from R1 to reach R6 the cost is 13 (5+3+5) [R1->R3->R4->R6], then R1 send the packet to R3, but on R3 the packet is send trought its intra-area route (R3 sent the packet to R4)... then the traffic travels R1-R3-R5-R6.
R1 escoge a R3 por RID que sería 3.3.3.3 sería el más alto, R3 no conoce la área 1 por R4 ya que solo tiene conexión al área 0, en cambio R3 si tiene conexión a area 1 y por ello escoje R5 por que tiene prioridad mas baja que es 20, y por ultimo llega a R6
RESPUESTA CORRECTA : C: R1 -> R3 -> R5 -> R6
intra-routes are preferred over inter-routes. This is a tricky question and there are two parts to consider with. First is which metric are the ABRs are advertise to area 0 the R6 routes, in other words is the metric to reach R6 in area 1. R3 has metric 25 and R4 metric is 5! Second part is the R1 to reach the ABR with the best metric advertisement for R6 which is R4. So the path will be R1->R2->R4->R6. So correct one is A. I reproduce the scenario in my lab.
I also reproduced it in my lab and the correct answer is C for me. R1 has two ASBRs informing how to reach R6, one through R3 and the other through R4. For R1, the path with the best metric is the one that uses R4, but this path has R3 in the middle (R1-R3-R4 metric is 8 while R1-R2-R4 metric is 10). Once the packet reaches R3 on its way to R6, R3 uses its own routing table to forward the packet (6.6.6.6 is the lo0 address of R6 and ge-0/0/3 the interface facing R5). Here is the tricky part of the question and when "intra-routes are preferred over inter-routes" enters into action:
root@R3> show route 6.6.6.6
inet.0: 21 destinations, 21 routes (21 active, 0 holddown, 0 hidden)
+ = Active Route, - = Last Active, * = Both
6.6.6.6/32 *[OSPF/10] 00:08:34, metric 25
> to 60.0.0.5 via ge-0/0/3.0
D is correct.
R3 summary route of R6 and send to area 0 with the cost of 25.
R3 to R1 take the cost of 5.
>>> If R1 choose to reach to R6 via R3 >>> cost = 25+5 = 30
R4 summary route of R6 and send to area 0 with the cost of 5.
R4 to R1 take the cost of 8 (=3+5) via R3.
>>> If R1 choose to reach to R6 via R4 >>> cost = 8+5 = 13
>>> R1 prefer route R1->R3->R4->R6
>>> D is correct.
this should be D. An ABR would give you the best cost to reach the destination, area details are hidden. Here, packets go from R1 to R3, R3 will choose the intra-area to R4, because R4 is advertising a better metric.
No, you only have the best cost to an ABR providing a Type-3 summary, which is R3. Once on R3 the Intra-Area route is preferred. There is no link between R3 and R4 in area 1.
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