Ace Test Series: Computer Networks - Ethernet

Question

Answer 1

1+20+20=41 B

but minimum payload in ethernet is 46B.

so add 5B i.e. 41+5=46B

Ethernet frame size= 46+26(ethernet header)

                             =72 Byte

Comments

I'm not getting your answer ...can u please explain..

I did ...1024-20-20-41====943

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 @ priyanka manwani   to reach message of 1 byte of application layer  to data link layer ,it will come through transport layer then network then it will reach at data link layer.since every layer add its own header to message ,so 1 byte(application layer data)+20(tcp header)+20(ip header)=41 byte of packet will reach at data link layer and it will be payload for frame but we know payload of frame should be  greater than 46 and less than 1500 byte.but we are getting 41 byte so we have to add padding byte to it (41+5=46 byte) .so ethernet frame size will be 46+26=72 Byte.

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Thank you got it now! Misunderstood the question