On the Equilibrium in a Queuing System with Retrials and Strategic Arrivals

Abstract

This paper considers a callback single-server system with an orbit and a First-Come First-Served (FCFS) service discipline. Customers (users, clients) that encounter a busy server are sent into orbit and then have the option to retry service after an exponential period of time. In addition, each customer entering the system uses a strategy and must independently decide when to arrive in the system within a fixed admission period of time so that the expected sojourn time is minimal. We interpret the arrival process as a Nash equilibrium solution of a noncooperative game when the arrival intensity is completely described by an unknown distribution function, and then we propose a way to find an equilibrium for the case when the client’s waiting time for service is obviously limited. The analytical solution for the equilibrium is illustrated numerically for two-person and three-person games.

1. Introduction

As a rule, in the classical sense, the queuing theory assumes a nonfixed infinite flow of requests to the system that do not have a strategy, and the traditional steady-state analysis is applicable. The paper considers a queuing system with an orbit when the request is given the opportunity to repeatedly try to get to the service. If a customer entering the system encounters a busy server, they are queued into orbit and can retry to occupy the server. Since queues with retrials are of great importance in modeling modern wireless telecommunication systems, there are many works that offer a steady-state performance analysis of such queues. In particular, retrial queues with orbit are well suited to service industry models where clients are allowed multiple service attempts. An overview of such models has been prepared, for example, in [1,2,3]. Important results on stability conditions based on the regenerative approach can be found in detail in [4,5].

For the first time, a strategic approach to defining a queue as a population of a fixed size of incoming customers is demonstrated in [6] for the one-server $?/M/1$ model. Customers are accepted by the server during a specified open period of time $[0,T]$, the total number of arrivals is assumed to be Poisson distributed, and customers make decisions independently of each other. The main direction of research given by Glazer and Hassin in [6] concerns the conclusion that the arrival rate is not constant at equilibrium over the facility open period, as is usually assumed in the classical queuing theory.

Following the statement of the problem from [7], we will consider a system with a finite number of users (players), which certainly guarantees the stability of the system. Let the same requests be served on the First-Come First-Served principle. All the same, contrary to this work, we will consider another objective function. Indeed, now, instead of observing the average sojourn time of the player, we will pay attention to the case when the player cannot stay in the system for more than the fixed time V. As the payoff function of the player, consider the value ${(V-\mathbb{E}W)}^{+}$, where W is their average sojourn time. The objective of a player is to choose a strategy that will maximize their gain or, in other words, minimize the total time that they spend in the system.

Let us focus on homogeneous customers and look for a symmetrical equilibrium solution in which all customers use the same arrival strategy F. This formulation of the problem is quite natural when the population is large and anonymous, and no identification of individual clients is required. In particular, in systems with a strategy, in many cases, a symmetric equilibrium is the only possible outcome (see, for instance, the review in [8,9,10,11]).

From the practice of research in this area, it is obvious that finding a symmetric equilibrium distribution is a rather cumbersome task and sometimes is not trivially solved even numerically.

Thus, we will define the queuing game, where all players arrive according to a symmetric continuous arrival strategy $F\left(t\right)$ (cdf) with the probability density function $f\left(t\right)$ (pdf), where the player still arrives into the system either at initial moment $t=0$ with probability p or at instant $t\in [t_e,T]$ following the distribution function, defined as

$$F\left(t\right)=\left\{\begin{array}{c}p,t\in [0,t_e)\hfill \\ p+\underset{t_e}{\overset{t}{\int }}f\left(\theta \right)d\theta ,t\in [t_e,T].\hfill \end{array}\right.$$

(1)

where $p\in (0,1)$, $t_e>0$, and $T<\infty$ are predefined, and the time T is commonly referred to as closing time.

Suppose also that the stability condition $\mu >\gamma$ holds exponential service with parameter $\mu$ and exponential seeking time with parameter $\gamma$; thus, $1/\mu$ and $1/\gamma$ are the average service and seeking times, respectively. Next, we consider a queuing game for two and three players using the new payoff function and compare the derived equilibrium with the results illustrated in our initial paper [7].

2. Two-Player Case

Without loss of generality, we will assume that one of the players, here the second one, chooses the arrival time with cdf (1). This means that the second player enters the system either at instant $t=0$ with probability p or at instant $t\in [t_e,T]$.

Further, our aspiration is to find the best response of the first player to the strategy of the second one defined by Rule (1).

Let the first player enter the system at instant $t=0$, then the payoff function is:

$$C\left(0\right)=(1-p){\int }_0^V\mu e^{-\mu \theta }(V-\theta )d\theta +p\frac{1}{2}{\int }_0^V\mu e^{-\mu \theta }(V-\theta )d\theta +p\frac{1}{2}{I}_0,$$

(2)

where $I_0$ denotes the time profit of the first player until limit V for the case when the second player must be serviced first. Then, the first player arrives at the server from orbit and is served

$$\begin{array}{c}\hfill I_0={\int }_0^V\mu e^{-\mu \theta }{\int }_0^{V-\theta }\gamma e^{-\gamma \tau }{\int }_0^{V-(\theta +\tau )}\mu e^{-\mu r}(V-r)drd\tau d\theta .\end{array}$$

(3)

In other words, following our reasoning, Formula (2) reflects that if the second player arrives at the same instance $t=0$ as the first one with probability p, then the server selects the player with probability $1/2$, i.e., either the first one immediately gets into the server, or it enters orbit prior to service. With probability $1-p$, the second player arrives after the first one, then the first can be served immediately.

If the first player joins orbit, meaning before limit time V, they must wait for the second to leave the server, come from orbit in exponential time, and be served; otherwise, their payoff will be zero.

So, it is fair that

$$\begin{array}{cc}\hfill & I_0=V-\frac{1}{\mu }-e^{-\mu V}\left[V-\frac{1}{\mu }+\frac{-V^2\gamma {\mu }^2+V{\gamma }^2\mu -2V{\gamma }^2+4V\gamma \mu -2V{\mu }^2+2\mu }{{(\gamma -\mu )}^2}\right]\hfill \\ \hfill & +(e^{-\mu V}-e^{-\gamma V})\frac{\mu \gamma }{{(\gamma -\mu )}^3}-e^{-\gamma V}\left(V-\frac{1}{\mu }\right)\frac{{\mu }^2}{{(\gamma -\mu )}^2}.\hfill \end{array}$$

(4)

Note that the second player was served on the interval $[0,t]$ with probability ${\int }_0^t\mu e^{-\mu \theta }d\theta =(1-e^{-\mu t}).$ For convenience, we also predefine the expression

$$\begin{array}{cc}\hfill & W={\int }_0^V\mu e^{-\mu \theta }(V-\theta )d\theta =V-\frac{1}{\mu }+\frac{1}{\mu }{e}^{-\mu V}.\hfill \end{array}$$

(5)

Thus,

$$\begin{array}{cc}\hfill & C\left(0\right)=\left(1-\frac{p}{2}\right)\underset{0}{\overset{V}{\int }}\mu e^{-\mu \theta }(V-\theta )d\theta +p\frac{1}{2}{I}_0\hfill \\ \hfill & =\left(1-p\frac{1}{2}\right)\left(V(1-e^{-\mu V})-\left(\frac{1}{\mu }-\frac{1}{\mu }{e}^{-\mu V}(1+\mu V)\right)\right)+p\frac{1}{2}{I}_0\hfill \\ \hfill & =(1-p\frac{1}{2})W+p\frac{1}{2}{I}_0.\hfill \end{array}$$

(6)

If the first player arrived at $t\in (0,t_e)$, the payoff function is:

$$\begin{array}{cc}\hfill & C\left(t\right)=(1-p)\underset{0}{\overset{V}{\int }}\mu e^{-\mu \theta }(V-\theta )d\theta +p(1-e^{-\mu t})\underset{0}{\overset{V}{\int }}\mu e^{-\mu \theta }(V-\theta )d\theta \hfill \\ \hfill & +p{e}^{-\mu t}{I}_0=(1-p{e}^{-\mu t})\underset{0}{\overset{V}{\int }}\mu e^{-\mu \theta }(V-\theta )d\theta +p{e}^{-\mu t}{I}_0\hfill \\ \hfill & =(1-p{e}^{-\mu t})W+p{e}^{-\mu t}{I}_0.\hfill \end{array}$$

(7)

Thus, if the second player arrived at $t=0$ with probability p, then either they were served before the time t (then the first one immediately goes to the service) or they must be served in the interval $[t,t+V]$. Then, the first player must arrive from orbit and be served no later than $t+V$. Note that in the limit $t\to 0+$, it is true that

$$C(0+)=(1-p)W+p{I}_0<C\left(0\right),$$

thus the function $C\left(t\right)$ increases in t. Moreover, it is also true that $W>I_0$.

Further, in equilibrium, we require the fulfillment of the condition $C\left(0\right)=C\left(t_e\right)$, so

$$\begin{array}{c}\hfill (1-\frac{p}{2})W+\frac{p}{2}{I}_0=(1-p{e}^{-\mu t_e})W+p{e}^{-\mu t_e}{I}_0,\end{array}$$

(8)

and it is obvious that

$$\begin{array}{c}\hfill t_e=\frac{ln2}{\mu }.\end{array}$$

(9)

It should be noted that in the two-person queuing game with retrials and average sojourn time as a cost function, the same formula was obtained for $t_e$ in [7].

If the first player arrival at $t>t_e$, the payoff function is:

$$\begin{array}{cc}\hfill & C\left(t\right)=p(1-e^{-\mu t})\underset{0}{\overset{V}{\int }}\mu e^{-\mu \theta }(V-\theta )d\theta +p{e}^{-\mu t}{I}_0\hfill \\ \hfill & +\underset{t_e}{\overset{t}{\int }}\left((1-e^{-\mu (t-\theta )})\underset{0}{\overset{V}{\int }}\mu e^{-\mu \tau }(V-\tau )d\tau +e^{-\mu (t-\theta )}{I}_0\right)dF\left(\theta \right)\hfill \\ \hfill & +\underset{t}{\overset{T}{\int }}\underset{0}{\overset{V}{\int }}\mu e^{-\mu \tau }(V-\tau )d\tau dF\left(\theta \right)\hfill \\ \hfill & =p(1-e^{-\mu t})W+p{e}^{-\mu t}{I}_0+W\underset{t_e}{\overset{t}{\int }}(1-e^{-\mu (t-\theta )})f\left(\theta \right)d\theta \hfill \\ & +I_0\underset{t_e}{\overset{t}{\int }}{e}^{-\mu (t-\theta )}f\left(\theta \right)d\theta +W\underset{t}{\overset{T}{\int }}f\left(\theta \right)d\theta .\hfill \end{array}$$

It is necessary that the condition

$$\begin{array}{c}\hfill \underset{t_e}{\overset{T}{\int }}f\left(\theta \right)d\theta =1-p\end{array}$$

(10)

is fulfilled, then the following expression can be obtained

$$\begin{array}{cc}\hfill & C\left(t\right)=W(1-p{e}^{-\mu t})+p{e}^{-\mu t}{I}_0-(W-I_0)e^{-\mu t}\underset{t_e}{\overset{t}{\int }}{e}^{\mu \theta }f\left(\theta \right)d\theta .\hfill \end{array}$$

(11)

To find the equilibrium density, we use condition $C^{\prime }\left(t\right)=0$, where $C^{\prime }\left(t\right)$ is expressed in the following way

$$\begin{array}{cc}\hfill & C^{\prime }\left(t\right)=p\mu (W-I_0)e^{-\mu t}-(W-I_0)f\left(t\right)+\mu (W-I_0)e^{-\mu t}\underset{t_e}{\overset{t}{\int }}{e}^{\mu \theta }f\left(\theta \right)d\theta ,\hfill \end{array}$$

thus the condition $C^{\prime }\left(t\right)=0$ implies

$$\begin{array}{c}\hfill p\mu (W-I_0)-(W-I_0)e^{\mu t}f\left(t\right)+\mu (W-I_0)\underset{t_e}{\overset{t}{\int }}{e}^{\mu \theta }f\left(\theta \right)d\theta =0.\end{array}$$

(12)

Further, we denote

$$\begin{array}{cc}\hfill & y\left(t\right)=\underset{t_e}{\overset{t}{\int }}{e}^{\mu \theta }f\left(\theta \right)d\theta ;y\left(t\right)\equiv 0,t<t_e,\hfill \end{array}$$

(13)

and after reduction by $W-I_0$, Equation (12) becomes

$$\begin{array}{c}\hfill y^{\prime }\left(t\right)=\mu y\left(t\right)+p\mu ,\end{array}$$

(14)

and then we find the general solution of the inhomogeneous linear differential equation by the Bernoulli method in the form $y\left(t\right)=u\left(t\right)v\left(t\right)$, where $A\left(t\right)=\mu$ and $B\left(t\right)=p\mu$, thus

$$\begin{array}{cc}\hfill u\left(t\right)& =e^{\int A\left(t\right)dt}=e^{\mu t},\hfill \\ \hfill v\left(t\right)& =\int \frac{B\left(t\right)}{u\left(t\right)}dt=-p{e}^{-\mu t}+K,\hfill \\ \hfill y\left(t\right)& =-p+K{e}^{\mu t},\hfill \end{array}$$

(15)

where K is an unknown constant. So, we obtain an analytical solution for equilibrium pdf $f\left(t\right)$ in the following form:

$$\begin{array}{c}\hfill f\left(t\right)=y^{\prime }\left(t\right)e^{-\mu t}=\mu K,\end{array}$$

(16)

and immediately

$$\begin{array}{c}\hfill C\left(t\right)=W(1-p{e}^{-\mu t})+p{e}^{-\mu t}{I}_0-K(W-I_0)(1-e^{-\mu (t-t_e)}).\end{array}$$

(17)

Thus, the constant K can be fined due to Definition (13)

$$\begin{array}{c}\hfill K=p{e}^{-\mu t_e},f\left(t\right)=\mu p{e}^{-\mu t_e}.\end{array}$$

(18)

Next, it is required in equilibrium that there be equality $C\left(t\right)=C\left(t_e\right)=const$, and Condition (10) also holds, then eventually, the system of equations can be written as:

$$\left\{\begin{array}{cc}\hfill & {\displaystyle t_e=\frac{ln2}{\mu },}\hfill \\ \hfill & {\displaystyle K=p{e}^{-\mu t_e},}\hfill \\ \hfill & {\displaystyle p=\frac{1}{1+\mu e^{-\mu t_e}(T-t_e)}.}\hfill \end{array}\right.$$

(19)

Summing up all the above considerations, we can formulate the following statement.

Theorem 1. 

An equilibrium in the two-person queuing game with retrials is represented by cdf (1), where $f\left(t\right)=K=const$ for $t\in [t_e,T]$ and parameters p, $t_e$, and K satisfy Condition (19).

To be able to compare the results with those obtained earlier in [7] for the case when the average sojourn time of a customer is used as an objective function, we take the same experiment parameters while varying stay limit V.

Example 1. 

Suppose the system arrivals occur on the interval $[0,2]$, the threshold value for the first player waiting $V=1$, and assume the service and retrieval rates $\mu =2$ and $\gamma =1$, respectively.

Thus, having calculated the values $I_0=0.16$ and $W=0.57$ from (19), we obtain the values

$$t_e=0.347,p=0.377,K=0.188,$$

and the expression for equilibrium pdf $f\left(t\right)$ and payoff function $C\left(t\right)$ in the following form:

$$f\left(t\right)=0.377,t\ge t_e,$$

$$C\left(0\right)=C\left(t\right)=0.491,t\ge t_e.$$

Figure 1 shows the pdf $f\left(t\right)$ and the $C\left(t\right)$ shape on the interval $[t_e,T]$.

Note that for sojourn time target function $C\left(t\right)$ from [7], the corresponding values are:

$$t_e=0.347,p=0.377,C\left(0\right)=C\left(t\right)=0.783\mathrm{for}{t}_e>t,$$

which is the same arrival strategy.

Example 2. 

Define $V=10$ outside the interval $[t_e,T]$, and let $T=2$, $\mu =2$, $\gamma =1$. Then, by carrying out the corresponding calculations, we obtain $I_0=9.49$, $W=9.5$. It can be obtained from System (19) that

$$t_e=0.347,p=0.377,K=0.188,$$

and

$$f\left(t\right)=0.377,t\ge t_e,$$

$$C\left(0\right)=C\left(t\right)=9.499,t\ge t_e.$$

The corresponding curves are shown in Figure 2.

Example 3. 

Now, let the expectation limit V coincide with T, namely $T=2$, $V=T$, $\mu =2$, and $\gamma =1$. Then, $I_0=0.97$ and $W=1.51$. As in Examples 1 and 2, the key values of the model remain unchanged and equal

$$t_e=0.347,p=0.377,K=0.188,$$

also

$$f\left(t\right)=0.377,t\ge t_e,$$

However, the expression for the target function $C\left(t\right)$ changes as follows

$$C\left(0\right)=C\left(t\right)=1.407,t\ge t_e.$$

The behavior of both functions on the interval $[t_e,T]$ is shown in Figure 3.

It is important to summarize that despite the different objective function the arrival strategy for the two-player game is the same as derived in [7]. As expected, it is noticeable in the examples that the player’s payoff significantly depends on stay limit V.

3. Three-Player Case

Consider the case when the second and third players enter the system according to the strategy when the interarrival times correspond to the distribution $F\left(t\right)$. Let the players be considered identical, so if they arrive at the same time in orbit or service, then they are selected equally likely. Next, we look for the optimal strategy for the first player.

If the first player arrives at the moment $t=0$ and assuming that $V\le T$, then the payoff is

$$\begin{array}{cc}\hfill C\left(0\right)=& {(1-p)}^{2}W+2p(1-p)\left(\frac{1}{2}W+\frac{1}{2}{I}_1^1\right)+p^2\left(\frac{1}{3}W+\frac{2}{3}\left(\frac{1}{2}{I}_0^2+\frac{1}{2}{I}_1^2\right)\right)\hfill \\ \hfill =& {(1-p)}^{2}W+p(1-p)(W+I_1^1)+\frac{p^2}{3}(W+I_0^2+I_1^2)\hfill \end{array}$$

(20)

where, similarly with the previous analysis, we predetermine the auxiliary values W, $I_0^2$, $I_1^2$, and $I_1^1$:

$$\begin{array}{cc}\hfill & W=\underset{0}{\overset{V}{\int }}\mu e^{-\mu \theta }(V-\theta )d\theta =V-\frac{1}{\mu }+\frac{1}{\mu }{e}^{-\mu V},\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill & {\displaystyle I_0^2=\underset{0}{\overset{V}{\int }}\mu e^{-\mu \theta }d\theta \underset{0}{\overset{V-\theta }{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-(\theta +\tau )}{\int }}\mu e^{-\mu r}(V-(\theta +\tau +r))dr}\hfill \\ \hfill & =V+\left(\frac{2}{\mu }+\frac{1}{\gamma }\right)(e^{-\mu V}-1)+\frac{V\gamma }{\gamma -\mu }{e}^{-\mu V}+\frac{{\mu }^2}{\gamma {(\gamma -\mu )}^2}(e^{-\gamma V}-e^{-\mu V})\hfill \\ \hfill & =W+\left(\frac{1}{\mu }+\frac{1}{\gamma }\right)(e^{-\mu V}-1)+\frac{V\gamma }{\gamma -\mu }{e}^{-\mu V}+\frac{{\mu }^2}{\gamma {(\gamma -\mu )}^2}(e^{-\gamma V}-e^{-\mu V}).\hfill \end{array}$$

(22)

If the second and third players arrived later than $t_e$, then the first player immediately goes to the server, and their payoff corresponds to W.

$$\begin{array}{cc}\hfill & {\displaystyle I_1^2=\underset{0}{\overset{V}{\int }}\mu e^{-\mu \theta }d\theta \underset{0}{\overset{V-\theta }{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-(\theta +\tau )}{\int }}\mu e^{-\mu \xi }d\xi \underset{0}{\overset{V-(\theta +\tau +\xi )}{\int }}\gamma e^{-\gamma \eta }d\eta }\hfill \\ \hfill & \times \underset{0}{\overset{V-(\theta +\tau +\xi +\eta )}{\int }}\mu e^{-\mu r}(V-(\theta +\tau +\xi +\eta +r))dr\hfill \\ \hfill & {\displaystyle =V-\frac{3}{\mu }-\frac{2}{\gamma }+e^{-\gamma V}\left[-\frac{3{\mu }^3}{{\left(\gamma -\mu \right)}^4}-\frac{{\mu }^2\left(V\mu +2\right)}{{\left(\gamma -\mu \right)}^3}+\frac{2\mu }{{\left(\gamma -\mu \right)}^2}-\frac{2}{\gamma -\mu }+\frac{2}{\gamma }\right]}\hfill \\ \hfill & {\displaystyle +e^{-\mu V}[\frac{3{\mu }^3}{{\left(\gamma -\mu \right)}^4}-\frac{2{\mu }^2\left(V\mu -1\right)}{{\left(\gamma -\mu \right)}^3}+\frac{\mu \left(V^2{\mu }^2-4V\mu -4\right)}{2{\left(\gamma -\mu \right)}^2}+\frac{V^2{\mu }^2+2V\mu +2}{\gamma -\mu }}\hfill \\ \hfill & +\frac{V^2{\mu }^2+4V\mu +6}{2\mu }],\hfill \end{array}$$

(23)

If only one of them came after $t_e$, and the other came at the same time as the first one, then at instant $t=0$, there are two players in the system who claim service. Then, with a probability $1/2$, the first goes to the server, and the other with the probability $1/2$ joins orbit.

$$\begin{array}{cc}\hfill & I_1^1=\underset{t_e}{\overset{V}{\int }}[\underset{0}{\overset{\theta }{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{\theta -s}{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-(s+\tau )}{\int }}\mu e^{-\mu r}(V-(s+\tau +r))dr\hfill \\ \hfill & +\underset{0}{\overset{\theta }{\int }}\mu e^{-\mu s}ds\underset{\theta -s}{\overset{\infty }{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-\theta }{\int }}\mu e^{-\mu \xi }d\xi \underset{0}{\overset{V-(\theta +\xi )}{\int }}\gamma e^{-\gamma \eta }d\eta \hfill \\ \hfill & \times \underset{0}{\overset{V-(\theta +\xi +\eta )}{\int }}\mu e^{-\mu r}(V-(\theta +\xi +\eta +r))dr\hfill \\ \hfill & +\underset{\theta }{\overset{V}{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{V-s}{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-(s+\tau )}{\int }}\mu e^{-\mu r}(V-(s+\tau +r))dr]dF\left(\theta \right)\hfill \\ \hfill & +I_0^2{\int }_V^{T}dF\left(\theta \right).\hfill \end{array}$$

(24)

If the first player goes into orbit, then the third player can arrive in the system while servicing the second one. Thus, the third player can take the server before the player from the orbit, so in that case, the payoff of the first player corresponds to $I_1^1$.

And finally, if three players arrived simultaneously at time $t=0$, then with the probability $1/3$ the first player will be served straight away, and with the probability $2/3$ the first and other one will go into orbit. Then, two options are equally probable, either the first player leaves the orbit earlier (his payoff corresponds to $I_0^2$), or the first player leaves the orbit last (his payoff corresponds to $I_1^2$).

After additional manipulations, we obtain the final expression in the form

$$\begin{array}{cc}\hfill I_1^1& =\underset{t_e}{\overset{V}{\int }}dF\left(\theta \right)[V+\left(\frac{2}{\mu }+\frac{1}{\gamma }\right)(\frac{e^{-\mu (V-\theta )}(\gamma e^{-\mu \theta }-\mu e^{-\gamma \theta })}{\gamma -\mu }-1)\hfill \\ \hfill & +\left(V-\theta -\frac{2}{\mu }-\frac{1}{\gamma }\right)\frac{\mu (e^{-\mu \theta }-e^{-\gamma \theta })}{\gamma -\mu }\hfill \\ \hfill & +\frac{(e^{-\mu \theta }-e^{-\gamma \theta })(\frac{\mu }{\gamma }+1+\mu \theta -V\mu )+\gamma \theta e^{-V\mu }}{\gamma -\mu }\hfill \\ \hfill & +\frac{\gamma (V-\theta )e^{-\mu (V-\theta )}(\gamma e^{-\mu \theta }-\mu e^{-\gamma \theta })}{{(\gamma -\mu )}^2}+\frac{\gamma e^{-V\mu }(e^{-\theta (\gamma -\mu )}-1)}{{(\gamma -\mu )}^2}\hfill \\ \hfill & -\frac{{\mu }^2(\gamma e^{-\mu \theta }-\mu e^{-\gamma \theta })(e^{-\mu (V-\theta )}-e^{-\gamma (V-\theta )})}{\gamma {(\gamma -\mu )}^3}]+I_0^2{\int }_V^{T}dF\left(\theta \right).\hfill \end{array}$$

(25)

We note separately that in the case $V>T$, the following expression holds

$$\begin{array}{cc}\hfill & I_1^1=\underset{t_e}{\overset{T}{\int }}[\underset{0}{\overset{\theta }{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{\theta -s}{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-(s+\tau )}{\int }}\mu e^{-\mu r}(V-(s+\tau +r))dr\hfill \\ \hfill & +\underset{0}{\overset{\theta }{\int }}\mu e^{-\mu s}ds\underset{\theta -s}{\overset{\infty }{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-\theta }{\int }}\mu e^{-\mu \xi }d\xi \underset{0}{\overset{V-(\theta +\xi )}{\int }}\gamma e^{-\gamma \eta }d\eta \hfill \\ \hfill & \times \underset{0}{\overset{V-(\theta +\xi +\eta )}{\int }}\mu e^{-\mu r}(V-(\theta +\xi +\eta +r))dr\hfill \\ \hfill & +\underset{\theta }{\overset{V}{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{V-s}{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-(s+\tau )}{\int }}\mu e^{-\mu r}(V-(s+\tau +r))dr]dF\left(\theta \right).\hfill \end{array}$$

(26)

However, for the sake of brevity, we omit further calculations for this case, since they are performed in the same way.

If the customer arrives at instant $t=0+$, then the payoff is:

$$\begin{array}{c}\hfill C(0+)={(1-p)}^{2}W+p(1-p)I_1^1+\frac{p^2}{2}(I_0^2+I_1^2)<C\left(0\right).\end{array}$$

(27)

Next, consider the case when the first player arrived at the time $t\in (0,t_e)$. Then, the first player is served immediately if either the other two have been served before instant t, or only one customer has been served before t, and the other has not managed to come from orbit. Otherwise, the first player enters the orbit.

So, for the case when $t+V\le T$, the first player payoff is:

$$\begin{array}{cc}\hfill & C\left(t\right)={(1-p)}^{2}W+2p(1-p)(1-e^{-\mu t})W\hfill \\ \hfill & {\displaystyle +2p\left(\underset{t_e}{\overset{t+V}{\int }}\right[\underset{t}{\overset{\theta }{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{\theta -s}{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{t+V-(s+\tau )}{\int }}\mu e^{-\mu r}(t+V-s-\tau -r)dr}\hfill \\ \hfill & +\underset{t}{\overset{\theta }{\int }}\mu e^{-\mu s}ds\underset{\theta -s}{\overset{\infty }{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{t+V-\theta }{\int }}\mu e^{-\mu \xi }d\xi \underset{0}{\overset{t+V-(\theta +\xi )}{\int }}\gamma e^{-\gamma \eta }d\eta \hfill \\ \hfill & \times \underset{0}{\overset{t+V-(\theta +\xi +\eta )}{\int }}\mu e^{-\mu r}(t+V-\theta -\xi -\eta -r)dr\hfill \\ \hfill & +\underset{\theta }{\overset{t+V}{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{t+V-s}{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{t+V-(s+\tau )}{\int }}\mu e^{-\mu r}(t+V-s-\tau -r)dr]f\left(\theta \right)d\theta \hfill \\ \hfill & +((1-e^{-\mu t})W+e^{-\mu t}{I}_0^2)\underset{t+V}{\overset{T}{\int }}f\left(\theta \right)d\theta )\hfill \\ \hfill & +p^2(\underset{0}{\overset{t}{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{t-s}{\int }}\gamma e^{-\gamma \tau }(1-e^{-\mu (t-(s+\tau \left)\right)})d\tau W+\underset{0}{\overset{t}{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{t-s}{\int }}\gamma e^{-\gamma \tau }{e}^{-\mu (t-(s+\tau \left)\right)}d\tau I_0^2\hfill \\ \hfill & +W\underset{0}{\overset{t}{\int }}\mu e^{-\mu s}{e}^{-\gamma (t-s)}ds+e^{-\mu t}{I}_1^2).\hfill \end{array}$$

Thus, after intermediate calculations, we obtain that $t+V<T$ holds:

$$\begin{array}{cc}\hfill C\left(t\right)=& W{(1-p)}^2+2p(1-p)(W(1-e^{-\mu t})+I_1^1)\hfill \\ \hfill & +p^{2}W(1-e^{-\mu t})+p^{2}W(\frac{\mu (e^{-\mu t}-e^{-\gamma t})}{{(\gamma -\mu )}^2}-\frac{\gamma }{\gamma -\mu }t{e}^{-\mu t}+\frac{\mu (e^{-\mu t}-e^{-\gamma t})}{\gamma -\mu })\hfill \\ \hfill & +p^2{I}_0^2(\frac{\gamma }{\gamma -\mu }t{e}^{-\mu t}-\frac{\gamma (e^{-\mu t}-e^{-\gamma t})}{{(\gamma -\mu )}^2})+p^2{e}^{-\mu t}{I}_1^2.\hfill \end{array}$$

(28)

Finally, let us consider the case when the first player enters the system at instant $t\in [t_e,T]$. Let the current player still not be able to wait more than the given time V, and also suppose first that $t+V<T$. For that t, the payoff is:

$$\begin{array}{cc}\hfill & C\left(t\right)=p^2\left[W+(I_1^2-W)e^{-\mu t}+(I_0^2-W)\gamma \mu \frac{e^{-\gamma t}-e^{-\mu t}(\mu t-\gamma t+1)}{{(\mu -\gamma )}^2}\right]\hfill \\ \hfill & +2p\underset{t_e}{\overset{t}{\int }}f\left(\theta \right)d\theta [W+(I_1^2-I_0^2)e^{-\mu t}\hfill \\ \hfill & +(I_0^2-W)\frac{\gamma \mu e^{-\gamma (t-\theta )}{e}^{-\mu \theta }-\gamma \mu e^{-\mu t}(1+(\mu -\gamma )(t-\theta ))+{(\mu -\gamma )}^2{e}^{-\mu (t-\theta )}}{{(\mu -\gamma )}^2}]\hfill \\ \hfill & +2p\underset{t}{\overset{t+V}{\int }}f\left(\theta \right)d\theta \left[(1-e^{-\mu t})W+D\left(\theta \right)\right]+2p\underset{t+V}{\overset{T}{\int }}f\left(\theta \right)d\theta (W+(I_0^2-W)e^{-\mu t})\hfill \\ \hfill & +\underset{t_e}{\overset{t}{\int }}\underset{t_e}{\overset{t}{\int }}f\left(\theta \right)f\left(\eta \right)d\theta d\eta [W+(e^{-\mu (t-\theta )}-e^{-\mu (t-\theta )}{e}^{-\mu (t-\eta )}+e^{-\mu (t-\eta )})(I_0^2-W)\hfill \\ \hfill & +e^{-\mu (t-\theta )}{e}^{-\mu (t-\eta )}(I_1^2-I_0^2)]+2\underset{t_e}{\overset{t}{\int }}(W+e^{-\mu (t-\theta )}(I_0^2-W))f\left(\theta \right)d\theta \underset{t+V}{\overset{T}{\int }}f\left(\eta \right)d\eta \hfill \\ \hfill & +2\underset{t_e}{\overset{t}{\int }}f\left(\theta \right)d\theta \underset{t}{\overset{t+V}{\int }}(W+e^{-\mu (t-\theta )}(D\left(\eta \right)-W))f\left(\eta \right)d\eta +W\underset{t+V}{\overset{T}{\int }}\underset{t+V}{\overset{T}{\int }}f\left(\theta \right)f\left(\eta \right)d\theta d\eta ,\hfill \end{array}$$

(29)

where $D\left(\theta \right)$ does not depend on t and is denoted as:

$$\begin{array}{cc}\hfill & D\left(\theta \right)=V+\left(\frac{2}{\mu }+\frac{1}{\gamma }\right)(\frac{e^{-\mu (V-\theta )}(\gamma e^{-\mu \theta }-\mu e^{-\gamma \theta })}{\gamma -\mu }-1)\hfill \\ \hfill & +\left(V-\theta -\frac{2}{\mu }-\frac{1}{\gamma }\right)\frac{\mu (e^{-\mu \theta }-e^{-\gamma \theta })}{\gamma -\mu }+\frac{(e^{-\mu \theta }-e^{-\gamma \theta })(\frac{\mu }{\gamma }+1+\mu \theta -V\mu )+\gamma \theta e^{-V\mu }}{\gamma -\mu }\hfill \\ \hfill & +\frac{\gamma (V-\theta )e^{-\mu (V-\theta )}(\gamma e^{-\mu \theta }-\mu e^{-\gamma \theta })}{{(\gamma -\mu )}^2}+\frac{\gamma e^{-V\mu }(e^{-a(\gamma -\mu )}-1)}{{(\gamma -\mu )}^2}\hfill \\ \hfill & -\frac{{\mu }^2(\gamma e^{-\mu \theta }-\mu e^{-\gamma \theta })(e^{-\mu (V-\theta )}-e^{-\gamma (V-\theta )})}{\gamma {(\gamma -\mu )}^3}.\hfill \end{array}$$

Expression (29) can be rewritten in the form:

$$\begin{array}{cc}\hfill & C\left(t\right)=p^{2}W+p^2(I_1^2-W)e^{-\mu t}+p^2(I_0^2-W)\frac{\gamma \mu }{{(\mu -\gamma )}^2}{e}^{-\gamma t}-p^2(I_0^2-W)\frac{\gamma \mu }{(\mu -\gamma )}{e}^{-\mu t}t-p^2(I_0^2-W)\frac{\gamma \mu }{{(\mu -\gamma )}^2}{e}^{-\mu t}\hfill \\ \hfill & +2p(1-p)W+2p(I_1^2-I_0^2)e^{-\mu t}\underset{t_e}{\overset{t}{\int }}f\left(\theta \right)d\theta +2p\frac{\gamma \mu }{{(\gamma -\mu )}^2}(I_0^2-W)e^{-\gamma t}\underset{t_e}{\overset{t}{\int }}{e}^{-\theta (\mu -\gamma )}f\left(\theta \right)d\theta -2p(I_0^2-W)\frac{\gamma \mu }{{(\gamma -\mu )}^2}{e}^{-\mu t}\underset{t_e}{\overset{t}{\int }}f\left(\theta \right)d\theta \hfill \\ \hfill & -2p(I_0^2-W)\frac{\gamma \mu }{\gamma -\mu }\underset{t_e}{\overset{t}{\int }}(t-\theta )f\left(\theta \right)d\theta +2p(I_0^2-W)e^{-\mu t}\underset{t_e}{\overset{t}{\int }}{e}^{\mu \theta }f\left(\theta \right)d\theta -2p(I_0^2-W)e^{-\mu t}\underset{t}{\overset{t+V}{\int }}f\left(\theta \right)d\theta +2p\underset{t}{\overset{t+V}{\int }}D\left(\theta \right)f\left(\theta \right)d\theta \hfill \\ \hfill & +2p(I_0^2-W)e^{-\mu t}\underset{t+V}{\overset{T}{\int }}f\left(\theta \right)d\theta +{(1-p)}^{2}W+2(I_0^2-W)\underset{t_e}{\overset{t}{\int }}{e}^{-\mu (t-\theta )}f\left(\theta \right)d\theta \underset{t_e}{\overset{t}{\int }}f\left(\eta \right)d\eta +(W-2I_0^2+I_1^2){\left[\underset{t_e}{\overset{t}{\int }}{e}^{-\mu (t-\theta )}f\left(\theta \right)d\theta \right]}^2\hfill \\ \hfill & +2(I_0^2-W)\underset{t_e}{\overset{t}{\int }}{e}^{-\mu (t-\theta )}f\left(\theta \right)d\theta \underset{t+V}{\overset{T}{\int }}f\left(\eta \right)d\eta +2\underset{t_e}{\overset{t}{\int }}{e}^{-\mu (t-\theta )}f\left(\theta \right)d\theta \underset{t}{\overset{t+V}{\int }}(D\left(\eta \right)-W)f\left(\eta \right)d\eta \hfill \end{array}$$

(30)

Then, for the next step, it is necessary to solve the system of equations:

$$\begin{array}{c}\hfill C\left(0\right)=C\left(t_e\right)=C\left(t\right)=\mathrm{const},t\in [t_e,T].\end{array}$$

(31)

From Formulas (20) and (28), it follows that the equation $C\left(0\right)=C\left(t_e\right)$ has the form

$$\begin{array}{cc}\hfill & p(1-p)(W+I_1^1)+\frac{p^2}{3}(W+I_0^2+I_1^2)\hfill \\ \hfill & =2p(1-p)(W(1-e^{-\mu t_e})+I_1^1)\hfill \\ \hfill & +p^{2}W(1-e^{-\mu t_e})+p^{2}W(\frac{\mu (e^{-\mu t_e}-e^{-\gamma t})}{{(\gamma -\mu )}^2}-\frac{\gamma }{\gamma -\mu }{t}_e{e}^{-\mu t_e}+\frac{\mu (e^{-\mu t_e}-e^{-\gamma t_e})}{\gamma -\mu })\hfill \\ \hfill & +p^2{I}_0^2(\frac{\gamma }{\gamma -\mu }{t}_e{e}^{-\mu t_e}-\frac{\gamma (e^{-\mu t_e}-e^{-\gamma t_e})}{{(\gamma -\mu )}^2})+p^2{e}^{-\mu t_e}{I}_1^2,\hfill \end{array}$$

or

$$\begin{array}{cc}\hfill & W(p-p^2+\frac{p^2}{3})+(p-p^2)I_1^1+\frac{p^2}{3}{I}_0^2+\frac{p^2}{3}{I}_1^2\hfill \\ \hfill & =W(2p(1-p)(1-e^{-\mu t_e})+p^2(1-e^{-\mu t_e}+\frac{\mu (e^{-\mu t_e}-e^{-\gamma t})}{{(\gamma -\mu )}^2}-\frac{\gamma }{\gamma -\mu }{t}_e{e}^{-\mu t_e}+\frac{\mu (e^{-\mu t_e}-e^{-\gamma t_e})}{\gamma -\mu }))\hfill \\ \hfill & +2p(1-p)I_1^1+p^2{I}_0^2(\frac{\gamma }{\gamma -\mu }{t}_e{e}^{-\mu t_e}-\frac{\gamma (e^{-\mu t_e}-e^{-\gamma t_e})}{{(\gamma -\mu )}^2})+p^2{e}^{-\mu t_e}{I}_1^2,\hfill \end{array}$$

(32)

Also, the required derivative can be calculated and the equation

$$C^{\prime }\left(t\right)=0,$$

(33)

is composed based on Expression (30).

As it already becomes clear, this solution is rather cumbersome for a numerical solution algorithm. Let us consider another way to write the payoff $C\left(t\right)$, which seems to be simpler, where in terms of the system of Kolmogorov’s differential inverse equations at instant $t\in [t_e,T]$ for state probabilities $p_{i,j,k}\left(t\right)$, at the moment t, the system received i customers, j customers are in the server, and k ones in the orbit:

$$\begin{array}{cc}\hfill & p_{0,0,0}^{\prime }\left(t\right)=-{\lambda }_0\left(t\right)p_{0,0,0}\left(t\right),\hfill \\ \hfill & p_{1,0,0}^{\prime }\left(t\right)=-{\lambda }_1\left(t\right)p_{1,0,0}\left(t\right)+\mu p_{1,1,0}\left(t\right),\hfill \\ \hfill & p_{1,1,0}^{\prime }\left(t\right)=-(\mu +{\lambda }_1\left(t\right))p_{1,1,0}\left(t\right)+{\lambda }_0\left(t\right)p_{0,0,0}\left(t\right),\hfill \\ \hfill & p_{2,0,1}^{\prime }\left(t\right)=-\gamma p_{2,0,1}\left(t\right)+\mu p_{2,1,1}\left(t\right),\hfill \\ \hfill & p_{2,1,0}^{\prime }\left(t\right)=-\mu p_{2,0,1}\left(t\right)+\gamma p_{2,0,1}\left(t\right)+{\lambda }_1\left(t\right)p_{1,0,0}\left(t\right),\hfill \\ \hfill & p_{2,1,1}^{\prime }\left(t\right)=-\mu p_{2,1,1}\left(t\right)+{\lambda }_1\left(t\right)p_{1,1,0}\left(t\right),\hfill \\ \hfill & p_{2,0,0}^{\prime }\left(t\right)=\mu p_{2,1,0}\left(t\right),\hfill \end{array}$$

(34)

where the arrival rate at instant t depends on the number of customers there are in the system, so

$$\begin{array}{cc}\hfill & {\displaystyle {\lambda }_0\left(t\right)=\frac{2f\left(t\right)}{1-F\left(t\right)},}\hfill \\ \hfill & {\displaystyle {\lambda }_1\left(t\right)=\frac{f\left(t\right)}{1-F\left(t\right)}.}\hfill \end{array}$$

Due to (34), the response for the player entering the system at instant t is

$$\begin{array}{cc}\hfill & C\left(t\right)=(p_{0,0,0}\left(t\right)+p_{1,0,0}\left(t\right)+p_{2,0,1}\left(t\right)+p_{2,0,0}\left(t\right))W\hfill \\ \hfill & +p_{1,1,0}\left(t\right)\left[\underset{t}{\overset{t+V}{\int }}f\left(\theta \right)d\theta \right(\underset{0}{\overset{\theta -t}{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{\theta -(t+s)}{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-(s+\tau )}{\int }}\mu e^{-\mu r}dr(V-(s+\tau +r))\hfill \\ \hfill & +\underset{0}{\overset{\theta -t}{\int }}\mu e^{-\mu s}ds\underset{\theta -(t+s)}{\overset{\infty }{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{t+V-\theta }{\int }}\mu e^{-\mu \xi }d\xi \underset{0}{\overset{t+V-(\theta +\xi )}{\int }}\gamma e^{-\gamma v}dv\hfill \\ \hfill & \times \underset{0}{\overset{t+V-(\theta +\xi +v)}{\int }}\mu e^{-\mu r}dr(t+V-(\theta +\xi +v+r))\hfill \\ \hfill & +\underset{\theta -t}{\overset{V}{\int }}\mu e^{-\mu s}ds\underset{0}{\overset{V-s}{\int }}\gamma e^{-\gamma \tau }d\tau \underset{0}{\overset{V-(s+\tau )}{\int }}\mu e^{-\mu r}(V-(s+\tau +r))dr+I_0^2\underset{t+V}{\overset{T}{\int }}f\left(\theta \right)d\theta \left)\right]\hfill \\ \hfill & +p_{2,1,0}\left(t\right)I_0^2+p_{2,1,1}\left(t\right)I_1^2.\hfill \end{array}$$

(35)

After carrying out similar calculations as in Formula (25), we obtain the following expression:

$$\begin{array}{cc}\hfill & C\left(t\right)=(p_{0,0,0}\left(t\right)+p_{1,0,0}\left(t\right)+p_{2,0,1}\left(t\right)+p_{2,0,0}\left(t\right))W\hfill \\ \hfill & +p_{1,1,0}\left(t\right)\left(\underset{t}{\overset{t+V}{\int }}f\left(\theta \right)d\theta \right[V+(\frac{2}{\mu }+\frac{1}{\gamma })(\frac{e^{-\mu (V-\theta )}(\gamma e^{-\mu \theta }-\mu e^{-\gamma \theta })}{\gamma -\mu }-1)\hfill \\ \hfill & +(V-\theta -\frac{2}{\mu }-\frac{1}{\gamma })\frac{\mu (e^{-\mu \theta }-e^{-\gamma \theta })}{\gamma -\mu }\hfill \\ \hfill & +\frac{(e^{-\mu \theta }-e^{-\gamma \theta })(\frac{\mu }{\gamma }+1+\mu \theta -V\mu )+\gamma \theta e^{-V\mu }}{\gamma -\mu }\hfill \\ \hfill & +\frac{\gamma (V-\theta )e^{-\mu (V-\theta )}(\gamma e^{-\mu \theta }-\mu e^{-\gamma \theta })}{{(\gamma -\mu )}^2}+\frac{\gamma e^{-V\mu }(e^{-\theta (\gamma -\mu )}-1)}{{(\gamma -\mu )}^2}\hfill \\ \hfill & -\frac{{\mu }^2(\gamma e^{-\mu \theta }-\mu e^{-\gamma \theta })(e^{-\mu (V-\theta )}-e^{-\gamma (V-\theta )})}{\gamma {(\gamma -\mu )}^3}]+I_0^2\underset{t+V}{\overset{T}{\int }}f\left(\theta \right)d\theta )\hfill \\ \hfill & +p_{2,1,0}\left(t\right)I_0^2+p_{2,1,1}\left(t\right)I_1^2.\hfill \end{array}$$

(36)

It is necessary that the normalization condition be fulfilled

$$\begin{array}{c}\hfill \underset{t_e}{\overset{T}{\int }}f\left(\theta \right)d\theta =1-p,\end{array}$$

(37)

and also boundary conditions for the Cauchy problem for Equation (34) can be fined as the system of state probabilities for $t=t_e$ in [7], namely:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{\displaystyle p_{0,0,0}\left(t_e\right)}\hfill & ={(1-p)}^2,\hfill \\ {\displaystyle p_{1,0,0}\left(t_e\right)}\hfill & =2p(1-p)(1-e^{-\mu t_e}),\hfill \\ {\displaystyle p_{1,1,0}\left(t_e\right)}\hfill & =2p(1-p)e^{-\mu t_e},\hfill \\ {\displaystyle p_{2,0,1}\left(t_e\right)}\hfill & =p^2\underset{0}{\overset{t_e}{\int }}\mu e^{-\mu s}{e}^{-\gamma (t_e-s)}ds=p^2\frac{\mu }{\mu -\gamma }(e^{-\gamma t_e}-e^{-\mu t_e}),\hfill \\ {\displaystyle p_{2,1,0}\left(t_e\right)}\hfill & =p^2\underset{0}{\overset{t_e}{\int }}\mu e^{-\mu s}\underset{0}{\overset{t_e-s}{\int }}\gamma e^{-\gamma \tau }{e}^{-\mu (t_e-(s+\tau ))}d\tau ds=p^2\mu \gamma \frac{e^{-\gamma t_e}-e^{-\mu t_e}(\mu t_e-\gamma t_e+1)}{{(\mu -\gamma )}^2},\hfill \\ {\displaystyle p_{2,1,1}\left(t_e\right)}\hfill & =p^2{e}^{-\mu t_e},\hfill \\ {\displaystyle p_{2,0,0}\left(t_e\right)}\hfill & =p^2\underset{0}{\overset{t_e}{\int }}\mu e^{-\mu s}\underset{0}{\overset{t_e-s}{\int }}\gamma e^{-\gamma \tau }(1-e^{-\mu (t_e-(s+\tau ))})d\tau ds\hfill \\ \hfill & {\displaystyle =p^2(1-e^{-\mu t_e}-\mu \frac{e^{-\gamma t_e}-e^{-\mu t_e}}{\mu -\gamma }-\gamma \frac{e^{-\gamma t_e}-e^{-\mu t_e}(\mu t_e-\gamma t_e+1)}{{(\mu -\gamma )}^2}).}\hfill \end{array}\right.\end{array}$$

Then, for the next step, it is necessary to solve the system of equations:

$$\begin{array}{c}\hfill C\left(0\right)=C\left(t_e\right)=C\left(t\right)=\mathit{const},t\in [t_e,T].\end{array}$$

(38)

Since explicit calculations for the case of three players in an equilibrium distribution are rather complex analytically, we will use a numerical approach that is quite common for problems of this type (see for instance [8]). A discrete approximation technique based on simulation on a grid with a small rank on a set $[0,1]\times [0,T]$ was proposed for a queuing game with retrials in [7]. The same algorithm is used below for the case of three players.

As before, the purpose of the study was to compare the player’s strategy for different objective functions, thus we need to choose the same parameters as in [7].

Example 4. 

Suppose the system arrivals occur on the interval $[0,T]$, $T=4$, and the threshold value for the first player waiting is $V=2$. Assume the service and retrieval rate $\mu =2$, $\gamma =1$, respectively. So, having calculated the parameters numerically from (32), (33), and (37) and apply the approximation algorithm on the grid for equilibrium pdf we obtain

$$t_e=0.483,p=0.421,C\left(0\right)=C\left(t\right)=0.810$$

and Figure 4 shows the pdf $f\left(t\right)$ and the $C\left(t\right)$ shape on the interval $[t_e,T]$. For $V=3.5$, the response value is $C\left(t\right)=0.850$, and for case $V=10$, we obtain $C\left(t\right)=1.756$, without changes in equilibrium strategy.

An alternative technique of the numerical solution of Kolmogorov’s system of differential Equations (34) and Formulas (32) and (37) gives the following solutions $t_e=0.471$, $p=0.398$, $C\left(0\right)=C\left(t\right)=0.797$ for $V=2$.

It should be noted that both methods give consistent results for the equilibrium arrival strategy for a three-player game with retrials. In contrast to the case of two players, now the equilibrium pdf $f\left(t\right)$ is not a constant and decreases in the interval $[0.471,4]$. Moreover, just as in the case of two players, where it was easy to obtain analytical solutions, in the case of three players, the strategies found numerically agree with the results of calculations in the paper [7], where another objective function was considered for the same system.

All numerical tests were realized in Python3 language using modules for symbolic calculation and executed on the HP ENVY Ultrabook with an Intel(R) Core(TM) i3 7100U 2.4 GHz processor with 4 GB of RAM, running Windows 10.

4. Conclusions

This paper deals with a game-theoretical approach to the analysis of a single-server queuing system with retrials from the orbit. In contrast to the classical approach of the queuing theory, it is assumed that the customers (players) are guided by the arrival strategy and so the arrival rate is not constant in general. We investigate the arrival strategy in Nash equilibrium, where the goal is to maximize the expected reward per player when the total time that he spends in the system is limited by constant V.

For games with two and three persons, it is shown that the optimal arrival behavior is such that the player either enters the system with a nonzero probability at the initial time, or there will be an arrival pause along a positive interval, after which the arrival rate is controlled by the equilibrium distribution density. In the case of three players, the equilibrium pdf is not a constant and decreases over the interval $[t_e,T]$.

In the further direction of the analysis, it is necessary to generalize the results and the numerical algorithm to the case of an arbitrary number of players for a system with retrials and optimal timing of arrival to a queue. It is supposed to consider the problem in the asymptotic setting as the number of players tends to infinity.