2 votes 2 votes Consider a logical address space of $8$ pages of $1024$ words mapped with memory of $32$ frames. How many bits are there in the physical address ? $9$ bits $11$ bits $13$ bits $15$ bits Operating System ugcnetcse-dec2011-paper2 operating-system memory-management + – makhdoom ghaya asked Aug 19, 2016 • recategorized Oct 10, 2018 by Pooja Khatri makhdoom ghaya 1.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 1 votes 1 votes Taking page size same as frame size Physical address = frame offset + word offset No of frames = 32 so needs 5 bits Frame size = page size =1k words = 10 bits So ans should be 5 + 10 = 15bits papesh answered Aug 19, 2016 • selected Sep 14, 2017 by sourav. papesh comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Ans: 15 bits (10 bits + 5 bits)= (Size of each frame + No. of frame) rishu_darkshadow answered Sep 14, 2017 rishu_darkshadow comment Share Follow See all 0 reply Please log in or register to add a comment.