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

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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. Consider the sliding window algorithm with SWS = 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 receiver
(and hence that Max SeqNum = 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].
Transcribed Image Text:. Consider the sliding window algorithm with SWS = 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 receiver (and hence that Max SeqNum = 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].
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