1) General
At IP-transmissions the data stream consists of
| • | Payload (pure audio data e.g. one or more MPEG frame) |
| • | IP Overhead |
2) IP Overhead in Bytes
The IP overhead in Byte is at :
2.1 UDP
46 Byte consisting of:
| • | IEEE802.3: 18 Byte |
| • | IP: 20 Byte |
| • | UDP: 8 Byte |
2.2 RTP
58 Byte consisting of:
| • | IEEE802.3: 18 Byte |
| • | IP: 20 Byte |
| • | UDP: 8 Byte |
| • | RTP: 12 Byte |
Note:
The RTP header is 12 Byte and not 16 Byte long since no contribution source identifiers are used.
2.3 MPEG TS
50 Byte consisting of:
| • | IEEE802.3: 18 Byte |
| • | IP: 20 Byte |
| • | UDP: 8 Byte |
| • | MPEG TS Header: 4 Byte |
3) IP Overhead percentage calculation
3.1 UDP
If the packet size mode is set variable then the payload is also rounded to the next full audio frame (MPEG) or audio block.
Here some examples for IP overhead calculation at UDP connections with variable packet size:
The following diagram shows the percentaged IP overhead at
- Algorithm: MPEG L3
- Sample rate: 48 kHz
- UDP
- packet size = 180 Byte (default value)
- packet size mode: var (default value)
3.2 RTP
Similar to UDP calculation but it must be considered that the size of all headers is 58 Byte instead of 46 Byte.
Example:
Algorithm: MPEG L2
Sample rate: 48 kHz
Bit rate: 256 kbit/s
Frame length = 256/48 * 144 Byte = 768 Byte
IP Overhead = ((768 Byte + 58 Byte)/768 Byte –1) * 100 % = approx. 7.6%
3.3 MPEG TS
Each MPEG TS packet is 188 Byte long and consist of
| • | payload: 184 Byte |
| • | MPEG TS Header: 4 Byte |
Since 
uses MPEG TS with UDP the whole overhead of one packet is 50 Byte (see item 2.3).
3.3.1 Theoretical optimum overhead
The optimum (i.e minimum) overhead is reached if
| • | every MPEG frame just needs one packet (and therefore the overhead just must be sent once for every frame) |
| • | frame length = 184 Byte |
Then the IP overhead is calculated to:
IP Overhead = ((184 Byte + 50 Byte)/184 Byte -1 ) * 100 % = approx. 27.2 %
3.3.2 Real overhead
Since the real MPEG frame length is usually different to 184 Byte and cannot transport part of two frames in a MPEG TS frame, the real overhead is usually much bigger then the theoretically optimum one.
Example 1 (bad overhead):
Algorithm: MPEG L3
Sample rate: 48 kHz
Bit rate: 128 kbit/s
Frame length: 128/48 * 144 Byte = 384 Byte
Frame length of 384 Byte means that 3 MPEG TS packets are necessary to transport one frame. This means:
IP Overhead = ((3*(MPEG TS packet + All other overheads))/384 Byte) – 1) * 100 %
= ((3*(188 Bytes + 46 Byte)/384 Byte –1) + 100% = approx. 82.8%
Example 2 (good overhead):
Algorithm: MPEG L3
Sample rate: 44.1 kHz
Bit rate: 56 kbit/s
Frame length: 56/44.1 * 144 Byte = 183 Byte
Frame length of 183 Byte means that just one MPEG TS packets is necessary to transport one frame. This means:
IP Overhead = (((MPEG TS packet + All other overheads))/183 Byte) – 1) * 100 %
= (((188 Bytes + 46 Byte)/183 Byte –1) + 100% = approx. 27.9 %