CCIE Collaboration Exam: Bandwidth Allocation for VoIP Calls | Ethernet Network

# Calculate Bandwidth Allocation for VoIP Calls with G.722 Codec | CCIE Collaboration Exam

### Question

Assume 18 bytes for the Layer 2 header and a 10-millisecond voice payload, how much bandwidth should be allocated to the strict priority queue for three VoIP calls that use a G.722 codec over an Ethernet network?

### Explanations

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A. B. C. D. E.

A.

References: http://www.cisco.com/en/US/tech/tk652/tk698/technologies_tech_note09186a0080094ae2.shtml.

To calculate the amount of bandwidth required for three VoIP calls that use a G.722 codec over an Ethernet network, we need to consider the following:

• Layer 2 header size: 18 bytes
• Voice payload size: 10 milliseconds
• Codec used: G.722
• Number of calls: 3

First, let's calculate the size of the voice payload:

G.722 codec uses a sample rate of 16 kHz and a bit rate of 64 kbps. This means that for every millisecond of audio, we have:

64 kbps / 8 bits = 8 KB of audio data

For a 10-millisecond voice payload, the size of the audio data is:

8 KB * 10 = 80 KB

Next, let's calculate the total packet size:

The total packet size is the sum of the voice payload size, Layer 2 header size, and any additional Layer 3 and Layer 4 overhead. Assuming no additional overhead, the total packet size is:

80 KB + 18 bytes = 80.018 KB

To convert this to kilobits, we multiply by 8:

80.018 KB * 8 = 640.144 kbps

This is the bandwidth required for one VoIP call. To calculate the bandwidth required for three VoIP calls, we simply multiply by 3:

640.144 kbps * 3 = 1920.432 kbps

However, we need to allocate this bandwidth to the strict priority queue. This means that we need to subtract any bandwidth allocated to other traffic classes. Assuming that no bandwidth is allocated to other traffic classes, the bandwidth allocated to the strict priority queue is:

1920.432 kbps = 1920.432 / 1000 = 1.920432 Mbps

Finally, we need to convert this to kilobits per second (kb/s):

1.920432 Mbps * 1000 = 1920.432 kb/s

Rounding this to the nearest tenth, we get:

1920.432 kb/s ≈ 1920.4 kb/s

Therefore, the answer is option B, 347.8 kb/s.