##### Disk head request sequence for worst case

Consider a disk with the 100 tracks numbered from 0 to 99 rotating at 3000 rpm. The number of sectors per track is 100 and the time to move the head between two successive tracks is 0.2 millisecond.

1. Consider an initial set of 100 arbitrary disk requests and assume that no new disk requests arrive while servicing these requests. If the head is initially at track 0 and the elevator algorithm is used to schedule disk requests, what is the worse case time to complete all the requests?
##### 5Comments
karmjit joshi
12 Jan 2017 01:45 pm

3000 rpm

i.e. 50 rps

1 r = 1000 /50 msec = 20 msec

So rotational latency = 20/2 = 10 msec per access.

In worst case we need to go from tracks 0-99. I.e. 99 seeks

Total time = 99  * 0.2 + 10*100  = 1019.8msec = 1.019 sec

not sure

Shubham Malaviya
12 Jan 2017 01:57 pm

Sequence for the worst case should be. 0,99,1,98,..... isn't it?  If this is true then total seeks will be more

1+2+3...+99   and so total seeks is 99*100/2.

What do you say?

Arul S
12 Jan 2017 06:22 pm

Here is my try.

The time for 3000 rotations = 1 minute

then, The time for one rotation = 20 ms

In the worst case, there can be disk requests in such a way that the disk head has to move all the way to the other end of the disk (i.e., move consecutively 99 tracks at one go), after reaching the other end, the disk might need to do one full rotation to reach particular block for data transfer.

Total seek time per request = (99 tracks * 0.2 ms) + (20 ms for full rotation) = 39.8 ms

Rotational delay per request = half of the disk rotation time = 10 ms

Data transfer time is not given, so we may assume it to be negligible.

Total time taken to service 100 requests = 100 * $(T_{seek})$  + 99 * $(T_{rotat\_delay})$= 100 * (39.8 ms) + 99 * (10 ms) = 3980 + 990 = 4970 ms

rachapalli vinay kumar
21 Jan 2017 12:19 pm

Rotational delay per request-- why there is a delay?

can you explain what it is please.

rachapalli vinay kumar
21 Jan 2017 12:22 pm

R