Computer Networks - Transport Layer - Sequence number wrap around

Question

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?

Comments

 40 Gbps link.
1 sec = 4Gb = 5GByte
so 1 byte = 1/5G
sequence number field in TCP = 32 bits 
so number of segment can numbers  = 2³²

So 2³² bits to take exhaust = 2³² /5G = 0.858sec  = 858.99 msec approx 

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@Prashant  every "Byte" consumes a sequence but not a "bit".

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sry for that.

Answer 1

sequence number is made of 32 bits, so there are total $2^{32}$ sequences possible.
BW = 40Gbps = 5GBps

Each byte consumes 1 sequence number. Therefore

$\frac{2^{32}}{5*2^{30}}$ = $\frac{4}{5}$ = 0.8 Second will be the wrap around time.

Comments

Thanks for the solution. @manu00x

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Is Bandwidth in powers of 10 or powers of 2 ?

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10