Advanced Data Representation and File Organisation: Question 4

Worked solution

Part (a): Normalising +22

Step 1, convert 22 to binary. Using bit values 16, 8, 4, 2, 1:

16 + 4 + 2 = 22

So 22 = 10110 in binary.

Step 2. Write this as a normalised fraction multiplied by a power of 2. The integer 10110 is the same value as the binary point shifted 5 places to the left:

10110 = 0.10110 x 2^5

Check: 0.10110 in binary = 1×2⁻¹ + 0×2⁻² + 1×2⁻³ + 1×2⁻⁴ + 0×2⁻⁵ = 0.5 + 0.125 + 0.0625 = 0.6875, and 0.6875 × 2⁵ = 0.6875 × 32 = 22. ✓

Since 22 is positive, the mantissa’s sign bit is 0, and the fraction 0.10110 already starts with a 1 immediately after the point, so it is already normalised.

Step 3, fit the mantissa into 8 bits (1 sign bit + 7 fraction bits) and the exponent into 4 bits. Pad the fraction 10110 with zeros on the right to fill 7 bits: 1011000.

Mantissa = 0 1011000  ->  01011000
Exponent = 5 in 4-bit two's complement = 0101

So Mantissa = 01011000, Exponent = 0101.

[3 marks]: [1] for the correct binary conversion of 22 (10110), [1] for correctly expressing it as a normalised fraction × 2⁵, [1] for the correct final 8-bit mantissa and 4-bit exponent.

Part (b): Normalising -22

The magnitude of -22 is the same as +22, so the representation is found by taking the two’s complement of the mantissa from part (a). The exponent does not change, because the size of the number has not changed, only its sign.

Step 1. Invert every bit of the mantissa 01011000:

01011000  ->  10100111

Step 2. Add 1:

  10100111
+ 00000001
-----------
  10101000

So the mantissa for -22 is 10101000, with the exponent unchanged at 0101.

Check: mantissa 10101000 has sign bit 1 (negative) and fraction bits 0101000, giving -1 + (0×2⁻¹ + 1×2⁻² + 0×2⁻³ + 1×2⁻⁴) = -1 + (0.25 + 0.0625) = -0.6875. Multiplying by 2⁵ = 32 gives -0.6875 × 32 = -22. ✓

[3 marks]: [1] for the correct method (two’s complement of the mantissa, exponent unchanged), [1] for correctly inverting and adding 1, [1] for the correct final mantissa and exponent.

Part (c): Confirming normalisation

A binary floating-point mantissa is normalised when it has no wasted leading bit:

• If the mantissa is positive (sign bit 0), the bit immediately after the sign bit must be 1, giving the form 0.1....
• If the mantissa is negative (sign bit 1), the bit immediately after the sign bit must be 0, giving the form 1.0....

This rule ensures the maximum number of significant bits are used to represent the value for a given exponent, which gives the greatest possible precision.

The mantissa found in part (b) is 10101000: the sign bit is 1, and the very next bit is 0, so it satisfies the negative-mantissa normalisation rule and is correctly normalised.

[2 marks]: [1] for correctly stating the normalisation rule (for both signs, or at least the negative case), [1] for correctly confirming it against the mantissa from part (b).

Final answers

• (a) +22 → Mantissa 01011000, Exponent 0101
• (b) -22 → Mantissa 10101000, Exponent 0101
• (c) Normalised: positive mantissas start 0.1..., negative mantissas start 1.0...; 10101000 satisfies this (sign bit 1 followed by 0).