I am still stuck with a problem with the M/G/1 queue: not quite the same as my original problem (discussed here) as I understand that now – but the next stage really – involving some manipulation of Laplace transforms.

I won’t post all the details here, because you can read them here instead and, if this sort of thing matters to you (and why wouldn’t it?) pick up the bounty I am offering on the maths Stack Exchange.

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## Puzzle about an M/G/1 queue

I am deeply puzzled by a question about the behaviour of an M/G/1 queue – i.e., a queue with a Markovian distribution of arrival times, a General distribution of service times and 1 server. I have asked about this on the Math Stackexchange (and there’s now a bounty on the question if you’d like to answer it there – but as I am getting nowhere with it, I thought I’d ask it here too.

(This is related to getting a more rigorous presentation on thrashing into my PhD thesis.)

Considering an M/G/1 queue with Poisson arrivals of rate λ – this comes from Cox and Miller’s (1965) “The Theory of Stochastic Processes” (pp 240 – 241) and also Cox and Isham’s 1986 paper “The Virtual Waiting-Time and Related Processes“.

My question is what is the difference between (using the authors’ notation) $p_0$ and $p(0,t)$? The context is explained below…

In the 1965 book (the 1986 paper presents the differentials of the same equations), $X(t)$ is the “virtual waiting time” of a process and the book writes of “a discrete probability $p_0(t)$ that $X(t)=0$, i.e., that the system is empty, and a density $p(x,t)$ for $X(t)>0$“.

The system consumes virtual waiting time in unit time, i.e., if $X(t)\leq\Delta t$ and there are no arrivals in time $\Delta t$ then $X(t + \Delta t) = 0$.

The distribution function of $X(t)$ is then given by: $F(x,t)=p_0(t)+\int_{0}^{\infty}p(z,t)dz$

They then state: $p(x, t+ \Delta t) = p(x + \Delta t, t)(1 - \lambda \Delta t) +p_0(t)b(x)\lambda\Delta t + \int_{0}^{x}p(x - y, t)b(y)dy\lambda\Delta t + o(\Delta t)$

I get all this – the first term on the RHS is a run-down of $X(t)>0$ with no arrivals, the second is adding $b(x)$ of service time when the system is empty at $t$ and the third, convolution-like, term is adding $b(y)$ of service time from an arrival when it’s not empty at $t$. (The fourth accounts for their being more than one arrival in $\Delta t$ but it tends to zero much faster than $\Delta t$ so drops out as $\Delta t$ approaches the limit.)

And … and this is where I have the problems … $p_0(t+\Delta t)=p_0(t)(1-\lambda\Delta t) +p(0,t)\Delta t(1 - \lambda\Delta t) + o(\Delta t)$

The first term of the RHS seems clear – the probability that the system is empty at $t$ multiplied by the probability there will be no arrivals in $\Delta t$, but the second is not clear to me at all.

I assume this term accounts for the probability of the system “emptying” during $\Delta t$ but I don’t see how that works, is anyone able to explain?

In other words, how does $p(0,t)\Delta t(1 - \lambda\Delta t)$ represent this draining? Presumably $(1 - \lambda\Delta t)$ again represents the possibility of zero arrivals in $\Delta t$, so how does $p(0, t)\Delta t$ represent the $X(t) \leq \Delta t$ situation?

If we take the equilibrium situation where $p_0(t) = p_0$ and $p(x, t) = p(x)$ then, if we differentiate and as $p^{\prime}_0 = 0$, we get $p_0 = \lambda p(0)$ – so, again, what does $p(0)$ represent?

## In praise of StackExchange

I have been ill – flu I think – for the last few days and so “blogging has been light”. But I have discovered StackExchange in the meantime.

Of course, I was always aware that StackExchange was there, but I suppose I thought of it as a parallel to “Experts-Exchange” – the awful, pay for a solution, site. But actually, once I overcame my loathing for having to login, I discovered it was a series of communities based on trust and free exchange of information, with acceptable behaviour policed through user activism and a really good “karmic” points system.

I have even been able to answer a couple of questions!

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