Strategic bidding in an accumulating priority queue: equilibrium analysis

Abstract

We study the strategic purchasing of priorities in a time-dependent accumulating priority M/G/1 queue. We formulate a non-cooperative game in which customers purchase priority coefficients with the goal of reducing waiting costs in exchange. The priority of each customer in the queue is a linear function of the individual waiting time, with the purchased coefficient being the slope. The unique pure Nash equilibrium is solved explicitly for the case with homogeneous customers. A general characterisation of the Nash equilibrium is provided for the heterogeneous case. It is shown that both avoid the crowd and follow the crowd behaviors are prevalent, within class types and between them. We further present a pricing mechanism that ensures the order of the accumulating priority rates in equilibrium follows a $C\mu$ type rule and improves overall efficiency.

Appendix: proofs

Proof

(Proof of Lemma 1 )

• 1. The proof of the recursive waiting time formula (2) follows the same arguments used in pp. 126–131 of Kleinrock (1976) for the discrete case. We outline the proof without going into all the details. If the AP rate of a singled out customer is a then her expected waiting is

  $$\begin{aligned} \begin{aligned} \mathrm {W}(a;\mathcal {F})&= \mathrm {W}_0+\mathrm {E}\sum _{i=1}^N\left[ \sum _{j=1}^{N_i(a)}X_{ij}+\sum _{j=1}^{M_i(a)}X_{ij}\right] \\&= \mathrm {W}_0+\sum _{i=1}^N\overline{x}_i(\mathrm {M}_i(a)+\mathrm {N}_i(a)), \end{aligned} \end{aligned}$$

  (10)

  where $\mathrm {W}_0$ is the expected remaining time of the customer in service, $X_{ij}$ is the service time of the j’th type i customer, $\mathrm {M}_i(a)$ is the expected number of type i customers who will overtake her, and $\mathrm {N}_i(a)$ is the expected number of type i customers who were present in the queue upon her arrival and will be admitted into service before her. The second equality in (10) requires a more cautious consideration because service times and the number of overtaking customers are not independent. The condition for the Wald identity,

  $$\begin{aligned} \mathrm {E}\sum _{j=1}^{M_i(a)}X_{ij}=\mathrm {E}M_i(a)\mathrm {E}X_{i1}, \end{aligned}$$

  to hold can be stated as (see p. 158 of Durrett (2010))

  $$\begin{aligned} \mathrm {E}X_{im}\mathbbm {1}_{\{M_i(a)\ge m\}}=\mathrm {E}X_{im}\mathop {\mathrm {P}}(M_i(a)\ge m), \ \forall m\ge 1. \end{aligned}$$

  This is indeed correct because the event $\{M_i(a)\ge m\}$ simply means that the m’th arrival of type i has overtaken the tagged customer, and this only depends on the service times prior to m (recall there is no preemption), and is therefore independent of $X_{im}$. In a similar manner, if we order the customers present in the queue by their arrival times then the event $\{N_i(a)\ge n\}$ does not depend on $X_{in}$, but only on the service times $X_{i1},\ldots ,X_{i(n-1)}$. If the customer arrived at time 0 and waited for $w\ge 0$ time in the queue then any type i customer with rate $b>a$ that arrived at time t such that $b(w-t)>aw$ has overtaken her. In other words, all arrivals in the interval $\left( 0,w\left( 1-\frac{a}{b}\right) \right)$ overtake the tagged customer. The arrival rate of such customers is $\lambda _i$, hence by the splitting splitting and superposition properties of the Poisson process and iterating on conditional expectation on the waiting time, we obtain

  $$\begin{aligned} \mathrm {M}_i(a)=\lambda _i \mathrm {W}(a;\mathcal {F})\int _a^\infty \left( 1-\frac{a}{b}\right) \ dF_i(b),\quad i=1,\ldots ,N. \end{aligned}$$

  (11)

  We are left with computing the number of type i customers in the queue (upon arrival) that the customer bidding a will not overtake. Clearly she will not overtake any customers with rate $b\ge a$, and the expected number of such customers is $\lambda _i\int _a^\infty \mathrm {W}(b;\mathcal {F})\ dF_i(b)$. A customer with priority $b<a$ who arrived at $t-s$ and waits v time in the queue will still be in the queue if $s<v$ and will not be overtaken if $(v-s)a<vb$. Thus, the probability of a customer with rate b who arrived at $t-s$ being overtaken by the tagged customer who arrived at t is then

  $$\begin{aligned} \mathop {\mathrm {P}}\left( s<W(b;\mathcal {F})<\frac{a}{a-b}s\right) =\mathop {\mathrm {P}}\left( W(b;\mathcal {F})>s\right) -\mathop {\mathrm {P}}\left( W(b;\mathcal {F})>\frac{a}{a-b}s\right) . \end{aligned}$$

  Again, we use the properties of the Poisson process, namely splitting and superposition, to obtain the expected number of such customers,

  $$\begin{aligned}&\lambda _i\int _0^\infty \left[ \mathop {\mathrm {P}}\left( W(b;\mathcal {F})>s\right) -\mathop {\mathrm {P}}\left( W(b;\mathcal {F})>\frac{a}{a-b}s\right) \right] ds\ dF_i(b) \\&\quad =\lambda _i\left[ \mathrm {W}(b;\mathcal {F})-\left( 1-\frac{b}{a}\right) \mathrm {W}(b;\mathcal {F})\right] \ dF_i(b)=\lambda _i\mathrm {W}(b;\mathcal {F})\frac{b}{a}\ dF_i(b) . \end{aligned}$$

  By integrating on all customer types we get

  $$\begin{aligned} \mathrm {N}_i(a)=\lambda _i\left[ \int _0^{a-} \mathrm {W}(b;\mathcal {F})\frac{b}{a}\ dF_i(b)+\int _a^\infty \mathrm {W}(b;\mathcal {F})\ dF_i(b)\right] , \end{aligned}$$

  (12)

  where $a-$ indicates that the integral does include the atom, $dF_i(a)$, if it exists. For more detailed analysis and justification of the above computations the reader is referred to Kleinrock (1976), Stanford et al. (2014). Combining (11) and (12) we get

  $$\begin{aligned}&\mathrm {W}(a;\mathcal {F})=\mathrm {W}_0\\&\quad +\sum _{i=1}^N\rho _i\left[ \int _0^{a-} \mathrm {W}(b;\mathcal {F})\frac{b}{a}\ dF_i(b){+}\int _a^\infty \left( \mathrm {W}(a;\mathcal {F})\left( 1{-}\frac{a}{b}\right) {+}\mathrm {W}(b;\mathcal {F})\ \right) dF_i(b)\right] , \end{aligned}$$

  or equivalently

  $$\begin{aligned} \mathrm {W}(a;\mathcal {F})=\frac{\mathrm {W}_0+\sum _{i=1}^N\rho _i\left[ \int _0^{a-} \mathrm {W}(b;\mathcal {F})\frac{b}{a}\ dF_i(b)+\int _a^\infty \mathrm {W}(b;\mathcal {F})\ dF_i(b)\right] }{1-\sum _{i=1}^N\rho _i\int _a^\infty \left( 1-\frac{a}{b}\right) \ dF_i(b)}. \end{aligned}$$

  By the work conservation property we have that

  $$\begin{aligned} \sum _{i=1}^N\rho _i\int _a^\infty \mathrm {W}(b;\mathcal {F})\ dF_i(b)=\frac{\mathrm {W}_0\rho }{1-\rho }-\sum _{i=1}^N\rho _i\int _0^{a-} \mathrm {W}(b;\mathcal {F})\ dF_i(b), \end{aligned}$$

  which leads to the general recursive formula (2). Note that $a-$ can be replaced by a because if there is a point mass at a the value inside the integral is zero. Furthermore, since the waiting time is a decreasing function of the bid and $\rho <1$, all above integrals are finite. Specifically, for every $i=1,\ldots ,N$ we have

  $$\begin{aligned} \int _a^\infty \mathrm {W}(b;\mathcal {F})\ dF_i(b) < \mathrm {W}(a;\mathcal {F})\int _a^\infty \ dF_i(b) < \mathrm {W}(a;\mathcal {F})<\infty , \end{aligned}$$

  and

  $$\begin{aligned} \int _a^\infty \left( 1-\frac{a}{b}\right) \ dF_i(b) < \int _a^\infty \ dF_i(b)<1. \end{aligned}$$

• 2. We can rewrite (2) as

  $$\begin{aligned} \mathrm {W}(a,\mathcal {F})=\frac{\frac{\mathrm {W}_0}{1-\rho }-\sum _{i=1}^N\rho _i H_i(a)}{1-\sum _{i=1}^N\rho _i J_i(a)}, \end{aligned}$$

  where

  $$\begin{aligned} H_i(a):=\int _0^{a}\mathrm {W}(b;\mathcal {F})\left( 1-\frac{b}{a}\right) dF_i(b) \end{aligned}$$

  and

  $$\begin{aligned} J_i(a):=\int _a^{\infty }\left( 1-\frac{a}{b}\right) dF_i(b). \end{aligned}$$

  The only possible points of discontinuity of $H_i(a)$ and $J_i(a)$ are ones such that there is a jump in the measure of integration, i.e. a point $\tilde{a}$ such that $F_i(\tilde{a}-)<F_i(\tilde{a})$ (and $dF_i(\tilde{a})>0$). Observe however, that in these points the value inside the integral is zero. Thus, we have established the continuity of $\mathrm {W}(a,\mathcal {F})$. The derivative is continuous at $\tilde{a}$ such that $F_i(\tilde{a}-)=F_i(\tilde{a})$ for some $i=1,\ldots ,N$ if

  $$\begin{aligned} \lim _{a\uparrow \tilde{a}}\frac{d}{da}\mathrm {W}(a,\mathcal {F})=\lim _{a\downarrow \tilde{a}}\frac{d}{da}\mathrm {W}(a,\mathcal {F}). \end{aligned}$$

  Let $H(a):=\sum _{i=1}^N \rho _i H_i(a)$ and $J(a):=\sum _{i=1}^N \rho _iJ_i(a)$. The first derivative at any point a such that $F_i(a-)=F_i(a), \ \forall i=1,\ldots ,N$ is

  $$\begin{aligned} \begin{aligned} \frac{d}{da}\mathrm {W}(a,\mathcal {F})&= \frac{-H'(a)(1-J(a))+J'(a)\left( \frac{\mathrm {W}_0}{1-\rho }-H(a)\right) }{(1-J(a))^2} \\&= \frac{\mathrm {W}(a,\mathcal {F})J'(a)-H'(a)}{1-J(a)}. \end{aligned} \end{aligned}$$

  We have already established that H(a) and J(a) are continuous. Hence, the denominator is continuous. Moreover, the numerator is continuous if

  $$\begin{aligned} K(a) {:=} \mathrm {W}(a,\mathcal {F})J'(a)-H'(a), \end{aligned}$$

  is continuous. With some caution we can apply the derivative chain rule to both integral terms. Using Assumption 1 that F is defined as a combination of point masses and intervals with positive density we have that

  $$\begin{aligned} \begin{aligned} K(a)&= \sum _{i=1}^N \rho _i\left[ \mathrm {W}(a,\mathcal {F})\frac{d}{da}\int _a^{\infty }\left( 1{-}\frac{a}{b}\right) dF_i(b){-}\frac{d}{da}\int _0^{a}\mathrm {W}(b,\mathcal {F})\left( 1{-}\frac{b}{a}\right) dF_i(b)\right] \\&= -\sum _{i=1}^N \rho _i\left[ \int _a^\infty \mathrm {W}(a,\mathcal {F})\frac{1}{b}dF_i(b)+\int _0^a\frac{b}{a^2}\mathrm {W}(b,\mathcal {F})dF_i(b)\right] . \end{aligned} \end{aligned}$$

  By the right continuity of the cdf we have that at any discontinuity point $\tilde{a}$, the term $\mathrm {W}_F(\tilde{a})\frac{1}{\tilde{a}}dF(\tilde{a})$ moves from the left integral to the right integral. Thus, we can conclude that $\lim _{a\uparrow \tilde{a}}K(a)=\lim _{a\downarrow \tilde{a}}K(a)$, for any discontinuity point $\tilde{a}$.

• 3. First we observe that $K(a)<0$ and therefore $\mathrm {W}(a;\mathcal {F})$ is a monotone decreasing function, as expected. It can further be verified that $J'(a)<0$ and $K'(a)>0$ for all $a>0$, hence

  $$\begin{aligned} \frac{d^2}{da^2}\mathrm {W}(a;\mathcal {F})=\frac{\rho K'(a)(1-\rho J(a))+K(a)\rho J'(a)}{(1-\rho J(a))^2}>0. \end{aligned}$$

  Therefore the expected waiting time of a single customer is strictly convex with respect to a change in her own AP rate, regardless of F.$\square$

Proof

(Proof of Proposition 2 ) We prove the monotonicity properties of

$$\begin{aligned} \tilde{\mathrm {W}}(b_i;\mathbf {b})=\frac{1-\sum _{k=i}^N\rho _k\left( 1-\frac{b_i}{b_k}\right) -\frac{C_i}{b_i^2}\sum _{k=1}^{i-1}\rho _k b_k \mathrm {W}(b_k;\mathbf {b})}{C_i\sum _{k=i}^N\frac{\rho _k}{b_k}}, \quad i=1,\ldots ,N, \end{aligned}$$

where

$$\begin{aligned} \mathrm {W}(b_i;\mathbf {b})=\frac{\frac{\mathrm {W}_0}{(1-\rho )}-\sum _{k=1}^{i-1}\rho _k(1-\frac{b_k}{b_i})\mathrm {W}(b_k;\mathbf {b})}{1-\sum _{k=i}^N \rho _k(1-\frac{b_i}{b_k})}, \ 1 \le i \le N. \end{aligned}$$

Throughout the proof we assume that $\mathbf {b}$ is ordered and only allow changes of single coordinates that maintain the order, i.e., $b_i\in (b_{i-1},b_{i+1})$ for $i=1,\ldots ,N$.

• 1. We will first show that $\tilde{\mathrm {W}}(b_i;\mathbf {b})$ is monotone increasing w.r.t. $b_i$. The result is immediate for $i=1$, as the sum in the numerator is empty. For $i>1$ it suffices to show that

  $$\begin{aligned} \frac{1}{b_i^2}\mathrm {W}(b_j;\mathbf {b})=\frac{\frac{\mathrm {W}_0}{(1-\rho )}-\sum _{k=1}^{j-1}\rho _k(1-\frac{b_k}{b_j})\mathrm {W}(b_k;\mathbf {b})}{b_i^2\left( 1-\sum _{k=j}^N\rho _k\right) +\sum _{k=j}^N\frac{b_j b_i^2}{b_k}}, \end{aligned}$$

  is monotone decreasing w.r.t. $b_i$ for all $j<i$. The denominator clearly increases with $b_i$. The expected waiting time $\mathrm {W}(b_k;\mathbf {b})$ increases with any $b_j$ such that $j\ne k$, and as this is the only element of the numerator dependent on $b_i$ we have that the numerator is decreasing.

• 2. Next we will show that $\tilde{\mathrm {W}}(b_i;\mathbf {b})$ decreases with $b_j$ if $j<i$. By taking derivative we have that

  $$\begin{aligned} \frac{d}{db_j}\tilde{\mathrm {W}}(b_i;\mathbf {b}) \propto -\rho _j\frac{d}{db_j}\left[ b_j\mathrm {W}(b_j;\mathbf {b})\right] -\sum _{\{k\ne j,k<i\}}\rho _k b_k\frac{d}{db_j}\mathrm {W}(b_k;\mathbf {b}). \end{aligned}$$

  The sum on the right-hand side is positive because $\frac{d}{db_j}\mathrm {W}(b_k;\mathbf {b})$ is positive for all $k\ne j$. It therefore remains to be shown that

  $$\begin{aligned} b_j\mathrm {W}(b_j;\mathbf {b})=\frac{\frac{\mathrm {W}_0}{1-\rho }-\sum _{k=1}^{j-1}\left( 1-\frac{b_k}{b_j}\right) \mathrm {W}(b_k;\mathbf {b})}{\frac{1}{b_j}\left( 1-\sum _{k=j+1}^N\rho _k\right) +\sum _{k=j+1}^N\frac{\rho _k}{b_k}}, \end{aligned}$$

  is an increasing function w.r.t. $b_j$. For $j=1$ this is clearly true, as the sum in the numerator is empty. Furthermore, the the convexity of the waiting time implies that $\mathrm {W}(b_j;\mathbf {b})$ is decreasing at a decreasing rate, and hence cannot decrease faster than $b_j>b_1$.

• 3. The last part of the Lemma is a negative result, that can be verified by examples. Numerical examples such as those in Fig.  2 show that the $\tilde{W}_i$ can be both increasing or decreasing with $b_j$ such that $j>i$. Obviously, explicit numbers can be plugged into the equations for this verification.$\square$