Consider the sliding window algorithm with SWS = RWS = 3, with no out-oforder arrivals, and with infinite-precision sequence numbers.

(a) Show that if DATA[6] is in the receive window, then DATA[0] (or in general any older data) cannot arrive at the receiver (and hence that MaxSeqNum = 6 would have sufficed).

(b) Show that if ACK[6] may be sent (or, more literally, that DATA[5] is in the sending window), then ACK[2] (or earlier) cannot be received. These amount to a proof of the formula given in Section 2.5.2, particularized to the case SWS = 3. Note that part (b) implies that the scenario of the previous problem cannot be reversed to involve a failure to distinguish ACK[0] and ACK[5]