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December 2nd, 2018, 05:00 PM   #1
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Question Multiclass M/M/1 queue which can simultaneously have one customer per class

Suppose there is a queue with exponential service time $\displaystyle \frac{1}{\mu}$ which accepts customers from K classes with poisson distribution and rate $\displaystyle \lambda_k$ but if a customer from any class arrives at the queue and there is already another customer from the same queue either waiting or receiving service, leaves.

I have performed simulation to find the total service time $\displaystyle S = W_q + \frac{1}{\mu}$ but failed miserably.

These are what I have found out from simulations so far:

1. service time $\displaystyle T_k$ is diffrent in every class.

2. $\displaystyle \lambda_k$ changes while leaving the queue server and becomes $\displaystyle \lambda_k'=\frac{\lambda_k}{1+\lambda_k*T_k}$ beacause when one customer arrives and stays in the queue for $\displaystyle T_k$, $\displaystyle \lambda_k*T_k$ customers arrive and leave immediatly without service. So one out of every $\displaystyle 1+\lambda_k*T_k$ gets service from the queue, hence $\displaystyle \lambda_k'=\lambda_k*\frac{1}{1+\lambda_k*T_k}$.

3. $\displaystyle \sum_k (\lambda_k'T_k)=(\sum_k \lambda_k')T$ in which T is the total mean service time of all customers regardless of class.

__________________________________________________ _________________________

So I found out that if we consider this queue as K M/M/1/1 queues with $\displaystyle T_k$ 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 $\displaystyle T_k$ that we are looking for. $\displaystyle P_0$ is the probability of each of these queues being empty which is by M/M/1/1 standards:

$\displaystyle P_0^k = \frac{1}{1+\lambda_kT_k}$

Average number of customers of K in the system is equal to $\displaystyle 1-P_0^k$, because every one of them could only have one customer:

$\displaystyle P_1^k=n_k=\frac{\lambda_kT_k}{1+\lambda_kT_k}$

The total number of customers in the system is the sum of $\displaystyle n_k$:

$\displaystyle N=\sum_kn_k
$
***By this point these are all verified by simulations***. After this I tried to find $\displaystyle T_k$s and failed. I thought $\displaystyle T_k$ should be like this:

$\displaystyle T_k=(N-n_k+1)*\frac{1}{\mu}$

or

$\displaystyle T_k=(1-P_1^k)(N-n_k+1)*\frac{1}{\mu}$

I thought that every customer, by entering the queue will have to wait for each one of customers from other classes be served for $\displaystyle \frac{1}{\mu}$ and then be served itself. None of the above work.

Could anyone Tell me why?
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class, customer, m or m or 1, multiclass, poisson, probabibility, queue, queueing, simultaneously, statistics, stochastic process



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