**5.04-3. Bellman Ford Algorithm - a change in DV (1, part 3)**Consider the network below, and suppose that at t=0, the link between nodes g and h goes down. And so at t=0, nodes g and h recompute their distance vectors (DVs). Following this recomputation, to which nodes will h send its new distance vector? (Note: to answer this question, you’ll need to know some of the DV entries at g and h at t=0, but hopefully they’ll be obvious by inspection).**Diagram Explanation:**The diagram is a network of nodes labeled from 'a' to 'i'. Each line represents a link between two nodes with an associated cost. For instance, the link between node 'a' and node 'b' has a cost of 8. At the center, there is a broken link between nodes 'g' and 'h' with a cost of 6, indicating that this link has gone down at t=0. Nodes 'g' and 'h' are in a state of computation which is illustrated with a “compute” label and yellow explosion graphics around them.**Answer Options:**- node i only- node e only- all nodes- node h does not send out its distance vector, since none of the least costs have changed to any destination.- nodes i and e only- nodes i and e and g onlyThe correct choice would involve analyzing the impact of the link failure on node h's distance vector and to which nodes the updated vector should be sent.

The Bellman-Ford algorithm is a graph traversal algorithm used to find the shortest paths from a single source vertex to all other vertices in a weighted directed graph. It can handle graphs with both positive and negative edge weights, as well as detect negative weight cycles.The algorithm iteratively relaxes the edges of the graph, gradually improving the estimated distances from the source vertex to each vertex. It starts by initializing the distance to the source vertex as 0 and the distances to all other vertices as infinity. Then, in each iteration, it considers all edges in the graph and updates the distance of each vertex if a shorter path is found.(1) The Correct answer is option(4): node i and e only.Explanation: The node h sends its new Distance Vector to the nodes i and e. As both i and e nodes are connected to node h and also both nodes are of the same cost from node h. So it will send the new Distance Vector to nodes i and e.All other incorrect options are:Option: Node i only Explanation: This option is incorrect because when the link between nodes g and h goes down, node h will not send its new distance vector to node i alone. The link failure between g and h does not directly affect the least cost to reach node i.Option: Node e only Explanation: This option is incorrect because when the link between nodes g and h goes down, node h will not send its new distance vector to node e alone. The link failure between g and h does not directly affect the least cost to reach node e.Option: Nodes i, e, and g only Explanation: This option is incorrect because when the link between nodes g and h go down, node h will not send its new distance vector to nodes i, e, and g,only.Option: All nodes Explanation: This option is incorrect because When the link between nodes g and h goes down, node h will not send its new distance vector to all nodes. The link failure between g and h does not result in any changes to the least costs to any destination, so there is no need for node h to send out its new distance vector.2. The correct option (1) is An essentially infinite amount of time; this is the count-to-infinity problem.Explanation: In the given scenario, when the link between nodes b and c goes down at t=0, the Bellman-Ford algorithm is used to update the distance vectors (DVs) of the affected nodes and propagate the changes to their neighbors. However, in certain cases, such as in a network with cycles or when there are indirect paths with lower costs, the count-to-infinity problem can occur.Here all other incorrect options are:Option 2 is Incorrect because This option suggests that the last time a DV calculation will take place is at time 2. However, in the given network, the count-to-infinity problem occurs due to the link failure between nodes b and c. As a result, the DV calculations will continue indefinitely, and there is no fixed last time for the calculations.Option 3 is Incorrect because This option suggests that the last time a DV calculation will take place is at time 3. However, as explained in option 2, the count-to-infinity problem causes the DV calculations to continue indefinitely. Therefore, there is no specific last time for the calculations.Option 4 is Incorrect because This option suggests that the last time a DV calculation will take place is at time 4. However, as mentioned before, the count-to-infinity problem leads to an infinite loop of DV exchanges. Therefore, the calculations will not stop at a specific time.Conclusion:(1) Node i and e only option (4) is the correct answer.(2) In summary, options 2, 3, and 4 are incorrect because they assume a fixed last time for DV calculations, while in reality, the count-to-infinity problem option (1) is correct causing the calculations to continue indefinitely.

Answered: 5.04-3. Bellman Ford Algorithm - a…