Referring to the OSPF link metrics in the exhibit, which path will traffic from R6 take to reach R1?
A.
R6, R3, R2, R4, R1
B.
R6, R3, R2, R1
C.
R6, R5, R4, R1
D.
R6, R5, R3, R2, R4, R1
Click the Exhibit button.
Referring to the OSPF link metrics in the exhibit, which path will traffic from R6 take to reach R1?
Referring to the OSPF link metrics in the exhibit, which path will traffic from R6 take to reach R1?
A.
R6, R3, R2, R4, R1
B.
R6, R3, R2, R1
C.
R6, R5, R4, R1
D.
R6, R5, R3, R2, R4, R1
I think that the corret answer is A) What do you think
i think that correct is B, R2 will not sent to R4, because have link to area 1.
B is correct.
When R1 advertises a subnet intto Area 1, R2 and R4 will receive it. R4 will generate a LSA 3 and advertise into Area 0. So, from R6 to R2, R2 will have two path to reach a subnet in Area 1. One is O and one is OIA, and O is prefered (although its cost is higher).
Agree that B is correct. Nguyen Duc Hoa has the detailed explanation.
Thank you for the explanation,
can someone please elaborate why R4 will generate a LSA 3 and advertise into Area 0, but R2 won’t do the same?
Cheers
rawmaterial has a good point. I was sold on Ducs explanation until I read raws question. Anyone care to detail why the same decision that influenced the path to use R2 according to the answer being B as opposed to R4? As raw says, wouldn’t they both generate an LSA 3? I don’t think enough information is provided here.
My understanding based on Moy’s OSPF Anatomy is that that both R2 and R4 would send a Type 3 (Summary) LSA into area 0 with the same cost which is the cost to the most distant component = 25. The LSA 3 is sent to R3 and R5. R3 creates a new LSA 3 for the path thru R2 = 25 + 5 = 30, and sends it to R6. R5 creates a new LSA 3 for the path thru R3 & R2 = 25 + 10 + 5 = 40 and also send it to R6. R6 will pick the R6-R3-R2-R1 path (B).
I am trying to follow Nguyen Duc Hoa’s explanation and broke it down to 5 steps below.
I don’t understand how he came to determine Answer B is correct on point (5) below.
(1) B is correct.
(2) When R1 advertises a subnet intto Area 1, R2 and R4 will receive it.
(3) R4 will generate a LSA 3 and advertise into Area 0.
(4) So, from R6 to R2, R2 will have two path to reach a subnet in Area 1.
(5) One is O (intra-area) and one is OIA (inter-area), and O (intra-area) is prefered (although its cost is higher).
Since R2 has two paths to R1, one directly through R1 (OIA – inter-area) and other one through R4 (O – intra-area)
And we know that O – Intra area path is prefered, then R2 will prefer the path through R4.
Which makes answer A correct.
I believe B is the answer
Please see RFC 2328 below and how it defines intra-area path.
If intra area paths are destinations belonging to one of the routers attached areas, then since R2 is attached to area 0 and area 1, then the path from R2 to R1 is an intra-area path. And the route from R1 advertised through R4 is coming in as LSA 3 (summary LSA) hence becomes an inter-area path.
This is how answer B becomes correct.
RFC 2328 section 11
http://www.ietf.org/rfc/rfc2328.txt
Path-type
There are four possible types of paths used to route traffic to
the destination, listed here in decreasing order of preference:
intra-area, inter-area, type 1 external or type 2 external.
Intra-area paths indicate destinations belonging to one of the
router’s attached areas. Inter-area paths are paths to
destinations in other OSPF areas. These are discovered through
the examination of received summary-LSAs. AS external paths are
paths to destinations external to the AS. These are detected
through the examination of received AS-external-LSAs.
also see explanation in this link
https://supportforums.cisco.com/thread/2239499
See this webside
http://packetcorner.wordpress.com/2012/08/20/ospf-route-selection-process/