A byte-addressable computer has a memory capacity of 2m Kbytes and can perform…
2014
A byte-addressable computer has a memory capacity of 2m Kbytes and can perform 2n operations. An instruction involving 3 operands and one operator needs a maximum of
- A.
3m bits
- B.
m + n bits
- C.
3m + n bits
- D.
3m + n + 30 bits
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Correct answer: D
Answer: 3m + n + 30 bits
Explanation:
Memory capacity of 2^m Kbytes means 2^m × 1024 bytes = 2^(m+10) addressable bytes.
Each byte address therefore, requires (m + 10) bits.
There are 3 operands, so the operand address fields together need 3 × (m + 10) = 3m + 30 bits.
The operator is chosen from 2^n operations, so the operator field needs n bits.
Total bits = (3m + 30) + n = 3m + n + 30 bits.
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