1 votes 1 votes average process size is 128KB per page entry requires 8 bytes, then optimal page size? 64B 128B 7254B none gate_forum asked Jan 17, 2016 gate_forum 3.7k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Optimal page size= sqrt(2.S.E) S=process size or fragment size. E=page table entry size . So page size P=Sqrt(2.128K.8)= 1K so optimal page size is 1K, none of the above is right answer. A. Dalal answered Jul 10, 2016 • edited Jul 11, 2016 by A. Dalal A. Dalal comment Share Follow See all 10 Comments See all 10 10 Comments reply asu commented Jul 10, 2016 reply Follow Share s= logical address space ??? 0 votes 0 votes Kapil commented Jul 11, 2016 reply Follow Share I think that the formula you have used has E = number of page entries and not PTE size. plz check again... 0 votes 0 votes A. Dalal commented Jul 11, 2016 reply Follow Share No, E is not number of page entries, E is a page table entry size. 0 votes 0 votes Kapil commented Jul 11, 2016 reply Follow Share can u provide a reference? 0 votes 0 votes A. Dalal commented Jul 11, 2016 reply Follow Share Dear Kapil boss for which exam u r preparing 0 votes 0 votes asu commented Jul 11, 2016 reply Follow Share @KAPIL E= PAGE TABLE ENTRY SIZE...REFER GALVIN BUT @ IITJEE S= LOGICAL ADDRESS SPACE HOW U TAKEN PROCESS SIZE 0 votes 0 votes Kapil commented Jul 11, 2016 reply Follow Share @asu http://dysphoria.net/OperatingSystems1/4_page_size.html 0 votes 0 votes asu commented Jul 11, 2016 reply Follow Share yes e = page table entry size in ur reference also 0 votes 0 votes Sanjay Sharma commented Oct 1, 2017 i edited by Sanjay Sharma Apr 4, 2020 reply Follow Share ... 0 votes 0 votes rajputashish657 commented Apr 4, 2020 reply Follow Share @asu bro Logical Adrees Space =Process Size U don't knw this simple thing 0 votes 0 votes Please log in or register to add a comment.
