Abandonment versus blocking in many-server queues: asymptotic optimality in the QED regime

Abstract

We consider a controlled queueing system of the $G/M/n/B+GI$ type, with many servers and impatient customers. The queue-capacity $B$ is the control process. Customers who arrive at a full queue are blocked and customers who wait too long in the queue abandon. We study the tradeoff between blocking and abandonment, with cost accumulated over a random, finite time-horizon, which yields a queueing control problem (QCP). In the many-server quality and efficiency-driven (QED) regime, we formulate and solve a diffusion control problem (DCP) that is associated with our QCP. The DCP solution is then used to construct asymptotically optimal controls (of the threshold type) for QCP. A natural motivation for our QCP is telephone call centers, hence the QED regime is natural as well. QCP then captures the tradeoff between busy signals and customer abandonment, and our solution specifies an asymptotically optimal number of trunk-lines.

Appendix

1.1 Proof of Lemma 3.2

Let $x \le 0$ and consider the diffusion process $Y$ characterized by

$$\begin{aligned} Y(t)=x + \sigma W(t) - \mu \int _{0}^{t}(\beta +Y(s))\mathrm{{d}}s, \quad t \ge 0, \end{aligned}$$

(5.1)

where $W$ is a standard Brownian motion. Next, we introduce the stopping time $\tau _0$ by

$$\begin{aligned} \tau _0= \inf \{ t \ge 0: Y(t)=0 \}. \end{aligned}$$

(5.2)

We intend to show that the function $F_\infty$ of (3.15) and (3.16) has the stochastic representation $F_\infty (x)=E[\mathrm{{e}}^{ -\gamma \tau _0}| Y(0)=x]$, for every $x \le 0$. To this end, introduce

$$\begin{aligned} \tilde{F}_\infty (x)=E[\mathrm{{e}}^{- \gamma \tau _0}| Y(0)=x], \end{aligned}$$

(5.3)

for all $x \le 0$. For each $N \ge |x|$, we also introduce the stopping time

$$\begin{aligned} \tau _N= \inf \{ t \ge 0: Y(t)=-N \}. \end{aligned}$$

(5.4)

To construct the function $F_\infty$, we begin with a sequence of functions $(F_n)$. Let $F_n$ be the unique solution to the boundary value problem

$$\begin{aligned} \frac{\sigma ^2}{2}F_n^{\prime \prime }(x)-(\beta \mu +\mu x) F^{\prime }_n(x)-\gamma F_n(x)=0, \quad \mathrm{for} \quad x<0,\\ F_n(0)=1, \quad \mathrm{and} \quad F_n(-n)=0. \end{aligned}$$

Then, with the use of It$\hat{o}$’s lemma, one can easily verify that

$$\begin{aligned} F_n(x)=E_x[\mathrm{{e}}^{- \gamma \tau _0} I_{[\tau _0 < \tau _n]}] \quad \mathrm{for} \quad -n \le x \le 0. \end{aligned}$$

Using the scale function associated with the diffusion $Y$, it follows that (see [17], Sect. 4.1 of Chapter 5)

$$\begin{aligned} P[ \tau _n < \tau _0 | Y(0)=x]= \frac{\int _{x}^{0} \mathrm{{e}}^{\frac{\mu }{\sigma ^2}(y^2 +2 \beta y)}\mathrm{{d}}y}{\int _{-n}^{0} \mathrm{{e}}^{\frac{\mu }{\sigma ^2}(y^2 +2 \beta y)}\mathrm{{d}}y}. \end{aligned}$$

Therefore, $P[ \tau _n < \tau _0 | Y(0)=x]$ is decreasing to zero as $n$ tends to infinity. Consequently, the sequence $(F_n(x))$ is increasing to the function $\tilde{F}_\infty$. We fix the interval $[x,0]$ and integrate the differential equation for $F_n$ twice on this interval to obtain an integral equation for $F_n$. Then we use $0\le F_n(y) \le \tilde{F}_\infty (y) \le 1$, on $[-n,0]$, and $\lim \limits _{n \rightarrow \infty }F_n(y)=\tilde{F}_\infty (y)$, together with the bounded convergence theorem, to conclude that the function $\tilde{F}_\infty$ also satisfies the same integral equation. By differentiating it twice, we observe that $\tilde{F}_\infty$ also satisfies (3.15), together with the boundary condition $\tilde{F}_\infty (0)=1$.

The stochastic representation (5.3) also implies that $\tilde{F}_\infty$ is increasing and $\tilde{F}^{\prime }_\infty (x) \ge 0$ on the interval $(-\infty ,0]$. Consequently, $\lim \limits _{x \rightarrow -\infty } \tilde{F}_\infty (x) =L_0$ exists, with $0 \le L_0 <1$. Furthermore, if $\tilde{F}^{\prime }_\infty (\xi )=0$ for some $\xi <0$, then by (3.15), $\tilde{F}^{\prime \prime }_\infty (\xi )>0$. Hence, $x =\xi$ is a strict local minimum. This is a contradiction since $\tilde{F}_\infty$ is increasing. Hence $\tilde{F}^{\prime }_\infty (x) > 0$, for all $x<0$ and as a consequence, $\tilde{F}_\infty$ is strictly increasing on $(-\infty ,0]$.

Our next step is to prove that $\lim \limits _{x \rightarrow -\infty } \tilde{F}_\infty (x) =0$. We consider the process $Z$ defined by $Z(t)=Y(t)+ \beta$, for all $t \ge 0$, where $Y$ is given in (3.17). Then $Z$ is an Ornstein-Uhlenbeck process that satisfies

$$\begin{aligned} Z(t)=(x+\beta )+\sigma W(t) - \mu \int _{0}^{t}Z(s)\mathrm{{d}}s, \quad t \ge 0. \end{aligned}$$

(5.5)

For each $y>x+\beta$, we introduce the stopping time $\tilde{\tau }_y$ by

$$\begin{aligned} \tilde{\tau }_y=\inf \{ t \ge 0 : Z(t) \ge y \}. \end{aligned}$$

(5.6)

Then $\tau _0$, defined in (5.2), is identical to $\tilde{\tau }_\beta$, and $\tilde{F}_\infty (x)= E[\mathrm{{e}}^{- \gamma \tilde{\tau }_\beta } | Z(0)=x + \beta ]$, for all $x<\min \{ - \beta , 0 \}$. Using the strong Markov property, we obtain

$$\begin{aligned} E[\mathrm{{e}}^{- \gamma \tilde{\tau }_0}|Z(0)=x+ \beta ]= E[\mathrm{{e}}^{- \gamma \tilde{\tau }_\beta }|Z(0)=x+ \beta ] E[\mathrm{{e}}^{- \gamma \tilde{\tau }_0}|Z(0)= \beta ] \quad \mathrm{if} \quad \beta <0, \end{aligned}$$

and

$$\begin{aligned} E[\mathrm{{e}}^{- \gamma \tilde{\tau }_\beta }|Z(0)=x+ \beta ]= E[\mathrm{{e}}^{- \gamma \tilde{\tau }_0}|Z(0)=x+ \beta ] E[\mathrm{{e}}^{- \gamma \tilde{\tau }_\beta }|Z(0)= 0] \quad \mathrm{if} \quad \beta >0. \end{aligned}$$

Therefore, to reach the desired conclusion, it suffices to demonstrate that

$$\begin{aligned} \lim \limits _{ y \rightarrow -\infty }E[\mathrm{{e}}^{- \gamma \tilde{\tau }_0}|Z(0)=y]=0. \end{aligned}$$

For this, consider the process $Z$ that is characterized by $Z(t)=y+\sigma W(t) - \mu \int _{0}^{t}Z(s)\mathrm{{d}}s$. Then it is well known that, via a random time change, one can write

$$\begin{aligned} Z(t)= \mathrm{{e}}^{ -\mu t}\left[ y+ B\left( \frac{\sigma ^2}{2 \mu }(\mathrm{{e}}^{ 2 \mu t}-1)\right) \right] , \quad t \ge 0, \end{aligned}$$

where $B$ is another Brownian motion. Next, we introduce a collection of stopping times $(T_y)$ with respect to this Brownian motion. Specifically, for each $y<0$, let

$$\begin{aligned} T_y= \inf \{ t \ge 0: B(t)=-y\}. \end{aligned}$$

This enables one to derive a relationship between $\tilde{\tau }_0$ and $T_y$, namely

$$\begin{aligned} \tilde{\tau }_0=\frac{1}{2 \mu }\log \left[ \left( \frac{2 \mu }{\sigma ^2}\right) T_y +1\right] . \end{aligned}$$

The distribution of the Brownian stopping time $T_y$ is well known (see [3]), and it will help us compute the limit $\lim \limits _{ y \rightarrow -\infty }E[\mathrm{{e}}^{- \gamma \tilde{\tau }_0}|Z(0)=y]$.

Observe that $\mathrm{{e}}^{- \gamma \tilde{\tau }_0}=[(\frac{2 \mu }{\sigma ^2})T_y +1]^{ - \frac{\gamma }{2 \mu }}$. It follows that

$$\begin{aligned} \lim \limits _{ y \rightarrow -\infty }E[\mathrm{{e}}^{- \gamma \tilde{\tau }_0}|Z(0)=y] =\lim \limits _{ y \rightarrow -\infty }E \left[ \frac{1}{[(\frac{2 \mu }{\sigma ^2})T_y +1]^{\frac{\gamma }{2 \mu }}} |B(0)=0\right] . \end{aligned}$$

Using the bounded convergence theorem, we notice that the limit on the right-hand side vanishes since $\lim \limits _{ y \rightarrow -\infty }T_y=\infty$ almost surely.

Consequently, $\lim \limits _{x \rightarrow -\infty } \tilde{F}_\infty (x) =0$. Now it is clear that $\tilde{F}_\infty$ satisfies (3.15) and (3.16). The uniqueness of solutions to (3.15) and (3.16) can be established by the fact that the difference of two solutions to (3.15) cannot have any positive local maxima. Therefore, $\tilde{F}_\infty$ is identical to $F_\infty$, as characterized via the initial value problem (3.15) and (3.16). It also has the stochastic representation (5.3).

In our next step, we show that $\lim \limits _{x \rightarrow -\infty } F^{\prime }_\infty (x) =0$ and the function $F_\infty$ is strictly convex. First we extend the function $F_\infty$ to $(-\infty ,\infty )$, so that it satisfies the differential equation (3.15) everywhere on $(-\infty ,\infty )$. Since $F_\infty$ is strictly increasing on $(- \infty , 0]$, by (3.16) it is clear that $\liminf \limits _{x \rightarrow -\infty } F^{\prime }_\infty (x) =0$. Thus, we can choose a sequence $(y_n)$ strictly decreasing to $-\infty$, such that $y_{n+1}<y_n<0$ and $0<F^{\prime }_\infty (y_{n+1})< F^{\prime }_\infty (y_{n})$, for all $n$. Consequently, there is a point $\xi _n$ such that $y_{n+1}<\xi _n<y_n$ and $F^{\prime \prime }_\infty (\xi _n)>0$. Note that the sequence $(\xi _n)$ is also strictly decreasing to $-\infty$.

Let $z=\inf \{ x \ge 0:F^{\prime }_\infty (x)\le 0 \}$. If $z$ is finite, then $F^{\prime }_\infty (z)=0$ and $F^{\prime }_\infty (x)>0$, for all $x <z$, and consequently, $F_\infty (z) \ge F_\infty (0)=1$. By (3.15), $Y_n$ has paths of bounded variation $F^{\prime \prime }_\infty (z)>0$, and hence $F_\infty$ has a strict local minimum at the point $x=z$, which is a contradiction. Therefore, $z$ cannot be finite and $F^{\prime }_\infty (x)>0$, for all $x$ in $(-\infty ,\infty )$.

Next we consider any point $x_1>0$, such that $x_1+\beta >0$. Then $F_\infty (x_1)>F_\infty (0)=1$ and $F^{\prime }_\infty (x_1) >0$. Using (3.15), we also obtain $F^{\prime \prime }_\infty (x_1) >0$. Now introduce the function $H(x)=F^{\prime \prime }_\infty (x)$, on the interval $[\xi _n, x_1]$. Then, by differentiating (3.15), we see that

$$\begin{aligned} \frac{\sigma ^2}{2}H^{\prime \prime }(x)-(\beta \mu +\mu x) H^{\prime }(x)-(\gamma +2\mu ) H(x)=0,\quad \mathrm{on}\, [\xi _n, x_1], \end{aligned}$$

$H(\xi _n)>0$ and $H(x_1)>0$. Let $\xi _n \le c \le x_1$ so that $H(c)=\min \limits _{[\xi _n, x_1]}H(x)$. Suppose that $H(c)\le 0$, then $\xi _n <c<x_1$ and $H^{\prime }(c)=0$. If $H(c)=0$, then by the uniqueness of the solution to the above differential equation, it follows that $H$ is identically zero. If $H(c)<0$, again by the above differential equation, we have $H^{\prime \prime }(c)<0$ and $x=c$ is a strict local maximum; this is a contradiction. Hence $H(c)>0$ and, consequently, $F^{\prime \prime }_\infty (x)>0$ on the interval $[\xi _n, x_1]$. But one can choose $x_1$ arbitrarily large and the sequence $(\xi _n)$ is decreasing to $-\infty$. We thus conclude that $F^{\prime \prime }_\infty (x)>0$, on the interval $(-\infty , \infty )$. This, together with the fact that $\liminf \limits _{x \rightarrow -\infty } F^{\prime }_\infty (x) =0$, implies $\lim \limits _{x \rightarrow -\infty } F^{\prime }_\infty (x) =0$. Hence, $F_\infty$ is a strictly convex function that satisfies all the conclusions of Lemma 3.2. This completes the proof.