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Part 2

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... distributed route computation algorithms discussed in class and shown in the textbook and notes. ... 2nd vector from F ... v]: c[C] = 5 + 7 = 12 ... – PowerPoint PPT presentation

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Title: Part 2


1
Part 2 Simulation Question   Given the graph
below, where weights are assigned on the edges,
simulate the first three updates which will be
carried out on the routing table information for
node 6 according to each of the distributed route
computation algorithms discussed in class and
shown in the textbook and notes. (a)
Distance Vector Routing 1) Give the initial
Routing Table for node A in the graph.
2) Show what happens when A receives the first
Distance Vector from B. 3) Show what happens
when A receives the second Distance Vector from
F. Explain your answers. (b) Link-State Routing
(SPF) from Dijkstras algorithm 1) Give the
initial Routing Table for node A in the graph.
2) Show one complete iteration of algorithm
13.1 and the resulting routing table for A.
3)Show a second complete iteration of algorithm
13.1 and the resulting routing table for A.
Explain your answers.
2
Distance Vector
  • Initial routing table for A

Algorithm 13.2
Dest. Next-hop Distance
A - 0
3
Distance Vector
  • 2. A receives 1st vector from B

B (A,6), (F,2),(C,7)
7
C
B
6
5
1
8
D
A
2
2
Dest. Next-hop Distance
A - 0
B B 6
C B 13
F B 8
6
3
F
E
7
Routes introduced by B in RED
4
Distance Vector
  • 3. A receives 2nd vector from F

F(A,3), (B,2), (C,8), (D,2), (E,7)
7
C
B
6
5
Dest. Next-hop Distance
A - 0
B F 5
C F 11
D F 5
E F 10
F F 3
1
8
D
A
2
2
6
3
F
E
7
Routes introduced by F in GREEN
5
Link-State (Algorithm 13.1)
  • 1. Initial

7
C
B
Set S
6
5
1
B C D E F
8
D
A
2
2
6
3
F
E
7
Routing-table
B B
C 0
D 0
E 0
F F
Distance-table
B 6
C -
D -
E -
F 3
6
Link-State
  • 2. 1st iteration

Choose F since DF is shortest. Remove F from
S. S c D(F) weight(F, v) for every node
v still in S. Then compare c with Dv cB 3
2 5 lt DB 6 cC 3 8 11 lt DC
inf cD 3 2 5 lt DD inf cE 3 7
10 lt DE inf Thus, update tables with shorter
values.
B C D E
7
C
B
6
5
1
8
D
A
2
2
6
3
E
F
7
R-table
B F
C F
D F
E F
F F
D-table
B 5
C 11
D 5
E 10
F 3
7
Link-State
  • 3. 2nd iteration

Choose B (D is fine too, because both DB and
DD are shortest) Remove B from S. S c
D(B) weight(B, v) for every node v still in S.
Then compare c with Dv cC 5 7 12 gt
DC 11 ? no update for C cD 5 4 9 gt
DD 5 ? no update for D cE 5 9 14 gt
DE 10 ? no update for E Thus, the tables
remains same as step 2.
C D E
7
C
B
6
5
1
8
D
A
2
2
6
3
F
E
7
Alternatively, If D is chosen instead of B at
this step, the table is
Dest. Next-hop Distance
B F 5
C D 10
D F 5
E F 10
F F 3
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