I have a concern for you regarding your lecture for Bellman-Ford. In the first ITERATION, THE LEAST OR MINIMUM cost IS 1 HOP WITH A PATH OF 7-4. HOW COME YOU NEED TO USE IT AS THE MINIMUM IN YOUR SECOND ITERATION. RATHER YOU TOOK 2 WITH A PATH OF 7-4-6. PLEASE, I NEED CLARIFICATION VERY URGENTLY.
In a traditional network this algo would be better. But what about in a Software Defined Network (SDN), where the Controller knows the topology of the network up front? In that case, Dijkstra's algo would be better. Right?
Why does this algorithm require less information? I am still confused how you would calculate all the values for V7 if you don't have all the edge connectivity/cost information.
If router A is not connected directly to router C, but router C is connected to B which in turn is connected to A. The routing Table of B will have connection information about C. So when B shares its routing table with A, A indirectly comes to know that C is connected to B. So A will simply add the Distance AB + BC to get distance AC. So the process continues and A gets the distance between itself and all other routers through its neighbors only.
What happens to the table when h = 0? Does that mean that because the number of hops is 0, the node that you are working from doesn't go off and find a path? Would we just put in the Lh(N) Path columns an infinity sign and a dash to indicate nothing is happening?
Shouldn't the cost of each hop that you calculate be the one going from the other nodes towards v7? Because the cost you fill in the matrix is what v7 needs to reach the other nodes and not the other way around.I think that is how Dijstra works and not Bellman Ford
Very well explained ! The first lesson on Bellman-Ford that I have understood , well done !
Great video. I especially appreciate the time spent to sneakily edit out the handwriting
I have a concern for you regarding your lecture for Bellman-Ford. In the first ITERATION, THE LEAST OR MINIMUM cost IS 1 HOP WITH A PATH OF 7-4. HOW COME YOU NEED TO USE IT AS THE MINIMUM IN YOUR SECOND ITERATION. RATHER YOU TOOK 2 WITH A PATH OF 7-4-6. PLEASE, I NEED CLARIFICATION VERY URGENTLY.
In a traditional network this algo would be better. But what about in a Software Defined Network (SDN), where the Controller knows the topology of the network up front? In that case, Dijkstra's algo would be better. Right?
I dont understand why in row 2 below L(3) is 2 and L (4) is 1 as well as why their respective paths
Why does this algorithm require less information? I am still confused how you would calculate all the values for V7 if you don't have all the edge connectivity/cost information.
If router A is not connected directly to router C, but router C is connected to B which in turn is connected to A. The routing Table of B will have connection information about C. So when B shares its routing table with A, A indirectly comes to know that C is connected to B. So A will simply add the Distance AB + BC to get distance AC. So the process continues and A gets the distance between itself and all other routers through its neighbors only.
why the route for V3 and V4 will not change? in row 2
What happens to the table when h = 0? Does that mean that because the number of hops is 0, the node that you are working from doesn't go off and find a path? Would we just put in the Lh(N) Path columns an infinity sign and a dash to indicate nothing is happening?
Correct. They would all be infinite distance with no paths
You saved me 320 Dollars ! passed CCNA and Bellman- Ford was one of the exam tests . Thanks xoxo
Shouldn't the cost of each hop that you calculate be the one going from the other nodes towards v7? Because the cost you fill in the matrix is what v7 needs to reach the other nodes and not the other way around.I think that is how Dijstra works and not Bellman Ford
Nice video . Thank you sir :)
Best explanation !
Hello