Name: Anonymous 2007-03-08 6:53 ID:LVfE4/xi
Assuming a process can be carried out in an infinite number of ways, and assuming that a function
f: (A € N) --> (B € R) maps a process to a numerical ranking (i.e, higher is better), then a process can be optimal
without step i. ?????????
and with step j. Profit!
Proof.
Profiting is better than not profiting. Therefore, of two procedures differing only in the inclusion of a profit step, the profit one will be better.
Even if there are an infinite number of ways to do something, there can be a point of maximum efficacy, where including an extra step will only reduce efficiency. Thus, if the inclusion of ?????? is this extra step, then reduction from optimal to sub-optimal performance may result.
f: (A € N) --> (B € R) maps a process to a numerical ranking (i.e, higher is better), then a process can be optimal
without step i. ?????????
and with step j. Profit!
Proof.
Profiting is better than not profiting. Therefore, of two procedures differing only in the inclusion of a profit step, the profit one will be better.
Even if there are an infinite number of ways to do something, there can be a point of maximum efficacy, where including an extra step will only reduce efficiency. Thus, if the inclusion of ?????? is this extra step, then reduction from optimal to sub-optimal performance may result.