# Queuing Theory problem, M/M/1/K queue with twist

Printable View

• Jul 3rd 2011, 05:49 PM
Abonimation
Queuing Theory problem, M/M/1/K queue with twist
The problem statement
Customers arrive to a register as a poisson process with arrival rate λ and are serviced with an exponential distribution with service rate μ. When a customer arrives he'll decide to join the line or not depending on how many people are currently in the system (not including himself). In other words, if at a given time n people are in the system then a new customer will join the line with probability βn, and not join the line and not take the service with probability 1-βn. However, 0 ≤ βn ≤ 1 (0 ≤ n ≤ N), βn = 0 (n > N).

(1) Find the balance equation expressed with the balance probability state pn (where n is the number of customers in the system).

(2) Solve the equation in (1).

My attempt at a solution

(1) So, my problem here is where the β will enter the problem. I'm thinking that the events "Customer stays" and "Customer arrives" are independent, resulting in the following:
With this I would get:
λ β0 p0 = μ p1
λ βn pn + μ pn = μ pn+1 + λ β[sub]n-1/SUB] pn-1 for 1 ≤ n ≤ N-1
λ βN-1 pN-1 = μ pN

Is this correct? If not, any hint?

(2) If the above is correct, how to solve it? Any hint is appreciated.
• Jul 7th 2011, 07:20 PM
Abu-Khalil
Re: Queuing Theory problem, M/M/1/K queue with twist
You are right. Those events are independent.

To solve that, just use the recursion to express any $p_n$ in terms of $p_0$ and then remember $\sum_{n=0}^N p_n=1$.
• Jul 12th 2011, 10:05 PM
Abonimation
Re: Queuing Theory problem, M/M/1/K queue with twist
Thanks! I calculated it and got a pretty messy, but hopefully correct, expression. :)