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A-LevelComputer ScienceProcessor fundamentals9618/12 · Oct/Nov 2022 · Q712 Marks

The following table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC), and an Index Register (IX). [Table: Instruction (Opcode, Operand, Explanation)] <address> can be an absolute or symbolic address # denotes a denary number, e.g. #123 B denotes a binary number, e.g. B01001101 The following table shows another part of the instruction set for the processor. [Table: Instruction (Opcode, Operand, Explanation)] The contents of memory addresses 50 and 51 are shown: [Table: Memory address, Data value (50: 01001101, 51: 10001111)]

  1. (a)Trace the program currently in memory using the trace table, stopping when line 90 is executed for a second time. [Table of program instructions and memory contents for trace table][5]
  2. (b)(i)The current contents of the ACC are: 0 1 0 1 0 0 1 1 Show the contents of the ACC after the execution of the following instruction. XOR B00011111[1]
  3. (b)(ii)The current contents of the ACC are: 0 1 0 1 0 0 1 1 Show the contents of the ACC after the execution of the following instruction. AND 50[1]
  4. (b)(iii)The current contents of the ACC are: 0 1 0 1 0 0 1 1 Show the contents of the ACC after the execution of the following instruction. LSL #3[1]
  5. (b)(iv)The current contents of the ACC are: 0 1 0 1 0 0 1 1 Show the contents of the ACC after the execution of the following instruction. OR 51[1]
  6. (c)Write the register transfer notation for each of the stages in the fetch-execute cycle described in the table. [Table: Description, Register transfer notation][3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

  1. [1] For ACC=0, IX=1 (after 76)
  2. [1] For ACC=2, 101=2 (after 85)
  3. [1] For ACC=0, ACC=1, 100=1, IX=1 (after 89 - first loop iteration complete)
  4. [1] For IX=4 (after 76 - second loop, first instruction)
  5. [1] For ACC=6, 101=6 (after 82)
  6. [1] For ACC=1, ACC=2, 100=2, IX=2 (after 89 - second loop iteration complete, stop after 90)
  7. [1] 0100 1100
  8. [1] 0100 0001
  9. [1] 1001 1000
  10. [1] 1101 1111
  11. [1] MAR ← [PC]
  12. [1] PC ← [PC] + 1
  13. [1] CIR ← [MDR]

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About this question

Topic
Processor fundamentals, A-Level Computer Science (syllabus 9618). Structured question, 12 marks.
Source
A-Level Computer Science, Oct/Nov 2022 Paper 1 Variant 2, question 7.

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