# GATE2007-IT-60

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For the network given in the figure below, the routing tables of the four nodes $A$, $E$, $D$ and $G$ are shown. Suppose that $F$ has estimated its delay to its neighbors, $A$, $E$, $D$ and $G$ as $8$, $10$, $12$ and $6$ msecs respectively and updates its routing table using distance vector routing technique.

$$\overset{\textbf{Routing Table of A}}{\begin{array}{|c|r|}\hline \text{A}&0\\ \hline \text{B}&40\\ \hline \text{C}&14\\ \hline \text{D}&17\\ \hline \text{E}&21\\ \hline \text{F}&9\\ \hline \text{G}&24\\ \hline \end{array}}\qquad \overset{\textbf{Routing Table of D}}{\begin{array}{|c|r|}\hline \text{A}&20\\ \hline \text{B}&8\\ \hline \text{C}&30\\ \hline \text{D}&0\\ \hline \text{E}&14\\ \hline \text{F}&7\\ \hline \text{G}&22\\ \hline \end{array}} \qquad \overset{\textbf{Routing Table of E}}{\begin{array}{|c|r|}\hline \text{A}&24\\ \hline \text{B}&27\\ \hline \text{C}&7\\ \hline \text{D}&20\\ \hline \text{E}&0\\ \hline \text{F}&11\\ \hline \text{G}&22\\ \hline \end{array}}\qquad \overset{\textbf{Routing Table of G}}{\begin{array}{|c|r|}\hline \text{A}&21\\ \hline \text{B}&24\\ \hline \text{C}&22\\ \hline \text{D}&19\\ \hline \text{E}&22\\ \hline \text{F}&10\\ \hline \text{G}&0\\ \hline \end{array}}$$

1. $$\begin{array}{|c|r|} \hline \text {A} & \text{8} \\\hline \text {B} & \text{20} \\\hline \text{C} & \text{17} \\\hline \text{D} & \text{12} \\\hline \text {E} & \text{10} \\\hline \text {F} & \text{0} \\\hline \text{G} & \text{6} \\\hline \end{array}$$
2. $$\begin{array}{|c|r|} \hline \text {A} & \text{21} \\\hline \text {B} & \text{8} \\\hline \text{C} & \text{7} \\\hline \text{D} & \text{19} \\\hline \text {E} & \text{14} \\\hline \text {F} & \text{0 } \\\hline \text{G} & \text{22} \\\hline \end{array}$$
3. $$\begin{array}{|c|r|} \hline \text {A} & \text{8} \\\hline \text {B} & \text{20} \\\hline \text{C} & \text{17} \\\hline \text{D} & \text{12} \\\hline \text {E} & \text{10} \\\hline \text {F} & \text{16} \\\hline \text{G} & \text{6} \\\hline \end{array}$$
4. $$\begin{array}{|c|r|} \hline \text {A} & \text{8} \\\hline \text {B} & \text{8} \\\hline \text{C} & \text{7} \\\hline \text{D} & \text{12} \\\hline \text {E} & \text{10} \\\hline \text {F} & \text{0} \\\hline \text{G} & \text{6} \\\hline \end{array}$$

edited
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how we get 17 for c?

Distance from $F$ to $F$ is $0$ which eliminates option (C).

Using distance vector routing protocol, $F \to D \to B$ yields distance as $20$ which eliminates options (B) and (D).

edited by
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@Rajarshi Can you please explained how you are calculated Distance F -> D -> B is 20.
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^F To D is given in question as 12 and D to B can be found from table of D i.e. 8. So total F-D-B is 12+8 =20.

ANS: A

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cool :)
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Can u explain it ??
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he already did, what else u want..?
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nice
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D to F is given 7. why are we getting F to D 12.
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@vishalshrm539 I found your comment very rude and distasteful. Everyone has varying comprehensible capabilities, and if you can genuinely help, please do so, otherwise, let others help and refrain from making unnecessary remarks.

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@vishar Consider all distances as edge weights in a bidirectional graph. D to F being $7$ does not necessarily mean that F to D will be $7$ as well. In fact, F to D is given as $12$ in the question.

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Should be best solution
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@Puja Mishra Just take any 1 option and verify others. In above question he took example of option A, considered its neighbour and verified for other nodes. Hope it helps.

ans a)

Basically it is a bidirectional weighted graph. The directed weights are obtained from the routing tables of the respective source to destinations. We need to calculate the shortest path from F to each node based on the distances from the routing tables.

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