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Pipeline and Vector Processing Q.44

0. Consider a hard disk with 16 recording surfaces (0-15) having 16384 cylinders (0-16383) and each cylinder contain 64 sectors (0-63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and addressing format is . A file of size 42797 KB is stored in the disk and the starting location of the file is . What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?

  • Option : D
  • Explanation : Since starting on disk starts from <1200, 9, 40> so no of sectors left on 9th surface is 24
    so, on 9th surface total storage of “12288” B is possible Now, a part from 9th surface, on cylinder no. 1200 only 6 surface is left.
    To storage possible on these 6 surface are
    = 6 * 26 * 29 ￫ storage on each sector.
    ᛎ
    = 196608 B
    So total on cylinder no. 1200, storage possible
    = 196608 + 12288 = 208896 B
    So Since file size is 42797 kB and out of which 208896 B are stored on cylinder no. 1200. So we are left only with only 43615232 B.
    Since in 1 cylinder, Storage Possible is
    = 24 * 26 * 29 B = 524288 B
    So we need about
    = 43615232B / 524288B
    = 83.189 more cylinders.
    Hence we’ll need the 1284th Cylinder to completely store the file. C02 after 1283rd cylinder we will left with data which will need 189
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