UGC NET CSE | June 2014 | Part 3 | Question: 43

Question

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 

• A. $3m$ bits
• B. $m + n$ bits
• C. $3m + n$ bits
• D. $3m + n + 30$ bits 

Answer 1

D. should be the ans...

n(operator)

m+10(operand1)

m+10(operand2)

m+10(operand3)

So ans = n + 3m + 30

Comments

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

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because of 2^m K bytes (1024 =2^10)

Answer 2

Memory capacity  is 2^m Kbytes= 2^m+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

Answer 3

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

Comments

Good question asked in UGC first time. :D