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Official PYQ · 2007Q11I/O SchedulingMCQ1 mark

GATE CS 2007 Q11 — I/O Scheduling

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Consider a disk pack with 1616 surfaces, 128128 tracks per surface and 256256 sectors per track. 512512 bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively:

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Answer: A

To find the capacity of the disk pack and the number of bits required to address a specific sector, we can perform the following step-by-step calculations:

1. Capacity of the Disk Pack

The total capacity is the product of the number of surfaces, tracks per surface, sectors per track, and bytes per sector:

Capacity=Surfaces×Tracks/Surface×Sectors/Track×Bytes/Sector\text{Capacity} = \text{Surfaces} \times \text{Tracks/Surface} \times \text{Sectors/Track} \times \text{Bytes/Sector}

Expressing each value as a power of 22:

  • Number of surfaces = 16=2416 = 2^4
  • Tracks per surface = 128=27128 = 2^7
  • Sectors per track = 256=28256 = 2^8
  • Bytes per sector = 512=29512 = 2^9 bytes
Capacity=24×27×28×29 bytes=228 bytes\text{Capacity} = 2^4 \times 2^7 \times 2^8 \times 2^9 \text{ bytes} = 2^{28} \text{ bytes}

Since 1 Mbyte=220 bytes1\text{ Mbyte} = 2^{20}\text{ bytes}:

Capacity=28×220 bytes=256 Mbytes\text{Capacity} = 2^8 \times 2^{20} \text{ bytes} = 256 \text{ Mbytes}

2. Bits Required to Specify a Particular Sector

A sector is uniquely identified by specifying its surface, track, and sector offset. The total number of sectors in the entire disk pack is:

Total Sectors=Surfaces×Tracks/Surface×Sectors/Track\text{Total Sectors} = \text{Surfaces} \times \text{Tracks/Surface} \times \text{Sectors/Track} Total Sectors=24×27×28=219 sectors\text{Total Sectors} = 2^4 \times 2^7 \times 2^8 = 2^{19} \text{ sectors}

The number of bits required to uniquely address 2192^{19} sectors is:

log⁡2(219)=19 bits\log_2(2^{19}) = 19 \text{ bits}

Alternatively, we can sum the bits needed for each addressing component:

  • Bits for surface: log⁡2(16)=4\log_2(16) = 4 bits
  • Bits for track: log⁡2(128)=7\log_2(128) = 7 bits
  • Bits for sector within track: log⁡2(256)=8\log_2(256) = 8 bits
Total bits=4+7+8=19 bits\text{Total bits} = 4 + 7 + 8 = 19 \text{ bits}

Thus, the capacity is 256256 Mbyte and the number of bits required is 1919 bits.

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