Fragmentation of packet

Question

Which of the following statements are True ?

Comments

I got your point brother but can't we take the floor of that decimal offset and store it ?

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Yes we have to take the floor not ceil

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I think S2 and S3 are correct

Answer 1

Answer. S2 and S3.

From transport layer data is 1100Bytes. Transport layer will add header(size given 10Bytes) and will send to network layer.

Therefore total data send to network layer will be 1110 Bytes,

Now MTU = 206 including ip header therefore max. data that can be send is 206-20 = 186 Byte

since 186B is not divisible by 8, so fragment size will be 184B.

S1)#Fragements = Ceil(1110/184) = 7

So, Fragment 1 to fragment 6 size will 184B

and fragment 7 size will be 8B(although remaining is 6B, but to make it divisible by 8 we add padding to make it 8B)

S2) $184/8 = 23$

offset of 4$^{th}$ fragment is $3*23 = 69$

 

S3) 6$^{th}$ fragment is not the last fragment , therefore MF bit will 1. for 7$^{th}$ fragment mf bit will be 0.

Comments

Good answer, just a minor glitch.

Use No. of Fragements = Ceil(1110/184) = 7

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Corrected

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Great!

Answer 2

S2,S3   are TRUE

S1 FALSE