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The minimum frame length for $10$ Mbps Ethernet is ________ bytes and maximum is _______ bytes.

  1. $64$ & $128$
  2. $128$ & $1518$ 
  3. $1518$ & $3036$
  4. $64$ & $1518$ 
asked in Computer Networks by Boss (39.9k points)
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2 Answers

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Ans is D: 64 and 1518

The minimum frame size is 64B as in ethernet csma/cd protocol is used which says Tt>=2Tp...

now as in ethernet 802.3 frame size (EXCLUDING DATA) is 18B....therefore minimum data is =64-18=46B

Now the maximum data allowed is 1500B as the MTU of ethernet is 1500B....

therefore maximum frame size =1500+18=1518B
answered by Active (4k points)
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ans is D. 64 & 1518 

The minimum size of an Ethernet frame is 64 bytes. The breakup of this size between the fields is: Destination Address (6 bytes) + Source Address (6 bytes) + Frame Type (2 bytes) + Data (46 bytes) + CRC Checksum (4 bytes). The minimum number of bytes passed as data in a frame must be 46 bytes. If the size of the data to be passed is less than this, then padding bytes are added.

 

The maximum size of an Ethernet frame is 1518 bytes. The breakup of this size between the fields is: Destination Address (6 bytes) + Source Address (6 bytes) + Frame Type (2 bytes) + Data (1500 bytes) + CRC Checksum (4 bytes). The maximum number of bytes of data that can be passed in a single frame is 1500 bytes.

answered by Active (1.1k points)


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