Poisson arrivals followed by locking

following is my problem:

Pulses arrive at a processor according to a Poisson process of rate λ.
Suppose each arriving pulse that is processed by the processor locks the processor for a fixed time T, so that pulses arriving during the locked period are not processed, and pulses arriving after the locked period are possible to be processed subject to the same locking rule by the following processed pulses.

Let $X (t)$ be the number of all pulses having been (being) processed up to time $t$. Then try to compute:

  1. $P (X (t) ≥ n+1)$ for $t ≥ nT$. Can we also compute the distribution of $X(t)$?
  2. the distribution of the
    inter-arrival time between every two
    processed pulses.
  3. the probability that the processor
    is free at an arbitrarily given time $t > 0$.

The farthest I can imagine is that $X$ seems to be a renewal process (although not sure how to show that), and I don’t know how to go further.

Also any reference that might be helpful?

Thanks in advance!

Update:

As Stijn commented, this a M/D/1 queue, which I have not learned yet and can’t find answers to my questions in the file he linked. Can someone give more clues?

Solutions Collecting From Web of "Poisson arrivals followed by locking"

Let $u_n(t) = P(X(t) \ge n)$, where the process starts at $t=0$. Condition on the time $Y$ of the first arriving pulse, which will be processed. Since there is a delay of time $T$ before another pulse can be processed,
$P(X(t) \ge n | Y = y) = P(X(t – y – T) \ge n-1)$ (and 0 if $t – y – T \le (n-1) T$).
Thus $u_n(t) = \int_0^{t-nT} \lambda \exp(-\lambda y) u_{n-1}(t-y-T)\, dy$.
I get $u_n(t) = 1 – q_n(t-nT) \exp(-\lambda (t-nT))$ where
$q_n(s) = \sum_{j=0}^{n-1} (\lambda s)^j/j!$.