A byte addressable computer has a memory capacity of 2m KB( kbytes ) and can…
2018
A byte addressable computer has a memory capacity of 2m KB( kbytes ) and can perform 2m operations. An instruction involving 3 operands and one operator needs maximum of
- A.
3m bits
- B.
3m + n bits
- C.
m + n bits
- D.
none of the above
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Correct answer: D
First, determine the address bits required. The memory capacity is 2^m KB. Since 1 KB equals 2^10 bytes, the total capacity is 2^(m+10) bytes. Thus, m + 10 bits are needed for addressing. The instruction contains one operator and three operands. Each operand requires m + 10 bits. If the opcode size is n bits, the total instruction length is n + 3(m + 10). Option B (3m+n) fails to account for the 10-bit conversion factor from KB.