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+23 votes

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 a set of disk requests to read data from tracks 32, 7, 45, 5 and 10. Assuming that the elevator algorithm is used to schedule disk requests, and the head is initially at track 25 moving up (towards larger track numbers), what is the total seek time for servicing the requests?
  2. 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?
asked in Operating System by Veteran (68.8k points)
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3 Answers

+19 votes
Best answer
Answer A)

We are using SCAN - Elevator algorithms.

We will need to go from 25->99->5. (As we will move up all the way to 99,servicing all request, then come back to 5.)

So total seeks = 74+94= 168

Total time = 168*0.2 = 33.60000

Answer for B)

We need to consider rotational latency too - >

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
answered by Veteran (48.6k points)
selected by

@Ahwan,you mentioned

Nothing is mentioned about the sector, for each track, it can be found anywhere, either in the next or in last. 
So take average rotational latency. Simple. 

So in the worst case,it can be in the last. By worst time it mean ,that my system should not take any time more than worst case time under all scenarios.

Now if i have such a request ,where we always need to do one complete rotation for each request so we should not consider average rotational latency.

@rahul sharma 5 
No, I did not mean worst case as total time in the worst case.
In worst case where can be the sector, in best case where can be the sector.
So I took the average of them.

I guess the answer will be same if you find total time if all sectors are last and total time if all sectors are in current position..if you will take average of them you will get same answer.

@rahul @joshi

Generally we do 100*(10+0.2)

But if we do such way, will ans be wrong?

or it could be like this 0,1,1,1,1.....1(now here seek time will be 1 since there is one track movmnt, but inter-track roatioanl time will be same as above),

@reena here only one track movement?

I am not getting, what u mean in this example??

could u elaborate it a bit?

what is the worse case time to complete all the requests?

Hi Guys,

I think instead of average rotational latency we should take time taken in full rotation. What is your opinion ?

If we are taking rotational latency because of "we do not know which sector is to be selected in those 100 random track request", then also in part - a) if request completion time asked instead of "seek time" then we have to consider the same avg. rotational latency in addition?
@bruv depends on question
One thing to notice here

request should go from 25 ->99 ->10

isnot it?

why upto 5 track request is taking?

Is there calculation mistake or am I missing somewhere?

Plz chk

@srestha ji, 

Is there calculation mistake 


am I missing somewhere?



25 ->99 ->5 .Means staring from 25 goto 99 and service all request and come back to 5 ans service all request on the way.
–3 votes


end track: 99


total seeks=(99-25)+(99-5)=74+94=168

total seek time = 0.2 * 168 ms


from 0 to 99

worst case time = 0.2 * 99 ms
answered by Boss (8.5k points)
You missed rotational latency
–3 votes

Question is about how much time required to satisfied 100  requests...

1 request time is = Tseek +TrotLatency + data transfer time

Datatransfer time = 0(as not given any info)

99 seeks and 100 rotations required in worst case...

So 99 * seek time + 100*rotational latency...
answered by Junior (601 points)

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