SOLUTION FOR ASSIGNMENT #3                                       Spring 1999
--------------------------

For 70% utilization we get (with a simulation run time of 1.0e8 time units):

  3 buffers: 3.41% loss
  4 buffers: 1.67% loss

So, 4 packet buffers, or 4 * 53 bytes = 212 bytes is needed

For 80% utilization we get (with a simulation run time of 1.0e8 time units):

  5 buffers: 2.19% loss
  6 buffers: 1.38% loss

So, 6 packet buffers, or 6 * 53 bytes = 318 bytes is needed

For 90% utilization we get (with a simulation run time of 1.0e8 time units):

  8 buffers: 2.12% loss
  9 buffers: 1.66% loss

So, 9 packet buffers, or 9 * 53 bytes = 477 bytes is needed

============================================================================

EXTRA CREDIT:

We use the Pollaczek-Khinchin (P-K) formula for an M/G/1 queue.  See any
textbook on queueing to find a derivation and discussion of the P-K formula.
The P-K formula is,

  N = rho + (lambda^2 * (second moment of service time)) / (2 * (1 - rho))

where rho is the utilization, lambda is the arrival rate, and the second moment
of the service time is just that.  For deterministic service times, the second
moment is simply the first moment (the mean) squared.

Using 1.0 as the service time, the arrival rate is then equal to rho.  We compute,

70% utilization:

  N = 0.70 * ((0.70^2) * (1^2)) / (2 * (1 - 0.70)) = 1.517 packets in the system

80% utilization:

  N = 2.400 packet in the system

90% utilization

  N = 4.950 packets in the system

An excellent undergraduate level text for performance modeling is:

  M. Molloy, "Fundamentals of Performance Modeling," Macmillan Publishing
  Company, New York, 1989.

---