Information heterogeneity in a retrial queue: throughput and social welfare maximization

Abstract

We consider an M/M/1 queue with retrials. There are two streams of customers, one informed about the server’s state upon arrival (idle or busy) and the other not informed. Both informed and uninformed customers decide whether to join the system or not upon arrival. Upon joining, customers who are faced with a busy server will retry several times until the server is idle to acquire service. The interval of retrials is exponentially distributed. We investigate equilibrium strategies for the customers and study the impact of information heterogeneity on the system throughput and social welfare. We find that social welfare is increasing in the fraction of informed customers and the maximum social welfare is reached when all customers are informed about the state of the server. On the other hand, we find that when the workload is low (or high), the throughput-maximizing server should conceal (or disclose) the state of the server to customers. When the workload falls in an intermediate range, information heterogeneity in the population (i.e., revealing the information to a certain portion of customers) leads to more efficient outcomes. Finally, numerical analyses are presented to verify our results and illustrate the impact of the retrial behavior on the system performance.

Appendices

Appendix A

In this section, we present some numerical examples to verify our results (for example, the equilibrium joining strategy, the throughput, social welfare, and the individual’s benefit for customers), and illustrate the impact of the retrial rate $\theta$ on the throughput and social welfare. Unless otherwise noted, we let $\mu =\theta =2$ and $\nu =2.5$ for all of the examples in this section.

1.1 A.1: Equilibrium joining strategy $(q_{U},q_{I})$

Fig. 5

Equilibrium joining probability for customers versus $\gamma$

Based on Proposition 2, the explicit equilibrium joining strategies for both customer types with respect to $\gamma$ are given in Fig. 5. Specifically, four different arrival rates ($\Lambda =1.3$, $\Lambda =2.5$, $\Lambda =4$, and $\Lambda =10$) are compared. In (a) of Fig. 5, we have $\gamma ^{*}<0$ and $\gamma ^{o}\in (0,1]$, which corresponds to the region II of Figs. 2 and 3. We can observe that $q_{I}=0$ when $\gamma \in [0,\gamma ^{o})$ and $q_{I}$ strictly increases with $\gamma$ when $\gamma \in [\gamma ^{o},1]$. On the other hand, notice that $\gamma \ge \gamma ^{*}$ always holds for any $\gamma \in [0,1]$ when $\gamma ^{*}<0$; thus, we have $q_{U}=1$, which is consistent with the results of Corollaries 1–2. (b) and (c) of Fig. 5 correspond to region III of Figs. 2 and 3. For uninformed customers, their equilibrium joining probability is increasing in $\gamma \in [0,\gamma ^{*})$ and we have $q_{U}=1$ when $\gamma \in [\gamma ^{*},1]$. For informed customers who find a busy server, they always balk when $\gamma \in [0,\gamma ^{o})$ and become more inclined to join when $\gamma \in [\gamma ^{o},1]$, i.e., $q_{I}$ is strictly increasing in $\gamma \in [\gamma ^{o},1]$. Finally, (d) of Fig. 5 corresponds to the region V. In this case, informed customers who find a busy server always balk and the joining probability for uninformed customers is decreasing in $\gamma$. By comparing the trend of $q_{U}$ in (c) and (d) of Fig. 5, we observe that $q_{U}$ is non-decreasing in $\gamma$ when $\Lambda$ is small, but is non-increasing in $\gamma$ when $\Lambda$ is large. In particular, when the demand $\Lambda$ is small (for example, $\Lambda =1.3,2.5,4$), the system is not too congested. As the expected utility for joining uninformed customers is higher than that of joining informed customers who find a busy server, the uninformed customers are more inclined to join the system when $\gamma$ increases. However, when $\Lambda$ is large (for example, $\Lambda =10$), the service capacity of system cannot meet the demands of customers. Because all informed customers who find an idle server will definitely join the system, when $\gamma$ increases, the total arrival rate to the system increases. Then the uninformed customers become more hesitant to join. That is to say, differently from the informed customers, the uninformed customer would take a different strategy as the best response to the increase of $\gamma$ under the different workloads.

1.2 A.2: Illustration of throughput and social welfare

We illustrate the impact of $\gamma$ on $q_{e}$ in Fig. 6, where $q_{e}=\lambda _{e}/\Lambda$, and $\lambda _{e}$ is the throughput of the system. Similarly to Fig. 5, four cases are presented ($\Lambda =1.3$, $\Lambda =2.5$, $\Lambda =4$, and $\Lambda =10$). In particular, we examine $q_{e}$ in three cases as follows: (1) The “unobservable” case depicts $q_{e}$ when the information is unavailable to all customers, independently of $\gamma$. (2) The “observable” case depicts $q_{e}$ when all arriving customers are given the information about the status of server, independently of $\gamma$. (3) The “heterogeneous” case depicts $q_{e}$ when $\gamma$ varies in [0, 1]. Therefore, we can observe that $q_{e}$ is constant $\gamma$ in cases (1)–(2), while $q_{e}$ varies with $\gamma$ in case (3).

Fig. 6

Effective joining probability versus $\gamma$

Numerical experiments show that when the arrival rate is relatively low ($\Lambda =1.3$), the service provider needs to hide the information to maximize the effective joining probability (i.e., $\gamma ^{t}=0$); see (a) of Fig. 6. This is because some customers join the system with strictly positive utility if they are informed, but all customers tolerate lower utility if they are uninformed. Thus, when the surplus of informed customers decreases (i.e., $\gamma$ decreases), the uninformed customers can have a higher probability of joining the system, which can result in a higher throughput. When the arrival rate is high ($\Lambda =10$), the joining probability is rather low in the unobservable case because all customers flinch at their arrivals by considering the externalities of others. However, in the observable case, customers will join the system with large potential arrival rate if the server is idle at arrivals. Thus, the disclosure of information is an effective strategy for the provider to attract arriving customers in this case (i.e., $\gamma ^{t}=1$); see (d) of Fig. 6.

Fig. 7

Social welfare versus $\gamma$ when $\mu =2$, $\theta =2$, $\nu =2.5$

However, when $\Lambda$ drops into an intermediate range, it is optimal for the provider to publish the information to a certain fraction of customers (i.e., $\gamma =\gamma ^{t}\in (0,1)$). That is, the throughput is unimodal in $\gamma \in [0,1]$. For example, $\gamma ^{t}\thickapprox 0.5$ when $\Lambda =2.5$ and $\gamma ^{t}\thickapprox 0.75$ when $\Lambda =4$; see (b) and (c) of Fig. 6. This shows that information heterogeneity is favorable for the service provider. By the way, we can observe that the throughput maximizer $\gamma ^{t}$ is increasing in the potential arrival $\Lambda$, which is consistent with our result in Proposition 3.

The corresponding social welfare is presented in Fig. 7. From Fig. 7, we can observe that no matter how the potential arrival rate $\Lambda$ changes, the social welfare is always non-decreasing in $\gamma$. When $\gamma =1$ (all customers are given the information about the server), the maximal social welfare can be attained. Recall that the optimal fraction of informed customers for throughput $\gamma ^{t}$ is increasing in $\Lambda$. When $\Lambda$ is small, we have $\gamma ^{t}=0$, but when $\Lambda$ is large, $\gamma ^{t}=1$. However, the social welfare maximizer $\gamma ^{\mathrm{soc}}$ is independent of $\Lambda$, and we always have that $\gamma ^{\mathrm{soc}}=1$.

1.3 A.3: Individual’s benefit and information fee

Fig. 8

Individual utility for customers versus $\gamma$

We illustrate the individual utility for customers and their difference between the two streams in Fig. 8 for $\Lambda =1.3$, $\Lambda =2.5$, $\Lambda =4$, and $\Lambda =10$. In Fig. 8, we can see that the utility for informed customers is higher than that for uninformed customers, i.e., $D(\gamma )\ge 0$ for $\gamma \in [0,1]$, which verifies our intuition that the value of information is nonnegative. Furthermore, the differences of utility are all non-increasing in $\gamma \in [0,1]$ for each case, which is consistent with our Lemma 7. In particular, when $\Lambda$ is small, informed customers and uninformed customers are more inclined to join this system; thus, the difference is too low (see (a) of Fig. 8). When $\Lambda$ is large, the utility for uninformed customers is 0; thus, the difference equals the utility of informed customers (see (d) of Fig. 8). Otherwise, when $\Lambda$ drops into an intermediate region, the changes of difference can be divided into three parts. Specifically, $D(\gamma )$ is decreasing in $\gamma$ slowly when $\gamma \in [0,\gamma ^{*})$ and declines rapidly when $\gamma \in [\gamma ^{*},\gamma ^{o})$. When $\gamma \in [\gamma ^{o},1]$, we have $D(\gamma )=0$.

In Fig. 8, we have that $\gamma ^{*}<0$ and $\gamma ^{*}>1$ in (a) and (d), respectively. Thus, the optimal strategy for the service provider is to conceal or reveal the information to arriving customers to maximize the throughput. In (b) and (c) of Fig. 8, $\gamma ^{*}$ drops into (0, 1). In this case, the throughput maximizer is attained at $\gamma ^{t}=\gamma ^{*}\in (0,1)$, in which we have ${\overline{S}}_{I}(q_{U},q_{I})\vert _{\gamma =\gamma ^{*}}>{\overline{S}}_{U}(q_{U},q_{I})\vert _{\gamma =\gamma ^{*}}=0$, i.e., $D(\gamma ^{*})>0$. This means that the information can bring a positive benefit to customers under the optimal fraction $\gamma ^{t}$.

1.4 A.4: The effect of $\theta$ on throughput and social welfare

Finally, we illustrate the throughput and social welfare in Fig. 9 when $\Lambda =2.5$. For $\gamma =0.1, 0.5, 0.7, 0.9$, we observe that the throughput is monotonically increasing in $\theta$. However, the social welfare is not monotone in $\theta$. When $\gamma =0.1$ and 0.9, we observe that the social welfare is decreasing in $\theta$ although it reduces the retrial times. This is because when $\theta$ is large, the throughput increases, which makes the system more congested, and the increased delay cannot compensate for the decreased delay on retrials. Furthermore, a different trend is presented when $\gamma =0.5$. Specifically, when $\theta$ is small ($\theta \in [1,2]$) or $\theta$ is large $(\theta \in [4,10])$, the social welfare is decreasing in $\theta$. When $\theta$ drops into the intermediate interval, the social welfare is increasing in $\theta$. That is to say, the improvement of $\theta$ will be not necessary to induce a higher social welfare even if it results in a shorter retrial time.

Fig. 9

Throughput and social welfare versus $\theta$ when $\Lambda =2.5$

In summary, although the social welfare is increasing in $\gamma$, it can be non-monotone in $\theta$. That is, in a retrial queueing system, the rate which maximizes the social welfare should be carefully selected. This observation also discloses the difference between our retrial queueing system and the classic queueing system.

Appendix B

Proof of Proposition 1

Multiplying equations (4.1)–(4.2) by $z_{1}^{j}z_{2}^{k}$ and summing over all j, k, we derive the following basic equations after some manipulations:

$$\begin{aligned}&\lambda _{0}\Pi _{0}(z_{1},z_{2})+\theta \left( \frac{\partial \Pi _{0}(z_{1},z_{2})}{\partial z_{1}}z_{1}+\frac{\partial \Pi _{0}(z_{1},z_{2})}{\partial z_{2}}z_{2}\right) = \mu \Pi _{1}(z_{1},z_{2}), \end{aligned}$$

(8.1)

$$\begin{aligned}&((1-z_{1})\Lambda _{U}q_{U}+(1-z_{2})\Lambda _{I}q_{I}+\mu )\Pi _{1}(z_{1},z_{2}) \nonumber \\&= \lambda _{0}\Pi _{0}(z_{1},z_{2}) + \theta \left( \frac{\partial \Pi _{0}(z_{1},z_{2})}{\partial z_{1}}+\frac{\partial \Pi _{0}(z_{1},z_{2})}{\partial z_{2}}\right) . \end{aligned}$$

(8.2)

Letting $z_{1}=1$ and substituting $\mu \Pi _{1}(1,z_{2})$ of (8.1) into (8.2), we can obtain

$$\begin{aligned} \Lambda _{I}q_{I}\Pi _{1}(1,z_{2})=\theta \frac{\partial \Pi _{0}(1,z_{2})}{\partial z_{2}}. \end{aligned}$$

(8.3)

By letting $z_{2}=1$, in a similar way, we have

$$\begin{aligned} \Lambda _{U}q_{U}\Pi _{1}(z_{1},1)=\theta \frac{\partial \Pi _{0}(z_{1},1)}{\partial z_{1}}. \end{aligned}$$

(8.4)

Letting $z_{1}=z_{2}=z$ and noticing that $\frac{\partial \Pi _{0}(z_{1},z_{2})}{\partial z_{1}}|_{z_{1}=z_{2}=z}+\frac{\partial \Pi _{0}(z_{1},z_{2})}{\partial z_{2}}|_{z_{1}=z_{2}=z}=\frac{\mathrm{d}\Pi _{0}(z,z)}{\mathrm{d}z}$, we have the following equation by substituting $\mu \Pi _{1}(z_{1},z_{2})$ of (8.1) into (8.2):

$$\begin{aligned} \theta \frac{d\Pi _{0}(z,z)}{dz}=\lambda _{1}\Pi _{1}(z,z). \end{aligned}$$

(8.5)

After eliminating $\Pi _{1}(z,z)$ by substituting (8.5) into (8.1), we get

$$\begin{aligned} \lambda _{1}\lambda _{0}\Pi _{0}(z,z)+\theta \frac{d\Pi _{0}(z,z)}{dz}(z\lambda _{1}-\mu )=0, \end{aligned}$$

with the solution

$$\begin{aligned} \Pi _{0}(z,z)=\frac{\mathrm{Const}}{\left( 1-\frac{\lambda _{1}}{\mu }z\right) ^{\frac{\lambda _{0}}{\theta }}}. \end{aligned}$$

(8.6)

Substituting (8.6) into (8.5) and using the normalizing condition $\Pi _{0}(1,1)+\Pi _{1}(1,1)=1$, we have

$$\begin{aligned} \Pi _{0}(z,z)= & {} \frac{\mu -\lambda _{1}}{\mu +\lambda _{0}-\lambda _{1}}\left( \frac{\mu -\lambda _{1}}{\mu -\lambda _{1}z}\right) ^{\frac{\lambda _{0}}{\theta }} \end{aligned}$$

(8.7)

$$\begin{aligned} \Pi _{1}(z,z)= & {} \frac{\lambda _{0}}{\mu +\lambda _{0}-\lambda _{1}}\left( \frac{\mu -\lambda _{1}}{\mu -\lambda _{1}z}\right) ^{2}\left( \frac{\mu -\lambda _{1}}{\mu -\lambda _{1}z}\right) ^{\frac{\lambda _{0}-\theta }{\theta }}. \end{aligned}$$

(8.8)

Thus, we can obtain (4.3) and (4.4) immediately. Noticing that $\Pi _{1}=\Pi _{1}(z_{1},1)|_{z_{1}=1}=\Pi _{1}(1,z_{2})|_{z_{2}=1}$, we substitute (4.4) into (8.3) and (8.4) to obtain the expected number of informed and uninformed customers when the server is idle, respectively. That is,

$$\begin{aligned} N_{0,I}= & {} \frac{\lambda _{0}\Lambda _{I}q_{I}}{\theta (\mu +\lambda _{0}-\lambda _{1})}, \end{aligned}$$

(8.9)

$$\begin{aligned} N_{0,U}= & {} \frac{\lambda _{0}\Lambda _{U}q_{U}}{\theta (\mu +\lambda _{0}-\lambda _{1})}. \end{aligned}$$

(8.10)

Next, we will derive the expected number of informed and uninformed customers when the server is busy, which are denoted by $N_{1,I}$ and $N_{1,U}$, respectively.

Letting $z_{1}=1$ and taking the derivative with respect to $z_{2}$ in (8.2), and taking $z_{2}=1$, we can obtain

$$\begin{aligned} -\Lambda _{I}q_{I}\Pi _{1}+\mu \frac{\partial \Pi _{1}(1,z_{2})}{\partial z_{2}}|_{z_{2}=1}= & {} \lambda _{0}\frac{\partial \Pi _{0}(1,z_{2})}{\partial z_{2}}|_{z_{2}=1}+\theta \frac{\partial ^{2}\Pi _{0}(z_{1},z_{2})}{\partial z_{1}\partial z_{2}}|_{z_{1}=z_{2}=1} \nonumber \\&+\,\theta \frac{\partial ^{2}\Pi _{0}(1,z_{2})}{\partial z_{2}^{2}}|_{z_{2}=1}. \end{aligned}$$

(8.11)

Differentiating with respect to $z_{2}$ in (8.3) and taking $z_{2}=1$, we have

$$\begin{aligned} \frac{\partial ^{2}\Pi _{0}(1,z_{2})}{\partial z_{2}^{2}}|_{z_{2}=1}=\frac{\Lambda _{I}q_{I}}{\theta }\frac{\partial \Pi _{1}(1,z_{2})}{\partial z_{2}}|_{z_{2}=1}. \end{aligned}$$

(8.12)

Substituting (8.12) into (8.11) and eliminating $\frac{\partial ^{2}\Pi _{0}(1,z_{2})}{\partial z_{2}^{2}}|_{z_{2}=1}$, we get

$$\begin{aligned} \frac{-\Lambda _{I}q_{I}\lambda _{0}}{\mu +\lambda _{0}-\lambda _{1}}+(\mu -\Lambda _{I}q_{I})N_{1,I}=\lambda _{0}N_{0,I}+\theta \frac{\partial ^{2}\Pi _{0}(z_{1},z_{2})}{\partial z_{1}\partial z_{2}}|_{z_{1}=z_{2}=1}. \end{aligned}$$

(8.13)

Analogously, we can obtain

$$\begin{aligned} \frac{-\Lambda _{U}q_{U}\lambda _{0}}{\mu +\lambda _{0}-\lambda _{1}}+(\mu -\Lambda _{U}q_{U})N_{1,U}=\lambda _{0}N_{0,U}+\theta \frac{\partial ^{2}\Pi _{0}(z_{1},z_{2})}{\partial z_{1}\partial z_{2}}|_{z_{1}=z_{2}=1}. \end{aligned}$$

(8.14)

Combining (8.13) and (8.14), we can eliminate $\frac{\partial ^{2}\Pi _{0}(z_{1},z_{2})}{\partial z_{1}\partial z_{2}}|_{z_{1}=z_{2}=1}$ and get that

$$\begin{aligned} \lambda _{0}(N_{0,I}-N_{0,U})= & {} \frac{(\Lambda _{U}q_{U}-\Lambda _{I}q_{I})\lambda _{0}}{\mu +\lambda _{0}-\lambda _{1}}+(\mu -\Lambda _{I}q_{I})N_{1,I}\nonumber \\&-(\mu -\lambda _{N}q_{N})N_{1,U}. \end{aligned}$$

(8.15)

Notice that $N_{1}=N_{1,I}+N_{1,U}$ and $N_{1}=\frac{d\Pi _{1}(z,z)}{dz}|_{z=1}$.

By differentiating with respect to z in (8.8), we have

$$\begin{aligned} N_{1,I}+N_{1,U}=\frac{(\lambda _{0}+\theta )\lambda _{1}\lambda _{0}}{\theta (\mu +\lambda _{0}-\lambda _{1})(\mu -\lambda _{1})}. \end{aligned}$$

(8.16)

Combining (8.9), (8.10), (8.15) and (8.16), we can derive $N_{1,U}$ and $N_{1,I}$ as follows:

$$\begin{aligned} N_{1,U}= & {} \frac{(\lambda _{0}+\theta )\Lambda _{U}q_{U}\lambda _{0}}{\theta (\mu +\lambda _{0}-\lambda _{1})(\mu -\lambda _{1})}, \end{aligned}$$

(8.17)

$$\begin{aligned} N_{1,I}= & {} \frac{(\lambda _{0}+\theta )\Lambda _{I}q_{I}\lambda _{0}}{\theta (\mu +\lambda _{0}-\lambda _{1})(\mu -\lambda _{1})}. \end{aligned}$$

(8.18)

So far, we can obtain the expected number of informed and uninformed customers as $N_{I}=N_{0,I}+N_{1,I}$ and $N_{U}=N_{0,U}+N_{1,U}$, as Eqs. (4.5) and (4.6) indicated. Because the effective arrival rate for uninformed customers is $\Lambda _{U}q_{U}$, the expected waiting time for them is $w_{U}=N_{U}/(\Lambda _{U}q_{U})$ from Little’s law (see Eq. (4.7)). For informed customers, their effective arrival rate to orbit is $\Pi _{1}\Lambda _{I}q_{I}$; with the help of Little’s law, we can obtain the expected waiting time $w_{I}$ similarly (see Eq. (4.8)). $\square$

Proof of Lemma 1

It is sufficient to prove that $w_{U}$ and $w_{I}$ are both increasing in $q_{U}$ and $q_{I}$ for $\Lambda _{U}q_{U}+\Lambda _{I}q_{I}<\mu$ and $0\le q_{I},q_{U}\le 1$. Notice that $\lambda _{0}=\Lambda _{U}q_{U}+\Lambda _{I}$ and $\lambda _{1}=\Lambda _{U}q_{U}+\Lambda _{I}q_{I}$; $w_{U}$ and $w_{I}$ can be written as

$$\begin{aligned} w_{U}= & {} \frac{\Lambda _{U}q_{U}+\Lambda _{I}}{\theta (\mu +\Lambda _{I}(1-q_{I}))}\cdot \frac{\mu +\Lambda _{I}(1-q_{I})+\theta }{\mu -(\Lambda _{U}q_{U}+\Lambda _{I}q_{I})}, \\ w_{I}= & {} \frac{1}{\theta }\cdot \frac{\mu +\Lambda _{I}(1-q_{I})+\theta }{\mu -(\Lambda _{U}q_{U}+\Lambda _{I}q_{I})}. \end{aligned}$$

Differentiating with respect to $q_{U}$ for a given $q_{I}$, we find that

$$\begin{aligned} \frac{\mathrm{d}w_{U}}{\mathrm{d}q_{U}}= & {} \frac{\mu +\Lambda _{I}(1-q_{I})+\theta }{\theta (\mu +\Lambda _{I}(1-q_{I}))} \frac{\Lambda _{U}(\mu +\Lambda _{I}(1-q_{I}))}{(\mu -(\Lambda _{U}q_{U}+\Lambda _{I}q_{I}))^{2}}>0, \\ \frac{\mathrm{d}w_{I}}{\mathrm{d}q_{U}}= & {} \frac{\mu +\Lambda _{I}(1-q_{I})+\theta }{\theta } \frac{\Lambda _{U}}{(\mu -(\Lambda _{U}q_{U}+\Lambda _{I}q_{I}))^{2}}>0. \end{aligned}$$

Similarly, by differentiating with respect to $q_{U}$ for a given $q_{I}$, we have

$$\begin{aligned} \frac{\mathrm{d}w_{U}}{\mathrm{d}q_{I}}= & {} \frac{\Lambda _{I}(\Lambda _{U}q_{U}+\Lambda _{I})}{\theta (\mu -(\Lambda _{U}q_{U}+\Lambda _{I}q_{I}))^{2}}\\&\left( \frac{\theta (\mu -(\Lambda _{U}q_{U}+\Lambda _{I}q_{I}))}{(\mu +\Lambda _{I}(1-q_{I}))^{2}}+\frac{\theta +\mu +\Lambda _{I}(1-q_{I})}{\mu +\Lambda _{I}(1-q_{I})}\right)>0, \\ \frac{\mathrm{d}w_{I}}{\mathrm{d}q_{I}}= & {} \frac{\Lambda _{I}}{\theta }\frac{\Lambda _{U}q_{U}+\Lambda _{I}+\theta }{(\mu -(\Lambda _{U}q_{U}+\Lambda _{I}q_{I}))^{2}}>0; \end{aligned}$$

thus, $w_{U}$ and $w_{I}$ are increasing in $q_{U}$ and $q_{I}$.

To show the ATC-type behavior, it is sufficient to prove that the best response functions $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} U_{U}({\hat{q}};q_{U},q_{I})$ and $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} U_{I}({\hat{q}};q_{U},q_{I})$ are both non-increasing in $q_{U}$ and $q_{I}$, respectively. Recall that $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} U_{U}({\hat{q}};q_{U},q_{I})=\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})$. If $S_{U}(q_{U},q_{I})>0$, we have $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})=1$. If $S_{U}(q_{U},q_{I})<0$, we have $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})=0$. Otherwise, if $S_{U}(q_{U},q_{I})=0$, we have $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})=q'\in [0,1]$. In summary, we have

$$\begin{aligned} \mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})=\left\{ \begin{array}{ll} 1, &{} \hbox {if}~{S_{U}(q_{U},q_{I})>0;} \\ q', &{} \hbox {if}~ {S_{U}(q_{U},q_{I})=0;} \\ 0, &{} \hbox {if}~{S_{U}(q_{U},q_{I})<0,} \end{array} \right. \end{aligned}$$

where $q'\in [0,1]$. Recall that $S_{U}(q_{U},q_{I})$ is decreasing in both $q_{U}$ and $q_{I}$, for a fixed $q_{I}\in [0,1]$; we consider the following three cases: (1) If $S_{U}(x,q_{I})<0$ for all $x\in [0,1]$, then $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})=0$, which is non-increasing in $q_{U}$. (2) If $S_{U}(x,q_{I})>0$ for all $x\in [0,1]$, then $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})=1$, which is also non-increasing in $q_{U}$. (3) Otherwise, if there exists an ${\hat{x}}\in [0,1]$ such that $S_{U}({\hat{x}},q_{I})=0$, then $S_{U}(q_{U},q_{I})>0$ if and only if $q_{U}\in [0,{\hat{x}})$. Thus, we have $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})=1$ for $q_{U}\in [0,{\hat{x}})$, $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})=q'\in [0,1]$ for $q_{U}={\hat{x}}$, and $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})=0$ for $q_{U}\in ({\hat{x}},1]$. As $1\ge q'\ge 0$, we can have that $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})$ is non-increasing in $q_{U}$ for any fixed $q_{I}\in [0,1]$. Similarly, we can show that $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} {\hat{q}}S_{U}(q_{U},q_{I})$ is non-increasing in $q_{I}$ for any fixed $q_{U}\in [0,1]$.

On the other hand, the monotonicity of $\mathop {\arg \max }_{{\hat{q}}\in [0,1]} U_{I}({\hat{q}};q_{U},q_{I})$ can be derived analogously, which completes this proof. $\square$

Proof of Lemma 2

It is not difficult to verify that $S_{U}(q_{U},q_{I})>S_{I}(q_{U},q_{I})$ for any strategy $(q_{U},q_{I})$ and fraction $\gamma \in [0,1]$, which can be used to reduce the range of possible equilibria. Based on that relation, we must have $U_{U}({\hat{q}};q_{U},q_{I})={\hat{q}}S_{U}(q_{U},q_{I})>{\hat{q}}S_{I}(q_{U},q_{I})=U_{I}({\hat{q}};q_{U},q_{I})$. When $(q_{U},q_{I})$ is an equilibrium strategy and $q_{I}>0$, based on the equilibrium conditions (3.3)–(3.4) we have that $S_{I}(q_{U},q_{I})\ge 0$ and $q_{U}S_{U}(q_{U},q_{I})>q_{U}S_{I}(q_{U},q_{I})\ge 0$. Thus, we can get $q_{U}=1$. (Otherwise, $S_{U}(q_{U},q_{I})$ can be decreased further to let more uninformed customers join.) That is to say, the equilibrium strategy pair $(q_{U},q_{I})$ should satisfy $q_{I}>0 \Rightarrow q_{U}=1$, i.e., $q_{U}<1 \Rightarrow q_{I}=0$. Also, notice that when $q_{U}=1$, we can have $q_{I}\in [0,1]$. Then we can focus our attention on the region $\Omega _{e}=\{(1,q_{I})\vert 0\le q_{I}\le 1\}\bigcup \{(q_{U},0)\vert q_{U}<1\}$ when we are seeking for the equilibrium strategy. $\square$

Proof of Proposition 2

It is readily found that $S_{U}(q_{U},q_{I})$ and $S_{I}(q_{U},q_{I})$ are strictly decreasing when $\Lambda _{U}q_{U}+\Lambda _{I}q_{I}<\mu$ from Lemma 1. We first consider the case that $\Lambda <\mu$. Notice that when $\Lambda <\mu$, we must have $\Lambda _{U}<\mu$. Then we consider the following cases:

• 1. If $S_{U}(1,0)>0$, we can obtain $q_{U}=1$ immediately because the behavior of an informed customer who finds a busy server is determined by the behavior of uninformed customers. Due to the monotonicity of $S_{I}(1,q)$ in q, we have the following three subcases:    (i) If $S_{I}(1,0)<0$, the maximum net benefit for an informed customer would be negative, the best response for an them is balking, and thus $q_{I}=0$ is the unique equilibrium.    (ii) If $S_{I}(1,0)\ge 0>S_{I}(1,1)$, there exists a unique solution of the equation $S_{I}(1,q)=0$ which lies in the interval (0, 1) and gives the second branch of (4.9).    (iii) If $S_{I}(1,1)\ge 0$, the best response for an informed customer is joining, and thus $q_{I}=1$, which derives the third branch of (4.9).

• 2. If $S_{U}(1,0)\le 0$, as we analyzed before, under equilibrium, we must have that $S_{U}(q_{U},q_{I})=0$, which implies that $q_{U}\le 1$ and $q_{I}=0$. Then the equilibrium can be attained when $S_{U}(q_{U},0)=0$. Because $S_{U}(q,0)$ is decreasing in $q\in [0,1]$, there exists a unique solution ${\hat{q}}=\frac{(\mu \nu \theta -\Lambda _{I})(\mu +\Lambda _{I})-\Lambda _{I}\theta }{\Lambda _{U}((\nu \theta +1)(\mu +\Lambda _{I})+\theta )}$ such that $S_{U}({\hat{q}},0)=0$. If the solution ${\hat{q}}$ lies in [0, 1], it characterizes an equilibrium strategy, i.e., $q_{U}={\hat{q}}$. Otherwise, if ${\hat{q}}<0$, we have $S_{U}(0,0)<0$, i.e., $q_{U}=0$. Thus, we derive the result of (4.11).

When $\Lambda \ge \mu$, from similar arguments we can get the following results:

$\bullet$ When $\Lambda _{U}<\mu$,

• 1. If $S_{U}(1,0)>0$, we have $q_{U}=1$ and

  $$\begin{aligned} q_{I}=\left\{ \begin{array}{ll} 0, &{} \hbox {if}~{S_{I}(1,0)<0;} \\ \frac{(\mu -\Lambda _{U})\theta \nu -(\mu +\Lambda _{I}+\theta )}{\Lambda _{I}(\theta \nu -1)}, &{} \hbox {if}~{S_{I}(1,0)\ge 0.} \end{array} \right. \end{aligned}$$

• 2. If $S_{U}(1,0)\le 0$, we have $q_{I}=0$ and

  $$\begin{aligned} q_{U}=\max \left\{ 0,\frac{(\mu \nu \theta -\Lambda _{I})(\mu +\Lambda _{I})-\Lambda _{I}\theta }{\Lambda _{U}((\nu \theta +1) (\mu +\Lambda _{I})+\theta )}\right\} . \end{aligned}$$

$\bullet$ When $\Lambda _{U}\ge \mu$, we have $q_{I}=0$ and

$$\begin{aligned} q_{U}=\max \left\{ 0,\frac{(\mu \nu \theta -\Lambda _{I})(\mu +\Lambda _{I})-\Lambda _{I}\theta }{\Lambda _{U}((\nu \theta +1)(\mu +\Lambda _{I})+\theta )}\right\} . \end{aligned}$$

Next, we will show that $q_{I}$ is non-decreasing in $\gamma$. When $q_{I}=0$ or $q_{I}=1$, it is obviously non-decreasing (constant) in $\gamma$. When $q_{I}=\frac{(\mu -\Lambda _{U})\theta \nu -(\mu +\Lambda _{I}+\theta )}{\Lambda _{I}(\theta \nu -1)}$, by taking the derivative of it with respect to $\gamma$, we have

$$\begin{aligned} \frac{\mathrm{d}q_{I}}{\mathrm{d}\gamma }=-\frac{(\mu -\Lambda )\theta \nu -(\mu +\theta )}{\gamma ^2(\theta \nu -1)\Lambda }. \end{aligned}$$

If $\Lambda \ge \mu$, we have $\frac{\mathrm{d}q_{I}}{\mathrm{d}\gamma }>0$. Otherwise, if $\Lambda <\mu$, notice that $S_{I}(1,1)<0\Leftrightarrow (\mu -\Lambda _{U})\theta \nu -(\mu +\Lambda _{I}+\theta )<0$, which also induces $\frac{\mathrm{d}q_{I}}{\mathrm{d}\gamma }>0$. By combining the results for $\Lambda <\mu$ and $\Lambda \ge \mu$, we can obtain Proposition 2, which completes this proof. $\square$

Proof of Lemma 3

Firstly, it is not difficult to verify that ${\overline{\mu }}^{o}>{\underline{\mu }}^{o}$ and ${\overline{\mu }}>{\underline{\mu }}$ when $\nu >1/\mu +1/\theta$. Then it is sufficient to prove that ${\overline{\mu }}^{o}>{\overline{\mu }}$ and ${\underline{\mu }}^{o}>{\underline{\mu }}$.

Let $x=\nu \theta -1>0$. We have ${\overline{\mu }}^{o}=\Lambda +\frac{\Lambda +\theta }{x}$ and ${\overline{\mu }}=\frac{\Lambda \big (x+2+\sqrt{(x+2)^{2}+\frac{4\theta (x+1)}{\Lambda }}\big )}{2(x+1)}$. Because $(x+2)^{2}+\frac{4\theta (x+1)}{\Lambda }=(x+2B)^2-4B^{2}+4B<(x+2B)^2$, where $B=1+\frac{\theta }{\Lambda }>1$, we have

$$\begin{aligned} {\overline{\mu }}<\frac{\Lambda (x+B+1)}{x+1}=\Lambda +\frac{\Lambda +\theta }{x+1}<\Lambda +\frac{\Lambda +\theta }{x}={\overline{\mu }}^{o}. \end{aligned}$$

On the other hand, notice that ${\overline{\mu }}={\underline{\mu }}+\Lambda$ and ${\overline{\mu }}^{o}={\underline{\mu }}^{o}+\Lambda$, and we have ${\underline{\mu }}^{o}>{\underline{\mu }}$, too.

When $\nu \rightarrow \infty$, we have ${\underline{\mu }}^{o}=\lim \limits _{x\rightarrow \infty }\frac{\Lambda +\theta }{x}=0$ and ${\overline{\mu }}{=}\lim \limits _{x\rightarrow \infty }\frac{\Lambda \big (x+2+\sqrt{(x+2)^{2}+\frac{4\theta (x+1)}{\Lambda }}\big )}{2(x+1)}=\Lambda$. Thus, this gives ${\underline{\mu }}^{o}<{\overline{\mu }}$, which completes our proof. $\square$

Proof of Lemma 4

From (4.4), we have $\lambda _{e}=\Pi _{1}\mu =\frac{\Lambda _{I}+\Lambda _{U}q_{U}}{\mu +\Lambda _{I}(1-q_{I})}\mu$. When $\gamma$ drops into different intervals, $q_{I}$ and $q_{U}$ are determined in Proposition 2. By substituting $q_{I}$ and $q_{U}$ into $\frac{\Lambda _{I}+\Lambda _{U}q_{U}}{\mu +\Lambda _{I}(1-q_{I})}$, we can derive $\lambda _{e}$ immediately. $\square$

Proof of Proposition 3

If $\gamma ^{o}<0$, $\lambda _{e}$ is a constant; see (5.1). Otherwise, from Lemma 4, when $\gamma \in [0,\gamma ^{*})$, the effective arrival rate in (5.3) is increasing in $\gamma$; when $\gamma ^{*}<\gamma <\gamma ^{o}$, the effective arrival rate is the first branch of (5.2), which is decreasing in $\gamma$. However, when $\gamma \ge \gamma ^{o}$, the effective arrival rate is a constant. Therefore, the maximal effective arrival rate can be attained at $\gamma ^{*}$. When $\gamma ^{*}\le 0$, the optimal fraction $\gamma ^{t}=0$; when $\gamma ^{*}\in (0,1)$, the optimal fraction $\gamma ^{t}=\gamma ^{*}$; when $\gamma ^{*}\ge 1$, the optimal fraction is $\gamma ^{t}=1$. To show that $\gamma ^{t}$ is non-decreasing in $\Lambda$, it is sufficient to prove that $\gamma ^{*}$ is increasing in $\Lambda$. By taking the derivative of $\gamma ^{*}$ with respect to $\Lambda$, we can get that

$$\begin{aligned} \frac{\mathrm{d}\gamma ^{*}(\Lambda )}{d\Lambda }= \frac{\mu -\frac{\theta }{\sqrt{\frac{4 \theta ^2 \nu }{\Lambda }+(1+\theta \nu )^2}}}{\Lambda ^2}. \end{aligned}$$

Notice that $\frac{\mathrm{d}\gamma ^{*}(\Lambda )}{d\Lambda }>0\Leftrightarrow \mu -\frac{\theta }{\sqrt{\frac{4 \theta ^2 \nu }{\Lambda }+(1+\theta \nu )^2}}>0\Leftrightarrow \frac{4 \theta ^2 \nu }{\Lambda }+(1+\theta \nu )^2\mu ^2>\theta ^2\Leftrightarrow \frac{1}{\Lambda }>\frac{\theta ^2/\mu ^2-(1+\theta \nu )^2}{4\theta ^2\nu }$. Recalling that $\nu >1/\mu +1/\theta$ (see the assumption at the beginning of Sect. 4.2), we have $\frac{\theta ^2/\mu ^2-(1+\theta \nu )^2}{4\theta ^2\nu }<0$, which implies that $\frac{1}{\Lambda }>0>\frac{\theta ^2/\mu ^2-(1+\theta \nu )^2}{4\theta ^2\nu }$. Therefore, we have $\frac{\mathrm{d}\gamma ^{*}(\Lambda )}{d\Lambda }>0$ for all $\Lambda >0$, which completes this proof. $\square$

Proof of Lemma 5

Similarly to Lemma 4, the social welfare can be easily derived by substituting $q_{I}$ and $q_{U}$ into $S_{\mathrm{soc}}=\Lambda _{I}{\overline{S}}_{I}(q_{U},q_{I})+\Lambda _{U}{\overline{S}}_{U}(q_{U},q_{I})$. $\square$

Proof of Proposition 4

If $\gamma ^{o}<0$, the $S_{\mathrm{soc}}$ is constant in $\gamma$; see (5.6). We will discuss the monotonicity of $\gamma$ in [0, 1] as follows: When $\gamma \in [0,\gamma ^{*})$, the $S_{\mathrm{soc}}$ in (5.8) is obviously increasing in $\gamma$ because its numerator and denominator are increasing and decreasing in $\gamma$, respectively. When $\gamma \in [\gamma ^{*},1]$ and $\gamma \ge \gamma ^{o}$, $S_{\mathrm{soc}}$ is a constant; see the second branch of (5.7). Thus, it is sufficient to show that $S_{\mathrm{soc}}$ is increasing in $\gamma \in [\gamma ^{*},\gamma ^{o})$.

When $\gamma \in [\gamma ^{*},\gamma ^{o})$, we define $x=\mu +\Lambda \gamma$ and $S_{\mathrm{soc}}=f_{3}(x)=\Lambda C\left[ \frac{\mu \nu }{x}-\frac{(\mu +\Lambda -x)(x+\theta )}{\theta x(x-\Lambda )}\right]$, where $x\in [\mu +\mu \gamma ^{*},\mu +\Lambda \gamma ^{o})$. Taking the derivative of $f_{3}(x)$ with respect to x, we have

$$\begin{aligned} \frac{\mathrm{d}f_{3}(x)}{\mathrm{d}x}=\frac{\Lambda C[-x^{2}(\theta -\mu +\theta \mu \nu )+2x\theta (\Lambda +\mu +\Lambda \mu \nu )-\theta \Lambda (\Lambda +\mu +\Lambda \mu \nu )]}{x^{2} \theta (x-\Lambda )^{2}}. \end{aligned}$$

It is sufficient to prove $\frac{\mathrm{d}f_{3}(x)}{\mathrm{d}x}>0$. It is obvious that the denominator is positive; then, we just need to discuss the sign of numerator when $x\in [\mu +\Lambda \gamma ^{*},\mu +\Lambda \gamma ^{o})$. Considering the quadratic function $g_{3}(x)=-x^{2}(\theta -\mu +\theta \mu \nu )+2x\theta (\Lambda +\mu +\Lambda \mu \nu )-\theta \Lambda (\Lambda +\mu +\Lambda \mu \nu )$, we have $\theta -\mu +\theta \mu \nu >0$ because $\nu >1/\mu +1/\theta$; then, this is a parabola pointing downwards, which is illustrated in Fig. 10. Its two roots are given as follows:

$$\begin{aligned} x_{1}= & {} \frac{\theta (\Lambda +\mu +\Lambda \mu \nu )-\sqrt{\theta (\theta +\Lambda )(\Lambda +\mu +\Lambda \mu \nu )\mu }}{\theta -\mu +\theta \mu \nu }, \\ x_{2}= & {} \frac{\theta (\Lambda +\mu +\Lambda \mu \nu )+\sqrt{\theta (\theta +\Lambda )(\Lambda +\mu +\Lambda \mu \nu )\mu }}{\theta -\mu +\theta \mu \nu }. \end{aligned}$$

Fig. 10

Illustration of $g_{3}(x)$

Because $\theta -\mu +\theta \mu \nu >0$ and $\Lambda +\mu +\Lambda \mu \nu >0$, the symmetric axis of $g_{3}(x)$ is on the right side of plane, which implies that $x_{2}>x_{1}>0$. Thus, it is sufficient to prove that $\mu +\Lambda \gamma ^{*}\ge x_{1}$ and $\mu +\Lambda \gamma ^{o}\le x_{2}$, in which case $g_{3}(x)>0$ when $x\in [\mu +\Lambda \gamma ^{*},\mu +\Lambda \gamma ^{o})$. Based on the definition of $\gamma ^{*}$ and $\gamma ^{o}$, we have

$$\begin{aligned} \mu +\Lambda \gamma ^{*}= & {} \mu +\frac{\Lambda }{2}+\frac{\Lambda }{2\theta \nu }+\frac{\Lambda }{2}\sqrt{(1+\frac{1}{\theta \nu })^{2}+\frac{4}{\nu \Lambda }}, \end{aligned}$$

(8.19)

$$\begin{aligned} \mu +\Lambda \gamma ^{o}= & {} \frac{\theta (\nu \Lambda +1)}{\nu \theta -1}. \end{aligned}$$

(8.20)

Notice that $x_{1}=\frac{\Lambda \sqrt{\theta (\Lambda +\mu +\Lambda \mu \nu )}}{\sqrt{\theta (\Lambda +\mu +\Lambda \mu \nu )}+\sqrt{\mu (\theta +\Lambda )}}<\Lambda<\frac{\Lambda }{2}+\frac{\Lambda }{2}\sqrt{(1+\frac{1}{\theta \nu })^{2}+\frac{4}{\nu \Lambda }}<\mu +\Lambda \gamma ^{*}$ (see (8.19)); thus, $\mu +\Lambda \gamma ^{*}> x_{1}$. Next, we will prove that $x_{2}>\mu +\Lambda \gamma ^{o}$.

Because $x_{2}=\frac{\Lambda \sqrt{\theta (\Lambda +\mu +\Lambda \mu \nu )}}{\sqrt{\theta (\Lambda +\mu +\Lambda \mu \nu )}-\sqrt{\mu (\theta +\Lambda )}}$, $\mu +\Lambda \gamma ^{o}=\Lambda +\frac{\Lambda +\theta }{\theta \nu -1}$ (see Eq. (8.20)), it is sufficient to prove that

$$\begin{aligned}&\Lambda \sqrt{\frac{(\theta +\Lambda )\mu }{\theta (\Lambda +\mu +\Lambda \mu \nu )}}>\frac{\Lambda +\theta }{\theta \nu -1}(1-\sqrt{\frac{(\theta +\Lambda )\mu }{\theta (\Lambda +\mu +\Lambda \mu \nu )}}) \nonumber \\&\Longleftrightarrow \frac{\theta (\nu \Lambda +1)}{\theta \nu -1}\sqrt{\frac{(\theta +\Lambda )\mu }{\theta (\Lambda +\mu +\Lambda \mu \nu )}}>\frac{\Lambda +\theta }{\theta \nu -1} \nonumber \\&\Longleftrightarrow \theta \mu \nu (\nu \Lambda +1)>\Lambda +\mu (\nu \Lambda +1)+\theta \nonumber \\&\Longleftrightarrow (\nu \Lambda +1)(\theta \nu -1)\mu -(\Lambda +\theta )>0. \end{aligned}$$

(8.21)

It is easy to verify that $(\nu \Lambda +1)(\theta \nu -1)\mu -(\Lambda +\theta )$ is increasing in $\nu$. Because $\nu >1/\mu +1/\theta$, by taking $\nu =1/\mu +1/\theta$ in (8.21), we find that $(\Lambda +{\Lambda \theta }/{\mu }+\theta )-(\Lambda +\theta )={\Lambda \theta }/{\mu }>0$, which verifies inequality (8.21); thus, we complete this proof. $\square$

Proof of Lemma 6

Under equilibrium, an individual uninformed customer receives a nonzero utility only if $q_{U}=1$. When $q_{U}<1$, we have $q_{I}=0$ and ${\overline{S}}_{U}(q_{U},0)=0$; thus, it is easy to derive that ${\overline{S}}_{I}(q_{U},0)>{\overline{S}}_{U}(q_{U},0)=0$. When $q_{U}=1$, the gap $D(1,q_{I})={\overline{S}}_{I}(1,q_{I})- {\overline{S}}_{U}(1,q_{I})$ can be written as

$$\begin{aligned} D(1,q_{I})=\frac{(1-q_{I})(\Lambda _{I}+\Lambda _{U})}{\mu +(1-q_{I})\Lambda _{I}}\cdot \left( \frac{\mu +(1-q_{I})\Lambda _{I}+\theta }{\theta [\mu -\Lambda +(1-q_{I})\Lambda _{I}]}C-R\right) . \end{aligned}$$

• 1. When $q_{I}=1$, we obviously have $D(1,1)=0$.

• 2. When $0\le q_{I}< 1$, it is sufficient to prove that $D(1,q_{I})\ge 0$, i.e.,

  $$\begin{aligned} \nu \le \frac{\mu +(1-q_{I})\Lambda _{I}+\theta }{\theta (\mu -\Lambda +(1-q_{I})\Lambda _{I})}. \end{aligned}$$

  (8.22)

• (i) If $q_{I}=0$, the equilibrium strategy for uninformed and informed customers is $q_{U}=1$ and $q_{I}=0$, respectively. Notice that the equilibrium condition that $(q_{U},q_{I})=(1,0)$ is $S_{I}(1,0)\le 0\Leftrightarrow \nu \le \frac{\mu +\Lambda _{I}+\theta }{\theta (\mu -\Lambda _{U})}$ (see Proposition 2), which satisfies the inequality (8.22).

• (ii) If $0<q_{I}<1$, we have $S_{I}(1,0)>0>S_{I}(1,1)$, thus we have $\nu =\frac{\mu +(1-q_{I})\Lambda _{I}+\theta }{\theta (\mu -\Lambda +(1-q_{I})\Lambda _{I})}$, then $D(1,q_{I})=0$, which also satisfies inequality (8.22). To sum up , when $q_{U}=1$, we always have $D(1,q_{I})\ge 0$, which completes our proof. $\square$

Proof of Lemma 7

When $\gamma ^{*}\in (0,1)$, two cases with respect to $\gamma ^{o}$ need to be discussed to investigate the monotonicity of $D(\gamma )$ in $\gamma$:

• 1. If $\gamma ^{o} \ge 1$, $\gamma$ can be divided into two subcases:    (i) When $\gamma \in [0,\gamma ^{*})$, $q_{U}=\max \big \{0,\frac{(\mu \nu \theta -\Lambda _{I})(\mu +\Lambda _{I})-\Lambda _{I}\theta }{\Lambda _{U}((\nu \theta +1)(\mu +\Lambda _{I})+\theta )}\big \}$ and $q_{I}=0$. By substituting them into (5.4) and (5.5), we have

  $$\begin{aligned} D(\gamma )=\left\{ \begin{array}{ll} \frac{(\mu +\Lambda \gamma +\theta )R}{(\nu \theta +1)(\mu +\Lambda \gamma )+\theta }, &{} \hbox {if}~ {(\mu \nu \theta -\Lambda _{I})(\mu +\Lambda _{I})\ge \Lambda _{I}\theta ;} \\ \frac{\mu R}{\mu +\Lambda \gamma }, &{} \hbox {if} ~{(\mu \nu \theta -\Lambda _{I})(\mu +\Lambda _{I})<\Lambda _{I}\theta .} \end{array} \right. \end{aligned}$$

  By taking the derivative with respect to $\gamma$ in the two branches above, respectively, we can obtain

  $$\begin{aligned} \frac{dD(\gamma )}{\mathrm{d}\gamma }=\left\{ \begin{array}{ll} -\frac{\nu \Lambda }{((\nu \theta +1)(\mu +\Lambda \gamma )+\theta )^{2}}<0, &{} \hbox {if}~{(\mu \nu \theta -\Lambda _{I})(\mu +\Lambda _{I})\ge \Lambda _{I}\theta ;}\\ -\frac{\Lambda \mu R}{(\mu +\Lambda \gamma )^{2}}<0, &{} \hbox {if}~{(\mu \nu \theta -\Lambda _{I})(\mu +\Lambda _{I})<\Lambda _{I}\theta .} \end{array} \right. \end{aligned}$$

  Thus, it is obvious that $D(\gamma )$ is decreasing in $[0,\gamma ^{*})$.    (ii) When $\gamma \in \left[ \gamma ^{*},1\right]$, we have $q_{U}=1$ and $q_{I}=0$, and then $D(\gamma )=\frac{C\Lambda }{\mu +\Lambda \gamma }\big (\frac{\mu +\Lambda \gamma +\theta }{\theta (\mu +\Lambda \gamma -\Lambda )}-\nu \big )$. It is intuitive that $\frac{C\Lambda }{\mu +\Lambda \gamma }$ is decreasing in $[\gamma ^{*},1]$, and then it is sufficient to prove that $\frac{\mu +\Lambda \gamma +\theta }{\theta (\mu +\Lambda \gamma -\Lambda )}-\nu$ is decreasing in $\gamma \in [\gamma ^{*},1]$ because $\frac{\mu +\Lambda \gamma +\theta }{\theta (\mu +\Lambda \gamma -\Lambda )}-\nu >0$. Define $g(\gamma )=\frac{\mu +\Lambda \gamma +\theta }{\theta (\mu +\Lambda \gamma -\Lambda )}-\nu$, and notice that $\frac{dg(\gamma )}{\mathrm{d}\gamma }=-\frac{(\Lambda +\theta )\Lambda }{(\mu +\Lambda \gamma -\Lambda )^{2}}<0$, and we have that $D(\gamma )$ is decreasing in $(\gamma ^{*},1]$.

• 2. If $\gamma ^{o} <1$, $\gamma$ can be divided into three subcases:    (i) When $\gamma \in [0,\gamma ^{*})$, the monotonicity of $D(\gamma )$ can be derived by the same argument as in (i) of case (1).    (ii) When $\gamma \in [\gamma ^{*},\gamma ^{o})$, we have $q_{U}=1$ and $q_{I}=0$, and the monotonicity of $D(\gamma )$ can be derived in the same way as (ii) of case (1).    (iii) When $\gamma \in [\gamma ^{o},1]$, we have $q_{U}=1$ and $q_{I}>0$. Through the analysis in Lemma 6, we have $D(1,q_{I})=0$ for any $q_{I}>0$. Thus, in this case, $D(\gamma )$ is constant in $\gamma$. In summary, if $\gamma ^{o}\ge 1$, $D(\gamma )$ is strictly decreasing in $\gamma \in [0,1]$. If $\gamma ^{o}<1$, $D(\gamma )$ is strictly decreasing in $\gamma \in [0,\gamma ^{o})$ and constant in $[\gamma ^{o},1]$.$\square$

Proof of Proposition 5

With the information fee, we first need to prove the proportion $\gamma ^{*}$ who buy information which is chosen by customers is an equilibrium. Notice that ${\overline{S}}_{U}(q_{U},q_{I})\vert _{\gamma =\gamma ^{*}}=0$, when $f={\overline{S}}_{I}(q_{U},q_{I})\vert _{\gamma =\gamma ^{*}}$, we have ${\overline{S}}_{I}(q_{U},q_{I})\vert _{\gamma =\gamma ^{*}}-f={\overline{S}}_{U}(q_{U},q_{I})\vert _{\gamma =\gamma ^{*}}=0$, the left-hand side is the real utility for informed customers and the right-hand side is the utility for uninformed customers. That is, when all others follow the strategy $\gamma ^{*}$ to buy information, it is indifferent for the individual to buy information or not because his utility is $\gamma [{\overline{S}}_{I}(q_{U},q_{I})\vert _{\gamma =\gamma ^{*}}-f]+(1-\gamma ){\overline{S}}_{U}(q_{U},q_{I})\vert _{\gamma =\gamma ^{*}}=0$. Thus, all strategies $\gamma \in [0,1]$ for the individual are the best response to others, and then $\gamma ^{*}$ is an equilibrium strategy for customers.

Next, we should prove that under this information fee, the equilibrium is unique. If there exists $\gamma ^{'}<\gamma ^{*}$, recalling that $\gamma ^{*}<\gamma ^{o}$, then $D(\gamma )$ is strictly decreasing in $[0,\gamma ^{*}]$ (see Lemma 7) and $D(\gamma ^{*})-f={\overline{S}}_{I}(q_{U},q_{I})\vert _{\gamma =\gamma ^{*}}-f=0$; we have $D(\gamma ^{'})-f>D(\gamma ^{*})-f=0$. Since customers are rational, the uninformed customers have an incentive to buy information and to deviate from $\gamma ^{'}$ to $\gamma ^{*}$, so $\gamma ^{'}$ is not an equilibrium strategy. Similarly, if $\gamma ^{'}>\gamma ^{*}$, we have $D(\gamma ^{'})-f<D(\gamma ^{*})-f=0$; thus, the informed customers have an incentive to give up buying information and to deviate from $\gamma ^{'}$ to $\gamma ^{*}$, too. Therefore, $\gamma ^{*}$ is the unique equilibrium strategy for customers. $\square$

Appendix C

This section is a supplement to Sect. 7. In this section, we first provide the proof of Proposition 6. Next, we will demonstrate properties of the equilibrium strategy and the expected individual benefits for both uninformed and informed customers. Finally, numerical examples are given to verify that the key insights of our base model can still hold in the generalized case.

Proof of Proposition 6

Denote by $w_{U}^{o}$ and $w_{I}^{o}$ the expected orbiting times of uninformed and informed customers who join the orbit, respectively. Notice that the arrival rate of uninformed customers to the orbit is $\Lambda _{U}q_{U}$. Thus, for a joining uninformed customer, her expected waiting time in the orbit is

$$\begin{aligned} (1-\Pi _{K})\cdot 0+\Pi _{K}\cdot w_{U}^{o}=\frac{N_{U}}{\Lambda _{U}q_{U}}. \end{aligned}$$

Similarly, for a joining informed customer, her expected waiting time in the orbit is

$$\begin{aligned} (1-\Pi _{K})\cdot 0+\Pi _{K}\cdot w_{I}^{o}=\frac{N_{I}}{\Lambda _{I}q_{I}}. \end{aligned}$$

Notice that since the customers (both uninformed and informed) in orbit are symmetric (share the same retrial rate $\theta$), then we have that $w_{U}^{o}=w_{I}^{o}$ based on their definitions. This gives that $\frac{N_{U}}{\Lambda _{U}q_{U}}=\frac{N_{I}}{\Lambda _{I}q_{I}}$, which implies that $\frac{N_{U}}{\Lambda _{U}q_{U}}=\frac{N_{I}}{\Lambda _{I}q_{I}}=\frac{N_{U}+N_{I}}{\Lambda _{U}q_{U}+\Lambda _{I}q_{I}}$. Therefore, to derive the expected orbiting time for a joining customer, it is sufficient to investigate the total number of customers in orbit, i.e., $N_{U}+N_{I}$. Then we can establish a two-dimensional continuous-time Markov chain {(L(t), N(t)); $t\ge 0$}, where L(t) and N(t) are the numbers of customer in the waiting line and in the orbit, respectively. We still use $\lambda _{0}\equiv \Lambda _{I}+\Lambda _{U}q_{U}$ and $\lambda _{1}\equiv \Lambda _{I}q_{I}+\Lambda _{U}q_{U}$ to denote the arrival rates of customers to the system when the waiting line is available and unavailable, respectively. Let $p_{(i,j)}$ be the steady-state probability that $P(L(\infty )=i,N(\infty )=j)$; we have the following balance equations for $1\le i\le K-1$ and $j\ge 0$:

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

(8.23)

$$\begin{aligned} (\lambda _{0}+\mu +j\theta )p_{(i,j)}= & {} \mu p_{(i+1,j)}+(j+1)\theta p_{(i-1,j+1)}+\lambda _{0}p_{i-1,j}, \end{aligned}$$

(8.24)

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

(8.25)

The following generating functions are defined to solve the equations above:

$$\begin{aligned} \Pi _{i}(z) = \sum _{j=0}^{\infty }z^{j}p_{(i,j)},~~~0\le i\le K. \end{aligned}$$

To simplify the notation, we define $\Pi _{i}\equiv \Pi _{i}(1)$ for $0\le i\le K$. Multiplying Eqs. (8.23)–(8.25) by $z^{j}$ and summing over all $j\ge 0$, we can derive the following basic equations after some manipulations:

$$\begin{aligned} \lambda _{0}\Pi _{0}(z)+z\theta _{0}\Pi _{0}'(z)= & {} \mu \Pi _{1}(z), \end{aligned}$$

(8.26)

$$\begin{aligned} (\lambda _{0}+\mu )\Pi _{i}(z)+z\theta _{0}\Pi _{i}'(z)= & {} \mu \Pi _{i+1}(z)+\theta _{0}\Pi _{i-1}'(z)+\lambda _{0}\Pi _{i-1}(z), \end{aligned}$$

(8.27)

$$\begin{aligned} (\lambda _{1}-\lambda _{1}z+\mu )\Pi _{K}(z)= & {} \theta \Pi _{K-1}'(z)+\lambda _{0}\Pi _{K-1}(z). \end{aligned}$$

(8.28)

From (8.26), we can obtain that

$$\begin{aligned} \Pi _{0}'(1)=\frac{\mu \Pi _{1}-\lambda _{0}\Pi _{0}}{\theta }. \end{aligned}$$

(8.29)

Multiplying equations in (8.27) and (8.28) by $z^{i}$ for $1\le i\le K-1$ and $z^{K}$, respectively, and then summing over (8.26)–(8.28), we have

$$\begin{aligned}&\lambda _{0}\Pi _{0}(z)+(\lambda _{0}+\mu )\sum _{i=1}^{K-1}\Pi _{i}(z)z^{i}+(\lambda _{1}-\lambda _{1}z+\mu )\Pi _{K}(z)z^{K} \nonumber \\&=\mu \Pi _{1}(z)+\sum _{i=1}^{K-1}(\mu \Pi _{i+1}(z)+\lambda _{0}\Pi _{i-1}(z))z^{i}+\lambda _{0}\Pi _{K-1}(z)z^{K} \nonumber \\&\Leftrightarrow \lambda _{0}(1-z)\Pi _{0}(z)+\sum _{i=1}^{K}(\lambda _{0}z^{i}(1-z)\Pi _{i}(z)-\mu z^{i-1}(1-z)\Pi _{i}(z)).\nonumber \\&\quad +\,(\lambda _{1}-\lambda _{0})z^{K}(1-z)\Pi _{K}(z)=0 \nonumber \\&\Leftrightarrow \lambda _{0}\Pi _{0}(z)+\sum _{i=1}^{K}(\lambda _{0}z-\mu )z^{i-1}\Pi _{i}(z)+(\lambda _{1}-\lambda _{0})z^{K}\Pi _{K}(z)=0. \end{aligned}$$

(8.30)

By letting $z=1$ in (8.30), we can derive the relationship between $\Pi _{0}$ and $\Pi _{K}$:

$$\begin{aligned} \Pi _{K}=\frac{\lambda _{0}-\mu +\mu \Pi _{0}}{\lambda _{0}-\lambda _{1}}. \end{aligned}$$

Furthermore, by taking the derivative with respect to z in (8.30) and letting $z=1$, we get

$$\begin{aligned}&\lambda _{0}\Pi _{0}'(1)+\sum _{i=1}^{K}((i\lambda _{0}-(i-1)\mu )\Pi _{i}+(\lambda _{0}-\mu )\Pi _{i}'(1))+(\lambda _{1}-\lambda _{0})(K\Pi _{K}+\Pi _{K}'(1))=0 \nonumber \\&\Leftrightarrow \Pi _{K}'(1)=\frac{\mu \Pi _{0}'(1)+\sum _{i=1}^{K}(i\lambda _{0}-(i-1)\mu )\Pi _{i}+(\lambda _{0}-\mu )\sum _{i=0}^{K-1}\Pi _{i}'(1)+ (\lambda _{1}-\lambda _{0})K\Pi _{K}}{\mu -\lambda _{1}}.\nonumber \\ \end{aligned}$$

(8.31)

Also, by taking the derivative with respect to z in (8.28) and letting $z=1$, we have

$$\begin{aligned} \mu \Pi _{K}'(1)=\lambda _{1}\Pi _{K}+\lambda _{0}+\Pi _{K-1}'(1)+\theta \Pi _{K-1}''(1). \end{aligned}$$

(8.32)

Similarly, by taking the derivative with respect to z in (8.26)–(8.27) and letting $z=1$, after some algebra we can obtain that

$$\begin{aligned} (\Pi _{1}''(1)-\Pi _{0}''(1))\theta= & {} \mu \Pi _{2}''(1)+\lambda _{0}\Pi _{0}''(1)-(\lambda _{0}+\mu +\theta )\Pi _{1}''(1), \\&\vdots&\\ (\Pi _{K-1}''(1)-\Pi _{K-2}''(1))\theta= & {} \mu \Pi _{K}''(1)+\lambda _{K-2}\Pi _{0}''(1)-(\lambda _{0}+\mu +\theta )\Pi _{K-1}''(1); \end{aligned}$$

thus, this gives

$$\begin{aligned} \mu \Pi _{K}'(1)=\theta \sum _{i=0}^{K-1}\Pi _{i}'(1)+\lambda _{0}+\Pi _{K-1}'(1)+\theta \Pi _{K-1}''(1). \end{aligned}$$

(8.33)

Combining (8.32) and (8.33), we can get $\lambda _{1}\Pi _{K}=\theta \sum _{i=0}^{K-1}\Pi _{i}'(1)$. This is intuitive because under steady state, the effective arrival rate to the orbit equals the output of the orbit itself. Therefore, combining (8.29), the $\Pi _{K}'(1)$ in (8.31) can be rewritten as

$$\begin{aligned} \Pi _{K}'(1)=\frac{\mu (\mu \Pi _{1}-\lambda _{0}\Pi _{0})+\theta \sum _{i=1}^{K}(i\lambda _{0}-(i-1)\mu )\Pi _{i}+[(\lambda _{0}-\mu )\lambda _{1}+\theta (\lambda _{1}-\lambda _{0})K]\Pi _{K} }{\theta (\mu -\lambda _{1})}. \end{aligned}$$

Then the total number of customers in the orbit can be expressed by $N_{\mathrm{orbit}}=\sum _{i=0}^{K}\Pi _{i}'(1)=\Pi _{K}'(1)+\lambda _{1}\Pi _{K}/\theta$, i.e.,

$$\begin{aligned} N_{\mathrm{orbit}}=\frac{\mu (\mu \Pi _{1}-\lambda _{0}\Pi _{0})+\theta \sum _{i=1}^{K}(i\lambda _{0}-(i-1)\mu )\Pi _{i}+[(\lambda _{0}-\lambda _{1})\lambda _{1}+\theta (\lambda _{1}-\lambda _{0})K]\Pi _{K} }{\theta (\mu -\lambda _{1})}. \end{aligned}$$

Thus, we can get $w_{U}^{o}=w_{I}^{o}=\frac{N_{\mathrm{orbit}}}{(\Lambda _{I}q_{I}+\Lambda _{U}q_{U})\Pi _{K}}$. Denote by $w_{L}$ the expected waiting time in line for customers; this should satisfy $w_{L}=\Pi _{0}\cdot \frac{0}{\mu }+\Pi _{1}\cdot \frac{1}{\mu }+\cdots +\Pi _{K-1}\cdot \frac{K-1}{\mu }+\Pi _{K} w_{L}$, which gives $w_{L}=(\sum _{i=0}^{K-1}i\cdot \Pi _{i})/(\mu \sum _{i=0}^{K-1}\Pi _{i})$. Therefore, the expected waiting times for uninformed customers are $w_{U} = \Pi _{K}\cdot \left( \frac{N_{\mathrm{orbit}}}{(\Lambda _{I}q_{I}+\Lambda _{U}q_{U})\Pi _{K}}+w_{L}\right) +\sum _{i=0}^{K-1}i\cdot \Pi _{i}/\mu =\frac{N_{\mathrm{orbit}}}{\Lambda _{I}q_{I}+\Lambda _{U}q_{U}}+\frac{\sum _{i=0}^{K-1}i\cdot \Pi _{i}}{\mu \sum _{i=0}^{K-1}\Pi _{i}}$, and the expected waiting time for informed customers who find a full waiting line (i.e., $L(t)=K$) is $w_{I}=\frac{N_{\mathrm{orbit}}}{(\Lambda _{I}q_{I}+\Lambda _{U}q_{U})\Pi _{K}}+\frac{\sum _{i=0}^{K-1}i\cdot \Pi _{i}}{\mu \sum _{i=0}^{K-1}\Pi _{i}}$, which completes this proof. $\square$

Notice that the expected net benefits for an uninformed customer and an informed customer who find a full waiting line and decides to enter the orbit are

$$\begin{aligned} S_{U}(q_{U},q_{I})= & {} R-w_{U}C, \\ S_{I}(q_{U},q_{I})= & {} R-w_{I}C, \end{aligned}$$

respectively. It is intuitive to find that $S_{U}(q_{U},q_{I})>S_{I}(q_{U},q_{I})$ because $w_{U}<w_{I}$. Denote by $(q_{U},q_{I})$ the equilibrium pair of customers; we have the following lemma immediately.

Lemma 8

Under equilibrium, we have $q_{U}\ge q_{I}$. In particular, if $q_{I}>0$, then $q_{U}=1$.

Proof

Because $S_{U}(q_{U},q_{I})>S_{I}(q_{U},q_{I})$ for any strategy $(q_{U},q_{I})$ and fraction $\gamma \in [0,1]$, $U_{U}({\hat{q}};q_{U},q_{I})={\hat{q}}S_{U}(q_{U},q_{I})>{\hat{q}}S_{I}(q_{U},q_{I})=U_{I}({\hat{q}};q_{U},q_{I})$. When $(q_{U},q_{I})$ is an equilibrium strategy, if $q_{I}>0$, based on the equilibrium conditions (3.3)–(3.4), we have that $S_{I}(q_{U},q_{I})\ge 0$ and $q_{U}S_{U}(q_{U},q_{I})>q_{U}S_{I}(q_{U},q_{I})\ge 0$. Thus, we can get $q_{U}=1$. (Otherwise, the expected utility can be decreased further to let more uninformed customers join.) That is to say, the equilibrium strategy pair $(q_{U},q_{I})$ should satisfy $q_{I}>0 \Rightarrow q_{U}=1$, i.e., $q_{U}<1 \Rightarrow q_{I}=0$. It is obvious that $q_{U}\ge q_{I}$. $\square$

Lemma 8 is similar to Lemma 2 in our base model, which helps to reduce the possible region of equilibria. Also, it should be noted that the system can be stable if and only if $\mu >\lambda _{1}$, i.e., $\mu >\Lambda _{U}q_{U}+\Lambda _{I}q_{I}=\Lambda [\gamma (q_{I}-q_{U})+ q_{U}]$. Next, we provide a numerical method to derive the equilibrium strategy: (1) When $\gamma =0$, the equilibrium strategy $q_{U}$ can be determined by $q_{U}=\max \{q\vert qS_{U}(q,0)\ge 0\}$. (2) When $\gamma =1$, the equilibrium strategy $q_{I}$ can be determined by $q_{I}=\max \{q\vert qS_{I}(0,q)\ge 0\}$. (3) When $\gamma \in (0,1)$, from Lemma 8, if $q_{U}<1$, we have $q_{I}=0$, which gives $q_{U}<\frac{\mu }{\Lambda (1-\gamma )}$. If $q_{U}=1$, we have $q_{I}<\frac{\mu -(1-\gamma )\Lambda }{\gamma \Lambda }$. Therefore, if $\Lambda _{U}<\mu$, it is sufficient to search the equilibrium pair $(q_{U},q_{I})$ in the region $\Omega _{(q_{U},q_{I})}^{(1)}=\{(q_{U},0)\vert q_{U}<\min \{\frac{\mu }{\Lambda (1-\gamma )},1\}\}\bigcup \{(1,q_{I})\vert q_{I}<\min \{\frac{\mu -(1-\gamma )\Lambda }{\gamma \Lambda },1\}\}$. Otherwise, if $\Lambda _{U}\ge \mu$, the possible region is $\Omega _{(q_{U},q_{I})}^{(2)}=\{(q_{U}, 0)\vert q_{U}<\min \{\frac{\mu }{\Lambda (1-\gamma )},1\}\}$. Thus, we consider the following three subcases to determine the equilibrium strategy $(q_{U},q_{I})$:

• 1. If $S_{U}(q,0)<0$ for $q<\min \big \{\frac{\mu -(1-\gamma )\Lambda }{\gamma \Lambda },1\big \}$, the equilibrium strategy is $(q_{U},q_{I})=(0,0)$.

• 2. If there exists ${\hat{q}}_{U}\in \big [0,\min \big \{\frac{\mu -(1-\gamma )\Lambda }{\gamma \Lambda },1\big \}\big )$ such that $S_{U}({\hat{q}}_{U},0)=0$, the equilibrium strategy is $(q_{U},q_{I})=({\hat{q}}_{U},0)$.

• 3. If $S_{U}(q,0)\ge 0$ for $q<\min \big \{\frac{\mu -(1-\gamma )\Lambda }{\gamma \Lambda },1\big \}$, the equilibrium joining probability for uninformed customers is $q_{U}=1$, and $q_{I}=\max \{q\vert qS_{I}(1,q)\ge 0\}$.

The existence of an equilibrium strategy is intuitive, but it is difficult to establish the uniqueness of the equilibrium strategy because the monotonicity of $S_{U}(q_{U},q_{I})$ and $S_{I}(q_{U},q_{I})$ cannot be proven analytically. However, the equilibrium strategy can be determined numerically by examining the three cases above. Under equilibrium, the expected individual benefits for informed and uninformed customers are given as follows:

Fig. 11

Throughput and social welfare versus $\gamma$ when $\Lambda =1.5$

Fig. 12

Throughput and social welfare versus $\gamma$ when $\Lambda =3$

Fig. 13

Throughput and social welfare versus $\gamma$ when $\Lambda =6$

$$\begin{aligned} {\overline{S}}_{I}(q_{U},q_{I})= & {} \Pi _{0}R+\Pi _{1}\left[ R-\frac{C}{\mu }\right] +\cdots +\Pi _{K-1}\left[ R-\frac{C(K-1)}{\mu }\right] \\&+\Pi _{K}q_{I}S_{I}(q_{U},q_{I})\\= & {} (1-\Pi _{K})R-\sum _{i=0}^{K-1}\frac{Ci\Pi _{i}}{\mu }+\Pi _{K}q_{I}S_{I}(q_{U},q_{I}), \\ {\overline{S}}_{U}(q_{U},q_{I})= & {} q_{U}S_{U}(q_{U},q_{I}). \end{aligned}$$

Then the total social welfare can be expressed as $S_{\mathrm{soc}}=\Lambda _{I}{\overline{S}}_{I}(q_{U},q_{I})+\Lambda _{U}{\overline{S}}_{U}(q_{U},q_{I})$. When the orbit size is truncated by $M=100$, Figs. 11 and 13 illustrate the throughput and social welfare of the system when the capacity of the waiting line is $K=5$, $\mu =\theta =2$ and $\nu =2.5$.

When $\Lambda =1.5$ (see Fig. 11), we can observe that the throughput can be maximized at $\gamma =0$. When $\Lambda =3$ (see Fig. 12), the throughput is unimodal in $\gamma$, and it can be maximized at $\gamma \approx 0.7$. That is, the service provider is better off in the presence of information heterogeneity. When $\Lambda =6$ (see Fig. 13), the maximal throughput can be attained at $\gamma =1$, i.e., all customers are informed of the state of waiting line. On the other hand, for $\Lambda =1.5,3,6$, Figs. 11 and 13 show that the social welfare is increasing in $\gamma$, and it can be optimized at $\gamma =1$, which coincides with the result in our base model (i.e., $K=1$). That is to say, in a retrial queue, when a finite waiting line is allowed before the server, our results in the retrial queues can still hold.