How can average queuing delay exceed 1 if the $La/R \leq 1$?

Consider a router with a packet arrival rate of $a$ where each packet has $L$ bits, and a transmission rate (in bits/s) of $R$. My textbook describes the traffic intensity as $\frac{La}{R}$ and states that as $\frac{La}{R} \to 1$, the average queueing delay approaches $\infty$. I'm just struggling to understand why this is the case. Is it because this increases the chance that $La$ exceeds $R$ during periods of time, and that long queues can accumulate?

It depends on the r.v's of the arrival the service process.

For example, suppose the arrival and service process are all deterministic process. For example, a packet of one bit ($L=1$) arrives in every second ($a=1$), and the router transmits a bit in every second ($R=1$). Of course that in such case the average queuing delay equals to 1.

However, if for example in a system where the packet arrivals are determined by a Poisson process with parameter of $\lambda=a$ seconds and job service times have an exponential distribution (M/M/1) with average of $1/\mu=\frac{L}{R}$ seconds (which describes a system with packet length $L$ and average transmission rate of $R$), i.e., $\mu=\frac{R}{L}$ the service time equals to $\frac{1}{\mu-\lambda}- \frac{1}{\mu}$. As $\frac{La}{R}$ approaches $1$, then $\mu-\lambda$ approaches $\infty$ the service time approaches infinity.

Now for the strange behavior of $\frac{La}{R}=1$ for M/M/1 queue: When $\frac{La}{R}=1$ that means that the probability a packet enters the router equals the probability that a packet exits the router. That means, the probability that the router contains $i$ packets equals the probability that the router contains $j$ packets, for every $i$ and $j$. Thus, the number of packets in router approaches $\infty$, as well as the queuing delay.

StackExchange Computer Science Q#96940, answer score: 3