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Question 5: A file of size K = 200 KBytes is to be sent from Node 1 to Node 5 traversing four identical intermediate links as shown in Figure below. The data link protocol operating on node 1 requires that V = 100 bits be added as overhead to any frame transmitted on the link. Assume the link transmission rate is R = 100 kb/s while the link propagation time, Tprop is equal to 1 msec. Node processing delay is ignored. An intermediate node can start transmitting the received frame immediately when it is fully received. Link 1 Link 2 Link 3 Link 4 1 2 3 5 R R R R Tprop Tprop Tprop Tprop a) If the file is NOT segmented at node 1, but rather sent as a whole together with the overhead V, compute the total transmission time, Ttotal1, till the file is completely received at Node 5. b) For this part assume the file at node 1 is divided into frames each of size F bits. Of course, as before V overhead bits are added to F data bits at node 1 before transmission on link 1. The figure below shows the case of dividing the frame into m frames. Note that Ty is the time to Link 1 Link 2 Link 4 transmit the frame on the link, F1 F2 while Tprop is the time it signal to propagate and reach the other end of the link. If the frame size F is chosen to be very small, then m is going to be large and these links are going to be transmitting overhead bits most of the time. The other extreme is when the frame size is equal to the original file size as in part (a) of this problem, then m is equal to 1. In this case the file transmission time is still not optimal. It is required to determine the best frame size that results in minimum transmission time. Hint: Let the frame size F vary from 1 bit all the way to original file size in bits, and for each value compute the corresponding T_total2. Link 3 Timele F3 F1 F2 F3 F1 F2 F3 T! Fm F1 F2 F3 Total Tprop Fm Fm Fm Time received time