Abstract
In standard queues, when there are waiting customers, service completions are followed by service commencements. In retrial queues, this is not the case. In such systems, customers try to receive service at a time of their choosing, or the server seeks the next customer for a non-negligible time. In this note, we consider a hybrid model with both a finite standard queue and an orbit. While in the orbit, customers try to join the standard queue in their own time. We assume that the retrial rate is a decision variable, and study both the Nash equilibrium and the socially optimal retrial rates, under a cost model that considers both waiting costs and retrial costs.
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Notes
Alternatively, one may assume that service is associated with a significantly large reward, and hence this prohibition is actually a result.
Elcan’s notation is a bit different. He denotes the steady-state probability by \(\mu \) (here it is p), the service by \(\alpha \) (here it is \(\mu \)) and retrial rate by \(\gamma \) (here it is \(\eta \)). Also, one needs to set \(r=1\) to fit the result to the one we quote. In the case where \(r=1\), the sums can be simplified: see [12].
In practice, we solve the equations by truncating the state space and the capacity of the orbit to a large k. Alternatively, the partial generating functions can be evaluated numerically.
Kulkarni presented the parameters in a normalized mode: He set the retrial cost to be 1 (here it is r). He set the waiting cost to be per service time and not per time unit, and hence his C is our \(c\mu \). Kulkarni also assumed general service time, but in the exponential service times his \(\frac{\tau }{S}=\frac{1}{\sqrt{2}}\).
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Acknowledgements
This research was supported by the Israel Science Foundation, grant #950/18, and by the Israeli Ministry of Science, Technology and Space, grant #8767171.
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Appendices
Appendices
A Solution of steady-state equations
While seeking for a solution to the infinite set of steady-state equations, we transformed the equations to a set of partial generating functions, \(g_i(z)\). By getting \(g_i(z)\), we can compute \(p_{i,j}\). The transformation is to multiply every equation by \(z^j\) and sum it from 0 to \(\infty \). We present an example of transforming Eq. (1).
\(\lambda g_0(z)+\eta z g_0'(z) = \mu g_1(z)\).
So, finally we get a system of linear differential equations:
In matrix form, we have a system of the type \(G'(z)=A(z)*G(z)\). This system cannot be solved analytically, because the coefficients (the A(z) matrix) are not constant, i.e., they depend on z. As can be seen in any textbook on partial differential equations, unless A(z) is a periodic function, no explicit solution exists.
B Computational steps
In this appendix, we provide further details regarding the computational procedures that were used in Sects. 3 and 4.
In both sections, the computational schemes started with truncating the state space to be large enough. Then, the search for the socially optimal retrial rate was simply by searching in a grid and then refining it. We have that, for every n, \(\eta ^*=\eta ^*(n)\) is bounded from above by \(\eta ^*(1)\). So, we looked at the value of the objective function at values of \(\eta \) separated by a distance of 0.1. Then, after identifying the region where \(\eta ^*\) lies, we refined the steps to much smaller ones.
The search for the equilibrium was done similarly. We looked at a two-dimensional grid of \((\gamma ,\eta )\). We constructed a matrix of the values of \(Cost(\gamma ,e)\) for each pair in the grid. Going back to the definitions, the best response is \(\gamma ^*(\eta )\), which minimizes \(Cost(\gamma ,\eta )\). Thus, the symmetric Nash equilibrium is in a column such that its minimal value is in the diagonal. Again, after obtaining the region that \(\eta _e\) lies in, we refined the grid.
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Kerner, Y., Shmuel-Bittner, O. Strategic behavior and optimization in a hybrid M/M/1 queue with retrials. Queueing Syst 96, 285–302 (2020). https://doi.org/10.1007/s11134-020-09672-w
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DOI: https://doi.org/10.1007/s11134-020-09672-w