>That first formula should work, where the last bins get an extra value, >but the second formula is not correct. The min function does not give >the desired result there. Anyway, it would be > >if VALUE/(Q+1) <= P, then bin number = VALUE/(Q+1) >else bin number = P + (VALUE-P*(Q+1))/Q. > >I think. Looks reasonable. It has the two linear pieces with the slopes of the number per bin in each section. One goes through the origin, and the other is offset to meet at the changeover point. This formulation does an extra division for the if-test, but it could be reformulated by comparing VALUE against a pre-calculated value (something like VALUE <= P*(Q+1)+Q). So the efficiency looks similar to the current rule. That satisfies my concern. Steven Sharp sharp@cadence.com -- This message has been scanned for viruses and dangerous content by MailScanner, and is believed to be clean.Received on Tue Oct 23 16:48:02 2007
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