A Quick Video Bandwidth question

[mayank.sharma]

Hello All,

I have a question.... And was hoping for some clarification on video bandwidth used by a VC call.

If you see the attached screen shot the Call rate from a end point for the call is 384kbps. But if you notice that the Transmit (199kbps) Video and Receive (221kbps) Video which equals to 420kbps which is well above 384kbps limit for this call. This would imply that you will need to provision 384kbps X 2 = 768kbps  (+ Overheads) for this call on the network from the endpoint. Correct me if I am wrong here?

This is my understanding as I have read somewhere previously Bandwidth is always symmetric for VC. If you need 1Mbps download for a call, then you also need 1Mbps upload. And that would explain why we see more then 384kbps for Transmit + Receive below.

Thanks in advance...

Regards,

Mayank

[Wayne DeNardi]

To try to put it really simply, a "384k" call is "384k up and 384k" down simultaneously, as you put it, a "symmetric" bandwidth, and this rate is the maximum that it will use and it will burst up and down within this limit.

So, in your case per your screenshot, at that point in time you did the capture, it was only using 199/384 of the transmit, and 211/384 of the receive, so well within the limit of the call.

Wayne
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[mayank.sharma]

Hi Wayne,

Thanks for the response.

That is my understanding too. "384k" call is "384k up and 384k" down.

To elaborate on my understanding..

A 384k channel is established to send the stream captured from the camera out from endpoint and A 384k channel is established to Receive the stream from the other end and display on screen connect to endpoint. Of course the presentation will eat into this if required. It will go the max of 384k max for Transmit and max of 384k for Receive but never above this but it can always use less than that depending on the fact if there is more or less motion/movement in the VC call itself.    

Hence from the networking prospective this would imply to me that I would need to provision on the network 384 X 2 = 768k (+ Overheads) for a call from this endpoint. Correct?

Regards,

Mayank

[Wayne DeNardi · Accepted solution]

No, a 384k call needs 384k of bandwidth.  It's 384k up and down simultaneously, you don't add the two numbers together.

Similar to when talking about a network switch, a 10Mb switch port does 10Mb in each direction at the same time - it's not a 20Mb switch port.

To try to put it in a physical world example, imagine the "bandwidth" is a window... you can shine a light through the window, and you could also shine a light through it in the opposite direction.  If you shine a light through it in both directions as the same time, doesn't make the window twice as much.

Wayne

[mayank.sharma]

All Good Wayne... : )

You are right... Just got my networking concepts a bit mixed up with the VC concepts..

Thanks for your help.

Regards,

Mayank