GATE CS 2024 Set 2 Q.14: ALU, Datapath and Control Unit

The format of a single-precision floating-point number as per the IEEE 754 standard is:

Sign	Exponent	Mantissa
(1 bit)	(8 bits)	(23 bits)

Choose the largest floating-point number among the following options.

Sign: 0, Exponent: 0111 1111, Mantissa: 1111 1111 1111 1111 1111 111

Sign: 0, Exponent: 1111 1110, Mantissa: 1111 1111 1111 1111 1111 111

Sign: 0, Exponent: 1111 1111, Mantissa: 1111 1111 1111 1111 1111 111

Sign: 0, Exponent: 0111 1111, Mantissa: 0000 0000 0000 0000 0000 000

Official answer and solution

Solution

Exclude 1111 1111 (reserved for Infinity/NaN) to identify the maximum normalized exponent 1111 1110 with an all-ones mantissa.

In the IEEE 754 single-precision format:

An exponent field of all ones (1111 1111 $= 255$ ) is reserved for special values: if the mantissa is zero, it represents $\pm\infty$ ; if the mantissa is non-zero (as in Option C), it represents NaN (Not a Number).
The largest valid biased exponent for a normalized finite number is 1111 1110 ( $254_{10}$ , giving true exponent $E = 254 - 127 = 127$ ).
With Sign $= 0$ (positive), Exponent $= \texttt{1111 1110}$ , and Mantissa $=$ all 1s, Option (B) represents $(2 - 2^{-23}) \times 2^{127}$ , which is the largest finite single-precision floating-point number (whereas Options A and D have biased exponent $127$ , i.e., $E = 0$ ).

Final answer

Option (B)

Common trap

Selecting Option (C) without realizing that Exponent = 1111 1111 with a non-zero mantissa encodes NaN, not a numeric value.

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