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.
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Appendix. Proofs
Appendix. Proofs
Proof of Lemma 1
In the birth–death process, solving the flow balance equations yields
Also, using the property that \(\sum _{n=0}^{{\overline{N}}} \pi _{n,\mu ,\psi ,q} = 1\), we get
and therefore,
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 \(0 \le n \le k-1\),
Then,
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
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:
When we expand the terms in the aforementioned expression, we need to obtain
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):
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
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):
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):
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):
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
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
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
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):
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
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
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):
Simplifying and rearranging the terms,
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., \(b_{q}(\lceil {\overline{n}} \rceil ,{\overline{N}}_{1}) = \lceil {\overline{n}} \rceil \)). \(\square \)
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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
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DOI: https://doi.org/10.1007/s11134-020-09671-x