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Published equation contexts

s=max⁡(∣x∣)2 b−1−1s = \frac{\max(|x|)}{2^{\,b-1} - 1}

Why this formula appears here

For weights, a common simplification is symmetric quantization around zero, with z=0 and the scale set directly by the largest magnitude present: s=max⁡(∣x∣)2 b−1−1s = \frac{\max(|x|)}{2^{\,b-1} - 1}. 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

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bb

Symbol b

b occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

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2 b−1−12^{\,b-1} - 1

Denominator: 2^b-1 - 1

The complete quantity below the fraction bar; it must be nonzero for this division.

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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.

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Published contexts (1)

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s=max⁡(∣x∣)2 b−1−1s = \frac{\max(|x|)}{2^{\,b-1} - 1}

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This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

For weights, a common simplification is symmetric quantization around zero, with z=0 and the scale set directly by the largest magnitude present: s=max⁡(∣x∣)2 b−1−1s = \frac{\max(|x|)}{2^{\,b-1} - 1}. 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

Meanings in this article

  • ss: the scale.
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