average time in disk system

Q . Ahard disk  system has the following has the following parameters:

number of tracks= 500

number of sector per track  =100

number of bytes per sector= 500

time taken by the head  to move from one track to adjacent track=1ms

rotation speed= 600 rpm

what is the average   time taken for tranferring 250 bytes from the disk?

option :

A. 300.5 ms

B.  255.5 ms

C. 255  ms

D. 300 ms

4Comments
Pritam Prasun @pritam
11 Jun 2016 04:07 pm

Important: On an average, to serve each request, head has to move half the number of tracks and  half the rotation.

So, time taken to do the above work = 250 ms (for half tracks) + (1/600)*(60)*(1/2)*1000 ms (for half rotation)

= 300 ms

Now, calculate the time needed to read 250 bytes.
= (250/500)*(1/100)*(1/10)*1000
= 0.5 ms

Hence, total time = 300.5 ms

Hradesh @hradeshpatel
11 Jun 2016 08:18 pm

sir...check this solution..

given that read/write head one track to adjacent track=1ms

so that no. of track=500

so   seek time=(1+2+3+4....+499)/500

   ( 499*500)/(500*2) =249.5 ms 

Avg. time to transfer = Avg. seek time + Avg. rotational delay + Data transfer time

So, rotational delay = 60 / 600 = 0.1 s

In 1 rotations we can transfer the whole data in a track which is equal to number of sectors in a track * bytes per track

= 100 * 500 = 50,000

i.e., in 0.1 s, we can transfer 50,000 bytes.
Hence time to transfer 250 bytes = 0.1 * 250 / 50,000 = 0.5 ms

Avg. rotational delay = 0.5 * rotational delay = 0.5 * 0.1s = 50 ms

So, average time for transferring 250 bytes = 249.5 + 50 + 0.5 = 300 ms

Pritam Prasun @pritam
11 Jun 2016 08:58 pm

Hradesh, This is a great catch. Your answer is more precise and correct.

Hradesh @hradeshpatel
11 Jun 2016 09:01 pm

Given that read/write head one track to adjacent track=1ms

so that no. of track=500

so   seek time=(1+2+3+4....+499)/500

   ( 499*500)/(500*2) =249.5 ms 

Avg. time to transfer = Avg. seek time + Avg. rotational delay + Data transfer time

So, rotational delay = 60 / 600 = 0.1 s

In 1 rotations we can transfer the whole data in a track which is equal to number of sectors in a track * bytes per track

= 100 * 500 = 50,000

i.e., in 0.1 s, we can transfer 50,000 bytes.
Hence time to transfer 250 bytes = 0.1 * 250 / 50,000 = 0.5 ms

Avg. rotational delay = 0.5 * rotational delay = 0.5 * 0.1s = 50 ms

So, average time for transferring 250 bytes = 249.5 + 50 + 0.5 = 300 ms

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