0 votes 0 votes Computer Networks ace-test-series computer-networks ethernet + – Sonali Rangwani asked Jan 23, 2017 edited Mar 6, 2019 by ajaysoni1924 Sonali Rangwani 445 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes 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 Rajnish Kumar answered Jan 23, 2017 Rajnish Kumar comment Share Follow See all 3 Comments See all 3 3 Comments reply priyanka manwani commented Dec 16, 2018 reply Follow Share I'm not getting your answer ...can u please explain.. I did ...1024-20-20-41====943 0 votes 0 votes Prateek Raghuvanshi commented Dec 16, 2018 reply Follow Share @ 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. 1 votes 1 votes priyanka manwani commented Dec 17, 2018 reply Follow Share Thank you got it now! Misunderstood the question 0 votes 0 votes Please log in or register to add a comment.