Skip to main content
Log in

Equilibrium in a finite capacity M/M/1 queue with unknown service rates consisting of strategic and non-strategic customers

  • Published:
Queueing Systems Aims and scope Submit manuscript

Abstract

We consider an \(M/M/1/{\overline{N}}\) observable non-customer-intensive service queueing system with unknown service rates consisting of strategic impatient customers who make balking decisions and non-strategic patient customers who do not make any decision. In the queueing game amongst the impatient customers, we show that there exists at least one pure threshold strategy equilibrium in the presence of patient customers. As multiple pure threshold strategy equilibria exist in certain cases, we consider the minimal pure threshold strategy equilibrium in our sensitivity analysis. We find that the likelihood ratio of a fast server to a slow server in an empty queue is monotonically decreasing in the proportion of impatient customers and monotonically increasing in the waiting area capacity. Further, we find that the minimal pure threshold strategy equilibrium is non-increasing in the proportion of impatient customers and non-decreasing in the waiting area capacity. We also show that at least one pure threshold strategy equilibrium exists when the waiting area capacity is infinite.

This is a preview of subscription content, log in via an institution to check access.

Access this article

Price excludes VAT (USA)
Tax calculation will be finalised during checkout.

Instant access to the full article PDF.

Institutional subscriptions

Fig. 1
Fig. 2
Fig. 3

Similar content being viewed by others

References

  1. Anand, K.S., Paç, M.F.: Quality–speed conundrum: trade-offs in customer-intensive services. Manag. Sci. 57(1), 40–56 (2011)

    Article  Google Scholar 

  2. Boudali, O., Economou, A.: The effect of catastrophes on the strategic customer behavior in queueing systems. Naval Res. Logist. 60(7), 571–587 (2013)

    Article  Google Scholar 

  3. Debo, L.G., Toktay, L.B., Van Wassenhove, L.N.: Queuing for expert services. Manag. Sci. 54(8), 1497–1512 (2008)

    Article  Google Scholar 

  4. Debo, L.G., Parlour, C., Rajan, U.: Signaling quality via queues. Manag. Sci. 58(5), 876–891 (2012)

    Article  Google Scholar 

  5. Debo, L., Veeraraghavan, S.: Equilibrium in queues under unknown service times and service value. Oper. Res. 62(1), 38–57 (2014)

    Article  Google Scholar 

  6. Economou, A., Kanta, S.: Optimal balking strategies and pricing for the single server Markovian queue with compartmented waiting space. Queueing Syst. 59(3–4), 237 (2008)

    Article  Google Scholar 

  7. Economou, A., Kanta, S.: Equilibrium customer strategies and social-profit maximization in the single-server constant retrial queue. Naval Res. Logist. 58(2), 107–122 (2011)

    Article  Google Scholar 

  8. Guo, P., Zipkin, P.: The effects of the availability of waiting-time information on a balking queue. Eur. J. Oper. Res. 198(1), 199–209 (2009)

    Article  Google Scholar 

  9. Hassin, R., Haviv, M.: Equilibrium threshold strategies: the case of queues with priorities. Oper. Res. 45(6), 966–973 (1997)

    Article  Google Scholar 

  10. Hassin, R., Haviv, M.: To Queue or Not to Queue: Equilibrium Behavior in Queueing Systems, vol. 59. Springer, Berlin (2003)

    Book  Google Scholar 

  11. Hassin, R.: Rational Queueing. CRC Press, Boca Raton (2016)

    Book  Google Scholar 

  12. Hassin, R., Roet-Green, R.: The impact of inspection cost on equilibrium, revenue, and social welfare in a single-server queue. Oper. Res. 65(3), 804–820 (2017)

    Article  Google Scholar 

  13. Haviv, M.: When to arrive at a queue with tardiness costs? Perform. Eval. 70(6), 387–399 (2013)

    Article  Google Scholar 

  14. Iravani, F., B Balciog̃lu, : On priority queues with impatient customers. Queueing Syst. 58(4), 239 (2008)

  15. Juneja, S., Shimkin, N.: The concert queueing game: strategic arrivals with waiting and tardiness costs. Queueing Syst. 74(4), 369–402 (2013)

    Article  Google Scholar 

  16. Kakutani, S.: A generalization of Brouwer’s fixed point theorem. Duke Math. J. 8(3), 457–459 (1941)

    Article  Google Scholar 

  17. Kostami, V., Rajagopalan, S.: Speed-quality trade-offs in a dynamic model. Manuf. Serv. Oper. Manag. 16(1), 104–118 (2013)

    Article  Google Scholar 

  18. Kulkarni, V.G.: On queueing systems by retrials. J. Appl. Probab. 20(2), 380–389 (1983)

    Article  Google Scholar 

  19. Li, L., Wang, J., Zhang, F.: Equilibrium customer strategies in markovian queues with partial breakdowns. Comput. Ind. Eng. 66(4), 751–757 (2013)

    Article  Google Scholar 

  20. Liu, Y., Cooper, W.L.: Optimal dynamic pricing with patient customers. Oper. Res. 63(6), 1307–1319 (2015)

    Article  Google Scholar 

  21. Liu, Y., Shang, W.: Follow the crowd to avoid congestion with unknown service rate. In: The 10th POMS-HK International Conference 2019 : Operations Excellence for a Better World ; Conference date: 05-01-2019 Through 06-01-2019. http://www.cb.cityu.edu.hk/ms/pomshk2019/conferenceprogram.htm

  22. Nair, S.: ’Virar crowd’ loses unity on packed local trains. The Times of India.https://timesofindia.indiatimes.com/city/mumbai/Virar-crowd-loses-unity-on-packed-local-trains/articleshow/53440493.cms (2016). Accessed 26 September 2019

  23. Naor, P.: The regulation of queue size by levying tolls. Econom. J. Econom. Soc. 37, 15–24 (1969)

    Google Scholar 

  24. Sumita, U., Masuda, Y., Yamakawa, S.: Optimal internal pricing and capacity planning for service facility with finite buffer. Eur. J. Oper. Res. 128(1), 192–205 (2001)

    Article  Google Scholar 

  25. Wolff, R.W.: Poisson arrivals see time averages. Oper. Res. 30(2), 223–231 (1982)

    Article  Google Scholar 

  26. Xiao, G., Dong, M., Li, J., Sun, L.: Scheduling routine and call-in clinical appointments with revisits. Int. J. Prod. Res. 55(6), 1767–1779 (2017)

    Article  Google Scholar 

Download references

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Rahul R. Marathe.

Additional information

Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Appendix. Proofs

Appendix. Proofs

Proof of Lemma 1

In the birth–death process, solving the flow balance equations yields

$$\begin{aligned}&\pi _{n,\mu ,\psi ,q} \\&\quad = \left\{ \begin{array}{ll} \begin{aligned}&{}\left( [\psi _{n-1}q + (1-q)] \dfrac{\Lambda }{\mu }\right) \pi _{n-1,\mu ,\psi ,q} = \cdots \\ {} &{}\quad = \prod _{j=0}^{n-1} \left( [\psi _{j}q + (1-q)]\dfrac{\Lambda }{\mu }\right) \pi _{0,\mu ,\psi ,q}, \end{aligned}&{} n \in \{1,2,\ldots ,n_{b}\}, \\ \begin{aligned}&{}\left( \dfrac{(1-q)\Lambda }{\mu }\right) \pi _{n-1,\mu ,\psi ,q} = \cdots \\ {} &{}\quad = \left( \dfrac{(1-q)\Lambda }{\mu }\right) ^{n-n_{b}} \prod _{j=0}^{n_{b}-1} \left( [\psi _{j}q + (1-q)]\dfrac{\Lambda }{\mu }\right) \pi _{0,\mu ,\psi ,q},\end{aligned} &{} n \in \{n_{b}+1,\ldots , {\overline{N}}\}. \end{array}\right. \end{aligned}$$

Also, using the property that \(\sum _{n=0}^{{\overline{N}}} \pi _{n,\mu ,\psi ,q} = 1\), we get

$$\begin{aligned} \pi _{0,\mu ,\psi ,q}= & {} \left[ 1 + \sum _{n=1}^{n_{b}} \prod _{j=0}^{n-1} \left( [\psi _{j}q+(1-q)]\dfrac{\Lambda }{\mu }\right) \right. \\&\left. + \prod _{j=0}^{n_{b}-1} \left( [\psi _{j}q+(1-q)]\dfrac{\Lambda }{\mu }\right) \sum _{n=n_{b}+1}^{{\overline{N}}} \left( \dfrac{(1-q)\Lambda }{\mu }\right) ^{n-n_{b}}\right] ^{-1} \end{aligned}$$

and therefore,

$$\begin{aligned}&\pi _{n,\mu ,\psi ,q} \\&\quad = {\left\{ \begin{array}{ll} \pi _{0,\mu ,\psi ,q} \prod _{j=0}^{n-1} \left( [\psi _{j}q+(1-q)]\dfrac{\Lambda }{\mu }\right) , &{} n \in \{1,2,\ldots ,n_{b}\}, \\ \pi _{0,\mu ,\psi ,q} \left( \dfrac{(1-q)\Lambda }{\mu }\right) ^{n-n_{b}} \prod _{j=0}^{n_{b}-1} \left( [\psi _{j}q+(1-q)]\dfrac{\Lambda }{\mu }\right) , &{} n \in \{n_{b}+1,n_{b}+2,\ldots , {\overline{N}}\}, \end{array}\right. } \end{aligned}$$

which completes the proof. \(\square \)

Proof of Lemma 2

For the pure threshold strategy to be in equilibrium, the customer’s utility \(U(n,k,q) \ge 0\) for all \(0 \le n \le k-1\) and \(U(n,k,q) \le 0\) for all \(k \le n \le n_{b}\). Hence, it follows that, for all \(0 \le n \le k-1\),

$$\begin{aligned} p \pi _{n,{\overline{\mu }},k,q} \left( V - \dfrac{n+1}{{\overline{\mu }}}c\right) + (1-p) \pi _{n,{\underline{\mu }},k,q} \left( V - \dfrac{n+1}{{\underline{\mu }}}c\right) \ge 0. \end{aligned}$$

Then,

$$\begin{aligned} \pi _{n,{\underline{\mu }},k,q} \left( \dfrac{n+1}{{\underline{\mu }}}c - V\right) \le \dfrac{p}{1-p} \pi _{n,{\overline{\mu }},k,q} \left( V - \dfrac{n+1}{{\overline{\mu }}}c\right) . \end{aligned}$$

Simplifying the expression further completes the proof. \(\square \)

Proof of Lemma 3

(i) Let \({\underline{\rho }} = \dfrac{\Lambda }{{\underline{\mu }}}\) and \({\overline{\rho }} = \dfrac{\Lambda }{{\overline{\mu }}}\) where \({\overline{\rho }} < {\underline{\rho }}\). We first write \(\theta _{q}({\tilde{k}},{\overline{N}})\) as

$$\begin{aligned} \theta _{q}({\tilde{k}},{\overline{N}}) = \dfrac{\sum _{j=0}^{{\tilde{k}}}{\underline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}}(1-q)^{j-{\tilde{k}}}{\underline{\rho }}^{j}}{\sum _{j=0}^{{\tilde{k}}}{\overline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}}(1-q)^{j-{\tilde{k}}}{\overline{\rho }}^{j}}. \end{aligned}$$

It is evident that \(\theta _{q}({\tilde{k}},{\overline{N}})\) is independent of \({\tilde{k}}\) at \(q=0\) and therefore, \(\dfrac{d\theta _{q}({\tilde{k}},{\overline{N}})}{d{\tilde{k}}} = 0\). At \(q=1\), \(\theta _{q}({\tilde{k}},{\overline{N}})\) in our case is the inverse of \(\Phi ({\tilde{k}})\) in Debo and Veeraraghavan [5]. Therefore, at \(q=1\), it is easy to see that \(\theta _{q}({\tilde{k}},{\overline{N}})\) is monotonically increasing in \({\tilde{k}}\) and \(\dfrac{d\theta _{q}({\tilde{k}},{\overline{N}})}{d{\tilde{k}}} > 0\).

We are now interested in the behavior of \(\theta _{q}({\tilde{k}},{\overline{N}})\) in \({\tilde{k}}\) when \(q \in (0,1)\). Consider two queue lengths \(z \in \{1,2,\ldots \}\) and \(z-1 \in \{0,1,2,\ldots \}\). Our aim is to prove that \(\theta _{q}(z,{\overline{N}}) - \theta _{q}(z-1,{\overline{N}}) > 0\). In order to prove that \(\theta _{q}(z,{\overline{N}}) - \theta _{q}(z-1,{\overline{N}}) > 0\), it suffices to show that Eq. (14) is satisfied:

$$\begin{aligned} \begin{aligned}&\left( \sum _{j=0}^{z}{\underline{\rho }}^{j} + \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j}\right) \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j} + \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\overline{\rho }}^{j}\right) \\&- \left( \sum _{j=0}^{z}{\overline{\rho }}^{j} + \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j}\right) \left( \sum _{j=0}^{z-1}{\underline{\rho }}^{j} + \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\underline{\rho }}^{j}\right) > 0. \end{aligned} \end{aligned}$$
(14)

When we expand the terms in the aforementioned expression, we need to obtain

$$\begin{aligned} \begin{aligned}&\left( \sum _{j=0}^{z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - \sum _{j=0}^{z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j} \right) + \left( \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - \sum _{j=0}^{z}{\overline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}} (1-q)^{j-z+1}{\underline{\rho }}^{j} \right) \\ {}&\quad +\, \left( \sum _{j=0}^{z}{\underline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}( 1-q)^{j-z+1}{\overline{\rho }}^{j} - \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z} {\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j} \right) \\&\quad +\, \left( \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\overline{\rho }}^{j} - \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\underline{\rho }}^{j} \right) > 0. \end{aligned} \end{aligned}$$
(15)

Consider \(\sum _{j=0}^{z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - \sum _{j=0}^{z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j}\) from Eq. (15):

$$\begin{aligned} \begin{aligned} \sum _{j=0}^{z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - \sum _{j=0}^{z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j}&= \left( {\underline{\rho }}^{z} + \sum _{j=0}^{z-1}{\underline{\rho }}^{j}\right) \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - \left( {\overline{\rho }}^{z} + \sum _{j=0}^{z-1}{\overline{\rho }}^{j}\right) \sum _{j=0}^{z-1}{\underline{\rho }}^{j} \\&= {\underline{\rho }}^{z} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - {\overline{\rho }}^{z} \sum _{j=0}^{z-1}{\underline{\rho }}^{j} \\&= {\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j-z} - \sum _{j=0}^{z-1}{\underline{\rho }}^{j-z} \right) \\&= {\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \dfrac{1}{{\overline{\rho }}^{z}} - \dfrac{1}{{\underline{\rho }}^{z}} + \dfrac{1}{{\overline{\rho }}^{z-1}} - \dfrac{1}{{\underline{\rho }}^{z-1}} +\cdots + \dfrac{1}{{\overline{\rho }}} - \dfrac{1}{{\underline{\rho }}}\right) . \end{aligned} \end{aligned}$$
(16)

We know that \({\underline{\rho }} > {\overline{\rho }} \implies \dfrac{1}{{\underline{\rho }}}< \dfrac{1}{{\overline{\rho }}} \implies \dfrac{1}{{\underline{\rho }}^{z}} < \dfrac{1}{{\overline{\rho }}^{z}}\) for all \(z > 0\). It follows that

$$\begin{aligned} \sum _{j=0}^{z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - \sum _{j=0}^{z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j} > 0. \end{aligned}$$
(17)

Now consider \(\sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - \sum _{j=0}^{z}{\overline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\underline{\rho }}^{j}\) from Eq. (15):

$$\begin{aligned}&\sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z} {\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - \sum _{j=0}^{z}{\overline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\underline{\rho }}^{j} = \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} \nonumber \\&\qquad -\, \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j} + {\overline{\rho }}^{z} \right) \left( (1-q){\underline{\rho }}^{z} + (1-q) \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \right) \nonumber \\&\quad = \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - (1-q) {\underline{\rho }}^{z} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - (1-q) {\underline{\rho }}^{z} {\overline{\rho }}^{z} \nonumber \\&\qquad -\,(1-q) \sum _{j=0}^{z-1}{\overline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} - {\overline{\rho }}^{z} (1-q) \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \nonumber \\&\quad = q \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j} - (1-q) {\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j - z} + 1 + \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j-z} \right) .\nonumber \\ \end{aligned}$$
(18)

Consider \(\sum _{j=0}^{z}{\underline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\overline{\rho }}^{j} - \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j}\) from Eq. (15):

$$\begin{aligned}&\sum _{j=0}^{z}{\underline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\overline{\rho }}^{j} - \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j} \nonumber \\&\quad = \left( \sum _{j=0}^{z-1}{\underline{\rho }}^{j} + {\underline{\rho }}^{z}\right) \left( (1-q) {\overline{\rho }}^{z} + (1-q) \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \right) \nonumber \\&\qquad -\, \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j} = (1-q) {\overline{\rho }}^{z} \sum _{j=0}^{z-1}{\underline{\rho }}^{j} + (1-q) {\overline{\rho }}^{z} {\underline{\rho }}^{z}\nonumber \\&\qquad +\, (1-q) \sum _{j=0}^{z-1}{\underline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} + (1-q) {\underline{\rho }}^{z} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \nonumber \\&\qquad -\, \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j} = (1-q) {\overline{\rho }}^{z} {\underline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\underline{\rho }}^{j - z} \right. \nonumber \\&\qquad \left. +\, 1 + \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j - z} \right) - q \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j}. \end{aligned}$$
(19)

Consider \(\sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\overline{\rho }}^{j} - \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\underline{\rho }}^{j}\) from Eq. (15):

$$\begin{aligned}&\sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\overline{\rho }}^{j} - \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=z}^{{\overline{N}}}(1-q)^{j-z+1}{\underline{\rho }}^{j} \nonumber \\&\quad = \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \left( (1-q){\overline{\rho }}^{z} + \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z+1}{\overline{\rho }}^{j} \right) \nonumber \\&\qquad -\, \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \left( (1-q){\underline{\rho }}^{z} + \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z+1}{\underline{\rho }}^{j} \right) \nonumber \\&\quad = (1-q){\overline{\rho }}^{z} \sum _{j=z+1}^{{\overline{N}}} (1-q)^{j-z}{\underline{\rho }}^{j} - (1-q){\underline{\rho }}^{z} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j}\nonumber \\&\qquad +\, \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z + 1} {\overline{\rho }}^{j} \nonumber \\&\qquad -\, \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z + 1} {\underline{\rho }}^{j} \nonumber \\&\quad = (1-q) {\overline{\rho }}^{z} {\underline{\rho }}^{z} \left( \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j-z} - \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j-z} \right) \nonumber \\&\qquad +\, (1-q) \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \nonumber \\&\qquad -\, (1-q) \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \nonumber \\&\quad = (1-q) {\overline{\rho }}^{z} {\underline{\rho }}^{z} \left( \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j-z} - \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j-z} \right) . \end{aligned}$$
(20)

It is apparent that the sum of \((1-q) {\overline{\rho }}^{z} {\underline{\rho }}^{z} \left( \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j-z} - \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j-z} \right) \) from Eq. (20), \(- (1-q) {\underline{\rho }}^{z} {\overline{\rho }}^{z} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j-z}\) from Eq. (18) and \((1-q) {\overline{\rho }}^{z} {\underline{\rho }}^{z} \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j - z}\) from Eq. (19) is equal to 0.

From Eqs. (16), (18) and (19), we find the sum of \({\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j-z} - \sum _{j=0}^{z-1}{\underline{\rho }}^{j-z} \right) \), \(-(1-q) {\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j - z} + 1 \right) \) and \((1-q) {\overline{\rho }}^{z} {\underline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\underline{\rho }}^{j - z} + 1 \right) \), which is equal to

$$\begin{aligned}&{\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j-z} - \sum _{j=0}^{z-1}{\underline{\rho }}^{j-z} \right) + (1-q) {\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\underline{\rho }}^{j - z} - \sum _{j=0}^{z-1}{\overline{\rho }}^{j - z} \right) \nonumber \\&\quad = {\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j-z} - \sum _{j=0}^{z-1}{\underline{\rho }}^{j-z} \right) - (1-q) {\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( -\sum _{j=0}^{z-1}{\underline{\rho }}^{j - z} + \sum _{j=0}^{z-1}{\overline{\rho }}^{j - z} \right) \nonumber \\&\quad = q {\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j-z} - \sum _{j=0}^{z-1}{\underline{\rho }}^{j-z} \right) > 0. \end{aligned}$$
(21)

From Eq. (17), it follows that \(q {\underline{\rho }}^{z} {\overline{\rho }}^{z} \left( \sum _{j=0}^{z-1}{\overline{\rho }}^{j-z} - \sum _{j=0}^{z-1}{\underline{\rho }}^{j-z} \right) > 0\) as \(q \in (0,1)\). Finally, consider the terms \(q \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \sum _{j=0}^{z-1}{\overline{\rho }}^{j}\) and \(-q \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \sum _{j=0}^{z-1}{\underline{\rho }}^{j}\) from Eqs. (18) and (19), respectively. Here, we know that \(\sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} > \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j}\). Adding these equations to Eq. (21), we obtain

$$\begin{aligned} \begin{aligned} \left( q {\underline{\rho }}^{z} + q \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\underline{\rho }}^{j} \right) \sum _{j=0}^{z-1} {\overline{\rho }}^{j} - \left( q {\overline{\rho }}^{z} + q \sum _{j=z+1}^{{\overline{N}}}(1-q)^{j-z}{\overline{\rho }}^{j} \right) \sum _{j=0}^{z-1} {\underline{\rho }}^{j} > 0. \end{aligned}\nonumber \\ \end{aligned}$$
(22)

From Eq. (22), we conclude that \(\theta _{q}({\tilde{k}},{\overline{N}})\) is always monotonically increasing in \({\tilde{k}}\) if \(q > 0\).

(ii) We consider \(\omega ({\tilde{n}})\), which is written as

$$\begin{aligned} \omega ({\tilde{n}}) = \left( \dfrac{{\overline{\mu }}}{{\underline{\mu }}}\right) ^{{\tilde{n}}} \dfrac{\dfrac{{\tilde{n}}+1}{{\underline{\mu }}}c - V}{V - \dfrac{{\tilde{n}}+1}{{\overline{\mu }}}c} = \left( \dfrac{{\overline{\mu }}}{{\underline{\mu }}}\right) ^{{\tilde{n}}+1} \dfrac{{\tilde{n}} - {\underline{n}}}{{\overline{n}} - {\tilde{n}}}. \end{aligned}$$

As \({\tilde{n}} \in [{\underline{n}},{\overline{n}})\) and \({\overline{\mu }} > {\underline{\mu }}\), it is straightforward to observe that \(\left( \dfrac{{\overline{\mu }}}{{\underline{\mu }}}\right) ^{{\tilde{n}}+1}\) is increasing in \({\tilde{n}}\). Also, \({\tilde{n}} - {\underline{n}}\) is increasing in \({\tilde{n}}\) and \({\overline{n}} - {\tilde{n}}\) is decreasing in \({\tilde{n}}\). We know that \(\omega ({\tilde{n}}) \ge 0\) for \({\tilde{n}} \in [{\underline{n}},{\overline{n}})\). As the function \(\omega ({\tilde{n}})\) is comprised of a product of two positive and increasing functions of \({\tilde{n}}\) divided by a decreasing function of \({\tilde{n}}\), we infer that \(\omega ({\tilde{n}})\) is monotonically increasing in \({\tilde{n}}\). \(\square \)

Proof of Lemma 4

Let \(\theta _{q}({\tilde{k}},{\overline{N}}) = \dfrac{\sum _{j=0}^{{\tilde{k}}}{\underline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}}(1-q)^{j-{\tilde{k}}}{\underline{\rho }}^{j}}{\sum _{j=0}^{{\tilde{k}}}{\overline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}}(1-q)^{j-{\tilde{k}}}{\overline{\rho }}^{j}}\). We consider q and \(q - \epsilon \) and then aim to show that \(\theta _{q - \epsilon }({\tilde{k}},{\overline{N}}) - \theta _{q}({\tilde{k}},{\overline{N}}) > 0\).

We can denote \(\sum _{j={\tilde{k}}+1}^{{\overline{N}}}(1-q+\epsilon )^{j-{\tilde{k}}}\rho ^{j} = \alpha \sum _{j={\tilde{k}}+1}^{{\overline{N}}}(1-q)^{j-{\tilde{k}}}\rho ^{j}\), where \(\alpha > 1\). Then, in order to prove that \(\theta _{q - \epsilon }({\tilde{k}},{\overline{N}}) - \theta _{q}({\tilde{k}},{\overline{N}}) > 0\), it is enough to consider and prove Eq. (23):

$$\begin{aligned}&\left( \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} + \alpha \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j}\right) \left( \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j}\right) \nonumber \\&\quad -\, \left( \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j}\right) \left( \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} + \alpha \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j}\right) > 0.\nonumber \\ \end{aligned}$$
(23)

From Eq. (23), we can easily show that \(\sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} - \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} = 0\) and \(\alpha \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j} \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j} - \alpha \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j} \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j} = 0\). The remaining terms satisfy the following conditions: \(\sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j} \left( \alpha - 1\right) > 0\) and \(\sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j} \left( 1 - \alpha \right) < 0\).

To show that \(\theta _{q - \epsilon }({\tilde{k}},{\overline{N}}) - \theta _{q}({\tilde{k}},{\overline{N}}) > 0\), it is sufficient to show for all \({\tilde{k}} \in \{0,1,2,\ldots \}\) that

$$\begin{aligned} \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j} - \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j} > 0. \end{aligned}$$
(24)

We will use the principle of mathematical induction to prove this. At \({\tilde{k}} = 0\), Eq. (24) is simplified to \(\sum _{j=1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j} - \sum _{j=1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j} > 0\) (\(\because {\underline{\rho }} > {\overline{\rho }}\)). We now assume that Eq. (24) is satisfied for \({\tilde{k}} = z\) and show that it is true for \({\tilde{k}} = z + 1\). At \({\tilde{k}} = z+1\), Eq. (24) becomes

$$\begin{aligned}&\sum _{j=0}^{z + 1} {\overline{\rho }}^{j} \sum _{j=z+2}^{{\overline{N}}} (1-q)^{j-z+1} {\underline{\rho }}^{j} - \sum _{j=0}^{z+1} {\underline{\rho }}^{j} \sum _{j=z+2}^{{\overline{N}}} (1-q)^{j-z+1} {\overline{\rho }}^{j} \nonumber \\&\quad = {\overline{\rho }}^{z} (1-q) \sum _{j=z+2}^{{\overline{N}}} (1-q)^{j-z} {\underline{\rho }}^{j} + (1-q) \sum _{j=0}^{z} {\overline{\rho }}^{j} \sum _{j=z+2}^{{\overline{N}}} (1-q)^{j-z} {\underline{\rho }}^{j} \nonumber \\&\qquad -\, {\underline{\rho }}^{z} (1-q) \sum _{j=z+2}^{{\overline{N}}} (1-q)^{j-z} {\overline{\rho }}^{j} - (1-q) \sum _{j=0}^{z} {\underline{\rho }}^{j} \sum _{j=z+2}^{{\overline{N}}} (1-q)^{j-z} {\overline{\rho }}^{j} \nonumber \\&\quad = {\overline{\rho }}^{z} (1-q) \sum _{j=z+1}^{{\overline{N}}} (1-q)^{j-z} {\underline{\rho }}^{j} - (1-q)^{2} {\overline{\rho }}^{z} {\underline{\rho }}^{z+1} \nonumber \\&\qquad +\, (1-q) \sum _{j=0}^{z} {\overline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}} (1-q)^{j-z} {\underline{\rho }}^{j} - (1-q)^{2} \sum _{j=0}^{z} {\overline{\rho }}^{j} {\underline{\rho }}^{z+1} \nonumber \\&\qquad -\, {\underline{\rho }}^{z} (1-q) \sum _{j=z+1}^{{\overline{N}}} (1-q)^{j-z} {\overline{\rho }}^{j} + {\underline{\rho }}^{z} (1-q)^{2} {\overline{\rho }}^{z+1} \nonumber \\&\qquad -\, (1-q) \sum _{j=0}^{z} {\underline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}} (1-q)^{j-z} {\overline{\rho }}^{j} + (1-q)^{2} \sum _{j=0}^{z} {\underline{\rho }}^{j} {\overline{\rho }}^{z+1}. \end{aligned}$$
(25)

From Eq. (25), we infer that \(-(1-q)^{2} {\overline{\rho }}^{z} {\underline{\rho }}^{z+1} + {\underline{\rho }}^{z} (1-q)^{2} {\overline{\rho }}^{z+1} < 0\). However, \({\overline{\rho }}^{z} (1-q) \sum _{j=z+1}^{{\overline{N}}} (1-q)^{j-z} {\underline{\rho }}^{j} - {\underline{\rho }}^{z} (1-q) \sum _{j=z+1}^{{\overline{N}}} (1-q)^{j-z} {\overline{\rho }}^{j}> |-(1-q)^{2} {\overline{\rho }}^{z} {\underline{\rho }}^{z+1} + {\underline{\rho }}^{z} (1-q)^{2} {\overline{\rho }}^{z+1}| > 0\). Similarly, \(-(1-q)^{2} \sum _{j=0}^{z} {\overline{\rho }}^{j} {\underline{\rho }}^{z+1} + (1-q)^{2} \sum _{j=0}^{z} {\underline{\rho }}^{j} {\overline{\rho }}^{z+1} < 0\). However, \((1-q) \sum _{j=0}^{z} {\overline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}} (1-q)^{j-z} {\underline{\rho }}^{j} - (1-q) \sum _{j=0}^{z} {\underline{\rho }}^{j} \sum _{j=z+1}^{{\overline{N}}} (1-q)^{j-z} {\overline{\rho }}^{j}> |-(1-q)^{2} \sum _{j=0}^{z} {\overline{\rho }}^{j} {\underline{\rho }}^{z+1} + (1-q)^{2} \sum _{j=0}^{z} {\underline{\rho }}^{j} {\overline{\rho }}^{z+1}| > 0\). We now observe that Eq. (24) is satisfied for \({\tilde{k}} = z+1\). Hence, \(\theta _{q}({\tilde{k}},{\overline{N}})\) is monotonically decreasing in q. \(\square \)

Proof of Lemma 5

Let \(\theta _{q}({\tilde{k}},{\overline{N}}) = \dfrac{\sum _{j=0}^{{\tilde{k}}}{\underline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}}(1-q)^{j-{\tilde{k}}}{\underline{\rho }}^{j}}{\sum _{j=0}^{{\tilde{k}}}{\overline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}}(1-q)^{j-{\tilde{k}}}{\overline{\rho }}^{j}}\). We aim to show that \(\theta _{q}({\tilde{k}},{\overline{N}}+1) - \theta _{q}({\tilde{k}},{\overline{N}}) > 0\) as presented in Equation (26):

$$\begin{aligned}&\left( \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}} + 1} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j}\right) \left( \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j}\right) \nonumber \\&\quad -\, \left( \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j}\right) \left( \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}+1} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j}\right) > 0.\nonumber \\ \end{aligned}$$
(26)

Simplifying and rearranging the terms,

$$\begin{aligned}&\left( \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j} + (1-q)^{{\overline{N}} + 1 -{\tilde{k}}} {\underline{\rho }}^{{\overline{N}}+1}\right) \left( \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j}\right) \nonumber \\&\qquad -\, \left( \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j}\right) \left( \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} + \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j} + (1-q)^{{\overline{N}} + 1 -{\tilde{k}}} {\overline{\rho }}^{{\overline{N}}+1}\right) \nonumber \\&\quad = (1-q)^{{\overline{N}} + 1 -{\tilde{k}}} {\underline{\rho }}^{{\overline{N}}+1} \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j} + (1-q)^{{\overline{N}} + 1 -{\tilde{k}}} {\underline{\rho }}^{{\overline{N}}+1} \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\overline{\rho }}^{j} \nonumber \\&\qquad -\, (1-q)^{{\overline{N}} + 1 -{\tilde{k}}} {\overline{\rho }}^{{\overline{N}}+1} \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j} - (1-q)^{{\overline{N}} + 1 -{\tilde{k}}} {\overline{\rho }}^{{\overline{N}}+1} \sum _{j={\tilde{k}}+1}^{{\overline{N}}} (1-q)^{j-{\tilde{k}}} {\underline{\rho }}^{j}\nonumber \\&\quad = (1-q)^{{\overline{N}} + 1 -{\tilde{k}}} {\overline{\rho }}^{{\overline{N}}+1} {\underline{\rho }}^{{\overline{N}}+1} \left( \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j- {\overline{N}}-1} - \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j-{\overline{N}}-1} \right) \nonumber \\&\qquad +\, (1-q)^{{\overline{N}} + 1 -2{\tilde{k}}} {\overline{\rho }}^{{\overline{N}}+1} {\underline{\rho }}^{{\overline{N}}+1} \left( \sum _{j=0}^{{\tilde{k}}} \left( (1-q){\overline{\rho }}\right) ^{j-{\overline{N}}-1} - \sum _{j=0}^{{\tilde{k}}} \left( (1-q){\underline{\rho }}\right) ^{j-{\overline{N}}-1} \right) .\nonumber \\ \end{aligned}$$
(27)

We know that \({\underline{\rho }} > {\overline{\rho }} \implies \dfrac{1}{{\underline{\rho }}}< \dfrac{1}{{\overline{\rho }}} \implies \dfrac{1}{{\underline{\rho }}^{z}} < \dfrac{1}{{\overline{\rho }}^{z}}\) for all \(z > 0\) and \({\tilde{k}} < {\overline{N}}+1\). It follows that \((1-q)^{{\overline{N}} + 1 -{\tilde{k}}} {\overline{\rho }}^{{\overline{N}}+1} {\underline{\rho }}^{{\overline{N}}+1} \left( \sum _{j=0}^{{\tilde{k}}} {\overline{\rho }}^{j- {\overline{N}}-1} - \sum _{j=0}^{{\tilde{k}}} {\underline{\rho }}^{j-{\overline{N}}-1} \right) + (1-q)^{{\overline{N}} + 1 -2{\tilde{k}}} {\overline{\rho }}^{{\overline{N}}+1} {\underline{\rho }}^{{\overline{N}}+1} \left( \sum _{j=0}^{{\tilde{k}}} \left( (1-q){\overline{\rho }}\right) ^{j-{\overline{N}}-1} - \sum _{j=0}^{{\tilde{k}}} \left( (1-q){\underline{\rho }}\right) ^{j-{\overline{N}}-1} \right) > 0\). Hence, \(\theta _{q}({\tilde{k}},{\overline{N}})\) is monotonically increasing in \({\overline{N}}\). \(\square \)

Proof of Proposition 4

From Lemma 5, we know that \(\theta _{q}(k,{\overline{N}}_{1}) > \theta _{q}(k,{\overline{N}}_{2})\) when \({\overline{N}}_{1} > {\overline{N}}_{2}\). It follows that \(b_{q}(k,{\overline{N}}_{1}) \ge b_{q}(k,{\overline{N}}_{2})\) for a given k. As a pure threshold strategy equilibrium is guaranteed to exist (Proposition 1), let the equilibrium for \({\overline{N}}_{2}\) occur at k, i.e., \(b_{q}(k,{\overline{N}}_{2}) = k\). If multiple equilibria exist for \({\overline{N}}_{2}\), consider the highest pure threshold strategy equilibrium. Now, as \(b_{q}(k,{\overline{N}}_{1}) \ge b_{q}(k,{\overline{N}}_{2})\), we can look at two cases:

Case 1. If \(b_{q}(k,{\overline{N}}_{1}) = b_{q}(k,{\overline{N}}_{2})\), then the equilibrium for \({\overline{N}}_{1}\) also occurs at k.

Case 2. If \(b_{q}(k,{\overline{N}}_{1}) > b_{q}(k,{\overline{N}}_{2})\), the equilibrium for \({\overline{N}}_{1}\) does not occur at k. From FTC behavior, we know that \(b_{q}(k+1,{\overline{N}}_{1}) \ge b_{q}(k,{\overline{N}}_{1})\). Now, if \(b_{q}(k+1,{\overline{N}}_{1}) = k+1\), the pure threshold strategy equilibrium for \({\overline{N}}_{1}\) occurs at \(k+1\). Otherwise, there is no pure threshold strategy equilibrium at \(k+1\) for \({\overline{N}}_{1}\). Then, we can consider \(k+2, k+3,\ldots ,\lceil {\overline{n}} \rceil \). In general, \(b_{q}(k + \varphi ,{\overline{N}}_{1})\) is bounded above by \(\lceil {\overline{n}} \rceil \) as \(\lim _{n \rightarrow {\overline{n}}} \omega (n) = +\infty \). As \(b_{q}(k + \varphi + 1,{\overline{N}}_{1}) \ge b_{q}(k + \varphi ,{\overline{N}}_{1})\), we find that \(b_{q}(k + \varphi + 1,{\overline{N}}_{1}) = \lceil {\overline{n}} \rceil \) if \(b_{q}(k + \varphi ,{\overline{N}}_{1}) = \lceil {\overline{n}} \rceil \). Therefore, if there exists no pure threshold strategy equilibrium from \(k+1\) (i.e., \(b_{q}(k+1,{\overline{N}}_{1}) \ne k+1\)) to \(\lceil {\overline{n}} \rceil - 1\) (i.e., \(b_{q}(\lceil {\overline{n}} \rceil - 1,{\overline{N}}_{1}) \ne \lceil {\overline{n}} \rceil - 1\)), there will always exist a pure threshold strategy equilibrium for \({\overline{N}}_{1}\) at \(\lceil {\overline{n}} \rceil \) (i.e., \(b_{q}(\lceil {\overline{n}} \rceil ,{\overline{N}}_{1}) = \lceil {\overline{n}} \rceil \)). \(\square \)

Rights and permissions

Reprints and permissions

About this article

Check for updates. Verify currency and authenticity via CrossMark

Cite this article

Srivatsa Srinivas, S., Marathe, R.R. Equilibrium in a finite capacity M/M/1 queue with unknown service rates consisting of strategic and non-strategic customers. Queueing Syst 96, 329–356 (2020). https://doi.org/10.1007/s11134-020-09671-x

Download citation

  • Received:

  • Revised:

  • Accepted:

  • Published:

  • Issue Date:

  • DOI: https://doi.org/10.1007/s11134-020-09671-x

Keywords

Mathematics Subject Classification

Navigation