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Equation 15 · The Network Is the Computer Again

What does this equation mean?

Tring(n,p)=2(p−1) α+2(p−1)p n β+p−1p n γ.T_{\mathrm{ring}}(n,p) = 2(p-1)\,\alpha + \frac{2(p-1)}{p}\,n\,\beta + \frac{p-1}{p}\,n\,\gamma .

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Start with2(p-1)
Divide byp
This relates toT_ring(n,p)
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This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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TringT_{\mathrm{ring}}

Symbol T_ring

TrT_ring is part of the quantity the equation computes from the expression on the right.

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nn

Symbol n

n is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

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pp

Symbol p

p is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

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α\alpha

Symbol α

the writing.

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β\beta

Symbol β

β is one of the signed contributions combined to compute the quantity on the left.

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γ\gamma

Symbol gamma

gamma is one of the signed contributions combined to compute the quantity on the left.

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=

=

The expressions on both sides represent the same quantity under the stated assumptions.

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fraction

fraction

Divide the expression above the line by the one below it.

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addition

addition

Add the term after the plus sign to the term or group before it.

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subtraction

subtraction

Subtract the following term or group from the preceding one. A leading minus marks a negative quantity.

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subscript

subscript

The lower label selects a particular version, component, or indexed member of the quantity. For example, x₀ and xₜ can be values at different positions.

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2(p−1)2(p-1)

Numerator: 2(p-1)

The complete quantity above the fraction bar.

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p−1p-1

Numerator: p-1

The complete quantity above the fraction bar.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction. Read it with the definitions, units, and assumptions supplied by the article.

What the article says around this equation

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: Tring(n,p)=2(p−1) α+2(p−1)p n β+p−1p n γT_{\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 2nβ\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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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: Tring(n,p)=2(p−1) α+2(p−1)p n β+p−1p n γT_{\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 2nβ\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 factor of (p-1)/p , while broadcast and reduce get a factor of one because everything must pass through a single root [ 3 ] . The point of the correction is that bus bandwidth should stay roughly constant as ranks increase, so it can be compared against hardware peak.

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