Questions 4: A disk has 42 recording surface has 6000 cylinders. There is an average of 500 sectors per track. Each sector contains 512 Bytes of data. a. What is the maximum number of bytes that can be stored? b. What is the data transfer rate in bytes per second at a rotational speed of 4200rpm? C. Suggest a a a suitable addressing scheme for the disk, assuming that there are 512 bytes/ Sector.

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Answer:

Part (a): Maximum Number of Bytes That Can Be Stored

Given:

  • Number of recording surfaces: 42
  • Number of cylinders: 6000
  • Number of sectors per track: 500
  • Number of bytes per sector: 512

To find the maximum number of bytes that can be stored, we need to calculate the total number of sectors and then multiply by the bytes per sector.

  1. Tracks Per Cylinder:

    • Each cylinder has one track per recording surface.
    • Total tracks per cylinder = 42 (one for each surface).

  2. Total Tracks:

    • Total tracks = Number of cylinders * Tracks per cylinder
    • Total tracks = 6000 cylinders * 42 tracks/cylinder
    • Total tracks = 252,000 tracks.

  3. Total Sectors:

    • Total sectors = Total tracks * Sectors per track
    • Total sectors = 252,000 tracks * 500 sectors/track
    • Total sectors = 126,000,000 sectors.

  4. Total Bytes:

    • Total bytes = Total sectors * Bytes per sector
    • Total bytes = 126,000,000 sectors * 512 bytes/sector
    • Total bytes = 64,512,000,000 bytes.

So, the maximum number of bytes that can be stored is 64,512,000,000 bytes.

Part (b): Data Transfer Rate in Bytes Per Second

Given:

  • Rotational speed: 4200 rpm (revolutions per minute)
  • Number of sectors per track: 500
  • Number of bytes per sector: 512

To find the data transfer rate in bytes per second, we need to calculate how many bytes can be read in one second.

  1. Rotational Speed in Revolutions per Second:

    • Rotational speed in rpm = 4200 rpm.
    • Rotational speed in rps (revolutions per second) = 4200 rpm / 60 seconds/minute = 70 rps.

  2. Bytes Per Revolution:

    • Bytes per revolution = Sectors per track * Bytes per sector
    • Bytes per revolution = 500 sectors/track * 512 bytes/sector
    • Bytes per revolution = 256,000 bytes.

  3. Data Transfer Rate:

    • Data transfer rate = Bytes per revolution * Revolutions per second
    • Data transfer rate = 256,000 bytes/revolution * 70 revolutions/second
    • Data transfer rate = 17,920,000 bytes/second.

So, the data transfer rate is 17,920,000 bytes per second.

Part (c): Suitable Addressing Scheme for the Disk

A suitable addressing scheme for the disk should allow us to uniquely identify each sector on the disk. Given the structure of the disk (cylinders, surfaces, tracks, and sectors), the typical addressing scheme used is Cylinder-Head-Sector (CHS) addressing. Here's how it can be defined:

  1. Cylinder (C):

    • Range: 0 to 5999 (since there are 6000 cylinders).

  2. Head (H):

    • Range: 0 to 41 (since there are 42 recording surfaces).

  3. Sector (S):

    • Range: 1 to 500 (sectors are usually numbered starting from 1).

Each sector can be addressed by a combination of these three values (C, H, S).

Example of CHS Addressing:

  • Cylinder (C): 1234
  • Head (H): 21
  • Sector (S): 305

In this example, the specific sector is located on:

  • Cylinder 1234
  • Head 21
  • Sector 305

Summary

  • The maximum number of bytes that can be stored is 64,512,000,000 bytes.
  • The data transfer rate is 17,920,000 bytes per second.
  • A suitable addressing scheme for the disk is the CHS (Cylinder-Head-Sector) addressing scheme.

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