VoIP performance experts should be familiar with - ITU-T Y.1541 concerning IPDV (IP packet delay variation = jitter); there are many sections on IPDV, particularily also on "aggregated PDV"; - IETF PMOL, ... IP performance metrics - IETF AVT, ... RFC's for RTCP XR (RFC 3611, 5093) and ietf drafts on work in progress RTCP XR metrics -Albrecht

-----Original Message----- From: voiceops-bounces at voiceops.org [mailto:voiceops-bounces at voiceops.org] On Behalf Of Jim Dalton Sent: Freitag, 29. Oktober 2010 14:15 To: 'David Knell'; voiceops at voiceops.org Subject: Re: [VoiceOps] Question about Packet Jitter

...

Jitter is an average measure of what's essentially noise (random variations in the arrival times of packets from a mean), and so I'd expect - for two jitter sources in series producing average jitter of j1 and j2 respectively - the overall jitter to be sqrt(j1^2 + j2^2).

Intuitively, a packet which is late as a result of the first jitter source might then be further delayed or delivered early by the second - so just summing the average jitter values isn't appropriate.

--Dave

[JD wrote:] Very good point. The relationship of overall jitter to be sqrt(j1^2 + j2^2) makes sense. However, I am having trouble working through a proof of the math. Any suggestions of a reference on jitter that might address this?

Many thanks for your insight,

Jim D.

...

-----Original Message----- From: voiceops-bounces at voiceops.org [mailto:voiceops- bounces at voiceops.org] On Behalf Of David Knell Sent: Friday, October 29, 2010 1:37 AM To: voiceops at voiceops.org Subject: Re: [VoiceOps] Question about Packet Jitter

On Thu, 2010-10-28 at 16:26 -0400, Scott Berkman wrote:

...

The only portion missing is the jitter through the SBC itself, which really should be negligible. Assuming that is 0, your equation below would be correct.

I think I'd disagree with this. Jitter is an average measure of what's essentially noise (random variations in the arrival times of packets from a mean), and so I'd expect - for two jitter sources in series producing average jitter of j1 and j2 respectively - the overall jitter to be sqrt(j1^2 + j2^2).

Intuitively, a packet which is late as a result of the first jitter source might then be further delayed or delivered early by the second - so just summing the average jitter values isn't appropriate.

--Dave

...

-----Original Message----- From: voiceops-bounces at voiceops.org [mailto:voiceops- bounces at voiceops.org] On Behalf Of Jim Dalton Sent: Thursday, October 28, 2010 8:34 AM To: voiceops at voiceops.org Subject: [VoiceOps] Question about Packet Jitter

I have a question about calculating jitter. Consider the call diagram below. A SIP call flows from the source device to the Session Border Controller (SBC) to the destination device. All RTP packets are proxied through the SBC.

+========+ +=====+ +=============+ | Source | Ingress Call Leg | SBC | Egress Call Leg | | Destination | | Device |------------------->| |------------------>| Device | +========+ +=====+ +=============+ Jitter Src to SBC ------------------->

Jitter Source to Destination --------------------------------------------->

If the packet jitter is known for the Ingress Call Leg from the source to SBC and for end to end packet flow from the source to the destination, is it possible to calculate jitter for the Egress Call Leg from the SBC to the destination device?

I do not think the following relationship is accurate. (jitter Source to Destination) less (jitter Src to SBC) = (jitter SBC to Destination)

Can anyone provide some guidance on this question?

Thank you,

Jim Dalton VoIP Least Cost Routing, Analysis, Billing 1.404.526.6053 www.TransNexus.com

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