Currently, the "turn on a reactor button" currently seeks (or at least is supposed to seek) the most resource effecient reactor when it turns on, keeping in mind the poor efficiency due to multiple reactors on same planet effect. However, what if there is a tie for most efficient, and due to poor efficiency shenanigans, it happens that the tie is between a Mk. III or a poor efficiency (when turned on) Mk. II. Which one should it choose?
Well, IMO, the tie breaker between two power reactors of the same efficiency should be the reactor that costs less per second to keep on. In the case mentioned above, the poor efficiency Mk. II reactor would be turned on under this logic. Possibly, if ties with efficiency are very rare, maybe if the efficiencies are within 5% (or something) of each other, just call it a tie and use the tie breaker logic.
On a related note, an oversight in the "find reactor to turn off" logic. Currently it seeks to turn off the least efficient reactor. Problem is, if it can't afford to turn it off (like a Mk. III and turning it off would bring energy into the negatives), it doesn't try the second least efficient reactor, rather, it just gives up.