# How to find the average downtime of a queuing system (and something more)?

Good afternoon, colleagues. How to get a couple of formulas from the questions. Here we are given a queuing system with initial parameters: $$\begin{matrix}\text{intake intensity} & \lambda\\ \text{service channels} & m\\ \text{service intensity} & \mu\\ \text{maximum queue size} &n\end{matrix}$$

And we know how to get these parameters:

a) Probability of denial of service $$p_{denial} = p_{n+m}$$ b) Relative intensity of service $$q=1-p_{n+m}$$ c) Absolute intensity of service $$A = q\cdot\lambda$$ d) Average queue length $$L_{queue}=\sum^n_{i=1}ip_{m+i}$$ e) Average time in queue $$T_{queue}=\sum^{n-1}_{i=0}\frac{i+1}{m\mu}p_{m+i}$$ f) Average number of busy channels $$N_{channels}=\sum^{m}_{i=1}ip_i+\sum^{m+n}_{i=m+1}mp_i$$

But with these, I did not find the necessary information in the relevant literature. Maybe they are derived from the above, or do they need additional calculations?

g) Probability that an incoming request will not wait at queue

h) Average downtime of the queuing system

i) Average time when there is no queue in the system

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• why you downvoted? the question seems to be relevant and non-trivial – RoyalGoose yesterday
• Presumably everything is Markov and your $p_i$ denote the stationary distribution? For (i), do you just mean the stationary probability that the system is in state 0? i.e. $p_0$? What is it that you're asking for here? (g) should be the probability that the system contains fewer than $m$ tasks, i.e. $p_0+p_1+\dots+p_{m-1}$. As for (h), what do you mean by "downtime"? Just the same as (i) again? – James Martin yesterday
• @JamesMartin Yes, a steady state is implied. g) we know the probability of a new claim arriving, and the probability of service, as I understand it, we need to find the probability that the claim will get into exactly one of $m$ states (not including $n$ - queue) h) downtime means the probability that no requests arrive at all, but we need to find exactly the average time i) similar to g. but mean time – RoyalGoose 23 hours ago
• Sorry, I do not understand any of those explanations. What do you mean by "probability that no requests arrive at all"? What is the distinction between "average time" and "mean time"? I still don't see why (g) wouldn't be $p_0+p_1+\dots+p_{m-1}$. An incoming claim starts service immediately if and only if there are fewer than $m$ claims already present in the system? – James Martin 23 hours ago
• @JamesMartin You may be right about g), and I wanted to confirm this. As for downtime, I'm not sure myself, perhaps it means the likelihood of the absence of claims in terms of time – RoyalGoose 23 hours ago