Question

Suppose the size of process is 10000 bytes and the relocation register is loaded with value 5000 which of the following memory address this process can access

Answer 1

Answer:

The page field is 8-bit wide, then the page size is 256 bytes.

Using the subdivision above, the first level page table points to 1024 2nd level page tables, each pointing to 256 3rd page tables, each containing 64 pages. The program's address space consists of 1024 pages, thus we need we need 16 third-level page tables. Therefore we need 16 entries in a 2nd level page table, and one entry in the first level page table. Therefore the size is: 1024 entries for the first table, 256 entries for the 2nd level page table, and 16 3rd level page table containing 64 entries each. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 2 (one second-level paget table) + 16 * 64 * 2 (16 third-level page tables) = 4608 bytes.