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The passage around this formula

Consider the two canonical all-reduce algorithms. A ring arranges participants in a cycle and performs a reduce-scatter followed by an all-gather, each in p-1 steps carrying n/p bytes: $T_{\mathrm{ring}}(n,p) = 2(p-1)\,\alpha + \frac{2(p-1)}{p}\,n\,\beta + \frac{p-1}{p}\,n\,\gamma$. The bandwidth term converges to 2n$\beta$ as p grows — it stops depending on the number of participants — which is why the ring is the bandwidth-optimal choice for large messages. NVIDIA’s own performance documentation encodes exactly this factor, defining the bus bandwidth of an all-reduce by applying a correction of 2(p-1)/p to the naive size-over-time figure, on the reasoning that an all-reduce requires 2(p-1) data transfers across p links; all-gather, reduce-scatter and all-to-all get a…

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