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?

Answer
Comment
Array

Option : D
Explanation : Since starting on disk starts from <1200, 9, 40> so no of sectors left on 9^th surface is 24
so, on 9^th surface total storage of “12288” B is possible Now, a part from 9^thsurface, 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 1284^th Cylinder to completely store the file. C₀₂ after 1283^rd cylinder we will left with data which will need 189

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