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The passage around this formula

But look at the latency term: 2(p-1)$\alpha$ grows linearly in the number of participants. A recursive-doubling or tree scheme instead completes in $\lceil$ $\log_2$ p $\rceil$ rounds: $T_{\mathrm{rd}}(n,p) = \lceil \log_2 p \rceil \,\alpha + \lceil \log_2 p \rceil \, n\,(\beta + \gamma)$. paying a logarithmic latency term but sending the full message in every round, so its bandwidth cost grows with $\log$ p rather than staying flat.

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