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+25 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 (59.5k points)
edited by | 1.8k views

3 Answers

+20 votes
Best answer

Answer for (A):

We are using SCAN - Elevator algorithms.

We will need to go from $25\rightarrow 99 \rightarrow  5$. (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 $= 168 \times 0.2 = 33.60000$

Answer for (B):

We need to consider rotational latency too $\rightarrow $

$3000$ 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 $= 99  \times 0.2 + 10 \times 100  = 1019.8 \ msec = 1.019 \ sec$

answered by Boss (42.5k points)
edited by

@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.
In the worst case, can the requests come in the form - 0, 99, 0, 99 .... so on. In that case for every two requests 99 seek would be needed. Correct me If I am wrong here...
–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 Loyal (8.2k 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 (599 points)

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