Question

An ISA specifies a word size of 8 bytes, byte addressability, and an address space of 256 K; it uses single-word instructions (i.e. each instruction is a single 8 byte word). What is the size of the MAR

Answer 1

2 bytes

Step-by-step explanation:

Size of the MAR ( memory address register ) = 18 bits = 2 bytes

Given that

address space = 256 K = 2^8

number of address location= 2^8 * 2^10 ( where 1K = 2^10 )

= 2^18

Given the number of address location = 2^18 ;

Hence 18 bits are required to store the address of the instruction