Consider bandwidth of line is 100 Mbps and sequence number of field consists 32 bits on a TCP machine. #techtud

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Consider bandwidth of line is 100 Mbps and sequence number of field consists 32 bits on a TCP machine. How much time it will take to cover all possible sequence numbers? ________________ sec (integer value only).
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is it 343.5 sec
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I think you are right. Please provide the answer. Here is the question and its answer, which I think is wrong. http://www.techtud.com/example/consider-bandwidth-line-100-mbps-and-sequence

As Tcp is byte streamed protocol it means we numbered each byte transmitted.

Here Sequence field is 32 bits, So we can uniquely numbered or identify =232 Bytes

Bandwidth is 100 Mbps.

Means 100 Mb   data is transmitted in 1 sec

(108 )/8 byte     data is transmitted in 1 sec

1 byte data is transmitted in 8/108 sec

232 byte data is transmitted in  (232 *8)/108 sec=235/108 seconds

Solving it you will get 343.597 seconds

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Hi Guys, I know 1 new sequence number is consumed consumed for SYN, FIN or 1 byte of Data (means for 100 Byte 100 sequence number will be used). But sometimes TCP sends Dummy packet for example when receivers window size is zero sender keeps polling receiver about ... is also know as Silly window Syndrome problem). So in such kind of cases does new sequence number is used for every dummy packet ?
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