COURSE DATA AND COMPUTER COMMUNICATIONS Suppose that we have

COURSE: DATA AND COMPUTER COMMUNICATIONS

Suppose that we have a point-to-point communication system using the Stop and Wait flow control. The round trip time (RTT) for this system is 20ms and we’ve measured the transmission time as 2ms. Recall that RTT is the total amount of time taken to send one frame successfully to another station (and, of course, receiving and acknowledgement). What is the propagation time for this system? Draw a timing diagram to support your answer.

ANY HELP PLEASE?

Solution

public ArrayList roundRobinJarjestus(ArrayList pstlst)whereas (i < pstlst.size())method is taken
pworkTime = p.getTooaeg(); //execute time
time = time + pworkTime;

if (((Protsess) pstlst.get(i+1)).getSaabumisaeg() < time)point in time is under time passed
queue.add(pstlst.get(i+1));
}

if (pworkTime-kvant > 0)over zero and still left one thing to execute
p.setTooaeg(pworkTime-kvant);
queue.add(p);
}
uuspst.add(queue.get(i));
i = i+1;
}
come back uuspst;
}