**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 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).Description of the network:- **Nodes:** a, b, c, d, e, f, g, h, i- Links with costs: - a to b: 8 - b to c: 1 - a to d: 1 - d to e: 1 - e to f: 1 - f to i: 1 - e to h: 1 - g to h: originally 6 (now down) - g to d: 1 - h to i: 1- At t=0, the link (with a cost of 6) between nodes g and h goes down.- Nodes g and h recompute their distance vectors.**Diagram explanation:**- Nodes are represented as circles with labels from "a" to "i".- Nodes g and h each display a "compute" action, showing they are recalculating.- The link between g and h is indicated as down, represented by a dashed line and an infinity symbol for its cost.**Question:**Which nodes will node h send its new distance vector to?**Options:**- ○ node i only- ○ nodes i and e only- ○ node h does not send out its distance vector, since none of the least costs have changed to any destination.- ○ nodes i and e and g only- ○ node e only- ○ all nodes

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.Conclusion:Node i and e only option (4) is the correct answer.

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Computer Science

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 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). a 1 1 at t=0 the link (with a cost of 6) between nodes g and h goes down b. compute ∞ g all nodes 8 1 6 1 e 1 compute h 1 1 1 1 node i only nodes i and e only O node h does not send out its distance vector, since none of the least costs have changed to any destination. O nodes i and e and g only node e only

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 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). a 1 1 at t=0 the link (with a cost of 6) between nodes g and h goes down b. compute ∞ g all nodes 8 1 6 1 e 1 compute h 1 1 1 1 node i only nodes i and e only O node h does not send out its distance vector, since none of the least costs have changed to any destination. O nodes i and e and g only node e only

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

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Concept explainers

Heuristic System

The term heuristic is derived from the Greek word heurisko, which means “I find, discover”. Heuristics can be mental shortcuts or approaches designed from previous experiences. The main aim of the heuristic approach is to find a good enough solution to solve the problem. Test and mistake, rule of thumb, or an educated estimate are all examples of heuristics. These methods are sufficient for achieving an immediate, brief aim. When a heuristic is used in practice, it works as predicted. It can, however, produce systematic errors. Trial and error is an important heuristic that may be used in a variety of situations, from cross-linking nuts and bolts to determining the values of given input variables in mathematical puzzles. Additional hypotheses, visual illustrations, simplification, and forward or backward reasoning are all common heuristics in mathematics.

Question

**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 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).

Description of the network:

- **Nodes:** a, b, c, d, e, f, g, h, i
- Links with costs:
- a to b: 8
- b to c: 1
- a to d: 1
- d to e: 1
- e to f: 1
- f to i: 1
- e to h: 1
- g to h: originally 6 (now down)
- g to d: 1
- h to i: 1

- At t=0, the link (with a cost of 6) between nodes g and h goes down.
- Nodes g and h recompute their distance vectors.

**Diagram explanation:**
- Nodes are represented as circles with labels from "a" to "i".
- Nodes g and h each display a "compute" action, showing they are recalculating.
- The link between g and h is indicated as down, represented by a dashed line and an infinity symbol for its cost.

**Question:**
Which nodes will node h send its new distance vector to?

**Options:**
- ○ node i only
- ○ nodes i and e only
- ○ node h does not send out its distance vector, since none of the least costs have changed to any destination.
- ○ nodes i and e and g only
- ○ node e only
- ○ all nodes

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