On Tue, 26 Oct 2021 22:21:23 -0400
Steven Rostedt [off-list ref] wrote:

I'm sure there's an algorithm somewhere that can give as the real max.

You got me playing with this more ;-)

OK, I added the rounding in the wrong place. I found that we can make
the max_div to be the same as the shift! The bigger the shift, the
bigger the max!

	mult = (1 << shift) / div;
	max_div = (1 << shift)

But the rounding needs to be with the mult / shift:

	return (val * mult + ((1 << shift) - 1)) >> shift;


When val goes pass 1 << shift, then the error will be off by more than
one.

-- Steve