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Algorithms for computing Maximally Redundant Trees for IP/LDP Fast-Reroute draft-enyedi-rtgwg-mrt-frr-algorithm-01. A lia Atlas (Juniper Networks) Gabor Enyedi , Andras Csaszar (Ericsson) IETF 83, Paris, France. Agenda. Briefly about the algorithm Problem Avoid using a node

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slide1

Algorithms for computing Maximally Redundant Trees for IP/LDPFast-Reroutedraft-enyedi-rtgwg-mrt-frr-algorithm-01

Alia Atlas (Juniper Networks)

Gabor Enyedi, Andras Csaszar (Ericsson)

IETF 83, Paris, France

agenda
Agenda
  • Briefly about the algorithm
  • Problem
  • Avoid using a node
  • Non-2-connected networks
agenda1
Agenda
  • Briefly about the algorithm
  • Problem
  • Avoid using a node
  • Non-2-connected networks
adag and partial order1
ADAG and partial order
  • Almost DAG (ADAG)
  • A<<B if there is a path from A to B
  • Root is both the shortest and the greatest

E

G

D

Root

C

S

B

A

H

adag and partial order2
ADAG and partial order
  • S<<E
    • Blue path: increasing [S, E]
    • Red path: decreasing [Root, S] and [E, Root]

E

G

D

Root

C

S

B

A

H

adag and partial order3
ADAG and partial order
  • S>>A
    • Blue path: increasing [S,Root] and [Root, A]
    • Red path: decreasing [A, S]

E

G

D

Root

C

S

B

A

H

adag and partial order4
ADAG and partial order
  • S and C are not ordered
    • Blue path: [S, E] and [C, E]
    • Red path: [A, S] and [A, C]

E

G

D

Root

C

S

B

A

H

agenda2
Agenda
  • Briefly about the algorithm
  • Problem
  • Avoid using a node
  • Non-2-connected networks
three trees
Three trees
  • We have tree trees
    • SPT
    • Two MRTs
  • There is no connection between SPT and MRTs
  • Impossible to find a redundant pair for SPT
  • Example:

Shortest path

1

C

D

10

1

Dest

S

No redundant pair for that!

1

10

1

B

A

1

agenda3
Agenda
  • Briefly about the algorithm
  • Problem
  • Avoid using a node
  • Non-2-connected networks
total order
Total order
  • Partial order can compare any X only with S
    • We need to compare any two nodes
  • Make a total order as well
    • If A<<B, let A<B
    • If A and B are not ordered select either A<B or B<A
    • This can be done with a topological oder after converting the ADAG into a DAG
  • Results:
      • If A<B, either A<<B or A and B are not ordered
a possible total order
A possible total order
  • Numbers are written next to nodes

8

7

E

G

6

D

Root

C

S

4

5

0

3

B

A

H

1

2

possible cases
Possible cases
  • If dst>>src, failed node F

If S<<F<E, it may be on the BLUEpath

8

7

E

G

6

D

Root

C

S

4

5

0

3

B

Otherwise: it may be on the REDpath

A

H

1

2

possible cases1
Possible cases
  • If dst<<src, failed node F

Otherwise: it may be on the BLUEpath

8

7

E

G

6

D

Root

C

S

4

5

0

3

If A<F<<S, it may be on the RED path

B

A

H

1

2

possible cases2
Possible cases
  • If dst and src are not ordered
    • There are four sub-paths

If F>>src, it may be on the first part of the RED path

If F and src are not ordered, and F>dest, it may be on the second part of the RED path

7

8

E

G

6

D

4

Root

C

S

5

0

3

F and src are not ordered, and F<dest, it may be on the second part of BLUE path

B

If F<<src, it may be on the first part of the BLUE path

A

H

1

2

agenda4
Agenda
  • Briefly about the algorithm
  • Problem
  • Avoid using a node
  • Non-2-connected networks
non 2 connected problem
Non-2-connected problem
  • In this case we don’t have a single order
    • Neither a partial order
    • Nor a total order
  • Convert the GADAG into an ADAG!

C

A

Non-root block

X

Local root block

D

B

non 2 connected problem1
Non-2-connected problem
  • In this case we don’t have a single order
    • Neither a partial order
    • Nor a total order
  • Convert the GADAG into an ADAG!

C

A

X1

Non-root block

Local root block

X2

D

B