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When reading through the Ethernet frame format in IEEE 802.3 , the EtherType was explained as below:

  1. 0 - 1500 (Decimal) comes under packet data payload length.
  2. 1536(0x600) and above it means the value is determining the type of the frame.(Eg. 0x800 stands for IPV4)

What about the values in between 1501 to 1535? Why these values been left off?

Note: On the Wiki link it is been mentioned it is not defined. But not finding as explanation in any standardized documents.

Arun Chettoor
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1 Answers1

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These values are reserved to avoid ambiguity (e.g. when 802.1Q tagging is used, slightly increasing the frame size).

Using the Ethertype field for the frame length is obsolete though, Ethernet II framing (using the Ethertype field to indicate the protocol carried as payload) far outnumbers any other frame type. Instead, the length of the frame is indicated by carrier loss or a special end-of-frame symbol, depending on the PHY in use.

Zac67
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  • Could you please elaborate how Ethertype field will change the 802.3 with the 802.1Q protocol frame size. – Arun Chettoor May 23 '18 at 04:54
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    The optional 802.1Q tag is inserted in front of the Ethertype field. Its four bytes increase the maximum frame size to 1522 bytes. (Two bytes are the tag marker 0x8100, four bits are used for priority codes, twelve bits are used for VLAN identification.) 802.1ad introduced double tagging ("QinQ"), so a standard frame might grow some more. – Zac67 May 23 '18 at 05:29
  • So as per your comments 802.1Q tagging and 802.1ad double tagging the space is left inorder to make it compatible with the 802.3. – Arun Chettoor May 29 '18 at 06:16