I came to your channel looking for an explanation on Ising models and now my TH-cam 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!
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?
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: th-cam.com/video/kWvG0_p4wO8/w-d-xo.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 th-cam.com/video/BdOB8x3SVAE/w-d-xo.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
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: th-cam.com/video/JZnzQ8YzVZg/w-d-xo.html. Finding the stationary distribution for the continuous case is easier as is explained here: th-cam.com/video/tbA2DnKTRxM/w-d-xo.html. I hope this helps
@@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 for the transition matrix take a look at this video th-cam.com/video/1meaW5GxUbY/w-d-xo.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 th-cam.com/video/tbA2DnKTRxM/w-d-xo.html. I hope this helps
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?
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.
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?
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 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.
Thank you for posting this. Very hard to find a good video on the subject and yours is very well explained.
I came to your channel looking for an explanation on Ising models and now my TH-cam 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!
very helpful thank you. This is the clearest explanation I found in youtube
please keep uploading.youtube needs ppl like you
Hello. Thanks for your comment. It made my day.
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?
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: th-cam.com/video/kWvG0_p4wO8/w-d-xo.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 th-cam.com/video/BdOB8x3SVAE/w-d-xo.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
hello! thanks for the help! i’m a bit unsure - so what exactly is the stationary distribution?
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: th-cam.com/video/JZnzQ8YzVZg/w-d-xo.html. Finding the stationary distribution for the continuous case is easier as is explained here: th-cam.com/video/tbA2DnKTRxM/w-d-xo.html. I hope this helps
@@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?
@@gtribello also, sorry to disturb again, but what is the transition matrix for?
@@ethelchew7367 for the transition matrix take a look at this video th-cam.com/video/1meaW5GxUbY/w-d-xo.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 th-cam.com/video/tbA2DnKTRxM/w-d-xo.html. I hope this helps
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?
Something is missing from the slides.
He actually burped at 19:39 :) Nice explanation
Ty😊
Thank you. Is there any video on transient solution
Thanks. I haven't made a video on the transient solution sorry. I am not even sure what the transient solution would be.
How many times this going to be explained!
awesome job bro!
why are those limits equal to lambda and mu respectively?
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.
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?
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
@@gtribello I want to contact you, can I get your number
@@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.
I don't know any thing (: , realy
this time pk pkk la
this is super difficult..............damn