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# Reply To: Packet-switched data

#32382
Rommel
Guest

You are designing your backbone connection between two routers. You know that you will generally see about 600 packets per second and 200 bytes per packet or 1600 bits per packet. The result of multiplying 600 pps by 1600 bits per packet gives us the amount of bandwidth we will need to support, 960,000 bps. We know that we can buy circuits in increments of 64,000 bps. This will give us the amount of data necessary to keep the circuit busy for 1 second. How many circuits will we need to keep delay under 10 milliseconds?
To get the traffic load we will need to:
960,000 bps/64,000 bps = 15 Erlangs of traffic load
To get the average transmission time we will need to multipy bytes per packet by 8 to get bits per packet. Then divide by 64000 bps (circuit speed) to get the average transmission time per packet
200 bytes/packet * 8 bits = 1600 bits/packet /64000 bps = 0.025 seconds to transmit or 25 milliseconds.
Delay factor 10 ms/25 ms = 0.4 delay factor

By using an erlang C table you will get a result of ~17 64K circuits