3 votes 3 votes 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? Computer Networks computer-networks tcp sequencenumber transport-layer wrap-around-time + – shane.126 asked Oct 13, 2017 • retagged Dec 15, 2019 by KUSHAGRA गुप्ता shane.126 5.2k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Prashant. commented Oct 13, 2017 i edited by Prashant. Oct 13, 2017 reply Follow Share 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 2 votes 2 votes Manu Thakur commented Oct 13, 2017 reply Follow Share @Prashant every "Byte" consumes a sequence but not a "bit". 1 votes 1 votes Prashant. commented Oct 13, 2017 reply Follow Share sry for that. 2 votes 2 votes Please log in or register to add a comment.
Best answer 7 votes 7 votes 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. Manu Thakur answered Oct 13, 2017 • selected Oct 13, 2017 by sourav. Manu Thakur comment Share Follow See all 5 Comments See all 5 5 Comments reply Prashant. commented Oct 13, 2017 reply Follow Share calculation will be reverse. :) its 4/5 1 votes 1 votes Manu Thakur commented Oct 13, 2017 reply Follow Share oh! yes, typo! 0 votes 0 votes shane.126 commented Oct 13, 2017 reply Follow Share Thanks for the solution. @manu00x 0 votes 0 votes Harsh Kumar commented Jun 14, 2018 reply Follow Share Is Bandwidth in powers of 10 or powers of 2 ? 0 votes 0 votes rewrihitesh commented Aug 9, 2018 reply Follow Share 10 0 votes 0 votes Please log in or register to add a comment.
