## 7.5 Steady-State Behavior of the M/M/1 Model

Henceforth, we assume that the calling population is infinite, the arrivals are assumed to follow a Poisson process with rate $$\lambda$$ arrivals per time unit - that is, the inter-arrival times are assumed to be exponentially distributed with mean $$1/\lambda$$. We also assume that service times are exponentially distributed with mean $$1/\mu$$. The queue discipline will be FIFO. Because of the exponential distribution assumption on the arrival process, these models are called Markovian.

A queuing system is said to be in statistical equilibrium or in steady state if the probability that the system is in a given state is not time-dependent: that is $P(L(t)=n)= P_n(t) = P_n$ is independent of time $$t$$. Two properties of such systems:

• they approach statistical equilibrium from any starting state,

• they remain in statistical equilibrium once it is reached.

For the models that we will consider next the steady-state parameter $$L$$, the average number of customers in the system, can be computed as $L= \sum_{n=0}^\infty nP_n$ where $$\{P_0,P_1,\dots\}$$ are the steady-state probabilities of finding $$n$$ customers in the system. Once $$L$$ is given, the other steady-state parameters can be computed easily from Little’s law: \begin{align*} w &= \frac{L}{\lambda}\\ w_Q &= w - \frac{1}{\mu}\\ L_Q &= \lambda w_Q, \end{align*} where $$\lambda$$ is the arrival rate and $$\mu$$ is the service rate.

Recall that for queues to reach statistical equilibrium a necessary and sufficient condition is that $/c<1$.

For the M/M/1 model the steady-state parameters can be shown to be equal to: \begin{align*} \rho &= \frac{\lambda}{\mu} \\ L &= \frac{\lambda}{\mu-\lambda}\\ w &= \frac{1}{\mu-\lambda}\\ w_Q &= \frac{\lambda}{\mu(\mu-\lambda)}\\ L_Q &= \frac{\lambda^2}{\mu(\mu-\lambda)}\\ P_n &= \left(1-\frac{\lambda}{\mu}\right)\left(\frac{\lambda}{\mu}\right)^n \end{align*}

Let’s consider an example. Suppose that the interarrival times and service times of a single-chair unisex hair-styling shop have been shown to be exponentially distributed. The values of $$\lambda$$ and $$\mu$$ are 2 per hour and 3 per hour, respectively: that is, the time between arrivals averages 1/2 hour and the service time averages 20 minutes. The server utilization and the probabilities for 0, 1, 2, 3, and 4 or more customers in the shop are computed as follows: \begin{align*} \rho &= \frac{\lambda}{\mu}=2/3\\ P_0 &= 1- \frac{\lambda}{\mu} = 1/3\\ P_1 &= \left(\frac{1}{3}\right) \left(\frac{2}{3}\right) =\frac{2}{9}\\ P_2 &= \left(\frac{1}{3}\right) \left(\frac{2}{3}\right)^2 =\frac{4}{27}\\ P_3 &= \left(\frac{1}{3}\right) \left(\frac{2}{3}\right)^3 =\frac{8}{81}\\ P_{\geq 4}&= 1-\sum_{n=0}^3=\frac{16}{81} \end{align*} From the above calculations the probability that the hair stylist is busy is $$1-P_0=0.67$$ and thus the probability that he is free is $$0.33$$. The average number of customer in the system is $L=\frac{\lambda}{\mu-\lambda}=\frac{2}{3-2}= 2 \mbox{ customers}$ The average time in the system is $w = \frac{L}{\lambda}=\frac{1}{\mu-\lambda} = \frac{2}{2}=1 \mbox{ hour}$ The average time the customer spends in the queue is $w_Q = w - \frac{1}{\mu}= 1- \frac{1}{3}= \frac{2}{3} \mbox{ hour}$ and the number of customer in the queue is $L_Q = \frac{\lambda^2}{\mu(\mu-\lambda)}=\frac{4}{3} \mbox{ customers}$

Let’s consider another example. Arrivals occur at rate $$\lambda = 10$$ per hour and management has a choice of two servers, one who works at rate $$\mu_1=11$$ customers per hour and the second at rate $$\mu_2 = 12$$ customers per hour. The utilizations are $$\rho_1 = 10/11=0.909$$ and $$\rho_2 = 10/12 = 0.833$$. Assume a M/M/1 model, then $L_1 = \frac{\lambda}{\mu_1-\lambda}=\frac{10}{11-10}=10$ whilst $L_2 = \frac{\lambda}{\mu_2-\lambda}=\frac{10}{12-10}=5$ Thus an increase in service rate from 11 to 12 customers per hour, a mere 9.1% increase, would result in a decrease in average number in system from 10 to 5, which is a 50% decrease!