Customers’ abandonment strategy in an M / G / 1 queue

Abstract

We consider an M / G / 1 queue in which the customers, while waiting in line, may renege from it. We show the Nash equilibrium profile among customers and show that it is defined by two sequences of thresholds. For each customer, the decision is based on the observed past (which determines from what sequence the threshold is taken) and the observed queue length (which determines the appropriate element in the chosen sequence). We construct a set of equations that has the Nash equilibrium as its solution and discuss the relationships between the properties of the service time distribution and the properties of the Nash equilibrium, such as uniqueness and finiteness.

Appendix A

Proof

$$\begin{aligned}&\mathbb {P}(Q \le y|Q \le R(0,a)\wedge A_2\wedge (I(T_1-a)+(1-I)(A_1-W_1) )) \\&\quad = \frac{\mathbb {P}(Q \le y,Q \le R(0,a)\wedge A_2\wedge (I(T_1-a)+(1-I)(A_1-W_1) ))}{\mathbb {P}(Q \le R(0,a)\wedge A_2\wedge (I(T_1-a)+(1-I)(A_1-W_1)))}. \end{aligned}$$

We give explicit expressions for both the numerator and the denominator.

The numerator is

$$\begin{aligned}&\mathbb {P}(Q \le y,Q \le R(0,a)\wedge A_2\wedge (I(T_1-a)+(1-I)(A_1-W_1) ))\\&\quad =\int _{0}^{A_1} \mathbb {P}(I=1)\mathbb {P}(Q \le y,Q \le R(0,a)\wedge A_2\wedge (T_1-a))\\&\qquad +P(I{=}0)\int _{w_1=0}^{A_2\wedge (A_1{-}w_1) }\mathbb {P}(Q \le y,Q \le R(0,a){\wedge } A_2{\wedge }(A_1-w_1) ) p(0,a,w_1)\mathrm{d}w_1, \end{aligned}$$

and the denominator is

$$\begin{aligned}&\mathbb {P}(Q \le R(0,a)\wedge A_2\wedge (I(T_1-a)+(1-I)(A_1-W_1)))\\&\quad =\int _{0}^{A_1} \mathbb {P}(I=1)\mathbb {P}(Q \le R(0,a)\wedge A_2\wedge (T_1-a))\\&\qquad +\,\mathbb {P}(I=0)\mathbb {P}(Q \le R(0,a)\wedge A_2\wedge (A_1-w_1) ) p(0,a,w_1)\mathrm{d}w_1. \end{aligned}$$

After taking the derivative, we obtain the p.d.f. of $Q|Q \le R(0,a)\wedge A_2\wedge (I(T_1-a)+(1-I)(A_1-W_1))$, and hence of $Y|N=2,A=a$ as well. $\square$