0 votes 0 votes 64MB of MM is divivded into 8 partitions... if processes are 6MB and 4MB is loaded then what is the percentage of internal fragmentation Aboveallplayer asked Nov 26, 2016 Aboveallplayer 255 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes SIze of one partitions $= \frac{64}{8} = 8$ MB internal fragmentation in 6 MB process = 8 - 6 = 2MB internal fragmentation in 4 MB process = 8 - 4 = 4MB therefore, % of internal fragmentation $= \frac{2 + 4}{16} \times 100 = 37.5 %$ Lokesh . answered Nov 26, 2016 Lokesh . comment Share Follow See all 0 reply Please log in or register to add a comment.
