Calculating size of pages and page tables

Question

A physical address is 32 bits and the virtual address is split as 10 bits 'off-set',2 bits 'byte addressability', 20 bits being left for identifying the page.

What size are the pages and the page tables?

I believe its:

2^10 = 1024.

The ones that are reserved pages are 2^2 = 4

I am not sure on how to use the addressbility though...

Thanks again guys :)

Answer 1

let me tell you:

In the architecture that you use, you will use 2 ²⁰ pages, and offset will help you identify the word location in a page. So page size is 2 ¹⁰ . What is the 2 bits of addressability? It is the addressability of bytes in a word => you have 4 bytes in a word and these 2 bits maps to the byte locations in your word.

Answer 2

Addressing mechanism is is similar to looking for a word in the dictionary, first, find the chunk of words with just the 1st letter matching and then keep narrowing down. Similarly, 1st 20 bits to locate the page , next 10 bits to locate the word the particular page and last 2 bits for the byte in the word.