Calculating Page Table Size

memory-management paging
mino picture mino · May 1, 2013 · Viewed 97.3k times · Source

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?

Answer

mino picture mino · May 1, 2013

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.

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