2 votes 2 votes n a 32 bit processor the virtual address is 22 bit:12 bit(Table index : Offset). What is the size of the page frame and the process page table assuming that each entry in the page table is 4 Bytes? Operating System operating-system paging memory-management discrete-mathematics programming-in-c + – Sneha bhusare asked Aug 6, 2017 Sneha bhusare 1.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Question is ambiguous as virtual address + index=34 not 32 So assuming a 34 bit system Offset=page frame size= 4KB Page table size = number of entries in page table* size of each entire =$2^{22}*2^{2}$=16MB Tesla! answered Aug 6, 2017 • edited Jan 19, 2018 by Tesla! Tesla! comment Share Follow See all 4 Comments See all 4 4 Comments reply Avinash Singh 1 commented Aug 21, 2017 reply Follow Share Page table size =number of entries in page table*size of each entire number of entries in page table=total number of pages in the process no of pages=virtual address space/page size VAS=2^22 page size=2^12 therefore no of pages=2^22/2^12 no of pages=2^10 and now Page table size=2^10*2^2 –1 votes –1 votes Tesla! commented Aug 21, 2017 reply Follow Share Total virtual address space is of 32 it off which 22 is for page number and 12 for offset 0 votes 0 votes Avinash Singh 1 commented Aug 21, 2017 reply Follow Share yes bro you r right i thought something else...thank you 0 votes 0 votes pankaj choudhary commented Jan 19, 2018 reply Follow Share how it is possible that 32 bit = 22 + 12? 1 votes 1 votes Please log in or register to add a comment.
