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The passage around this formula

The repair is to restructure the loops so that the working set of the innermost computation is small enough to sit in a fast store and is used many times before being evicted. Partition the operands into b $\times$ b blocks and compute block by block. Three such blocks — two inputs and an accumulator — must be resident simultaneously, so the tile size is bounded by the fast-store capacity M in words: $Q(b) \;\approx\; \frac{2N^3}{b} + N^2, \qquad \text{subject to} \quad 3b^2 \le M$. Taking b as large as the constraint allows gives

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