Question

The RISC-V 32-bit architecture supports virtual memory with 32-bit virtual addresses mapping to 32-bit physical addresses. The page size is 4Kbytes, and page table entries (PTEs) are 4 bytes each.

Translation is performed using a 2-level page table structure. Bits 31:22 of a virtual address index the first-level page table. If the selected first-level PTE is valid, it points to a second-level page table. Bits 21:12 of the virtual address then index that second-level page table. If the selected second-level PTE is valid, it points to the physical page.

How many bytes are required for the first-level page table?

How many bytes are required for a second-level page table?

If the entire virtual address space were allocated, how many bytes would be required for page tables?

If the following ranges of virtual addresses are allocated:

• 0x00000000 to 0x011FFFFF
• 0x10000000 to 0x1FFFFFFF
• 0x7FC00000 to 0x7FFFFFFF

How many bytes would be required for page tables?

Answer 1

as we can see that first level page table has 10 bits so 2^(10) possible locations so we can see that it is 1024 and each entry is of 4 bytes so the total requirements is 1024*4= 4KBytes

Similarly for second level page table 10 bits are used and 4 byte per entry so 4KBytes

Size of pages = 4KB

and memory size as there are 2^20 pages possible with this address

210 * 210 * 4KB 4GB

as we can see that

• 0x00000000 to 0x011FFFFF
  • the number of addresses are 0x011FFFFF = 1,88,74,367‬ addresses and 1 page table contains 2^10 = 1024 of such entries so
  • page tables required will be 1,88,74,367‬/1024= 18432 tables and
  • for these tables we need 18432/1024 = 18 page directories
  • So total requirement will be 18432+18 =18450 tables or 18450*4KB=73800 KB
• 0x10000000 to 0x1FFFFFFF similarly here the number of addresses are 0x0FFFFFFF=26,84,35,455‬
  • Page tables=2,62,144
  • first level page table = 256
  • Hence the number of pages will be 262144+256 = 262400 or 1049600 KB
• 0x7FC00000 to 0x7FFFFFFF
  • here it will be 3FF FFFF‬ = 6,71,08,863‬
  • Page tables = 6,71,08,863‬ /1024 = 65536
  • Page directory tables= 65536/1024 = 64
  • Total memory requirement = 65600 tables or 262400KB

Do give a thumbs up and in case there are doubts leave a comment.