Strategic behaviour in a tandem queue with alternating server

Abstract

This paper considers an unobservable two-site tandem queueing system attended by an alternating server. We study the strategic customer behaviour under two threshold-based operating policies, applied by a profit-maximizing server, while customers’ strategic behaviour and server’s switching costs are taken into account. Under the Exact-N policy, in each cycle the server first completes service of N customers in the first stage ($Q_1$), then switches to the second stage ($Q_2$) and then serves those N customers before switching back to $Q_1$ to start a new cycle. This policy leads to a mixture of Follow-the-Crowd and Avoid-the-Crowd customer behaviour. In contrast, under the N-Limited policy, the server switches from $Q_1$ to $Q_2$ also when the first queue is emptied, making this regime work-conserving and leading only to Avoid-the-Crowd behaviour. Performance measures are obtained using matrix geometric methods for both policies and any threshold N, while for sequential service ($N=1$) explicit expressions are derived. It is shown that the system’s stability condition is independent of N, and of the switching policy. Optimal performances in equilibrium, under each of these switching policies, are analysed and compared through a numerical study.

Appendix A: Calculating \(\vec {P_o}\)

In this appendix, we elaborate the process of calculating $\vec {P}_0$ by replacing one of the non-repeating matrix balance equations with the normalizing equation and by that obtaining a system of equations with a unique solution.

1.1 A.1 Exact-N scenario

For notational simplicity let

$$\begin{aligned} \begin{aligned}&\phi =B_0+RA_2, \\&\psi =(I-R)^{-1}\vec {e} \ , \end{aligned} \end{aligned}$$

where $\phi$ is a matrix of size $(2N)\times (2N)$ and $\psi$ is a column vector of size (2N).

Thus (13) and (15) become

$$\begin{aligned} {\left\{ \begin{array}{ll} \vec {P}_0\phi =\vec {0} \\ \vec {P}_0\psi =1 \end{array}\right. } \ . \end{aligned}$$

Denote $\phi _j$ as the jth column of the matrix $\phi$ and expand the first equation:

$$\begin{aligned} \vec {P}_0 \Big [\phi _1 \ \ \phi _2 \ \ \ldots \ \ \phi _{2N} \Big ] = \langle 0,0,\ldots ,0 \rangle \ . \end{aligned}$$

Now replace the first column of the matrix $\phi$ by the second equation:

$$\begin{aligned} \vec {P}_0 \Big [\psi \ \ \phi _2 \ \ \ldots \ \ \phi _{2N} \Big ] = \langle 1,0,\ldots ,0 \rangle \ . \end{aligned}$$

This system has a unique solution for $\vec {P}_0$.

1.2 A.2 N-Limited scenario

The notation in this case is as follows:

$$\begin{aligned} \begin{aligned}&\phi = \begin{pmatrix} B_0 &{} C_1 \\ B_1 &{} A_1+RA_2 \\ \end{pmatrix}, \\&\psi =\langle \vec {e},(I-R)^{-1}\vec {e} \rangle \ , \end{aligned} \end{aligned}$$

where $\phi$ is a matrix of size $(3N+1)\times (3N+1)$ and $\psi$ is a column vector of size $3N+1$, in which the first $N+1$ entries are ones. Similarly to the Exact-N scenario, (19), (23) and (24) become

$$\begin{aligned} {\left\{ \begin{array}{ll} \langle \vec {P}_0,\vec {P}_1 \rangle \phi =\vec {0}, \\ \langle \vec {P}_0,\vec {P}_1 \rangle \psi =1, \end{array}\right. } \end{aligned}$$

and with a similar outcome:

$$\begin{aligned} \langle \vec {P}_0,\vec {P}_1 \rangle \Big [\psi \ \ \phi _2 \ \ \ldots \ \ \phi _{2N} \Big ] = \langle 1,0,\ldots ,0 \rangle \ . \end{aligned}$$

This gives a solution for $\vec {P}_0$ and $\vec {P}_1$.