Another improvement to this strategy is to read as many blocks as possible of the
smaller relation, R say, into the database buffer, saving one block for the inner relation, and
one for the result relation. If the buffer can hold nBuffer blocks, then we should read
(nBuffer − 2) blocks from R into the buffer at a time, and one block from S. The total
number of R blocks accessed is still nBlocks(R), but the total number of S blocks read is
reduced to approximately [nBlocks(S)*(nBlocks(R)/(nBuffer − 2))]. With this approach,
the new cost estimate becomes: