Network Layer Routing Algorithm

1. Network Layer

2.  Shortest Path Algorithm
 Distance Vector Routing
- Count to Infinite Problem
 Link State Routing
 Hierarchical Routing

3. Shortest Path Algorithm
 The first 5 steps used in computing the shortest path from A to D.
The arrows indicate the working node.

4.  Shortest path is the fastest path rather than the path
with fewest edges or kilometers
 Initially no path is known so all nodes are labeled with
infinity
 Labels on the edges could be computed as a function
of the distance ,bandwidth, average traffic
,communication cost ,measured delay and other
factors

5.  Initially, no path is known – All nodes are labeled with
infinity.
2. As the algorithm progresses labels are changed
reflecting the better path. The label is called
as tentative label.
3. As it is found that the label represents the shortest
path then it is made permanent and never changed
thereafter.

6. Distance Vector Routing
(a) A subnet. (b) Input from A, I, H, K, and the new
routing table for J.

7.  Distance Vector routing is called Routing information
Protocol(RIP)
 A Route with the less number of hops is considered as
the best route
 Every Router advertises its best route to the other
routers
 Convergence : setting of routers to the best paths
across the network
 It reacts rapidly to good news & converge slowly to bad
news .

8. Possible route from J to B
JA+AB=8+12 =20
JI+IB=10+36 = 46
JH+HB=12+31=43
JK+KB = 6+28 =34
Min =20 when taken ‘A’ Route , So 20 is updated in J new
table

9. Find the route from J to C
 JA+AC =8+25=33
 JI+IC=10+18=28
 JH+HC=12+19=31
 JK+KC=6+36=42
 Shortest distance is 28 when C Route taken

10. Find the Route from J to D
 JA+AD =8+40=48
 JI+ID=10+27=37
 JH+HD=12+8=20
 JK+KD=6+24=30
 Shortest distance is 20 through H route

11.  Find the Route from J to E
 JA+AE =8+14=22
 JI+IE=10+7=17
 JH+HE=12+30=42
 JK+KE=6+22=28
 17 is the shortest distance from the I route

12.  Find the Route from J to F
 JA+AF=8+23=31
 JI+IF=10+20=30
 JK+KF=6+40
 JH+HF=12+19=13
 30 is the shortest distance through I Route

13.  Find the Route from J to G
 JA+AG=8+18=26
 JH+HG=12+6=18
 JK+KG=6+31=37
 JI+IG=10+31=41
 18 is the shortest distance through H Route

14.  Find the Route from J to I
 JA+AI=8+21=29
 JH+HI=12+14=26
 JK+KI=6+22=28
 JI+IJ=10+0=10
10 is the shortest distance from I Route

15.  Find the Route from J TO K
 JA+AK=8+24=32
 JH+HK=12+22=34
 JK+KK=6+0=6
 JI+IK=10+22=34
 6 is the shortest distance from the K Route

16.  Find the route from J To L
 JA+AL=8+29=37
 JH+HL=12+9=21
 JK+KL=6+9=15
 JI+IL=10+33=43
 15 is the shortest distance from K Route

17. 3 Distance Vector Routing (2)
The count-to-infinity problem.

18. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
Initialization

19. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
10
2
2
2
2
2
2
1
Direct
Neighbours

20. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
10
2
2
2
2
2
2
1
3
3
11
3
3
11
Neighbours
of neighbours

21. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
10
2
2
2
2
2
2
1
3
3
11
3
3
11
13
13
13
13
Neighbours
of neighbours
of neighbours

22. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
10
2
2
2
2
2
2
1
3
3
11
3
3
11
13
13
13
13
Stable
convergence

23. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
10
2
2
2
2
2
2
1
3
3
11
3
3
11
13
13
13
13
1
Good news:
A new link!

24. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
10
2
2
2
2
2
2
1
3
3
11
3
3
11
1
13
1
13
1
Direct
endpoints
know

25. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
4
2
2
2
2
2
2
1
3
3
3
3
3
3
1
3
1
3
1
Neighbours
know

26. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
4
1
4
2
2
2
2
2
2
1
3
3
3
3
3
3
1
3
1
3
1
Neighbours
of neighbours
know

27. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
4
1
4
2
2
2
2
2
2
1
3
3
3
3
3
3
1
3
1
3
1
A happy and
stable network

28. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
4
1
4
2
2
2
2
2
2
1
3
3
3
3
3
3
1
3
1
3
Bad news:
Link crash!!

29. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
4
1
2
2
2
2
2
2
1
3
3
3
3
3
3
Direct
endpoints
know

30. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
4
1
2
2
2
2
2
2
1
3
3
3
3
3
3
10

31. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
4
1
2
2
2
2
2
2
1
3
3
3
3
3
3
10
5
13
11
13
Get help
from
neighbours

32. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
4
1
2
2
2
2
2
2
1
3
3
3
3
7
7
10
5
13
11
13
Routing loop
(due to
inconsistent
state info)

33. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
8
1
2
2
2
2
2
2
1
3
3
3
3
7
7
10
9
13
11
13

34. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
8
1
2
2
2
2
2
2
1
3
3
3
3
11
11
10
9
13
11
13
Counting
to infinity…

35. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
2
2
2
2
2
2
1
3
3
3
3
11
11
10
13
13
11
13

36. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
2
2
2
2
2
2
1
3
3
3
3
11
13
10
13
13
11
13

37. 1
10 2
2
2
A
B C
E
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
A
C
E
B
D
0
0
0
0
0
10
1
2
2
2
2
2
2
1
3
3
3
3
11
13
10
13
13
11
13
Stability
again

38. Link State Routing
 Each router must do the following:
• Discover its neighbors, learn their network address.
• Measure the delay or cost to each of its neighbors.
• Construct a packet telling all it has just learned.
• Send this packet to all other routers.
• Compute the shortest path to every other router.

39. Link State Routing : Learning about
the Neighbors
(a) Nine routers and a LAN.
(b) A graph model of (a).

40.  States of lines of all the routers in a network is
considered to build a common graph of the entire
network .
 New artificial node ‘N’ to which A,C & F are connected.
Possibility of visiting A to C on the LAN is represented
by ANC

41. Link State Routing (3): Measuring
Line Cost
 A subnet in which the East and West parts are
connected by two lines.

42. Setting a Link Cost

43.  Delay = Round trip time of ECHO Packet / 2
 Sending router get a reasonable estimate of the delay
by measuring round trip time & dividing it by two

44. Link State Routing : Building Link
State Packets
 (a) A subnet. (b) The link state packets for this
subnet.

45. Link State Routing : Distributing the
Link State Packets

46. The Packet buffer for router B
 Packet from A arrives directly . So it must be sent to C
& F and Acknowledgement to A , as indicated by the
Flag bits
 E is different ,It arrives twice one via EAB and Vis EFB .
IT must be sent only to C and it must be acknowledged
to both A & F

47. Duplicate arrives scenario
 If a copy of C’s state arrives from F before the fourth
entry in the table has been forwarded. The sixth bit
will be changed to 100011 to indicate packet must be
Acknowledged to F but not sent there

48. Hierarchical Routing
 Hierarchical routing.

49.  The routers are divided into regions . Each router has
complete details about how to route packets to
destination with in its own region. But it does not have
any idea about the internal structure of other regions