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Strategic customer behavior and optimal policies in a passenger–taxi double-ended queueing system with multiple access points and nonzero matching times

Abstract

This paper considers an observable double-ended queueing system of passengers and taxis, where matching times follow an exponential distribution. We assume that passengers are strategic and decide to join the queue only if their expected utility is nonnegative. We show that the strategy of passengers is represented by a unique vector of thresholds corresponding to different cases of the number of taxis observed in the system upon passenger arrival. Furthermore, we develop a heuristic algorithm to find an optimal range of fees to be levied on passengers to maximize social welfare or revenues.

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Acknowledgements

The research of Hung Q. Nguyen was supported by JST SPRING, Grant Number JPMJSP2124. The research of Tuan Phung-Duc was supported by JSPS KAKENHI Grant Numbers 18K18006 and 21K11765.

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Appendices

Appendix A Proof of Proposition 1

We will prove Proposition 1 by induction on p. The statement is equivalent to

$$\begin{aligned} T(p,j) \le T(p+1,j), \end{aligned}$$
(A1)

for any fixed values of j.

Since \(T(0,j) = 0\), it is obviously implied from the recursive formulas that \(T(0,j) \le T(1,j)\); thus, (A1) holds with \(p=0\). Assuming that (A1) holds with \(p=q-1\) for any integer \(q\ge 1\), which indicates, for any fixed value of j,

$$\begin{aligned} T(q-1,j) \le T(q,j). \end{aligned}$$
(A2)

We show that it holds with \(p=q\), which indicates that we need to prove that, for any fixed value of j,

$$\begin{aligned} T(q,j) \le T(q+1,j), \end{aligned}$$

by considering the following five cases.

  • When \(j=K\), from (2), we have

    $$\begin{aligned} T(q,j)=\frac{1}{S\mu } + T(q-1,K-1), \end{aligned}$$
    (A3)

    and

    $$\begin{aligned} T(q+1,j)=\frac{1}{S\mu } + T(q,K-1). \end{aligned}$$
    (A4)

    Since \(T(q-1,K-1) \le T(q,K-1)\) by assumption (A2), from (A3) and (A4), we obtain

    $$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ j=K. \end{aligned}$$
    (A5)
  • When \(S<j<K\), from (2), we have

    $$\begin{aligned} T(q,j) = \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,j+1) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,j-1), \end{aligned}$$

    and

    $$\begin{aligned} T(q+1,j) =\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q+1,j+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,j-1). \end{aligned}$$

    Now, due to (A5), it is seen that the inequality \(T(q,j) \le T(q+1,j)\) holds for \(j=K-1\) because

    $$\begin{aligned} T(q,K-1)&= \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,K) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2)\\&\le \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q+1,K)+\frac{S\mu }{\lambda _t+S\mu }T(q,K-2)\\&= T(q+1,K-1). \end{aligned}$$

    Then, it is easily obtained by induction on j, that

    $$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ S<j<K. \end{aligned}$$
    (A6)
  • When \(j=0\), from (1), we have

    $$\begin{aligned} T(q,0) =\frac{1}{\lambda _t} + T(q-1,1), \end{aligned}$$
    (A7)

    and

    $$\begin{aligned} T(q+1,0) =\frac{1}{\lambda _t} + T(q,1). \end{aligned}$$
    (A8)

    Since \(T(q-1,1) \le T(q,1)\) because of the inductive assumption, from (A7) and (A8), we obtain

    $$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ j=0. \end{aligned}$$
    (A9)
  • When \(0<j<S\), from (2), we have

    $$\begin{aligned} T(q,j) =\frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }T(q-1,j+1) +\frac{j\mu }{\lambda _t+j\mu }T(q,j-1), \end{aligned}$$

    and

    $$\begin{aligned} T(q+1,j) =\frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }T(q,j+1) +\frac{j\mu }{\lambda _t+j\mu }T(q+1,j-1), \end{aligned}$$

    Now, due to (A9), it is seen that the inequality \(T(q,j) \le T(q+1,j)\) holds for \(j=1\) because

    $$\begin{aligned} T(q,1)&= \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) +\frac{\mu }{\lambda _t+\mu }T(q,0) \\&\le \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q,2) +\frac{\mu }{\lambda _t+\mu }T(q+1,0) \\&= T(q+1,1). \end{aligned}$$

    Then, it is easily obtained by induction on j that

    $$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ 0<j<S. \end{aligned}$$
    (A10)
  • Last, when \(j=S\), from (2), we have

    $$\begin{aligned} T(q,j)=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1), \end{aligned}$$
    (A11)

    and

    $$\begin{aligned} T(q+1,j)=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q+1,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q+1,S-1). \end{aligned}$$
    (A12)

    However, note that \(T(q,S+1) \le T(q+1,S+1)\) and \(T(q,S-1) \le T(2,S-1)\) (implied from results (A6) and (A10)). Therefore, from (A11) and (A12), we obtain

    $$\begin{aligned} T(q,j) \le T(q+1,j) \ \text {for} \ j=S. \end{aligned}$$
    (A13)

Equations (A5), (A6), (A9), (A10), (A13) complete our proof.

Appendix B Proof of Lemma 1

We prove Lemma 1 by induction on j.

First, note that

$$\begin{aligned} T(1,0)=\frac{1}{\lambda _t} +\frac{\lambda _t}{\lambda _t+S\mu }T(0,1)=\frac{1}{\lambda _t}, \end{aligned}$$

and

$$\begin{aligned} T(1,K)=\frac{1}{S\mu } +\frac{S\mu }{\lambda _t+S\mu }T(0,K-1)=\frac{1}{S\mu }. \end{aligned}$$

By induction on j, we have

$$\begin{aligned} T(1,j)&=\frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }T(0,j-1) +\frac{j\mu }{\lambda _t+j\mu }T(1,j-1) \\&= \frac{1}{\lambda _t+j\mu } +\frac{j\mu }{\lambda _t+j\mu }.\frac{1}{\lambda _t}\\&= \frac{1}{\lambda _t}, \end{aligned}$$

for \(1\le \ j\le S-1\); and

$$\begin{aligned} T(1,j)&=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(1,j+1) +\frac{S\mu }{\lambda _t+S\mu }T(0,j-1)\\&= \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }.\frac{1}{S\mu } \\&= \frac{1}{S\mu }, \end{aligned}$$

for \(S+1\le \ j\le K-1\).

Finally,

$$\begin{aligned} T(1,S)&=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(1,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(1,S-1)\\&= \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }.\frac{1}{S\mu } +\frac{S\mu }{\lambda _t+S\mu }.\frac{1}{\lambda _t} \\&= \frac{\lambda _t^2+\left( S\mu \right) ^2+\lambda _t\left( S\mu \right) }{\lambda _t\left( S\mu \right) \left( \lambda _t+S\mu \right) }. \end{aligned}$$

It can also be noted that

$$\begin{aligned} T(1,S)-T(1,S-1)&=\frac{\lambda _t^2+\left( S\mu \right) ^2+\lambda _t\left( S\mu \right) }{\lambda _t\left( S\mu \right) \left( \lambda _t+S\mu \right) } - \frac{1}{\lambda _t}\\&= \frac{S\mu }{\lambda _t(\lambda _t+S\mu )} > 0, \end{aligned}$$

and

$$\begin{aligned} T(1,S)-T(1,S+1)&=\frac{\lambda _t^2+\left( S\mu \right) ^2+\lambda _t\left( S\mu \right) }{\lambda _t\left( S\mu \right) \left( \lambda _t+S\mu \right) } - \frac{1}{S\mu }\\&= \frac{\lambda _t}{S\mu (\lambda _t+S\mu )} > 0. \end{aligned}$$

Appendix C Proof of Lemma 2

We prove this lemma by induction on p. We can easily see that it holds with \(p=0\) and \(p=1\) due to Lemma 1. Assume that it holds with \(p=q-1\) for any integer \(q\ge 2\). Additionally, assume that when \(p=q-1\), the inequality holds in the case of \(j=S\). (We show that under the same inductive assumptions, it also holds when \(p=q\) and \(j=S\) later in Proposition 2.) Then, from assumptions, we have

$$\begin{aligned} T(q-1,j) \le \frac{1}{j\mu }+T(q-1,j-1),\text { for } 1\le j\le S. \end{aligned}$$
(C14)

We show that the inequality holds with \(p=q\), which indicates that we need to prove that

$$\begin{aligned} T(q,j) \le \frac{1}{j\mu } + T(q,j-1), \text { for}\ 1 \le j \le S-1. \end{aligned}$$
(C15)

Assume there exists \(1\le j\le S-1\) such that

$$\begin{aligned} T(q,j) > \frac{1}{j\mu }+T(q,j-1). \end{aligned}$$
(C16)

From (2), we have

$$\begin{aligned} T(q,j)&=\frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }T(q-1,j+1) +\frac{j\mu }{\lambda _t+j\mu }T(q,j-1) \\&< \frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }T(q-1,j+1) +\frac{j\mu }{\lambda _t+j\mu }\left( T(q,j)-\frac{1}{j\mu }\right) \\&= \frac{\lambda _t}{\lambda _t+j\mu }T(q-1,j+1) +\frac{j\mu }{\lambda _t+j\mu }T(q,j), \end{aligned}$$

which is equivalent to

$$\begin{aligned} T(q,j) < T(q-1,j+1). \end{aligned}$$
(C17)

On the other hand, we also have

$$\begin{aligned} T(q-1,j+1) < \frac{1}{(j+1)\mu }+T(q-1,j), \end{aligned}$$
(C18)

according to the inductive assumption. From (C16), (C17) and (C18), we obtain

$$\begin{aligned} \frac{1}{j\mu }+T(q,j-1) < \frac{1}{(j+1)\mu }+T(q-1,j+1), \end{aligned}$$

which implies

$$\begin{aligned} T(q,j-1) < T(q-1,j). \end{aligned}$$
(C19)

Additionally, from (2), we have

$$\begin{aligned}&T(q,j-1) \\&\quad =\frac{1}{\lambda _t+(j-1)\mu } +\frac{\lambda _t}{\lambda _t+(j-1)\mu }T(q-1,j) +\frac{(j-1)\mu }{\lambda _t+(j-1)\mu }T(q,j-2) \\&\quad > \frac{1}{\lambda _t+(j-1)\mu } +\frac{\lambda _t}{\lambda _t+(j-1)\mu }T(q,j-1) +\frac{(j-1)\mu }{\lambda _t+(j-1)\mu }T(q,j-2) \end{aligned}$$

(due to (C19)). This implies

$$\begin{aligned} T(q,j-1) > \frac{1}{(j-1)\mu } + T(q,j-2). \end{aligned}$$

By induction on j, it finally implies

$$\begin{aligned} T(q,1) > \frac{1}{\mu } + T(q,0). \end{aligned}$$
(C20)

However,

$$\begin{aligned} T(q,1)-T(q,0)&= \left( \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) +\frac{\mu }{\lambda _t+\mu }T(q,0) \right) - T(q,0) \\&= \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) - \frac{\lambda _t}{\lambda _t+\mu }T(q,0) \\&= \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) - \frac{\lambda _t}{\lambda _t+\mu }\left( \frac{1}{\lambda _t}+ T(q-1,1) \right) \\&= \frac{\lambda _t}{\lambda _t+\mu }\left( T(q-1,2)-T(q-1,1)\right) \\&\le \frac{\lambda _t}{\lambda _t+\mu }.\frac{1}{2\mu } \qquad \text {(due to the inductive assumption)} \\&< \frac{1}{\mu }, \end{aligned}$$

which contradicts (C20). This indicates that (C15) holds and thus completes the proof.

Appendix D Proof of Lemma 3

We prove this lemma by induction on p. We can easily see that it holds with \(p=0\) and \(p=1\) due to Lemma 1. Assume that it holds with \(p=q-1\) for any integer \(q\ge 2\). Additionally, assume that when \(p=q-1\), the inequality holds in the case of \(j=S\). (We will show that, under the same inductive assumptions, it also holds when \(p=q\) and \(j=S\) later in Proposition 2.) Then, from assumptions, we have

$$\begin{aligned} T(q-1,j-1) \le \frac{1}{\lambda _t}+T(q-1,j),\ \text {for } S+1\le j\le K. \end{aligned}$$
(D21)

We show that the inequality holds with \(p=q\), which indicates that we need to prove that

$$\begin{aligned} T(q,j-1) \le \frac{1}{\lambda _t}+T(q,j),\ \text {for } S+2\le j\le K. \end{aligned}$$
(D22)

First, notice that

$$\begin{aligned}&T(q,K-1) - T(q,K) \\&\quad = \left( \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,K)+\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2)\right) -T(q,K)\\&\quad = \frac{1}{\lambda _t+S\mu } - \frac{S\mu }{\lambda _t+S\mu }T(q,K)+\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2)\\&\quad = \frac{1}{\lambda _t+S\mu } - \frac{S\mu }{\lambda _t+S\mu }\left( \frac{1}{S\mu }+T(q-1,K-1) \right) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2)\\&\quad = \frac{S\mu }{\lambda _t+S\mu }\left( T(q-1,K-2)-T(q-1,K-1) \right) \\&\quad \le \frac{S\mu }{\lambda _t+S\mu }.\frac{1}{\lambda _t} \text {(due to the inductive assumptions)} \\&\quad < \frac{1}{\lambda _t}, \end{aligned}$$

which indicates that (D22) holds with \(j=K\). Now, we make an inductive assumption on j; and for any \(S+2 \le j\le K-1\), consider the following:

$$\begin{aligned}&T(q,j-1) - T(q,j) \\&\quad = \left( \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,j)+\frac{S\mu }{\lambda _t+S\mu }T(q-1,j-2)\right) \\&\quad - \left( \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,j+1)+\frac{S\mu }{\lambda _t+S\mu }T(q-1,j-1)\right) \\&\quad =\frac{\lambda _t}{\lambda _t+S\mu }\left( T(q,j)-T(q,j+1) \right) +\frac{S\mu }{\lambda _t+S\mu }\left( T(q-1,j-2)-T(q-1,j-1) \right) \\&\quad \le \frac{\lambda _t}{\lambda _t+S\mu }.\frac{1}{\lambda _t}+\frac{S\mu }{\lambda _t+S\mu }.\frac{1}{\lambda _t}\ \text {(due to the inductive assumptions)} \\&\quad = \frac{1}{\lambda _t}. \end{aligned}$$

Appendix E Proof of Proposition 2

We prove Proposition 2 by induction on p. The statement is equivalent to the following inequalities.

$$\begin{aligned} T(p,j) \le T(p,j+1), \text {for}\ 0 \le j \le S-1, \end{aligned}$$
(E23)

and

$$\begin{aligned} T(p,j) \ge T(p,j+1), \text {for}\ S \le j \le K-1. \end{aligned}$$
(E24)

We already showed that (E23) and (E24) hold with \(p=0\) and \(p=1\) in Lemma 1. Assuming that (E23) and (E24) hold with \(p=q-1\) for any integer \(q\ge 2\), which indicates that

$$\begin{aligned} T(q-1,j) \le T(q-1,j+1), \text {for}\ 0 \le j \le S-1, \end{aligned}$$

and

$$\begin{aligned} T(q-1,j) \ge T(q-1,j+1), \text {for}\ S \le j \le K-1. \end{aligned}$$

We show that it holds with \(p=q\), which indicates that we need to prove that

$$\begin{aligned} T(q,j) \le T(q,j+1), \text {for}\ 0 \le j \le S-1, \end{aligned}$$

and

$$\begin{aligned} T(q,j) \ge T(q,j+1), \text {for}\ S \le j \le K-1. \end{aligned}$$

by considering the following five cases.

  • When \(j=0\), consider the following

    $$\begin{aligned} T(q,1)-T(q,0)&= \left( \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) +\frac{\mu }{\lambda _t+\mu }T(q,0) \right) - T(q,0) \\&= \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) - \frac{\lambda _t}{\lambda _t+\mu }T(q,0) \\&= \frac{1}{\lambda _t+\mu } +\frac{\lambda _t}{\lambda _t+\mu }T(q-1,2) - \frac{\lambda _t}{\lambda _t+\mu }\left( \frac{1}{\lambda _t}+ T(q-1,1) \right) \\&= \frac{\lambda _t}{\lambda _t+\mu }\left( T(q-1,2)-T(q-1,1)\right) \\&\ge 0 \qquad \text {(due to the inductive assumption)}, \end{aligned}$$

    which indicates that

    $$\begin{aligned} T(q,0) \le T(q,1). \end{aligned}$$
    (E25)
  • When \(1\le j \le S-2\), from (2) we have

    $$\begin{aligned} T(q,j) =\frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }T(q-1,j+1) +\frac{j\mu }{\lambda _t+j\mu }T(q,j-1), \end{aligned}$$
    (E26)

    and

    $$\begin{aligned} T(q,j+1)&= \frac{1}{\lambda _t+(j+1)\mu } +\frac{\lambda _t}{\lambda _t+(j+1)\mu }T(q-1,j+2) \nonumber \\&\quad +\, \frac{(j+1)\mu }{\lambda _t+(j+1)\mu }T(q,j). \end{aligned}$$
    (E27)

    We prove \(T(q,j) \le T(q,j+1)\) by contradiction. Assuming \(\exists j,\ T(q,j) > T(q,j+1)\), combining with (E27), we have

    $$\begin{aligned} T(q,j) > \frac{1}{\lambda _t+(j+1)\mu } +\frac{\lambda _t}{\lambda _t+(j+1)\mu }T(q-1,j+2) +\frac{(j+1)\mu }{\lambda _t+(j+1)\mu }T(q,j), \end{aligned}$$

    which is equivalent to

    $$\begin{aligned} T(q,j) > \frac{1}{\lambda _t}+T(q-1,j+2). \end{aligned}$$

    However, we also have \(T(q-1,j+2) \ge T(q-1,j+1)\) (due to the inductive assumption), so

    $$\begin{aligned} T(q,j) > \frac{1}{\lambda _t}+T(q-1,j+1). \end{aligned}$$
    (E28)

    From (E26) and (E28), we obtain

    $$\begin{aligned} T(q,j)&< \frac{1}{\lambda _t+j\mu } +\frac{\lambda _t}{\lambda _t+j\mu }\left( T(q,j)-\frac{1}{\lambda _t}\right) +\frac{j\mu }{\lambda _t+j\mu }T(q,j-1), \end{aligned}$$

    which is equivalent to

    $$\begin{aligned} T(q,j) < T(q,j-1). \end{aligned}$$

    By induction on j (by repeating the same procedure), it finally implies

    $$\begin{aligned} T(q,1) < T(q,0), \end{aligned}$$

    which contradicts (E25) that is proved above. This contradiction indicates that

    $$\begin{aligned} T(p,j) \ge T(p,j+1), \text {for}\ 1 \le j \le S-2. \end{aligned}$$
    (E29)
  • When \(j=K-1\), consider the following

    $$\begin{aligned}&T(q,K-1)-T(q,K) \\&\quad = \left( \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,K) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2) \right) - T(q,K) \\&\quad = \frac{1}{\lambda _t+S\mu } +\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2) - \frac{S\mu }{\lambda _t+S\mu }T(q,K) \\&\quad = \frac{1}{\lambda _t+\mu } +\frac{S\mu }{\lambda _t+S\mu }T(q-1,K-2) - \frac{S\mu }{\lambda _t+\mu }\left( \frac{1}{S\mu }+ T(q-1,K-1) \right) \\&\quad = \frac{S\mu }{\lambda _t+S\mu }\left( T(q-1,K-2)-T(q-1,K-1)\right) \\&\quad \ge 0 \qquad \text {(due to the inductive assumption)}, \end{aligned}$$

    which indicates that

    $$\begin{aligned} T(q,K-1) \ge T(q,K). \end{aligned}$$
    (E30)
  • When \(S+1\le j \le K-2\), from (2), we have

    $$\begin{aligned} T(q,j) =\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,j+1) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,j-1), \end{aligned}$$
    (E31)

    and

    $$\begin{aligned} T(q,j+1) =\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,j+2) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,j). \end{aligned}$$
    (E32)

    Due to the inductive assumption, we have \(T(q-1,j-1) \ge T(q-1,j)\); therefore, due to (E30), the inequality \(T(q,j) \ge T(q,j+1)\) holds for \(j=K-2\). By induction on j, we obtain

    $$\begin{aligned} T(q,j) \ge T(q,j+1)\ \text {for}\ S+1 \le j \le K-2. \end{aligned}$$

    Next, we prove that \(T(q,S) \ge T(q,S+1)\). From (2), we have

    $$\begin{aligned} T(q,S)=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1), \end{aligned}$$
    (E33)

    and

    $$\begin{aligned} T(q,S+1) = \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+2) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,S). \end{aligned}$$

    To prove the above inequality, we show that

    $$\begin{aligned} T(q,S-1)\ge T(q-1,S). \end{aligned}$$
    (E34)

    From (2), we have

    $$\begin{aligned}&T(q,S-1) \\&\quad =\frac{1}{\lambda _t+(S-1)\mu } +\frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) +\frac{(S-1)\mu }{\lambda _t+(S-1)\mu }T(q,S-2) \\&\quad \ge \frac{1}{\lambda _t+(S-1)\mu } +\frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) \\&\qquad +\, \frac{(S-1)\mu }{\lambda _t+(S-1)\mu }\left( T(q,S-1) - \frac{1}{(S-1)\mu }\right) \\&\qquad \quad \text {(due to}\, \mathrm{Lemma\, 2}) \\&\quad = \frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) +\frac{(S-1)\mu }{\lambda _t+(S-1)\mu }T(q,S-1), \end{aligned}$$

    which implies that (E34) is true. Therefore,

    $$\begin{aligned} T(q,S) \ge T(q,S+1). \end{aligned}$$
    (E35)

    From (E33) and (E35), we have

    $$\begin{aligned} T(q,S) \le \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1), \end{aligned}$$

    which implies

    $$\begin{aligned} T(q,S-1) +\frac{1}{S\mu }\ge T(q,S), \end{aligned}$$

    and this also completes the proof of Lemma 2.

  • Finally, we prove that \(T(q,S) \ge T(q,S-1)\). To show the above equality, first note that

    $$\begin{aligned}&T(q,S-1) \\&\quad =\frac{1}{\lambda _t+(S-1)\mu } +\frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) +\frac{(S-1)\mu }{\lambda _t+(S-1)\mu }T(q,S-2) \\&\quad \le \frac{1}{\lambda _t+(S-1)\mu } +\frac{\lambda _t}{\lambda _t+(S-1)\mu }T(q-1,S) +\frac{(S-1)\mu }{\lambda _t+(S-1)\mu }T(q,S-1), \end{aligned}$$

    due to (E29), and this implies

    $$\begin{aligned} T(q,S-1) - T(q-1,S) \le \frac{1}{\lambda _t}. \end{aligned}$$
    (E36)

    Now, due to (E36), Lemma 3 and the inductive assumptions, we have

    $$\begin{aligned}&T(q,S) - T(q,S+1) \\&\quad =\left( \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1) \right) \\&\qquad -\, \left( \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+2) +\frac{S\mu }{\lambda _t+S\mu }T(q-1,S) \right) \\&\quad = \frac{\lambda _t}{\lambda _t+S\mu }\left( T(q,S+1)-T(q,S+2)\right) +\frac{S\mu }{\lambda _t+S\mu }\left( T(q,S-1)-T(q-1,S)\right) \\&\quad \le \frac{\lambda _t}{\lambda _t+S\mu }.\frac{1}{\lambda _t} +\frac{S\mu }{\lambda _t+S\mu }.\frac{1}{\lambda _t}\\&\quad = \frac{1}{\lambda _t}, \end{aligned}$$

    which indicates that

    $$\begin{aligned} T(q,S) \le \frac{1}{\lambda _t} + T(q,S+1). \end{aligned}$$
    (E37)

    (Note that this conclusion also completes the proof of Lemma 3).

    Now, due to (E37), we have

    $$\begin{aligned} T(q,S)&=\frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }T(q,S+1) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1) \\&\ge \frac{1}{\lambda _t+S\mu } +\frac{\lambda _t}{\lambda _t+S\mu }\left( T(q,S) - \frac{1}{\lambda _t} \right) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1) \\&= \frac{\lambda _t}{\lambda _t+S\mu }T(q,S) +\frac{S\mu }{\lambda _t+S\mu }T(q,S-1), \end{aligned}$$

    which implies \(T(q,S) \ge T(q,S-1)\).

Appendix F Proof of Proposition 3

First, we prove that

$$\begin{aligned} T(p,0) \ge \frac{p}{\lambda _t}, \end{aligned}$$
(F38)

for all \(p=1,2,\ldots .\)

This inequality holds for \(p=1\) because \(T(1,0)=\frac{1}{\lambda _t}\). Assume that it also holds for \(p=q \ge 1\), indicating that \(T(q,0) \ge \frac{q}{\lambda _t}\). We have

$$\begin{aligned} T(q+1,0)&= \frac{1}{\lambda _t} + T(q,1)\\&\ge \frac{1}{\lambda _t} + T(q,0) \qquad \text {(due to}\, \mathrm{Proposition\, 2})\\&\ge \frac{q+1}{\lambda _t}. \end{aligned}$$

Therefore, by induction on p, we obtain that (F38) is true. \(\lim _{p \rightarrow +\infty } \frac{p}{\lambda _t} = +\infty \), which implies

$$\begin{aligned} \lim _{p \rightarrow +\infty } T(p,0) = +\infty . \end{aligned}$$

By induction on j using formula (2), we can easily obtain

$$\begin{aligned} \lim _{p \rightarrow +\infty } T(p,j) = +\infty . \end{aligned}$$

for all \(j=0,1,2,\ldots ,K\).

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Nguyen, H.Q., Phung-Duc, T. Strategic customer behavior and optimal policies in a passenger–taxi double-ended queueing system with multiple access points and nonzero matching times. Queueing Syst (2022). https://doi.org/10.1007/s11134-022-09786-3

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  • DOI: https://doi.org/10.1007/s11134-022-09786-3

Keywords

  • Double-ended queueing system
  • Strategic queueing
  • Matching queue