On Mon, 10 Nov 1997, Jon Mansey wrote:
>
> Hi,
>
> Please excuse a newbie question, but where does the 0.6 derive from in
> these calcs? Is it the latency introduce by the trip up and back to the
> satellite? If so it seems rather high, my calcs give me more like 250ms.
>
0.6 sec (or more accurate between 534ms and 580ms depending on the
location of the ground station and your workstation) is coming from the
double-hop path; from your workstation --> satellite --> destination
and then destination -->satellite --> sender(your workstation)
Each way is about 250ms to 270ms. So doubling that it gets you approx.
0.6s. This is actually the total RTT which is used for the calculation of
the BDP.
Pambos
> Puzzled,
>
> > to 65535 x 8/0.6s = 870Kbps. Solaris TCP/IP stack is able to open both
> > the transmit window/buffer and receive window/buffer upto 65535 bytes.
> >
> > So the maximum achievable throughputs will look like the following:
> >
> > Transmit_TCP_stack Maximum Throughput
> > Microsoft TCP: approx. 109Kbps
> > Solaris TCP approx. 870Kbps
> >
>
> [email protected] Chief Science Officer
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