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

1. $3m$ bits
2. $3m + n$ bits
3. $m + n$ bits
4. none of the above

### 1 comment

How we can add byte + bits . Why byte is not converted into bits ?

Memory capacity = $2^{m}KB$ = $2^{m+10}B$

Since byte addressable memory is given $1$ word = $1$ byte

Memory capacity = $2^{m+10}$ words

No. of bits required to point 1 memory location  = $m+10$ bits

Total no. of operations = $2^{n}$

No. of bits to define 1 operation = $n$bits

Given an instruction have 3 operands and 1operator.

No. of bits needed by instructions = $3*(m+10) + n$

=$3m + n + 30$ bits

Option d) is the correct answer

### 1 comment

why 3*(m+10) done???
3m+n.... B

ans should be D
now isro copied question from UGC Net.

Ans D) none

Memory capacity is actually $2^{m+10} B$.

operations: n

Operand:     m +10

So, n + 3*( m + 10 ) = n + 3m + 30

Step-1: Memory capacity is 2^m KB = 2^(m+10) Bytes
Step-2: Total number of operations 2^n
Step-3: Every instruction involving 3 operands and 1 operator
Step-4: Number of bits needed by instruction is= 3*(m+10)+n
=3m+n+30 bits

so answer is (d) none of the above
by