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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

1. $3m$ bits
2. $m + n$ bits
3. $3m + n$ bits
4. $3m + n + 30$ bits

recategorized | 3.3k views

D. should be the ans...

 n(operator) m+10(operand1) m+10(operand2) m+10(operand3)

So ans = n + 3m + 30

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why 10 bits for each operands ??????????
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because of 2^m K bytes (1024 =2^10)
Memory capacity  is 2Kbytes= 2m+10bytes.  Whole memory can be represented by m+10 bits.
2n operations can be represented by n bits.
Now instruction of form
Instruction op1,op2,op3. Will take  (3m+30+n) bits
Option D

edited
+1 vote

option is D

To specify a particular operation, out of the 2n possible operations, one needs n bits. As the machine is byte addressable, to specify a particular byte we need (m+10) bits (2(n+10) bytes are there).

So 3 addressable and 1 operations needs 3(m+10)+n=3m+n+30 bits

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Good question asked in UGC first time. :D