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

Answer:

2 bytes

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