The general area of mathematics that treats this is queueing theory,
often covered in operations research books and classes. The
information presented does not entirely allow for a computation, so
with due respect to myoarin-ga's concern, I suspect this is motivated
by either a practical or theoretical interest in the subject. Similar
problems must be addressed in scheduling of a multiprocessor
computer's CPUs.
There can be different "protocols" when more than one queue (tool) is
involved, but generally speaking if the queues (tools) are shared
among requestors, then a more efficient utilization (less wait time)
obtains. If the 100 tools are designated each to a group of three
workers, then we have done nothing to improve efficiency over of the 3
workers sharing 1 tool, as indeed we've merely multiplied that
situation a hundredfold.
Many queuing models will treat the case where the "service time", here
designated as 15 minutes, is itself a probabilistic and not
deterministic interval. That is, 15 minutes might be an average value
and not a fixed one.
In the present circumstance that three workers will each need to use
the tool only once during an eight-hour shift, for exactly 15 minutes,
one can take the "arrival times" of the three requests to be randomly
and uniformly distributed in the 8 hour interval. You can work out
the exact answer for the 3 on 1 case. Clearly the probability of
"collision" is low but not negligibly small, as the first request
removes up to half an hour from the shift, during which another
request would create a "tool" conflict.
regards, mathtalk-ga |
