This is a simple paging system of 224 bytes of physical memory, 256 pages of logical address space, and a page size of 210 bytes.

1. How many bits are in a logical address?

2. How many bytes are in a page frame?

3. How many bits of the physical address specify the page frame?

4. How many entries are in the page table?

5. How many bits are needed to store an entry in the page table (It is assumed that each entry in the page table contains a presence / absence bit in addition to the page frame number)

 

1) Logical address space contains = 256 pages= 2^8

Page size = 2^10 bytes

So, total logical address space =2^8 * 2^10 = 2^18 bytes

For 2^18 byte address space - 18 bit address is required.

2) Number of page frames = logical address / size of page

Page size = 2^10 bytes

Logical address space size = 2^18 bytes 

Number of page frames = 2^18/2^10 = 2^8 

so, 8 bytes are present in the page frames basically

this is the basic idea 

3) Physical Address is of size 24 bit as there are 

2^24 address in physical memory 

size of each page is = 2^10 bytes 

So, Number of frames are = physical address/page size

             = 2^24/2^10

             =2^14

so, 14 bytes are reserved for frames in physical address 

4) Number of entries in a page table 

there are 256 pages that number of entries are there in

page tables 

256 = 2^8 

so 256 entries are present at most 

5) number of bits stored in page table is basically

page frame size in physical memory + extra bit 

14 bits + 1 bits (for presence absence )

so 15 bits is the required answer