Fixed and floating point, errors, normalisation, underflow and overflow – A-level Computer Science exam questions with mark scheme

1Standard5 marks

A computer stores real numbers in floating point form using a mantissa of 8 bits and an exponent of 4 bits. Both are two's complement. The binary point of the mantissa is immediately after the sign bit.

(a) Calculate the denary value of the number with mantissa 01101000 and exponent 0011. [3]

(b) Calculate the denary value of the number with mantissa 01010000 and exponent 1110. [2]

Exponent 0011 = 3 (so the binary point moves 3 places right)(1)
Mantissa 0.1101000 becomes 110.1(1)
6.5(1)
Exponent 1110 = −2 and mantissa 0.101 (= 0.625) shown, or point moved 2 places left to give 0.00101(1)
0.15625 (allow 5/32)(1)
Answer: (a) 6.5 (b) 0.15625
Examiner tip: Say which way the point moves: a negative exponent moves it left, so the number gets smaller.

2Hard5 marks

A computer stores real numbers in floating point form using an 8-bit mantissa followed by a 4-bit exponent. Both are in two's complement and the binary point of the mantissa is immediately after the sign bit.

(a) Represent −3.25 as a normalised floating point number in this format. Show your working. [4]

(b) Explain why floating point numbers are normalised. [1]

3.25 = 11.01 in binary (allow equivalent method: −3.25 = −4 + 0.75 written as two's complement fixed point 100.11) (the question asks for working: a correct final answer, mantissa 10011000 and exponent 0010, with no working scores 2 marks only)(1)
Exponent 0010 (point moved 2 places, 0.1101 × 2²)(1)
Positive mantissa 01101000 (allow, by the alternative method, normalising 100.11 to 1.0011 by moving the point 2 places)(1)
Two's complement of the mantissa (invert and add 1) giving 10011000 (normalised as the first two bits differ) (alternative method: award for the final mantissa 10011000 obtained from 1.0011)(1)
To maximise the precision for the number of bits available / so each value has a unique representation (allow: avoids wasting bits on leading 0s or 1s)(1)
Answer: (a) Mantissa 10011000, exponent 0010 (b) maximises precision / unique representation
Examiner tip: For a negative mantissa a normalised form starts 10, not 11; check this after negating.

3Hard6 marks

A floating point system uses an 8-bit two's complement mantissa (point after the sign bit) and a 4-bit two's complement exponent. All numbers are stored normalised.

(a) Calculate the smallest positive normalised value that can be stored. Give your answer as a fraction or decimal. [3]

(b) Calculate the largest positive value that can be stored. [2]

(c) The designer changes the format to a 6-bit mantissa and a 6-bit exponent. State the effect on range and on precision. [1]

Smallest positive normalised mantissa is 01000000 (= 0.5)(1)
Most negative exponent 1000 = −8(1)
0.5 × 2⁻⁸ = 2⁻⁹ = 1/512 = 0.001953125(1)
Largest mantissa 01111111 (= 127/128) with largest exponent 0111 (= 7)(1)
127 (127/128 × 128)(1)
Range increases and precision decreases (both needed)(1)
Answer: (a) 1/512 = 0.001953125 (b) 127 (c) greater range, less precision
Examiner tip: The smallest positive normalised mantissa is 0.1000000 (0.5), not 0.0000001.

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