Question

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

Answer 1

$\text{Memory size}=2^{m}K,Bytes=2^m{.2}^{10}Bytes=2^{(m+10)}Bytes$

Address size = ( m + 10) bits

Number of operations $=2^n$

opcode bits = n-bits

Instruction format:

$\begin{array}{cccc}\text{opcode}& \text{add 1}& \text{add 2}& \text{add 3}\end{array}$
n (m+10) (m+10) (m+10)

(3m+n+30) bits