Consider a disk drive with a disk speed of 4400 rpm and an average seek time of 8 milliseconds. Each track has 64 sectors, each sector is 512 bytes and each page is 4KB. How long will it take in milliseconds to read 4 arbitrary pages (say 2, 70, 7, 90) in a file without knowing much else about the placement of those pages on the disk?

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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Would you like more details or have any questions?

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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.

Learn how to calculate disk read time with a 4400 RPM disk drive and an 8 ms average seek time. This example shows how to compute rotational latency, seek time, and transfer time for reading 4 arbitrary pages. Ideal for computer science students exploring disk performance metrics.

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