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Equation 31 · Part 4 · How a Model Actually Gets Small Enough to Run on a Phone

Symbol s^2

E ⁣[(x−x^)2]≈s212\mathbb{E}\!\left[(x - \hat{x})^2\right] \approx \frac{s^2}{12}
s2s^2

What this part means

the square of s; the scale.

Its job in the formula

s2s^2 occurs above the fraction bar. The numerator is divided by the entire denominator below it.

Where the article explains it

The standard affine mapping takes a real value x , a scale s and a zero-point z , and produces an integer xqx_q = clip\mathrm{clip}(⌊xs⌉+z, qmin⁡, qmax⁡)\left(\left\lfloor \frac{x}{s} \right\rceil + z,\ q_{\min},\ q_{\max}\right), \qquad x^\hat{x} = s(xq−z)\left(x_q - z\right) where ⌊\lfloor ⋅\cdot ⌉\rceil is round-to-nearest and x^\hat{x} is the dequantised approximation actually used in arithmetic.

The passage around this formula

for a signed integer of b bits. An 8-bit integer has 2^8=256 representable levels; a 4-bit integer has 2^4=16 . That sixteen-fold reduction in available codes is the entire cost of quantization in one number. Treating the rounding error as approximately uniform over one quantization step s , its expected squared magnitude is the classical result E ⁣[(x−x^)2]≈s212\mathbb{E}\!\left[(x - \hat{x})^2\right] \approx \frac{s^2}{12}. so halving the number of bits, which roughly doubles s at fixed range, roughly quadruples the expected squared error per weight. That is why INT8 is usually described as close to free and INT4 is not: the error a network has to absorb does not grow gently as bits are removed, it grows quadratically in the step size, and every bit…

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

Open the illustrated exponents: repeated multiplication and powers guide →

See this notation across published equations →

Sources cited in the article section

These citations provide research context; check each source for the exact claim it supports.