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Suppose you are designing a sliding window protocol for a $1Mbps$ point to point link which has one way latency of $1.25s$. Assuming that each frame carries $1kB$ of data, determine the minimum number of bits required for the sequence number.

Answer provided - $9$ bits.
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$Sender Window size = 1+2a.$

$Propagation Time (one way latency ) = 1.25sec$

$L = 1*1024*8 bits$

$Bandwidth = 1mbps = 10^6b$

$Transmission Time = L/B = 1024*8/10^6 = 0.008192sec$

$a=Propagation time / Transmission Time = 152.5$

$Sender Window size = 1+2a = 306.17$

If we have to 306.17 pkts then sequence no should have

$Sequence no field = log(306.17) = 9$

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