Question

consider a logical address space of 64 pages of 1,024 words each, mapped onto a physical memory of 32 frames. a. how many bits are there in the logical address? b. how many bits are there in the physical address?

Answer 1

Explanation:

When we address within a 1024-word page, it requires 10 bits because 1024 = 210. Now, since the logical address space is made up of 64 or 26 pages long. Therefore, the logical addresses must be 10+6 which is 16 bits. In the same way, since there are 32 or 25 physical frames, the physical addresses will be 5 + 10, i.e.,  15 bits long. We can see both the addresses being found out in this way.