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Consider the subnet given in the below diagram. Here, distance vector routing algorithm is used.

The following vectors have just come in to router $C$:

from $B$: $\left ( 3, 0, 6, 5, 3, 8, 3 \right )$;  

from $D$: $\left ( 1,5, 3, 0, 8, 6, 3 \right )$;  

from $E$: $\left ( 4, 3, 1, 1, 0, 4, 6 \right )$; and

from $F$: $\left (3, 8, 3, 6, 4, 0,9 \right )$.   

 

The measured delays to $B$, $D$, $E$ and $F$ are $2, 3, 1$ and $6$, respectively.

What is $C's$ new routing table?

  1.      $\left (5, 2, 0, 2, 5, 5, 5 \right )$
  2.      $\left (4, 4, 0, 3, 1, 6, 5\right )$
  3.      $\left (5, 4, 0, 3, 5, 6, 6\right )$
  4.      $\left (4, 2, 0, 2, 1, 5, 5\right )$
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Given routing vectors coming into router C are:
From B: (3, 0, 6, 5, 3, 8, 3);
From D: (1, 5, 3, 0, 8, 6, 3);
From E: (4, 3, 1, 1, 0, 4, 6);
From F: (3, 8, 3, 6, 4, 0, 9);
Given that the distance metric (delay) from C to B is 2; C to D is 3; C to E is 1 and C to F is 6


So the distance vector at C if going via B is = Distance from C to B + Distance from B to other routers

  •  Going via B gives (2+3, 2+0, 2+6, 2+5, 2+ 3, 2+8, 2+3) = (5, 2, 8,  7,  5, 10,  5 )
  •  Going via D gives (3+1, 3+5, 3+3, 3+0, 3+8, 3+6, 3+3) = (4,  8, 6,  3, 11, 9,   6 )
  •  Going via E gives (1+4, 1+3, 1+1, 1+1, 1+0, 1+4, 1+6) = (5,  4,  2,  2,  1,  5,  7 )
  •  Going via F gives (6+3, 6+8, 6+3, 6+6, 6+4, 6+0, 6+9) = (9, 14, 9, 12, 10 , 6, 15)

So, taking minimum for each destination except C from the above calculations,

we get the distance vector at C as (4, 2, 0, 2, 1, 5, 5)

Hence option D is correct .

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