Question

Consider a hard disk with 16 recording surfaces (0-15) having 32768 cylinders and each track contains 128 sectors. Data storage capacity in each sector is 2KB. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no. >. A file of size 48688 KB is stored in the disk and the last sector of the file is stored at sector with address <8692, 5, 30>. The file is stored in a contiguous manner. What is the address of starting location sector of the file?

Answer 1

option (a)
$\text{No. of sector per cylinder = 16 * 128 = 2048}\text{Sector no.of last sector}\Rightarrow \text{(8692 * 2048)+(5*128)+30= 17801886}\text{No. of sectors to stores file}=\frac{48688KB}{2K}=24344\text{First sector no.of file = 17801886 - 24344 + 1 = 17777543}\text{cylinder no. of first sector = 17777543/2048 = 8680}\text{Surface no. of first sector = (17777543\%2048)/128 =7}\text{Sector no. of first sector= (17777543\%2048) \%128 =7}\text{address = <8680, 7, 7>}$