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My question has a queue M/M/1/2, that is, a system with exponential interarrival and service times, one server and having a room only for 2 customers (including the one in service, and another that is waiting). Let the number of customers in the room at time $t$ be $X(t)$. I model this as a birth and death process with transition rate matrix being

$$\begin{bmatrix} -\lambda & \lambda & 0 \\\\ \mu & -\lambda - \mu & \lambda \\\\ 0&\mu&-\mu \end{bmatrix}.$$

I have trouble to understand and compute:

  1. "the average number of customers in the room in the long run"

    Is this to compute $\lim_{t \rightarrow \infty} X(t)/t$, or expectation of the limit distribution of $X(t)$ as $t \rightarrow \infty$?

  2. "the proportion of potential customers that enter the room in the long run"

    How is this represented in formula?

  3. "If the server works twice faster, how much more business it will do"

    How is "how much more business" represented in formula?

For each part, are there some theorems that can help me get started?

Thank you so much!

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Maybe business means the number of customers served? If a served customers also pays $n$ amount of monetary units, I guess it's $\text{(#customers served)}n$, no? – Stijn May 2 '11 at 13:24
@Stijn: Thank you! The first sentence makes sense. I was wondering how I shall tackle the questions? – Jessie May 2 '11 at 13:26

2 Answers 2

up vote 2 down vote accepted

If we look at $N(t)$ to be the number of people in the room, then $S = \left\{ 0,1,2 \right\}$ with transitions of $\lambda$ from $0 \rightarrow 1$ and $1 \rightarrow 2$, and transitions of $\mu$ from $2 \rightarrow 1$ and $1 \rightarrow 0$. By using Cuts in the rate diagram, we can find the stationary distribution since the rate diagram is irreducible. We get the equations: $$\lambda \pi_0 = \mu \pi_1$$ $$\lambda \pi_1 = \mu \pi_2$$ $$\pi_0 + \pi_1 + \pi_2 = 1$$ where $\pi = (\pi_0 , \pi_1 , \pi_2)$ is the steady state. $$\Rightarrow \pi = \left(\frac{1}{1+ \lambda/\mu + (\lambda/\mu)^2},\frac{\lambda/\mu}{1+ \lambda/\mu + (\lambda/\mu)^2},\frac{(\lambda/\mu)^2}{1+ \lambda/\mu + (\lambda/\mu)^2}\right)$$ Thus the solution to number 1 is: $$E[x] = 0 \cdot \frac{1}{1+ \lambda/\mu + (\lambda/\mu)^2} + 1 \cdot \frac{\lambda/\mu}{1+ \lambda/\mu + (\lambda/\mu)^2} + 2 \cdot \frac{(\lambda/\mu)^2}{1+ \lambda/\mu + (\lambda/\mu)^2}$$ $$= \frac{\lambda/\mu + 2(\lambda/\mu)^2}{1+ \lambda/\mu + (\lambda/\mu)^2}$$ Problem number 2 is a littler trickier and I'll have to think about it some more (I want to say that it's just $\pi_0 + \pi_1$), but Problem number 3 is recalculating the stationary distribution for the new $\mu'$ and finding out the proportion of potential customers that enter the room in the long run in comparison to the solution to number 2 when we used the old $\mu$.

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Thanks! I think I made some error when writing down the rate matrix. Each of its row should be summed up to be 0. Just corrected. – Jessie May 2 '11 at 18:20

I took the liberty of expanding on my comment in this (partial) answer. Maybe it will help you out.

1) The reader I use for my Queueing Theory course denotes by $P(L=n)$ the probability of the number of customers when the system is in equilibrium. That is; $P(L=n) = \lim_{t\to \infty} P(L(t) = n)$ where $L(t)$ denotes the number in the system at time $t$.

2) The way I interpret the question is that a potential customer can join the room and then leave if the room is full or join the queue if it is not. According to the PASTA property (=Poisson Arrivals See Time Averages, see, for example,, this is equal to $p_c$, the probability of $c$ customers in the system (for a $M/M/c/c$ queue).

The reader I use is available online:

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