Question

Suppose that we have a computer system using 32-bit logical address and 46-bit physical address. It also uses paging for memory management with a single-level page table organization. The page size is 4K bytes and each page table entry is 32 bits or 4 bytes in size.

Calculate the number of bits in each field in the logical address, the size in bytes of the page table, and the number of frames.

Answer 1

Page Size = 4K bytes = 2^12. Thus offset = 12 bits. And page number = 32-12 = 20 bits.

So fields of logical address: Page Number = 20 bits and offset = 12 bits.

Page Table Size = Number of entries * size of each entry = 2^20 * 4 (each page table entry is 4 bytes in size)

Page Table Size = 2^24 = 4MB.

Frame Size = Page Size = 4KB = 2^12 bytes

Number of frames = Size of physical memory / frame size = 2^46 / 2^12 = 2^34.