Consider TCP data transfer over a 1 Gbps link with maximum segment lifetime (MSL) = 60 seconds. The

Q: What is GATE-CS?

GATE-CS is a competitive examination conducted in India. This question is from the 2022 GATE-CS Computer Networks paper.

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This GATE-CS 2022 Computer Networks question has a medium difficulty level and is marked as an important question.

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Question

Consider TCP data transfer over a 1 Gbps link with maximum segment lifetime (MSL) = 60 seconds. The minimum number of bits required for the sequence number field to prevent wrap-around during the MSL is ___

Accepted Answer

Answer: 33Concept -- TCP Sequence Number Space (from Kurose and Ross, Computer Networking; RFC 793):To prevent sequence number wrap-around, the sequence number space must be large enough that no sequence number is reused within 2*MSL (the maximum time a segment can remain in the network).However, wrap-around must be prevented within the MSL (maximum segment lifetime). Bytes transmitted in one MSL = bandwidth * MSL = 1 Gbps * 60 s = 10^9 bits/s * 60 s = 60 * 10^9 bits = 60 * 10^9 / 8 bytes = 7.5 * 10^9 bytes.Number of sequence numbers needed = 7.5 * 10^9. Bits needed = log2(7.5 * 10^9) = log2(7.5) + log2(10^9) = approximately 2.9 + 29.9 = 32.8. Round up to 33 bits.Minimum bits = ceil(log2(7.5 * 10^9)) = 33 bits.

Consider TCP data transfer over a 1 Gbps link with maximum segment lifetime (MSL) = 60 seconds. The minimum number of bits required for the sequence number field to prevent wrap-around during the MSL is ___

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Consider TCP data transfer over a 1 Gbps link with maximum segment lifetime (MSL) = 60 seconds. The minimum number of bits required for the sequence number field to prevent wrap-around during the MSL is ___

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