What does this equation mean?

separate integrations, because nothing built for one pair transfers to any other. A protocol that both sides implement exactly once instead collapses this to $C_{\text{standardized}} = M + N$. The assumption a shared protocol is betting on is that the fixed cost of building and maintaining the standard itself is smaller than the integrations it removes, and that those removed integrations were genuinely redundant rather than differently necessary. When M and N are both large and still growing, the gap between M $\times$ N and M + N dominates every other consideration; when either is small — a single application talking to a handful of tools it controls — a bespoke integration can still be cheaper than adopting a shared protocol at all, and treating standardization as free in that case would be cargo-culting rather than economizing.