Computer System Architecture - Central Processing Unit

27. Consider the following sequence of micro operations.
MBR ⇽ PC
MAR ⇽ X
PC ⇽ Y
Memory ⇽ MBR
Which one of the following is a possible operation performed by this sequence

  • Option : D
  • Explanation : Given: A sequence of micro operations
    MBR ⇽ PC
    MAR ⇽ X
    PC ⇽ Y
    Memory ⇽ MBR
    To find: What these micro operation represents?
    Analysis: On conver ting these operations to normal english
    1. First micro operation stores the value of PC into memory Base register (MBR)
    2. Second micro operation stores value of X into Memory Address register (MAR)
    3. Third micro operation stores value of Y into PC
    4. Fourth micro operation stores value of MBR to memory.
    So before execution of these instructions PC holds the value of next instruction to be executed. We first stores the value of PC to MBR and then through MBR to memory i.e. we are saving the value of PC in memory and then load PC with a new value. This can be done only in two types of operations conditional branch and interrupt service. As we are not checking here for any conditions so it is interrupt service.
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Consider the following program segment for a hypothetical CPU having three user registers R1, R2 and R3.

30. Consider that the memory is word addressable with size 32 bits and the program has been loaded starting from memory location 1000 (decimal). If an interrupt occurs during the ADD instruction, what will be the return address pushed on to the stack.

  • Option : B
  • Explanation : If memory is word addressable then 1st instruction require two words space (1000 – 1001) 2nd instruction requires only 1 word space ( 1002) and instruction 3rd requires only 1 word space (1003) hence 3rd is add instruction and interrupt occurs during that so after executing that instruction CPU handles the interrupt so return address is 1004
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