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Bellman-Ford Algorithm
Consider computations for one destination d
Initialization
Each node table has 1 row for destination d Distance of node d to itself is zero: Dd=
Distance of other node j to d is infinite: Dj= , for j d
Next hop node nj = -1 to indicate not yet defined for j d
Send Step
Send new distance vector to immediate neighbors across local link
Receive Step
At node j , find the next hop that gives the minimum distance to d , Minj { Cij + Dj } Replace old (nj, Dj(d)) by new (nj, Dj(d)) if new next node or distance Go to send step
Bellman-Ford Algorithm
Now consider parallel computations for all destinations d
Initialization
Each node has 1 row for each destination d Distance of node d to itself is zero: Dd(d)=
Distance of other node j to d is infinite: Dj(d)= , for j d
Next node nj = -1 since not yet defined
Send Step
Send new distance vector to immediate neighbors across local link
Receive Step
For each destination d , find the next hop that gives the minimum distance to d , Minj { Cij+ Dj(d) } Replace old (nj, Di(d)) by new (nj, Dj(d)) if new next node or distance found Go to send step
Initial (-1, ) (-1, ) (-1, ) (-1, ) (-1, )
1 (-1, ) (-1, ) (6,1) (-1, ) (6,2) 2 3
Root Node
D 6 =
D 3 =D 6 + n 3 =
3 1
5
4 6
2
2
3
4
2
1
1
2
3
5
D^ D^6 = 5 =D 6 + n 5 =
Initial (-1, ) (-1, ) (-1, ) (-1, ) (-1, )
1 (-1, ) (-1, ) (6, 1) (-1, ) (6,2) 2 (3,3) (5,6) (6, 1) (3,3) (6,2) 3
Root Node
3 1
5
4 6
2
2
3
4
2
1
1
2
3
5 0
Initial (3,3) (4,4) (6, 1) (3,3) (6,2)
1 (3,3) (4,4) (4, 5) (3,3) (6,2)
2
3
Root Node
3 1
5
4 6
2
2
3
4
2
1
1
2
3
5 0
Network disconnected; Loop created between nodes 3 and 4
Initial (3,3) (4,4) (6, 1) (3,3) (6,2)
1 (3,3) (4,4) (4, 5) (3,3) (6,2)
2 (3,7) (4,4) (4, 5) (5,5) (6,2)
3
Root Node
3 1
5
4 6
2
2
3
4
2
1
1
2
3
5 0
Node 4 could have chosen 2 as next node because of tie
3
5
4 6
2
2
3
4
2
1
1
2
3
5
1
Root Node
Node 1 could have chose 3 as next node because of tie
1 2 3 4 1 1 1
1 2 3 4 1 1
X
(a)
(b)
Update Node 1 Node 2 Node 3 Before break (2,3) (3,2) (4, 1) After break (2,3) (3,2) (2,3) 1 (2,3) (3,4) (2,3) 2 (2,5) (3,4) (2,5) 3 (2,5) (3,6) (2,5) 4 (2,7) (3,6) (2,7) 5 (2,7) (3,8) (2,7)
Counting to Infinity Problem
Nodes believe best path is through each other (Destination is node 4)
