1 votes 1 votes consider a system where each file is associated with a 16 bit number.for each file ,each user should have the read and write capability.how much memory is needed to store each user's access data?A) 16KB B) 32KB C) 64KB D) 128KB Shivi Patel asked Jun 18, 2016 Shivi Patel 656 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes is the ans A 16KB Sanjay Sharma answered Jun 18, 2016 Sanjay Sharma comment Share Follow See all 5 Comments See all 5 5 Comments reply Catalan 1920 commented Jun 21, 2016 reply Follow Share how ? please explain 0 votes 0 votes Sanjay Sharma commented Jun 22, 2016 reply Follow Share 16 bits so 2^16 =64kbit =8KB now each user has read and write capability so need 1 more bit so 8x2=16 KB 0 votes 0 votes mcjoshi commented Oct 9, 2016 reply Follow Share each file is associated with a 16 bit number. this means that no. of files $= 2^{16}$ and $16bit$ are required for read access and $16-bits$ for write access. Am i correct @sanjay sir, and @Kapilp 0 votes 0 votes Kapil commented Oct 9, 2016 reply Follow Share Each file requires 2 bits (read and write) Associated bits with each file are 00,01,10,11 . 0 votes 0 votes mcjoshi commented Oct 9, 2016 reply Follow Share means no. of files $= 2^{16}$ 1-bit for reading, 1 for writing, which add to total $2-bits$ memory needed $= 2^{16}*\frac{2}{8} = 16KB$ 1 votes 1 votes Please log in or register to add a comment.