...
Steady state operation implies a distribution function of Ni that is the same for all i, thus:
RHS Term 2
which implies:
RHS Term 4
...
If the faster computer is used, the arrival rate of new jobs is λ. If the ten slower computers are used, each is responsible for 10% of the jobs arriving by a Poisson distribution at rate λ/10. Envision 10 separate queues leading to 10 separate slow computers. Each queue is "magically" assigned 10% of the customers. A job is never split and handled by multiple of the 10 computers in parallel.
Determine the fraction of time each computer is busy and the average total delay for both options (1) and (2)
One Fast Computer
So:
and:
Ten Slow Computers
and:
...
for j =1, 2, ..., P, where:
Example 1: How does the mean waiting time for class 1 customers compare to that of class 2 customers?
Given there are two classes of traffic (P = 2), let class 1 be standard data and class 2 be urgent data.
where:
As we would expect, note that:
Example 2: What are N, Nq, T, and W for each class of customer and for the system as a whole?
Consider a single-server queueing system with a nonpreemptive priority queueing discipline and two classes of customers. The arrivals for each class are Poisson with rates λ1 = 2 arrivals/second, and λ2 = 1 arrival/second, respectively.
Class 1 customers have exponential service times with an average service time of 1/4 seconds, while class 2 customers have deterministic service with a service time of 1/4 seconds. This implies μ1 = 4 services/second and μ2 = 4 services/second.
First note that:
As such, the queueing system is stable.
Class 1 Customers (w/exponential service times)
Class 2 Customers (w/deterministic service times)
So:
and:
and:
Putting it all together
Thus, we can use Little's Law to find the Nq for each class of customer:
and:
Likewise, we can solve for T and N of each class:
To analyze the entire system (comprised of both customer classes), we can solve for W, T, N, and Nq:
