http://gotitsolutions.org/2008/01/31/the-effect-of-latency-and-packet-loss-on-effective-throughput.html Get Information Technology Conjecture... Sun, 14 Nov 2010 23:34:36 +0000 hourly 1 http://wordpress.org/?v= http://gotitsolutions.org/2008/01/31/the-effect-of-latency-and-packet-loss-on-effective-throughput.html/comment-page-1#comment-499 Greg Schmitz Fri, 19 Sep 2008 12:19:36 +0000 http://www.gotitsolutions.org/2008/01/31/the-effect-of-latency-and-packet-loss-on-effective-throughput.html#comment-499 I'm a little curious about the math behind this. While I believe I understand the methodology for the equation (though I'm not sure on the 1.2), the values for throughput seem to be an order of magnitude off. No matter how I do the math, at .00001% I see a throughput of 379,473,319bps or 361.89Mbps, and by 1% I see 379,473bps or .36Mbps. Judging from the image, it appears that this test yielded a 3,800Mbps throughput at .00001% (aprox.) The way I see it, in the case of the .00001% probability of packet loss, the math would be: Throughput in bits/sec = ((1,250 * 8)/(100 / 1000)) * (1.2 / (.00001 / 100)^.5) So is there a discrepancy in the image, or in my math? I’m a little curious about the math behind this. While I believe I
understand the methodology for the equation (though I’m not sure on
the 1.2), the values for throughput seem to be an order of
magnitude off. No matter how I do the math, at .00001% I see a
throughput of 379,473,319bps or 361.89Mbps, and by 1% I see
379,473bps or .36Mbps. Judging from the image, it appears that this
test yielded a 3,800Mbps throughput at .00001% (aprox.) The way I
see it, in the case of the .00001% probability of packet loss, the
math would be: Throughput in bits/sec = ((1,250 * 8)/(100 / 1000))
* (1.2 / (.00001 / 100)^.5) So is there a discrepancy in the image,
or in my math?

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