Click the Exhibit button.
You have the multi-area OSPF network design shown in the exhibit
Which path will traffic from R1 transit to reach R4 if the R2-R4 link fails?
A. R1-R2-R5-R3-R4
B. R1-R2-R3-R5-R4
C. R1-R2-R3-R4
D. R1-R2-R5-R4
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Says Multi-Area OSPF, but the configuration is not with Multi-Area adyacency! (there´s no “secondary” statement on area 0.0.0.10)
If is a real Multi-Area design, the correcta answer will be C
If is like the configuration on the exhibit, the correct answer will be D.
The correct is C
Is true that OSPF prefeers Intra-Area, but the exercise says that OSPF design is Multi-Area. So, will be a secondary PtToPt link between R2 and R3 when the Multi-Area configuration is applyed. Please see page 4-16 in AJSPR book.
And the path will be: R1-R2-R3-R4
from R2 perspective. LSAs from R3 that containing R4 routes (or any route from are 10) are masrked as (O IA) <R2>R5>R4
Many thanks for this clarification
D is correct answer.
D is the correct answer.
I’ve implemented the same topology in the lab.
D
Intra-Area LSA wins to Inter-Area LSA even for a sub-optimal path.
OSPF will use cost as the metric to choose the shortest path for each destination, this is true but it’s not entirely correct. OSPF will first look at the “type of path” to make a decision and secondly look at the metric. This is the prefered path list that OSPF uses:
Intra-Area (O)
Inter-Area (O IA)
External Type 1 (E1)
NSSA Type 1 (N1)
External Type 2 (E2)
NSSA Type 2 (N2)
After the path selection it will look at the lowest cost path
https://tools.ietf.org/id/draft-ietf-ospf-multi-area-adj-09.xml
“In the current OSPF specification, intra-area paths are preferred over inter-area paths.”
A is the correct answer in a standard OSPF implementation.
Intra-Area LSA wins to Inter-Area LSA even for a sub-optimal path
C looks right to me.
Under no R2-R4 failure:
R2 injects R4 LSA3 with cost of 5.
In the same time R3 injects R4 LSA3 with cost 5 as well.
R1’s cost to reach out R2 is 1 and cost to R4 through R2 is 1+5=6. R2 win, R1-R2-R4.
R1’s cost to reach out R3 is 2 and cost to R4 through R3 is 2+5=7.
Under the R2-R4 failure:
R2 injects R4 LSA3 with cost of 5+5=10 (R2-R5-R4).
In the same time R3 injects R4 LSA3 with cost 5.
R1’s cost to reach out R2 is 1 and cost to R4 through R2 is 1+10=11.
R1’s cost to reach out R3 is 2 and cost to R4 through R3 is 2+5=7. R4 win, R1-R2-R3-R4.