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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
asked in Computer Networks by Veteran (14.8k points) 15 152 316
reopened by | 206 views

http://gateoverflow.in/59997/mtu Follow this link to know how the fragmentation occured. On destination host the packets are recieved as size of 480 and 440. So length will be 440+20(IP header)=460 and offset = 480/8 =60. So I think third option

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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
answered by Boss (9k points) 6 56 152
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since in first link 954 bytes are transferred,why are we considering only 920 for next link??

and while you are calculating last fragment,you are writing that last fragment is 440+20+12 in which data is (440+20) =460 bytes,but data part here is just 440 (i am confused that whether it should be 460 or 440)and while calculating offset,you took just 480 not (480 + 20).pls explain this

and one more doubt i have,when divind further i should consider data + header into consideration or just data part??

 

Thanks for the help
@Akriti

Fragmentation is done at network layer so we are considering only data at IP layer, hence for offset also we have taken IP data part only.. 954 is the frame size at Data link layer after adding frame header

data at n/w layer is 440, 20 is IP header and 12 is Frame header so we are not taking header part for fragmentation

while doing fragmentation we consider pnly data part not the header because MTU tells size of maximum data that can go in the given network
thanks @cse23
in third link,there will be no fragmentation as its MTU is large enough to holdprevious packets

480 + 20 + 12 <=512

440 + 20 + 12 <=512
Could you please explain this diagramatically? Didn't quite get the explaination.

Also as soon as the packet reaches R1 its contents are : 14+20+900 , no ? Why not ? Kindly help.
@the.brahmin.guy,as the packet reaches R1,its size is 920 bytes .so MTU of first router is 1024 bytes,even if we subtract 20 bytes of IP header and 14 bytes of DLL header,then also we can accomodate 920 bytes of payload in it..hence,no fragmentation.
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