This is probably a known result, but I couldn't find any resource pointing directly to the issue I'm trying to solve.
Suppose you start a logistic mission that needs that during its time $T_m$ a given object is always working. At the beginning of your mission, you have an instance of this object working plus $k$ spare items that could replace the object if a random failure (that happens with a given rate $\lambda$) occurs. Furthermore, after the occurrence of a failure, the broken instance is repaired in a fixed time $T_r$.
A mission is successful if there is at least one item working for its entire time.
What I want to calculate is the probability that the mission is successful. It is obvious that this problem is a queue problem. In fact:
- The failures are the calls (or the customers) arriving at the desk;
- the spare items are the number of operators;
- $T_r$ is the duration of the call (which is fixed, in this case).
However, I'm not interested in the probability at the equilibrium that I'm running out of items nor in the average number of items under repair, which are the quantities that are most often found in queue theory. I'm interested in the following:
What is the probability that there is no queue in a M/D/k system before a fixed $T_m$ time?
We can say that the system is initially empty, since the $k$ spare items are all brand-new and ready to replace a broken item.
This is where I'm in at the moment:
- If $T_m<T_r$ than the mission is successful if $k$ or less failures occur, so we have:
$$P(\text{success}) = \sum_{i=0}^k e^{-\lambda T_m} \frac{(\lambda T_m)^i}{i!}$$
- For $k=1$, after doing some calculations, I found that:
$$P(\text{success}) = e^{-\lambda T} \sum_{j=0}^{n+1} \frac{\lambda^j}{j!} (T_m-(j-1)T_r)^j$$
where $n=\lfloor\frac{T_m}{T_r}\rfloor$.
The last result matches with a simulation that covers the general case (with arbitrary $k$) that I wrote in R (if requested, I could provide it). I'm not arriving at the general result.