Joseph W. Doherty Sat, 08/09/2008 - 04:14
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If you thoughput you're thinking of is data throughput, the maximum throughput would be the ratio between the data payload and all overhead. Or, payload usage divided by the total bandwidth consumption.


For instance, if you were using TCP and sending payload in 1460 segments, there's minimally the 40 bytes of IP and TCP overhead, 18 bytes of Ethernet L2 overhead (assuming no VLAN tag, etc.) and 20 bytes of Ethernet L1 overhead. Or 1400/1538 * 100 Mbps = 91 Mbps. If you were only sending 1 byte of payload per frame, then you would have 1/84 * 100 Mbps = 1.19 Mbps throughtput.

mahesh-gohil Sun, 08/10/2008 - 18:28
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Dear


can you provide me any link to such document which should be inline with IUT standatrd


--Mahesh

mahesh-gohil Sun, 08/10/2008 - 20:00
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Dear Joseph,


please provide link to formula or any rfc for that


it is urgent


mahesh

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