Continuous Markov chains, arriving pairs

Question

I have been trying to sort out this exercise but really stuck on this. Preparing myself for exams and found many exercise on continuous Markov chains but I am always stuck when it comes to transition , rate and detailed balances equations (note that this is not an assignment or a homework just a preparation for my June exam). I have also upload my results so if you could help me with the answers by correcting and give me hints and suggest me how to deal with transition, rate and detailed balance equations

Regards

Consider the continuous time Markov chain model for a single server queue with a modification that allows for arrivals in pairs. The arrival rate is $\lambda$ and the service rate is $\mu$, but each arrival, with probability $\theta$, is of a pair of individuals rather than a single individual. The corresponding continuous time Markov chain model for the queue size has as its state space the set of all non-negative integers and the following transition rates:

transition

$ i \rightarrow i+1 \space \space rate \space \space \lambda(1-\theta) \space \space (i \geq 0) $

$ i\rightarrow i+2 \space \space rate \space \space \lambda\theta \space \space (i \geq 0) $

$ i\rightarrow i-1 \space \space rate \space \space \mu \space \space (i \geq 1) $

1. We need to find an expression for the mean number of individuals who arrive per unit time.
2. Traffic intensity
3. The balance equation and explain why the detailed balance equations cannot be valid for this model, except in the special case $\theta = 0$.

My answer is:

1. The mean is $\lambda \theta$
2. the traffic intensity is

$\lambda(1+\theta)/\mu$

Why is $\lambda(1+\theta)$

I think just because the arrivals are iid and we have

$2\lambda \theta + \lambda(1-\theta) = \lambda(1+\theta)$

1. ?

Answer 1

1. We wish to determine the rate of arriving customers, i.e. the number of customers that arrive per unit time. With probability $1-\theta$ a single customer arrives. With probability $\theta$ two customers arrive. Because the arrival process is a Poisson process (I assume this, based on the text you gave me) and we are randomly splitting this Poisson process with probability $\theta$, we now have two arrival Processes with rates $\lambda (1-\theta)$ ($1$ customer per arrival) and $\lambda \theta$ ($2$ customers per arrival), respectively. So, we have that the rate of arriving customers is $\lambda(1-\theta) \cdot 1 + \lambda\theta \cdot 2 = \lambda(1+\theta)$.

2. The service rate is $\mu$. At most, we can handle $\mu$ customers per time unit. So, the traffic intensity $\rho := \lambda(1+\theta)/\mu$, which should be smaller than $1$.

3. Let $p_i := \lim_{t \to \infty} \mathbb{P}(X(t) = i)$, where $X(t)$ is the number of customers in the system at time $t$. Thus, $p_i$ is the equilibrium probability of having $i$ customers in the system. The balance equations are now as follows. \begin{align} \lambda p_0 &= \mu p_1, \\ (\lambda + \mu)p_1 &= \lambda (1-\theta) p_0 + \mu p_2, \\ (\lambda + \mu)p_i &= \lambda \theta p_{i-2} + \lambda (1-\theta) p_{i-1} + \mu p_{i+1}, \quad i \ge 2. \end{align} And this can be solved by looking at the approach in Chapter 6 of these lecture notes. I do not see why one cannot solve this as you mentioned, but I might have overlooked something.

Answer 2

For every $\mu\gt(1+\theta)\lambda$, the generating function $g(s)=\sum\limits_{n\geqslant0}\pi_ns^n$ of the stationary distribution $(\pi_n)$ is given by $$ g(s)=\frac{\mu-(1+\theta)\lambda}{\mu-(1+\theta s)\lambda s}. $$ This follows from the balance equation, valid for every $n\geqslant0$, $$ (\lambda+\mu\mathbf 1_{n\ne0})\pi_n=\mu\pi_{n+1}+(1-\theta)\lambda\pi_{n-1}+\theta\lambda\pi_{n-2}, $$ with the convention that $\pi_{-1}=\pi_{-2}=0$. For example, the mean queue size is $$ g'(1)=\frac{(1+\theta)\lambda}{\mu-(1+\theta)\lambda}=\frac{\rho}{1-\rho}, $$ where $\rho\lt1$ is the traffic intensity, defined as $$ \rho=\frac{(1+\theta)\lambda}{\mu}. $$ Decomposing the rational fraction $g(s)$ as $$ g(s)=(1+\theta)\frac{1-\rho}{\rho\nu}\left(\frac{\alpha_+}{1-\alpha_+s}+\frac{\alpha_-}{1+\alpha_-s}\right), $$ where $$ \alpha_\pm=\frac{\rho(\nu\pm1)}{2(1+\theta)},\qquad\nu=\sqrt{1+4\theta(1+\theta)/\rho}, $$ one sees that, for every $n\geqslant0$, $$ \pi_n=(1+\theta)\frac{1-\rho}{\rho\nu}\left(\alpha_+^{n+1}+(-1)^n\alpha^{n+1}_-\right). $$ For example, $\pi_0=1-\rho$, $\pi_1=\rho(1-\rho)/(1+\theta)$, and so on.