Question 2. (b) [Seems to be an interesting question]

We think the answer is R/M-(N/(N+1)D) where in the best case scenario (i.e., for maximum throughput) the last packet for every window is dropped, and we do not lose any more packets till the window moves. Therefore, on an average, we waste 1 packet, i.e., an efficiency of N/N+1 for every time we are forced to drop a packet. This happens D times every second and the maximum number of packets sent every second is R/M, hence the result.