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Official PYQ · 2009Q51I/O SchedulingMCQ1 mark

GATE CS 2009 Q51 — I/O Scheduling

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A hard disk has 6363 sectors per track, 1010 platters each with 22 recording surfaces and 10001000 cylinders. The address of a sector is given as a triple ⟨c,h,s⟩\langle c, h, s \rangle, where cc is the cylinder number, hh is the surface number and ss is the sector number. Thus, the 0th^{th} sector is addresses as ⟨0,0,0⟩\langle 0, 0, 0 \rangle, the 1st^{st} sector as ⟨0,0,1⟩\langle 0, 0, 1 \rangle, and so on

The address ⟨400,16,29⟩\langle 400, 16, 29 \rangle corresponds to sector number:

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Answer: C

To find the sector number corresponding to the address ⟨c,h,s⟩=⟨400,16,29⟩\langle c, h, s \rangle = \langle 400, 16, 29 \rangle, we first identify the disk parameters from the given data:

  • Number of cylinders (CC): 10001000 (indexed from 00 to 999999)
  • Number of surfaces per cylinder (HH): 10 platters×2 surfaces/platter=2010 \text{ platters} \times 2 \text{ surfaces/platter} = 20 (indexed from 00 to 1919)
  • Number of sectors per track (SS): 6363 (indexed from 00 to 6262)

Since the 0th0^{\text{th}} sector is addressed as ⟨0,0,0⟩\langle 0, 0, 0 \rangle, the addressing follows the standard Cylinder-Head-Sector (CHS) mapping. The total number of sectors preceding the address ⟨c,h,s⟩\langle c, h, s \rangle is given by:

Sector Number=(c×H×S)+(h×S)+s\text{Sector Number} = (c \times H \times S) + (h \times S) + s

Substituting the given values c=400c = 400, h=16h = 16, and s=29s = 29:

Sector Number=(400×20×63)+(16×63)+29=(400×1260)+1008+29=504,000+1008+29=505,037\begin{aligned} \text{Sector Number} &= (400 \times 20 \times 63) + (16 \times 63) + 29 \\ &= (400 \times 1260) + 1008 + 29 \\ &= 504,000 + 1008 + 29 \\ &= 505,037 \end{aligned}

Thus, the address corresponds to sector number 505037505037.

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