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Clarification

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Sylvain Jeaugey 2018-01-30 09:17:29 -08:00
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@ -54,7 +54,7 @@ Considering that each rank has a bandwidth to the outside world of _B_, the time
`t = (S*2*(n-1)) / (n*B)`
Indeed, we have _S_ elements, 2*(n-1) operations per element, and _n_ links of bandwidth _B_ to perform them.
Reordering the elements, we find that
Reordering the equation, we find that
`t = (S/B) * (2*(n-1)/n)`