be repaired immediately, while the other type is a complex breakdown, which requires repair by the OEM. Moreover, it is assumed that the delay time and repair times for both simple and complex breakdowns are generally distributed. We derive the conditional expectations of the residual delay time and repair times, which are essential figures for customers to make decisions. Subsequently, we consider the fully observable queue and obtain the equilibrium strategies of customers. Finally, we conduct two numerical examples. One is a Markovian case and its results are an extension of those in Wang and Zhang [14] and Economou and Kanta [22]. Another example supports the conclusion that the equilibrium threshold strategy is not unique.

Several issues remain to be addressed. In this paper, we merely consider the fully observable case. It is an interesting direction for future work to explore the unobservable cases. In addition, homogeneous customers are assumed in our work. Therefore, an intriguing question is to investigate the equilibrium strategies with heterogeneous customers. Moreover, another interesting topic is to study the difference in balking probabilities between consecutive stages.