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Got a doubt while studying about LLC layer. From Network layer I came to know that IP fragmentation will be done, For L4 TCP will segment the data.

Here, the question is Protocols starting from Layer-2 (For eg : STP , RSTP), who will fragment the data.

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    And I'll just ask it once more: could you please look at the preview before posting your questions? Once again the markup was a mess. Thank you.
    – Teun Vink
    Dec 23 '13 at 13:02
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    Layer 2 protocols that support fragmentation in Layer 2 (that I know about): Multilink PPP, ATM, by default fragments Layer 3 traffic into 53-bytes cells, via the AAL ATM Adaptation Layer.
    – aseaudi
    Dec 31 '13 at 22:46
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Layer 2 does not do any fragmentation. It is up to L3 to pass data to L2 in a packet/frame size that will already match L2's MTU. In fact, the reason L3 does fragmentation in the first place is is because of the limitations of the L2 Protocol.

IP doesn't care if the packet size is 1500 bytes of 9999 bytes, it just knows what its underlying L2 protocol can handle. The "Packet Length" field in the IP header is a 16 bit value, which means IP can create a packet as big as 65536 bytes if necessary, but only if the underlying L2 fabric can support it.

Now specifically speaking for "L2 protocols" like CDP, STP, VTP, (etc...), those by design send very small sized frames that should never need to be broken up into smaller frames. So what you said is, in a way, correct, that the protocol itself handles the fragmentation (by means of not needing to send frames that would require any additional fragmentation).

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Layer-2 (For eg : STP , RSTP) , who will fragment the data .

STP, RTSP, etc... are Ethernet payloads, and Ethernet does not support fragmentation.

Offhand, the only common layer2 protocol I can come up with that does support fragmentation is MLPPP; MLPPP fragmentation is commonly used to make L2 Frames small enough so voice traffic on E / T-carrier circuits doesn't see too much delay.

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  • So .. Protocol it self handles the fragmentation ? Dec 23 '13 at 12:40
  • STP & RSTP don't need fragmentation
    – This
    Dec 23 '13 at 13:06
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    Beacause it send small amount of data ? Dec 23 '13 at 13:09
  • yes, the application normally doesn't need much and if it does, it knows how to break the frames into MTU-sized chunks
    – This
    Dec 23 '13 at 13:19
  • Any protcol is there to miplemnt this fragmentation on DataLink layer ? Dec 24 '13 at 14:32
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Actually breaking up of PDUs takes place only at two layers:1)At transport layer called SEGMENTATION. 2)At network layer called FRAGMENTATION.

The need for fragmentation as mentioned in earlier answers are as per the restrictions upon on L-2 interfaces(E.g:for Ethernet it is 1500 bytes).This value is called MTU (Maximum Transmission Value).If Ethernet interfaces receive frames larger than this default MTU value they will drop those frames.Hence it has to be decided at L-3 level whether or not to fragment the IP datagram as per L-2 requirement.All the process of fragmentation and re-segmentation are done using multiple fields in IP HEADER. And to answer your question there is no fragmentation done at L-2 level for L-1 layer requirement.

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Every interface has an MTU(MAx Transmision unit). For Ethernet it is 1500 bytes(usually). So the payload needs to be fragmanted (at Layer 3 ) so that it can fit into this max 1500 bytes. This 1500 bytes is the payload for the Layer 2.

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  • Yep, he was asking about Layer 2. So i explained that the fragmenting takes place at layer 3.
    – mihai
    Dec 24 '13 at 7:20
  • STP,RSTP are Layer 2 technologies that make sure there is no loop on the topology. They have nothing to do data fragmentation. Fragmentation takes place at layer 3.
    – mihai
    Dec 27 '13 at 9:17
  • Of course the layer 2 protocol decides how much data it can handle. In case of STP these are BPDU, and these are control packets. Data traffic is still ehternet.
    – mihai
    Dec 27 '13 at 9:24

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