Sample-path analysis of general arrival queueing systems with constant amount of work for all customers

Abstract

We consider a discrete-time queueing system where the arrival process is general and each arriving customer brings in a constant amount of work which is processed at a deterministic rate. We carry out a sample-path analysis to derive an exact relation between the set of system size values and the set of waiting time values over a busy period of a given sample path. This sample-path relation is then applied to a discrete-time $G/D/c$ queue with constant service times of one slot, yielding a sample-path version of the steady-state distributional relation between system size and waiting time as derived earlier in the literature. The sample-path analysis of the discrete-time system is further extended to the continuous-time counterpart, resulting in a similar sample-path relation in continuous time.

Appendix

1.1 Lemmas and proofs

In this appendix, we state and prove a few lemmas that are needed for the proof of Theorem 1 in Sect. 2.

Lemma 1

For the discrete-time $(c,k)$-system, let $Q=\{(t_1,d_1),\dots ,(t_n,d_n)\}$ be a $B(n)$ set, and let $P=\{(t_1,d_1),\dots ,(t_n,d_n),(t_{n+1},d_{n+1})\}$ with $t_{n+1}=d_n$ and $d_{n+1}=t_1+\lceil k(n+1)/c\rceil -1$ $($which is a $B(n+1)$ set $)$. If $S^{[c]}_Q=W^{[k]}_Q$, then $S^{[c]}_P=W^{[k]}_P$.

Proof

Consider the setting where the $(c,k)$-system is empty right before $t_1$, and $n+1$ customers $\mathcal C _1,\dots ,\mathcal C _{n+1}$ enter the system at $t_1,\dots ,t_{n+1}$, respectively. Since $P$ is a $B(n+1)$ set, the $\mathcal C _i$ depart at $d_i=t_1+\lceil ki/c\rceil -1$, $i=1,\dots ,n+1$, and exactly $c$ units of work is performed at each of the slots $t_1,\dots ,d_{n+1}-1$. Let $\ell =N_Q(d_n)=|\{i\le n:d_i=d_n\}|$, the number of those in $\mathcal C _1,\dots ,\mathcal C _n$ who are in system at $d_n$ (and also depart at the end of slot $d_n$), and $v=$ the number of units of work remaining for these $\ell$ customers at (the beginning of) slot $d_n$. We have (cf. (2.3)),

$$\begin{aligned} (\ell -1)k+1\le v\le \min \{\ell k,c\} \text{ and } \ell =\lceil v/k\rceil . \end{aligned}$$

(6.1)

Also $N_P(t)=N_Q(t)$ for $t_1\le t<d_n$, since $t_{n+1}=d_n$. We consider the three cases ($c\ge 1,k=1$), ($c=1,k\ge 1$) and ($c\ge 2, k\ge 2$) separately.

Case (i): $c\ge 1,k=1$. By (6.1), $\ell =\lceil v/k\rceil =v$. Write $n=qc+r$ with $q=\lceil n/c\rceil -1\ge 0$ and $0<r\le c$. With $k=1$, the total number of work units for customers $\mathcal C _1,\dots ,\mathcal C _n$ is $n$. Since exactly $c$ units of work is performed at $t\in \{t_1,\dots ,d_n-1\}$ and since $\mathcal C _n$ departs at $d_n$, we must have $r=v=\ell$. Two subcases are considered below.

Subcase (i.1): $0<r<c$. In this case, the total number of work units (including the work of $\mathcal C _{n+1}$) at $d_n$ equals $r+1\le c$, so that $d_{n+1}=d_n$. It follows that

$$\begin{aligned} S_Q&= \{N_Q(t):t_1\le t< d_n\}\uplus \{r\},\\ S_P&= \{N_Q(t):t_1\le t<d_n\}\uplus \{r+1\},\\ W_P&= W_Q\uplus \{1\}. \end{aligned}$$

Since $\{r\}^{[c]}=\{1_{(r)}\}$ and $\{r+1\}^{[c]}=\{1_{(r+1)}\}$ and $S^{[c]}_Q=W_Q^{[k]}=W_Q$,

$$\begin{aligned} S^{[c]}_P=S^{[c]}_Q\uplus \{1\}=W_Q\uplus \{1\}=W_P. \end{aligned}$$

Subcase (i.2): $r=c$. The total number of work units (including the work of $\mathcal C _{n+1}$) at $d_n$ equals $c+1$, so that $d_{n+1}=d_n+1$. It follows that

$$\begin{aligned} S_Q&= \{N_Q(t):t_1\le t< d_n\}\uplus \{c\},\\ S_P&= \{N_Q(t):t_1\le t<d_n\}\uplus \{1,c+1\},\\ W_P&= W_Q\uplus \{2\}. \end{aligned}$$

Since $\{1,c+1\}^{[c]}=\{1_{(c)},2\}$ and $\{c\}^{[c]}=\{1_{(c)}\}$ and $S^{[c]}_Q=W_Q$,

$$\begin{aligned} S^{[c]}_P=S^{[c]}_Q\uplus \{2\}=W_Q\uplus \{2\}=W_P. \end{aligned}$$

This completes the proof for case (i).

Case (ii): $c=1,k\ge 1$. Since $c=1$, all of $\mathcal C _1,\dots ,\mathcal C _{n-1}$ must have departed before $d_n$, and $\mathcal C _n$ has just one unit of work remaining at $d_n$. So $\ell =v=1$. When $\mathcal C _{n+1}$ enters the system at $t_{n+1}=d_n$, it takes $k$ additional slots to complete the work of $\mathcal C _{n+1}$. So $d_{n+1}=d_n+k$. It follows that

$$\begin{aligned} S_Q&= \{N_Q(t):t_1\le t< d_n\}\uplus \{1\},\\ S_P&= \{N_Q(t):t_1\le t<d_n\}\uplus \{1_{(k)},2\},\\ W_P&= W_Q\uplus \{k+1\}. \end{aligned}$$

Since $\{k+1\}^{[k]}=\{1_{(k-1)},2\}$ and $S_Q=S_Q^{[c]}=W^{[k]}_Q$,

$$\begin{aligned} W^{[k]}_P&= W^{[k]}_Q\uplus \{1_{(k-1)},2\}\\&= S_Q\uplus \{1_{(k-1)},2\}\\&= \{N_Q(t):t_1\le t<d_n\}\uplus \{1\}\uplus \{1_{(k-1)},2\} \\&= S_P, \end{aligned}$$

completing the proof for case (ii).

Case (iii): $c\ge 2,k\ge 2$. We further consider the following three subcases.

Subcase (iii.1): $d_{n+1}=d_n$. Necessarily, $c\ge v+k$ (the amount of work at $d_n$, including the work of $\mathcal C _{n+1}$), which together with (6.1) implies $c>\ell k\ge 2\ell$. Also we have $W_P=W_Q\uplus \{1\}$, $S_Q=\{N_Q(t):t_1\le t<d_n\}\uplus \{\ell \}$, and $S_P=\{N_Q(t):t_1\le t<d_n\}\uplus \{\ell +1\}$. Since $\ell +1\le 2\ell <c$, we have $\{\ell \}^{[c]}=\{1_{(\ell )}\}$, $\{\ell +1\}^{[c]}=\{1_{(\ell +1)}\}$, so that

$$\begin{aligned} S^{[c]}_P=S^{[c]}_Q\uplus \{1\}=W^{[k]}_Q\uplus \{1\}=W^{[k]}_P. \end{aligned}$$

Subcase (iii.2): $d_{n+1}=d_n+1$. Necessarily, $v+k\le 2c$, which together with (6.1) implies $2c\ge v+k>\ell k\ge 2\ell$, i.e. $c>\ell$. Then $W_P=W_Q\uplus \{2\}$ (implying $W^{[k]}_P=W^{[k]}_Q\uplus \{1_{(2)}\}$),

$$\begin{aligned} S_Q=\{N_Q(t):t_1\le t<d_n\}\uplus \{\ell \},S_P=\{N_Q(t):t_1\le t<d_n\}\uplus \{1,\ell +1\}, \end{aligned}$$

from which and $c>\ell$ it follows that

$$\begin{aligned} S^{[c]}_P=S^{[c]}_Q\uplus \{1_{(2)}\}=W^{[k]}_Q \uplus \{1_{(2)}\}=W^{[k]}_P. \end{aligned}$$

Subcase (iii.3): $d_{n+1}\ge d_n+2$. Necessarily $\ell =1$. Since $\mathcal C _{n+1}$ is the only customer in system after $d_n$ and since exactly $c$ units of work is performed at each of the slots $d_n+1,\dots ,d_{n+1}-1$, we have $k\ge c(d_{n+1}-d_n-1)+1\ge d_{n+1}-d_n+1$. So $W_P=W_Q\uplus \{d_{n+1}-d_n+1\}$ (implying $W^{[k]}_P=W^{[k]}_Q\uplus \{1_{(d_{n+1}-d_n+1)}\}$),

$$\begin{aligned} S_Q=\{N_Q(t):t_1\le t<d_n\}\uplus \{\ell \},S_P=\{N_Q(t):t_1\le t<d_n\}\uplus \{\ell +1,1_{(d_{n+1}-d_n)}\}. \end{aligned}$$

It follows that

$$\begin{aligned} S^{[c]}_P=S^{[c]}_Q\uplus \{1_{(d_{n+1}-d_n+1)}\} =W^{[k]}_Q\uplus \{1_{(d_{n+1}-d_n+1)}\}=W^{[k]}_P. \end{aligned}$$

This completes the proof for case (iii). The proof of Lemma 1 is complete. $\square$

Lemma 2

For the discrete-time $(c,k)$-system, let $Q=\{(t_1,d_1),\dots ,(t_n,d_n)\}$ be a $B(n)$ set with $t_n<d_n$. Let $j$ satisfy $t_n\le j<j+1\le d_n$, and let $P=\{(t_1,d_1),\dots ,(t_n,d_n),(j,d_{n+1})\}$ and $P^{\prime }=\{(t_1,d_1),\dots ,(t_n,d_n),(j+1,d_{n+1})\}$ $($with $d_{n+1}=t_1+\lceil (n+1)k/c\rceil -1)$, which are both $B(n+1)$ sets. Then

$$\begin{aligned} S^{[c]}_P=W^{[k]}_P\;if\,and\,only\,if \; S^{[c]}_{P^{\prime }}=W^{[k]}_{P^{\prime }}. \end{aligned}$$

Proof

Consider the setting where the $(c,k)$-system is empty right before $t_1$, and $n+1$ customers $\mathcal C _1,\dots ,\mathcal C _{n+1}$ enter the system at $t_1,\dots ,t_n,t_{n+1}$, respectively, where $t_{n+1}=j$ or $j+1$. Since $P$ and $P^{\prime }$ are both $B(n+1)$ sets, the $\mathcal C _i$ depart at $d_i=t_i+\lceil ki/c\rceil -1$, $i=1,\dots ,n+1$, and exactly $c$ units of work is done at each of the slots $t_1,\dots ,d_{n+1}-1$. Let $\ell =N_Q(j)$, the number of those in $\mathcal C _1,\dots ,\mathcal C _n$ who are in system at $j$, and $v=$ the number of units of work remaining at $j$ for the $\ell$ customers $\mathcal C _{n-\ell +1},\dots ,\mathcal C _n$. Since none of $\mathcal C _{n-\ell +2},\dots ,\mathcal C _n$ begins service before time $j$, we have $(\ell -1)k+1\le v\le \ell k$, implying $\ell =\lceil v/k\rceil$ (cf. (2.3)). On the other hand, with $t_{n+1}=j$, $v+k$ units of work (including $\mathcal C _{n+1}$’s work) needs to be completed at $d_{n+1}$, so $(d_{n+1}-j)c+1\le v+k\le (d_{n+1}-j+1)c$, implying $d_{n+1}-j+1=\lceil \frac{v+k}{c}\rceil$. We have shown

$$\begin{aligned} \ell +1=\left\lceil \frac{v}{k}\right\rceil +1 =\left\lceil \frac{v+k}{k}\right\rceil \text{ and } d_{n+1}-j+1=\left\lceil \frac{v+k}{c}\right\rceil . \end{aligned}$$

(6.2)

Since $\mathcal C _{n+1}$ departs at $d_{n+1}$ for both cases $t_{n+1}=j$ and $t_{n+1}=j+1$, the waiting time of $\mathcal C _{n+1}$ is $d_{n+1}-j+1$ for the former case and $d_{n+1}-j$ for the latter. As the waiting times of $\mathcal C _1,\dots ,\mathcal C _n$ do not depend on $t_{n+1}$, we have

$$\begin{aligned} W_P\uplus \{d_{n+1}-j\}=W_{P^{\prime }}\uplus \{d_{n+1}-j+1\}. \end{aligned}$$

(6.3)

Also, $\mathcal C _{n+1}$ begins service at the same time for the two cases $t_{n+1}=j$ and $t_{n+1}=j+1$. In particular, for $t_{n+1}=j$, $\mathcal C _{n+1}$ cannot begin service at $j (<d_n)$. It follows that

$$\begin{aligned} N_P(t)=N_{P^{\prime }}(t)\quad \text{ for }\; t\ne j, \; N_P(j)=\ell +1, \; N_{P^{\prime }}(j)=\ell . \end{aligned}$$

So,

$$\begin{aligned} S_P\uplus \{\ell \}=S_{P^{\prime }}\uplus \{\ell +1\}. \end{aligned}$$

(6.4)

By (6.2)–(6.4),

$$\begin{aligned} S_P\uplus \left\{ \left\lceil \frac{v+k}{k}\right\rceil -1\right\}&= S_{P^{\prime }}\uplus \left\{ \left\lceil \frac{v+k}{k}\right\rceil \right\} ,\end{aligned}$$

(6.5)

$$\begin{aligned} W_P\uplus \left\{ \left\lceil \frac{v+k}{c}\right\rceil -1\right\}&= W_{P^{\prime }}\uplus \left\{ \left\lceil \frac{v+k}{c}\right\rceil \right\} . \end{aligned}$$

(6.6)

By Lemma 3 below, $\lceil \lceil (v+k)/k\rceil /c\rceil =\lceil \lceil (v+k)/c\rceil /k\rceil$, which is denoted by $\rho$. We consider $\rho =1$ and $\rho >1$ separately.

Case (i): $\rho =1$. We have $\lceil (v+k)/k\rceil \le c$ and $\lceil (v+k)/c\rceil \le k$, so

$$\begin{aligned} \left\{ \left\lceil \frac{v+k}{k}\right\rceil \right\} ^{[c]} =\left\{ \left\lceil \frac{v+k}{k}\right\rceil -1\right\} ^{[c]} \uplus \{1\}, \end{aligned}$$

(6.7)

$$\begin{aligned} \left\{ \left\lceil \frac{v+k}{c}\right\rceil \right\} ^{[k]} = \left\{ \left\lceil \frac{v+k}{c}\right\rceil -1\right\} ^{[k]}\uplus \{1\}, \end{aligned}$$

(6.8)

By (6.5)–(6.8),

$$\begin{aligned} S^{[c]}_P=S^{[c]}_{P^{\prime }}\uplus \{1\} \text{ and } W^{[k]}_P=W^{[k]}_{P^{\prime }}\uplus \{1\}. \end{aligned}$$

It follows that $S^{[c]}_P=W^{[k]}_P$ if and only if $S^{[c]}_{P^{\prime }}=W^{[k]}_{P^{\prime }}$.

Case (ii): $\rho >1$. It is readily seen that $\{\lceil \frac{v+k}{k}\rceil \}^{[c]}$ has one more copy of $\rho$ and one fewer copy of $\rho -1$ than $\{\lceil \frac{v+k}{k}\rceil -1\}^{[c]}$, and $\{\lceil \frac{v+k}{c}\rceil \}^{[k]}$ has one more copy of $\rho$ and one fewer copy of $\rho -1$ than $\{\lceil \frac{v+k}{c}\rceil -1\}^{[k]}$. It follows from (6.5) and (6.6) that

$$\begin{aligned} S^{[c]}_P\uplus \{\rho -1\}=S^{[c]}_{P^{\prime }}\uplus \{\rho \} \text{ and } W^{[k]}_P\uplus \{\rho -1\}=W^{[k]}_{P^{\prime }}\uplus \{\rho \}, \end{aligned}$$

which implies that $S^{[c]}_P=W^{[k]}_P$ if and only if $S^{[c]}_{P^{\prime }}=W^{[k]}_{P^{\prime }}$. This completes the proof. $\square$

Lemma 3

For positive integers $i,j$ and $\ell$, we have

$$\begin{aligned} \lceil \lceil \ell /i\rceil /j\rceil =\lceil \lceil \ell /j\rceil /i\rceil . \end{aligned}$$

Proof

Let $\rho =\lceil \ell /i\rceil$. Then $\ell \le i\rho$, and $\lceil \ell /j\rceil \le \lceil i\rho /j\rceil \le i\lceil \rho /j\rceil$, implying $\lceil \lceil \ell /j\rceil /i\rceil \le \lceil \rho /j\rceil =\lceil \lceil \ell /i\rceil /j\rceil$. By symmetry, $\lceil \lceil \ell /i\rceil /j\rceil \le \lceil \lceil \ell /j\rceil /i\rceil$. This completes the proof. $\square$