Hard disk

Question

A certain hard disk rotates at 6000rpm. It has 1KB data per sector and 64 sectors per track and has seek time = 5sec. What is the time required to read data from 800 random sectors ??

Please solve this

Answer 1

Tavg = Tseek + Trot + Ttrans 

Tseek=Avg seek time

Trot=Avg rotational time

Ttrans=Transfer Time of 1 sector

 6000 rotations in 60 sec

1 rotaion can complete in = 60sec/6000 = 10ms

now in 10ms can read 64 * 1KB data 

To read one sector of size 1KB =(10ms * 1KB / 64KB) = (10/64)ms

Here condition is given that all sectors are reads from random location so we need seek time , rotation time for each sector plus data read time.

Tseek=5sec = 5000ms given

Trot=10/2 =5ms

Tavg= (5000+5+(10/64) ) * 800ms = 4004125ms

 

Comments

@papesh , if it is successive then what will be the ans ?

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is it transfer time *800 + seek time + rotational latency ?

Answer 2

is the answer 2004125 millisec ?