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Here the thing to keep in mind is : 

For every byte in TCP , we have a sequence number assigned .

Hence for wraparound of all bytes possible , we need to cover maximum number of sequence numbers i.e. we have to start from sequence number 0 and continue till (232 - 1) as sequence number field in TCP is of 32 bits . After (232 - 1) , the numbering will begin from 0 again which is known as "wraparound of sequence numbers" .

Hence data involved before wraparound of sequence numbers = 232 B  (as 1 sequence number is for 1 byte)

                                                                                           = 235 bits

   Given data rate of the network                                            = 200 Mbps

                                                                                           = 200 * 106 bps

  Hence time taken for wraparound of sequence numbers         = Data  size  /  Data rate

                                                                                           = 235 / (200 * 106)  s

                                                                                           = 171.799 s

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Number of sequence number possible=$2^{32} \text{byte} =2^{32} \times 2^3 \text{bits}=2^{35} \text{bits}$

Bandwidth$ =200 Mbps$

$\Rightarrow 200 \times 10^{6} \text{bits}\,\text{in } 1 \text{second}$

$\Rightarrow 1 \text{bits}\,\text{in } \frac{1}{200 \times 10^6} \text{second}$

$\Rightarrow 2^{35} \text{bits}\,\text{in } \frac{2^{35} }{200 \times 10^6} \text{second}$

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shane.126 asked Oct 13, 2017
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TCP operates over a 40 Gbps link. If TCP uses the full bandwidth continuously, how long (in msec) would it take the sequence number to wrap around completely?