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Suppose that host A is connected to a router R 1, R 1 is connected to another router, R2, and R2 is connected to host B. Suppose that a TCP message that contains 900 bytes of data and 20 bytes of TCP header is passed to the IP code at host A for delivery to B.Show the Total length, Identification, DF, MF, and Fragment offset fields of the IP header in each packet transmitted over the three links.

Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frame header,

link R1-R2 can support a maximum frame size of 512 bytes, including an 8-byte frame header,

and link R2-B can support a maximum frame size of 512 bytes including a 12-byte frame header.

in Computer Networks by Active (2.8k points)
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2 Answers

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Link A-R1

Length= (920Bdata+20Bheader)=940B   

Link R1-R2(here the packet is being fragmented into 2 packets)

Packet1:Length=(504B data+ 8B header)=512B  [It is correct bcz data's length i.e 504 is divisible by 8]

Packet 2:Length=(416B data + 8B header)=424B [It is correct bcz data's length i.e 416 is divisible by 8]

Link R2-B( here packet 1 from link R1-R2 is fragmented into to two packets say packet1,1 and packet 1,2)

Packet1.1 Length=(496 B data+ 12 bytes header)=508 B [It is correct bcz data's length i.e 496 is divisible by 8]

Packet 1.2 Length=(8 B data+ 12 bytes header)=20 B [It is correct bcz data's length i.e 8 is divisible by 8]

Packet 2 Length=(416 B data+ 12 bytes header)=428 B [It is correct bcz data's length i.e 416 is divisible by 8]

So Receiver receives three packets.

PACKET ID MF HL=Headersize/4 TL  OFFSET
PACKET 1,1 X 1 3 508 0
PACKET 1,2 X 1 3 20 496/8=62
PACKET 2 X 0 3 428 (496+8)/8=63
by Active (3.3k points)
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Here I doubt that ,in the link R1-R2, Total length must be 512Bytes (504+8) for 1st packet and TL=424B (416+8) .

by Active (2.8k points)
0
512B contain 8 B as frame header = this routing is done at router ryt?

so MTU at router = 512- 8 = 504 only.

not remove 20 B of network header = 504- 20 = 484 which is not divisible by 8 so 480 B size packet is sent

hence total size for 1st packet = 480 +20 = 500

for second 920- 480= 440.B

so total size = 440+20 = 460B

hope you got it.
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