Hello this question is related to the network management I hope to answer it wit

Question

Hello

this question is related to the network management

I hope to answer it with explaination, but don't use your handwriting, please...

Consider the figure below: suppose that host A is connected to router RI, router RI is connection to router R2. router R2 is connected to R3. and R3 is connected to host B. Suppose that a TOP massage that contains 900 bytes of data and 20 bytes of Tcp header is passed to the lp layer at host A for delivery host B. show header length Total Length, Don't More Fragments, and Fragment offset fields of the lP header in each packet transmitted over all the four links. Assume that link A-RI can support a maximum frame size of 1024 a bytes including a 14 bytes frame header, link RI-R2 can support a maximum frame size of 512 bytes including an 8 byte frame header link R2-R3 can support a maximum of 256 bytes including a 12 tes frame header. and link R3-B can support a maximum frame size of 1024 bytes including a 14 bytes frame header MTU-1024 MTU 512 MTU 256 MTU-1024 R1 R2 R3

Explanation / Answer

TCP Packet = 900 (data) + 20 (IP header) = 920 Bytes.
IP Packet = 920 (data) + 20 (IP header) = 940 Bytes

Link A-R1: Can support 1024 bytes including 14-byte frame header (e.g., 1010-byte IP data packet) thus no fragmentation is needed.
  
Length = 940, ID = x, DF = 0, MF = 0, Offset = 0.

Link R1-R2: Can support 512 bytes including 8-byte header (e.g., 512-8=504 bytes of IP data packet and 504 – 20 (IP header) = 484). Since IP data in a frame must be multiple of 8 bytes:
8*x 484 => x 60.5 bytes. Use x = 60 bytes
IP data 1 = 60*8 = 480 Bytes
IP packet size 1 = 480 + 20 = 500 Bytes.
IP data 2 = 920 – 480 = 440 Bytes
IP packet size 2 = 440 + 20 = 460 Bytes.

[1]   Length = 500, ID = x, DF = 0, MF = 1, Offset = 0.
[2]   Length = 460, ID = x, DF = 0, MF = 0, Offset = 60.

Link R2-R3: Can support 256 bytes including 12-byte header (e.g., 256-12=244 bytes of IP data packet and 244 – 20 (IP header) = 224). Since IP data in a frame must be multiple of 8 bytes:
8*x 224 => x 28 bytes. Use x = 28 bytes

Packet 1 of R1 arrives at R2,
480/224=2.14 (So there will be 3 packets.)

IP data 1 = 28*8 = 224 Bytes
IP packet size 1 = 224 + 20 = 244 Bytes.
IP data 2 = 480 – 224 = 256 Bytes (Only 236 bytes can be sent).
IP packet size 2 = 236 + 20 = 256 Bytes.
Ip data 3 = 480-(224+236)=20 Bytes.
IP packet size 3 = 20 + 20 = 40 Bytes

[1]   Length = 244, ID = x, DF = 0, MF = 1, Offset = 0.
[2]   Length = 256, ID = x, DF = 0, MF = 1, Offset = 0.
[3] Length = 40, ID = x, DF = 0, MF = 0, Offset = 28.

Packet 2 of R1 arrives at R2,
440/224=1.96 (So there will be 2 packets.)

IP data 1 = 28*8 = 224 Bytes
IP packet size 1 = 224 + 20 = 244 Bytes.
IP data 1 = 440 – 224 = 216 Bytes
IP packet size 1 = 216 + 20 = 236 Bytes.

[1]   Length = 244, ID = x, DF = 0, MF = 1, Offset = 0.
[2]   Length = 236, ID = x, DF = 0, MF = 0, Offset = 28.

Link R3-B: Can support 1024 bytes including 14-byte frame header (e.g., 1010-byte IP data packet) thus no fragmentation is needed.

Length = 244, ID = x, DF = 0, MF = 0, Offset = 0.

Length = 256, ID = x, DF = 0, MF = 0, Offset = 0.

Length = 40, ID = x, DF = 0, MF = 0, Offset = 0.

Length = 244, ID = x, DF = 0, MF = 0, Offset = 0.

Length = 236, ID = x, DF = 0, MF = 0, Offset = 0.

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