TCP HEADER , FRAME SIZE

Question

Answer 1

Given that, for R1-R2 the maximum packet size including 8 B  header is 512 B. It means that a packet in R1-R2 can carry maximum data of  512-8 =504 B only.

We need to transfer 900 B data and 20 B of TCP header, total 920  B which is beyond the limit 504 B. Hence the packet will be fragmented.

• First packet will carry first 504 B
• Second packet will carry 920-504 =416 B

Total length of second packet = 416+ 8 (header) = 424 B

Comments

Given answer is 460

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424B seems correct to me.

Answer 2

Maximum frame size allowed = 512 B (include frame header)
Frame Header Size = 8 B
So maximum packet size allowed = 512 B - 8 B = 504 B

Maximum data size=504-20=484 ≈ 480 // multiply of 8

• First packet will carry =480 data+20 IP header+8 frame header=508B
• Second packet will carry 920-480 =440 B

Total length of IP header = 440+ 20 (include IP header) = 460 B