Question

If there are 32 segments, each of size 1kb, then the logical address should have how many bits (assume that word size is 32-bits)?

Answer 1

Answer:

The answer will be 15 bits.

There are 32 segments, to represent these segments, 5 bits are required (since 2^5 = 32 ). Having selected a page, to select a particular byte one needs 10 bits (since 2^10 = 1K byte) or we can say,

block offset= 1 Kbytes= 2^10 bytes = 10 bits require for offset

So, totally 5 + 10 =15 bits are needed.