Queuing Exercises Sheet 1
For a G/G/1 system. (Traffic Intensity: A measure of the utilization of a station. )
For an M/M/1 system
Given: response time (r):0.050 [s]
We are looking for the throughput :
(see book p. 556)
- Queue length (n) = 10
- Maximum response time () = 2 [seconds]
- Maximum request rate ()
The maximum service time possible is .
Therefore is: [request per second].
The implication holds for
From this 2 implications we can retrieve
Both solutions are equivalent, only the notation differs.
- expected value of the queue size () = 100
- interarrival time() = 1/5 milliseconds = 0.2 milliseconds = 0.0002 seconds
We search the time until a given job is "trough the system"
Mean arrival rate = 1/0.0002 = 5000 mails per second
As the queue has always the same size, the interarrival time has to be equal to the service time. And the Mean arrival rate has to be the same as the mean service rate therefore = 5000 mails per second.
The expected value of the interarrival time (and therefore also the service time) has to be = 0.0002 seconds.
This give us = 0.0202 [seconds].
This solution is based on the official given solution during the course of 2009.
Here the PDF-2009
Using Little's Law we find:
--Giulio 14:31, 31. Jan. 2012 (CET)
Where we did not use . I guess it does not make sense to say that this still holds, even if the traffic intensity is 0.4. I would rather use the formulas for M/M/1, which indicate that , and since holds, can't be.
Given: and (from the previous exercise)
From the previous exercise (the alternative solution) we know that , therefore
--Giulio 14:33, 31. Jan. 2012 (CET)