At 03:06 PM 2/11/01 -0500, you wrote:
    > >If processes A and B are on the ready queue (A is running), and C arrives
    > >(external event causing A to be interrupted before the end of its
    > >timeslice).  After processing, the ready queue looks like:
    >
    > >B C A (with B running).
    >
    >
    >Essentially this means that the arrival event pre-empts the reinsertion
    >of A into the ready queue. On a display event however, the interruption
    >of the running process and resultant reinsertion into the ready queue
    >is processed first before the display correct?
    
    It doesn't matter where you show it.
    
    
    >And finally, in the example above, if C was arriving at the end of A's
    >timeslice I assume the ready queue would look like:
    >
    >B A C (with B running)
    
    Another coin flip, yes.
    
    
    
    
    >Because we process the end of the timeslice first sending A to the end
    >of the ready queue, then process the arrival of C. Because of what you
    >said in class about external events at the same time as internal events not
    >causing the process to leave the cpu, B which just arrived on the CPU as a
    >result
    >of A's timeslice ending does not get interrupted, and C simply get's stuck
    >on the back of the ready queue.