MAXIMUM (Y): lose P2's subtraction entirely by having both additions apply on top of each other with P2's write squeezed in between and then overwritten: P1 reads 100; P2 reads 100, writes 50 (based on its own read); P1 (using its earlier stale read of 100) writes 120, discarding P2's update; P3 reads 120, writes 130. Y=130. MINIMUM (X): lose BOTH additions by having P2 read the ORIGINAL 100 early, then P1 and P3 both apply in sequence (100→120→130), and finally P2 writes its stale result 100-50=50 LAST, overwriting everything. X=50 (this is also the theoretical floor, since P2's own read value can never be below the initial 100 given only additions could have preceded it). Y-X = 130-50 = 80.