I'm reading through an example of page tables and just found this:
Consider a system with a 32-bit logical address space. If the page size in such a system is 4 KB (2^12), then a page table may consist of up to 1 million entries (2^32/2^12). Assuming that each entry consists of 4 bytes, each process may need up to 4 MB of physical address space for the page table alone.
I don't really understand what this 4MB result represents. Does it represent the space the actual page table takes up?
This question is related to
memory-management
paging
My explanation uses elementary building blocks that helped me to understand. Note I am leveraging @Deepak Goyal's answer above since he provided clarity:
We were given a logical 32-bit address space (i.e. We have a 32 bit computer)
Consider a system with a 32-bit logical address space
We were also told that
each page size is 4 KB
As Depaak said, we calculate the number of pages in the page table with this formula:
Num_Pages_in_PgTable = Total_Possible_Logical_Address_Entries / page size
Num_Pages_in_PgTable = 2^32 / 2^12
Num_Pages_in_PgTable = 2^20 (i.e. 1 million)
The authors go on to give the case where each entry in the page table takes 4 bytes. That means that the total size of the page table in physical memory will be 4MB:
Memory_Required_Per_Page = Size_of_Page_Entry_in_bytes x Num_Pages_in_PgTable
Memory_Required_Per_Page = 4 x 2^20
Memory_Required_Per_Page = 4 MB (Megabytes)
So yes, each process would require at least 4MB of memory to run, in increments of 4MB.
Now if a professor wanted to make the question a bit more challenging than the explanation from the book, they might ask about a 64-bit computer. Let's say they want memory in bits. To solve the question, we'd follow the same process, only being sure to convert MB to Mbits.
Let's step through this example.
- Logical address space: 64-bit
- Page Size: 4KB
- Entry_Size_Per_Page: 4 bytes
Recall: A 64-bit entry can point to one of 2^64 physical page frames - Since Page size is 4 KB, then we still have 2^12 byte page sizes
- 1 KB (kilobyte) = 1 x 1024 bytes = 2^10 bytes
- Size of each page = 4 x 1024 bytes = 2^2 x 2^10 bytes = 2^12 bytes
`Num_Pages_in_PgTable = Total_Possible_Logical_Address_Entries / page size
Num_Pages_in_PgTable = 2^64 / 2^12
Num_Pages_in_PgTable = 2^52
Num_Pages_in_PgTable = 2^2 x 2^50
Num_Pages_in_PgTable = 4 x 2^50 `
Memory_Required_Per_Page = Size_of_Page_Entry_in_bytes x Num_Pages_in_PgTable
Memory_Required_Per_Page = 4 bytes x 8 bits/byte x 2^52
Memory_Required_Per_Page = 32 bits x 2^2 x 2^50
Memory_Required_Per_Page = 32 bits x 4 x 2^50
Memory_Required_Per_Page = 128 Petabits
[2]: Operating System Concepts (9th Ed) - Gagne, Silberschatz, and Galvin
Since the Logical Address space is 32-bit long that means program size is 2^32 bytes i.e. 4GB. Now we have the page size of 4KB i.e.2^12 bytes.Thus the number of pages in program are 2^20.(no. of pages in program = program size/page size).Now the size of page table entry is 4 byte hence the size of page table is 2^20*4 = 4MB(size of page table = no. of pages in program * page table entry size). Hence 4MB space is required in Memory to store the page table.
In 32 bit virtual address system we can have 2^32 unique address, since the page size given is 4KB = 2^12, we will need (2^32/2^12 = 2^20) entries in the page table, if each entry is 4Bytes then total size of the page table = 4 * 2^20 Bytes = 4MB
Suppose logical address space is **32 bit so total possible logical entries will be 2^32 and other hand suppose each page size is 4 byte then size of one page is *2^2*2^10=2^12...* now we know that no. of pages in page table is pages=total possible logical address entries/page size so pages=2^32/2^12 =2^20 Now suppose that each entry in page table takes 4 bytes then total size of page table in *physical memory will be=2^2*2^20=2^22=4mb***
Source: Stackoverflow.com