why 10 bits for each operands ??????????

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+5 votes

A byte addressable computer has a memory capacity of $2^{m}$ Kbytes and can perform $2^{n}$ operations. An instruction involving $3$ operands and one operator needs a maximum of

- $3m$ bits
- $m + n$ bits
- $3m + n$ bits
- $3m + n + 30$ bits

+4 votes

Best answer

+7 votes

Memory capacity is 2^{m }Kbytes= 2^{m+10}bytes. Whole memory can be represented by m+10 bits.

2^{n} operations can be represented by n bits.

Now instruction of form

Instruction op1,op2,op3. Will take (3m+30+n) bits

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