2 2 votes CO & Architecture dma + – Kriss Singh 2.3k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
Best answer 14 14 votes In 1s CPU makes 300 million ticks, so time for a tick (clock time) is 1/(300 * 106 ) = 3.3 ns. For 900 cycles, CPU time will be 900*3.3 = 2970 ns = 2.97µs Now, time for data transfer = 200/5000s = 40ms = 40000 µs So, percentage of CPU time for transfer = 2.97/(40000+ 2.97) = 0.0074% sanjeev_zerocode answered Sep 14, 2014 • selected Dec 4, 2016 by Tendua sanjeev_zerocode comment Share Follow See all 3 Comments 3 3 Comments reply Arjun commented Oct 27, 2014 reply Follow flag In 1s CPU makes 300 million ticks, so time for a tick (clock time) is 1/(300 * 106 ) = 3.3 ns. For 900 cycles, CPU time will be 900*3.3 = 2970 ns = 2.97µs Now, time for data transfer = 200/5000s = 40ms = 40000 µs So, percentage of CPU time for transfer = 2.97/(40000+ 2.97) = 0.0074% 3 3 replyShare sanjeev_zerocode commented Oct 27, 2014 reply Follow flag Thanks for clearing 1 1 replyShare phelps18 commented Jan 7, 2018 reply Follow flag sir since hard disk is directly sending data to memory doesn't it mean it is burst mode? if so then cpu overhead will be 100%. 1 1 replyShare Please log in or register to add a comment.