Appendix A – Calculation of Emax and bias

Except for the calculation of E_limit, these calculations are general for any format where the coefficient and exponent are both integers.

1. Let p be the precision (total length of the coefficient) of a format, in digits.
2. Let E_max be the maximum positive exponent for a value in a format, as defined in IEEE 854. That is, the largest possible number is
   (10^p–1) / (10^(p–1)) × 10^E_max
   For example, if p=7 this is 9.999999 × 10^E_max.
3. The exponent needed for the largest possible number is E_max–(p–1) (because, for example, the largest coefficient when p=7 is 9999999, and this only needs to be multiplied by 10^E_max / 10^(p–1) to give the largest possible number).
4. E_min=–E_max (as defined by IEEE 854 for base 10 numbers). That is, the smallest normal number is 10^E_min. The exponent needed for this number is E_min (its coefficient will be 1).
5. The number of exponents, E_normals, used for the normal numbers is therefore 2 × E_max – p + 2. (The values –E_max through –1, 0, and 1 through E_max–(p–1).)
6. Let E_tiny be the exponent of the smallest possible (tiniest, non-zero) subnormal number when expressed as a power of ten. This is E_min – (p–1).
   For example. if p=7 again, the smallest subnormal is 0.000001 × 10^E_min, which is 10^E_tiny.
   The number of exponents needed for the subnormal numbers, E_subnormals, is therefore E_min – E_tiny, which is p – 1.
7. Let E_range be the number of exponents needed for both the normal and the subnormal numbers; that is, E_normals + E_subnormals. This is (2 × E_max + 1).
8. Place E_tiny so its encoded exponent (the exponent with bias added) is 0 (the encoded exponent cannot be less than 0, and we want an all-zeros number to be valid – hence an encoded exponent of 0 must be valid).
9. Let E_limit be the maximum encoded exponent value available. For the formats in the specification, this is 3 × 2^ecbits – 1, where ecbits is the length of the exponent continuation in bits (for example, E_limit is 191 for the 32-bit format).
10. Then, the number of exponent values available is E_limit + 1, which is 3 × 2^ecbits.
11. Now, to maximize E_max, E_range = E_limit + 1
    That is, 2 × E_max + 1 = 3 × 2^ecbits.
12. Hence: E_max = (3 × 2^ecbits – 1)/2 = E_limit/2
    Note that the divisions by 2 must be truncating integer division.
13. If E_limit is odd (always the case in these encodings), one value of exponent would be unused. To make full use of the values available, E_min remains as the value just calculated, negated, and E_max is increased by one.^[1] 
    Hence: E_min = –E_limit/2
    and: E_max = E_limit/2 + 1
    (where the divisions by 2 are truncating integer division).
14. And: bias = –E_tiny = –E_min + p – 1

For example, let E_limit = 191 and p = 7 (the 32-bit format). Then:

  E_max = 191/2 + 1 = 96
  E_min = –95
  E_tiny = –101
  bias = 101

The parameters and derived values for all three formats are as follows:

Format ecbits Elimit p Emax Emin bias 32-bit 6 191 7 96 –95 101 64-bit 8 767 16 384 –383 398 128-bit 12 12287 34 6144 –6143 6176

Note that it is also possible to consider the coefficients in these formats to have a decimal point after the first digit (instead of after the last digit). With this view, the bit patterns for the layouts are identical, but the bias would be decreased by p–1, resulting in the same value for a given number.