Sixteen Bits, Split Two Ways

A team wants to halve a model's memory by storing its numbers in 16 bits instead of 32. Two 16-bit formats are on the table, and they split their bits differently.

Every format below stores a number as

\pm\,(1 + f) \times 2^{e}

$e$ is a whole-number exponent, held in the exponent bits. Each format allows only a fixed range of $e$ .
$f$ is a fraction held in the $M$ mantissa bits, so it can only be one of $0,\ \tfrac{1}{2^M},\ \tfrac{2}{2^M},\ \dots,\ \tfrac{2^M-1}{2^M}$ .

A value that falls between two storable numbers is rounded to the nearer one. A value bigger than the largest storable number becomes infinity, which is called an overflow.

format	total bits	exponent bits	allowed $e$	mantissa bits $M$
FP32	32	8	$-126$ to $127$	23
FP16	16	5	$-14$ to $15$	10
BF16	16	8	$-126$ to $127$	7
FP8 (E5M2)	8	5	$-14$ to $15$	2

Two values from a training run have to be stored:

An activation of $90{,}000$ , produced by one unusually loud layer.
A weight that was exactly $3.0$ , just after the optimizer adds $0.005$ to it. Its true new value is $3.005$ .

What does each 16-bit format actually store for these two values?

Sixteen Bits, Split Two Ways

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