Timeline for How to prove positive recurrence of a queue server system that stops for maintenance?
Current License: CC BY-SA 3.0
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| Mar 21, 2017 at 11:24 | comment | added | Serguei Popov | "if we consider the states as a tuple of queue size and of a binary variable that indicates if server is maintained" - here we can kind of "join" all the maintenance states into one (as indicated in the response). But you can alsouse the Lyapunov function $f(1,n)=n$, $f(0,n)=n+c$ for suitable $c>0$ ($1$ means server works, $0$ means it's under maintenance). | |
| Mar 21, 2017 at 4:37 | comment | added | kon psych | Yes $V$ is $f$ in the book you sent (I am used to see Lyapunov function denoted by $V$). If we consider the states as a tuple of queue size and of a binary variable that indicates if server is maintained, then the set of states for which $f$ is 0 is infinite. | |
| Mar 21, 2017 at 1:00 | comment | added | Serguei Popov | For continuous time it's difficult to speak about "several steps". But if your transition rates are uniformly bounded, then you can switch to discrete time (the embedded chain)?.. | |
| Mar 21, 2017 at 0:58 | comment | added | Serguei Popov | If $V$ is $f$, then it is finite (at least in one-dimensional case). | |
| Mar 21, 2017 at 0:52 | comment | added | Serguei Popov | Who is $V(\cdot)$? | |
| Mar 21, 2017 at 0:06 | comment | added | kon psych | The last theorem is what I was looking for but if I am not mistaken, is for discrete time only? Also your first argument is not valid because the set of states $s$ for which $V(s)$ < constant should be finite. Example 6.16 in the book of my question highlights why this is necessary. I can see nonetheless that including maintenance state in function would work. | |
| Mar 20, 2017 at 13:29 | history | edited | Serguei Popov | CC BY-SA 3.0 |
added 75 characters in body
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| Mar 20, 2017 at 13:24 | history | answered | Serguei Popov | CC BY-SA 3.0 |