I think, that it consists of (includes); Design, Performance; etc.
Answer:
Please see the attached file for complete Answer.
Explanation:
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
