Consider a variant of the stop and wait protocol, where the sender instead of sending a single frame and waits for an ack, it sends 2 frames and waits for a single ack. When an ack arrives, the sender sends the next two frames and so on (so each ack arriving to the sender indicates that two frames have been received by the receiver). If one or both frames do not make it to the receiver, the receiver would not ack any frames. If the sender does not hear from the receiver after a timeout value, it will retransmits the last two frames. Assume that each data frame is equal to 1500 Bytes and each ack frame is equal to 40 Bytes. The link between the sender and the receiver has a 10 Mbps bandwidth and each one-way propagation delay of 10 msec. Answer the following questions: [2 pts each] (a) What is the total round-trip time (RTT) from sending the ?rst bit of the ?rst frame and receiving an ack back? (b) What is the total throughput for this protocol, in Mbps when no packets get lost? What percentage of the link bandwidth is wasted due to this protocol? (c) What is the minimum number of sequence numbers we need on packets for this protocol to work successfully? How many bits would you dedicate in headers to carry the sequence number? (d) Draw the time-line diagram (include times at each event) between the sender and the receiver until the ?rst two frames make it to the receiver. Assume that the second frame was lost the ?rst time it was transmitted and assume that the timeout value used by the sender is chosen to be twice the RTT (obtained from part (a)). (e) What is the total throughput for this protocol, in Mbps, if the channel always drop one of the two frames?