Originally Posted by
Kacee
It shows that if there's IN/I, there will be R. But that's a given. If that's not true, it's an inventory error.
There will be many, many times, however, when flights are R>0 and IN0, I0. Using your theory, you'll miss a very high percentage of upgradeable flights.
Example:
Yup.
I never watch waitlisted flights when P=0. When PN > 0, I start to get interested. If PN=9, an UG is almost a certainty.
P opens before PN which opens before R which opens well before I(N).
So, it is absolutely true that the existence of I space in predictive of R, but the converse is not true. Consider IN a "square" and R a "rectangle". If you're only searching for squares, you're gonna miss a lot of rectangles. Focus on the parallelograms (PN) instead (which you can't search for, I don't think) for better indicators.