Customer equilibrium in a single-server system with virtual and system queues

Abstract

Consider a non-preemptive M/M/1 system with two first-come first-served queues, virtual (VQ) and system (SQ). An arriving customer who finds the server busy decides which queue to join. Customers in the SQ have non-preemptive priority over those in the VQ, but waiting in the SQ is more costly. We study two information models of the system. In the unobservable model, customers are notified only whether the server is busy, and in the observable model they are also informed about the number of customers currently waiting in the SQ. We characterize the Nash equilibrium of joining strategies in the two models and demonstrate a surprising similarity of the solutions.

Appendix: Notation and proofs

Notation and definitions

SQ	System queue
VQ	Virtual queue
$C_\mathrm{s}, C_\mathrm{v}$	Waiting costs in the SQ and VQ
$\varphi$	The cost ratio, $\varphi = \frac{C_\mathrm{v}}{C_\mathrm{s}}$
$\lambda$	Mean arrival rate of customers to the system
$\mu$	Mean service rate
$r_\mathrm{s}$	Probability of joining the SQ in the unobservable case
$S(r_\mathrm{s})$	Expected net benefit for a customer following the strategy $r_\mathrm{s}$
$L_\mathrm{s}(t), L_\mathrm{v}(t)$	The number of customers in the SQ and the VQ at time t
E(L), $E(W_\mathrm{s})$, $E(W_\mathrm{v})$	Expected waiting time in the system, SQ and VQ, respectively
$\hat{E}[W|l_\mathrm{s}]$	The expected waiting time in the VQ in time units per customer
$E(L_\mathrm{s})$, $E(L_\mathrm{v})$	Expected number of customers in the SQ and VQ
$\rho , \rho _\mathrm{s}, \rho _\mathrm{v}$	Occupation rates in the entire system, the SQ and the VQ
P	Stationary probabilities
$l_\mathrm{s}$	Number of customers in the SQ
$s(l_\mathrm{s})$	Threshold strategy of joining the SQ
T	Threshold strategy, $T=n+r$ $(r \in [0,1], n \in \mathbb {N})$
f	Number of unoccupied places in the SQ
b(f)	Mean busy period when there are $n+1-f$ customers in the SQ
$b'(f)$	Normalized b(f)

Proof of Proposition 4.1

From the transition rate diagram,

$$\begin{aligned} (\lambda + \mu )P_{n+1,0}= \lambda r P_{n0}, \end{aligned}$$

and therefore

$$\begin{aligned} P_{n+1,0} =\frac{\rho r}{\rho +1}P_{n0} . \end{aligned}$$

(23)

A cut around nodes $0',00,10,\ldots ,j0$ gives

$$\begin{aligned} \lambda P_{j0} = \mu P_{01} + \mu P_{j+1,0} , \quad j=0,1,\ldots ,n , \end{aligned}$$

or, after reindexing,

$$\begin{aligned} P_{j0}=\rho P_{j-1,0}-P_{01}. \end{aligned}$$

We now substitute $P_{01}$ from the upper horizontal cut equation

$$\begin{aligned} \lambda (1-r) P_{n0} + \lambda P_{n+1,0}= \mu P_{01} \, \end{aligned}$$

and $P_{n+1,0}$ from (23) and obtain

$$\begin{aligned} P_{j0}=\rho P_{j-1,0}-\frac{1+ \rho -r}{1+\rho } \rho P_{n0} , \quad j=1,\ldots ,n+1 . \end{aligned}$$

A recursive application of this equation gives

$$\begin{aligned} P_{j0} = \rho ^j P_{00} - \frac{1+\rho - r}{1+\rho } P_{n0} \sum \limits _{k=1}^{j} \rho ^k ,\quad j=1,\ldots ,n+1 . \end{aligned}$$

(24)

Specifically, for $j=n$,

$$\begin{aligned} P_{n0} = \rho ^n P_{00} - \frac{1+\rho - r}{1+\rho } P_{n0} \sum \limits _{k=1}^{n} \rho ^k , \end{aligned}$$

giving

$$\begin{aligned} P_{n0} = \frac{(1+\rho )\rho ^n}{(1+ \rho ) \sum \nolimits _{k=0}^{n}\rho ^k -r \sum \nolimits _{k=1}^{n}\rho ^k} P_{00} . \end{aligned}$$

(25)

Substituting $P_{n0}$ in (23) gives (4).

From Eqs. (25) and (24), we obtain

$$\begin{aligned} P_{j0} = \rho ^j P_{00} - \frac{1+ \rho - r}{1 + \rho } \frac{(1+\rho )\rho ^n}{(1+ \rho ) \sum \nolimits _{k=0}^{n}\rho ^k -r \sum \nolimits _{k=1}^{n}\rho ^k} \sum \limits _{k=1}^{j} \rho ^k P_{00}, \end{aligned}$$

which gives (3).

Proof of Proposition 4.2

Equation (8) follows from the horizontal cut between the rows $i,i-1$:

$$\begin{aligned} \lambda (1-r) P_{ni} + \lambda P_{n+1,i}= \mu P_{0,i+1} , \quad i=0,1,2,\ldots . \end{aligned}$$

(26)

A cut that contains the nodes $0i,1i,\ldots ,ji$ gives

$$\begin{aligned} \lambda P_{ji} + \mu P_{0i} = \mu P_{j+1,i} + \mu P_{0,i+1}\quad \begin{array}{lr} i=1,2,\ldots , \\ j=0,1,2,\ldots , n-1 . \end{array} \end{aligned}$$

(27)

By Eq. (26) and reindexing, we have

$$\begin{aligned} P_{ji} = P_{0i}-(1-r)\rho P_{ni} -\rho P_{n+1,i}+ \rho P_{j-1,i}\quad \begin{array}{lr} i=1,2,\ldots , \\ j=1, 2, . . . ,n. \end{array} \end{aligned}$$

A recursive application of this relation leads to Eq. (7).

We now find an expression for $P_{ni}$ and $P_{n+1,i}$. Equation (7) for $P_{ni}$ yields

$$\begin{aligned} \left[ \sum \limits _{k=0}^{n}\rho ^k -r\sum \limits _{k=1}^{n}\rho ^k\right] P_{ni} = \sum \limits _{k=0}^{n}\rho ^k P_{0i} - \sum \limits _{k=1}^{n}\rho ^k P_{n+1,i}, \end{aligned}$$

(28)

and the cut around the node $(n+1)i$ is

$$\begin{aligned} (1+\rho )P_{n+1,i} = \rho P_{n+1,i-1} + \rho rP_{ni} . \end{aligned}$$

(29)

By Eqs. (8), (28) and (29) we have

$$\begin{aligned} P_{ni} = \frac{ (1+\rho )(1-r) \sum \nolimits _{k=1}^{n+1}\rho ^k}{\sum \nolimits _{k=1}^{n+1}\rho ^k + \sum \nolimits _{k=0}^{n}\rho ^k - r\sum \nolimits _{k=1}^{n}\rho ^k} P_{n,i-1} + \frac{\sum \nolimits _{k=1}^{n+2}\rho ^k}{\sum \nolimits _{k=1}^{n+1}\rho ^k + \sum \nolimits _{k=0}^{n}\rho ^k - r\sum \nolimits _{k=1}^{n}\rho ^k} P_{n+1,i-1} . \end{aligned}$$

(30)

With this result and Eq. (29), we also get

$$\begin{aligned} P_{n+1,i}&= \left[ 1+\frac{\rho r\sum \nolimits _{k=1}^{n+2}\rho ^k}{\sum \nolimits _{k=1}^{n+1}\rho ^k + \sum \nolimits _{k=0}^{n}\rho ^k - r\sum \nolimits _{k=1}^{n}\rho ^k} \right] \frac{\rho }{\rho + 1} P_{n+1,i-1} \nonumber \\&\quad +\frac{(1+\rho )(1-r) r \sum \nolimits _{k=2}^{n+2}\rho ^k}{\sum \nolimits _{k=1}^{n+1}\rho ^k + \sum \nolimits _{k=0}^{n}\rho ^k - r\sum \nolimits _{k=1}^{n}\rho ^k} P_{n,i-1} . \end{aligned}$$

(31)

In terms of X, Y, Z, as defined in Eqs. (9)–(11), we have

$$\begin{aligned} P_{ni} = YP_{n,i-1} + XP_{n+1,i-1} , \quad P_{n+1,i} = ZP_{n+1,i-1} + \frac{\rho r}{1+\rho } YP_{n,i-1}, \end{aligned}$$

therefore

$$\begin{aligned} P_{ni}= & {} Y^iP_{n0} + X\sum \limits _{k=0}^{i-1}Y^kP_{n+1,i-1-k} ,\\ P_{n+1,i}= & {} Z^iP_{n+1,0} + \frac{\rho r}{1+\rho } Y\sum \limits _{k=0}^{i-1}Z^kP_{n,i-1-k}, \end{aligned}$$

and we get (5) and (6).