floating point representation

Question

I faced a question that asked to choose the Correct Number Representation for Summation of Float32 × Int32

$$ A = \sum_{n=0}^{127} b[n] \cdot c[n] $$

Given:

• b is an array of 32-bit IEEE-754 floating-point values (float32).
• c is an array of 32-bit signed integers (int32, using two's complement).
• both b and c are natural numbers (i.e. positive integers).

My Understanding So Far:

• b[n] can range approximately between:

  $$ (1.0 \times 2^{0}) \quad \text{to} \quad (1.111\dots1_2 \times 2^{127}) \approx 2^{128} - 2^{104} $$

• c[n] can range from \$1\$ to \$2^{31} - 1\$.

• Therefore, the maximum possible product \$b[n] \cdot c[n]\$ can be roughly:

  $$ (2^{128} - 2^{104}) \cdot (2^{31} - 1) \approx 2^{159} $$

• Summing 128 such terms could reach as high as:

  $$ 128 \cdot 2^{159} \approx 2^{166} $$

Now, considering representations:

1. A 192-bit signed integer (int192) can represent values up to \$2^{191} - 1\$, which should cover the entire possible result exactly.
2. A 64-bit float (float64) has 11 bits exponent one sign the rest mantissa

Question:

Choices

• (A) int64 will always give the exact result
• (B) int192 will always give the exact result
• (C) float32 will give a correct result, but possibly with rounding
• (D) float64 will give a correct result, but possibly with rounding
• (E) Both B and D are correct
• (F) None of the above allow a correct result, even with rounding

The correct answer was (E). I did not understand why the exponent bits are enough and why we only have a problem with the rounding, so I would appreciate an explanation regarding the floating point representation range.

Answer 1

Exponent bits are 11 for float64, so the signed number range for the exponent does appear to be adequate. It's going to be approximately \$2^{1024}\$ maximum for the multiplier and you only need \$2^{166}\$ .

As far as rounding- there are 52 bits in the mantissa of a float64 number and you are generating a result with 32+23 = 55 significant bits, so there are not enough to exactly represent all possible results.