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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 
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3 Answers

+4 votes
Best answer

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 ??????????
0
because of 2^m K bytes (1024 =2^10)
+7 votes
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
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edited by
+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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0
Good question asked in UGC first time. :D
Answer:

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