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 λ arrivals per time unit - that is, the inter-arrival times are assumed to be exponentially distributed with mean 1/λ. We also assume that service times are exponentially distributed with mean 1/μ. 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)=Pn(t)=Pn 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=∞∑n=0nPn where {P0,P1,…} 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: w=LλwQ=w−1μLQ=λwQ, where λ is the arrival rate and μ 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: ρ=λμL=λμ−λw=1μ−λwQ=λμ(μ−λ)LQ=λ2μ(μ−λ)Pn=(1−λμ)(λμ)n
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 λ and μ 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: ρ=λμ=2/3P0=1−λμ=1/3P1=(13)(23)=29P2=(13)(23)2=427P3=(13)(23)3=881P≥4=1−3∑n=0=1681 From the above calculations the probability that the hair stylist is busy is 1−P0=0.67 and thus the probability that he is free is 0.33. The average number of customer in the system is L=λμ−λ=23−2=2 customers The average time in the system is w=Lλ=1μ−λ=22=1 hour The average time the customer spends in the queue is wQ=w−1μ=1−13=23 hour and the number of customer in the queue is LQ=λ2μ(μ−λ)=43 customers
Let’s consider another example. Arrivals occur at rate λ=10 per hour and management has a choice of two servers, one who works at rate μ1=11 customers per hour and the second at rate μ2=12 customers per hour. The utilizations are ρ1=10/11=0.909 and ρ2=10/12=0.833. Assume a M/M/1 model, then L1=λμ1−λ=1011−10=10 whilst L2=λμ2−λ=1012−10=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!