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.