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 December 2nd, 2018, 04:00 PM #1 Newbie   Joined: Dec 2018 From: Australia Posts: 1 Thanks: 0 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? Tags class, customer, m or m or 1, multiclass, poisson, probabibility, queue, queueing, simultaneously, statistics, stochastic process Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post life24 Computer Science 4 December 18th, 2016 02:54 AM MMath New Users 0 July 6th, 2016 05:04 AM flapalot Algebra 1 January 30th, 2012 05:06 AM Dld3 Economics 3 November 2nd, 2011 11:17 AM

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