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edited | 680 views

@ParimalAndhalkar take IP header 20 Bytes than try this .

is ans 1700 B ??
1680
answer should be 1700.if u want i can post the solution..i took ip header as 20 B..
ceil(1600/480)  = 4 fragments

each fragment take 20B header  ie extra 80bytes

1600+80 = 1680
yeah 1680 is answer,ur approach is wrong ..i made mstk previously..i fragmented the packet is A also,that why 1700 is coming..
@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.

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

How did you get 5 fragments?

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.

+1 vote

mtu is 480.

A sends payload 1600 B to B

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

ohh thanx

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

1700 is the actual size of the data