2
$\begingroup$

In queueing theory, for M/G/1 queue, there is Pollaczek-Khinchine formula to easily calculate the expected number of customers in the system by combining it with Little's law. I would like to know if I can use this approach to calculate the expected number of customers in the system of M/M/1/K queue (Poisson arrival with rate $\lambda$, Exponential service time with mean $\frac{1}{\mu}$, single server, finite capacity K).
I have tried to divide into two cases, where there are less than K customers in the system and where there are exactly K customers in the system. Then compute the expected number of customers in each case and weight average them using probabilities $p_k$ and $1-p_k$, where $p_k$ is the long-run probability there are K customers in the system. By this approach, I don't seem to get the solution which is $\frac{\rho[1-(K+1)\rho^K+K\rho^{K+1}]}{(1-\rho)(1-\rho^{K+1})}$, where $\rho$ is $\frac{\lambda}{\mu}$.
Can someone suggest which direction I should look into?

Edit: As suggested by Mick, I can calculate it directly. However, I am particularly interested in using Pollaczek-Khinchine formula and Little's law to get the expected number of customers.

$\endgroup$
2
$\begingroup$

A more straightforward way is to treat the system as a continuous time Markov chain with statespace $\{ 0,1,\ldots,K\}$, where $i$ denotes the number of customers in the system. Let $p_k(t) = \mathbb P(X_t = k)$, where $X_t$ denotes the state that the Markov chain is in at time $t$. Now from the theory we know that $$ \dot p (t) = p(t) \cdot Q $$ where $$ Q_{kl} = \begin{cases} -\lambda & k=l=0, \\ -(\lambda+\mu) & 0<l=k<K, \\ \lambda & l= k+1\leq K, \\ \mu & l=k-1, \\ -\mu & k=l=K. \end{cases} $$ We also know that the stationary distribution $\pi=(\pi_0,\pi_1,\ldots,\pi_K)$ satisfies $$ \pi\cdot Q = 0, $$ which gives us the recursion $$ \lambda\pi_{k-1}-(\mu+\lambda)\pi_k + \mu \pi_{k+1} = 0. $$ The solution of is recursion is given by $$ \pi_k = q^k \frac{q-1}{q^{K+1}-1}, \quad q = \frac{\lambda}{\mu}. $$ Now the long-term behavior of $X_t$ is described by the distribution $\pi$, hence if $N$ denotes the number of customers in the system after some time, then $$ \mathbb E N = \sum_{k=1}^K k\pi_k =\frac{q-1}{q^{K+1}-1} \sum_{k=1}^K kq^k = \frac{q(1-(K+1)q^K+Kq^{K+1})}{(q^{K+1}-1)(q-1)} $$

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. Yes, I am aware of the method to calculate the expected number of customers directly. I am trying to do it with another method. $\endgroup$
    – lovemath
    Aug 8 '20 at 20:28

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.