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Suppose customers join a queue with a poisson arrival rate . If a customer is not served within a unit of time, she abandons the queue. Customers are served by two servers: one of the servers runs the first-come-first-served (FCFS) policy and the other one the last-come-first-served (LCFS) policy. The FCFS server has a service time that is iid exponentially with mean $\lambda m$, where $\lambda<1$. The LCFS server has a service time that is iid exponentially with mean $\delta m$, where $\delta<1-\lambda$. A customer departs the queue after being served by either of the servers. I would like to show that the average length of the queue is at least $(1-\delta)m -o(m)$. Any input will be appreciated!

P.S. A special case of this problem when $\delta=0$ has been solved here Average queue length with impatient customers while the intuition seems similar, I have not been able to adapt this approach.

afshi7n
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The system you are describing is the simple $M/M/2/\infty$ queue with heterogeneous servers (one with service rate $1/(\lambda m)$, the other with the service rate $1/(\delta m)$) and the arrival rate $m$.

This queue can be solved analytically. Although the formulas will be cumbersome. The full solution for your case you can find here https://www.jstor.org/stable/167292. Although there are also matrix-geometric solutions available out there, which may be somewhat simpler.

Since you are interested only in the queue length, i would suggest the following answer. Let $b$ be the minimum service rate among the two i.e. $b=\max(1/(\lambda m), 1/(\delta m))$. Then the average queue length (I mean here TOTAL number of jobs in the system i.e. queue + servers) in your queue will be $\ge $ than the average queue length in the similar queue but with identical servers working at rate $b$. But for identical servers it is known (see, for example, page 152 here https://pdfs.semanticscholar.org/848f/a1f48ad9d3edb24b05667f15cfc633eb8f69.pdf ) that the average queue length is equal to $$ {2 {m \over 2 b} \over 1 - ({m \over 2 b})^2}. $$

Now, probably, some algebra will help you get the result you are looking for.

rrv
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  • I'm probably missing something, but (1) what does the $\infty$ in $M/M/2/\infty$ represent? And (2) how does this model incorporate the "impatient customer" aspect of the OP, i.e. "If a customer is not served within a unit of time, she abandons the queue."? – antkam Sep 19 '19 at 18:23
  • $M/M/2$ is the notation describing the specific queueing system. Please check what is "Kendall notation" in Wikipedia. I was answering your question and in your question there are no impatient customers. – rrv Sep 19 '19 at 20:39
  • sorry, I am not the OP so I dont know what you mean by "my question". :) The OP did include impatient customers (see 2nd sentence). Also, I did check Kendall notation in wiki and the $\infty$ just means no restriction on queue length, which is the default anyway, so I was surprised to see you write $M/M/2/\infty$ instead of simply $M/M/2$ and hence I was wondering if you meant something else by the $\infty$... – antkam Sep 19 '19 at 20:46
  • @rrv thanks for your note, but there are some differences with my setting and the settings you refer to. 1) in my setting there's a one FCFS and one LCFS server. 2) the impatience plays a significant role in the analysis. the M/M/2 queuing model does not include impatience – afshi7n Sep 20 '19 at 01:58
  • @afshi7n 1) in your setting the type of server does not matter. this follows from the fact that FCFS or LCFS policies are equivalent with respect to the average queue size. 2) in your setting there is no impatience. – rrv Sep 20 '19 at 14:00
  • the 2nd sentence of the OP: "If a customer is not served within a unit of time, she abandons the queue." models customer impatience. – antkam Sep 20 '19 at 19:44
  • Indeed i have missed this second sentence. My apologies. Anyway, the approach that I suggested above with the lower bound for the average queue length still works but. You have to use another formulae for the average queue length: $m p_{loss} (w+{1\over b})$, where $\tau$ is the constant impatience time and $w$ and $p_{loss}$ are the stationary mean waiting time in the queue and the stationary probability that a customer is lost, respectively ( $w$ and $p_{loss}$ can be found using expressions (1) in https://doi.org/10.1007/BF02564722 ). – rrv Sep 21 '19 at 09:55
  • It is incorrect that it doesn't matter if the FCFS policy is used or LCFS. Because customers abandon if they do not receive service after one unit of time, the type of policy does matter. To see an example, you can refer to the link I posted in the question. – afshi7n Sep 21 '19 at 16:07
  • since you are concerned only with the average queue size, it does not matter whether FCFS or LCFS is used. – rrv Sep 22 '19 at 08:23