1 votes 1 votes A paging system uses 32-bit address, each of which specifies 1 byte of memory. The system has a main memory unit of 1024 MB, and a page size of 32 KB. How many bits does the system use to maintain displacements (offset)? Operating System paging + – MIRIYALA JEEVAN KUMA asked Nov 10, 2017 MIRIYALA JEEVAN KUMA 2.3k views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments MIRIYALA JEEVAN KUMA commented Nov 11, 2017 reply Follow Share I know that.. how do you come know that displacement offset = no.of pages. any reference, please ? 0 votes 0 votes Surajit commented Nov 11, 2017 reply Follow Share It should be 15.It cannot be 17.page size 32kb = 2^15. Address size 32 bits.This is how the virtual address will be decoded. |17 bits --page nos for a process | 15 bits -- offset or displacement| 0 votes 0 votes Ankit Srivastava 7 commented Dec 29, 2017 reply Follow Share I think displacement offset is nothing but number of bits required to represent number of pages.. because displacement we usually calculate in page accessing and there is some limit like zero to 199 etc so to represent that 200 pages what will be number of bits require that is displacement offset.. In given question number of pages require (32-15) 17 bit so 17 is answer.. 0 votes 0 votes Please log in or register to add a comment.