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A-LevelComputer ScienceData representation9618/31 · May/June 2021 · Q114 Marks

Real numbers are stored in a computer system using floating-point representation with: • 10 bits for the mantissa • 6 bits for the exponent • Two's complement form for both the mantissa and the exponent.

  1. (a)Calculate the normalised floating-point representation of -7.25 in this system. Show your working. Mantissa Exponent[3]
  2. (b)Calculate the denary value of the given binary floating-point number. Show your working. Mantissa 1 0 1 1 0 0 0 1 1 1 Exponent 0 0 0 1 1 1[3]
  3. (c)The given binary floating-point number is not normalised. Normalise the floating-point number. Show your working. Mantissa 0 0 0 0 0 0 1 1 1 1 Exponent 1 0 0 1 1 1[3]
  4. (d)(i)The denary number 513 cannot be stored accurately as a normalised floating-point number in this computer system. (i) Explain the reason for this.[3]
  5. (d)(ii)(ii) Describe an alteration to the way floating-point numbers are stored to enable this number to be stored accurately using the same total number of bits.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

  1. [1] (a) Calculation of mantissa (e.g., 111.01 to 0111010000 or 1000110000 for -7.25 in two's complement)
  2. [1] (a) Calculation or use of exponent (e.g., 0.11101 × 2^3 or 1.00011 × 2^3 or appropriate shifting)
  3. [1] (a) Correct Answer: Mantissa 1000110000, Exponent 000011
  4. [1] (b) Working for exponent (e.g., identifies 7 from 000111)
  5. [1] (b) Working for mantissa (e.g., converts 1011000111 to 1.011000111 or similar, handles two's complement for negative)
  6. [1] (b) Correct Answer: -78.25
  7. [1] (c) Working (e.g., explaining shift of 6 places to the left for mantissa, -6 to exponent)
  8. [1] (c) Correct Mantissa: 0111000000
  9. [1] (c) Correct Exponent: 100001
  10. [1] (d)(i) Requires 11 bits / more than 10 bits to store (accurately) / reference to maximum (positive) number that can be stored = 511
  11. [1] (d)(i) Denary 513 in binary is 1000000001 // Normalised: 0.1000000001
  12. [1] (d)(i) Results in overflow
  13. [1] (d)(ii) The number of bits for the mantissa must be increased
  14. [1] (d)(ii) 11/12 bits mantissa and 5/4 bits exponent

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About this question

Topic
Data representation, A-Level Computer Science (syllabus 9618). Structured question, 14 marks.
Source
A-Level Computer Science, May/June 2021 Paper 3 Variant 1, question 1.

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