The M/M/1 Queue

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  • Опубликовано: 21 авг 2024

Комментарии • 30

  • @alexdamado
    @alexdamado 6 лет назад +6

    Thank you for posting this. Very hard to find a good video on the subject and yours is very well explained.

  • @Tom-kg6qe
    @Tom-kg6qe 4 года назад +2

    I came to your channel looking for an explanation on Ising models and now my RUclips feed is populated with your videos... I'm not complaining, I love this quantum statistical maths! And it's far better than the usual trash that populates my feed!

  • @achillesarmstrong9639
    @achillesarmstrong9639 5 лет назад +1

    very helpful thank you. This is the clearest explanation I found in youtube

  • @11amanie
    @11amanie Год назад +1

    Ty😊

  • @Rahul-uk4su
    @Rahul-uk4su 4 года назад +1

    please keep uploading.youtube needs ppl like you

    • @gtribello
      @gtribello  4 года назад +1

      Hello. Thanks for your comment. It made my day.

  • @exapsy
    @exapsy 6 лет назад +2

    I was searching for 2 days now regarding the M/M/1 queue and ... STILL I can't find anything related to my problem. Everything has Calculus when what I only want is a lecture with the simple formulas like LITTLE's law etc. Can anyone guide me?

  • @bowenzhang4471
    @bowenzhang4471 4 года назад +1

    Can I ask why the two limits (mu and lambda) indicates that the serving time and the interval of customer's arrival obey exponential distribution and Poison distribution respectively?

    • @gtribello
      @gtribello  4 года назад

      Hello. Thanks for your question first of all. When \lim_{t -> 0 } P_n(n+1)(t) is a constant you have a Poisson process as this is a definition of a Poisson process. I have an explanation for this in the video I made on the Poisson process here: ruclips.net/video/kWvG0_p4wO8/видео.html. A similar argument holds for \lim_{t -> 0 } P_n(n-1)(t). If this limit is a constant then the probability density function for an exponential random variable drops out when you solve the equations for the reasons explained in ruclips.net/video/BdOB8x3SVAE/видео.html I know this answer sounds a bit like I am saying that the answer to your question is "because it is." I hope that by watching the other videos helps. Good luck

  • @rayhey
    @rayhey 6 лет назад +3

    Something is missing from the slides.

  • @elcalifornio4797
    @elcalifornio4797 6 лет назад +1

    How many times this going to be explained!

  • @ethelchew7367
    @ethelchew7367 3 года назад +1

    hello! thanks for the help! i’m a bit unsure - so what exactly is the stationary distribution?

    • @gtribello
      @gtribello  3 года назад +2

      Hi Ethel. The stationary distribution tells us the long time probability of being in each of the various states in the chain. I have a video that explains this (for the discrete case) here: ruclips.net/video/JZnzQ8YzVZg/видео.html. Finding the stationary distribution for the continuous case is easier as is explained here: ruclips.net/video/tbA2DnKTRxM/видео.html. I hope this helps

    • @ethelchew7367
      @ethelchew7367 3 года назад

      @@gtribello oh okay thank you, sorry another question but at around 5:30, there’s the part of having a third limit where it can be rewritten as a sum of 3 limits. May i know the purpose of having this limit?

    • @ethelchew7367
      @ethelchew7367 3 года назад

      @@gtribello also, sorry to disturb again, but what is the transition matrix for?

    • @gtribello
      @gtribello  3 года назад +1

      @@ethelchew7367 for the transition matrix take a look at this video ruclips.net/video/1meaW5GxUbY/видео.html. This video tells you what the transition matrix is for in the context of Markov chains in discrete time. To go to the continuous case you need to take some limits, which is explained in this video ruclips.net/video/tbA2DnKTRxM/видео.html. I hope this helps

  • @mananwalia4720
    @mananwalia4720 7 лет назад +1

    awesome job bro!

  • @romanbudilovsky504
    @romanbudilovsky504 5 лет назад +2

    He actually burped at 19:39 :) Nice explanation

  • @bhupenderkumarsom1975
    @bhupenderkumarsom1975 5 лет назад +1

    Thank you. Is there any video on transient solution

    • @gtribello
      @gtribello  5 лет назад

      Thanks. I haven't made a video on the transient solution sorry. I am not even sure what the transient solution would be.

  • @tomasgonzales
    @tomasgonzales 6 лет назад +1

    why are those limits equal to lambda and mu respectively?

    • @gtribello
      @gtribello  6 лет назад +1

      Hello. They are equal to lambda and mu because we decide that is what they are equal to. These decisions are made in the same same way that we decide that many of the other limits are equal to zero. This may all seem rather arbitrary but setting the limits in this way gives the queue some useful properties. For example by choosing to set the limits in this way we ensure that customers arrive in accordance with a Poisson process with parameter lambda. Furthermore, the time taken to serve each customer is an exponentially distributed random variable with parameter mu because of the way we choose to set mu. I hope this helps.

    • @tomasgonzales
      @tomasgonzales 6 лет назад

      Hey thanks for the reply I have an assignment where I have to construct the Q matrix and I don't know how to explain that step. You reckon not justifying will be fine?

    • @gtribello
      @gtribello  6 лет назад

      Hello again. It depends on the question. There are a number of ways of formulating the theory of queues and the precise proof that your tutor will may be different. The following notes go into things in more detail than the video so they may help you: gtribello.github.io/mathNET/resources/jim-chap24.pdf

    • @sabermohammed1053
      @sabermohammed1053 4 года назад

      @@gtribello I want to contact you, can I get your number

    • @gtribello
      @gtribello  4 года назад

      @@sabermohammed1053 I am not sure I am comfortable putting my number publically in youtube. If you go searching online you should be able to find my email address quite easily. Drop me an email and let me know what you want to talk about and then we can work things out from there. Be well.

  • @user-ih6gb7li9i
    @user-ih6gb7li9i 7 лет назад +2

    I don't know any thing (: , realy

  • @EWang-yn5sy
    @EWang-yn5sy 5 лет назад +1

    this is super difficult..............damn

  • @LauDonThe5Facts
    @LauDonThe5Facts 5 лет назад +1

    this time pk pkk la