1
$\begingroup$

I have a two-server queue with Poisson arrival rate and $\lambda$ exponential services with $\mu$ ( first server service rate) and 2$\mu$ ( 2nd server service rate). Capacity is infinite.

Then why is the number of customers in the queue at time $t$ not a Markov Process?

Can you please help me out?

Thank You.

$\endgroup$
2
$\begingroup$

You need extra information to make the process a Markov process. The Markov property requires that the future depend only on the current state, but suppose you have only a single customer in the system. The future behaviour of the system depends on the history of the process (namely which server that job started service with) not just the current state. (It could experience service at rate either $\mu$ or $2\mu$.)

$\endgroup$
0
$\begingroup$

When there is one customer in the system, we need to know which server is serving that customer. So let the state space be $S=\{0,1_A,1_B\}\cup\{2,3,4,\ldots\}$, then $\{X(t):t\geqslant 0\}$ is a continuous-time Markov chain modelling the queueing system. Assume that customers arriving to an empty system always choose the faster server (denoted by state $1_B$. The global balance equations are given by \begin{align} \lambda\pi_0 &= \mu\pi_{1_A} + 2\mu\pi_{1_B}\\ (\lambda+\mu)\pi_{1_A} &= 2\mu\pi_2\\ (\lambda+2\mu)\pi_{1_B} &= \lambda\pi_0+\mu\pi_2\\ \lambda\pi_n &= 3\mu\pi_{n+1},\ \ n\geqslant 2. \end{align} The first three equations yield $$ \pi_{1_A} = \frac{\lambda^2}{\mu(2\lambda+3\mu)}\pi_0,\ \ \pi_{1_B} = \frac{\lambda ^2+3 \lambda \mu }{2 \mu (2 \lambda +3 \mu )}\pi_0, \ \ \pi_2 = \frac{\lambda ^2 (\lambda +\mu )}{2 \mu ^2 (2 \lambda +3 \mu )}\pi_0. $$ The rest of the balance equations yield the recursion $$\pi_n = \left(\frac\lambda{3\mu}\right)^{n-2}\pi_2 = \left(\frac\lambda{3\mu}\right)^{n-2} \frac{\lambda ^2 (\lambda +\mu )}{2 \mu ^2 (2 \lambda +3 \mu )}\pi_0,\ n\geqslant 3.$$ From $\sum_{n=0}^\infty \pi_n=1$ we have \begin{align} \pi_0 &= \left(1 + \frac{\lambda^2}{\mu(2\lambda+3\mu)}+\frac{\lambda ^2+3 \lambda \mu }{2 \mu (2 \lambda +3 \mu )} + \sum_{n=2}^\infty \left(\frac\lambda{3\mu}\right)^{n-2} \frac{\lambda ^2 (\lambda +\mu )}{2 \mu ^2 (2 \lambda +3 \mu )} \right)^{-1}\\ &= 1-\frac{9 \lambda (\lambda +\mu )}{5 \lambda ^2+15 \lambda \mu +18 \mu ^2}. \end{align} It follows that \begin{align} \pi_{1_A} &= \frac{\lambda^2}{\mu(2\lambda+3\mu)}\left(1-\frac{9 \lambda (\lambda +\mu )}{5 \lambda ^2+15 \lambda \mu +18 \mu ^2}\right) = \frac{2 \lambda ^2 (3\mu-\lambda)}{\mu \left(5 \lambda ^2+15 \lambda \mu +18 \mu ^2\right)}\\ \pi_{1_B} &= \frac{\lambda ^2+3 \lambda \mu }{2 \mu (2 \lambda +3 \mu )}\left(1-\frac{9 \lambda (\lambda +\mu )}{5 \lambda ^2+15 \lambda \mu +18 \mu ^2}\right) = \frac{9 \lambda \mu ^2-\lambda^3}{5 \lambda ^2 \mu +15 \lambda \mu ^2+18 \mu ^3}\\ \pi_n &= \left(\frac\lambda{3\mu}\right)^{n-2} \frac{\lambda ^2 (\lambda +\mu )}{2 \mu ^2 (2 \lambda +3 \mu )}\left(1-\frac{9 \lambda (\lambda +\mu )}{5 \lambda ^2+15 \lambda \mu +18 \mu ^2}\right) = \frac{3^{2-n} (3\mu-\lambda) (\lambda +\mu ) \left(\frac{\lambda }{\mu }\right)^n}{5 \lambda ^2+15 \lambda \mu +18 \mu ^2}. \end{align} The expected size in system is \begin{align} L &= \pi_{1_A}+\pi_{1_B}+\sum_{n=2}^\infty n\pi_n\\ &= \frac{2 \lambda ^2 (3\mu-\lambda)}{\mu \left(5 \lambda ^2+15 \lambda \mu +18 \mu ^2\right)} + \frac{9 \lambda \mu ^2-\lambda^3}{5 \lambda ^2 \mu +15 \lambda \mu ^2+18 \mu ^3} + \sum_{n=2}^\infty n\frac{3^{2-n} (3\mu-\lambda) (\lambda +\mu ) \left(\frac{\lambda }{\mu }\right)^n}{5 \lambda ^2+15 \lambda \mu +18 \mu ^2}\\ &= \frac{27 \lambda \mu (\lambda +\mu )}{-5 \lambda ^3+27 \lambda \mu ^2+54 \mu ^3}. \end{align} By Little's law, the expected sojourn time of a customer is $$ W = \frac L\lambda =\frac{27 \lambda \mu (\lambda +\mu )}{-5 \lambda ^4+27 \lambda^2 \mu ^2+54\lambda \mu ^3}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.