test-series

Question

Comments

@Bikram sir plz explain this

Answer 1

Network A has MTU of 1500 bytes
Network B has MTU of 620 bytes
we have to find the size of packets that is received at network C

Here is the thing , i think the data sent is 1380 , that is .. it does not include the header .

therefore we have the data as 1380 B.

now it will have no problem in network A .

but for network B , MTU = 600+20

thus , the data is broken into three segments .. (600+20) + (600+20)+(180+20)

thus the total data received will be 1440 B .

PS

made easy solution doesn't make much sense .. i think 1380 is the total data that station -1 is sending , otherwise they would have explicitly mentioned that it contains 20 B header , like they did for MTU of networks .

Answer 2

Network A has MTU of 1500 bytes
Netwoek B has MTU of 620 bytes
Network C has MTU of 1500 bytes ( MTU includes header size)

Segment Size = 1380 bytes, header size is 20B, data = 1360B

Network A has no problem, because segment size is 1380 Bytes
and MTU is 1500 Bytes.
Network B can carry 600B data + 20 B header
There will be 3 fragments each of size 600B, 600B, 160B respectively
Network C will receive total bytes
600 + 600+ 160 + 20+20+20 = 1420B

Comments

@rajoramanoj  any explanation is given?

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THE SOLUTION IS GIVEN AS

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may be it's given correct, i need to refresh my concepts!!