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M/M/c Queueing Model - YouTube
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this is a lecture 5 application of a
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continuous time Markov chain in simple
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Markovian queuing models the first
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lecture we have discussed the definition
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of a stochastic process in particular
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continuous-time markov chain then we
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have considered the Kolmogorov
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differential equation Chapman Kolmogorov
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equations the transient solutions for
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the ctmc the second lecture we have
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discussed the special case of continuous
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time Markov chain that is a birth-death
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process we have discussed in lecture 2
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lecture 3 the special case of
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birth-death process with a very
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important stochastic process that is a
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Poisson process is discussed in the
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lecture 3 the lecture 4 we have
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discussed the MM 1 queueing model that
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is a very special and important queueing
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model and the underlying stochastic
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process for the MM 1 queueing model that
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is a birth-death process with the birth
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rates are lambda and death rates are you
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in the fourth lecture we have discussed
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only the MM 1 queueing model in this
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lecture we are going to consider the
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other simple Markovian queuing models as
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an application of a continuous-time
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markov chain
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so in this lecture I am going to discuss
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other than the MM 1 queueing model I am
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going to discuss a simple Markovian
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queuing models starting with the m/m/c
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infinity queueing model then the finite
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capacity model Markovian setup emam 1
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yen queueing model then I am going to
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discuss the multi-server finite capacity
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model that is here moms CK queueing
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model after that I am going to discuss
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the last system that is here mom si si
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model for a infinite server model that
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is yummy infinity also I'm going to
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discuss at the end I'm going to discuss
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the finite source Markovian queuing
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models whereas the other five models the
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population is infinite source so the
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last one is a finite source Markovian
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queuing model also I'm going to discuss
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as the application of continuous time
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Markov chain the first model is a multi
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server infinite capacity Markovian
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queuing model the letter M denotes the
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inter arrival time is exponentially
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distributed with parameter lambda the
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service time by the each server that is
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exponentially distributed with the
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parameter mu and all we have more than
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one servers suppose you consider as a C
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where C is a positive integer and all
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the servers are identical and each
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server is doing the service which is
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exponentially distributed with the
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parameter mu which is independent of the
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all other service and the service time
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is independent with the inter arrival
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time also with these assumptions if you
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make a random variable X of T is the
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number of customers in the system at any
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time T that is a stochastic process
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since the possible values of number of
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customers in the system at any time T
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that is going to be 0 1 2 and so on
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therefore it is a discrete state and you
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are observing the queuing system at any
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time T therefore it is a continuous time
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so discrete state continuous times two
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stick process and if you observe the
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system is a keep moving into the
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different states because of either
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arrival or the service completion from
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the any one of the C servers so suppose
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there are no customer in the system and
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the system moves from the state is zero
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to one by one arrival so the inter
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arrival time is exponentially
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distributed therefore the rate in which
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the system is moving from the state 0 to
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1 is lambda like that you can visualize
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the rates for the system moving from 1
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to 2 2 to 3 and so on whereas whenever
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the system size is 1 2 and so on till C
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since we have a C number of the servers
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in the system who are entering into the
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system they will get they will start
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getting the service immediately suppose
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the system goes from state 1 to 0 that
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means that the customer entered into the
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system and he get the service
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immediately and the service time is
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exponentially distributed with the
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parameter mu therefore whenever the
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service is completed the system goes
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from the state 1 to 0 therefore the rate
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is mu whereas a from 2 to 1 there are 2
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customers in the system and both are
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under service at any time in if any one
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of the service if any one of the service
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completely service then the system moves
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from 2 to 1 so the service completion
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will be minimum offer the service time
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of the both the servers since each
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server is doing the service
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exponentially distributed with the
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parameter mu therefore the minimum of
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two exponential
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both are independent also therefore that
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is also going to be exponentially
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distributed with the sum of parameters
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so it is going to be parameter will be
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mu plus mu that is to Mew
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so the system moves from the state of
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two to one maybe the rate will be to me
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like that it will be keep going till the
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state from C to C minus 1 that means we
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have C servers therefore whenever the
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system size is also less than or equal
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to C that means all the customers are
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under service now we'll discuss the rate
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in which the system is moving from the
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state C plus 1 to C this is some state
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is a C plus 1 that means and the number
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of customers in the system that is C
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plus 1 we have C servers therefore one
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customer will be waiting for the service
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80 in the queue therefore the system is
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moving from C plus 1 to C that is
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nothing but one of the server completed
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the service out of C service therefore
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the sir the rate will be the service
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time completion service time leave me
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exponential distribution with the
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parameter C mu naught C plus 1 mu it is
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a we have only C service therefore the
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minimum of exponentially distributed
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with the parameters mu and so on with
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the C exponentially distributed random
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variables therefore that is going to be
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exponential distribution with the
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parameter mu plus mu plus receive news
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therefore it is going to be seeming like
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that the rate will be the death rate
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will be C mu after C plus 1 onwards
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whereas from 0 to C it will be mu 2 mu 3
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mu and so on till C mu after that it
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will be C mu from the state from C
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c + 1 - c c + 2 - c plus 1 and so on
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and if you see the state transition
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diagram you can observe that this is a
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birth-death process so before that let
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me explain what is a m/m/c infinity
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means whenever C customers or C service
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or any one of the C servers are
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available then the customers we get the
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service immediately he for all the C
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servers are busy then the customer has
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to wait till any one of the C servers
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are going to be completing their service
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so that is the way the system works
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therefore you will have the system size
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the system size the underlying
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stochastic process is going to be a
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birth-death process I said it's a
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special case of a continuous time Markov
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chain because the transitions are only
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the neighbors transition with the
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forward rates that is lambda and
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backward rates are the death rates are
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going to be mu 2 mu and so on
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therefore this is a special case of a
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continuous time Markov chain the
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underlying stochastic process for the
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m/m/c infinity model that is a
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birth-death process the birth rates are
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lambda whereas the death rates depends
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on the N the MU n is a function of Y n
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therefore it is called a state dependent
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and death rates it may not be the
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function n times mu it can be a function
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of Y and then we can use the word state
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dependent so here it is a linear
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function so state dependent death rates
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and that the thread sorry n times mu
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whenever yarn is lies between 1 - C
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and the MU n is going to be C times mu
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for n is greater than or equal to C that
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you can observe it from the state
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transition diagram also the death rates
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are going to be C mu here also C mu and
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so on therefore this is a birth-death
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process with the state dependent death
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rates now our interest is to find out
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the steady state or equilibrium solution
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since it is a infinite capacity model if
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you observe the birth-death process with
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the infinite state space then you need a
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condition so that the steady state
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probabilities exist so whenever lambda
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by C mu is less than 1 whenever lambda
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by C mean is less than 1 you can find
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out the limiting probabilities so
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sometimes I use the letter P and
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sometimes I use the word by n both are 1
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of the same so you find out the steady
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state probability by solving a PI Q is
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equal to P Q is equal to 0 and the
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summation of a PA is equal to 1 and if
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you recall the birth-death process the
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steady state the probabilities the PI
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not has the 1 divided by the series
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whenever the denominator series
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converges then you will get the PS so
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either I use a pea and sapiens both are
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on the same so here summation of a PA is
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equal to 1 and P if you make a vector P
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P times a Q is equal to 0 if you solve
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that equation and the denominator of P
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naught that expression that is going to
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be converges only if lambda divided by C
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mu is less than 1 so therefore whenever
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this condition is there the queueing
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system is stable also if you put C is
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equal to 1 you will get the mm1 queue so
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using the normalizing condition you are
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getting the P naught and the P naught is
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1 divided by this so this is a series so
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this series is going to be converges
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if this condition is satisfied so by
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solving that equation so you are getting
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P and C in terms of peanut and using
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normalizing and constant you are getting
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a V naught therefore this is a steady
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state also known as the equilibrium
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solution for the m/m/c infinity model so
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here we are using the birth-death
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process with the birth rates are lambda
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and the death rates are given in the
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this form and use the same logic of the
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stationary distribution for the birth
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death process using that we are getting
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the steady state or gabriel solution for
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the MMC model
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