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I have the following question: Suppose I have a product with $U$ users and $S$ seats (licences). Where $S < U$. A user may only use the given product if there are seats available, and during the using of the product the user will occupy one seat.

I have the following statistics on usage: The average usage time per session (per a user occupying one seat continuously), the average number of sessions per day.

I'm interested in knowing what are the odds of a user being denied a seat after requesting one. Also, I'd like to know the 'utilization' of the licences. In effect, what proportion of time are the licences unoccupied.

Thank you very much.

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    $\begingroup$ A couple of questions: (1) Are we assuming all seats are open at the beginning? (2) Once a user has occupied a seat, do they return to the pool of users seeking to occupy a seat? (3) Is everyone continuously requesting a license? $\endgroup$
    – Todd Burus
    Commented Feb 6, 2020 at 13:43
  • $\begingroup$ @ToddBurus Users only seek for seats several times a day, when they found a seat, they occupy it for some time then free it. During the day they may have to seek a seat again. At the beginning of the day all seats are vacant. Also, this is a practical question I don't need a 100% accurate result, a good estimate is enough. $\endgroup$
    – iliar
    Commented Feb 6, 2020 at 15:07
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    $\begingroup$ Okay, thanks. One more--are the seats vacant at the start of each day, or just at the start of the overall process? If it's at the start of each day, what defines a "day"? (I guess that's technically two). $\endgroup$
    – Todd Burus
    Commented Feb 6, 2020 at 15:10
  • $\begingroup$ @ToddBurus I'd say the situation is like this: At a certain time all users begin 'using'. They continue using for 9 hours which I call a 'day'. During these 9 hours they occasionally occupy a 'seat' (license). I have the average number of times this happens per day (number of sessions) and also the average time it takes them to stop occupying the seat (the time per session). Also, I think the problem is simpler if we assume a 'steady state', so it's also a good enough solution. $\endgroup$
    – iliar
    Commented Feb 6, 2020 at 18:31

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