# Breaktime fraction and tasks between breaks have reciprocal mean

Is your queue free one tenth of the time? Expect to work ten tasks between breaks.

If I have 1 worker dedicated to a stream of independently distributed tasks each with fixed unit time, and this stream averages $$\mu \in (0,1)$$ new tasks per unit time, then the distribution for freeing the queue after exactly $$x$$ units of work is

$\text{Borel}_\mu(x) = \frac{ e^{- x \mu} (x \mu)^{x-1}}{x!}$

Numerical summation of $$\sum_{x=1}^{\infty} \text{Borel}_\mu(x) = 1$$ for any $$\mu \in ( 0, 1)$$, which demonstrates a valid distribution. Performing a similar numerical summation in this same range to find the mean also converges:

$E(X) = \sum_{x=1}^{\infty} x \text{Borel}_\mu(x) \approx \frac{1}{1-\mu}$

So, quite simply, by this distribution, if my minion spends 1/5 of his time with a free queue, then I should expect him to work 5 tasks (on average) before the queue is again free. Also, by the exponential distribution, having 4/5 of his time busy, I should expect a free queue to last about 5/4 task units.

Fixed task lengths may make this result seem trivial, so I could assign an exponential random variable to the time on each task, with unit mean. Keeping the mean arrival rate of $$\frac{1}{\mu}$$ will leave $$\mu \in ( 0, 1)$$ time free on average just like $$\text{Borel}_\mu(x)$$. Specifically, I have the following:

Time to next decrement from queue (non-empty) has a distribution

$~ Exp(X,1) = e^{-x}$

Time to next increment to queue has a distribution

$~ Exp(x, \mu) = \mu e^{- x \mu}$

Because decrement and increment is memory-less, I can quickly explore the function of a working queue. Before I begin, at $$x = 0$$, I have an empty queue (0-job state). There is no chance of decrement, so work start is predicted by $$~ Exp(x,\mu)$$. At that moment the queue will be active on the first job; I'll call this the 1-job state. The 1-job state lasts until the job finishes, or a second job is queued. If a second job is queued, I have the 2-job state, which this will last until either a job finishes (returning to the 1-job state), or else a third job will be queued (3-job state). And so on.

Will this FRIFRO (First random-in, first random-out) look like the continuous variant of the next $$\text{Borel}_\mu(x)$$ distribution? I'm not expecting any surprises but I'm still working on it so in time I'll get around to it.