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Suppose that host A is connected to a router R 1, R 1 is connected to another router, R
2, and R 2 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.
asked in Computer Networks by Active (2.1k points) 10 45 | 40 views

2 Answers

0 votes

Here I doubt that ,in the link R1-R2, Total length must be 512Bytes (504+8) for 1st packet and TL=424B (416+8) .

answered by Active (2.1k points) 10 45
0 votes

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

 

answered ago by Active (1.9k points) 3 5 18


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