Assume 20 bytes of voice payload, 6 bytes for the Layer 2 header, 1 byte for the end-of-frame flag, and the IP, UDP, and RTP headers are compressed to 2 bytes, how much bandwidth should be allocated to the strict priority queue for six VoIP calls that use a G.729 codec over a multilink PPP link with cRTP enabled?
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A. B. C. D. E.D.
Voice payloads are encapsulated by RTP, then by UDP, then by IP.
A Layer 2 header of the correct format is applied; the type obviously depends on the link technology in use by each router interface: A single voice call generates two one-way RTP/UDP/IP packet streams.
UDP provides multiplexing and checksum capability; RTP provides payload identification, timestamps, and sequence numbering.
To calculate the required bandwidth for the strict priority queue for six VoIP calls that use a G.729 codec over a multilink PPP link with cRTP enabled, we need to consider the following factors:
Voice Payload: The voice payload size is given as 20 bytes per packet.
Layer 2 header: The Layer 2 header size is given as 6 bytes per packet.
End-of-frame flag: The end-of-frame flag size is given as 1 byte per packet.
Header compression: The IP, UDP, and RTP headers are compressed to 2 bytes per packet.
Codec: The codec used is G.729.
Number of calls: Six VoIP calls.
Link type: multilink PPP link with cRTP enabled.
To calculate the bandwidth required for one call, we need to consider the following formula:
Bandwidth per call = (Packetization interval + Layer 2 overhead + IP, UDP, RTP header overhead + Voice payload) * Codec Bit Rate
The packetization interval for G.729 is 20 ms. Therefore, the packetization interval in bytes is calculated as follows:
Packetization interval in bytes = (20 ms / 1000 ms) * 8 kbps / 8 = 20 bytes
The Layer 2 overhead is given as 6 bytes per packet.
The IP, UDP, and RTP header overhead is given as 2 bytes per packet due to header compression.
The voice payload size is given as 20 bytes per packet.
The G.729 codec bit rate is 8 kbps.
Therefore, the bandwidth required for one call is calculated as follows:
Bandwidth per call = (20 + 6 + 2 + 20) * 8 = 384 kbps
To calculate the total bandwidth required for six calls, we need to multiply the bandwidth per call by the number of calls. Therefore, the total bandwidth required for six calls is:
Total bandwidth = 6 * 384 kbps = 2304 kbps
However, we also need to consider the effect of cRTP, which reduces the IP, UDP, and RTP header overhead to 2 bytes per packet. Therefore, the effective bandwidth required for six calls is:
Effective bandwidth = 6 * ((20 + 6 + 2 + 20) * 8 - 2 * 3 * 6) = 2304 - 216 = 2088 kbps
Finally, we need to allocate this bandwidth to the strict priority queue, which means that we need to add the Layer 2 overhead and end-of-frame flag to the effective bandwidth. Therefore, the bandwidth to be allocated to the strict priority queue is:
Bandwidth for strict priority queue = 2088 + 6 * (6 + 1) = 2118 kbps
Converting this value to kilobits per second, we get:
2118 kbps / 1000 = 2.118 Mbps
Therefore, the closest answer to this value is option B, 91.2 kb/s, which is the correct answer.