GATE CS 2018 Q.35 (CS Q.25): OSI/TCP-IP Layering, Physical Layer and Switching

Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= $10^9$ 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 _______.

Your answer

Official answer and solution

Official answer: 34
From: Official GATE 2018 answer key

Solution

Compute the time needed to cycle through the full 32-bit TCP sequence number space at 1 Gbps.

TCP sequence numbers are 32 bits wide and count individual bytes transferred. The total sequence space consists of $2^{32} = 4,294,967,296$ sequence numbers.

The transmission bandwidth is 1 Gbps:

\text{Rate} = \frac{10^9\text{ bits/sec}}{8\text{ bits/byte}} = 1.25 \times 10^8\text{ bytes/sec}

The time required to wrap around the entire sequence number space is:

T = \frac{2^{32}\text{ bytes}}{1.25 \times 10^8\text{ bytes/sec}} = \frac{4,294,967,296}{125,000,000} \approx 34.3597\text{ seconds}

Rounding to the closest integer gives 34 seconds.

Final answer

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

Source

Paper: GATE 2018 Computer Science and Information Technology (CS)
Question: 35 of 65, printed as CS Q.25
Section: Computer Science and Information Technology

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