# TCP Warp Around time - Sequence number Problem

Consider a long-lived TCP session with an end-to-end bandwidth of 1Gbps. The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closet integer) before this sequence number can be used again is _________.

The solution given for above question was : [(2^32)*8]/(10^9) = 34.3597 sec

But what I think it should be : [(2^32)208]/(10^9) = 687.194 sec

I have multiplied it by 20 because minimum size of TCP packet is 20Bytes.

Please help me to clear the confusion which answer should be right and why. If there is another answer for the above question please let us know.

• Looking at your question, you probably don't know what "The minimum time (in seconds, rounded to the closet integer) before this sequence number can be used again" is theoretically . Commented Jul 8, 2022 at 11:26
• Also its the TCP header which is minimum 20 bytes. Commented Jul 8, 2022 at 11:27
• @RinkeshP Thanks for your comment. I searched around a little and got "The sequence number is a counter used to keep track of every byte sent outward by a host. If a TCP packet contains 1400 bytes of data, then the sequence number will be increased by 1400 after the packet is transmitted." Commented Jul 8, 2022 at 14:48