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Bandwidth Overhead over DSL connections

Internet Service Providers typically rate their connectivity services based on download and upload bandwidth. So for example you may subscribe to an aDSL connection rated at 2 Mbps (Megabits per Second) download and 1 Mbps upload.

The traffic reaching the provider’s network from the aDSL modem is then routed to its eventual destination by traversing an ATM (Asynchronous Transfer Mode) network, and to do this, packets are broken up into fixed-size cells, with routing information added to the cell in the cell header. In simple terms, this process requires additional data (routing headers and padding).

In many cases, the Internet Service Provider defines the bandwidth available as the bandwidth INCLUDING this overhead.

So it can be important to understand the real bandwidth utilized for each phone call, INCLUDING this overhead.

Example 1: A call using the G.711u codec

  • Each RTP packet contains 20ms of audio (typical)
  • Each 20ms of audio requires 160 bytes

Each second of audio will require 50 packets, each containing an audio payload of 160 bytes.

Before being transmitted over the network the IP packet will contain:

  • 160 bytes for the audio payload
  • 12 bytes for the RTP header
  • 8 bytes for the UTP header
  • 20 bytes for the IP header

…for a total of 200 bytes per packet

If the transmission medium is Ethernet, the IP packet is encapsulated in an Ethernet Frame which adds an 18-byte header, for a total of 218 bytes per frame * 50 packets per second * 8 bits per byte.

This equates to 87200 bits per second or 87.2 kbps.

However, on an ATM circuit, each IP packet of 200 bytes will first have a header and trailer added to it (typically an 8-byte header and an 8-byte trailer), and then the resulting 216 byte payload needs to be split into cells of 48 bytes each, and a 5-byte routing header added to each cell – meaning that each cell is 53 bytes in size. So the IP packet would be split across 5 cells, for a total of 53 bytes per cell * 5 cells per packet * 50 packets per second * 8 bits per byte.

This equates to 106000 bits per second, or 106 kbps.

Example 2: A call using the G.729 codec

  • Each RTP packet contains 20ms of audio (typical)
  • Each 20ms of audio requires 20 bytes

Each second of audio will require 50 packets, each containing an audio payload of 20 bytes.

Before being transmitted over the network the IP packet will contain:

  • 20 bytes for the audio payload
  • 12 bytes for the RTP header
  • 8 bytes for the UTP header
  • 20 bytes for the IP header

…for a total of 60 bytes per packet

If the transmission medium is Ethernet, the IP packet is encapsulated in an Ethernet Frame which adds an 18-byte header, for a total of 78 bytes per frame * 50 packets per second * 8 bits per byte.

This equates to 31200 bits per second or 31.2 kbps.

However, on an ATM circuit, each IP packet of 60 bytes will first have a header and trailer added to it (typically an 8-byte header and an 8-byte trailer), and then the resulting 76 byte payload needs to be split into cells of 48 bytes each, and a 5-byte routing header added to each cell – meaning that each cell is 53 bytes in size. So the IP packet would be split across 2 cells, for a total of 53 bytes per cell * 2 cells per packet * 50 packets per second * 8 bits per byte.

This equates to 42400 bits per second, or 42.4 kbps.

Example 3: A call using the GSM codec

  • Each RTP packet contains 20ms of audio (typical)
  • Each 20ms of audio requires 33 bytes

Each second of audio will require 50 packets, each containing an audio payload of 33 bytes.

Before being transmitted over the network the IP packet will contain:

  • 33 bytes for the audio payload
  • 12 bytes for the RTP header
  • 8 bytes for the UTP header
  • 20 bytes for the IP header

…for a total of 73 bytes per packet

If the transmission medium is Ethernet, the IP packet is encapsulated in an Ethernet Frame which adds an 18-byte header, for a total of 91 bytes per frame * 50 packets per second * 8 bits per byte.

This equates to 36400 bits per second or 36.4 kbps.

However, on an ATM circuit, each IP packet of 73 bytes will first have a header and trailer added to it (typically an 8-byte header and an 8-byte trailer), and then the resulting 89 byte payload needs to be split into cells of 48 bytes each, and a 5-byte routing header added to each cell – meaning that each cell is 53 bytes in size. So the IP packet would be split across 2 cells, for a total of 53 bytes per cell * 2 cells per packet * 50 packets per second * 8 bits per byte.

This equates to 42400 bits per second, or 42.4 kbps.

Conclusion

If the audio stream is exchanged over an internet connection that uses an ATM infrastructure, it is very likely that there will be no bandwidth saving when using G.729 instead of GSM as a codec.

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