computer network

Question

Suppose a router receives an IP packet containing 600 data bytes and has to forward the packet to a network with maximum transmission unit of 200 bytes. Assume that IP header is 20 bytes long. What are fragment offset values for divided packets?

(A) 22, 44, 66, 88
(B) 0, 22, 44
(C) 0, 22, 44, 66
(D) 22, 44, 66

Answer 1

DATA=600B, MTU = 200B ,IP HEADER = 20B.

So in each frame 20B is reserved for IP HEADER ==> remaining is 180B

180 is not divisible by 8 , so we take 176 (nearest ones divisible by 8) as data along with 20B header.

we will get four segments  like 

we will get fragment offset by dividing the datasent with 8 like 0 ,176 / 8 (22) , (176+176) / 8 (44) ,(176+176+176) / 8 (66).

Comments

why did u divide by 8?

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Fragment offset gives info about  Position of fragment of user data in original datagram

Fragment Offset contains  field contains 13 bits so it can represent 2^13 bytes .

So,if we multiply by 8 we get the data.

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packet =[20+580]              MTU =[20+180]

176              176             176              52+4(padding )

(0 to 21)    (22 to 43)    (44 to 65)    (66 to 72)

Range  [0, 22, 44 ,66]

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in last, it would be 54+20(header) not 72+20(header) plzz edit it