A2 · Practice questions · Normalisation, range and precision

Shift left, exponent down.

Six original Cambridge-style questions on normalisation, range, precision and the limits of floating point, every value checked, with mark points.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Two different bits.

Normalised: positive starts 01, negative starts 10.
Normalise: shift the mantissa left, subtract the shift count from the exponent; check the value is unchanged.
More mantissa bits: precision up. More exponent bits: range up.
Limits: recurring binary fractions give rounding errors; too big is overflow, too close to zero is underflow.

Explain

[2 marks]

Give two reasons why floating-point numbers are normalised.

To give the maximum precision for the number of bits in the mantissa (no bits wasted on leading 0s or 1s). ✓
So that each number has a unique representation (making comparison easier). ✓

Convert

[3 marks]

A floating-point number has an 8-bit mantissa and a 4-bit exponent, both two's complement. Normalise 00010110 0100, showing the new mantissa and exponent.

Mantissa shifted 2 places left: 01011000. ✓
Exponent 4 - 2 = 2: 0010. ✓
Value unchanged: 2.75. ✓

Convert

[3 marks]

Using the same format, normalise 11110100 0101 and give its denary value.

Mantissa shifted 3 places left until it starts 10: 10100000. ✓
Exponent 5 - 3 = 2: 0010. ✓
Value: -0.75 × 4 = -3. ✓

Calculate

[4 marks]

For the same format, write the binary and the denary value of (a) the largest positive number and (b) the smallest positive normalised number.

(a) Mantissa 01111111, exponent 0111. ✓
= 127/128 × 2⁷ = 127. ✓
(b) Mantissa 01000000, exponent 1000. ✓
= 0.5 × 2^-8 = 1/512 (0.001953125). ✓

Analysis

[3 marks]

A designer must use 12 bits for each floating-point number and is choosing between an 8-bit mantissa with a 4-bit exponent and a 6-bit mantissa with a 6-bit exponent. Explain the effect of each choice.

8-bit mantissa: more significant bits, so greater precision (smaller gaps between values). ✓
6-bit exponent: larger range (exponent -32 to +31, not -8 to +7), so much larger and smaller numbers can be stored. ✓
Trade-off: increasing one decreases the other for a fixed total number of bits. ✓

Explain

[4 marks]

The denary value 0.1 is stored using an 8-bit mantissa and a 4-bit exponent as 01100110 1101. Explain why this is not exactly 0.1, and describe one consequence for a program. Also explain the term overflow.

0.1 is a recurring fraction in binary (0.000110011...), so the mantissa can hold only the first bits. ✓
The stored value is 0.099609375, a rounding error of 0.000390625. ✓
Consequence: errors accumulate over repeated calculations / an equality test on REAL values may fail. ✓
Overflow: the result of a calculation is too large to be represented, because the exponent needed is greater than the maximum. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.