Performance analysis of service systems with priority upgrades

Abstract

In this paper, we study the performance of service systems with priority upgrades. We model the service system as a single-server two-class priority queue, with queue 1 as the normal queue and queue 2 as the priority queue. The queueing model of interest has various applications in healthcare services, perishable inventory and project management. We comprehensively examine the system’s stationary distribution, computational algorithm design and sensitivity analysis. We observe that when queue 2 is large, the conditional distribution of queue 1 approximates a Poisson distribution. The tail probability of queue 2 decays geometrically, while the tail probability of queue 1 decays much faster than queue 2’s. This helps us design an algorithm that computed the stationary distribution. Finally, by using the algorithm, we perform a sensitivity analysis on various system parameters, i.e., the arrival rates, service rates and the upgrade rate. The numerical study provides helpful insights into designing such service systems.

Appendices

Appendix 1

The current theory for the tail types and asymptotics of the stationary distributions is mainly for discrete-time processes. We first review the basic sufficient conditions for the discrete-time QBD process to have a stationary distribution whose tail decays geometrically and then tailor the theory to the continuous-time process.

The discrete-time QBD process is introduced as follows. Let $\{(X_{n}, Y_{n}), n = 0, 1, {\ldots } \}$ be a discrete-time Markov chain with countable state space S. Assume that $X_{n}$ is a nonnegative integer value, and $Y_{n}$ has the state space $S_{0}$ if $X_{n} = 0$, the state space $S_{1}$ if $X_{n} = 1$, etc. Thus, $S = (\{0\} \times S_{0}) \cup (\{1\} \times S_{1})$. We refer to $X_{n}$ and $Y_{n}$ as the level and background process, respectively. The transition probability matrix P of the Markov process is given by

$$\begin{aligned} P=\left( {{\begin{array}{ccccc} {B_0 }&{}\quad {B_1 }&{}\quad &{}\quad &{} \\ {B_{-1} }&{}\quad {A_0 }&{}\quad {A_1 }&{}\quad &{} \\ &{}\quad {A_{-1} }&{}\quad {A_0 }&{}\quad {A_1 }&{} \\ &{}\quad &{}\quad \ddots &{}\quad \ddots &{}\quad \ddots \\ \end{array} }} \right) , \end{aligned}$$

(24)

where the block size may be finite or infinite.

Lemma 2

(Neuts 1981) Suppose that P defined in Eq. 24 is ergodic. Let ${\varvec{v}} = ({\varvec{v}}_{n}, n \ge 0$) be its stationary distribution. There exists a minimal nonnegative solution R of the matrix equation:

$$\begin{aligned} R=R^{2}A_{-1} +RA_0 +A_1 , \end{aligned}$$

(25)

and the stationary distribution has the following matrix geometric form:

$$\begin{aligned}&{\varvec{\upnu }_n =\varvec{\upnu }_1 R^{n-1},} \quad {n>1}, \end{aligned}$$

(26)

$$\begin{aligned}&\varvec{\upnu }_0 B_0 +\varvec{\upnu }_1 B_{-1} =\varvec{\upnu }_0 , \end{aligned}$$

(27)

$$\begin{aligned}&\varvec{\upnu }_0 B_1 +\varvec{\upnu }_1 \left( {A_0 +RA_{-1} } \right) =\varvec{\upnu }_1 . \end{aligned}$$

(28)

Lemma 3

Under the assumption of Lemma 2, we define the matrix generating function $A^{*}(z) = z^{-1}A_{1} + A_{0}+ zA_{-1}$ . If there exist a positive row vector ${\varvec{x}}$, a positive column vector ${\varvec{y}}$ , and a real number $z \in (0, 1)$ that satisfy the following conditions:

$$\begin{aligned} \mathbf{x}A^{{*}}(z)= & {} \mathbf{x}, \end{aligned}$$

(29)

$$\begin{aligned} A^{{*}}(z)\mathbf{y}= & {} \mathbf{y}, \end{aligned}$$

(30)

$$\begin{aligned} \mathbf{xy}< & {} \infty , \end{aligned}$$

(31)

$$\begin{aligned} \varvec{\upnu }_1 \mathbf{y}< & {} \infty , \end{aligned}$$

(32)

then we have the finite limitation

$$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow \infty } z^{-n}\varvec{\upnu }_n =\frac{\varvec{\upnu }_1 \mathbf{r}}{z\mathbf{xr}}{} \mathbf{x}, \end{aligned}$$

(33)

where $\mathbf{r} = (I-A_{0}- { RA}_{-1}-{ zA}_{-1})$ y.

Proof

Lemma 3 follows by Theorem 2.1.1 and 2.2.1 in Sakuma and Miyazawa (2005). $\square$

Proof of Theorem 1

Denote with P the transition probability matrix of its corresponding embedded Markov chain. The transition probability matrix P has the QBD form of Eq. 24. We have

$$\begin{aligned} P=\left( {{\begin{array}{cccc} {D_0^{-1} }&{}\quad &{}\quad &{} \\ &{}\quad {D^{-1}}&{}\quad &{} \\ &{}\quad &{}\quad {D^{-1}}&{} \\ &{}\quad &{}\quad &{}\quad \ddots \\ \end{array} }} \right) Q+I, \end{aligned}$$

(34)

where $D_{0}$ = −diag{$C_{0}$} and D = −diag{$Q_{0}$}; and I is the identity matrix. Let ${\varvec{v}} = ({\varvec{v}}_{0}, {\varvec{v}}_{1}, {\ldots })$ be the stationary distribution of the embedded Markov chain, i.e., ${\varvec{v}}P = {\varvec{v}}$ and ${\varvec{ve}} = 1$. Then, we have $\varvec{\uppi }_{n}= \beta ^{-1}{\varvec{v}}_{n}D^{-1}$, for all $n\ge 1$, where

$$\begin{aligned} \beta =\varvec{\upnu }_0 D_0^{-1} \mathbf{e}+\sum _{n=1}^\infty {\varvec{\upnu }_n D^{-1}{} \mathbf{e}} =\varvec{\upnu }_0 D_0^{-1} \mathbf{e}+\varvec{\upnu }_1 (I-R)^{-1}D^{-1}{} \mathbf{e}. \end{aligned}$$

(35)

From the assumption, P is positive recurrent, and its invariant vector ${\varvec{v}}$ is given by Lemma 2. Let $z = \rho$, and assume that $x_{0} = 1$ and $y_{0} = 1$. By Eq. 29, we have

$$\begin{aligned}&\frac{x_0 }{\lambda _1 +\lambda _2 +\mu _2 }(\lambda _2 z^{-1}+\mu _2 z)+\frac{x_1 }{\lambda _1 +\lambda _2 +\mu _2 +\lambda _T }\lambda _T z^{-1}=x_0 , \end{aligned}$$

(36)

$$\begin{aligned}&\frac{x_{i-1} \lambda _1 }{\lambda _1 +\lambda _2 +\mu _2 +(i-1)\lambda _T }+\frac{x_i (\lambda _2 z^{-1}+\mu _2 z)}{\lambda _1 +\lambda _2 +\mu _2 +i\lambda _T }\nonumber \\&\quad +\frac{x_{i+1} (i+1)\lambda _T z^{-1}}{\lambda _1 +\lambda _2 +\mu _2 +(i+1)\lambda _T }=x_i ,\quad {i>0.} \end{aligned}$$

(37)

From Eqs. 36 and 37, we have

$$\begin{aligned} {x_i =\frac{\lambda _1 +\lambda _2 +\mu _2 +i\lambda _T }{\lambda _1 +\lambda _2 +\mu _2 }\times \frac{\lambda _1^i }{\prod \limits _{k=1}^i {k\lambda _T } },}\quad {i\ge 0.} \end{aligned}$$

(38)

By Eq. 30, we have

$$\begin{aligned} (\lambda _2 z^{-1}+\mu _2 z)y_0 +\lambda _1 y_1= & {} (\lambda _1 +\lambda _2 +\mu _2 )y_0 , \end{aligned}$$

(39)

$$\begin{aligned} i\lambda _T z^{-1}y_{i-1} +(\lambda _2 z^{-1}+\mu _2 z)y_i +\lambda _1 y_{i+1}= & {} (\lambda _1 +\lambda _2 +\mu _2 +i\lambda _T )y_i ,\quad {i>0}. \end{aligned}$$

(40)

From Eqs. 39 and 40, we have

$$\begin{aligned} {y_i =\rho ^{-i},}\quad {i\ge 0.} \end{aligned}$$

(41)

It is easy to verify Eq. 31. To verify Eq. 32, we use Lemma 4.1 in Xie et al. (2009). First, we have

$$\begin{aligned} \mathbf{v}_1 \mathbf{y}=\sum _{q_1 \ge 0} {v_{(1, q_1 )} \rho ^{-q_1 }} \le \sum _{(q_{2} ,q_{1} )} {v_{(q_{2} ,q_{1} )} \rho ^{-q_1 }} . \end{aligned}$$

(42)

In addition, we know the marginal distribution of queue 1 decays faster than any geometric distribution. Then, it is easy to verify that

$$\begin{aligned} \sum _{q_1 \ge 0}^\infty {\varvec{\uppi }(\cdot ,q_1 )\rho ^{-q_1 }} <\;\infty . \end{aligned}$$

(43)

By Lemma 4.1 in Xie et al. (2009), we have

$$\begin{aligned} \mathbf{v}=\varvec{\uppi }D_1 /(\varvec{\uppi }D_1 \mathbf{e}), \end{aligned}$$

(44)

where $D_1 =-diag(Q).$

Define the following matrix,

$$\begin{aligned} D_{00} =D_{11} =\left( {{\begin{array}{cccc} {d_0 }&{}\quad &{}\quad &{} \\ &{}\quad {d_1 }&{}\quad &{} \\ &{}\quad &{}\quad {d_2 }&{} \\ &{}\quad &{}\quad &{}\quad \ddots \\ \end{array} }} \right) ,D_2 =\left( {{\begin{array}{cccc} {D_{00} }&{}\quad &{}\quad &{} \\ &{}\quad {D_{11} }&{}\quad &{} \\ &{}\quad &{}\quad {D_{11} }&{} \\ &{}\quad &{}\quad &{}\quad \ddots \\ \end{array} }} \right) , \end{aligned}$$

(45)

where $d_i =(\lambda _1 +\lambda _2 +\mu _1 +\mu _2 +\lambda _{\mathrm{T}} )/\rho ^{i}.$

Let $\varvec{\uptheta }\hbox {= }\varvec{\uppi }\hbox {D}_2 /(\varvec{\uppi }D_1 \mathbf{e})$, it is easy to verify that

$$\begin{aligned} \sum _{(q_{2} ,q_{1} )} {\theta _{(q_{2} ,q_{1} )} \rho ^{-q_1 }} =\frac{\lambda _1 +\lambda _2 +\mu _1 +\mu _2 +\lambda _{\mathrm{T}} }{\varvec{\uppi }D_1 \mathbf{e}}\sum _{(q_{2} ,q_{1} )} {\pi _{(q_{2} ,q_{1} )} (\rho ^{2})^{-q_1 }} <\infty . \end{aligned}$$

(46)

Thus,

$$\begin{aligned} \mathbf{v}_1 \mathbf{y}\le \sum _{(q_{2} ,q_{1} )} {v_{(q_{2} ,q_{1} )} \rho ^{-q_1 }} <\infty . \end{aligned}$$

(47)

Therefore, by Lemma 3, we have

$$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow \infty } \rho ^{-n}{} \mathbf{v}_n =\frac{\mathbf{v}_1 \mathbf{r}}{\rho \mathbf{xr}}{} \mathbf{x}. \end{aligned}$$

(48)

Since $\varvec{\uppi }_{n}= \beta ^{-1}{} \mathbf{v}_{n}D^{-1}$, we have

$$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow \infty } \rho ^{-n}\varvec{\uppi }_n =\frac{\mathbf{v}_1 \mathbf{r}}{\beta \rho \mathbf{xr}}{} \mathbf{x}D^{-1}. \end{aligned}$$

(49)

The constants $\mathbf{v}_{1}{} \mathbf{r}$ and $\mathbf{xr}$ are positive and finite. Denote with $\mathbf{c} = (c_{0}, c_{1},{\ldots })$ the Poisson distribution with parameter $\lambda _{1}/\lambda _{T}$. Define

$$\begin{aligned} \alpha =\frac{\mathbf{v}_1 \mathbf{r}}{\mathbf{xr}}\frac{\mu _2^{-1} e^{{\lambda _1 }/{\lambda _T }}}{\beta \rho (\rho +1)}>0. \end{aligned}$$

(50)

Finally, we have $\mathop {\lim }\limits _{n\rightarrow \infty } \rho ^{-n}\varvec{\uppi }_n =\alpha \mathbf{c}$. $\square$

Appendix 2

Proof of Theorem 2

1) Apparently, $s_{1} >s_{2}$. The following equation holds:

$$\begin{aligned} \mathop {\lim }\limits _{k\rightarrow \infty } \frac{{\varvec{\upeta }}_1 (k)}{{\varvec{\upeta }}_2 (k)}= & {} \mathop {\lim }\limits _{k\rightarrow \infty } \frac{{\varvec{\upeta }}_1 (0)\lambda _1^k \prod _{i=1}^k {\frac{1}{s_1 +i\lambda _T }} }{{\varvec{\upeta }}_2 (0)\lambda _1^k \prod _{i=1}^k {\frac{1}{s_2 +i\lambda _T }} }=\mathop {\lim }\limits _{k\rightarrow \infty } \frac{{\varvec{\upeta }}_1 (0)\prod _{i=1}^k {(s_2 +i\lambda _T )} }{{\varvec{\upeta }}_2 (0)\prod _{i=1}^k {(s_1 +i\lambda _T )} } \nonumber \\= & {} \frac{{\varvec{\upeta }}_1 (0)}{{\varvec{\upeta }}_2 (0)}\mathop {\lim }\limits _{k\rightarrow \infty } \prod _{i=1}^k {\frac{s_2 +i\lambda _T }{s_1 +i\lambda _T }}. \end{aligned}$$

(51)

Let $\mu _{1 }=\upalpha \lambda _{T}, \upalpha > 0$,

$$\begin{aligned} \mathop {\lim }\limits _{k\rightarrow \infty } \prod _{i=1}^k {\frac{s_1 +i\lambda _T }{s_2 +i\lambda _T }}= & {} \mathop {\lim }\limits _{k\rightarrow \infty } \prod _{i=1}^k {\frac{\mu _1 +i\lambda _T }{i\lambda _T }} =\mathop {\lim }\limits _{k\rightarrow \infty } \prod _{i=1}^k {\frac{i+\alpha }{i}} \nonumber \\= & {} \mathop {\lim }\limits _{k\rightarrow \infty } \prod _{i=1}^k {(1+\frac{\alpha }{i})} \ge \mathop {\lim }\limits _{k\rightarrow \infty } \left( {1+\sum _{i=1}^k {\frac{\alpha }{i}} } \right) \rightarrow +\infty . \end{aligned}$$

(52)

Therefore,

$$\begin{aligned} \mathop {\lim }\limits _{k\rightarrow \infty } \frac{{\varvec{\upeta }}_1 (k)}{{\varvec{\upeta }}_2 (k)}=0. \end{aligned}$$

(53)

2) For any small $\theta > 0$, we have

$$\begin{aligned} \frac{{\varvec{\upeta }}_1 (k+1)}{{\varvec{\upeta }}_1 (k)}=\frac{{\varvec{\upeta }}_1 (0)\lambda _1^{k+1} \prod _{i=1}^{k+1} {\frac{1}{s_1 +i\lambda _T }} }{{\varvec{\upeta }}_1 (0)\lambda _1^k \prod _{i=1}^k {\frac{1}{s_1 +i\lambda _T }} }=\frac{\lambda _1 }{s_1 +(k+1)\lambda _T }. \end{aligned}$$

(54)

From Eq. 54, we know that when k goes to infinity, and ${\varvec{\upeta }}_{1}(k+1)/{\varvec{\upeta }}_{1}(k)$ goes to zero, which is smaller than any positive values. Therefore, there exists $k^{*}=\max \{0,\left\lceil {(\lambda _1 /\theta -s_1 )/\lambda _T -1} \right\rceil \}$ such that for any $k>k^{*}$, we have

$$\begin{aligned} \frac{{\varvec{\upeta }}_1 (k+1)}{{\varvec{\upeta }}_1 (k)}<\theta . \end{aligned}$$

(55)

Therefore, ${\varvec{\upeta }}_{1}(k)$ approaches 0 faster than any geometric decay. The proof for the modified queue with service rate $s_{2}$ follows similarly. $\square$

Proof of Lemma 1

For any sample path $\omega$, we sort customers who arrive at queue 1 before time t into the following categories:

• (1) Customers who leave queue 1 without accepting any services: These customers belong to neither $N_{1}(t)$ nor $L_{2}(t)$.

• (2) Customers who are served by the server with service rate $\mu _{1}$: These customers do not belong to $N_{1}(t)$ but may belong to $L_{2}(t)$.

• (3) Customers who have been transferred but were attended by the server: These customers belong to neither $N_{1}(t)$ nor $L_{2}(t)$.

• 4) Customers who are still in queue 1: These customers belong to $L_{2}(t)$ as well.

Therefore, for any time t and sample path $\omega$, we have $N_{1}(t, \omega ) \le L_{2}(t, \omega$). This implies $N_{1}(t) \le _{st} L_{2}(t)$. The inequality $L_{1}(t) \le _{st }N_{1}(t)$ can be proved analogously.

Proof of Theorem 3

By Lemma 1, it is easy to see that, for any $n \ge$ 0,

$$\begin{aligned} \Pr \left\{ {L_1>n} \right\} \le \Pr \left\{ {N_1>n} \right\} \le \Pr \left\{ {L_2 >n} \right\} . \end{aligned}$$

(56)

Then, we have

$$\begin{aligned} \sum _{k=n+1}^\infty {\varvec{\uppi }(\cdot ,k)}= & {} \Pr \left\{ {N_1>n} \right\} \nonumber \\\le & {} \Pr \left\{ {L_2 >n} \right\} =\sum _{k=n+1}^\infty {{\varvec{\upeta }}_2 (k)} \nonumber \\= & {} \left( {1+\sum _{k=1}^\infty {\lambda _1^k \prod _{i=1}^k {\frac{1}{s_2 +i\lambda _T }} } } \right) ^{-1}\sum _{k=n+1}^\infty {\lambda _1^k \prod _{i=1}^k {\frac{1}{s_2 +i\lambda _T }} }. \end{aligned}$$

(57)

The other direction can be proved in a similar manner.

Given any $\gamma > 0$, let $k^{*}=\max \{0,\left\lceil {(\lambda _1 /\gamma -s_2 )/\lambda _T -1} \right\rceil \}$. For any $k>k^{*}$, we have

$$\begin{aligned} \sum _{j=k+1}^\infty {{\varvec{\upeta }}_2 (j)} ={\varvec{\upeta }}_2 (k)\sum _{j=1}^\infty {\prod _{i=1}^j {\frac{\lambda _1 }{s_2 +(i+k)\lambda _T }} } \le {\varvec{\upeta }}_2 (k)\sum _{j=1}^\infty {\gamma ^{j}} ={\varvec{\upeta }}_2 (k)\frac{\gamma }{1-\gamma }. \end{aligned}$$

(58)

It follows that

$$\begin{aligned} \varvec{\uppi }(\cdot ,k)<\sum _{j=k}^\infty {\varvec{\uppi }(\cdot ,j)} \le \;\sum _{j=k}^\infty {{\varvec{\upeta }}_2 (j)} \le (1+\gamma ){\varvec{\upeta }}_2 (k). \end{aligned}$$

(59)

The proof for the other half of the theorem can be completed in a similar manner. $\square$

Appendix 3

1) Deriving G iteratively by successive substitution:

This method, described by Neuts (1981), uses

$$\begin{aligned} {G_{(n+1)} =-\left( {\bar{{Q}}_1 +G_{(n)}^2 \bar{{Q}}_{-1} } \right) \bar{{Q}}_0^{-1} ,} \quad {n\ge 0}, \end{aligned}$$

(60)

which is derived from Eq. 21. Starting with $G_{(0)} = 0$, successive approximations of G can be obtained by using Eq. 60. The iteration is repeated until two consecutive iterates of G differ by less than the predefined tolerance $\varepsilon$:

$$\begin{aligned} \left\| {G_{(n+1)} -G_{(n)} } \right\| <\varepsilon , \end{aligned}$$

(61)

where ${\vert }{\vert }\cdot {\vert }{\vert }$ is an appropriate matrix norm. The sequence $\{G_{(n)}\}$ is entry-wise non-decreasing that can be proven by induction:

$$\begin{aligned} G_{(1)} =-\left( {\bar{{Q}}_1 +G_{(0)}^2 \bar{{Q}}_{-1} } \right) \bar{{Q}}_0^{-1} =-\bar{{Q}}_1 \bar{{Q}}_0^{-1} \ge 0=G_{(0)} . \end{aligned}$$

(62)

The matrices $-\bar{{Q}}_0^{-1}$ and $\bar{{Q}}_{-1}$ are non-negative. For $\bar{{Q}}_{-1}$, this is readily seen considering the structure of $\bar{{Q}}$. $\bar{{Q}}_0^{-1}$is non-positive because $\bar{{Q}}_0^{-1}$ is diagonally dominant with negative diagonal and non-negative off-diagonal elements.

If $G_{(n+1)} \ge G_{(n)}$, we have

$$\begin{aligned} G_{(n+2)} =-\left( {\bar{{Q}}_1 +G_{(n+1)}^2 \bar{{Q}}_{-1} } \right) \bar{{Q}}_0^{-1} \ge -\left( {\bar{{Q}}_1 +G_{(n)}^2 \bar{{Q}}_{-1} } \right) \bar{{Q}}_0^{-1} =G_{(n+1)}. \end{aligned}$$

(63)

The monotone convergence of $\{G_{(n)}\}$ toward G is shown by Neuts (1981).

2) Deriving $\varvec{\uptheta }_{0 }$ and $\varvec{\uptheta }_{1}$:

Taking the boundary balance equations and normalization condition $\varvec{\uptheta } \mathbf{e} = 1$, we have:

$$\begin{aligned} \left( {\varvec{\uptheta }_0 ,\varvec{\uptheta }_1 } \right) \left( {{\begin{array}{lll} {\bar{{C}}_0 }&{}\quad {(\bar{{C}}_1 )^{{*}}}&{}\quad \mathbf{e} \\ {\bar{{Q}}_{-1} }&{}\quad {(\bar{{Q}}_0 +G\bar{{Q}}_{-1} )^{{*}}}&{}\quad {(\mathbf{I}-G)^{-1}{} \mathbf{e}} \\ \end{array} }} \right) =\left( {0,\ldots ,0,1} \right) , \end{aligned}$$

(64)

where $(\cdot )^{{*}}$ indicates that the last column of the included matrix is removed to avoid linear dependency. The removed column is replaced by the normalization condition. Therefore, Eq. 64 is solved for computing $\varvec{\uptheta }_{0 }$ and $\varvec{\uptheta }_{1}$.

$$\begin{aligned} \left( {\varvec{\uptheta }_0 ,\varvec{\uptheta }_1 } \right) =\left( {0,\ldots ,0,1} \right) \left( {{\begin{array}{lll} {\bar{{C}}_0 }&{}\quad {(\bar{{Q}}_1 )^{{*}}}&{}\quad \mathbf{e} \\ {\bar{{Q}}_{-1} }&{}\quad {(\bar{{Q}}_0 +G\bar{{Q}}_{-1} )^{{*}}}&{}\quad {(\mathbf{I}-G)^{-1}{} \mathbf{e}} \\ \end{array} }} \right) ^{-1}. \end{aligned}$$

(65)

3) Deriving $\varvec{\uptheta }$:

The steady-state probability vectors $\varvec{\uptheta }_{i}$ can be obtained quite easily by using Eq. 20. Of course not all $\varvec{\uptheta }_{i}$ can be computed due to their infinite number, but the elements of $\varvec{\uptheta }_{i}$ converge toward 0 for increasing i since sp($G)<$ 1.