Disk Access Time

$\text{Access Time} = \underbrace{\text{Seek Time} + \text{Rotational Latency}}_\text{OS can optimize this} + \text{Transfer Time}$

An average seek needs a half rotation.


Suppose a disk has 7200 RPM, an average seek time of 8ms, and a transfer rate of 100 MB/s. State the average time to read 4KB at random.

7200 rotations in 60 seconds $\rightarrow$ 1 rotation in $\frac{60}{7200}\times 1000\text{ms}=8.3\text{ms}$

$\Rightarrow$ average seek $4.15\text{ms}$

Transfer time for $4\text{KB}$ is $\frac{4\text{KB}}{100 \frac{\text{MB}}{\text{s}}} = 0.04 \text{ms}$

$\text{Access Time} = 8.3 \text{ms} + 4.15 \text{ms} + 0.04 \text{ms} = 12.2 \text{ms}$

Matthis Kruse
Masters Student

My research interests include programming languages and compilers.