Assume 6 bytes for the Layer 2 header, 1 byte for the end-of-frame flag, and a 40-millisecond voice payload, how much bandwidth should be allocated to the strict priority queue for five VoIP calls that use a G.729 codec over a multilink PPP link?
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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 bandwidth required for five VoIP calls using the G.729 codec over a multilink PPP link, we need to consider the size of the Layer 2 header, the end-of-frame flag, and the voice payload.
Let's break down the calculation step by step:
(20 bytes/packet) / (20 milliseconds/packet) = 1,000 bytes/second = 8,000 bits/second
Layer 2 header and end-of-frame flag: The Layer 2 header is 6 bytes, and there is an additional 1-byte end-of-frame flag. Therefore, the total size of the Layer 2 header and end-of-frame flag is 7 bytes.
Total bandwidth required for five VoIP calls: To calculate the total bandwidth required for five VoIP calls, we need to multiply the bandwidth required for one call (8,000 bits/second) by five (the number of calls) and add the bandwidth required for the Layer 2 header and end-of-frame flag (7 bytes).
Total bandwidth = (8,000 bits/second/call) x 5 + (7 bytes x 8 bits/byte) = 40,000 bits/second + 56 bits/second = 40,056 bits/second
Converting bits per second to kilobits per second (kb/s):
40,056 bits/second = 40,056/1000 kb/s ≈ 40 kb/s
Therefore, the answer is A. 87 kb/s.