Multiclass M/M/1 queue which can simultaneously have one customer per class
$begingroup$
Suppose there is a queue with exponential service time $frac{1}{mu}$ which accepts customers from K classes with Poisson distribution and rate $lambda_k$ but if a customer from any class arrives at the queue while there is already another customer from the same class either waiting or receiving service, leaves.
I have performed a simulation to find the total service time $T = W_q + frac{1}{mu}$ but failed.
These are what I have found out from simulations so far:
service time $T_k$ is diffrent in every class.
$lambda_k$ changes while leaving the queue server and becomes $lambda_k'=frac{lambda_k}{1+lambda_k*T_k}$ beacause when one customer arrives and stays in the queue for $T_k$, $lambda_k*T_k$ customers arrive and leave immediatly without service. So one out of every $1+lambda_k*T_k$ gets service from the queue, hence $lambda_k'=lambda_k*frac{1}{1+lambda_k*T_k}$.
$sum_k (lambda_k'T_k)=(sum_k lambda_k')T$ in which T is the total mean service time of all customers regardless of class.
Update:
So I found out that if we consider this queue as K M/M/1/1 queues with $T_k$ service times, this could lead to finding the average number of customers in the system.
The followings are verified by simulation.
Every class has a separate M/M/1/1 queue with service time equal to $T_k$ that we are looking for. $P_0$ is the probability of each of these queues being empty which is by M/M/1/1 standards:
$P_0^k = frac{1}{1+lambda_kT_k}$
The average number of customers of K in the system is equal to $1-P_0^k$, because every one of them could only have one customer:
$P_1^k=n_k=frac{lambda_kT_k}{1+lambda_kT_k}$
The total number of customers in the system is the sum of $n_k$:
$N=sum_kn_k$
By this point these are all verified by simulations. After this I tried to find $T_k$s and failed. I thought $T_k$ should be like this:
$T_k=(N-n_k+1)*frac{1}{mu}$
or
$T_k=(1-P_1^k)(N-n_k+1)*frac{1}{mu}$
These have very close results to $T_k$s but don't work. I thought that every customer, by entering the queue will have to wait for each one of the customers from other classes be served for $frac{1}{mu}$ and then be served itself. None of the above work.
Any suggestions?
probability stochastic-processes markov-chains queueing-theory
asked Nov 30 '18 at 5:20
AnaisAnais
437
$endgroup$
add a comment |
$begingroup$
Suppose there is a queue with exponential service time $frac{1}{mu}$ which accepts customers from K classes with Poisson distribution and rate $lambda_k$ but if a customer from any class arrives at the queue while there is already another customer from the same class either waiting or receiving service, leaves.
I have performed a simulation to find the total service time $T = W_q + frac{1}{mu}$ but failed.
These are what I have found out from simulations so far:
service time $T_k$ is diffrent in every class.
$lambda_k$ changes while leaving the queue server and becomes $lambda_k'=frac{lambda_k}{1+lambda_k*T_k}$ beacause when one customer arrives and stays in the queue for $T_k$, $lambda_k*T_k$ customers arrive and leave immediatly without service. So one out of every $1+lambda_k*T_k$ gets service from the queue, hence $lambda_k'=lambda_k*frac{1}{1+lambda_k*T_k}$.
$sum_k (lambda_k'T_k)=(sum_k lambda_k')T$ in which T is the total mean service time of all customers regardless of class.
Update:
So I found out that if we consider this queue as K M/M/1/1 queues with $T_k$ service times, this could lead to finding the average number of customers in the system.
The followings are verified by simulation.
Every class has a separate M/M/1/1 queue with service time equal to $T_k$ that we are looking for. $P_0$ is the probability of each of these queues being empty which is by M/M/1/1 standards:
$P_0^k = frac{1}{1+lambda_kT_k}$
The average number of customers of K in the system is equal to $1-P_0^k$, because every one of them could only have one customer:
$P_1^k=n_k=frac{lambda_kT_k}{1+lambda_kT_k}$
The total number of customers in the system is the sum of $n_k$:
$N=sum_kn_k$
By this point these are all verified by simulations. After this I tried to find $T_k$s and failed. I thought $T_k$ should be like this:
$T_k=(N-n_k+1)*frac{1}{mu}$
or
$T_k=(1-P_1^k)(N-n_k+1)*frac{1}{mu}$
These have very close results to $T_k$s but don't work. I thought that every customer, by entering the queue will have to wait for each one of the customers from other classes be served for $frac{1}{mu}$ and then be served itself. None of the above work.
Any suggestions?
probability stochastic-processes markov-chains queueing-theory
asked Nov 30 '18 at 5:20
AnaisAnais
437
$endgroup$
$begingroup$
I'm having trouble parsing this description. Can you clarify how this is different from $K$ different M/M/1 queues?
$endgroup$
– Brian Tung
Dec 1 '18 at 0:19
add a comment |
$begingroup$
Suppose there is a queue with exponential service time $frac{1}{mu}$ which accepts customers from K classes with Poisson distribution and rate $lambda_k$ but if a customer from any class arrives at the queue while there is already another customer from the same class either waiting or receiving service, leaves.
I have performed a simulation to find the total service time $T = W_q + frac{1}{mu}$ but failed.
These are what I have found out from simulations so far:
service time $T_k$ is diffrent in every class.
$lambda_k$ changes while leaving the queue server and becomes $lambda_k'=frac{lambda_k}{1+lambda_k*T_k}$ beacause when one customer arrives and stays in the queue for $T_k$, $lambda_k*T_k$ customers arrive and leave immediatly without service. So one out of every $1+lambda_k*T_k$ gets service from the queue, hence $lambda_k'=lambda_k*frac{1}{1+lambda_k*T_k}$.
$sum_k (lambda_k'T_k)=(sum_k lambda_k')T$ in which T is the total mean service time of all customers regardless of class.
Update:
So I found out that if we consider this queue as K M/M/1/1 queues with $T_k$ service times, this could lead to finding the average number of customers in the system.
The followings are verified by simulation.
Every class has a separate M/M/1/1 queue with service time equal to $T_k$ that we are looking for. $P_0$ is the probability of each of these queues being empty which is by M/M/1/1 standards:
$P_0^k = frac{1}{1+lambda_kT_k}$
The average number of customers of K in the system is equal to $1-P_0^k$, because every one of them could only have one customer:
$P_1^k=n_k=frac{lambda_kT_k}{1+lambda_kT_k}$
The total number of customers in the system is the sum of $n_k$:
$N=sum_kn_k$
By this point these are all verified by simulations. After this I tried to find $T_k$s and failed. I thought $T_k$ should be like this:
$T_k=(N-n_k+1)*frac{1}{mu}$
or
$T_k=(1-P_1^k)(N-n_k+1)*frac{1}{mu}$
These have very close results to $T_k$s but don't work. I thought that every customer, by entering the queue will have to wait for each one of the customers from other classes be served for $frac{1}{mu}$ and then be served itself. None of the above work.
Any suggestions?
probability stochastic-processes markov-chains queueing-theory
asked Nov 30 '18 at 5:20
AnaisAnais
437
$endgroup$
Suppose there is a queue with exponential service time $frac{1}{mu}$ which accepts customers from K classes with Poisson distribution and rate $lambda_k$ but if a customer from any class arrives at the queue while there is already another customer from the same class either waiting or receiving service, leaves.
I have performed a simulation to find the total service time $T = W_q + frac{1}{mu}$ but failed.
These are what I have found out from simulations so far:
service time $T_k$ is diffrent in every class.
$lambda_k$ changes while leaving the queue server and becomes $lambda_k'=frac{lambda_k}{1+lambda_k*T_k}$ beacause when one customer arrives and stays in the queue for $T_k$, $lambda_k*T_k$ customers arrive and leave immediatly without service. So one out of every $1+lambda_k*T_k$ gets service from the queue, hence $lambda_k'=lambda_k*frac{1}{1+lambda_k*T_k}$.
$sum_k (lambda_k'T_k)=(sum_k lambda_k')T$ in which T is the total mean service time of all customers regardless of class.
Update:
So I found out that if we consider this queue as K M/M/1/1 queues with $T_k$ service times, this could lead to finding the average number of customers in the system.
The followings are verified by simulation.
Every class has a separate M/M/1/1 queue with service time equal to $T_k$ that we are looking for. $P_0$ is the probability of each of these queues being empty which is by M/M/1/1 standards:
$P_0^k = frac{1}{1+lambda_kT_k}$
The average number of customers of K in the system is equal to $1-P_0^k$, because every one of them could only have one customer:
$P_1^k=n_k=frac{lambda_kT_k}{1+lambda_kT_k}$
The total number of customers in the system is the sum of $n_k$:
$N=sum_kn_k$
By this point these are all verified by simulations. After this I tried to find $T_k$s and failed. I thought $T_k$ should be like this:
$T_k=(N-n_k+1)*frac{1}{mu}$
or
$T_k=(1-P_1^k)(N-n_k+1)*frac{1}{mu}$
These have very close results to $T_k$s but don't work. I thought that every customer, by entering the queue will have to wait for each one of the customers from other classes be served for $frac{1}{mu}$ and then be served itself. None of the above work.
Any suggestions?
probability stochastic-processes markov-chains queueing-theory
probability stochastic-processes markov-chains queueing-theory
asked Nov 30 '18 at 5:20
AnaisAnais
437
asked Nov 30 '18 at 5:20
AnaisAnais
437
edited Dec 10 '18 at 23:44
Anais
asked Nov 30 '18 at 5:20
AnaisAnais
437
asked Nov 30 '18 at 5:20
AnaisAnais
437
asked Nov 30 '18 at 5:20
AnaisAnais
437
437
$begingroup$
I'm having trouble parsing this description. Can you clarify how this is different from $K$ different M/M/1 queues?
$endgroup$
– Brian Tung
Dec 1 '18 at 0:19
add a comment |
$begingroup$
I'm having trouble parsing this description. Can you clarify how this is different from $K$ different M/M/1 queues?
$endgroup$
– Brian Tung
Dec 1 '18 at 0:19
$begingroup$
I'm having trouble parsing this description. Can you clarify how this is different from $K$ different M/M/1 queues?
$endgroup$
– Brian Tung
Dec 1 '18 at 0:19
$begingroup$
I'm having trouble parsing this description. Can you clarify how this is different from $K$ different M/M/1 queues?
$endgroup$
– Brian Tung
Dec 1 '18 at 0:19
add a comment |
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$begingroup$
I'm having trouble parsing this description. Can you clarify how this is different from $K$ different M/M/1 queues?
$endgroup$
– Brian Tung
Dec 1 '18 at 0:19