Optimal staffing for ticket queues

Abstract

Ticket queues are popular in many service systems. Upon arrival, each customer is issued a numbered ticket and receives service on a first-come-first-served basis according to the ticket number. There is no physical queue; customers may choose to walk away and return later (before their numbers are called) to receive service. We study the problem of optimal staffing in such a system with two capacity levels, where the staffing decision can only be based on ticket numbers, as opposed to the physical queue length in a traditional system. Using renewal reward theorem, we first derive the long-run average total cost (including customer delay and abandonment costs, operating cost and cost for changing staffing levels) and then obtain the optimal solution using fractional programming. In addition, we pursue a random-walk analysis, which leads to some highly accurate approximations.

7 Appendix

Proof of Lemma 1

First we look at $\mathsf{E}T_{11}$. Let $\widetilde{\mathbf{D}}_1$ be the $N\times N$-dimensional matrix obtained by removing the entries in generator $\mathbf{D}$ associated with state $(1,N+1)$:

(7.1)

From the phase-type distribution theory,

$$\begin{aligned} \mathsf{E}T_1=(\underbrace{0,\dots ,0}_{L+1},1,0,\dots ,0)\times (-\widetilde{\mathbf{D}}_1^{-1})\times \mathbf {e'}, \end{aligned}$$

(7.2)

where ${\mathbf {e}}'$ is the transpose of the $N-$dimensional unit vector. It is direct to verify the inverse of $\widetilde{\mathbf{D}}_1$, denoted by $\widetilde{\mathbf{D}}_1^{-1}=(\widetilde{d}_{ij})_{N\times N}$, can be written as

(7.3)

$\mathsf{E}T_{11}$ directly follows (7.2)–(7.3). Note that the expectation of $T_{12}$ is $(1+\mathsf{E}X)$ multiplied by $\mathsf{E}T_{11}$ replacing $\theta _1$ and $\rho _1$ by $\theta _2$ and $\rho _2$, respectively. Hence, $\mathsf{E}T_{12}$ can be obtained by (7.2)–(7.3) replacing $\theta _1$ and $\rho _1$ by $\theta _2$ and $\rho _2$, respectively. $\square$

Proof of Lemma 3

First we consider the customer delay cost incurred by the period $T_{11}$. For the Markov chain $\{(S_1(t),S_2(t),Q(t)), t\ge 0\}$ given in (2.2) starting with state $(1,0,L+1)$, let V(1, 0, i) be the number of visits to state (1, 0, i) during $T_1$, $i=0,\cdots , N.$ In view of the definition of $\widetilde{\mathbf{D}}_1^{-1}$ given by (7.3). Let $f_{L+1,i}$ be the probability that the chain visits state (1, 0, i) from state $(1,0,L+1)$ and $f_{i,i}$ the probability that the chain revisits state (1, 0, i), $i=0,1,\cdots , N$. From the theory of absorbing property of the first passage probability of the transient Markov chain, we know that V(1, 0, i) is a geometric random variable with

$$\begin{aligned} \mathsf{Pr}\left( V(1,0,i)=n\right) =f_{L+1,i}\times (1-f_{ii})\times (f_{ii})^{n-1}, \ \ n=1, 2, \cdots , i\ne L+1, \\ \mathsf{Pr}\left( V(1,0,L+1)=n\right) =(1-f_{L+1,L+1})\times (f_{L+1,L+1})^{n-1}, \ \ n=1, 2, \cdots , i=L+1. \end{aligned}$$

The first passage probabilities $f_{L+1,i}$ and $f_{ii}$ can be computed by

$$\begin{aligned}&f_{L+1,i}=\frac{\sum _{n=1}^\infty p_{L+2,i+1}^n}{\sum _{n=0}^\infty p_{i+1,i+1}^n}, \ i\ne L+1,\\&f_{ii}=\frac{\sum _{n=1}^\infty p_{i+1,i+1}^n}{\sum _{n=0}^\infty p_{i+1,i+1}^n}, \ i=0,1,\cdots ,N. \end{aligned}$$

Here, $p^n_{ij}$ is the n-step transition probability for the transition probability matrix given by

$$\begin{aligned} \mathbf{P}=(p_{ij})_{(N+1)\times (N+1)}=\frac{1}{\lambda +\mu } \widetilde{\mathbf{D}}_1+I. \end{aligned}$$

Here $\widetilde{\mathbf{D}}_1$ is given by (7.1), and I is an identity matrix. This implies

$$\begin{aligned} \left( I-\mathbf {P}\right) ^{-1}=(\lambda +\mu _1) \left( -\widetilde{\mathbf{D}}_1\right) ^{-1}. \end{aligned}$$

In view of (7.3), we have

$$\begin{aligned} f_{L+1,i}=\frac{{{\widetilde{d}}}_{L+2,i+1}}{{{\widetilde{d}}}_{i+1,i+1}} \ \ \text{ and } \ \ f_{ii}=\frac{(\lambda +\mu _1){{\widetilde{d}}}_{i+1,i+1}+1}{(\lambda +\mu _1){{\widetilde{d}}}_{i+1,i+1}}. \end{aligned}$$

We have

$$\begin{aligned} \mathsf{E} V(1,0,i)=- (\lambda +\mu _1){{\widetilde{d}}}_{L+2,i+1}, \ i=0,\cdots , N. \end{aligned}$$

This implies that the expected sojourn times in states (1, i) are

$$\begin{aligned} \frac{1}{\lambda +\mu _1} \mathsf{E} V(1,0,i)=- {{\widetilde{d}}}_{L+2,i+1}, \ i=0,\cdots , N. \end{aligned}$$

(7.4)

Given the ticket queue length i, the corresponding number of waiting customers follows a binomial distribution with mean $(1-{\alpha }_1)i$. Then, the customer delay cost in $T_{11}$ is

$$\begin{aligned}&h\left[ (1-{\alpha }_1)(-{{\widetilde{d}}}_{L+2,3})+2(1-{\alpha }_1)(-\widetilde{d}_{L+2,4})+\cdots \right. \nonumber \\&\quad \left. +(N-1)(1-{\alpha }_1)(-\widetilde{d}_{L+2,N+1})\right] . \end{aligned}$$

(7.5)

The customer delay cost in $T_{12}$ can be obtained by (7.5) in which $\mu _1$ is replaced by $\mu _2$. Hence the proof of the lemma is completed. $\square$

Proof of Lemma 4

First, according to Theorem 1 of Omahen and Marathe [21],

$$\begin{aligned} \mathsf{E} \Big (W \Big | \text{ arriving } \text{ during } (T^{(s)}_{21}+T^{(s)}_{22})\Big )= \frac{\lambda \overline{{\alpha }}_2}{\mu (\mu -\lambda \overline{{\alpha }}_2)} + \frac{\mathsf{E}(T^{(s)}_{21})^2}{2\mathsf{E} T^{(s)}_{21}}. \end{aligned}$$

(7.6)

The Laplace–Stieltjes transform of $T^{(s)}_{21}$ is given by

$$\begin{aligned} \mathsf{E} e^{-s T^{(s)}_{21}}=\sum _{i=0}^{N-L^+}\genfrac(){0.0pt}1{N-L^+}{i}{\alpha }_2^{N-L^+-i}\overline{{\alpha }}_2^i \left( \frac{\mu }{\mu +s} \right) ^i =\left( \frac{\mu +s{\alpha }_2}{\mu +s} \right) ^{N-L^+}. \end{aligned}$$

This implies

$$\begin{aligned} \mathsf{E}T^{(s)}_{21}= & {} \frac{(N-L^+)\overline{{\alpha }}_2}{\mu }, \nonumber \\ \mathsf{E} \Big (T_{21}^{(s)}\Big )^2= & {} \frac{\overline{{\alpha }}_2^2(N-L^+)^2+(1-{\alpha }_2^2)(N-L^+)}{\mu ^2}. \end{aligned}$$

(7.7)

Using (7.6), we have

$$\begin{aligned}&\mathsf{E} \Big (W \Big | \text{ arriving } \text{ during } (T^{(s)}_{21}+T^{(s)}_{22})\Big )\nonumber \\&\qquad = \frac{\lambda \overline{{\alpha }}_2}{\mu (\mu -\lambda \overline{{\alpha }}_2)} + \frac{\overline{{\alpha }}_2(N-L^+)+1+{\alpha }_2}{2\mu }. \end{aligned}$$

(7.8)

By (2.6),

$$\begin{aligned} \mathsf{E}(T^{(s)}_{21}+T^{(s)}_{22})&=\Big [p_2 \mathsf{E}\tau _2 +p_3 \left( \mathsf{E} \tau _2+\mathsf{E} \tau _3\right) +\cdots + p_{N-L+1}\sum \limits _{i=1}^{N-L}\mathsf{E}\tau _{i+1}\Big ]\\&=\frac{ \overline{{\alpha }}_2(N-L^+)}{\mu -\lambda \overline{{\alpha }}_2}. \end{aligned}$$

Hence, the lemma for $L>-1$ follows from $T^{(s)}_{23}=0$, (2.14) and (7.8).

Now we consider the case $L=-1$. According to the definition of $T^{(s)}_{23}$ and $\tau _{11}$ and $\tau _{12}$ in (2.6), we have

$$\begin{aligned} \mathsf{E}\Big (W \Big |\hbox { arrivals during}\ T^{(s)}_{23}\Big )&=\frac{\mu _1}{\mu }\mathsf{E}\Big (W \Big |\hbox { arrivals during}\ \tau _{12}\Big )\nonumber \\&\quad + \frac{\mu _2}{\mu }\mathsf{E}\Big (W \Big |\hbox { arrivals during}\ \tau _{11}\Big ). \end{aligned}$$

(7.9)

Taking derivative with resect to “s" in (2.8)–(2.9), we have

$$\begin{aligned} \mathsf{E}\tau _{11}=\frac{1/(\lambda +\mu _1)}{1-\lambda \mu _2/[\mu (\lambda +\mu _1)]} +\frac{\lambda /(\lambda +\mu _1)}{1-\lambda \mu _2/[\mu (\lambda +\mu _1)]} \mathsf{E}\tau _2 +\frac{\lambda \mu _1/[\mu (\lambda +\mu _1)]}{1-\lambda \mu _2/[\mu (\lambda +\mu _1)]}\mathsf{E}\tau _{12},\\ \mathsf{E}\tau _{12}=\frac{1/(\lambda +\mu _2)}{1-\lambda \mu _1/[\mu (\lambda +\mu _2)]} +\frac{\lambda /(\lambda +\mu _2)}{1-\lambda \mu _1/[\mu (\lambda +\mu _2)]} \mathsf{E}\tau _2 +\frac{\lambda \mu _2/[\mu (\lambda +\mu _2)]}{1-\lambda \mu _1/[\mu (\lambda +\mu _2)]}\mathsf{E}\tau _{11}. \end{aligned}$$

Therefore, we can write down $\mathsf{E}\Big (W|\text { arriving during } \tau _{11} \Big )$ and $\mathsf{E}\Big (W|\text { arriving during } \tau _{12} \Big )$ as

$$\begin{aligned} \mathsf{E}\Big (W|\text { arriving during } \tau _{11} \Big )= & {} \frac{{\lambda }/{(\lambda +\mu _1)}}{1-\lambda \mu _2/[\mu (\lambda +\mu _1)]} \cdot \frac{\mathsf{E}\tau _2}{\mathsf{E}\tau _{11}} \cdot \mathsf{E} \Big (W|\text { arriving during } \tau _2 \Big ) \\&+\frac{\lambda \mu _1/[\mu (\lambda +\mu _1)]}{1-\lambda \mu _2/[\mu (\lambda +\mu _1)]} \cdot \frac{\mathsf{E}\tau _{12}}{\mathsf{E}\tau _{11}} \cdot \mathsf{E} \Big (W|\text { arriving during } \tau _{12} \Big ),\nonumber \\ \mathsf{E}\Big (W|\text { arriving during } \tau _{12}\Big )= & {} \frac{ \lambda /(\lambda +\mu _2)}{1-\lambda \mu _1/[\mu (\lambda +\mu _2)]} \cdot \frac{\mathsf{E}\tau _2}{\mathsf{E}\tau _{12}}\cdot \mathsf{E}\Big (W|\text { arriving during } \tau _2 \Big ) \nonumber \\&+\frac{\lambda \mu _2/[\mu (\lambda +\mu _2)]}{1-\lambda \mu _1/[\mu (\lambda +\mu _2)]} \cdot \frac{\mathsf{E}\tau _{11}}{\mathsf{E}\tau _{12}} \cdot \mathsf{E}\Big (W|\text { arriving during } \tau _{11} \Big ). \end{aligned}$$

By Theorem 2 of Omaben and Marathe (1978),

$$\begin{aligned} \mathsf{E}\Big (W|\text { arriving during } \tau _2 \Big )=\frac{1}{\mu -\lambda \overline{{\alpha }}_2}. \end{aligned}$$

Hence, we have

$$\begin{aligned}&\mathsf{E}\Big (W|\text { arriving during } \tau _{11} \Big )\nonumber \\&\qquad =\mathsf{E}\Big (W|\text { arriving during } \tau _{12} \Big ) =\frac{\lambda }{(\mu +\lambda {\alpha }_2)(\mu -\lambda \overline{{\alpha }}_2)}. \end{aligned}$$

(7.10)

Recalling from (2.6) that

$$\begin{aligned} \mathsf{E}T^{(s)}_{23}&={\alpha }_2^N\mathsf{E}\tau _{11}+(1-{\alpha }_2^N) \Big (\frac{\mu _1}{\mu }\mathsf{E}\tau _{12} +\frac{\mu _2}{\mu }\mathsf{E}\tau _{11} \Big ), \end{aligned}$$

(7.11)

we know that

$$\begin{aligned}&\mathsf{E}\Big (W|\text { arriving during } T^{(s)}_{23} \Big )\nonumber \\&\quad = \Big ({\alpha }_2^N+(1-{\alpha }_2^N) \frac{\mu _2}{\mu }\Big )\frac{\mathsf{E} \tau _{11}}{\mathsf{E}T^{(s)}_{23}} \mathsf{E}\Big (W|\text { arriving during } \tau _{11} \Big )\nonumber \\&\qquad +(1-{\alpha }_2^N) \frac{\mu _1}{\mu }\frac{\mathsf{E} \tau _{12}}{\mathsf{E}T^{(s)}_{23}}\mathsf{E}\Big (W|\text { arriving during } \tau _{12} \Big ). \end{aligned}$$

(7.12)

Combining (7.10)–(7.12) yields that

$$\begin{aligned}&\mathsf{E}T^{(s)}_{23} \times \mathsf{E}\Big (W|\text { arriving during } T^{(s)}_{23}\Big )\nonumber \\&\qquad =\frac{\lambda }{\mu -\lambda \overline{{\alpha }}_2}\frac{\alpha _2^N\mu _1(\mu _2-\mu _1)+\mu _1(\lambda +\mu _1)+\mu _2(\lambda +\mu _2) }{\mu _1\mu _2(\mu +2\lambda )(\mu -\lambda \overline{\alpha }_2)}. \end{aligned}$$

The lemma for $L=-1$ directly follows from (2.14) and (7.8). $\square$

Proof of Lemma 6

In view of Lemma 3, we know that

$$\begin{aligned} \mathsf{E}T_{10}=\frac{1}{\lambda +\mu _1}\mathsf{E} V(1,0,0). \end{aligned}$$

By (7.3)–(7.4), we have the lemma for $\mathsf{E}T_{10}$. Now consider $\mathsf{E}T_{20}$. Let $\tau ^{(0)}_{1i}$ is the accumulative time for one server idle during $\tau _{1i}, i=1,2$. Then we have

$$\begin{aligned} \mathsf{E} \tau ^{(0)}_{11}= & {} \frac{1}{\lambda +\mu _1}+\frac{\lambda }{\lambda +\mu _1} \Big [ \frac{\mu _2}{\mu }\mathsf{E} \tau ^{(0)}_{11}+\frac{\mu _1}{\mu }\mathsf{E} \tau ^{(0)}_{12}\Big ],\\ \mathsf{E} \tau ^{(0)}_{12}= & {} \frac{1}{\lambda +\mu _2}+\frac{\lambda }{\lambda +\mu _2} \Big [ \frac{\mu _2}{\mu }\mathsf{E} \tau ^{(0)}_{11}+\frac{\mu _1}{\mu }\mathsf{E} \tau ^{(0)}_{12}\Big ]. \end{aligned}$$

This gives that

$$\begin{aligned} \mathsf{E}T_{20}= & {} \Big [{\alpha }_2^{N}+\frac{\mu _2}{\mu }(1-{\alpha }_2^N) \Big ] \mathsf{E}\tau ^{(0)}_{11}+\frac{\mu _1}{\mu }(1-{\alpha }_2^N) \mathsf{E}\tau ^{(0)}_{12}\nonumber \\= & {} \frac{{\alpha }_2^N\mu _1(\mu _2-\mu _1)+\mu _2(\lambda +\mu _2)+\mu _1(\lambda +\mu _1) }{\mu _1\mu _2(\mu +2\lambda )}, \end{aligned}$$

which prove the lemmas for $\mathsf{E}T_{20}$. $\square$

Proof of Lemma 7

The first three results directly follow from the random-walk theory (see, for example, [24]). (iv), by noting that T is a stopping time for the sequence $\{Y_n, n\ge 1\}$, follows from Wald’s equation. Now we show (v). Let ${{{\mathcal {F}}}}_n$ be the sigma field generated by $\{(X_i,Y_i),i=1,\dots ,n \}$. Note both $\{S_n\}$ and $\{Y_n\}$ are adapted to the filtration $\{ {\mathcal {F}}_n, n\ge 1 \}$. Hence

$$\begin{aligned} {E}\Big (\sum _{i=1}^nS_{i-1}Y_i\Big |{\mathcal {F}}_{n-1}\Big )= \sum _{i=1}^{n-1}S_{i-1}Y_i+{E}(S_{n-1}Y_n|{\mathcal {F}}_{n-1}) = \sum _{i=1}^{n-1}S_{i-1}Y_i+S_{n-1}{E}Y_n, \end{aligned}$$

(7.13)

where we use the fact that $Y_n$ is independent of ${\mathcal {F}}_{n-1}$. Taking expectation on both sides of (7.13), we have

$$\begin{aligned} {E} \Big (\sum _{i=1}^nS_{i-1}Y_i\Big ) ={E} \Big ( \sum _{i=1}^{n-1}S_{i-1}Y_i \Big )+(\mathsf{E} S_{n-1})\times {E}Y_n =\cdots ={E}\Big (\sum _{i=1}^{n-1}S_i\Big )\times \mathsf{E} Y_1. \end{aligned}$$

Hence,

$$\begin{aligned} {E} \Big (\sum _{i=1}^TS_{i-1}Y_i\Big )= & {} \mathsf{E} Y_1 \times {E}\Big (\sum _{i=1}^{T-1}S_i\Big )= \mathsf{E} Y_1 \times \Big [ {E}\Big (\sum _{i=1}^{T}S_i\Big ) -{E}S_T \Big ]\nonumber \\= & {} \mathsf{E} Y_1 \times \Big [ {E}\Big (\sum _{i=1}^{T}S_i\Big ) -\gamma \times {E}T \Big ]. \end{aligned}$$

(7.14)

Now we consider ${E}\Big (\sum _{i=1}^{T}S_i\Big )$. Write $X_i=\xi _i+\gamma$, where $\xi _i$’s are i.i.d., and ${E}\xi _i=0$, $\textsf {Var}(\xi _i)=\sigma ^2$, we have

$$\begin{aligned} \sum _{i=1}^{T}S_i= \sum _{i=1}^{T}\sum _{n=1}^{i}X_n= & {} \sum _{n=1}^{T}(T-n+1)X_n=TS_T-\sum _{n=1}^{T}(n-1)(\xi _n+\gamma ) \\= & {} TS_T-\frac{\gamma }{2}T(T-1)-\sum _{n=1}^{T}(n-1)\xi _n. \end{aligned}$$

The last term above is a martingale. Hence

$$\begin{aligned} {E}\Big (\sum _{i=1}^{T}S_i\Big ) ={E}(TS_T)-\frac{\gamma }{2}{E}T^2 +\frac{\gamma }{2}{E}T. \end{aligned}$$

(7.15)

Applying optimal stopping theorem to the martingale $\{ (S_n-n\gamma )^2-n\sigma ^2 \}$ yields

$$\begin{aligned} {E}\Big (S_T-T\gamma \Big )^2=\sigma ^2{E}T ~~\Rightarrow & {} ~~{E}(S_T-T\gamma )^2=\sigma ^2{E}T \nonumber \\\Rightarrow & {} ~~ {E}S_T^2+\gamma ^2{E}T^2-2\gamma {E}(S_TT) =\sigma ^2{E}T \nonumber \\\Rightarrow & {} ~~ {E}(S_TT)=\frac{{E}S_T^2 }{2\gamma }-\frac{1-\gamma ^2}{2\gamma }{E}T +\frac{\gamma }{2}{E}T^2.\nonumber \\ \end{aligned}$$

(7.16)

Plug (7.16) into (7.15), and we have

$$\begin{aligned} {E}\Big (\sum _{i=1}^{T}S_i\Big ) =\frac{{E}S_T^2}{2\gamma }+\frac{2\gamma ^2-1}{2\gamma }{E}T. \end{aligned}$$

Since, by $\mathsf{E} S_T=\gamma \cdot \mathsf{E} T=A\pi _A-B(1-\pi _A)$,

$$\begin{aligned} {E}S_T^2=A^2\pi _A+B^2(1-\pi _A)=(A-B)\gamma {E}T+AB, \end{aligned}$$

we simplify ${E}\Big (\sum _{i=1}^{T}S_i \Big )$ as

$$\begin{aligned} {E}\Big (\sum _{i=1}^{T}S_i\Big ) = \Big (\frac{A-B}{2}+\gamma -\frac{1}{2\gamma } \Big ) {E}T +\frac{AB}{2\gamma }. \end{aligned}$$

Thus, by (7.14), we have

$$\begin{aligned} {E} \Big (\sum _{i=1}^TS_{i-1}Y_i\Big )= & {} \mathsf{E} Y_1 \times \Big [\Big (\frac{A-B}{2} +\gamma -\frac{1}{2\gamma } \Big ){E}T+\frac{A B}{2\gamma } -\gamma {E}T \Big ]\\= & {} \mathsf{E} Y_1 \times \Big [\Big (\frac{A-B}{2}-\frac{1}{2\gamma } \Big ){E}T +\frac{A B}{2\gamma } \Big ]. \end{aligned}$$

This gives (v). Therefore, the lemma is proved. $\square$

Proof of Lemma 8

Going along the line of the proof of Lemma 7, the lemma can be proved similarly. $\square$