It All Fits In RAM Now

Your team has been quoting the classic argument: a B-tree with fanout 509 finds a key among $10^9$ in 4 block reads, while a balanced binary tree needs about 30, so use the B-tree. Then the machine gets 512 GB of RAM and the entire index is resident. Someone proposes replacing the B-tree with a plain balanced binary tree, arguing the disk argument no longer applies.

What is the strongest response?

They are wrong, because a hash table strictly dominates any tree once the data is resident.

The argument survives with the units swapped: the transfer unit becomes the cache line rather than the disk block, so a wide node that fills whole cache lines still beats 30 dependent pointer chases through scattered memory.

The argument holds unchanged. The external memory model counts transfers of a block of size B, and that accounting says nothing about which medium the block lives on, so the 4 reads against 30 comparison carries over to RAM exactly as it stands.

They are right on the numbers. In RAM every access costs the same, so what matters is comparison count, and both structures do about 30 comparisons.