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Consider the idea behind Fig. $4-21,$ but now for a disk with a mean seek time of $6\: msec,$ a rotational rate of $15,000\: rpm,$ and $1,048,576$ bytes per track. What are the data rates for block sizes of $1\: KB, 2\: KB,$ and $4\: KB,$ respectively?


in Operating System by Veteran (59.5k points) | 15 views

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We have:

$15000\;\text{rpm} = 60 \;\text{seconds}$

$\therefore \;1\;\text{rpm} = \frac{60}{15000} = 0.004 \;\text{seconds} = 4\;\text{msec}$

We know that:

Disk Latency = average seek time + average rotational delay + transfer time + controlled overhead.

Now, average access time neede to read $\mathbf k$-bytes $=8 + \frac{4}{2} + \frac{\mathbf k}{262144}\times 4$

For $1-\text{KB}$ block access time $=10.015625\;\text{msec}$

For $2-\text{KB}$ block access time $=10.03125\;\text{msec}$

For $3-\text{KB}$ block access time $=10.0625\;\text{msec}$

So, there is not much difference between the access time of all three.

Answer: $\therefore $ Data rates are: $102,240\;\frac{\text{KB}}{\text {sec}}, \; 204, 162\;\frac{\text{KB}}{\text {sec}}, \;\text{and} \;407,056\;\frac{\text{KB}}{\text {sec}}$

by Boss (19.2k points)
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