GATE CS 2022 Q.60: Transport Layer: UDP, TCP Flow and Congestion Control, Sockets

Consider the data transfer using TCP over a 1 Gbps link. Assuming that the maximum segment lifetime (MSL) is set to 60 seconds, the minimum number of bits required for the sequence number field of the TCP header, to prevent the sequence number space from wrapping around during the MSL is____________.

Your answer

Official answer and solution

Official answer: 33
From: Official GATE 2022 answer key

Solution

Calculate the maximum number of bytes that can be transmitted within the Maximum Segment Lifetime and find the minimum number of bits needed to represent that space.

Link bandwidth $B = 1\text{ Gbps} = 10^9\text{ bits/second}$ .

Converting to bytes per second:

\text{Rate} = \frac{10^9}{8} = 1.25 \times 10^8\text{ bytes/s}

Maximum Segment Lifetime (MSL) = 60 seconds.

The total number of bytes transmitted during MSL is:

N = 1.25 \times 10^8 \times 60 = 7.5 \times 10^9\text{ bytes}

To prevent the sequence number space from wrapping around during MSL, the sequence number range $2^k$ must be at least $N$ :

2^k \ge 7.5 \times 10^9

Evaluating powers of 2:

$2^{32} = 4,294,967,296 \approx 4.29 \times 10^9 < 7.5 \times 10^9$
$2^{33} = 8,589,934,592 \approx 8.59 \times 10^9 \ge 7.5 \times 10^9$

Therefore, the minimum number of bits required is $k = 33$ .

Final answer

Common trap

Assuming the standard 32-bit TCP sequence number field without computing whether $2^{32}$ wraps around in 60 seconds at 1 Gbps.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2022 Computer Science and Information Technology (CS)
Question: 60 of 65
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source