Student Initials: 24. [8 marks] An Internet user would like to transfer a messag
ID: 3911417 • Letter: S
Question
Student Initials: 24. [8 marks] An Internet user would like to transfer a message from his computer (host A) to another computer (host B). The size of the message passed to the IP layer is 2400 bytes. NET2 MTU 640 NET3 MTU= 1 500 NET1 R1 R2 MTU 1500 The datagram(s) carrying the message will have to cross two routers (R1 and R2) and 3 networks (NET1, NET2, and NET3) as described in the above figure. Each of these networks has a specific MTU (for example the MTU of network 1 is 1500 bytes. MTU here stands for Maximum Transmission Unit, the largest data that a frame can carry.). a) Give the number of the IP datagram(s) needed, and the total length of each IP datagram at each For the cases where the datagram is fragmented, give the value of the fragment Offset field and the value of the More bit in the Flags field of each fragment. b)Explanation / Answer
Here we have to send message from Host A to Host B
Size of the message =2400 bytes
At Network 1:
MTU of this network =1500
Length of IP&TCP headers to be included in every packet=20+20
Total number of Fragments =upper seal(2400/1500)=2 packets
Packet 1:
Length=1460+40
More Bit=1
Fragement offset=0
Packet 2:
Length=940+40
More Bit=0
Fragment offset=1460/8=182
At Network 2:
MTU of this network=640
Packets that is recieved by network 2 =1500,980
Let us consider the first fragment,as the MTU of this network is 640 the datagram is again fragmented
No.of fragments =upperseal(1500/640)=3 packets
Packet 1:
Length=600+40
More bit=1
Fragment offset=0
Packet 2 :
Length=600+40
More bit=1
Fragement offset=640/8=80
Packet 3:
Length=260+40
More bit=1
Fragement Offset=160
Now Consider Second packet Of Size 980 from Network 1
This Fragement is again fragemented into 2 fragments
Packet 4:
Length =600+40
More bit=1
Fragment offset=182
Packet 5:
Length =340+40
Fragment offset=242
More bit=0
At Network 3:
As the MTU of this network is Greater than the all datagrams from network 2
So there is no fragmentation at Network 3
More bit of each packet =0