To calculate the total time to read 4 arbitrary pages from the disk, we need to consider several components:

1. Rotational Latency: The time to rotate the disk so that the desired sector is under the read head.
2. Seek Time: The time to move the read head to the correct track.
3. Transfer Time: The time to read the data once the read head is positioned on the correct track and sector.

Let’s go through the solution step-by-step:

Given Information

• Disk speed: 4400 RPM
• Average seek time: 8 milliseconds
• Each track has 64 sectors
• Each sector is 512 bytes
• Each page is 4 KB (4096 bytes)

Steps

1. Rotational Latency

The disk rotates at 4400 RPM (revolutions per minute), so we can calculate the time for one complete revolution: $\text{Time for one revolution} = \frac{60,000 \text{ ms}}{4400 \text{ RPM}} = 13.64 \text{ ms}$ The average rotational latency (the average time it takes for the desired sector to align under the read head) is half of this: $\text{Average rotational latency} = \frac{13.64}{2} = 6.82 \text{ ms}$

2. Seek Time

The average seek time to move the head to the desired track is given as 8 milliseconds.

3. Transfer Time

Each page is 4 KB, or $4096$ bytes. Since each sector is $512$ bytes, we need: $\frac{4096 \text{ bytes}}{512 \text{ bytes per sector}} = 8 \text{ sectors per page}$ Assuming sequential reading within each page, the time to transfer each page can be approximated as: $\text{Transfer time per page} = 8 \text{ sectors} \times \text{(time per sector)}$ To find the time per sector, we use the fact that each revolution reads 64 sectors: $\text{Time per sector} = \frac{13.64 \text{ ms}}{64} = 0.2131 \text{ ms}$ Therefore: $\text{Transfer time per page} = 8 \times 0.2131 = 1.7048 \text{ ms}$

Total Time Calculation

For each page read, we must consider:

• Seek time: 8 ms (assuming each page may be on a different track)
• Rotational latency: 6.82 ms
• Transfer time: 1.7048 ms

So, the total time per page is: $8 + 6.82 + 1.7048 = 16.5248 \text{ ms}$

Since there are 4 pages, the total time to read them is: $4 \times 16.5248 = 66.0992 \text{ ms}$

Answer

It will take approximately 66.1 milliseconds to read the 4 arbitrary pages.

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Related Questions

1. How would the total read time change if the disk speed increased to 7200 RPM?
2. What would the time be if all pages were located sequentially on the same track?
3. How does rotational latency affect performance in disk drives with faster RPMs?
4. If the sector size were larger (e.g., 1024 bytes), how would that impact transfer time?
5. How would caching affect the time required to read these arbitrary pages?

Tip: Understanding the components of disk read time (seek, rotational latency, transfer) is essential for optimizing file storage and retrieval.