. Consider the sliding window algorithm with SWS = RWS = 3,with no out-of-order arrivals and with infinite-precisionsequence 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, thatDATA[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) impliesthat the scenario of the previous problem cannot be reversed toinvolve a failure to distinguish ACK[0] and ACK[5].

Consider we use the sliding window algorithm with SWS= RWS=3 Sliding window algorithm is used where…

Answered: Consider the sliding window algorithm…

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

Problem 1PE

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Use of XOR function

XOR function or exclusive OR function was introduced in Excel 2013. This function is available under excel logical functions. For the given statement, the XOR function will return true only when either of the statements is true. It returns false when both the statements are true or when both the statements are false.

Question

. 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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