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Need help calculating pps in VOICE

MACattack

Hi I hope somebody can help me. I am reading Cisco press ONT and I am in calculating BW for VOICE. I cannot figure out how a 20 ms voice got 50 pps, Ido understand from adding a 40 bytes overhead for RTP, UDP and IP and data link depends what type.

Below is the mentioned information is Cisco press.

Calculate the packet rate. The packet rate is inversed packetization period (converted to seconds). For example, if the packetization period is 20 ms, which is equivalent to 0.020 seconds, the packet rate is equal to 1 divided by 0.020, resulting in a packet rate of 50 packets per second (pps).

In the book if we ave 160 bytes of packet + 40 bytes overhaed (RTP,UDP and IP) = 200 bytes
with out data link or any compression.

I want to know what is the pps of the 200 bytes.

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Comments

Steve10393

MACattack wrote:

In the book if we ave 160 bytes of packet

What? Explain what this statement means in more detail?

EdTheLad

I dont think you are understanding what you are reading.PPS stands for packets per second, if you are transmitting 50 packets per second how long does it take to transmit 1 packet.Answer 1sec/50(number of packets in interval) = .02seconds = 20ms
Reversing the question, if it takes 20ms to transmit a packet,how many packets can you transmit per second. 1/.02 = 50 packets per second.
Now your question "I want to know what is the pps of the 200 bytes" you have not given enough information as far as i can see, the size of the packet will effect the packetization, the larger the packet the larger the packetization time and hence the smaller the pps.
Maybe there is a formula you need but since i never studied voice i dont know it.

dtlokee

The number of packets per second (pps) is determined by the sampling interval. Typical sampling intervals will be 20 or 30ms which means each packet contains 20 or 30ms of voice. This results in a pps rate of 50 for a 20ms sample, or 33 for a 30ms sample. Depending on the codec used you can calculate the total BW required for each call including the overhead of the L2, IP, UDP and RTP headers.

for example a g7.11 codec will sample at a rate of 64,000 bits per second

so 64,000/8 = 8000 Bytes per second

if each packet contains 20ms of voice (.020s)

8,000 * .020 = 160 bytes of voice payload per packet.

also the packet rate will be 1/.02 = 50

so to figure the bandwidth take the size of the voice payload and add the header overhead (40 bytes)

160 + 40 = 200

then multiply by the pps rate

200 + 50 = 10,000 bytes per second

then multiply by 8 to get bits per second

10,000 * 8 = 80,000 bits per second.

That would be the required BW for each call, also you should add 10% for call processing overhead (RTCP and your signaling protocols, MGCP, H.323, SIP, H.24

and if you're using VAD you can typically reduce the amount by 25-23% (Cisco says 35% but 25 might be a better estimate if you don't want the call quality to suffer when there many calls)

now compare that to the g.729 codec (MOS score of 3.9 vs 4.1 for the g.711) at a codec rate of 8,000 bits per second.

8,000 / 8 = 1,000

1000 * .02 = 20 bytes of voice payload per packet

20 + 40 = 60 bytes with header overhead

60 + 50pps = 3,000 bytes total per second

3,000 * 8 = 24,000 bits per second.

Even though the g.729 MOS score is only slightly less it's required BW is much less (30%) of the g.711 codec. If you can throw in cRTP ir dtops even more.

Netstudent

So if you knew how much BW was needed for a call, wouldn't you need to know your erlangs to get a estimated total bandwidth capacity needed for a WAN connection? Or am I off on that one?

MACattack

Sorry guys for the incomplete information.

I got it, thanks Ed it's clear to me now.

Yeah it mentioned that it all depends on the packetizzation.

For normal Cisco by default used 20 ms of voice / packet

20 ms of voice using g.729 codec has 8 000 bps
RTP, UDP and IP got 12 bytes, 8 bytes and 20 bytes total of 40 bytes

8000 x 20 /1000 = 160 bits per secs / 8 total of 20 bytes

so my voice payload is 20 byte and 40 bytes for overhead RTP = 60 bytes since it is not multiples of 8 an additional 4 byte padding is added. (64 bytes)

64 bytes + ppp is my DLayer of 6 bytes is 80 bytes without using cRTP or tunneling.

So I got it
if I have 20 ms of voice / 1 = .05 or reverse 1/.02 = 50 packets per sec
30 ms of voice /1 = 33 pps
40 ms /1 = 25 pps

It is getitng decrease as more voice packet you put in IP packet.
Thanks

dtlokee

Netstudent wrote:

So if you knew how much BW was needed for a call, wouldn't you need to know your erlangs to get a estimated total bandwidth capacity needed for a WAN connection? Or am I off on that one?

This deals more with the total number of calls (or trunks) required to provide the desired probability the call will be completed (blocked calls). For the BW requirements you need to figure how many calls you want to support (from your erlangs calc) then multiply by the BW for each call (and add in a little for call processing overhead). This is how you typically figure the LLQ settings for your WAN connections.

So with 2 locations connected via a t-1 and you want to allow 4 calls with the g7.29 codec between them the total BW required will be 4 * 24,000 (this will vary slightly with the type of encapsulation) or 96Kb/s. The erlangs part comes in when you determine how many calls you want to allow at the same time.

MACattack

Thanks dtlokee, it's clear to me now to get the number of pps I need to divide the voice in ms.
1 month to go for ONT yeah