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Consider a scenario shown in below diagram. Suppose a TCP message that contains 900 bytes of data and 20 bytes of TCP header is transferring from Host H1 to Host H2. Assume H1 to R1 MTU includes 14-bytes of header, R1 to R2 MTU includes 8 bytes of frame header R2 to H2 link MTU includes 12 bytes of frame header. Then what will be the last fragment fields when the message reaches to Host H2?image:CN5/Q14a.PNG

  • Length = 440; ID = x; DF = 0; MF = 0; Offset = 60
  •   Length = 500; ID = x; DF = 0; MF = 1; Offset = 0
  •   Length = 460; ID = x; DF = 0; MF = 0; Offset = 60
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Answer will be (C)

Tcp segment size = 900 +20 is 920 bytes

when it will go to IP layer at A, A's network layer will receive 920 as whole IP data and it will add 20 bytes of header

so IP packet size = 920 +20

                          =940 bytes

now H1-----R1

MTU=1024 bytes

so no need of fragmentation and (940 + 14) bytes are transferred

for R1--------R2

WHEN 940 comes it should be fragmented

data part is 920 so it is fragmented into 480 and 440 bytes

now (480+20+8)and (440+20+8) is sent

for R2-------B

again 920 is fragmented into 480 and 440

so,

(480+20+12)and (440 +20+12) are sent

hencelast fragment is 440+20+12 in which data is (440+20) =460 bytes

offset(how much data is ahead of the fragment) = 480/8 =60
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