2 votes 2 votes In the given network system, station A needs to send a payload of 1600B from its network layer to station B. If fragmentation is done, then the actual data size to be transmitted is? Somoshree Datta 5 asked Jan 18, 2019 Somoshree Datta 5 1.2k views answer comment Share Follow See all 23 Comments See all 23 23 Comments reply Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share What is meant by this question? Can someone pls explain.? 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share What do we need to find out? 0 votes 0 votes eyeamgj commented Jan 18, 2019 reply Follow Share WHAT IS ANSWER GIVEN ? I AM THINKING 1640 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share Answer is given as 1680. 0 votes 0 votes eyeamgj commented Jan 18, 2019 reply Follow Share WHAT IS HEADER SIZE TAKEN IN SOLUTION? 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share this is the solution that is given. 0 votes 0 votes eyeamgj commented Jan 18, 2019 reply Follow Share ACTUALLY QUESTION IS ASKING THAT HOW MUCH TOTAL DATA WILL BE REACH TO B .....WHAT I FEELING INItIALLY IS THAT TRANSMITTED ONLY FROM NETWORK ONE BUT QUESTION IS ASKING FOR TO REACH TO B 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share Amount of data won't change right? Total bytes that are transmitted along with headers will only change right? 0 votes 0 votes balchandar reddy san commented Jan 18, 2019 reply Follow Share if 460 is not divisible by 8 then they should send 456B right instead of 464? 2 votes 2 votes eyeamgj commented Jan 18, 2019 reply Follow Share I USED WORD DATA FOR HEADER + PAYLOAD 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share But data means only payload part right?why are you including header also? 0 votes 0 votes eyeamgj commented Jan 18, 2019 reply Follow Share YA BUT IT IS CONCEPTUAL TERMINOLOGY DEPENDS ON QUESTION GENERALLY WHAT IS MEANS .....I USED TO GET IDEA FROM WORDINGS OF QUESTIONS 0 votes 0 votes himgta commented Jan 18, 2019 reply Follow Share @Somoshree Datta 5 It should be 456+20 ,Anyways it does not change the final answer! But my doubt is that Network 1 is transferring datagrams to Network 2, then network 2 has to transfer to destination....Why network 2 does not add any header here? 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share himgta How are you getting the same answer as that given in the solution? If the question is trying to ask the total number of bytes that are being sent to B, then I am getting answer as 1700 B. 1st fragment: 456B data+20B header 2nd fragment: 456B data+20B header 3rd fragment: 456B data+20B header 4th fragment: 112B data+20B header 5th fragment: 120B data+20B header So total number of bytes that are being sent= 1600+(20*5)= 1700B Shouldnt this be the answer? 1 votes 1 votes balchandar reddy san commented Jan 18, 2019 i reshown by balchandar reddy san Jan 18, 2019 reply Follow Share fourth fragment can contain 232B of data right? 0 votes 0 votes Kunal Kadian commented Jan 18, 2019 reply Follow Share @Somoshree Datta 5 Why are you taking 112 B and 120 B data separately, you can take 232 B data in 4th fragment itself. 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share Kunal Kadian Because we are already fragmenting the packet into 2 fragments of size 1480B+20B as the first fragment and 120B+20B while passing through the network having MTU of 1500B. As we dont recombine fragments in intermediate routers, so we cant combine 120B and 112B and form a fragment of 232B. Reassemby of all fragments take place at the final destination and not at the intermediate routers. 1 votes 1 votes Kunal Kadian commented Jan 18, 2019 reply Follow Share @Somoshree Datta 5 I think we are supposed to take bottleneck MTU. Here minimum MTU = 480 B 0 votes 0 votes gauravkc commented Jan 18, 2019 reply Follow Share 464+20=484 $\gt$ 480 (MTU). How are we sending these packets through that network? 0 votes 0 votes Kunal Kadian commented Jan 18, 2019 reply Follow Share @gauravkc The explanation they have provided is wrong. We cannot add more bytes to 460 1 votes 1 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share @gauravkc Solution provided by made easy is wrong. 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share @Kunal Kadian we consider the bottleneck MTU only when we discover the path MTU using ICMP packets. Otherwise we don't consider it. How is host A supposed to know the bottleneck MTU unless and until it uses path MTU discovery technique? 0 votes 0 votes Kunal Kadian commented Jan 18, 2019 reply Follow Share @Somoshree Datta 5 Correct. But I thought that it is always assumed that minimum MTU is known (using MTU discovery technique). Maybe I will have to look other questions of similar kind, to clear the doubt.Did you see any such other question? 0 votes 0 votes Please log in or register to add a comment.