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Consider Host A sends a UDP datagram containing 1100 bytes of user data to host B over an ethernet LAN. Maximum transmission unit (MTU) of the network is 206 bytes (including the network layer overhead). Size of UDP header is 10 bytes and size of IP header is 20 bytes. There is no option field in the IP header.
Consider the following statements:
S1 : Total 6 fragments will be there.
S2 : The offset of 4th fragment will be 69.
S3 : The MF flag of 6th fragment is set i.e. 1.
Which of the following options is correct?

asked in Computer Networks by (391 points) | 166 views
someone plzz no.of fragments are calculated... i got 6 fragments but ans is 7,,...
UDP Datagram total size = 1100 + 10 = 1110 bytes, as it is said in the question 1100 bytes are user data.

1110 bytes will be contained in an ip packet, 20 bytes header will be added, will be transferred to data link layer.

MTU = 206 bytes, 206-20 bytes= 186, 186 not divisible by 8, hence each fragmented segment will carry 184 bytes of udp data gram + 20 bytes of ip header.

1110/184 = 6.something, hence 7 is the answer.
Thanx a lott.. :) now my doubt is cleared.

only S1 is false right?

and thanks for the explanation Manu

why we are not subtracting one more 20 from mtu that we generally do???? please help stuck

1 Answer

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S1 false S2 true S3

answered by Active (2.8k points)

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