Calculating Page Table Size

Question

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

User · Answer

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

User · Answer

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

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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    This means that every memory address can be 32 bits long    A 32-bit entry can point to one of 2 32 physical page frames  2   stated differently    A 32-bit register can store 2 32 different values    We were also told that      each page size is 4 KB    1 KB  kilobyte    1 x 1024 bytes   2 10 bytes 4 x 1024 bytes   2 2 x 2 10 bytes    4 KB  i e  2 12 bytes   The size of each page is thus 4 KB  Kilobytes NOT kilobits     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   Example  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   Givens         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       How Many pages In Page Table    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       How Much Memory in BITS Per Page   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

User · Answer

Since we have a virtual address space of 2 32 and each page size is 2 12  we can store  2 32 2 12    2 20 pages  Since each entry into this page table has an address of size 4 bytes  then we have 2 20 4   4MB  So the page table takes up 4MB in memory

User · Answer

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

[memory-management] Calculating Page Table Size

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