M/M/S Queue, probability interpretation
$begingroup$
In an M/M/s queue, what does this expression mean? :
$sum_{n=0}^{s-1}{(s-n)P_n}$
Furthermore, is it possible that the following equation holds? :
$sum_{n=0}^{s-1}{(s-n)P_n} = (1-rho)s$
If so, how could I demonstrate it?
EDIT: An M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers.
Thanks,
Louis
probability queueing-theory
asked Nov 28 '18 at 19:25
Louis-Philippe NoëlLouis-Philippe Noël
123
$endgroup$
add a comment |
$begingroup$
In an M/M/s queue, what does this expression mean? :
$sum_{n=0}^{s-1}{(s-n)P_n}$
Furthermore, is it possible that the following equation holds? :
$sum_{n=0}^{s-1}{(s-n)P_n} = (1-rho)s$
If so, how could I demonstrate it?
EDIT: An M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers.
Thanks,
Louis
probability queueing-theory
asked Nov 28 '18 at 19:25
Louis-Philippe NoëlLouis-Philippe Noël
123
$endgroup$
$begingroup$
What is an M/M/S queue? What are the variables that you're using?
$endgroup$
– Tki Deneb
Nov 29 '18 at 17:56
$begingroup$
@TkiDeneb an M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers. Thanks for your time!
$endgroup$
– Louis-Philippe Noël
Nov 29 '18 at 21:08
$begingroup$
I'm afraid I can't help you then, I'm not familiar with this subject.
$endgroup$
– Tki Deneb
Nov 29 '18 at 22:23
add a comment |
$begingroup$
In an M/M/s queue, what does this expression mean? :
$sum_{n=0}^{s-1}{(s-n)P_n}$
Furthermore, is it possible that the following equation holds? :
$sum_{n=0}^{s-1}{(s-n)P_n} = (1-rho)s$
If so, how could I demonstrate it?
EDIT: An M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers.
Thanks,
Louis
probability queueing-theory
asked Nov 28 '18 at 19:25
Louis-Philippe NoëlLouis-Philippe Noël
123
$endgroup$
In an M/M/s queue, what does this expression mean? :
$sum_{n=0}^{s-1}{(s-n)P_n}$
Furthermore, is it possible that the following equation holds? :
$sum_{n=0}^{s-1}{(s-n)P_n} = (1-rho)s$
If so, how could I demonstrate it?
EDIT: An M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers.
Thanks,
Louis
probability queueing-theory
probability queueing-theory
asked Nov 28 '18 at 19:25
Louis-Philippe NoëlLouis-Philippe Noël
123
asked Nov 28 '18 at 19:25
Louis-Philippe NoëlLouis-Philippe Noël
123
edited Nov 29 '18 at 22:55
Louis-Philippe Noël
asked Nov 28 '18 at 19:25
Louis-Philippe NoëlLouis-Philippe Noël
123
asked Nov 28 '18 at 19:25
Louis-Philippe NoëlLouis-Philippe Noël
123
asked Nov 28 '18 at 19:25
Louis-Philippe NoëlLouis-Philippe Noël
123
123
$begingroup$
What is an M/M/S queue? What are the variables that you're using?
$endgroup$
– Tki Deneb
Nov 29 '18 at 17:56
$begingroup$
@TkiDeneb an M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers. Thanks for your time!
$endgroup$
– Louis-Philippe Noël
Nov 29 '18 at 21:08
$begingroup$
I'm afraid I can't help you then, I'm not familiar with this subject.
$endgroup$
– Tki Deneb
Nov 29 '18 at 22:23
add a comment |
$begingroup$
What is an M/M/S queue? What are the variables that you're using?
$endgroup$
– Tki Deneb
Nov 29 '18 at 17:56
$begingroup$
@TkiDeneb an M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers. Thanks for your time!
$endgroup$
– Louis-Philippe Noël
Nov 29 '18 at 21:08
$begingroup$
I'm afraid I can't help you then, I'm not familiar with this subject.
$endgroup$
– Tki Deneb
Nov 29 '18 at 22:23
$begingroup$
What is an M/M/S queue? What are the variables that you're using?
$endgroup$
– Tki Deneb
Nov 29 '18 at 17:56
$begingroup$
What is an M/M/S queue? What are the variables that you're using?
$endgroup$
– Tki Deneb
Nov 29 '18 at 17:56
$begingroup$
@TkiDeneb an M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers. Thanks for your time!
$endgroup$
– Louis-Philippe Noël
Nov 29 '18 at 21:08
$begingroup$
@TkiDeneb an M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers. Thanks for your time!
$endgroup$
– Louis-Philippe Noël
Nov 29 '18 at 21:08
$begingroup$
I'm afraid I can't help you then, I'm not familiar with this subject.
$endgroup$
– Tki Deneb
Nov 29 '18 at 22:23
$begingroup$
I'm afraid I can't help you then, I'm not familiar with this subject.
$endgroup$
– Tki Deneb
Nov 29 '18 at 22:23
add a comment |
1 Answer
1
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$begingroup$
If $p_n$ is the stationary distribution then the first expression is the expected number of free servers.
The equation is indeed correct and this can be argued using Little’s law: the arrival rate to each server is $frac{lambda}{s}$ and the expected time in service is $frac{1}{mu}$, so the probability each server is free is $1-frac{lambda}{smu}$. The servers are identical so the the sum is just $s$ times the probability of a free server.
answered Dec 2 '18 at 21:15
QQQQQQ
59645
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add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $p_n$ is the stationary distribution then the first expression is the expected number of free servers.
The equation is indeed correct and this can be argued using Little’s law: the arrival rate to each server is $frac{lambda}{s}$ and the expected time in service is $frac{1}{mu}$, so the probability each server is free is $1-frac{lambda}{smu}$. The servers are identical so the the sum is just $s$ times the probability of a free server.
answered Dec 2 '18 at 21:15
QQQQQQ
59645
$endgroup$
add a comment |
$begingroup$
If $p_n$ is the stationary distribution then the first expression is the expected number of free servers.
The equation is indeed correct and this can be argued using Little’s law: the arrival rate to each server is $frac{lambda}{s}$ and the expected time in service is $frac{1}{mu}$, so the probability each server is free is $1-frac{lambda}{smu}$. The servers are identical so the the sum is just $s$ times the probability of a free server.
answered Dec 2 '18 at 21:15
QQQQQQ
59645
$endgroup$
add a comment |
$begingroup$
If $p_n$ is the stationary distribution then the first expression is the expected number of free servers.
The equation is indeed correct and this can be argued using Little’s law: the arrival rate to each server is $frac{lambda}{s}$ and the expected time in service is $frac{1}{mu}$, so the probability each server is free is $1-frac{lambda}{smu}$. The servers are identical so the the sum is just $s$ times the probability of a free server.
answered Dec 2 '18 at 21:15
QQQQQQ
59645
$endgroup$
If $p_n$ is the stationary distribution then the first expression is the expected number of free servers.
The equation is indeed correct and this can be argued using Little’s law: the arrival rate to each server is $frac{lambda}{s}$ and the expected time in service is $frac{1}{mu}$, so the probability each server is free is $1-frac{lambda}{smu}$. The servers are identical so the the sum is just $s$ times the probability of a free server.
answered Dec 2 '18 at 21:15
QQQQQQ
59645
answered Dec 2 '18 at 21:15
QQQQQQ
59645
answered Dec 2 '18 at 21:15
QQQQQQ
59645
answered Dec 2 '18 at 21:15
QQQQQQ
59645
59645
add a comment |
add a comment |
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$begingroup$
What is an M/M/S queue? What are the variables that you're using?
$endgroup$
– Tki Deneb
Nov 29 '18 at 17:56
$begingroup$
@TkiDeneb an M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers. Thanks for your time!
$endgroup$
– Louis-Philippe Noël
Nov 29 '18 at 21:08
$begingroup$
I'm afraid I can't help you then, I'm not familiar with this subject.
$endgroup$
– Tki Deneb
Nov 29 '18 at 22:23