Data representation (two's complement, floating point, normalisation, error checking) – A-level Computer Science exam questions with mark scheme

1Core3 marks

An 8-bit register stores signed integers using two's complement.

(a) Show how the denary value −45 is represented in this register. [2]

(b) State the range of denary values that can be stored in an 8-bit two's complement register. [1]

Method: 45 = 00101101 and bits inverted to 11010010, or −128 + 83 approach shown (implied by a correct answer)(1)
11010011(1)
−128 to +127 (allow −2⁷ to 2⁷−1)(1)
Answer: (a) 11010011 (b) −128 to 127
Examiner tip: In (b) the range is asymmetric: there is one more negative value than positive.

2Core5 marks

Two numbers are stored as 8-bit two's complement integers:

A = 01101011 B = 11100110

(a) Convert B to denary. [2]

(b) Add A and B using binary addition, giving your answer as an 8-bit binary number. Show any carry that is discarded. [2]

Place values used: −128 + 64 + 32 + 4 + 2 (or inverting and adding 1 to get 26)(1)
−26 (correct answer with no working scores 2 marks)(1)
Column addition shown with carries, giving the 9-bit result 101010001(1)
01010001 with the carry out of the most significant bit (the ninth bit) discarded(1)
81 (allow follow-through from an incorrect 8-bit answer to (b), converted correctly as two's complement)(1)
Answer: (a) −26 (b) 01010001 (c) 81
Examiner tip: The most significant bit of a two's complement number has a value of −128, not +128.

3Standard5 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.

4Hard5 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.

5Hard6 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.

6Standard6 marks

(a) A byte uses even parity, with the most significant bit as the parity bit and the 7-bit ASCII code for 'C' (1000011) in the other bits. State the byte stored. [1]

(b) The byte 10110010 is received. A student says: "Even parity checking passed, so the byte must be correct." Explain why the student may be wrong. [2]

(c) The first 12 digits of an ISBN-13 are 978140883219. Calculate the check digit. Digits are multiplied alternately by 1 and 3 starting with 1, the products are summed, and the check digit makes the total a multiple of 10. [3]

11000011 (parity bit 1 because 1000011 has three 1s)(1)
Parity can only detect an odd number of bit errors (accept: it only detects single-bit errors)(1)
If two (or any even number of) bits are flipped, the number of 1s stays even, so the byte would pass the check even though it is corrupted (accept a worked example, e.g. two bits of the original changed)(1)
Weighted sum: 9×1+7×3+8×1+1×3+4×1+0×3+8×1+8×3+3×1+2×3+1×1+9×3 = 114(1)
Next multiple of 10 is 120 (or 10 − (114 mod 10))(1)
Check digit 6 (correct check digit with no working scores 3 marks)(1)
Answer: (a) 11000011 (b) parity cannot detect an even number of bit errors, so a corrupted byte can still pass (c) 6
Examiner tip: Parity detects an odd number of flipped bits only; never say it "corrects" errors.

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