A Discrete-Time Priority Queue with Train Arrivals

Abstract

We analyze a discrete-time priority queue with train arrivals. Messages of a variable number of fixed-length packets belonging to two classes arrive to the queue at the rate of one packet per slot. We assume geometrically distributed message lengths. Packets of the first class have transmission priority over the packets of the other class. By using probability generating functions, some performance measures such as the moments of the packet delay are calculated. The impact of the priority scheduling discipline and the correlation in the arrival process is shown by some numerical examples.

Transform-Domain Analysis of Packet Delay in Network Nodes with QoS-Aware Scheduling

“…Clearly, no significant differentiation in the per-class delay distribution is expected under any of these disciplines. On the other hand, the most extreme way of service differentiation is AP (Absolute Priority) or 'HOL priority' (Head of Line), either preemptive or non-preemptive, see [45,47,24,9,48] as well as the contribution of Walraevens et al in this volume. Under AP, the next scheduled packet is the one that (1) belongs to the set of queued packets with highest priority and (2) has the longest waiting time of the packets in the set.…”

“…As was the case with (48), this relation can be solved iteratively. However, still assuming that we know the sequence Ω n (α), it is possible to provide a direct solution of the expected values E[m [n] 1 ] from (59).…”

“…In order to do so, we assume that the quantities Ω n (α) are available. As we discussed, they can either be obtained analytically by iterating (48) and taking the limit z → α in each step, or they follow from the numerical inversion method discussed in the previous paragraph. Differentiating (48) to z and taking the limit z → 1 on both sides, we find after some straightforward manipulations and using (46):…”

“…Through (42) and (43), this directly provides the pgf of d 1 . Indeed, (48) can be solved in an iterative way since Ω N (z) appears only on the left-hand side while the other side only depends on Ω 1 (z) to Ω N −1 (z). However, the problem is that we also have to determine the constant Ω n (α) in step n (n = 1, .…”

“…After careful inspection of the expression (57), one can prove that its dominant pole can only originate from the factor (z − A T (z)) −1 appearing in the second term, but also present in each Ω n (z) through (46) and (48). Note that the multiplicity of this factor is equal to 1 in all of these terms.…”