Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= 109 bits-per- second)

Q. Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= 10⁹ 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    

Ans: 34

Sol:

As sequence number field of TCP is 32 bits, so there are total 2³² unique sequence number are possible (from 0 to 2³²-1), which is limit of TCP data.

But if you want to send data more than 2³² bytes in TCP, then you need to repeat this procedure after sending 2³² bytes of data or unique sequence numbers. This concept is known as wrap around which allow sending unlimited data using TCP.

Therefore, question is asking for wrap around time which is equal to pass all unique sequences first, i.e., 2³², TCP assigns 1 sequence number to each byte of data.T_{wrap−around} = (Total data) / (Bandwidth) = (2³² bytes) / (10⁹ bits per second) = (2³² * 8 bits) / (10⁹ bits per second) = 34.35 seconds = 34 (in seconds)

GATE’ answer is same as either ceiling value or floor value (i.e., 34 and 35 both are correct).