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O-LevelComputer ScienceData Transmission2210/12 · Oct/Nov 2016 · Q43 Marks

4 Nine bytes of data are transmitted from one computer to another. Even parity is used. An additional parity byte is also sent. The ten bytes arrive at the destination computer as follows: parity bit bit 2 bit 3 bit 4 bit 5 bit 6 bit 7 bit 8 byte 1 1 1 1 0 1 1 1 0 byte 2 0 0 0 0 0 1 0 1 byte 3 0 1 1 1 1 0 0 0 byte 4 1 1 0 0 0 0 0 0 byte 5 1 0 1 1 1 1 1 0 byte 6 0 1 0 1 1 0 0 1 byte 7 0 1 1 1 0 0 1 1 byte 8 0 0 1 1 0 1 1 0 byte 9 1 1 0 0 0 0 1 1 parity byte 0 0 1 0 0 0 1 0 One of the bits was corrupted during the data transmission. (a) Circle the corrupt bit in the corrupt byte in the table above. (b) Explain how the corrupted bit was found.

  1. (a)Circle the corrupt bit in the corrupt byte in the table above.[1]
  2. (b)Explain how the corrupted bit was found.[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.

  1. [1] (a) Intersection of Row 7 and column 4 circled
  2. [2] (b) Row (byte number) 7 has an odd number of 1s (five 1s)
  3. Column (bit number) 4 has an odd number of 1s (five 1s)

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About this question

Topic
Data Transmission, O-Level Computer Science (syllabus 2210). Structured question, 3 marks.
Source
O-Level Computer Science, Oct/Nov 2016 Paper 1 Variant 2, question 4.

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