Calculating Bandwidth Utilization
The amount of milliseconds (of voice) in a packet determines the number of packets required per
second. If the RTP is set to 20 milliseconds of voice per packet, you will need 50 packets per
second.
Formula = 1000ms / number of ms of voice per packet = packets / second required.
Example: 1000ms / 20ms = 50 packets / sec)
Each packet acquires various amounts of information depending on the network type. For
example, there is approximately a 40 byte header added for Ethernet packets. Each time a
packet is sent that header must be attached, which will increase the packet size, and by
extension, the amount of bandwidth required for the connection.
To estimate the per-call TCP/IP bandwidth utilization (Layer 3) is simple. Assuming one call with
no silence suppression enabled,
Bandwidth = ((packets / sec * 320 bits / packet) + CODEC bit rate) * 1.05
Example 1: G.711 @ 20ms/packet (50 packets/second)
Bw = ((50packets / sec * 320 bits / packet) + 64Kbps) * 1.05
Bw = ((16Kbps) + 64Kbps) * 1.05
Bw = ((80 Kbps) * 1.05
Bw = 84Kbps
Example 2: G.729a @ 30ms / packet (33.3 packets / second)
Bw = ((33.3packets / sec * 320 bits / packet) + 8Kbps) * 1.05
Bw = ((10.7Kbps) + 8Kbps) * 1.05
Bw = ((18.7 Kbps) * 1.05
Bw = 19.7Kbps
If Silence Suppression is on:
Peak Bandwidth is calculated in the manner above; however, the average bandwidth is
estimated at 50% of peak.
Example 3: G729a with 30ms/packet
Average Bandwidth = 19.7Kbps * 0.5 = 9.9Kbps
Xcelerator IP Per Connection Bandwidth Calculation
The RTP plus UDP plus IP headers will add on 40 bytes. The Ethernet header and trailer account
for another 18 bytes of overhead, for a total of at least 58 bytes of overhead before there are any
voice bytes in the packet. These headers, plus the Ethernet header, produce the overhead for
shipping the packets. This overhead can range from 20% to 80% of the bandwidth consumed
over the LAN and WAN.
2
July 2007
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