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Consider the sliding window algorithm with \( S W S=R W S=3 \), with no out-of-order arrivals and with infinite-precision sequence numbers.
(a) Show that if DATA \( [6] \) is in the receive window, then DATA[D] (or in general any older data) cannot arrive at the receiver (and hence that MaxSeqNum \( =6 \) would have sufficed).
(b) Show that if \( A C K[6] \) 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 \( \mathrm{ACK}[\mathrm{O}] \) and \( \mathrm{ACK}[5] \).
Consider the sliding window algorithm with SWS $=\mathrm{RWS}=3$, with no out-of-order 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 recelver (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 $\mathrm{ACK}[\mathrm{O}]$ and $\mathrm{ACK}[5]$.
(a) Show that if DATA \( [6] \) is in the receive window, then DATA[D] (or in general any older data) cannot arrive at the receiver (and hence that MaxSeqNum \( =6 \) would have sufficed).
(b) Show that if \( A C K[6] \) 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 \( \mathrm{ACK}[\mathrm{O}] \) and \( \mathrm{ACK}[5] \).
Consider the sliding window algorithm with SWS $=\mathrm{RWS}=3$, with no out-of-order 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 recelver (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 $\mathrm{ACK}[\mathrm{O}]$ and $\mathrm{ACK}[5]$.
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