Seventh Homework Assignment
Posted 7 Apr 2004 by herman

Due 21 April 2004 at 11:59pm.

Run a discrete-event simulation to compare serving policies. To do this, modify the program, which is a version of the program I ran in class (that one was A description of the homework is in homework7.pdf.

Confused?, posted 16 Apr 2004 by herman
There seems to be some confusion about the assignment, particularly with "queues". I think about the situation as there being one customer generator feeding into three queues: a queue for S0, a queue for S2, and a queue for S3 (actually, one could imagine there being a fourth queue, one for X, but the service time on X is so short that its queue is always empty --- because X is so fast it finishes dealing with each customer instantly).

Probably the easiest way to change the simulator in would be to put the logic for selection (S0, S1 or S2) in the fire() method of customerGenerate, since that is the event of a new customer arriving to the system. At the time the customer arrives, X chooses which server should get the new customer, so it makes sense to put this decision in fire(). Of course, that means that you need to simulate the new customer being sent to one of the servers. Study the logic of customerServe, and how it simulates a queue of customers for a single server. You'll actually need three queues, one for each server.

Still confused? Please ask questions (send email).

Solutions, posted 6 May 2004 by herman
Generally, the pattern is that the first method (random) is inferior to the second method (round robin), and the third method is worse than either of the others. Typical average queue lengths would be approximately 1.5, 1.1, and 2.2. Intuitively, the policy of the third method (choose the least busy server) should be best, but because it samples queue lengths only periodically --- as driven by the visualizer --- the information is "out of date" and it will repeat assigning to one server (the least busy at the last sampling period); this causes one server queue to build up, hence the higher average for this case.

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