DRDO CSE 2022 Paper 2 | Question: 2

Question

What is the size of the physical address space in a paging system, which has a page table containing $64$ entries of $11$ bit each (including valid and invalid bit) and a page size of $256$ bytes.

Answer 1

As nothing is mentioned assuming PTE  = Frame number bits + valid/invalid bit

Number of bits for frame number  = $11 -1$ = $10$

So bits in PA = Frame number bits + offset bits = $10$ $+$  $\log 2^{8}$ = $10$ $+$  $8$ = $18$ bits

Physical address space = $2^{18}$ = $256$ $KB$

Answer 2

The size of the physical address space in a paging system can be calculated by multiplying the number of entries in the page table by the page size.

In this case, the page table has 64 entries, and each entry is 11 bits long, including valid and invalid bits. The page size is 256 bytes. Therefore, the size of the physical address space is 64 * 256 bytes = 16384 bytes.