OS Scheduling

Question

We are given a computer system consisting of a CPU and a disk. We are told that, each user request has a compute time of 80 msec and an average generates 10 disk requests. We are further told that, the service time at the disk is 10 msec. What is the maximum number of user requests that can be satisfied per second?

Answer 1

Ans.  5.555 user reqest per second

Each User request requires 80 m sec of CPU Time and 10 disk request as IO Time.

total time required for a process = CPU Time + IO Time

                                                   = 80 ms + 10 disk request * ( 10 ms per disk request )

                                                  = 180 ms

Number of process per second = 1 sec / 180 ms = 1/ (180 * 10^-3 ) = 5.555

Comments

Ok, i have a doubt that why is overlapping of I/O request not considered i.e. when a process goes for I/O, cpu can still continue processing other available programs in memory.That way number of programs will be more and hence the number of users will be more. Only the last I/O operations would not be overlapped in that case.

Eg- process 1 executes 40ms and is in wait state for I/O. then process 2 can continue its execution for that 100 ms and then say process one returns in ready queue and is executed when processor is free...

pls explain....??

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i have not considered I/O overlapping because its not given in the question.

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but the question is asking the maximum number of user which could further be increased from 5 to (1000-100)/80 i.e. 900/80 i.e. 11.25  if we use overlapping I/o

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I agree with you point. But see this question once

https://gateoverflow.in/1843/gateoverflow.in

here they have exactly mention in which order processes will be schedule and I/O times are overlapped as many possible.

Answer 2

First process completes cpu at 80, cpu assigned to second and first process  do io, second process completes cpu at 160, first process completes io  at 180

Similarly 9 th process completes io at 980

Process 10 to 12 cannot complete io but ll finish cpu time  at 960

Coz only one disk  multiple io request at same time to same disk not available,  disk will have an io queue where processes ll be lined up

Ans must be 9

Comments

Yes.. that can be considered but then what is the correct approach for such questions

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Wats given ans?

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as given by @sandeep singh..

Answer 3

The answer is 9. 

If P1 comes and after 80 ms goes for I/O, it will complete its I/O at 180ms. 

Now after 80ms, CPU takes on P2. After 160 ms CPU takes on P3. This goes on and on, till P12, whose CPU burst will be over at 960ms. We won't consider processes after that as we have to find out how many within a second. 

But in this time of 1000 ms we have to find out, how many processes completed their I/O?

So after 80ms, I/O of P1 will start, which will be completed at 180 ms. Then P2, which will be completed at 280 ms. This will go on till P9, which will be completed at 980 ms. 

This calculation can also be done by forming an equation. If a Gantt Chart of I/O is drawn, starting at t=80 ms and I/O at every 100 ms, 

80+ x(100) <=1000
x=9 (integer value)