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Serialization Delay?

December 15, 2009

Serialization/Access-Rate is the physical clocking speed of the interface (ie 64-kbps/128-kbps etc), which determines the amount of data that can be encapsulated on to the wire.

Serialization Delay or Serialization Rate is a constant based on the access rate of the interface. It is the time needed to place data on the physical wire.

These values are set in hardware and cannot be changed.

A data frame can be sent onto the physical wire ONLY at the serialization rate of the interface. Thus serialization delay is the size of the frame in bits divided by the clocking speed of the interface.

Serialization Delay = Frame Size/Link Speed

For example, a 1500-byte frame (12000-bits/64000-bits) will take 187.5ms to serialize (put on the wire) on a 64-kbps circuit.

Link- Frame Size (Bytes)
Speed 64 128 256 512 1024 1500
64 kbps 8 ms 16 ms 32 ms 64 ms 128 ms 187 ms
128 kbps 4 ms 8 ms 16 ms 32 ms 64 ms 93 ms
256 kbps 2 ms 4 ms 8 ms 16 ms 32 ms 46 ms
512 kbps 1 ms 2 ms 4 ms 8 ms 16 ms 23 ms
768 kbps 0.640 ms 1.28 ms 2.56 ms 5.12 ms 10.4 ms 15 ms

For low-speed WAN connections (those with a clocking speed of 768kbps or below), it might be necessary to provide a mechanism for Link Fragmentation and Interleaving (LFI) when running delay sensitive application like voice.

LFI fragments large data frames into regularly sized pieces, to interleave small delay sensitive frames into the flow, so that the end-to-end delay can be predicted accurately. This helps prevent jitter (variable delay) that would have been caused when the delay sensitive packets was delayed behind large data frames being serialized.

Two LFI tools which are commonly implemented, are MLP (Multilink PPP) on serial links and FRF.12 on Frame-Relay links.

But the question is how much delay is acceptable and what size should the fragments be?
That will depend on how much you want to limit the jitter too. Decide on the max jitter delay and work out the fragment size with this formula:

Fragment_Size = (Max_Jitter_Delay * Link_Speed_in_kbps) / 8

For example, if  a maximum variable delay of 10ms on a 128-kbps circuit is needed, then
Fragment_Size = (10ms * 128) / 8
Fragment_Size = 160 bytes

From the previous formula we can see that a 160 bytes payload size will take 10ms to serialize. So the longest a delay sensitive packet (or any packet fragment) would have to wait in the transmit-queue is 10 ms.

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2 comments

  1. Hi,

    Thanks for sharing such a nice article. I’ve one doubt. Is there any relation between NIC interface speed and the WAN speed.? Suppose if we would have used T1 technology, then would the NIC interface clock speed equal to 1.544 mbps.?
    We can set diff values to the NIC speed. Now, if we set it to AUTO, will it be equal to 1.544 mbps.?


    • I am not understanding your question.
      The clockrate specified or default assumed defines the rate at which data is sent.
      If this does not answer your question please provide more detail…



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