Made Easy Gate Mock-2 Question 56 [closed]

Question

In the given network system, station A needs to send a payload of 1600 B from its network layer to station B. If fragmentation is done, then the actual data size to be transmitted is ______________

Comments

At network one we have to divide the 1600 byte data.

At network 1:==> divide the packet into two part  ( 1480  + 120 ) and add the 20 bytes of the head to each.

At network 2 :==> MTU = 480, here we have to divide 1480 into three parts + 20 byte of header to each. So, total packets we have ==== 4.

At network 3:==> No need to divide. the packes.

So, we got 4 packets.

1680 === 1600 (data) + 80 ( header for four packets).

1680.

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how are the packets sent?

is it 460B+20B, since 460B is not divisible by 8,456B+30B is sent?

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@kumar.dilip

Thank you :) I did a mistake which I realized from your solution :D

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What about the fragment containing 120B data and 20B header?why aren't u adding that?

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@Somoshree Datta 5 You are right.. @kumar.dilip then 20 B has to be added more right? 1700 B should be the ans then

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https://gateoverflow.in/296435/made-easy-mock#c296509

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Somoshree Datta 5

Yes, it should be 1700.

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I am getting as follows

network 1 : (1480B payload + 20B Header) + (120B payload + 20B Header)

network 2 : (460B payload + 20 Header)*3 + ((120B payload + 20B Header)) + (120B payload + 20B Header)

Hence total of 100B overhead.