Question

Consider a system which uses $2^{M}K$ bytes of memory with a byte addressable scheme. The system supports 3-address instructions. If there are total $2^i$ number instructions supported by the system then the size of an instruction in bytes, for M = 2 and i = 4 is

• A. 4
• B. 3
• C. 6
• D. 5

Answer 1

The correct option is D 5

option (c)

$\text{Memory size}=2^{M}Kbytes=2^{M-10}bytes$

Address size = M +10 bits

Number of instruction $=2^i$

Opcode-bits = i-bits

Instruction

i M+10 M+10 M+10

Opcode

add.1

add.2

add.3

Total size = (i + 3M + 30) bits
For i = 4 and M= 2
Instruction size = 4 + 3 * 2 + 30
= 40 - bits
= 5 bytes