Strategic customer behavior in an M/M/1 feedback queue

Abstract

We investigate the behavior of equilibria in an M/M/1 feedback queue where price- and time-sensitive customers are homogeneous with respect to service valuation and cost per unit time of waiting. Upon arrival, customers can observe the number of customers in the system and then decide to join or to balk. Customers are served in order of arrival. After being served, each customer either successfully completes the service and departs the system with probability q, or the service fails and the customer immediately joins the end of the queue to wait to be served again until she successfully completes it. We analyze this decision problem as a noncooperative game among the customers. We show that there exists a unique symmetric Nash equilibrium threshold strategy. We then prove that the symmetric Nash equilibrium threshold strategy is evolutionarily stable. Moreover, if we relax the strategy restrictions by allowing customers to renege, in the new Nash equilibrium, customers have a greater incentive to join. However, this does not necessarily increase the equilibrium expected payoff, and for some parameter values, it decreases it.

Appendices

The nonreneging case

$$\begin{aligned} P^{(x)}&= \begin{bmatrix} A_0^{(1)} &{} A_1^{(1)} &{} 0 &{} 0 &{} \cdots &{} \cdots &{} \cdots \\ A_{-1}^{(2)} &{} A_0^{(2)} &{} A_1^{(2)} &{} 0 &{} \cdots &{} \cdots &{} \cdots \\ 0 &{} A_{-1}^{(3)} &{} A_0^{(3)} &{} A_1^{(3)} &{} \cdots &{} \cdots &{} \cdots \\ 0 &{} 0 &{} A_{-1}^{(4)} &{} A_0^{(4)} &{} \cdots &{} \cdots &{} \cdots \\ \vdots &{}\vdots &{} \vdots &{} \vdots &{} \ddots &{} \ddots &{} \vdots \\ \vdots &{}\vdots &{}\vdots &{} \vdots &{} \vdots &{} A_{-1}^{(\lceil x \rceil + 1)} &{} A_0^{(\lceil x \rceil + 1)} \\ \end{bmatrix}, \, \end{aligned}$$

(A.1)

$$\begin{aligned} A_{-1}^{(k)}&= \begin{bmatrix} 0 &{} 0 &{} \cdots &{} 0\\ \frac{\mu q}{\lambda +\mu } &{} 0 &{} \cdots &{} 0 \\ 0 &{} \frac{\mu q}{\lambda +\mu } &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} \frac{\mu q}{\lambda +\mu } \end{bmatrix} \in {\mathbb {R}}^{k \times (k-1)} \, \qquad k = 2, \cdots , \lceil x \rceil + 1, \end{aligned}$$

(A.2)

$$\begin{aligned} A_{0}^{(k)}&= \begin{bmatrix} 0 &{} 0 &{} \cdots &{} \frac{\mu (1-q)}{\lambda +\mu }\\ \frac{\mu (1-q)}{\lambda +\mu } &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \ddots &{} \ddots &{} \vdots \\ 0 &{} \cdots &{} \frac{\mu (1-q)}{\lambda +\mu } &{} 0 \end{bmatrix} \in {\mathbb {R}}^{k \times k} \, \quad k = 1, \cdots , \lfloor x \rfloor -1, \end{aligned}$$

(A.3)

$$\begin{aligned} A_{0}^{(\lfloor x \rfloor )}&= \begin{bmatrix} \frac{\lambda (1-(x-\lfloor x \rfloor ))}{\lambda +\mu } &{} 0 &{} \cdots &{} 0\\ 0 &{} \frac{\lambda (1-(x-\lfloor x \rfloor ))}{\lambda +\mu } &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} \frac{\lambda (1-(x-\lfloor x \rfloor ))}{\lambda +\mu } \end{bmatrix} \nonumber \\&+ \begin{bmatrix} 0 &{} 0 &{} \cdots &{} \frac{\mu (1-q)}{\lambda +\mu }\\ \frac{\mu (1-q)}{\lambda +\mu } &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \ddots &{} \ddots &{} \vdots \\ 0 &{} \cdots &{} \frac{\mu (1-q)}{\lambda +\mu } &{} 0 \end{bmatrix} \in {\mathbb {R}}^{\lfloor x \rfloor \times \lfloor x \rfloor }, \, \end{aligned}$$

(A.4)

$$\begin{aligned} A_{0}^{(k)}&= \begin{bmatrix} \frac{\lambda }{\lambda +\mu } &{} 0 &{} \cdots &{} 0\\ 0 &{} \frac{\lambda }{\lambda +\mu } &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} \frac{\lambda }{\lambda +\mu } \end{bmatrix}\nonumber \\&+ \begin{bmatrix} 0 &{} 0 &{} \cdots &{} \frac{\mu (1-q)}{\lambda +\mu }\\ \frac{\mu (1-q)}{\lambda +\mu } &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \ddots &{} \ddots &{} \vdots \\ 0 &{} \cdots &{} \frac{\mu (1-q)}{\lambda +\mu } &{} 0 \end{bmatrix} \in {\mathbb {R}}^{k \times k} \qquad k = \lfloor x \rfloor +1, \, \lceil x \rceil +1, \end{aligned}$$

(A.5)

$$\begin{aligned} A_{1}^{(k)}&= \begin{bmatrix} \frac{\lambda }{\lambda +\mu } &{} 0 &{} \cdots &{} 0 &{} 0\\ 0 &{} \frac{\lambda }{\lambda +\mu } &{} \cdots &{} 0 &{} 0\\ \vdots &{} \vdots &{} \ddots &{} \vdots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} \frac{\lambda }{\lambda +\mu } &{} 0 \end{bmatrix} \in {\mathbb {R}}^{k \times (k+1)} \, \quad k = 1, \cdots , \lfloor x \rfloor -1, \end{aligned}$$

(A.6)

$$\begin{aligned} A_{1}^{(\lfloor x \rfloor )}&= \begin{bmatrix} \frac{\lambda p}{\lambda +\mu } &{} 0 &{} \cdots &{} 0 &{} 0\\ 0 &{} \frac{\lambda p}{\lambda +\mu } &{} \cdots &{} 0 &{} 0\\ \vdots &{} \vdots &{} \ddots &{} \vdots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} \frac{\lambda p}{\lambda +\mu } &{} 0 \end{bmatrix} \in {\mathbb {R}}^{\lfloor x \rfloor \times (\lfloor x \rfloor +1)} \qquad A_1^{(\lceil x \rceil )} = \varvec{0}_{\lceil x \rceil \times (\lceil x \rceil +1)} . \end{aligned}$$

(A.7)

.

1.1 Proof of Lemma 1

For integer $d \ge 1$, we first define

$$\begin{aligned} ^{d}\nu ^{(x)}_{i,j}: = \left( (P^{(x)})^d \varvec{e} \right) _{\frac{j(j-1)}{2}+i} \qquad 1 \le i \le j \le \lceil x \rceil , \end{aligned}$$

(B.1)

where $V^d$ is the dth power of V and $\varvec{v}_k$ is the kth element of a vector $\varvec{v}$. We now prove by mathematical induction that $^{d}\nu ^{(x)}_{j,j}$ and $^{d}\nu ^{(x)}_{i,j}$ are increasing in j for any d and $1 \le i \le j \le \lceil x \rceil$.

When $d =1$,

$$\begin{aligned}&^{1}\nu ^{(x)}_{1,j} \,= \,^{1}\nu ^{(x)}_{1,j+1} = 1-\frac{\mu q}{\lambda +\mu } \quad 1 \le j \le \lceil x \rceil , \end{aligned}$$

(B.2)

$$\begin{aligned}&^{1}\nu ^{(x)}_{i,j} \,= \,^{1}\nu ^{(x)}_{i,j+1} = \,^{1}\nu ^{(x)}_{j+1,j+1}= 1 \quad 1 < i \le j \le \lceil x \rceil . \end{aligned}$$

(B.3)

That is, $^{1}\nu ^{(x)}_{j+1,j+1} = \, ^{1}\nu ^{(x)}_{j,j} > \, ^{1}\nu ^{(x)}_{1,1}$ for $1 < j \le \lceil x \rceil$, and $^{1}\nu ^{(x)}_{i,j+1} = \, ^{1}\nu ^{(x)}_{i,j}$ for $1 \le i \le j \le \lceil x \rceil$.

Next suppose that the induction assumption is at the dth transition,

$$\begin{aligned} ^{d}\nu ^{(x)}_{j+1,j+1} \, \ge \, ^{d}\nu ^{(x)}_{j,j} \qquad ^{d}\nu ^{(x)}_{i,j+1} \, \ge \, ^{d}\nu ^{(x)}_{i,j} \qquad 1 \le i \le j \le \lceil x \rceil . \end{aligned}$$

(B.4)

Before proving that (B.4) holds for $d+1$, we first prove that $^{d}\nu ^{(x)}_{i,j} \, \ge \, ^{d+1}\nu ^{(x)}_{i,j}$. Since $^{d}\nu ^{(x)}_{i,j}$ represents the sum of probabilities of being in each state in ${\mathcal {S}}$ at the dth transition, that is, the probability that the tagged customer is still in the system in the dth transition, if the initial state is (i, j). It follows from this physical interpretation that $^{d}\nu _{i,j}^{(x)} = 1$ when $d < i$. Furthermore, since the event that the tagged customer is still in the system after $d+1$ transitions is a subset of the event that it is still in the system after d transitions, it must be the case that $^d\nu ^{(x)}_{i,j}$ is decreasing in d.

Then by expanding both $^{d+1}\nu ^{(x)}_{i,j+1}$ and $^{d+1}\nu ^{(x)}_{i,j}$ as in Eq. (3.1) and collecting identical terms together, for $1 \le i \le j \le \lceil x \rceil$, we have

$$\begin{aligned}&\left( ^{d+1}\nu ^{(x)}_{i,j+1}-\, ^{d+1}\nu ^{(x)}_{i,j} \right) \nonumber \\&\quad =\frac{\lambda }{\lambda +\mu } \left( \left( ^{d}\nu ^{(x)}_{i,j+2}-\,^{d}\nu ^{(x)}_{i,j+1} \right) \, \mathbb {1}_{\lbrace j <\lfloor x \rfloor -1 \rbrace } + p\, \left( ^{d}\nu ^{(x)}_{i,j+2}-\,^{d}\nu ^{(x)}_{i,j+1} \right) \, \mathbb {1}_{\lbrace j = \lfloor x \rfloor -1 \rbrace }\right. \nonumber \\&\qquad \left. + (1-p) \left( ^{d}\nu ^{(x)}_{i,j+1}-\, ^{d}\nu ^{(x)}_{i,j} \right) \mathbb {1}_{\lbrace j = \lfloor x \rfloor \rbrace } \right) \nonumber \\&\qquad + \frac{\mu }{\lambda +\mu } \left( (1-q) \, \left( \, ^{d}\nu ^{(x)}_{j+1,j+1}-\, ^{d}\nu ^{(x)}_{j,j} \right) \, \mathbb {1}_{\lbrace i =1 \rbrace } + \left( q \, \left( \,^{d}\nu ^{(x)}_{i-1,j} - \, ^{d}\nu ^{(x)}_{i-1,j-1} \right) \right. \right. \nonumber \\&\qquad \left. \left. + (1-q) \, \left( ^{d}\nu ^{(x)}_{i-1,j+1}- \,^{d}\nu ^{(x)}_{i-1,j} \right) \right) \mathbb {1}_{\lbrace i >1 \rbrace } \right) . \end{aligned}$$

(B.5)

Again, by expanding $^{d+1}\nu ^{(x)}_{j+1,j+1}$, using Eq. (3.1), and collapsing the term $\displaystyle \frac{\mu q}{\lambda +\mu } \, ^{d}\nu ^{(x)}_{j,j}$, we obtain, for $1 \le i \le j \le \lceil x \rceil$,

$$\begin{aligned}&\left( \, ^{d+1}\nu ^{(x)}_{j+1,j+1}- \,^{d+1}\nu ^{(x)}_{j,j} \right) \ge \left( ^{d+1}\nu ^{(x)}_{j+1,j+1}- \, ^{d}\nu ^{(x)}_{j,j}\right) \nonumber \\&\quad = \frac{\mu (1-q)}{\lambda +\mu }\left( ^{d}\nu ^{(x)}_{j,j+1}-\,^{d}\nu ^{(x)}_{j,j} \right) + \frac{\lambda }{\lambda +\mu } \left( \left( ^{d}\nu ^{(x)}_{j+1,j+2} - \,^{d}\nu ^{(x)}_{j,j} \right) \mathbb {1}_{\lbrace j < \lfloor x \rfloor -1 \rbrace } \right. \nonumber \\&\qquad \left. + \left( p \, \left( ^{d}\nu ^{(x)}_{j+1,j+2}- \, ^{d}\nu ^{(x)}_{j,j} \right) +(1-p)\, \left( ^{d}\nu ^{(x)}_{j+1,j+1}- \, ^{d}\nu ^{(x)}_{j,j} \right) \right) \mathbb {1}_{\lbrace j = \lfloor x \rfloor -1 \rbrace } \right. \nonumber \\&\qquad \left. +\left( ^{d}\nu ^{(x)}_{j+1,j+1}- \,^{d}\nu ^{(x)}_{j,j} \right) \mathbb {1}_{\lbrace j = \lfloor x \rfloor \rbrace } \right) , \end{aligned}$$

(B.6)

where the inequality in (B.6) holds strictly if and only if $d > j-1$. It follows from induction assumption (B.4) that $^{d}\nu ^{(x)}_{j+1,j+2} \ge \,^{d}\nu ^{(x)}_{j+1,j+1} \ge \,^{d}\nu ^{(x)}_{j,j}$, hence, for $1 \le i \le j \le \lceil x \rceil$,

$$\begin{aligned} ( ^{d+1}\nu ^{(x)}_{i,j+1} - \, ^{d+1}\nu ^{(x)}_{i,j} )\ge 0 , \end{aligned}$$

(B.7)

and

$$\begin{aligned} (^{d+1}\nu ^{(x)}_{j+1,j+1}- \, ^{d+1}\nu ^{(x)}_{j,j}) \ge {\frac{\mu (1-q)}{\lambda +\mu }} \, (^{d}\nu ^{(x)}_{j,j+1}-\, ^{d}\nu ^{(x)}_{j,j}) \ge 0 . \end{aligned}$$

(B.8)

Finally, from Eq. (3.2),

$$\begin{aligned} \varvec{w}^{(x)} = \frac{1}{\lambda +\mu } \, (I-P^{(x)})^{-1} \varvec{e} =\frac{1}{\lambda +\mu } \, \sum _{d=0}^{\infty } (P^{(x)})^d \varvec{e} . \end{aligned}$$

(B.9)

It follows that

$$\begin{aligned}&w_{i,j} = \frac{1}{\lambda +\mu } \left( 1+ \sum _{d = 1}^{\infty } \, ^{d}\nu ^{(x)}_{i,j} \right) , \end{aligned}$$

(B.10)

$$\begin{aligned}&w_{j,j} = \frac{1}{\lambda +\mu } \left( 1+ \sum _{d = 1}^{\infty } \, ^{d}\nu ^{(x)}_{j,j} \right) \end{aligned}$$

(B.11)

are increasing in j. $\square$

1.2 Proof of Lemma 2

• When $n< x_1 = n+ p_1 < n+ p_2 = x_2 \le n+1$, from Eq. (3.2),

  $$\begin{aligned}&\left( {\mathbf {I}}-P^{(x_1)} \right) \varvec{w}^{(x_1)} \, = \, \frac{1}{\lambda +\mu }\varvec{e}, \end{aligned}$$

  (B.12)
  $$\begin{aligned}&\left( {\mathbf {I}}-P^{(x_2)} \right) \varvec{w}^{(x_2)} \, = \, \frac{1}{\lambda +\mu }\varvec{e} , \end{aligned}$$

  (B.13)

  where the probability transition matrices $P^{(x_1)}$ and $P^{(x_2)}$ have the same dimension. Noting that $P^{(x_2)}$ and $P^{(x_1)}$ only differ in n rows, we have

  $$\begin{aligned} \left( I-P^{(x_2)} \right) \, (\varvec{w}^{(x_2)}-\varvec{w}^{(x_1)})= & {} (P^{(x_2)}-P^{(x_1)}) \varvec{w}^{(x_1)}\nonumber \\= & {} \frac{\lambda (p_2-p_1)}{\lambda +\mu } \, \begin{bmatrix} {\mathbf {0}}_{\frac{n(n-1)}{2} \times 1} \\ w^{(x_1)}_{1,n+1} - w^{(x_1)}_{1,n} \\ w^{(x_1)}_{2,n+1} - w^{(x_1)}_{2,n} \\ \vdots \\ w^{(x_1)}_{n,n+1} - w^{(x_1)}_{n,n} \\ {\mathbf {0}}_{(2n+3) \times 1} \end{bmatrix} . \end{aligned}$$

  (B.14)

  We know from Eq. (B.10) that $w^{(x_1)}_{i,n+1} - w^{(x_1)}_{i,n} > 0$. Also, since the $((i_1,j_1),(i_2,j_2))$th entry of $(I-P^{(x_2)})^{-1}$ is the tagged customer’s expected number of visits to state $(i_2,j_2)$ starting from state $(i_1,j_1)$ in ${\mathcal {S}}$ before she leaves the system,

  $$\begin{aligned} (I-P^{(x_2)})^{-1} > 0 . \end{aligned}$$

  Hence

  $$\begin{aligned} \varvec{w}^{(x_2)}-\varvec{w}^{(x_1)} = (I-P^{(x_2)})^{-1} \, (P^{(x_2)}-P^{(x_1)}) \, \varvec{w}^{(x_1)} > 0 . \end{aligned}$$

  (B.15)

• When $x_1 = n$ and $x_2 =n + p_2$ or $x_2 = n+1$, $P^{(x_1)}$ and $P^{(x_2)}$ have different sizes. However, we can write $P^{(x_2)}$ as

  $$\begin{aligned} P^{(x_2)} \quad&= \quad \left[ \begin{array}{c c c c c | c} A_0^{(1)} &{} A_1^{(1)} &{} \cdots &{} \cdots &{} \cdots &{} 0\\ A_{-1}^{(2)} &{} A_0^{(2)} &{} A_1^{(2)} &{} \cdots &{} \cdots &{} 0\\ \vdots &{} A_{-1}^{(3)} &{} A_0^{(3)} &{} A_1^{(3)} &{} \cdots &{} 0 \\ \vdots &{}\vdots &{} \vdots &{} \ddots &{} \ddots &{} \vdots \\ \vdots &{} \vdots &{} \vdots &{} A_{-1}^{(n+1)} &{} A_0^{(n+1)} &{} 0 \\ \hline 0 &{} \cdots &{} \cdots &{} \cdots &{} {A}_{-1}^{(n+2)} &{} {A}_0^{(n+2)} \end{array} \right] \end{aligned}$$

  (B.16)
  $$\begin{aligned}&= \quad \left[ \begin{array}{c | c} {\bar{P}}^{(x_2)} &{} \varvec{0}_{\frac{(n+1)(n+2)}{2} \times (n+2)}\\ \hline \varvec{0}_{(n+2) \times \frac{n(n+1)}{2}} \quad {A}_{-1}^{(n+2)}&{} {A}_0^{(n+2)} \end{array} \right] , \end{aligned}$$

  (B.17)

  and

  $$\begin{aligned} (I-P^{(x_2)})^{-1} = \left[ \begin{array}{c | c} ( I-{\bar{P}}^{(x_2)})^{-1} &{} {\mathbf {0}} \\ \hline N \, ( I-{\tilde{P}}^{(x_2)})^{-1} &{} ( I- {A}_0^{(n+2)} )^{-1} \end{array} \right] , \end{aligned}$$

  (B.18)

  where $N = \begin{bmatrix} \varvec{0}_{(n+2) \times \frac{n(n+1)}{2}}&( I- {A}_0^{(n+2)} )^{-1} \, {A}_{-1}^{(n+2)} \end{bmatrix}$. When $x_1 = n$, the position where the tagged customer can join is at most $n+1$, so we compare $\varvec{w}^{(x_1)}_k$ with $\varvec{w}^{(x_2)}_k$ for $k = 1, \ldots , \displaystyle \frac{(n+1)(n+2)}{2}$. If we define $\varvec{{\bar{w}}}^{(x_2)}: = \left[ {I}_{\frac{(n+1)(n+2)}{2}} \quad \varvec{0}\right] \varvec{w}^{(x_2)}$, then

  $$\begin{aligned} \varvec{{\bar{w}}}^{(x_2)} = \frac{1}{\lambda +\mu }( I-{\bar{P}}^{(x_2)})^{-1} \, \varvec{e} , \end{aligned}$$

  (B.19)

  thus

  $$\begin{aligned} \varvec{{\bar{w}}}^{(x_2)}-\varvec{w}^{(x_1)} = \frac{\lambda \, p_2}{\lambda +\mu } \, (I-{\bar{P}}^{(x_2)})^{-1}\, \begin{bmatrix} {\mathbf {0}}_{\frac{n(n-1)}{2} \times 1} \\ w^{(x_1)}_{1,n+1} - w^{(x_1)}_{1,n} \\ w^{(x_1)}_{2,n+1} - w^{(x_1)}_{2,n} \\ \vdots \\ w^{(x_1)}_{n,n+1} - w^{(x_1)}_{n,n} \\ {\mathbf {0}}_{(n+1) \times 1} \end{bmatrix} > 0 . \end{aligned}$$

  (B.20)

  In (B.20), with an abuse of notation, we include the case $p_2 = 1$.

• When $x_1 = n_1 +p_1, x_2= n_2 +p_2$, and $n_1 <n_2$, by comparing the expected sojourn time for all the consecutive integers between $x_1$ and $x_2$, it follows from the aforementioned reasoning that

  $$\begin{aligned}&w_{i,j}^{(x_1)} < w_{i,j}^{(\lceil x_1 \rceil )}&1&\le i \le j \le \lceil x_1 \rceil +1, \nonumber \\&\cdots&\end{aligned}$$

  (B.21)
  $$\begin{aligned}&w_{i,j}^{(\lceil x_2 \rceil -2)} < w_{i,j}^{(\lceil x_2 \rceil -1)}&1&\le i \le j \le \lceil x_2 \rceil -1, \end{aligned}$$

  (B.22)
  $$\begin{aligned}&w_{i,j}^{(\lceil x_2 \rceil -1)} < w_{i,j}^{(x_2)}&1&\le i \le j \le \lceil x_2 \rceil . \end{aligned}$$

  (B.23)

  Hence, $w_{i,j}^{(x_1)} < w_{i,j}^{(x_2)}, \, 1 \le i \le j \le \lceil x_1\rceil +1$.

The reneging case

1.1 .

$$\begin{aligned} {\hat{P}}^{(\lfloor x \rfloor +1,x)}= & {} \begin{bmatrix} A_0^{(1)} &{} A_1^{(1)} &{} \cdots &{} \cdots &{} \cdots &{} \cdots &{} 0\\ A_{-1}^{(2)} &{} A_0^{(2)} &{} A_1^{(2)} &{} \cdots &{} \cdots &{} \cdots &{} 0\\ \vdots &{} A_{-1}^{(3)} &{} A_0^{(3)} &{} A_1^{(3)} &{} \cdots &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} A_{-1}^{(4)} &{} A_0^{(4)} &{} A_1^{(4)} &{} \cdots &{} 0 \\ \vdots &{}\vdots &{} \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{}0 &{}0 &{} 0 &{} 0 &{} {\hat{A}}_{-1}^{(\lfloor x \rfloor +1)} &{} {\tilde{A}}_0^{(\lfloor x \rfloor +1)} \end{bmatrix} , \end{aligned}$$

(C.1)

$$\begin{aligned} {\hat{A}}_{-1}^{(\lfloor x \rfloor +1)}= & {} \begin{bmatrix} 0 &{} 0 &{} \cdots &{} 0\\ \frac{\mu q + \mu (1-q)(1-(x-\lfloor x \rfloor ))}{\lambda +\mu } &{} 0 &{} \cdots &{} 0 \\ 0 &{} \frac{\mu q + \mu (1-q)(1-(x-\lfloor x \rfloor ))}{\lambda +\mu } &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} \frac{\mu q + \mu (1-q)(1-(x-\lfloor x \rfloor ))}{\lambda +\mu } \end{bmatrix} \in {\mathbb {R}}^{(\lfloor x \rfloor + 1) \times \lfloor x \rfloor }, \nonumber \\\end{aligned}$$

(C.2)

$$\begin{aligned} {\tilde{A}}_0^{(\lfloor x \rfloor +1)}= & {} \begin{bmatrix} \frac{\lambda }{\lambda +\mu } &{} 0 &{} \cdots &{} 0\\ 0 &{} \frac{\lambda }{\lambda +\mu } &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} \frac{\lambda }{\lambda +\mu } \end{bmatrix} + \begin{bmatrix} 0 &{} 0 &{} \cdots &{} \frac{\mu (1-q)}{\lambda +\mu } \\ \frac{\mu (1-q)\, (x - \lfloor x \rfloor )}{\lambda +\mu } &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \ddots &{} \ddots &{} \vdots \\ 0 &{} \cdots &{} \frac{\mu (1-q) \, (x - \lfloor x \rfloor )}{\lambda +\mu } &{} 0 \end{bmatrix} \in {\mathbb {R}}^{(\lfloor x \rfloor +1) \times (\lfloor x \rfloor +1 )}, \nonumber \\ \end{aligned}$$

(C.3)

$$\begin{aligned} \varvec{g}= & {} \frac{\mu q \, R_0}{\lambda +\mu } \, \varvec{{e}}_k \, - \, \frac{1}{\lambda +\mu } \, \varvec{{e}} \qquad k = \frac{j(j-1)}{2}+1, \, j = 1, \ldots , \lfloor x \rfloor +1 . \end{aligned}$$

(C.4)

1.2 .

$$\begin{aligned}&{\hat{P}}^{(x,x)} = \begin{bmatrix} A_0^{(1)} &{} A_1^{(1)} &{} \cdots &{} \cdots &{} \cdots &{} \cdots &{} 0\\ A_{-1}^{(2)} &{} A_0^{(2)} &{} A_1^{(2)} &{} \cdots &{} \cdots &{} \cdots &{} 0\\ \vdots &{} A_{-1}^{(3)} &{} A_0^{(3)} &{} A_1^{(3)} &{} \cdots &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} A_{-1}^{(4)} &{} A_0^{(4)} &{} A_1^{(4)} &{} \cdots &{} 0 \\ \vdots &{}\vdots &{} \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{}0 &{}0 &{} 0 &{} 0 &{} {\hat{A}}_{-1}^{(\lfloor x \rfloor +1)} &{} {\hat{A}}_0^{(\lfloor x \rfloor +1)} \end{bmatrix} , \end{aligned}$$

(C.5)

$$\begin{aligned}&{\hat{A}}_{0}^{(\lfloor x \rfloor +1)} = \begin{bmatrix} \frac{\lambda }{\lambda +\mu } &{} 0 &{} \cdots &{} 0\\ 0 &{} \frac{\lambda }{\lambda +\mu } &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} \frac{\lambda }{\lambda +\mu } \end{bmatrix}\nonumber \\&+ \begin{bmatrix} 0 &{} 0 &{} \cdots &{} \frac{\mu (1-q) (x-\lfloor x \rfloor )}{\lambda +\mu } \\ \frac{\mu (1-q) (x-\lfloor x \rfloor )}{\lambda +\mu } &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \ddots &{} \ddots &{} \vdots \\ 0 &{} \cdots &{} \frac{\mu (1-q) (x-\lfloor x \rfloor )}{\lambda +\mu } &{} 0 \end{bmatrix} \in {\mathbb {R}}^{(\lfloor x \rfloor +1) \times (\lfloor x \rfloor +1)}, \end{aligned}$$

(C.6)

Proof of Lemma 4

Proof

It follows from the explanation at the beginning of Sect. 3.1 that the tagged customer will never renege after joining if she uses $x_{tag} \ge \lceil x \rceil$ when others uses x. Thus the comparison of $z^{(x)}_{i,j}$ and ${\hat{z}}_{i,j}^{(\lfloor x \rfloor +1,x)}$ is actually the comparison of the respective expected sojourn time $w^{(x)}_{i,j}$ and ${\hat{w}}_{i,j}^{(\lfloor x \rfloor +1,x)}$. As in Lemma 1, the tagged customer’s expected sojourn time is her expected total number of visits of different states until she leaves the system, times $\frac{1}{\lambda +\mu }$. Similar to the proof of Lemma 1, we first define

$$\begin{aligned} ^{d}{\hat{\nu }}^{(x)}_{i,j}: = \left( ({\hat{P}}^{(\lfloor x \rfloor +1, x)})^d \varvec{e} \right) _{\frac{j(j-1)}{2}+i} \qquad 1 \le i \le j \le \lfloor x \rfloor +1 , \end{aligned}$$

(D.1)

which is the probability that the tagged customer is still in the system at the dth transition, if her initial state is (i, j), she never reneges, and others join and renege with threshold x. We prove Eq. (4.6) by mathematical induction.

First, it follows from the physical interpretation that when $d \le \lfloor x \rfloor$, $^d \nu ^{(x)}_{i,j} = \, ^d{\hat{\nu }}^{(\lfloor x \rfloor +1,x)}_{i,j}$. This is because others’ reneging will not affect the tagged customer until she rejoins the queue and is in service for the second time, and the queue size has reached $\lfloor x \rfloor + 1$ before she is in service for the first time. This is possible only when $d > \lfloor x \rfloor$. Indeed,

$$\begin{aligned} ^{\lfloor x \rfloor +1} \nu ^{(x)}_{i,j} > \, ^{\lfloor x \rfloor +1}{\hat{\nu }}^{(\lfloor x \rfloor +1,x)}_{i,j} \qquad 1< i \le j = \lfloor x \rfloor + 1 . \end{aligned}$$

(D.2)

Figure 11 shows an example of Eq. (D.2) with $(i,j) = (2,3)$ in (a), $(i,j) = (3,3)$ in (b), and $2< x < 3$. The gray block represents the tagged customer, and the white ones represent other customers in the system. It shows the sample paths where the reneging of others affects the tagged customer at the earliest possible transition. The first row is when others use threshold 2, and the second row is when others use threshold 3. In both Fig. 11a, b, the reneging of others affects the queue size at the first transition, but will not affect the tagged customer until $d >3$.

Fig. 11

An example of the comparison between $^{d} \nu ^{(x)}_{i,\lfloor x \rfloor +1}$ and $^{d}{\hat{\nu }}^{(\lfloor x \rfloor +1,x)}_{i,\lfloor x \rfloor +1}$ when $2< x < 3$

Next, we assume that $^d \nu ^{(x)}_{i,j} \ge \,^d{\hat{\nu }}^{(\lfloor x \rfloor +1,x)}_{i,j}$. Then we write the difference between $P^{(x)}$ and ${\hat{P}}^{(\lfloor x \rfloor +1,x)}$ in the form

$$\begin{aligned} P^{(x)}= {\hat{P}}^{(\lfloor x \rfloor +1,x)} + E \varDelta , \end{aligned}$$

(D.3)

where

$$\begin{aligned} E = \begin{bmatrix} {\mathbf {0}}_{\left( \frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}+1\right) \times \lfloor x \rfloor } \\ {I}_n \end{bmatrix} \quad \varDelta = \begin{bmatrix} {\mathbf {0}}_{\lfloor x \rfloor \times \frac{\lfloor x \rfloor (\lfloor x \rfloor -1)}{2}}&-\frac{\mu (1-q)(1-p)}{\lambda +\mu } {I}_{\lfloor x \rfloor }&\frac{\mu (1-q)(1-p)}{\lambda +\mu } {I}_{\lfloor x \rfloor }&{\mathbf {0}}_{\lfloor x \rfloor \times 1} \end{bmatrix} .\qquad \end{aligned}$$

(D.4)

Hence,

$$\begin{aligned} (P^{(x)})^{d+1} \varvec{e} = P^{(\lfloor x \rfloor +1,x)}\,\left( (P^{(x)})^d \varvec{e} \right) + E \varDelta \,\left( (P^{(x)})^d \varvec{e} \right) \ge P^{(\lfloor x \rfloor +1,x)} \,\left( (P^{(x)})^d \varvec{e} \right) \ge (P^{(\lfloor x \rfloor +1,x)})^{d+1} \varvec{e}.\nonumber \\ \end{aligned}$$

(D.5)

The first inequality follows from the conclusion that $^{d}\nu ^{(x)}_{i,j}$ is increasing in j for any d and $1 \le i \le j \le \lceil x \rceil$ in Lemma 1. Since $^{d}\nu ^{(x)}_{i,\lfloor x \rfloor + 1} \ge ^{d}\nu ^{(x)}_{i,\lfloor x \rfloor }$ for $i = 1, \dots , \lfloor x \rfloor$,

$$\begin{aligned} \varDelta \,\left( (P^{(x)})^d \varvec{e} \right) = \frac{\mu (1-q)(1-p)}{\lambda +\mu } \begin{bmatrix} &{}^{d}\nu ^{(x)}_{1,\lfloor x \rfloor + 1} - ^{d}\nu ^{(x)}_{1,\lfloor x \rfloor } \\ &{} ^{d}\nu ^{(x)}_{2,\lfloor x \rfloor + 1} - ^{d}\nu ^{(x)}_{2,\lfloor x \rfloor } \\ &{} \vdots \\ &{} ^{d}\nu ^{(x)}_{\lfloor x \rfloor ,\lfloor x \rfloor + 1} - ^{d}\nu ^{(x)}_{\lfloor x \rfloor ,\lfloor x \rfloor } \end{bmatrix} \ge 0 . \end{aligned}$$

(D.6)

The second inequality is from the induction assumption. Hence,

$$\begin{aligned} ^{d+1} \nu ^{(x)}_{i,j} \ge \, ^{d+1} {\hat{\nu }}^{(\lfloor x \rfloor +1,x)}_{i,j} . \end{aligned}$$

(D.7)

This concludes the proof for inequality (4.6).

When $x = \lfloor x \rfloor$, ${\hat{P}}^{(\lfloor x \rfloor , \lfloor x \rfloor )} = P^{(\lfloor x \rfloor )}$, and so $\varvec{{\hat{z}}}^{(\lfloor x \rfloor ,\lfloor x\rfloor )} = \varvec{z}^{(\lfloor x \rfloor )}$. Also, $A_1^{(\lfloor x \rfloor )} = 0$, so

$$\begin{aligned}&\left[ I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \quad 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \right] P^{(\lfloor x \rfloor )} \nonumber \\&\quad = \left[ I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \quad 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \right] P^{(\lfloor x \rfloor )} \begin{bmatrix} I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \\ 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \end{bmatrix} \left[ I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \quad 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \right] . \end{aligned}$$

(D.8)

Multiplying Eq. (3.2) by $\left[ I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \quad 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \right]$ on both sides and applying Eq. (D.8), we have

$$\begin{aligned} (I - {\bar{P}}^{(\lfloor x \rfloor )}) \, \left[ I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \quad 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \right] \varvec{w}^{(\lfloor x \rfloor )} = \frac{1}{\lambda +\mu } \, \varvec{e}, \end{aligned}$$

(D.9)

where ${\bar{P}}^{(\lfloor x \rfloor )}:= \left[ I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \quad 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \right] P^{(\lfloor x \rfloor )} \begin{bmatrix} I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \\ 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \end{bmatrix}$. Hence,

$$\begin{aligned} (\varvec{w}^{(\lfloor x \rfloor )})_{1:\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} = \left( I - {\bar{P}}^{(\lfloor x \rfloor )} \right) ^{-1}\frac{1}{\lambda +\mu } \, \varvec{e} . \end{aligned}$$

(D.10)

Similarly,

$$\begin{aligned} (\varvec{{\hat{w}}}^{(\lfloor x \rfloor +1, \lfloor x \rfloor )})_{1:\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} = \left( I - \bar{{\hat{P}}}^{(\lfloor x \rfloor +1, \lfloor x \rfloor )} \right) ^{-1}\frac{1}{\lambda +\mu } \, \varvec{e} , \end{aligned}$$

(D.11)

where $\bar{{\hat{P}}}^{(\lfloor x \rfloor )}:= \left[ I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \quad 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \right] {\hat{P}}^{(\lfloor x \rfloor +1, \lfloor x \rfloor )} \begin{bmatrix} I_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}} \\ 0_{\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2} \times (\lfloor x \rfloor +1)} \end{bmatrix}$. Since $P^{(\lfloor x \rfloor +1, x )}$ and $P^{(x)}$ have the same first $\displaystyle \frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}$ lines and rows, it follows that $\varvec{w}_{1:\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}}^{(\lfloor x \rfloor )} = \varvec{{\hat{w}}}_{1:\frac{\lfloor x \rfloor (\lfloor x \rfloor +1)}{2}}^{(\lfloor x \rfloor +1, \lfloor x \rfloor )}$. $\square$

E

1.1 E.1 The derivative of function f(p)

(E.1)

The calculation of $f'(p)$ is implemented in Wolfram Mathematica.

1.2 E.2 Social welfare expressions

The social welfare in the N-case:

(E.2)

The social welfare in the $R-$case:

(E.3)