...
Thus, we can apply Little's Law scoped at the queue as:
Computing the
...
Utilization (i.e. Pr(server is busy))
The average service time is the inverse of the service rate, or 1/μ. Thus:
...
Thus, T = T1 + T2 + T3 = 0.883 + 0.367 + 0.867 = 2.067 seconds.
Utilization
Recall that utilization is the average fraction of time the system is busy. For a single server queueing system, the utilization is the average fraction of time the server is busy.
We can characterize the utilization in terms of the arrival rate λ and the service rate μ.
Method 1
The measured utilization during the interval [0, t'] is:
For a stable system operating in steady state:
Recall that:
where:
Method 2
Consider a system with a FCFS service and a single output channel.
We can remove the FCFS constraint, but analysis is more complex. So, let Ni denote the number of customers in the network just after the departure of the ith customer and let Ai+1 denote the number of arrivals that occur while customer i+1 is being served.
Case 1
Ni > 0, meaning the network is not empty when the ith customer departs.
The service for the (i+1)st customer begins immediately after the departure of the ith customer, so Ni+1 = Ni + Ai+1 - 1.
Case 2
Ni = 0, meaning the network is empty when the ith customer departs.
The network remains empty until the arrival of the (i+1)st customer, so Ni+1 = Ni + 1 + Ai+1 - 1 = 0 + 1 + Ai+1 - 1 = Ai+1.
Let's define an indicator function to denote whether or not the network is empty:
Combining the expressions for cases 1 and 2 into a single expression, we can write Ni+1 = Ni - U(Ni) + Ai+1, where Ni+1, Ni, and Ai+1 are random variables and U(Ni) is a random variable.
This means that E[Ni+1] = E[Ni] - E[U(Ni)] + E[Ai+1]. Note that E[U(Ni)] = Pr(Ni > 0).
We assume a steady state network, which means E[Ni] = E[Ni+1]. Thus, E[Ni] = E[Ni] - E[U(Ni)] + E[Ai+1] and E[U(Ni)] = E[Ai+1]. This result indicates the probability that the server is busy immediately after a departure is equal to the average number of arrivals that occur in one service time.
Our goal is to find the utilization (ρ), or the fraction of time the server is busy, or the probability the server is busy at an arbitrary time instant (all meaning the same thing).
When does ρ = E[U(Ni)]?
Deterministic Arrivals
Assume a D/D/1 queueing system (recall D = deterministic). The above model sports one arrival per second and has a service time of 1/2 second per customer.
Ni = 0 for all i, so E[U(Ni)] = E[0] = 0. But, ρ = 1/2. So, E[U(Ni)] != ρ for this D/D/1 queueing system.
Poisson Arrivals
However, it can be shown that for Poisson arrivals, E[U(Ni)] = ρ. Thus, for Poisson arrivals, ρ = E[U(Ni)] = E[Ai+1].
This means the utilization of the server is equal to the average number of arrivals during a service interval.
We will use this fact to evaluate the performance of queueing systems with Poisson arrivals.