Wow this one really my hurt my head… The video training I understood, but when I applied the concept to practice questions it never worked and I was going round in circles.
This is how I was first shown how to calculate bandwidth requirements:
- Payload Size (g.729 default payload = 20 bytes)
- Add L3 Overhead of 40 Bytes
- Add L2 overhead (Ethernet = 18, Frame Relay = 6 etc…)
- Multiply by packets per second value for codec (G7.29 = 50PPS)
- * 8 to go from bytes to bits
That all makes sense.. but it never worked when applied to actual examples.
So after much research, I found a great post from someone on Cisco who totally nailed it…:
*Turns out where I really going wrong was with cRTP taking the Layer 3 payload from 40 bytes to 2-4 bytes. Also the actual math approach is different here and it simply works for all real world scenarios or test questions…
I found it a little bit incorrect and too complicated to calculate the actual bandwidth used by a voice call as it is explained in the CCDA OCG:
- the WAN headers size total is 7 instead of 6: the OCG forgot the byte for the end-of-frame flag; I agree that’s only a slight difference on WAN but a bigger difference on Ethernet with 4 more bytes.
- the simple formula isVoice bandwidth (bps) = codec bit rate * (voice whole packet size) / (voice payload size)
You can find below the whole accurate information to calculate the voice bandwidth per call with 2 examples.
Use the following assumptions when calculating voice bandwidth:
- IP/UDP/RTP header uses 40 bytes (20+8+12).
- cRTP reduces the IP/UDP/RTP header to 2 or 4 bytes; cRTP is not available on Ethernet.
- The WAN Layer 2 header (MPPP or FRF.12) adds 7 bytes (6+1 byte for the end-of-frame flag) on a point-to-point circuit.
- Ethernet adds 18 bytes (14+4 bytes of Frame Check Sequence (FCS) or Cyclic Redundancy Check (CRC))
- G.711 uses a default 160 bytes payload size and G.729 20 bytes.
- Voice packet size = (Layer 2 header) + (IP/UDP/RTP header) + (voice payload).
- Voice bandwidth (bps) = codec bit rate * (voice packet size) / (voice payload size)
As an example, calculate the WAN bandwidth used at a site that will have 10 concurrent G.729 calls with cRTP and a default voice payload of 20 bytes.
- G.729 codec is used: 8 kbps codec bit rate.
- cRTP = 2-byte IP/UDP/RTP header.
- Default voice payload= 20 bytes.
- WAN header = 7 bytes.
- Voice packet size = 7 bytes + 2 bytes + 20 bytes = 29 bytes.
- BW per call = (8 * 29)/20 = 11.6 kbps.
- BW for 10 calls = 11.6 * 10 = 116 kbps.
Here is a second example: Calculate the WAN bandwidth used by a G.711 call with no cRTP and a default voice payload of 160 bytes.
- G.711 codec is used: 64 kbps codec bit rate
- IP/UDP/RTP header = 40 bytes
- Default voice payload= 160 bytes
- WAN header = 7 bytes
- Voice packet size = 7 bytes + 40 bytes + 160 bytes = 207 bytes
- BW per call = 64000 * 207 / 160 = 82.8 kbps