GATE 2018 CS Q.25: solution and answer

GATE 2018 · CSQ.25 · 1 mark · NAT

Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= $1 0^{9}$ bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is ______.

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Answer and solution

ANSWER

The sequence number can be reused after at least about 34.36 seconds, which rounds to 34 seconds.

Official answer: 34 to 35Source: Official answer key, GATE 2018 CS (IIT Guwahati)

TCP numbers every byte of the data stream, and the sequence number field is 32 bits wide. A sequence number can therefore take $2^{32}$ distinct values and wraps around after $2^{32}$ bytes.
The starting value 1234 does not matter. The same sequence number comes round again only after exactly $2^{32}$ bytes have been sent, whatever the starting point.
Data to be sent, in bits: $2^{32} \times 8 = 2^{35} = 34, 359, 738, 368$ bits.
The sender cannot push data faster than the link bandwidth, $1 0^{9}$ bits per second, so the minimum time is the one at full bandwidth.
$Minimum time = \frac{2 ^{35}}{1 0 ^{9}} = \frac{34 , 359 , 738 , 368}{1 0 ^{9}} \approx 34.36 seconds .$
Rounded to the closest integer, the time is 34 seconds.

KEY IDEA

TCP sequence numbers count bytes in a 32-bit field, so they wrap after $2^{32}$ bytes. The minimum reuse time is the time to send $2^{32}$ bytes at full link speed.

WATCH OUT FOR

Counting $2^{32}$ bits instead of $2^{32}$ bytes (forgetting the factor of 8) gives about 4.29 seconds. Another slip is subtracting the starting value 1234, which is irrelevant to the wrap-around time.

From the GATE 2018 CS paper. Every question of this paper · Download the full paper · All 12 TCP questions, with notes

GATE 2018 CS Q.25: solution and answer

Answer and solution

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