A main memory system has 8 Gbyte of RAM If the memory is add

A main memory system has 8 Gbyte of RAM. If the memory is addressed in 32-bit words, how wide does the memory address register need to be? If the memory is byte-addressable, how wide does the memory address register have to be? If the memory is bit-addressable, how wide does the memory address register have to be?

Solution

A) For 32-bit:

2^32 = 4*1024*1024*1024 = 4 GB, A 32-bit architecture is limited to 4 GB. For 8 GB RAM two 32 bit registers are needed.