operating system - Determine addresses and page table size -


i know question has been asked many times still feel struggle it.

given:  physical memory: 2^20 32-bit system page size: 2^10

i need determine physical , virtual addresses calculate page table size.

i not sure have correct have got far:

->virtual address (22 bits specifying page #)(10 bit specifying offset) ->physical address like** (10 bits specifying frame #)(10 bits specifying offset)

the number of page entries 2^20/2^10=2^10

how calculate page table size? confused since each pte has both virtual , physical addresses. number of page entries*(physical address+virtual address)? how different if have present bit set?

size of page table depends on metadata bits hold each entry (valid dirty, etc..)

basically size be: (num_of_pages)*(num_of_bits_for_frame_number + meta_bits). ex. (valid bit): 2^22*(10 + 1)


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