Fragmentation

Question

header size not given.

Comments

ceil(1600/480)  = 4 fragments

each fragment take 20B header  ie extra 80bytes

1600+80 = 1680

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yeah 1680 is answer,ur approach is wrong ..i made mstk previously..i fragmented the packet is A also,that why 1700 is coming..

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@resuscitate

I don't know why you changed your approach but you were right the first time. Answer must be 1700 bytes only. And number of fragments must be 5.

At A we fragment the main bigger datagram into 2 fragments where one fragment contains 1480 bytes of data (divisible by 8) and the second one contains 120 bytes.

Now at the first router R the first fragment is further fragmented into 4 fragments each of 456 data bytes (divisible by 8).

So at B we received total 5 packets each with 20B of header. Therefore extra bytes received is (5x20)=100B.

answer =1600+100=1700.

Answer 1

There are 5 fragments.

After first network  
1480+20 
120+20 
After second network 
1480 packet will be divided into 
456+20 
456+20 
456+20 
112+20 
And second fragment is  
120+20 
So total will be 
(120+20)+(456+20)+(456+20)(456+20)+(112+20)=1700 bytes...

so 1700 Bytes transmitted.

The almost same question also asked in gate 2016 set 1 http://quiz.geeksforgeeks.org/gate-gate-cs-2016-set-1-question-63/

also same type of problem you can found from Kuroe & ross book , chapter 4 , No-19p

https://www.chegg.com/homework-help/consider-sending-2400-byte-datagram-link-mtu-700-bytes-suppo-chapter-4-problem-19p-solution-9780133128093-exc  

Comments

How did you get 5 fragments?

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At A we fragment the main bigger datagram into 2 fragments where one fragment contains 1480 bytes  and the second one contains 120 bytes.

At the first router R the first fragment is further fragmented into 4 fragments each of 456 data bytes.

so in total 4 + 1 = 5 fragments.

Answer 2

mtu is 480.

payload is 460 B.

A sends payload 1600 B to B

so total data A sends to B = (460+20,460+20,460+20 ,220+20) = 1680 B.

Comments

ohh thanx

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5 fragments.

After first network 
1480+20
120+20
After second network
1480 packet will be divided into
456+20
456+20
456+20
112+20
And second fragment is 
120+20
So total will be
(120+20)+(456+20)+(456+20)(456+20)+(112+20)=1700 bytes...

so 1680 is  not the answer.

The almost same question also asked in gate 2016 set 1 http://quiz.geeksforgeeks.org/gate-gate-cs-2016-set-1-question-63/

also same type of problem you can found from Kuroe & ross book , chapter 4 , No-19p

https://www.chegg.com/homework-help/consider-sending-2400-byte-datagram-link-mtu-700-bytes-suppo-chapter-4-problem-19p-solution-9780133128093-exc  

 

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Here we took 456 (not 460) as it is closest divisible by 8 that is required in Offset field, thats the reason, So if you calculate through 460 and has 4 fragments than the answer is 1680 which is wrong.

 

Thanks

Answer 3

1700 is the actual size of the data