How do I derive the steady state probabilities for the $M/M/1/k$ queueing systems with finite system capacity $k$?

  • 2
    $\begingroup$ What you are describing is usually called an $M/M/1/k$ queuing system. (Sometimes a different letter is used in place of $k$.) Here, $k$ refers to the capacity. Most textbooks (and probably most websites) that derive steady-state probabilities for $M/M/1$ will also do the same for $M/M/1/k$. $\endgroup$ – LarrySnyder610 Sep 20 '20 at 13:01

There are a couple of ways to derive the steady state probabilities for a $M/M/1/k$ queuing system with Markovian* arrivals (the first $M$), exponential service time distribution (the second $M$), a single server (the 1), and a finite total system capacity of $k$. Note this implies the queue can be at most $k-1$.

*Recall a system with $M$ arrivals has exponential interarrivals with rate $\lambda$ and mean interarrival time of $\frac{1}{\lambda}$. Equivalently, arrivals occur according to a homogeneous Poisson process with rate $\lambda$.

One approach
The $M/M/1/k$ queuing system can be thought of as a Continuous Time Markov Chain (CTMC), $\{X(t), t>=0\}$, where $X(t)$ represents the total number of customers in the system. We can take advantage of the system capacity (constraint) being finite.

If the arrivals happen at rate $\lambda$ and the service rate is $\mu$, then the transition rate diagram is given below. Note the finite number of states, with the statespace $\mathcal S = \{0,1,2,3,\ldots,k\}$.

Transition rate diagram

The steady state probabilities, $\mathbf \pi = [\pi_0\; \pi_1\; \pi_2\; \ldots\; \pi_k]$, are obtained by the solution to the steady-state equations (flow out = flow in),

$$ \lambda \pi_0 = \mu \pi_1 \\ (\lambda+\mu) \pi_1 = \lambda \pi_0 + \mu \pi_2 \\ (\lambda+\mu) \pi_2 = \lambda \pi_1 + \mu \pi_3 \\ \vdots \\ \lambda \pi_{k-1} = \mu \pi_k $$

and the normalization equation $\sum_{i=0}^k \pi_k = 1$. The steady-state equations contain the dependency pattern of a Birth-Death Process, with the final equation being modified due to the finite state space (though this will not hamper the solution).

Ignoring the requirement for the $\pi$'s to sum to 1 for now, set $\pi_0 = 1$. Then $\hat \pi_j = \left(\frac{\lambda}{\mu}\right)^j$, or letting $\rho=\frac{\lambda}{\mu}$, then $\hat \pi = \rho^j$.

To normalize the solution, we need to divide the $\hat \pi$'s by the normalization constant $G = \hat \pi_0 + \hat \pi_1 + \hat \pi_2 +\hat \pi_3 + \cdots + \hat \pi_k$, which is a finite sum because we have a finite state space.

Rearranging, $$ \begin{align} G &= \hat \pi_0 + \hat \pi_1 + \hat \pi_2 +\hat \pi_3 + \cdots \hat \pi_k \\ &= 1 + \rho + \rho^2 + \rho^3 + \cdots + \rho^k \\ &= \left(\frac{1-\rho^{k+1}}{1-\rho} \right) \end{align} $$ which gives $$\frac{1}{G} = \frac{1-\rho}{1-\rho^{k+1}}$$ which is valid for any value of $\rho \in [0,1)$ with the special case of $\rho=1$ requiring the trivial $G = k+1$ yielding $$\frac{1}{G}=\frac{1}{k+1}$$ for this exception case.

This implies that $$ \begin{array}{ll} \pi_j = \rho^j\left(\frac{1-\rho}{1-\rho^{k+1}}\right) & \text{for } \rho\ne 1 \tag{Key Result} \\ \pi_j = \frac{1}{k+1} & \text{for } \rho = 1 \\ \end{array} $$


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