Robust queueing theory: an initial study using imprecise probabilities

Abstract

We study the robustness of performance predictions of discrete-time finite-capacity queues by applying the framework of imprecise probabilities. More concretely, we consider the Geo/Geo/1/L model with probabilities of arrival and departure that are no longer fixed, but are allowed to vary within given intervals. We distinguish between two concepts of independence in this framework, namely repetition independence and epistemic irrelevance. In the first approach, we assume the existence of time-homogeneous probabilities for arrival and departure, which leads us to consider a collection of stationary queues. In the second, the stationarity assumption is dropped and we allow the arrival and departure probabilities to vary from time point to time point; they may even depend on the complete history of queue lengths. We calculate bounds on the expected queue length, the probability of a particular queue length and the probability of turning on the server. For the expected queue length, both approaches coincide. For the other performance measures, we observe and discuss various differences between the bounds obtained for these two approaches. One of our observations is that ergodicity may break down due to imprecision: bounds on expected time averages of certain functions on the state space are not necessarily equal to the bounds on the expectation of that function at random instants in a steady-state queue.

Appendices

Appendix

Proofs

Proof of Theorem 1

First notice that

$$\begin{aligned} \underline{E}^{{\text {EI}}}_{1:n}{\left( f\vert X_{1:m}\right) }&=\min {{\left\{ E^{p_{A,D}}_{1:n}{\left( f\vert X_{1:m}\right) }:p_{A,D}\in \mathscr {T}^{{\text {EI}}}\right\} }}\\&=\min {{\left\{ E^{p_{A,D}}_{1:n-1}{\left( E^{p_{A,D}}_{1:n}{\left( f\vert X_{1:n-1}\right) }\vert X_{1:m}\right) }:p_{A,D}\in \mathscr {T}^{{\text {EI}}}\right\} }}\\&=\min {{\left\{ E^{p_{A,D}}_{1:n-1}{\left( E^{p'_{A,D}}_{1:n}{\left( f\vert X_{1:n-1}\right) }\vert X_{1:m}\right) }:p_{A,D}\in \mathscr {T}^{{\text {EI}}},p'_{A,D}\in \mathscr {T}^{{\text {EI}}}\right\} }}\\&=\min {{\left\{ E^{p_{A,D}}_{1:n-1}{\left( \underline{E}^{{\text {EI}}}_{1:n}{\left( f\vert X_{1:n-1}\right) }\vert X_{1:m}\right) }:p_{A,D}\in \mathscr {T}^{{\text {EI}}}\right\} }}\\&=\underline{E}^{{\text {EI}}}_{1:n-1}{\left( \underline{E}^{{\text {EI}}}_{1:n}{\left( f\vert X_{1:n-1}\right) }\vert X_{1:m}\right) }, \end{aligned}$$

where the crucial third equality holds because the local probabilities of the probability trees in $\mathscr {T}^{{\text {EI}}}$ are chosen independently of each other. By continuing in this way, we find that

$$\begin{aligned}&\underline{E}^{{\text {EI}}}_{1:n}{\left( f\vert X_{1:m}\right) }\\&\quad =\underline{E}^{{\text {EI}}}_{1:n-1}{\left( \underline{E}^{{\text {EI}}}_{1:n}{\left( f\vert X_{1:n-1}\right) }\vert X_{1:m}\right) }\\&\quad =\underline{E}^{{\text {EI}}}_{1:n-2}{\left( \underline{E}^{{\text {EI}}}_{1:n-1}{\left( \underline{E}^{{\text {EI}}}_{1:n}{\left( f\vert X_{1:n-1}\right) }\vert X_{1:n-2}\right) }\vert X_{1:m}\right) }\\&\quad =\dots \\&\quad =\underline{E}^{{\text {EI}}}_{1:m+1}{\left( \underline{E}^{{\text {EI}}}_{1:m+2}{\left( \dots \underline{E}^{{\text {EI}}}_{1:n-1}{\left( \underline{E}^{{\text {EI}}}_{1:n}{\left( f\vert X_{1:n-1}\right) }\vert X_{1:n-2}\right) } \dots \vert X_{1:m+1}\right) } \vert X_{1:m}\right) }. \end{aligned}$$

The result now follows because

$$\begin{aligned} \underline{E}^{{\text {EI}}}_{1:1}{\left( h\right) }&=\min {{\left\{ E^{p_{A,D}}_{1:1}(h):p_{A,D}\in \mathscr {T}^{{\text {EI}}}\right\} }}\\&=\min {{\left\{ \sum _{x_1\in \mathscr {X}}p(x_1)h(x_1):p_{X_1}\in \mathscr {Q}_1\right\} }} =\underline{Q}_1(h) \text { for all}\, h\in \mathscr {L}(\mathscr {X}), \end{aligned}$$

and because, for all $i\in \mathbb {N}$:

$$\begin{aligned} \underline{E}^{{\text {EI}}}_{1:i+1}{\left( h\vert X_{1:i}\right) }&=\min {{\left\{ E^{p_{A,D}}_{1:i+1}(h\vert X_{1:i}):p_{A,D}\in \mathscr {T}^{{\text {EI}}}\right\} }}\\&=\min \bigg \{\sum _{x_{i+1}\in \mathscr {X}}{q{\left( x_{i+1}\vert X_{i},a_{X_{1:i}},d_{X_{1:i}}\right) }}h(X_{1:i},x_{i+1}):\\&\quad \quad \quad \quad (\forall x_{1:i}\in \mathscr {X}^i)\, a_{x_{1:i}} \in [\underline{a},\overline{a}], (\forall x_{1:i}\in \mathscr {X}^i)\, d_{x_{1:i}} \in [\underline{d},\overline{d}]\bigg \}\\&=\underline{Q}_{i+1}(h\vert X_{1:i}) \text { for all}\, h\in \mathscr {L}(\mathscr {X}^{i+1}). \end{aligned}$$

Proof of Theorem 2

We provide the proof for $\underline{E}^{{\text {EI}}}_{n}{\left( h\right) }$; the proof for $\overline{E}^{{\text {EI}}}_{n}{\left( h\right) }$ is completely analogous. It follows from Eqs. (16) and (18) and Theorem 1 that

$$\begin{aligned} \underline{E}^{{\text {EI}}}_{n}{\left( h\right) }=\underline{Q}_1\underline{Q}^{n-1}h. \end{aligned}$$

Therefore, the result follows—by induction—if, for any function $h\in \mathscr {L}(\mathscr {X})$ that satisfies Eq. (17), we can show (a) that $\underline{Q}h$ also satisfies Eq. (17) and (b) that, for all $y\in \mathscr {X}$, the minimum in

$$\begin{aligned} \underline{Q}h(y) =\min {{\left\{ \sum _{x\in \mathscr {X}}{q{\left( x\vert y,a,d\right) }}h(x):a \in \{\underline{a},\overline{a}\}, d \in \{\underline{d},\overline{d}\}\right\} }} \end{aligned}$$

(19)

is obtained for $a=\underline{a}$ and $d=\overline{d}$.

For all $y\in \{1,\dots ,L\}$, let $m_y:= h(y)-h(y-1)\ge 0$, where the inequality follows from Eq. (17).

We first prove (b). For $y=0$, Eqs. (1) and (19) imply that

$$\begin{aligned} \underline{Q}h(0)&=\min {{\left\{ (1-a)h(0)+ah(1):a\in \{\underline{a},\overline{a}\},d\in \{\underline{d},\overline{d}\}\right\} }}\nonumber \\&=\min {{\left\{ h(0)+am_1:a\in \{\underline{a},\overline{a}\},d\in \{\underline{d},\overline{d}\}\right\} }} =h(0)+\underline{a}m_1, \end{aligned}$$

(20)

where the last step holds because $m_1\ge 0$. Similarly, for $y \in \{1,\dots ,L-1\}$, Eqs. (2) and (19) imply that

$$\begin{aligned} \underline{Q}h(y)&=\min \big \{[d(1-a)]h(y-1)+[(1-d)(1-a)+da]h(y)\nonumber \\& +[(1-d)a]h(y+1) :a\in \{\underline{a},\overline{a}\}, d \in \{\underline{d},\overline{d}\}\big \}\nonumber \\&=\min {{\left\{ h(y)-d(1-a)m_y+(1-d)am_{y+1}:a\in \{\underline{a},\overline{a}\},d\in \{\underline{d},\overline{d}\}\right\} }}\nonumber \\&= h(y)-\overline{d}(1-\underline{a})m_y+(1-\overline{d})\underline{a}m_{y+1}, \end{aligned}$$

(21)

where the last step holds because $m_y\ge 0$ and $m_{y+1}\ge 0$.

Finally, for $y=L$, Eqs. (3) and (19) imply that

$$\begin{aligned} \underline{Q}h(L)&=\min {{\left\{ [d(1-a)]h(L-1)+[1-d(1-a)]h(L):a\in \{\underline{a},\overline{a}\},d\in \{\underline{d},\overline{d}\}\right\} }}\nonumber \\&=\min {{\left\{ h(L)-d(1-a)m_L:a\in \{\underline{a},\overline{a}\},d\in \{\underline{d},\overline{d}\}\right\} }}= h(L)-\overline{d}(1-\underline{a})m_L, \end{aligned}$$

(22)

where the last step holds because $m_L\ge 0$. This concludes the proof of (b).

We now prove (a): $\underline{Q}h(y+1)-\underline{Q}h(y)\ge 0$ for all $y\in \{0,\dots ,L-1\}$. For $y=0$, this holds because it follows from Eqs. (20) and (21) that

$$\begin{aligned} \underline{Q}h(1)-\underline{Q}h(0)&= \left( h(1)-\overline{d}(1-\underline{a})m_1+(1-\overline{d})\underline{a}m_{2} \right) - \left( h(0)+\underline{a}m_1 \right) \\&= m_1-\overline{d}(1-\underline{a})m_1+(1-\overline{d})\underline{a}m_{2} -\underline{a}m_1\\&\ge m_1-\overline{d}(1-\underline{a})m_1 -\underline{a}m_1 =(1-\underline{a})(1-\overline{d})m_1\ge 0. \end{aligned}$$

For $y\in \{1,\dots ,L-2\}$, this holds because it follows from Eq. (21) that

$$\begin{aligned}&\underline{Q}h(y+1)-\underline{Q}h(y)\\&\quad =\left( h(y+1)-\overline{d}(1-\underline{a})m_{y+1}+(1-\overline{d})\underline{a}m_{y+2} \right) \nonumber \\&\qquad - \left( h(y)-\overline{d}(1-\underline{a})m_y+(1-\overline{d})\underline{a}m_{y+1} \right) \\&\quad = m_{y+1}-\overline{d}(1-\underline{a})m_{y+1}+(1-\overline{d})\underline{a}m_{y+2} +\overline{d}(1-\underline{a})m_y-(1-\overline{d})\underline{a}m_{y+1}\\&\quad \ge m_{y+1}-\overline{d}(1-\underline{a})m_{y+1}-(1-\overline{d})\underline{a}m_{y+1}=(1-\underline{a})(1-\overline{d})m_{y+1}\ge 0. \end{aligned}$$

For $y=L-1$, this holds because it follows from Eqs. (21) and (22) that

$$\begin{aligned} \underline{Q}h(L)-\underline{Q}h(L-1)&= \left( h(L)-\overline{d}(1-\underline{a})m_L\right) \\&\quad -\, \left( h(L-1)-\overline{d}(1-\underline{a})m_{L-1}+(1-\overline{d})\underline{a}m_{L} \right) \\&= m_L-\overline{d}(1-\underline{a})m_L +\overline{d}(1-\underline{a})m_{L-1}-(1-\overline{d})\underline{a}m_{L}\\&\ge m_L-\overline{d}(1-\underline{a})m_L -(1-\overline{d})\underline{a}m_{L}\\&=\left( (1-\overline{d})(1-\underline{a}) +\overline{d}\underline{a}\right) m_L \ge 0. \end{aligned}$$

Proof of Theorem 3

For all $n\in \mathbb {N}$, it follows from subadditivity [27, Chap. 2.6.1(e)] that

$$\begin{aligned}&\underline{E}^{{\text {EI}}}_{1:n}{\left( \frac{1}{n}\sum _{i=1}^{n}{\mathbb {I}_{k}(X_{i})}\right) } \ge \frac{1}{n}\sum _{i=1}^{n}\underline{E}^{{\text {EI}}}_{1:n}{\left( {\mathbb {I}_{k}(X_{i})}\right) } =\frac{1}{n}\sum _{i=1}^{n}\underline{E}^{{\text {EI}}}_{i}{\left( {\mathbb {I}_{k}}\right) }, \end{aligned}$$

whence

$$\begin{aligned} \liminf _{n\rightarrow \infty }\underline{E}^{{\text {EI}}}_{1:n}{\left( \frac{1}{n}\sum _{i=1}^{n}{\mathbb {I}_{k}(X_{i})}\right) } \ge \liminf _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\underline{E}^{{\text {EI}}}_{i}{\left( {\mathbb {I}_{k}}\right) } \ge \liminf _{n\rightarrow \infty }\underline{E}^{{\text {EI}}}_{n}{\left( {\mathbb {I}_{k}}\right) }, \end{aligned}$$

where the last inequality follows from the definition of the limit inferior. The proof for the upper expectations is completely analogous; the inequalities are reversed and subadditivity is replaced by superadditivity.