Hi Steven,
That is an option I have been looking at but that means I have ping the
server every 1 second or so in order to determine the timeout. I was
wondering if there is a way to determine this by looking at the time it took
to send the data, and am now wondering if the SendString function does not
introduce this? If I measure the time before calling the function and after
the function execution is complete, does this tell me how fast the link is?
If I send 70 bytes and it took 100ms to complete, does this mean I can sent
7000 bytes per second?
Regards,
W
On Mon, May 4, 2009 at 3:05 PM, Steven Kamradt <[email protected]>wrote:
> You could try getting ping times between the
>
> -Steven
>
> On May 4, 2009, at 4:28 AM, [email protected] wrote:
>
> > Hi Everyone,
> >
> > I am wondering if there is a way that synapse can help me determine
> > the speed of a TCP/IP Link between a client and server. I have a
> > scenario where some thin or unreliable links can be used to link a
> > TCP client and server and have to deal with the following:
> >
> > 1) If the link fails, data will be archived until the link returns
> > and then the data will be sent to the server
> > 2) If there is more data at the client than the link can handle, I
> > want to limit/ throttle the data that is sent.
> >
> > The data that is sent from the client is normal CSV string, each
> > only about 70 bytes long. As the client receives the data, it
> > processes it and passes it on to a central server. In a previous
> > implementation I sent the data from the client to the server and
> > then the server would respond with 1 byte to acknowledge the receipt
> > of the data. I then measured the time it took from the time the
> > message was send to the time the acknowledgment was received from
> > the server and measure the speed of the link using the number of
> > bytes that was sent vs the time it took to send the data. I am now
> > in a scenario where I have no control over the server and therefor
> > cannot rely on the acknowledgment packet coming back, and therefor
> > this question.
> >
> > Kind Regards,
> >
> > W
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