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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?

in Computer Networks by Active (5k points) | 201 views
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What is meant by this question? Can someone pls explain.?
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What do we need to find out?
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WHAT IS ANSWER GIVEN ?

I AM THINKING 1640
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Answer is given as 1680.
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WHAT IS HEADER SIZE TAKEN IN SOLUTION?
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this is the solution that is given.

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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
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Amount of data won't change right? Total bytes that are transmitted along with headers will only change right?
+2
if 460 is not divisible by 8 then they should send 456B right instead of 464?
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I USED WORD  DATA FOR HEADER + PAYLOAD
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But data means only payload part right?why are you including header also?
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YA BUT IT IS CONCEPTUAL TERMINOLOGY  DEPENDS ON QUESTION GENERALLY WHAT IS MEANS .....I USED TO GET IDEA FROM WORDINGS OF QUESTIONS
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@ 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?

+1

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?

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fourth fragment can contain 232B of data right?
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@Somoshree Datta 5 Why are you taking 112 B and 120 B data separately, you can take 232 B data in 4th fragment itself.

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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.

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@Somoshree Datta 5 I think we are supposed to take bottleneck MTU. Here minimum MTU = 480 B

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464+20=484 $\gt$ 480 (MTU). How are we sending these packets through that network?
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@gauravkc The explanation they have provided is wrong. We cannot add more bytes to 460

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@gauravkc

Solution provided by made easy is wrong. 

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@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? 

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@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?

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