GATE 2016, SET I, 1 Mark
A processor can support a maximum of 4GB, where the memory is word-addressable (a word consists of two bytes). The size of address bus of the processor is at least _________bits.
Sol.
First we will convert 4GB in to Bytes
4GB = 230 X 4 = 230 X 22 = 232 Bytes
Given in problem, a word consist 2 Bytes,
So, no of words = 232 / 2 = 231
The size of address bus of the processor is at least 31 bits.