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