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Each user request generates 10 disk requests on average and has 10 msec service time.
So, time required for disk I/O = 10×10 msec = 100 msec.
Total time = computational time + disk I/O time = 80 + 100 msec = 180 msec.
in 180msec-->1 user's rquests are serviced
in 1 sec how many user's can generate--> 1000/180=5.5=5 users
1 user-->10 requests
5users-->50 requests.
so answer shud be 50 requests 
answer given is 5
 

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Bro,you started to calculate user ..No need to do that

you got that in 180 ms ,1 user request is satisfied..thats it 

in 1s 1000/180=5 user request is satisfied

and it ends right here!!

so 5 is the correct answer

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Now, consider pipelining as above.

Now, the for the last process, the waiting time needs to be completed. Before, the waiting time for the last process, burst time for all the processes can be overlapped.

So, subtracting waiting time for last proces from 1000 ms, we get 900 ms.

Now, max no of processes = $\frac{total time}{burst time}$ = $\frac{900}{80}$ = 11.2 $\approx$ 11 processes/users

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Another approach as we do consider in TCP.

Here, Tt (transmission time)= 80 ms   (which correlates with the main part of CPU computation)

Tp (propagation time) = 100 ms (for a total of 10 disk accesses)

Let, 'N' be the frames that can be transmitted for  100% efficiency

So, (N-1)Tt = Tp

We get N = 9/4

Now, in Tt + Tp = 180ms we can transfer 9/4 frames.

In 1000 ms, how much ?

So, in 1000 ms, frames transmitted  =  $\frac{1000 * \frac{9}{4}}{180}$ = 12.5 $\approx$ 12

This answer is extremely close to above.

But above answer is more appropriate.

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