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Editorial Team
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Asked: 2026-06-18T01:34:47+00:00

Consider a hypothetical 32-bit microprocessor having 32-bit instructions composed of two fields: the first

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Consider a hypothetical 32-bit microprocessor having 32-bit instructions composed of two fields: the first byte contains the opcode and the remainder the immediate operand or an operand address.

  1. What is the maximum directly addressable space capacity(in bytes)?
    For the first question I got 3 bytes but I’m not entirely sure why that is the answer.

  2. what is the maximum number of instructions which can be supported?

3.What is the smallest size in bits needed for the program counter and instruction register?

I’m really confused as to where to even start with these questions. Any help would be greatl appreciated.

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1 Answer

  1. Editorial Team

    For the first part, your maxiumum addressable space will be 

    2^([instruction length]-[remain])  => 2^(32-8)
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