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GO Classes DPP | GATE CS, DA | DBMS | Disk Waiting Time

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A magnetic disk rotates at $4200\ \mathrm{rpm}$. Its average positioning time is $5\ \mathrm{ms}$

The average waiting time is defined as:

$$\mathrm{Average\ positioning\ time}+\mathrm{Average\ rotational\ waiting\ time}$$

Approximately what is the average waiting time?

  1. $7\ \mathrm{ms}$
     
  2. $10\ \mathrm{ms}$
     
  3. $12\ \mathrm{ms}$
     
  4. $14\ \mathrm{ms}$

1 Answer

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First, convert RPM to rotations per second $:4200/60=70\ \mathrm{rotations/sec}$

Time for one complete rotation $:1/70\ \mathrm{sec}=14.286\ \mathrm{ms}$

For a randomly located sector, average rotational latency is half a rotation:

${14.286}/{2}=7.143\ \mathrm{ms}$

Now add the average positioning time:

$5+7.143=12.143\ \mathrm{ms} \approx\boxed{12\ \mathrm{ms}}$

Answer : C

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