Wrap around time

Question

Given bandwidth is 128 Gbps. (use giga has 2 power 30), the total time for a complete wrap-around of sequence numbers in TCP

Answer 1

128 Gbps

Therefore,

1 sec -> 2^7 * 2^30 bits

1 sec -> 2^37 bits

1 sec -> 2^37 / 2^3  Bytes

1 sec -> 2^34 Bytes

As TCP is Byte stream protocol,

1 sec -> 2^34 sequence numbers.

But, we have only 32 bits in sequence no. field,

Therefore,

2^34 seq nos. ->  1 sec

2^32 seq nos -> Time for wrap around

Time for wrap around = 2^32 / 2^34 sec

Time for wrap around = 0.25 sec.