Networking - Individual Task

Running Head: NETWORKING 0 | P a g e
Networking
Individual Task
Student name

Networking 1 | P a g e
Answer: 1
1. Next hop address is 180.20.0.30
2. M2 interface is suitable for it
3. Three hops are required for this deliver R1, R3, and Host B
4. R1 send it to default router.
5. R2 used m2 interface for forwarding that packet
6. 180.0.0.10 is used for forwarding that packet to their destination
7. 133.0.10.0/24 is destination network
8. Routing table for Router R3:
130.24.0.0/16
Network address Next-hop address Interface
130.24.10.16 180.20.0.30/8 m0
230.10.50.0/24
Network address Next-hop address Interface
230.10.50.24 --------------------- m1
Default: 180.0.10.10/8 m0

Networking 2 | P a g e
Answer: 2
a. IP datagram size is 5000 bytes including header.
Next destination MTU size = 1500 bytes
IP header size is 20 byes minimum (Huitema, 2000).
Therefore, size of data is 5000-20 =4980 bytes, so total number of fragments are
4980/1500 = 3.32
We take it as four fragments for destination. The fragment offsets would be multiples of
1500-20= 1480
1480/8 = 185
Therefore, 0,185,370, and so on
Fragment Starting byte Ending byte
1 1 1500
2 1501 3000
3 3001 4500
4 4501 5060
b.
Fragment Total Bytes Header
bytes
Data size in
bytes
Flag for more
fragments
Fragments
offset
1 1500 20 1480 1 0
2 1500 20 1480 1 185
3 1500 20 1480 1 370
4 560 20 540 0 555
c. All the fragments are having an IP header with them. Therefore, it is having more bytes
as compared to packet arrived from the other layer (Forouzan, 2007). It is also depend on
the MTU size of destination. If MTU size is greater than packet size than there is no
change, only header is added with that packet (Cowley, 2012).