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:

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

  2. $\lambda_k$ 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 $\lambda_k'=\lambda_k*\frac{1}{1+\lambda_k*T_k}$.

  3. $\sum_k (\lambda_k'T_k)=(\sum_k \lambda_k')T$ 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 $T_k$ 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:


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


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:




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?

  • $\begingroup$ 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

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