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Re: Queuing theory...

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Hello,

I will correct the following part cause i have just
translate it myself from french and it contains
some mistakes...


Here is the matematical modeling of an M/M/1 queuing system:

We already know that to satisfy a Poisson process we must
have that N(t1)- N(t0), N(t2)- N(t1) etc. must be independant
that means the counting increments must be independant.

We have the following relation between the Poisson law
and Exponential law:

the expected value E(X exponential) = 1 / E(X poisson)

so if the arrival is poissonian then the interarrivals are
exponential..

Now i will calculate the expected mean waiting time and
mean number of custumers in the queuing system:

The Little law says that the waiting time in the queuing system:

Ws = Ls/lambda

And the waiting time in the queue of the queuing system is:

Wq = Ws - 1/Mu
       =  Ls/lambda - 1/Mu

And the Little law gives us the expected mean number of custumers in the
queue:

Lq = Wq*lambda =  Ls - Phi  and  Phi = lambda/Mu

That implies:

Ls - Lq = Phi


When the system is in a stationary state , the balance equations gives:

State                          rate in = rate out

state 0:                       Mu * P1 = lambda*P0
state 1:  lambda*P0 + Mu*P2 = lambda*P1 + Mu*P1
state 2:  lambda*P1 + Mu*P3 = lambda*P2 + Mu*P2
...
state n:  lambda*Pn-1 + Mu*Pn+1 = lambda*Pn + Mu*Pn


And that gives us the following balance equations:

lambda * P0 = Mu*P1   <=>  P1 = (lambda/Mu)*P0

lambda * P1 = Mu*P2    <=>  P1 = (lambda/Mu)^2*P0

lambda * P2 = Mu*P3    <=>  P1 = (lambda/Mu)^3*P0

...

lambda * Pn-1 = Mu*Pn    <=>  P_n = (lambda/Mu)^k*P0  [1]



Note: P0 means the probability of having zero custumer in the queuing 
system and ^ means power.


And we have also that:

Sum(n=0 -> infinity) (Pn) = 1  that means the sum of probabilities equal 1

And  [1] gives us:

Sum(n=0 -> infinity) (Phi^n*P0) =1

And by resolving the sum if a geometric series that gives us:

P0 * 1/(1-Phi) = 1 => P0 = 1 - Phi  and Phi < 1

And [1] gives us:

Pn = (1-Phi) Phi^n

And we have that means the mean expected number iof custumer in the 
queing system is:

Ls = Sum(n=0 -> infinity) (n*Pn)

That implies:

Ls = Sum(n=0 -> infinity) (n*(1-Phi)*Phi^n) = Phi/1-Phi   et Phi<1

and we have the mean expected number of custumer in the queue of the 
queing system is :

Lq = Ls -Phi = Phi^2/ (1-Phi)  et Phi<1


Little law gives us the mean waiting time in the system and in the queue:

Ws = 1/lambda * Phi/(1-Phi)  and Phi<1

et

Wq= 1/lambda * Phi^2/(1-Phi)  and Phi<1




Thank you,
Amine Moulay Ramdane.

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Thread

Queuing theory... aminer <aminer@toto.net> - 2013-12-17 16:48 -0800
  Re: Queuing theory... aminer <aminer@toto.net> - 2013-12-17 17:08 -0800
  Re: Queuing theory... aminer <aminer@toto.net> - 2013-12-17 17:19 -0800
  Re: Queuing theory... aminer <aminer@toto.net> - 2013-12-17 17:31 -0800

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