8.1 Legacy Frame Relay Traffic Shaping


Can someone please explain why the following command has also been configured ?

frame-relay fragment 320

I understand from the task wording that interleaving is needed because of large files being transferred, so this seems to be the reason for using this command, but, if so, why has been chosen the 320 value ? Is it random or is it based on something I don't seem to understand ?




  • No, this is not a random value. I always think like this:

    For FRF for every DS0 (64K) of link speed, you need 80 bytes for fragment size. I don't have the task handy right now, but I assume the bandwidth for this task/link is 256 Kbps.

    So for every 64K you need 80 bytes of fragment, which will total 320 in this case.


    Good luck!

  • JoeMJoeM ✭✭✭
    • To decrease the serialization delay on the circuit ensure that all the
      shaping intervals are the smallest possible, and that a single packet
      cannot take more than one interval to be transmitted. 


    I read this and I think,  

    smallest shaping intervals possible?  = smallest Tc possible    

    single packet needs only one interval?  = fragment everything to resulting Bc size (convert from bit to byte)


  • Just in case


    How to obtain the Fragment Size:


    Serialization Delay : Fragment Size(bit) /link  bw (bps)

    So in this case we are given the Serialization Delay which is 10 ms and the link bw (256kbp)

    So the formula will be :

    Fragment Size = Serialization Delay * Link BW

    X= 0,01 * 256000

    X= 2560

    That's the fragment Size that we can send over that link in 10 ms.

    Now, on the Frame-Relay configuration we MUST set it in bytes so we must turn the 2560 bits into Bytes:

    2560/8 = 320.


    That's how you get the value




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