Problem Breakdown:

Specifications:

• Number of cylinders: 18,500
• Number of heads: 16
• Number of sectors per track: 75
• Spin rate: 5400 RPM
• Seek time between adjacent tracks: 2 ms
• Sector size: 1000 bytes
• To calculate:
  • (a) Disk capacity
  • (b) Time to read the entire disk sequentially
  • (c) Time to transfer 350 bytes of data

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(a) Disk Capacity

The total capacity of the disk is calculated using the formula:

$\text{Disk Capacity} = \text{Cylinders} \times \text{Heads} \times \text{Sectors per Track} \times \text{Sector Size}$

Substitute the given values:

$\text{Disk Capacity} = 18,500 \times 16 \times 75 \times 1000 \, \text{bytes}$

$\text{Disk Capacity} = 22,200,000,000 \, \text{bytes} = 22.2 \, \text{GB}$

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(b) Time to Read Entire Disk Sequentially

The time to read the disk includes:

1. Rotational latency (average time for the disk to spin half a revolution): $\text{Rotational latency} = \frac{60}{\text{Spin rate}} \div 2 = \frac{60}{5400} \div 2 = 5.56 \, \text{ms}$

2. Data transfer time: The disk contains the following number of sectors: $\text{Total sectors} = 18,500 \times 16 \times 75 = 22,200,000 \, \text{sectors}$

   Total data size: $\text{Data size} = 22,200,000 \times 1000 = 22.2 \, \text{GB}$

   Time to read one revolution: $\text{Time per revolution} = \frac{60}{5400} = 11.11 \, \text{ms}$

   Total time to read the disk: $\text{Total time} = \text{Time per revolution} \times \text{Number of revolutions}$

   Since one revolution covers all the sectors per track, the number of revolutions required is proportional to the total number of tracks: $\text{Number of revolutions} = \text{Cylinders} \times \text{Heads} = 18,500 \times 16 = 296,000$

   $\text{Total time} = 11.11 \, \text{ms} \times 296,000 = 3,288,560 \, \text{ms} \, \text{or} \, 54.81 \, \text{minutes}$

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(c) Time to Transfer 350 Bytes of Data

To transfer data, we calculate the data transfer time based on the disk's rotational speed and the sector size. The transfer time for one sector (1000 bytes) is:

$\text{Time per sector} = \frac{\text{Time per revolution}}{\text{Sectors per track}}$ $\text{Time per sector} = \frac{11.11 \, \text{ms}}{75} = 0.148 \, \text{ms}$

For 350 bytes (partial sector): $\text{Time for 350 bytes} = \frac{350}{1000} \times \text{Time per sector} = 0.35 \times 0.148 = 0.0518 \, \text{ms}$

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Summary:

• (a) Disk Capacity: 22.2 GB
• (b) Time to read entire disk sequentially: 54.81 minutes
• (c) Time to transfer 350 bytes: 0.0518 ms

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Do you want further breakdown or detailed explanations for any part?

Related Questions:

1. How is rotational latency calculated for different RPMs?
2. What happens to transfer times if sector size increases?
3. How does seek time impact total read times in random access?
4. Can the time to read the entire disk decrease if the heads read in parallel?
5. How does RPM relate to overall disk performance?

Tip: Always consider the rotational latency when calculating access times, as it can significantly impact the overall performance.