A byte addressable computer has a memory capacity of 2ᵐ K bytes and can perform 2ⁿ operations. An instruction involving 3 operands and one operator needs a…
A byte addressable computer has a memory capacity of 2ᵐ K bytes and can perform 2ⁿ operations. An instruction involving 3 operands and one operator needs a maximum of
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Correct answer: (D) 3m + n + 30 bits
Explanation
Memory capacity is 2ᵐ Kbytes = 2^(m+10) bytes, so the whole memory can be addressed with m + 10 bits. 2ⁿ operations can be represented by n bits. An instruction of the form OP, operand₁, operand₂, operand₃ will therefore take 3(m + 10) + n = 3m + n + 30 bits. So the answer is (d).